Electrical technology. Volume II, Machines and measurements 9789332514416, 9789332517943, 9332514410, 9789332517950, 9332517959, 9789332524569, 9332524564

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ELECTRICAL TECHNOLOGY MACHINES AND MEASUREMENTS Volume II

S. P. Bali Former Faculty Member Military College of Electronics and Mechanical Engineering Secunderabad, India

Delhi • Chennai

Copyright © 2013 Dorling Kindersley (India) Pvt. Ltd. Licensees of Pearson Education in South Asia No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material in this eBook at any time. ISBN 9789332514416 eISBN 9789332517943 Head Office: A-8(A), Sector 62, Knowledge Boulevard, 7th Floor, NOIDA 201 309, India Registered Office: 11 Local Shopping Centre, Panchsheel Park, New Delhi 110 017, India

Dedicated to My late wife SUKSHAM BALI (10 April 1940 – 07 August 2007)

And departing leave behind us FOOTPRINTS on the sands of time

Preface

A celebrity is a person who works hard all his life to become well known, then wears dark glasses to avoid being recognised. With the rising momentum of development, the scope of text books dealing, in particular, with Electrical Engineering has expanded considerably. An important guiding objective in writing this book is to provide the students with a text book they can read, understand and study by themselves. Intended to serve as a text book for the subject of Electrical Engineering for BE/B.Tech Degree students it will also serve as a text-cum-reference for the students of Diploma Engineering. So also it will be useful to candidates appearing for AMIE, IETE, GATE, UPSC Engineering Services and IAS entrance examination. It will be equally helpful to practicing engineers to understand the theoretical aspects of their professions. Despite the publication of a large number of text books on this field, the students continue to remain perplexed. Keeping this fact in mind, this text book has been developed in a systematic manner, with emphasis on basic concepts. Written in a simple, easy to understand language, reinforced by illustrations which speak of themselves and are easy to understand and supplemented by selected worked examples based on step-by-step solutions the various chapters are interlinked, yet independent. The book can be read in the sequence in which it is written without facing any difficulty. The following features are intended to serve as learning aids: 1. More than enough worked examples are given in each chapter, wherever applicable to emphasize the practical utility of the results derived. 2. Detailed summary is given at the end of each chapter, as an aid to memory. 3. Multiple choice questions (MCQ) along with their answers are included in each chapter for the self-assessment of the student. 4. The illustrative method of treatment is used, each illustration bringing home a point. 5. Conventional questions are also given at the end of each chapter. Answers to numerical questions are also given. 6. Where possible, mechanical analysis is given. 7. Equivalent circuits are given for a better understanding of the problem. 8. The per unit system is discussed in detail with plenty of worked examples. The book has been designed in two volumes, Volume 1: Electrical Fundamentals and Volume 2: Machines and Measurements. Vol. 2: Machines and Measurements cconsists of 24 chapters divided in two parts: Part B on Electric Machines which comprises seventeen chapters discussing every aspect of electric machines and Part C on Electrical Measurements that comprises seven chapters discussing the measurement of voltage, current, resistance and power and their display. Part B on Electric Machines comprises seventeen chapters. Starting with Electromechanical Energy Conversion motor and generator action and their interdependence are explained. Then are explained d.c. Generators, their types and construction. The e.m.f equation is derived, followed by generator characteristics and their suitability for specific applications. Paralleling of generators is discussed next followed by worked examples. Development and measurement of torque is discussed next. Classification and characteristics of motors follow. Different methods of starting motors and reversing d.c. motors are then discussed followed by losses and efficiency of d.c. motors. Then motor control is discussed threadbare. Followed by topics like reversing the direction of motors, interlocking devices, retardation and stopping, then comes the most important topic of transformers. Mutual coupling dot convention and types of transformers and their construction features are discussed. After developing the e.m.f. equation the equivalent circuit of a transformer is developed and the idea of reflected impedance introduced. This is followed by losses, efficiency and maximum power transfer, then auxiliary topics like auto transformers, current transformers, potential transformers and transformer connections are discussed supplemented by worked examples. An introduction to three phase transformers and their connections is given. This is followed by synchronous generators and motors. Alternator windings, coil group connections, winding pitch and

vi Electrical Technology distribution factors are explained. Alternator synchronizing procedure is also discussed. Synchronous motors come next; their construction, operation, starting, followed by power factor control, the characteristics of these motors are discussed. The text is supplemented by worked examples. In three phase induction motors, the formation of a rotating magnetic field is discussed in details. After explaining its construction, slip and its effect on rotor frequency are discussed. The characteristics of polyphase induction motors and their starting techniques are then explained. The various types of polyphase induction motors, their construction and phase splitting are then discussed. Shaded pole motors and wound rotor induction motors, various types, and their characteristics are explained in detail. In the chapter on specialized motors, reluctance motors, hysteresis motors, stepper motors and linear induction motors, producing linear motion, are briefly explained. The next four chapters are also of an introductory type. Chapter 47 on per unit system gives a detailed description of per unit resistance, per unit impedance, and per unit system for transformer calculations supplemented by worked examples. Part C of the book on Electrical Measurements comprises seven chapters. Starting with measurement systems and errors associated with them various types of meter movements, controlling and deflecting torques are discussed in details, supplemented by worked examples. The constructional details of the various types of meter movements are also given. Exploiting these meter movements as ammeters, voltmeters and ammeters and the terms associated with their performance are explained followed by worked examples, commercial type multi function multi-range meters (volt Ohm milliammeters, Voms) are then explained. The internal circuits and working of a Simpson Multimeter and calculations of the various shunts and multipliers, along with their equivalent circuits are given. The various types of Wattmeters and energy meters, their construction and operation details are given. Measurement of power is explained next along with worked out examples. Cathode Ray Oscilloscope, their types, controls and construction are given next followed by detailed setting up procedure. Appendices A to M provide extremely useful information concluding with a brief Glossary of terms have been made available as online resources. The book also features —a Web-based circuit simulator, specially created to help students practice key circuits. The customized FREE version integrated with the book will enable students to build, analyze and learn the circuits. Besides being used as a practice/pre-lab tool by students, it can also serve as an exciting tool for instructors to teach the circuits. Last but not the least I would like to put on record the appreciation of the production and editorial staff at Pearson providing me unstained help in completing the project right from the day it was conceived till its completion. Suggestions for improvement of the book will be thankfully acknowledged.

S. P. Bali

Contents Preface

v

32.13 32.14 32.15 32.16

PART B: ELECTRIC MACHINES 31. Electromechanical Energy Conversion 31.1 31.2 31.3 31.4 31.5 31.6 31.7 31.8 31.9 31.10 31.11 31.12 31.13 31.14 31.15 31.16

Introduction 589 Energy Conversion Devices 589 Related Fundamental Laws 590 D.c. Motor and Generator Mechanical Construction 592 Elementary Generator 593 Average e.m.f. Generated in a Quarter Revolution 595 Fundamental d.c. Generator Equation for Average e.m.f. Between Brushes 595 Biot-Savart Relationship 595 Motor Action Versus Generator Action 599 Power Flow Diagrams 599 Conversion Process in a Machine 601 Linear and Rotary Motion 602 Methods of Analysis 603 Energy Balance 604 Rotary Motion 606 Doubly Excited Rotating Machines 606 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

32. D.C. Generators 32.1 32.2 32.3 32.4 32.5 32.6 32.7 32.8 32.9 32.10 32.11

589

Generator Losses 629 Polarity of Brushes 629 Voltage Regulation 630 Paralleling of Generators 630 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

33. D.C. Motors

33.1 Introduction 637 33.2 Development and Measurement of Torque 638 33.3 Prony Brake 639 33.4 Two-Scale Prony Brake 640 33.5 Dynamometer 640 33.6 Back Electromotive Force in a Motor 641 33.7 Classification of Motors 643 33.8 Construction 644 33.9 Characteristics of d.c. Motors 645 33.10 Relation between Torque and Speed of a Motor 648 33.11 Direct-Current, Motor Starting Problems 651 33.12 D.c. Starting Switch 652 33.13 D.c. Motor Reversing 655 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

34. Efficiency of Direct Current Machinery 608

Introduction 608 Mechanical Construction 609 Armature Structure 610 Commutator and Brushes 611 Armature Windings 612 Commutation 616 Armature Reaction 617 E.M.F. Equation 618 Separately Excited Generator 621 Basic Generator Types 621 Schematic Diagram and Equivalent Circuit 622 32.12 Characteristics of d.c. Generators 625

637

34.1 34.2 34.3 34.4 34.5 34.6

659

Introduction 659 Basic Efficiency Relationships 659 Types of Losses in d.c. Machines 660 Motor Enclosures 664 Maintenance and Accessibility 665 Cooling and Ventilation 665 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

35. D.C. Motor Control 35.1 Introduction 667 35.2 Control Devices 667

667

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35.3 Pilot Control Devices 669 35.4 Manual d.c. Motor Starters 671 35.5 Automatic Direct-Current Motor Starters 674 35.6 Comparison of Manual Versus Automatic Starter 674 35.7 Reversing Control of Direct Current Motors 674 35.8 Retardation and Stopping 676 35.9 Ward-Leonard System 683 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

37.3 Three-phase Transformers 727 37.4 Paralleling Three-phase Transformer Banks 728 37.5 Harmonic Suppression in Three-phase Connections 728 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

38.

36. Transformers—Single Phase 686 36.1 Introduction 686 36.2 Mutual Coupling 687 36.3 Series Connection of Mutually-Coupled Coils 688 36.4 Parallel Connection of Mutually Coupled Coils 689 36.5 The Dot Convention 690 36.6 Preventing Mutual Inducance 691 36.7 Transformer 692 36.8 Transformer Construction 692 36.9 Ideal Transformer 694 36.10 Transformation Ratio 696 36.11 General Transformer Equation 697 36.12 Practical Transformer 699 36.13 Transformer Ratings 701 36.14 Transformer Equivalent Circuits 703 36.15 Secondary Voltage Phasor Relations 707 36.16 Transformer Voltage Regulation 708 36.17 Maximum Power Transfer 708 36.18 Losses in Transformers 710 36.19 Short Circuit Test 711 36.20 Open Circuit Test 711 36.21 Transformer Efficiency 712 36.22 Autotransformer 713 36.23 Power Transformers 715 36.24 Instrument Transformers 716 36.25 Pulse Transformers 717 36.26 Transformer Connections 718 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

37.

Three-phase Transformers

723

37.1 Introduction 723 37.2 Three-phase Connections of Single-phase Transformers 723

Synchronous Generators— Alternators

732

38.1 Introduction 732 38.2 Physical Construction of d.c. Machines 733 38.3 Alternator Windings 735 38.4 Synchronous Alternator 737 38.5 Stator Windings 739 38.6 Distribution Factor 741 38.7 Basic Voltage Generation Formula 744 38.8 Alternator Performance 744 38.9 Alternator Percentage Regulation 745 38.10 Relation between Generated Voltage and Terminal Voltage of an Alternator at Various Load Power Factor 745 38.11 Equivalent Circuit 753 38.12 Parallel Operation 754 38.13 Alternator Synchronizing Procedure 757 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

39.

Synchronous Motors

763

39.1 39.2 39.3 39.4 39.5 39.6

Introduction 763 General 763 Synchronous Motor Construction 764 Synchronous Motor Operation 764 Starting Synchronous Motors 766 Synchronous Motor Power Factor Control 767 39.7 Synchronous Motor V Curve 768 39.8 Synchronous Capacitors 771 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

40.

Induction Motors (Three Phase) 40.1 Introduction 776 40.2 The Rotating Magnetic Field 777 40.3 Speed of the Rotating Magnetic Field

776

779

Contents ix

40.4 Direction of Rotation 783 40.5 Slip and its Effect on Rotor Frequency and Voltage 783 40.6 Construction of a Three-phase Induction Motor 786 40.7 Rotor Impedance and Current 787 40.8 Locus of the Current 789 40.9 Losses and Efficiency 790 40.10 Air Gap Power 792 40.11 Maximum Torque 793 40.12 Induction Motor Torque-Speed Characteristics 795 40.13 Wrim and Scim: A Comparison 796 40.14 Starting Techniques for Induction Motors 796 40.15 Determination of Induction Motor Parameters 798 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

41. Induction Motors (Single Phase)

805

41.1 41.2 41.3 41.4 41.5 41.6 41.7 41.8 41.9

Introduction 805 Classes of Induction Motors 806 Getting the Rotor Started 807 Construction 808 Phase Splitting 808 Locked-Rotor Torque 809 Resistance-Start Split-Phase Motors 811 Capacitor-Start Split-Phase Motors 813 Reversing Single-Phase Induction Motors 815 41.10 Dual-Voltage Operation 815 41.11 Shaded-Pole Motors 818 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

42.

Specialized Motors 42.1 42.2 42.3 42.4 42.5 42.6

822

Introduction 822 Reluctance-Start Induction Motor 822 Hysteresis Motors 824 Stepper Motors 826 LIM 830 Universal Motors 832 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

43.

Servos and Synchros 43.1 43.2 43.3 43.4 43.5 43.6

44.

836

Introduction 836 Operational Amplifiers 836 Practical Issues 840 D.c. Servomotors 840 A.c. Servomotors 842 Synchros 843 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

Open-Loop and Closed-Loop Systems 847

44.1 44.2 44.3 44.4 44.5 44.6 44.7 44.8 44.9 44.10 44.11 44.12 44.13

Introduction 847 Control System 847 Servomechanisms 848 Open-Loop and Closed-Loop Systems 849 Automation 849 Components of a Control System 850 Transfer Function 850 Regulators and Servomechanisms 851 Transient Periods 852 Damping 852 The Basic Feedback Control System 854 The Integrator 857 Steady-State Operation of a Compound Generator 860 44.14 Automatic Frequency Control 861 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

45.

Inverters and Converters 45.1 45.2 45.3 45.4 45.5 45.6

Introduction 864 Converting Machines 864 Rectifiers: a.c. to d.c. Conversion 865 Three-phase Full-wave Rectifiers 869 Uninterruptible Power Supplies (UPS) 870 Inverters Changing d.c. Voltage to a.c. Voltage 871 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

46. Controlled Rectifiers 46.1 46.2 46.3 46.4

864

Introduction 876 Direct Thyratron Control 877 Thyristor Control of Motors 877 Thyristor Control of Motor 881

876

x Electrical Technology

Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

46.5 Universal Motor Control 881 46.6 Complete Control System 882 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

47. Per-unit System

50. 884

47.1 47.2 47.3 47.4

Introduction 884 Resistance as Per Unit Quantity 885 Impedance as a Per-unit Quantity 885 Per-unit Quantities for Three-phase Circuits 886 47.5 Per-unit System-Transformer Calculations 887 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

PART C: ELECTRICAL MEASUREMENTS 48. Measurements and Error

903

48.1 Introduction 903 48.2 Definitions 904 48.3 Accuracy to Measure Electron Performance 905 48.4 Factors Affecting Accuracy 905 48.5 Measurement Systems 906 48.6 Calibration 908 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

49. Meter Movement 49.1 49.2 49.3 49.4 49.5 49.6 49.7 49.8 49.9 49.10 49.11 49.12 49.13 49.14

Ammeters, Voltmeters and Ohmmeters 50.1 50.2 50.3 50.4 50.5 50.6

51.1 51.2 51.3 51.4 51.5 51.6 51.7 51.8 51.9

51.11 51.12

Introduction 911 Design Principles 911 Meters 915 Moving-Coil Instruments 915 Classification of Measuring Instruments 917 Gravity Control 918 Moving-Iron Instruments 919 Polarized Moving-Iron Instrument 922 Dynamometer-Type Instruments 924 Induction-Type Instruments 926 Hot-Wire Instruments 930 Thermocouple Instruments 931 Galvanometers 932 The Electrostatic Voltmeter 932

Introduction 938 Special Features 938 Ammeters 940 Voltmeters 948 Ohmmeters 953 Measurement of Insulation Resistance Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

955

51. Watt Meters and Energy Meters 960

51.10

911

937

51.13

Introduction 960 Watt Meters 960 Dynamometer-Type Watt Meter 961 Compensating Coil 963 Induction-Type Watt Meters 964 Energy Meters 965 General Classification 965 Direct Current Types 966 Alternating Current Types (Induction Watt-Hour Meters) 969 Poly-Phase Induction Watt-Hour Meters 974 Errors in Energy Meters 974 Measurement of Power in Three-phase Circuits 975 Measurement of Three-phase Power 977 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

52. Multimeters—V.O.Ms 52.1 52.2 52.3 52.4 52.5 52.6 52.7 52.8 52.9

983

Introduction 983 Direct Current Ranges 984 Direct Voltage Ranges 985 Multi-Range Ohmmeters 986 Alternating Current Ranges 986 Alternating Voltage Ranges 988 V.O.M. Specifications 989 The Simpson Model 260 Multi-Meter 993 Digital Multi-Meters 997

Contents xi

Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

53. The Oscilloscope 53.1 53.2 53.3 53.4 53.5 53.6 53.7 53.8

Introduction 1003 CRT 1004 Intensity 1005 Fluorescent Screen 1006 Focusing 1007 Deflection 1008 Time Base 1009 Graticules 1010

53.9 Block Diagram 1011 Summary Multiple Choice Questions (MCQ) Conventional Questions (CQ)

1003 54. Oscilloscope Techniques

1014

54.1 Introduction 1014 54.2 Preliminary Checks 1014 54.3 Screen Pattern Obtained with Deflection Voltages 1018 54.4 Voltage and Current Measurements 1020 Summary

Index

1023

Part B

ELECTRIC MACHINES Eyebolt Cast iron frame

Rotor

Bearing cap

Stator Bracket pulley end

Ball bearing

Ball bearing Fan

Keyed shaft

Bearing cap

Grease plug Pan guard Cast iron conduit box

Lead hole opening

Leads

Electromechanical Energy Conversion

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Electromechanical Energy Conversion

31

OBJECTIVES In this chapter you will learn about:   Fundamental laws: Faraday’s law, Fleming’s right-hand and left-hand rule and Lenz’s law   Development of torque and motor action   Induced e.m.f. and generator action   Relation between motor action and generator action  Magnetic structure of d.c. machines Elementary generator and generated voltage    E.m.f. equation   Interaction of magnetic fields of current-carrying conductors   Magnitude of mechanical force   Generator and motor power flow  Conversion process in a machine  Linear and rotary motion  Methods of analysis  Energy balance—action and reaction  Singly excited and doubly excited machines Electric machines

31.1 INTRODUCTION Electrical energy does have an overriding advantage in the sense that it can be transmitted and controlled more easily than most forms of energy. Electricity is most often generated in large installations, transformed to appropriate voltage in units called transformers and carried overland for various distances. Then it is usually retransformed to lower voltages and finally converted to whatever form required. Devices that receive mechanical input energy and convert it to electrical energy, and those devices that r­ eceive electrical e­ nergy and convert it to appropriate mechanical force or motion are called ­electromechanical devices. These broad fields of ­energy conversion are handled at the mechanical-to-electrical end by generators and at the electrical-to-mechanical end by m ­ otors. A major reason that these devices are so very widely used is that they are relatively efficient and controllable; so much so that our entire industrialized civilization is inseparably involved with their use.

31.2  ENERGY CONVERSION DEVICES In motors and generators, the means of coupling between mechanical and electrical energy is through appropriately located and controlled magnetic fields. There are a few known and commercially practical electromechanical energy conversion phenomena that govern the function of devices other than motors and generators. A few of these have been enumerated below: 1. There are electrostatic forces between the plates of a capacitor which require very high voltages for comparatively small forces. 2. Piezoelectric transducer effects are those by which a crystal is deformed and voltages are produced. Here, the motion is exceedingly small even though the forces may be very substantial.

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590  Electrical Technology 3. Magnetostrictive effects are those by which certain magnetic materials change their dimensions minutely under the influence of a magnetic field. Here, as in the piezoelectric transducer, the motion is small—even microscopic—but the force may be large. These three phenomena are useful in instrumentation, communication and control. They are at least substantially reversible in the sense that they work both ways. However, they are not suited to handling either high power or continuous motion. There are other useful energy conversion processes: 4. In case of chemical-to-electrical energy, as in batteries, some cycles are reversible in the sense that energy can be put back in and stored. The processes so far are inefficient or heavy in relation to their power or both. Intensive development is underway. 5. Thermal energy may be directly converted to electrical energy by thermocouples but the voltage is low so the principal use is in measurement of temperature. 6. The entire field of energy release through chemical combustion of gaseous, liquid or solid hydrocarbon fuels is of enormous importance. These phenomena are not reversible in the sense that we cannot usually put in energy and receive back the original fuel. To create a fuel at the present time takes about 15 times the energy that the combustion of that fuel will release. The study of the release of the chemical energy in the form of heat and the conversion of the heat energy by the use of an intermediate fuel into mechanical energy is the science of thermodynamics. 7. In the last generation, the release of heat energy from the fission of heavy elements such as uranium and plutonium has made the whole new important science of atomic energy. There is promise that even more energetic energy release cycles may be perfected and controlled through nuclear fusion. Some practical means may be found whereby the fusion process may produce electricity directly without intervening thermodynamic cycles. The steadily worsening fuel energy, availability coupled with the steadily increasing load, national, and world wide used for usable energy reinforces the need for new fruitful research.

31.3  RELATED FUNDAMENTAL LAWS All rotating electric machines as well as the related special transducers operate on the same principles and obey the same fundamental laws. These are as follows. 1. Faraday’s law of induction 2. Kirchhoff’s voltage and current laws 3. Ampere’s circuital law of the magnetic field 4. Ohm’s law 5. Biot-Savart’ s law of force on a conductor in a magnetic field 6. Watt’s determination of the relation between force, work, time and power φ Lines of magnetic flux

Direction of motion

31.3.1  Faraday’s Law of Induction

When the magnetic flux linking a conductor is changing, an e.m.f. is induced whose magnitude is proportional to the rate of change of flux. Direction of In electrical work, there are many right-hand rules and/ generated S N or left-hand rules. They are convenient ways of remembervoltage and current ing interrelationships where polarities and directions make a + great difference. One of the first of this type is due to Fleming, who related Faraday’s work as follows (see Figure 31.1) If the magnetic field is considered as stationary in space, the conductor then is considered as moving orthogonally Moving conductor across it. The right-hand is extended with the thumb, index finger, and second finger extended at right angles to each other so that they are orthogonally arranged. With this arrangement, the magnetic field is represented by the index finger Figure 31.1  Faraday’s Relationship with the field represented as travelling from the North Pole to the South Pole in the direction that the finger points. If the thumb then is considered as pointing in the direction of motion of the conductor, the middle finger points in direction that conventional current will flow, as illustrated in Figure 31.2.

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Electromechanical Energy Conversion  591

Heinrich Lenz stated (two years after Faraday’s basic discovery) a relationship that turned out to be basic in electromechanical energy transformation. In all cases of electromagnetic induction, an induced voltage will cause a current to flow in a closed circuit in such a direction that the magnetic field caused by that current will oppose the change that produced the current. This relationship is really a form of stating the basic fact of the conservation of energy. It is basic to the operation of inductances, transformers, motors and generators and is illustrated in Figure 31.3. Figure 31.2  Fleming’s Right-hand Rule

,

Figure 31.3  Magnetic Forces on a Conductor

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592  Electrical Technology

31.4  D.C. MOTOR AND GENERATOR MECHANICAL CONSTRUCTION In a practical working machine, a functional arrangement must be used to support a relatively large number of coils in an arrangement that will allow the conductors of the coils to be moved in and out of a strong magnetic field. A dense or concentrated magnetic field is necessary to reach a working voltage in a generator with coils of as few a turns as possible. The fewer turns allow a large wire diameter in the limited available space. This usually means a flux density B sufficiently high so that it nearly saturates the magnetic field structure. This, in turn, means a relatively large and heavy magnetic structure of specially chosen ­magnetic alloys as shown in Figure 31.4. In a practical sense, this means that the bulk of the volume and weight of a motor or generator is composed of a structure that is arranged in such a way that the winding coils can be c­ ontrollably passed into and out of the magnetic flux. There are only a few types of motion that are continuous or cyclically continuous. Unless that motion is made to oscillate in a cyclic fashion, it is a ‘one-shot’ proposition. A straight line motion or an angular motion is both ‘one-shot’ situations. The only exception that gives continuous smooth motion is rotary motion. Rotary motion has no counterpart in nature. It is a man-conceived, mechanically-developed way of doing things. To build an effective magnetic structure that will allow rotary motion, a motor is configured around a cylindrical stack of magnetic alloy discs that are pressed or keyed to a shaft. The rotating magnetic structure or armature is then pierced or slotted to allow room for the coils that are mounted on it.

Figure 31.4  Magnetic Structure (Motor or Generator) The magnetic poles that the conductors are passed by during rotary motion must then face inwards towards the cylindrical surface of the rotating structure. They are, in fact, curved on their inner face to conform—at least in part—to the rotating shape. Since a magnetic line of force must be a closed loop, there must be a return path between the outer ends of the magnetic poles. This function is taken up by the heavy alloy structure of the main frame. This configuration is still incomplete until the end structures or end bells are provided, which serve to support the bearings which, in turn, allow the required rotary motion and yet confine that motion in such a way that no mechanical contact takes place between the armature structure and the field poles. In addition, the bearings are required to confine the axial position of the armature as a whole. Furthermore, the bearings must support any forces due to belts, gears, couplings, or other direct mounted drive or drive mechanisms. The magnetic field is usually provided by a set of multiturn field coils that surround the field poles in the space between the armature and the main frame. These coils are of various types, depending on where their required current comes from. The armature windings that are contained or mounted in the slots in the armature magnetic structure are proportioned particularly according to the relations in Eq. 31.2. However, many coils and turns per coil are provided. There must be sufficient space to accommodate them. The physical size of the armature is determined by the dimensions of the windings it must contain and the total pole flux within it. The windings are interconnected in a few practical configurations, but they always comprise a closed loop or loops of inter-connected coils.

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Electromechanical Energy Conversion  593

31.5  ELEMENTARY GENERATOR At the moment shown in Figure 31.5, each conductor of the single turn coil is moving at right angles to the magnetic field. It can be assumed that the whole gap between the north and south magnetic poles is filled with a straight uniform field. Fleming’s rule shows that the right-hand conductor is generating an electromotive force that causes the current to flow into the page. With the left-hand conductor, the opposite holds. These two opposite-direction voltages are actually connected in series by the back coil connection at the rear and by an external circuit that is connected across the ends of the coil labelled + and –. Back coil Flux

Cut away pole

nt

rre

Cu

S is

+

Ax

–

o

on

N

ati

t f ro

Motion

External connections

Figure 31.5  An Elementary Generator In many diagrams used for illustrative purposes, no such configuration actually exists, because the coil has no support and the magnetic field is across a very substantial pole-to-pole gap in open air. This type of diagram is used because the coils that are actually used cannot be readily seen. They are buried in the armature magnetic structure. After 90° of rotation, as illustrated in Figure 31.6(a), the single-turn coil is now moving parallel to the assumed uniform magnetic field. At this moment, the coil is momentarily not cutting any magnetic linkages and, therefore, not generating any voltage. On the next 90° of motion, the coil reverses the situation, and the induced voltage is of the opposite polarity from the original situation demonstrated in Figure 31.5. Figure 31.6(b) shows an idealized generator case of a uniform field flowing from pole to pole. In this situation, the voltage in any one turn will vary sinusoidally. This generation of an alternating current is true in a practical machine also, even though the waveform is not usually a sine wave since the magnetic field does not flow straight across from a remote pole to another remote pole.

360°

330°

300°

240°

120° 150°

90°

60°

30°

0

180° 210°

e

270°

1

θ

–1 (b)

Figure 31.6  (a) Instantaneous Angular Positions (b) Generated Voltage Versus Angular Position The rectification of this unwanted a.c. voltage is conventionally performed by the commutator. This is simply a rotating mechanical switch composed of insulated segments connected to the ends of the coils. Fixed brushes are arranged to contact the commutator segments as shown in Figure 31.7. A study of Figure 31.7 will show that the bottom brush is always positive in polarity, since it connects to whichever commutator segment is positive at the moment owing to its position and motion in the magnetic field.

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594  Electrical Technology – Brush

+1 S

N

e

(a)

360°

330°

300°

270°

240°

210°

180°

150°

90°

60°

120°

+ Brush

0

30°

0 Commutator

(b)

Figure 31.7 Generated Voltage Versus Angular Positions (a) Commutator and Brushes (b) Generated Voltage vs Angular Position (Same Basis) as Figure 31.6 The result of a commutated single coil elementary generator is then a full-wave rectified alternating current. In a practical d.c. machine, the magnetic field direction is not straight across from field pole to field pole, but radially inward or radially outward at the surface of the armature owing to the magnetic properties of the armature structure. Furthermore, the field will be relatively uniform in strength after an initial entering change from no field to full field as seen by the coil. The result is that the generated voltage per coil is more accurately represented by the flat top waveform shown in Figure 31.8.

Figure 31.8 Generated Voltage Versus Angular Position for a Single Coil (a) Coil and Commutator (b) Generated Voltage vs Angular Position for a Single Coil When more coils are added and spaced uniformly around the armature—as is always the case in a practical machine—there are always a number of coils generating voltage. These voltages are additive owing to the internal series connections of the coils. The result is a uniform direct current with only small voltage variations as coils are switched in and out. It is usual in a d.c. machine to have the field poles so proportioned that about 70 per cent of the outer surface of the armature surface is covered by—and therefore—influenced by the field poles. This is the practical maximum pole area without pole-to-pole flux leakage. For an actual magnetic circuit, it is necessary that there be at least two field poles, and that the poles exist by integer numbers of pairs. Thus, a motor or generator is identified as ‘two-pole’, ‘four-pole’, ‘six-pole’, and so on. Either by having multipoles or through the direct result of winding configuration, there will be some parallel paths in a d.c. armature. Example 31.1 A conductor moving at a velocity of 1.5 m/s has a length of 40 cms as it moves through a uniform field of 1 tesla (Wb/m2). Calculate the voltage induced in the conductor when it moves through the reference field at (1) an angle of 90° (2) an angle of 35° and (3) an angle of 40°. Solution: 1. E = Blv sin qV (31.1) = (1) (0.4) (1.5) sin 90° = 0.6 V

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2.

3.

E = Blv sin qV = 0.6 sin 35° = 0.344 V E = Blv sin qV = Blv sin 120° = 0.52 V

31.6  AVERAGE E.M.F. GENERATED IN A QUARTER REVOLUTION The e.m.f. between the brushes of multicoil armature is produced by numerous coils connected in series, in which each coil consists of many conductors of wire. In order to calculate the resultant e.m.f. between the brushes, it is first necessary to determine the average e.m.f. induced in a single conductor in one quarter of a revolution (i.e., from a position, where there are zero, to a position where there are maximum flux linkages). Let us assume that the total flux produced between the poles consists of f webers and that t is the time required for one quarter of a revolution (i.e., 90 electrical degrees). Since the flux linkages have gone from zero to maximum in one-quarter of a revolution, the average e.m.f. induced in a single turn of two coils during this period is Eav/coil = 4f NS (V/coil)

where, N is the number of turns per coil and S is the relative speed in revolutions per second (rps) between the coil of N turns and the magnetic field.

31.7  FUNDAMENTAL D.C. GENERATOR EQUATION FOR AVERAGE E.M.F. BETWEEN BRUSHES It is possible to calculate the average voltage rating of a single coil (having one or more turns) rotating at a given speed (rps) under a given pole of known field strength; this may be derived in the following manner. If Z is the total number of armature conductors or coil sides and if a is the number of parallel coil paths between brushes of opposite polarity, then the total number of turns N per armature circuit is Z/2a. The total average induced e.m.f. between brushes, therefore, is

φ Zω P volts(V) 2π .α  where, f is the flux per pole in webers (Wb) w is the angular velocity in radians/second (rad/s) Eg =

(31.2)

31.8  BIOT-SAVART RELATIONSHIP The law of the force on a conductor when in a magnetic field, which quantifies the effect in Lenz’s law, is named after Biot and Savart. This law relates the magnetic flux per unit area B, the length of the conductor that is immersed in the magnetic field L (or l), and the current I, which together produce a force in whichever units are consistent.

31.8.1  Motor Action When two or more sources of magnetic fields are arranged in a manner so that their fluxes—or a component of their fluxes—are parallel within a common region, a mechanical force will be produced that tends to either force the sources of flux together or force them apart. A force of repulsion will occur if the two magnetic sources have components of flux that are parallel and in the same direction; this will be indicated by a net increase in flux called flux bunching in the common region. A force of attraction will occur if the respective fluxes have components that are parallel and in opposite directions, this will be indicated by a net subtraction of flux in the common region. This has been represented in Figure 31.9. The use of Biot Savart law is not only confined to a motor but is essential to both a motor as well as a generator. Any conductor that is moving across a magnetic field and at the same time carrying a current will exert a force on its confining structure. This force is termed motor action. The arrangement of the conductors used in a motor or generator to produce

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596  Electrical Technology their generator action results in the conductors being in the influence of the magnetic field. They are also producing a force lateral to that field. The force is obviously proportional to the current in the conductors. The remaining conductors that are carrying the same current but are not in the influence of the magnetic field do not produce any motor action. We need to bear in mind that the conductor that is generating a voltage is also subject to a mechanical force. Both effects are present. Under severe short-circuit conditions, the forces between adjacent conductors can be high enough to physically crush the insulation of transformers, motors and generators, bend bus bars, tear switch boards apart, and cause switches and circuit breakers to come apart with explosive violence. Thus, in those applications, where the available short circuit current is of a magnitude that would cause destruction of apparatus if a fault occurred, special current limiting devices as well as mechanical bracing and conductor support must be installed.

31.8.2  Elementary Two-Pole Motor Figure 31.9 Interaction of Magnetic Fields of Adjacent Current-carrying Conductors (a) Currents in Opposite Directions (b) Currents in the Same Direction end of an arrow that represents the direction of current in conductor A. The dot in the centre of conductor B is the point of an arrow, indicating the direction of current in conductor B. The direction of flux around each conductor is determined by the right-hand rule. The broken lines show the paths of component fluxes, assuming the rotor and stator were energized at different times. The dotted line indicates the direction and path of the resultant flux with both rotor and stator energized at the same time. The net flux on top of conductor A, due to the magnet as also a result of the current in the conductor is additive (bunching), causing a downward mechanical force F. A similar action occurs at the bottom of conductor B, causing an upward mechanical force. The net result is a counter clockwise (CCW) turning movement or torque called motor action.

Figure 31.l0 shows a rotor core, containing two insulated conductors in rotor slots, and the rotor centred between the poles of a stationary magnet (called the stator). The + mark on the end of conductor A is the tail

Figure 31.10  Elementary Two-pole Motor

31.8.3  Magnitude of the Mechanical Force The magnitude of the mechanical force exerted on a straight conductor that is carrying an electric current and situated within and perpendicular to a magnetic field, as illustrated in Figure 31.11(a), is expressed by where,

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F = Bleff .I

(31.3)

F = mechanical force (N) B = flux density of stator field (T) I = current in rotor conductor (A) leff = effective length of rotor conductor (m)

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Figure 31.11 (a) Conductor Carrying Current, Situated within and Perpendicular to the B Field (b) Conductor Skewed b The effective length of a conductor is that component of its length that is immersed in and normal to the magnetic field. Thus, if the conductor is not perpendicular to the magnetic field as shown in Figure 31.11(b), then the effective length of the conductor is expressed by leff = l sin a (31.4) Angle b is called the skewing angle, which may range from 0 to 30 degrees in electrical machines. The direction of the mechanical force exerted on the conductor is determined by flux bunching. β°

31.8.4  Developed Torque Figure 31.12(a) shows a rotor coil made up of a single loop situated in a two-pole stator field of uniform flux density. The effective length of each conductor (coil side) does not include the end connections. The end connections, also called end turns, are used to connect the conductors in series, but because they are not immersed in the magnetic field, they do not develop torque. The distance d between the centre of the shaft and the centre of a conductor is the moment arm. The direction of developed torque may be determined from an end view of the conductors and magnet poles, as seen from the battery end in Figure 31.12(b). The direction of flux due to the known direction of current is determined by the right-hand rule, and the direction of the mechanical force on each conductor is determined by the flux bunching effect. The resultant torque, produced by the two-­conductor couple, is CCW and has a magnitude equal to TD  = 2F.d N.m    TD  = 2Bleff ×Id Nm

End-turn

Window

N

S

d

I

I

Coil side End-turn

+

–

F

(31.5)

31.8.5  Left-Hand Rule of Motor Action Fleming’s right-hand rule relates the direction of magnetic flux, the direction of motion, and the direction of induced voltage polarity. A similar mutually orthogonal relationship exists between the direction of the magnetic field flux, the direction of the imposed voltage and current, and the direction of the resulting mechanical force or motor action. The left-hand rule fits this force relationship if we again let the index finger point in the direction of the flux from north to south, and let the middle finger point in the direction of the

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Shaft

N

+

S

F

Figure 31.12 (a) Single-loop Rotor Coil Carrying an Electric Current, and Situated in a ­Two-pole Field (b) End View of Coil, Showing Direction of Developed Force

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598  Electrical Technology imposed voltage and its resulting conventional current flow. Under these conditions, the thumb will point in the direction of the force that is developed by the Biot-Savart relationship. This has been illustrated in Figure 31.l3. and may be compared Flux with Figure 31.2. t n In the usual situation, both rules are valid. In a e rr Cu generator, the current resulting from the generated Flux voltage flowing through the load circuit will proForce duce a force that opposes the motion that caused t the voltage generation. This opposing force will be n rre Force proportional to the load current and is the reason a Cu generator requires more driving force as its electriFigure 31.13  Left-hand Rule cal load is increased. In motor, the motor torque or turning effect is proportional to the current that is flowing from the electrical power source. At the same time, the motor is generating a voltage according to Fleming’s right hand rule, which opposes the line voltage direction. A motor will then run just fast enough to develop just enough opposing voltage or back e.m.f., that it will limit its current to just enough to supply the required torque. Example 31.2 Each armature winding conductor in a direct current generator is subject to the influence of the magnetic field for 0.254 m of length. If, as a result of generated voltage, a current of 42.5 A flows, how much force is felt by the conductor if the field is 1.581 Wb/m2? Solution: B = 1.581 Wb/m2, I = 42.5 A, l = 0.254 m, F = BIL newtons = (1.581) × (42.5) × (0.254) = 17.1 newtons Example 31.3 A d.c. machine is generating 125 V while delivering 8 A to a load. In its armature circuit; what voltage must be generated internally in the armature? Solution: Vi = Eg – (Ia Ra) Eg = Vi + (Ia Ra)   = 125 + (8 × 1.35) = 135.8 V Example 31.4 Assume that each coil in Figure 31.12(a) has a length of 0.30 m and a skew angle of 15°. The distance between the centre of each conductor and the centre of the shaft is 0.60 m. The combined resistance of the coil and its connections to a 36 V battery is 4.0 W. If the stator field has a uniform flux density of 0.23 T between the poles, determine the magnitude and direction of the developed torque. Solution: From Figure 31.12(b);  a = (90 – b )º = (90 – 15)º = 75º   I = Ebat /R = 36 / 4.0 = 9 A T = 2B.I (l sin a).d = 2' × 0.23 × 9(0.3 sin75º) × 0.60 = 0.72 N/m The direction of the developed torque is counter clockwise.

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31.9  MOTOR ACTION VERSUS GENERATOR ACTION If, whenever motor action occurs, generator action is also developed, the question can be raised whether the converse is also true. Generator action is shown in Figure 31.14(b), where a mechanical force moves a conductor in an upward direction inducing an e.m.f. in the direction shown. When a current flows as a result of this e.m.f., there is a currentcarrying conductor existing in a magnetic field; hence, the result is motor action. Shown as a dashed line in Figure 31.14(b), the force developed as a result of motor action opposes the motion that produced it. It may then be stated categorically that generator action and motor action occur simultaneously in all rotating machinery. Hence, the same dynamo may be operated either as a motor or a generator or both. Note: The dynamotor is a typical d.c. dynamo that combines motor and generator action in the armature conductors. Correspondingly, the synchronous converter is a typical a.c. dynamo that performs the same function. A more graphic representation in terms of rotational elements is presented in Figure 31.14, which compares the elementary motor and generator for the same direction of rotation and shows the electric circuits of each. It is the key to an understanding of electromechanical energy conversion. Given the direction of applied voltage and current shown in Figure 31.14(a), the motor action that results produces a clockwise rotational force on both conductors. The direction of induced counter e.m.f. is also shown as opposed to the applied voltage, both in Figure 31.14(a) and in the motor circuit of Figure 31.14(c). It should be observed that in order for the current to produce clockwise rotation and to have the direction shown in Figure 31.14(c), it is necessary that the applied armature terminal voltage Va must be greater than the developed counter e.m.f. Ec. Direction of applied voltage and current Direction of rotation and torque produced by current

Direction of induced voltage and current

N

S

N

S

Direction of induced e.m.f.

Retarding torque produced by current (a)

Ia

Va

+

Direction of armature rotation and driving torque

M

(b)

Va = Ec + IcRc + I a Ec φ Ra

Va > Ec (c)

Driving torque by motor action

Ia

Eg = Va + IaRa + Ia Eg φ Ra

Retarding torque Load

Va

G

Driving torque > retarding torque

Driving M torque

(d)

Figure 31.14 Elementary Motor Action Versus Generator Action (a) Elementary Motor (b) Elementary Generator (c) Motor Circuit (d) Generator Circuit

31.10  POWER FLOW DIAGRAMS A clean picture of the dynamo, operating as either a motor or a generator is shown by the combined power flow diagram in Figure 31.l5. On the left side of the diagram is mechanical power, and on the right side is electric power.

31.10.1  Generator Power Flow If mechanical power is applied to the shaft of a dynamo as the input, the power at the shaft is TS/5252 hp. A dynamo, driven mechanically as a generator, sustains certain rotational losses. The difference between the rotational losses and the input mechanical power represents the net mechanical power converted to electric power by electromechanical conversion (EgIa). But the generator also sustains some internal electric losses, which subtract from the electric power developed. The net electric power output is, therefore, EgIa minus the electric losses, or the terminal voltage twices the total current delivered to the load (Vt It, shown at the right side in Figure 31.15). To summarize, as shown is Figure 31.15, for a dynamo

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600  Electrical Technology operating as a generator or as an alternator: The generated E or Eg exceeds the armature terminal voltage Va applied across the load. Electric output power Developed electric power

Mechanical Rotational losses power input + Electric losses  =EgIa= Mechanical Power input – Rotational losses =Electric output power + Electric losses

(31.6) (31.7) (31.8)

Figure 31.15  Combined Power Flow Diagram for Motor or Generator Action

31.10.2  Motor Power Flow Electric power applied to the terminals of a motor (VtIt), shown at the right of Figure 31.15, is immediately reduced by certain electric losses within the motor. The difference appears in the form of internal electric power (EcIa) which is converted to mechanical power by electromechanical conversion. The mechanical power available, which has been produced by internal motor torque, (EcIa//746) sustains some internal mechanical losses. The difference between these mechanical losses and the mechanical power produced as a result of electromechanical conversion is the mechanical output power. In summary, as shown in Figure 31.15, for a dynamo operating as a motor; Mechanical output power

Electric     Electric losses power input + Rotational losses

Developed = Ecla = Electric power mechanical power input Mechanical + Rotational = output power losses

(Electric losses)

(31.9)

(31.10) (31.11)

So, the dynamo is in reality, really very simple and straight forward, as shown in Figure 31.l5. The area of mechanical power is on the left side of the dashed vertical lines, and the area of electric power is on the right side of the lines. The centre area is represented by the change of energy state, or electromechanical conversion (since energy can be neither created nor destroyed) where no loss occurs. Putting mechanical power into a dynamo involves mechanical power loss, change of state, electric power loss and electric output. Putting electric power into a dynamo involves electric loss, change of energy state, mechanical power loss and mechanical output. Example 31.5 A 10 kW, 230 V, 1750 rpm, shunt generator was run light (1750 rpm) as a motor to determine its rotational losses at its rated load. The applied voltage across the armature (Va) computed for the test was 245 V, and the armature current drawn was 2 A. The field resistance of the armature was 230 W, and the armature circuit resistance measured 0.2 W. Calculate (1) Rotational losses at full load; (2) full load armature circuit loss and field loss; and (3) efficiency of the generator at rated load.

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Solution: 1. Rotational loss   = Va Ia – I 2a  Ra    = (245 × 2) – (22 × 0.2)    2. At the rated load,

= 490 – 0.8 = 489.2 W

I= L



P 10, 000 W = = 43.5 A Vt 230 V

I a = It + I L =



Full load armature loss = I

2 a

230 V + 43.5 = 44.5 A 230 W

R a = (44.5) 2 0.2 = 396 W

Field loss = Vf × If = 230 V' 1 A = 230 W 3. Efficiency of a generator at any load



=

output at that load output at that load + rotational loss + copper loss s at rated load

=

rated output rated output + rotational loss + copperr loss at rated load

Efficiency at full load =

10000 6 + 230)’100 10000 + 489.2 + (396

= 90.0 per cent Example 31.6 An armature consists of 40 coils and each coil has 20 turns. When the armature is rotated at 200 rad/s in a four-pole field structure having a flux of 5 m Wb/pole, and there are four paths in the armature. Calculate (1) the number of conductors and (2) the voltage between brushes generated by the armature. Solution: 1. Z = (40 coils) × (2 conductors/turn) (20 turns/coil) = 1600 conductors 2. φ Zω P Eg = 2π a =

(5 × 10−3 )(1600)(200)(4) 2π × 4

= 254.6 V (between brushess)

31.11  CONVERSION PROCESS IN A MACHINE Much electrical energy is used to provide simple heat, either for cooking or for heating a room. Some heat conversion is used to produce light as in the common light bulb, but a lot of electrical energy is used to drive machines. All electrical machines operate on a common set of principles. These principles apply whether the device operates with alternating current or with direct current. The most simple, so far as construction is concerned, involve linear movement; these include the relay and the contactor. The former was by far the most common electrical machine of all, but now is being replaced by the electronic switch which has no moving parts. The contactor remains in common use for switching power circuits on and off. Simply made rotating machines have few applications. Most rotating machines are complicated to create. An electrical machine is one that links an electrical energy system to another energy system by providing a reversible means of energy flow in its magnetic field. The magnetic field is the coupling between the two systems and is the neutral link. The energy transferred from one system to the other is temporarily stored in the system and then released to the other system. Usually, the energy system coupled to the electrical energy is a mechanical one. Not all machines deal with large amounts of

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602  Electrical Technology e­ nergy—those operating at very low power levels are often termed transducers, particularly when ­providing signals with which to activate electronic control devices. In a mechanical energy system, a mechanical force is associated with the displacement of its point of action. An electromagnetic system can develop a mechanical force in two ways: (1) by alignment; and (2) by interaction. The force of alignment can be illustrated by the arrangements shown in Figure 31.16. In Figure 31.16(a) two poles are situated opposite one another; each is made of a ferromagnetic material and a flux passes from the one to the other. The surfaces through which the flux passes are said to be magnetized surfaces and they are attracted towards one another, as indicated in Figure 31.16(a). The force of alignment acts in any direction that will increase the magnetic energy stored in the ­arrangement. In the first case, it will try to bring the two poles together since this decreases the reluctance of the air gap in the magnetic circuit and, hence, will increase the flux and, consequently, the stored energy. This principle of increasFigure 31.16 Force of Alignment ing the stored magnetic energy is the most important one and is the (a) Force of Attraction key to machine theory. (b) Lateral Force of Alignment In the second case, as shown in Figure 31.16(b), the poles are not situated opposite one another. The resultant force tries to achieve stored magnetic energy by two component actions: (1) by attraction of the poles towards one another; and (2) by aligning the poles laterally. If the poles move laterally, the cross-sectional area of the air-gap is increased and the reluctance is reduced with a consequent increase in the stored magnetic energy, as before. Both actions attempt to align the poles to the point of maximum stored energy, i.e., when the poles are in contact with a maximum area of contact. The force of alignment necessarily acts in the direction of lines of flux.

31.12  LINEAR AND ROTARY MOTION Many devices demonstrate the principle of the force of alignment. The most familiar is that of the permanent magnet and its attraction for ferromagnetic materials. Electromagnetic devices, such as the relay (Figure 31.17), demonstrate the force of alignment giving rise to linear motion. When the coil is energized, a flux is set up in the relay core and the air gap. The surfaces adjacent to the air-gap become magnetized and are attracted, hence putting the armature plate in the direction indicated. The relay’s function is to operate switches and this is used extensively in telephone exchanges.

Figure 31.17  The Electromagnetic Relay-force of Alignment Giving Rise to Linear Motion

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The force of alignment can also be used to produce rotary motion, as in the reluctance motor shown in Figure 31.18. In this case, the rotating piece, termed the rotor, experiences radial forces in opposite directions, thereby cancelling each other out. The rotor also experiences torque due to the magnetized rotor and pole surfaces, attempting to align themselves. This alignment torque occurs in any rotating machine which does not have a cylindrical rotor, i.e., in a rotor that is salient, as in the case shown in Figure 31.18. The force of interaction has the advantage of simplicity in its application. To calculate or even to estimate the energy stored in the magnetic fields of many arrangements is difficult, if not impossible. Many of these cases, however, can be dealt with by the relationship F = Bli. This also includes the case of a beam of electrons being deflected by a magnetic field in, say, a cathode ray tube. There are numerous applications involving the force of interaction to give rise to rotary motion. These include the synchronous and induc- Figure 31.18 Reluctance Motor-force of tion m ­ achine as well as commutating machines. All are variations of the Alignment Giving Rise to same theme, each having a different characteristic suitable for the various ­Rotary Motion industrial drives required. A simple machine illustrating the principle ­involved is shown in N Figure 31.19. By passing a current through the coil, it experiences a force on each of the coil sides resulting in a torque about the axis of rotation. A practical machine requires many conductors in order to develop a sufficient torque, depending on the manner in which the conductors are arranged, and the various machines are created. Each is a system linking device. At one end is the electrical system; at the other end is the mechanical system. In between, there is a magnetic field forming a two S way-link between them. If there is to be a flow of energy, all three will be involved simultaneously. Figure 31.19 Rotary Machine Illustrating There are three methods of approach to analyze the energy conversion Force of Interaction process, each of which has to allow for the imperfections of a machine. No machine gives out as much energy as it takes in. The difference is termed the losses—the losses in the electrical system, the mechanical system and the magnetic system. Note: The reaction in the electrical system—apart from the flow of current—is the introduction of an e.m.f. into that system; the product of e.m.f. and current gives the rate of electrical energy conversion.

31.13  METHODS OF ANALYSIS The basic energy conversion procession involves the magnetic coupling field and its action and reaction on the electrical and mechanical systems. There are three methods of approach: (1) the classical approach; (2) the generalized machine approach; and (3) transient operation (not dealt with here). The classical approach is based on the assumption that the operation of a machine can be predicted from a study of the machine losses. Such an approach is simple, but it has two major disadvantages. First, it deals almost exclusively with the machine operating under steady-state conditions and thus transient response conditions are virtually ignored, i.e., when it is accelerating or decelerating. Second, the losses of each machine are different. As a consequence, each type of machine requires to be separately analyzed. Such a process is both tedious and repetitive but it is simple. The generalized approach depends on a full analysis of the coupling field as observed from the terminals of the machine windings. The losses are recognized as necessary digressions from the main line of the analysis. The coupling field is described in terms of mutual inductance. Such an approach, therefore, considers any machine to be merely an arrangement of coils, which are magnetically linked. No attention is paid as to the form of the machine initially and it exists as a box with terminals from which measurements may be made to determine the performance of the machine. The measured quantities are voltage, current, power, frequency, torque and rotational speed, from which may be derived the resistance and inductance values for the coils. In the light of the derived values, it is possible to analyze the performance, both under steady state as well as transient conditions. The generalized approach has been universally appreciated, since it has so much to offer. It has taken a long time (since the late 1950s) to attain recognition. The difficulty that produced its earlier use was the complexity of the mathematics

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604  Electrical Technology involved, but once the mathematical manipulations have been carried out, they need not be repeated. The results can be separately modified to analyze not only most types of machines but also the different modes of operation. The individual machine is considered by making simplifying assumptions at the end of the analysis, but the effect of the theory is to concentrate attention upon the properties common to all machines.

31.14  ENERGY BALANCE Let us consider the operation of a simple attracted-armature relay, such as the one shown in Figure 31.20. We may assume that initially the switch is open and that there is no stored field energy. These conditions are quite normal. After the switch is closed, the sequence of events falls into four distinct groups in the following manner.

Figure 31.20  Attracted-armature Relay 1. After the switch is closed, the current rises exponentially. If L1 is the inductance of the coil for the initial position of the armature, the initial rate of rise of current is given by V/L1. The electrical ­energy from the source is partly dissipated as i2R losses in the magnetizing coil while the remainder is converted into stored energy in the magnetic field. During this period, the armature experiences an attractive force, but the various mechanical restraints prevent it from moving. 2. At some appropriate value of current, the armature begins to move. This occurs when the force of attraction fE balances the mechanical force fM. During the motion of the armature, there are many changes of energy in the system. On the mechanical side, energy is required to stretch the spring, drive the external load and to supply the kinetic energy required by the moving parts. At the same time, the air gaps are being reduced with a consequent increase in the inductance of the arrangement. This causes a reaction in the electrical system in the form of an induced e.m.f. This induced e.m.f. tends to reduce the coil current and also permits the conversion of electrical energy. It is the reaction to the action. 3.  The armature cannot continue to move indefinitely but, instead, it hits an end stop. This causes the kinetic energy of the system to be dissipated as noise, deformation of the poles and vibration. 4. Now that there is no further motion of the system, the inductance becomes constant at a new higher value L2. The current increases exponentially to a value V/r. The rate of rise is less than the initial rate of rise since the inductance is now much greater. The energy flow processes are, therefore, many and yet they are typical of many machines; in rotating machines there is no sudden stop, but otherwise the processes are similar. To handle so many at one time, it is necessary to set up an energy balance convention. Since the conversion process can take place in either direction, let the energies be input energies to the system, WE and WM, respectively. In the internal system, let the stored magnetic field energy be Wf, the stored mechanical energy be Ws, and the nonuseful thermal energy, (due to i2R loss, friction, etc.), which is a loss, be Wl. The arrangement is illustrated in Figure 31.21. Between any two states of the system, the energy balance (Figure 31.22) may be expressed as Figure 31.21  Energy Balance Diagram

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WE +WM =W f +Ws +Wl



(31.12)

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Electromechanical Energy Conversion  605

Alternatively, the energy rates of flow may be expressed as dW f dWs dWl p E + pM = + + dt dt dt 

∆WE (pE = ei)

(31.13)

Increasing

∆Wf

∆WM (pM = −fEu)

Figure 31.22  Idealized Energy Balance Diagram It follows that the ideal and essential energy balance may be expressed as d WE + d WM = d Wf and, hence, the power balance may be expressed as

pE + pM =

dW f dt 

(31.14)

(31.15)

The actions and reactions are indicated in Figure 31.23. On the electrical side, the applied voltage is v and this is opposed by the reaction in the form of back e.m.f. The electrical power is pE = vi while the rate of conversion is ei. These two terms are equal only when the i 2R loss is either neglected or considered external to the conversion process, as in the idealized system of Figure 31.23 Actions and Reactions in a Figure 31.22. Practical Conversion System On the mechanical side, the mechanical input force fM acts towards the conversion system and moves in a similar direction, say, with a velocity u. The reaction to this is the magnetically developed force fE. These two forces are equal and opposite only when the mechanical system is at rest or moves with uniform velocity. The difference would otherwise give rise to acceleration and, hence, there could not be steady state conditions. There is also a slight difference between the forces when the mechanical system is moving due to friction. Example 31.7 An electromagnet is made using a horseshoe core as shown in Figure 31.24. The core has an effective length of 600 mm and a cross-sectional area of 500 mm. A rectangular block of steel is held by the electromagnets force of alignment and a force of 20 N is required to free it. The magnetic circuit through the block is 200 mm long and the effective cross-sectional area is again 500 mm2. The relative permeability of both core and block is 700. If the magnet is energized by a coil of 100 turns, estimate the coil current. Solution: There are two air gaps in the magnetic circuit; hence, the force to part the circuit is double that at any one air-gap. f m = 2.

B2 A B2 A = = 20 N 2 µ0 µ0

 20 × 4π × 10−7  B=  = 0.222 T −6  500 × 10  H =

B 0.222 = = 250 At /m µ0 µ r 4π × 10−7 × 700

F = Hl = 250 × (600 + 200) × 10−3 = 200 At = Ns = 1001 I =

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200 = 2.0 A 100

Figure 31.24  For Example 31.7

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31.15  ROTARY MOTION The principles of the linear machine also apply to the rotating machine by replacing x by λ (the angle of rotational distortion) and u by w cut the angular velocity of the rotor. Angular velocity → Symbol: wr Unit: radians per second (rad/s) Angular speed symbol: n or mn Unit: revolutions per second (r/s) The torque of a rotating machine is given by M E =

dWf dλ

Note: M rather than T is used to symbolize the torque of a rotating machine.

31.16  DOUBLY EXCITED ROTATING MACHINES The reluctance machine and also the relays are singly excited systems, i.e., only the stator or the rotor is excited by a current-carrying coil. In either case, the motion is caused by a movable part changing its position to reduce in reluctance of the magnetic circuit. Due to the physical construction in the rotary case, the axis of the rotor tries to align itself with the axis of the field. In order to strengthen the attraction towards alignment, both the rotor and the stator can be ­excited. A simple arrangement is shown in Figure 31.25. This is called a doubly excited system. With this arrangement, the stator and the rotor each have two magnetic poles. Such a machine is described as a two-pole machine. Machines can be made with greater number of poles. The windings and the magnetic circuit components give rise to clearly defined rotor and stator fields. By the symmetry of the fields, each has an axis. The axes indicate the directions of the mean magnetizing forces across the airgaps for each of the fields. Since the gap lengths are constant, the field axis can have complexor properties ascribed to them, their magnitudes being related to the respective magnetomotive force (m.m.f.s). It is now possible to describe the torque as being created by the m.m.f. axes trying to align themselves. The axes are functions of the Figure 31.25 Simple Doubly Excited Rotary coil construction and, hence, became less dependent on the construcSystem tion of the core members, which only serve to distribute the field although they are, of course, required to ­ensure that a sufficient field strength is available, and it follows that specially shaped rotors are no longer r­ equired. The total flux in the air gap is the result of the combination of the field due to the rotor winding as well as due to the stator winding. The torque is created by the desire for the alignment of the fields. There are three important families of doubly excited rotating machines. 1. Synchronous machines

Stator flux Rotor flux

— —

alternating current direct current

2. Asynchronous machines

Stator flux Rotor flux

— —

alternating current alternating current



3. Commutator machines

Stator flux — Rotor flux

direct current direct current

S UM M A RY 1.  Electrical energy can be transmitted and controlled more easily. 2.  Electricity is generated in large installations, transformed to appropriate voltage in transformers, carried overhead for various distances, and then it is usually

M31_AUTH_ISBN_C31.indd 606

retransformed to lower voltages and converted to whatever form required. 3.  The means of coupling between mechanical and electrical energy is through appropriately located and controlled magnetic fields.

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Electromechanical Energy Conversion  607

4. When the magnetic flux linking a conductor is changing an e.m.f. is induced whose magnitude is proportional to rate of change of flux. 5. Right-hand and left-hand rules are convenient ways of remembering interrelationships where ­ polarities and directions make a great difference. 6. In all cases of electromagnetic induction, an induced voltage will cause a current to flow in a closed circuit in such a direction that the magnetic field which is caused by the current will oppose the change that produced the current. 7. The bulk of the volume and weight of a motor or generator is composed of a structure that is arranged in such a way that the winding coils can be controllably passed into and out of the magnetic flux. 8. Rotary motion is a man-conceived mechanically developed way of doing things.

9. The armature is slotted to allow room for the coils that are mounted on it. 10. The magnetic poles are curved on their inner face to conform to the rotary shaft. 11. The armature windings always comprise a closed loop or loops of interconnected coils. 12. The commutator is a rotating mechanical switch composed of insulated segments connected to the ends of the coils. 13. Any conductor that is moving across a magnetic field and at the same time carrying a current will exert a force on its confining structure. This force is termed motor action. 14. The effective length of a conductor is that particular component of its length that is immersed in and normal to the magnetic field. 15. Generator action and its counterpart motor action occur simultaneously in all rotating machinery.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. The right-hand motor rule is usually used to determine (a) Flux density (b)  Flux direction (c) Direction of conductor movement (d) Induced current

2. Torque on a single loop of wire in a magnetic field is

(a) (b) (c) (d)

Constant The same as field flux The turning force Never at a maximum value

3. The amount of torque on a current-carrying conductor in a magnetic field depends on

(a) The amount of current in the conductor (b) The current direction in the conductor (c) The direction of rotation (d) The direction of magnetic field between the two poles

4. To obtain motor action, current is supplied to a loop of wire in a magnetic field by (a) Slip rings (b)  Brushes (c) Split rings d) Brushes and a commutator

5. The principle of motor action is

(a) A conversion of mechanical energy to electrical energy (b) A conversion of electrical energy to mechanical energy (c) An unpredictable phenomenon (d) A conversion of chemical energy to mechanical energy

6. The generator effect in a motor produces a

(a) (b) (c) (d)

High power factor Counter electromotive force High resistance Reduced line voltage

ANSWERS (MCQ) 1. (b)  2. (c)  3. (a)  4. (d)  5. (b)  6. (b).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. What is a reversible energy conversion process? 2. When is motor action present in a generator? 3. When a unit is acting as a generator, in which direction does the motor action take place?

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4. Explain why all electric machines, when operating, develop torque and generator voltage at the same time. 5.  Explain four electromechanical energy conversion effects.

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32

D.c. Generators OBJECTIVES In this chapter you will learn about:    Distinguish between the functions of a motor and generator    Distinguish between lap and wave windings    Understand shunt, series, and compound generators and armature winding    Describe types of d.c. generators and their characteristics    Calculate generated e.m.f. for a generator using E = V + IaRa    Understand current and voltage relations in the three types of generators    Understand the need for, and requirements of paralleling generators    Solve simple problems using the above relations

(c) Cumulative Compound

(c) Differential Compound

(c) Interpole Compound Compound generators

32.1 INTRODUCTION A dynamo is a rotating electric machine which is capable of converting electrical energy into mechanical energy when operating in the motor mode, and mechanical energy into electrical energy when operating in the generator mode. For the generator, rotary motion is supplied by a prime mover (a source of mechanical energy) in order to produce relative motion between the conductors of the armature and the magnetic field of the dynamo in order to generate electrical energy. For the motor, electrical energy is supplied to the conductors and the magnetic field winding of the dynamo as well in order to produce an electromagnetic force between them and thus produce mechanical energy. This gives rise to a number of interesting possibilities and choices in determining which shall be the rotor (the part of the dynamo that rotates) and which shall be the stator (the part of the dynamo that is stationary). The various types of dynamo possibilities are as follows. 1. The direct current (d.c.) dynamo which has a rotating armature and a stationary field. 2. The synchronous (a.c.) dynamo with a rotating armature and a stationary field.

D.c. Generators  609

3. The synchronous (a.c.) dynamo with a rotating field and a stationary armature. 4. The asynchronous (a.c.) dynamo which has both stationary and rotating armature windings. Direct current generators, however—once the mainstay of electric power for large and small industrial plants—are being increasingly replaced by solid-state devices that convert available a.c. to d.c. for d.c.-drive system and other d.c. applications.

32.2  MECHANICAL CONSTRUCTION (Refer to Figure 32.1). The mainframe is required to serve as a return path for all the circulating magnetic flux that passes from the field poles to the armature. This flux-carrying requirement determines the needed cross-section of magnetic material, usually carbon steel. As a result of the needed flux-carrying ability, there is usually far more metal than is needed for structural strength requirements. The most usual structure is a rolled ring one with the end closure butt welded in automatic machinery which, in turn, leaves a visible weld. This ring structure must be lathe turned so that it’s inside surface is a true cylinder and so that its ends are square with the bore. These steps are required to maintain the necessary geometric dimensions so that the finished machine will assemble without unnecessary adjustment and also so that the field pole shoes will fit concentrically around the armature. Some form of mounting feet is usually welded into this frame structure, as can be seen in Figure 32.2 (a representative mainframe unit). The alternative mounting is to use the end bells for bolting surfaces. This structure is seen at its simplest in medium sizes in automobile starting motors where the butt weld shows clearly. These bolts may appear as regularly spaced hexagon heads around the middle of the frame. Where space is critical, these fastenings may be countersunk. Again, this counter sinking is used on an automobile starter because, quite literally, external bolt heads would prevent mounting the unit as closely as is required. Typical line of magnetic’ flux

Main frame

N

Field pole

re atu m r c A eti gn ture a M uc str

Typical slot to allow for armature winding coil space needs (one of many)

aft

Sh

S Air gap between face of pole and armature surface

Field pole

Figure 32.1 Magnetic Structure (Motor or Generator) Repeated here for Convenience



Figure 32.2 Representative d.c. Mainframe Unit

There are other constructions seen on d.c. machine mainframes, depending upon the make, size and application. Some units are of cast iron or cast steel where the field poles and mounting feet are integral. This makes a very neat unit but the machining required largely cancels any advantages. Cast iron construction limits the flux density that can be achieved and is largely obsolete. On very large units, the mainframe is split into an upper and lower half with a bolted flange joint on the horizontal centreline. This construction appears when the armature is too large and heavy to insert without a hoist. On the largest sizes, the field poles and field coils also require crane handling and, therefore, a separable structure serves two main purposes.

610  Electrical Technology Small to medium units may have their mainframe structure of punched laminations assembled in stacks. This structure makes an integral unit of the frame and the field poles. It can produce an excellent and very sound structure but the punching die costs are high. Figure 32.3 illustrates a punched mainframe construction. Field poles are usually, but not always, made of thin laminations of highly magnetic steel alloys. Laminated construction is necessary on the inner or pole shoe end of the field poles. This is because of the pulsations of field strength that result when the notched armature rotor magnetic structure passes the pole shoe. Variations in field strength result in Figure 32.3  Punched Mainframe Construction internal eddy currents being generated in the magnetic structure. These eddy currents are losses. They may be largely prevented by having laminated magnetic structures. Laminated structures allow magnetic flux to pass along the length of the laminations but do not allow electric eddy currents to pass across the structure from one lamination to another. The assembled stack of laminations is held t­ ogether as a unit by appropriately placed rivets. The outer end of the laminated pole is curved to fit very closely into the inner surface of the mainframe. A typical laminated field pole and pole shoe are shown in Figure 32.4. Any break, whatever, in a magnetic structure, causes significant reluctance which is roughly analogous to resistance, so that more ampere turns are necessary to make a given magnetic flux flow in the structure as a whole. More ampere turns mean more heat, which is a loss, so that the pole to mainframe joint is usually quite tightly clamped by the field pole Figure 32.4 A Typical Laminated Field mounting bolts. Pole and Pole Shoe

32.3  ARMATURE STRUCTURE The armature structure serves a dual purpose in the sense that it is the support for the winding conductors that pass through the magnetic field and is also a substantial portion of the magnetic flux circuit. Since any part of the armature magnetic structure sees cyclic reversals in magnetic flux direction, it is potentially subject to more severe eddy current losses than the field pole shoes and is, therefore, invariably laminated. The usual construction from the very smallest sizes up into the integral horse power or kilowatt range is a stack of discs of magnetic alloy steel. These discs are notched or perforated in the outer periphery to accommodate and support the armature windings. The required stack size is held as a unit by appropriate rivets, which are parallel to the shaft. The rivets themselves are a path for eddy currents and are a short-circuit path for generated voltages, so they must be insulated. Either that or their function must be taken by adhesively bonding the lamination stack by using insulating varnish. However, the stack is originally held; the windings themselves reinforce the unit as a whole. On the larger sizes the joint ­between the laminations and shaft is keyed in order to transmit the torque forces. On smaller units, a press fit is sufficient. The very large units are complicated by the need to keep the lamination sizes within the range of reasonable punch and die tooling. In these cases, the core of the armature is called a spider. It is a structural member that fills the radial space between the shaft and the built-up lamination stack, which now takes the form of a magnetic ring. Practical considerations limit the size of armature laminations to a few modular sizes, and also limit the range of the number of available winding slots in any one diameter. Slots Commutator The armature winding coils are placed in these slots Shaft (Figure 32.5) in various arrangements. However, these coils are configured electrically; they must be so arranged mechanically that they may be mounted in lamination slots. This mounting must be mechanically secure against centrifugal and torque forces and must be adequately insulated and also allow the Bearing Armature simplest possible assembly. In the larger sizes, these coils become very difficult to form, handle and assemble in position. Figure 32.5  The Armature

D.c. Generators  611

32.4  COMMUTATOR AND BRUSHES The basic purpose of commutators is to change the alternating current to direct current. The windings terminate in the rotary switching unit known as the commutator. This unit is almost always made of wedgeshaped segments of hard-drawn copper. The copper segments are insulated from each other as well as from their end clamps by strips of mica. No better material has been found. The requirements of low resistance, excellent conductivity, and good wear resistance are conflicting. Figure 32.6 shows a typical armature rotor stack and commutator complete with coils, fan and bearings. A practical commutator does not switch just one coil but operates with many. These coils are connected in series to form a winding and the winding is a continuous one tapped at regular intervals by connections made to a commutator with many segments, as shown in Figures 32.7 and 32.8. The arrangement is called a commutator winding. The ­advantages of this form of connection are that there are now many coils contributing to the system, thereby making it bigger and more effective, and only two of the coils can be short circuited at any instant by the brushes, leaving the large majority of the coils unaffected.

Figure 32.6 Direct Current Armature Complete with Coils, Commutator Fan and Bearings

Coils in slots Commutator segment construction

Wedge Slot liner Main insulation Cross-section of coils in slot

Figure 32.7 A Simple Commutator Winding and Winding Construction Details

Separator Conductor

Figure 32.8  Commutator Segment

The switching function is shared by the commutator and the brushes. These brushes are made of a carbon or carbon graphite or a copper-filled carbon mixture. Here, low contact resistance, some controlled internal resistance and good wearing qualities are required. The actual contact surface is between the brush and a copper oxide surface on the commutator. If the rubbing friction and electrical sparking are extreme, the copper oxide is not renewed as fast as it wears and the result is poor life. Good life is achieved under proper design conditions. The commutator assembly is pressed or pressed and keyed to the shaft at a predetermined distance from the lamination stack. An armature is electrically completed by attaching the appropriate winding coil ends to the commutator segments. The joint is made with solder or even hightemperature bronze for heavy duty. Finally, the whole assembly is strapped for strength to resist centrifugal stresses, dipped in insulating material and baked, balanced and equipped with a fan for cooling.

612  Electrical Technology

32.5  ARMATURE WINDINGS The working part of a motor or generator, whether alternating or direct current, constitutes the armature windings. These are where the voltage is generated and where the force that results in turning effort or motor action is developed. The field windings serve to produce the magnetic field that is required; they carry only from 2 to 10 per cent of the current in the machine if they are shunt windings. Similarly, if the field is series wound, it will carry full armature current but will have only a few per cent of the voltage that is present, across the armature. Either way, the armature windings have a much larger wattage and are a more critical part of the design. Only two basic configurations of windings are used—lap winding and wave winding. Some larger machines use a combination of these two basic types known as frog-leg winding, because of the appearance of the coil before it is installed. There are further subdivisions of each type, having relation to the number of conductors that are brought along in parallel such that a winding may be said to be simplex if a single conductor, duplex if doubled, and triplex if tripled, and so on. An understanding of these winding configurations is necessary to determine the number of paths that exist as also to identify the currents that each conductor is required to carry. The armature windings—whether on the rotor or stator—are always of the non-salient type and are distributed equally in slots adjacent to the air gap around the periphery of the armature. Essentially, there are two kinds, depending on the type of closure or re-entrance of the winding: (1) closed-circuit windings used in d.c. dynamos and (2) open-circuit windings used in a.c. dynamos. Regardless of the type or application, most armature windings consist of diamond-shaped preformed coils as shown in Figure 32.9 (a), which are inserted into the armature slots and connected in a manner to produce a complete winding. Each coil consists of many turns of fine linen-covered, cotton-covered, or enamel-covered wire, individual taped, lacquerdipped, and insulated from the armature slot. The number of conductors (Z) in any given coil will be twice the number of turns making up the coil, i.e., two conductors per turn. In general, armature coils span 180 degrees, i.e., from the center of a given pole to the center of a pole of opposite polarity, which is physically adjacent, as shown in Figure [32.9 (b) and (c)]. If a coil covers a span of 180 electric degrees, it is called a full-pitch coil, whereas one that spans less than 180 degrees is called a fractional pitch coil. An armature wound with a fractional pitch is called a chorded winding. Chorded windings require the use of less copper than full-pitch coils

Figure 32.9 Types of d.c. Armature Coils and End Connections (a) Preformed Armature Coil (b) Lap-wound Coil (c) Wave-wound Coil (d) Two-layer Windings

D.c. Generators  613

but they have approximately the same characteristics because the shorter front-end and back-end are inactive. A coil that spans 150 electrical degrees would have a pitch factor p of 150°/180° = 8.33 or 83.3 per cent. In general pitch factor of less than 80 percent are avoided. Most windings are two-large windings, i.e., two coil sides are inserted in each slot. In winding a two-layer armature, as has been shown in Figure 32.9 (d), one coil side p is placed in the bottom of a slot and the other side is not. The second coil side is not inserted until all the other armature coils have been inserted in the bottom slots. When coil side x has been inserted, only then is coil side 1 inserted; when coil side y has been inserted in a bottom slot, only then is coil side 2 inserted, and so on. The purpose of this procedure is to assume both strength against centrifugal forces and nearly perfect equality in the size, shape and weight of all the coils. The same purpose behind all the various winding configurations to take a conductor path from one polarity of brush via a commutator segment, up through the magnetic field, around the back end of the armature structure, back the opposite way through the opposite field, and finally back to another commutator segment. This process is repeated around the commutator and the armature lamination stack until the winding is complete, symmetrical and balanced. Certain requirements must be carried out. 1. Every coil must be of such a configuration that when seen as going away from the commutator, it passes through the influence of one field polarity and when coming back it passes through the influence of the other field polarity. In this way, the voltages generated are additive. The coils may be of one or many turns (see Figure 32.9). 2. The coils must be interconnected at the commutator in such a way that all the conductors under the influence of one magnetic pole are connected so that their incremental voltages are additive. 3. The whole winding must be configured in such a way that regardless of I the angular position of the armature, the commutator-brush connection relation to the field pole influence is maintained in the same sense. V 4. The coil shapes and end configuration must make maximum use of the Ra copper so as to minimize resistance loss, cost and weight. 5. The windings should be as strong as possible and braced and/or supported E where required. Copper is a superior electrical conductor, but it is a poor structural material for a highly stressed location such as a rotating armature winding. 6. The finished armature assembly should be in good ­dynamic balance, and the balance must be maintained in service. All these requirements are met more or less in modern lap, wave, or frog-leg windings. In fact, an armature winding is a remarkable packaging job as far as density and ­efficiency are concerned. The equivalent network of a generator armature winding is given in Figure 32.10.

Figure 32.10 Equivalent Network of a Generator Armature Winding

32.5.1  Lap Windings The differences among winding types arise from the way the coil ends are configured. A lap winding may have one or more turns of an approximately trapezoidal shape, with the two ends close to each other so that they can be connected to adjacent commutator segments. There are small variations in a duplex or higher plex lap coil, but the coil ends are near each other. Figure [32.11 (a) and (b)] shows a basic lap winding with two turns and three active commutator segments. A brush is required for each pole of a lap-wound a­ rmature. Co

il 2 Co

il 1

(a)

(b)

Figure 32.11 (a) Lap Armature-winding Coil, (b) Lap-wound Armature

614  Electrical Technology A winding of this type can be clearly seen in Figure 32.12 starting from segment A, the conductor current path is 1, 2, 3 and 4 to complete the first loop at segment B. However, the second loop starts at B, and the path is 5, 6, 7 and 8 ending at segment C. In this way, each loop overlaps the next. Development of the lap winding can be seen from Figure 32.13. 1

12

2

11

3 F

10

E

A D

B

4

C

9

5 8

6

7 D

E

F

9 10 11 12

Figure 32.12 Simple Illustration of the Lap Winding in Figure 32.11

B

A

2

3

C

4

5

D

6

7

E

8

9 10

To 7

To 12

Figure 32.13  Development of the Lap Winding

32.5.2  Wave Windings The wave winding in contrast is designed for high voltage applications and it has two parallel current paths. Such a winding can use only two brushes regardless of the number of individual poles. The wave winding has the same basic trapezoidal appearance in the coil body, but the coils are spread apart (see Figures 32.14 and 32.15). The coil ends are connected to commutator segments that are one less or one more segment than the angular distance between two like polarity field poles. A wave winding must go at least twice around the armature before it closes back where it started.

Figure 32.14 Wave-wound Armature

Figure 32.15  Illustration of Wave Winding

Figure 32.16 shows a wave winding, where the conductor starts from the segment (1) and passes through a slot (2, 3) to the rear of the drum where it turns (3, 4) and completes the path in the other direction (4, 5). Path 2 to 3 is shown as a north pole, while path 4 to 5 is the opposite or a south pole. Figure 32.17 shows the partial winding laid out or its developed version. The wave winding obtains its name from the shape of the coils before they are placed on the armature core, as they look like a wave. Figure 32.17 shows this type of winding where the series-connected armature conductors are cutting flux under adjacent unlike poles at the same time. The winding, as illustrated, goes around the armature enough times to connect all the conductors in series. A wave winding of this type has an advantage over a lap winding for some uses because it causes a higher voltage for a given number of poles and armature conductors. The coils in each path are series connected, and the number of paths is always two for any number of poles. The voltage produced by the generator is the sum of the voltages produced around the entire armature.

D.c. Generators  615 A

B

C

N

11

1

2

D

E

F

N

S

3

4

5

G

6

7

8

H

I

J

K

S

9 10 11 12 13 14 15

Load

Figure 32.16 Development of the Wave Winding

Figure 32.17  Developed Wave Winding

For the lap winding, the various paths are parallel-connected. There are as many paths in a lap winding as there are poles in the machine. For a given number of poles and armature conductors, a lower voltage is produced; however, the load can be greater than with the wave winding.

32.5.3  Lap and Wave Winding—A Comparison Lap Winding: There are as many parallel paths as the product of multiplicity and the number of poles. Each coil, at any instant, contains a group of coils in series. The current carried by each armature coil is I/a. The winding requires as many brushes as there are poles. Lap windings are used in high current low voltage d.c. dynamos. Wave Winding: The number of paths in the armature is twice the multiplicity. Each coil, at any instant, contains a group of coils in series (similar to the lap winding). The current carried by each armature coil is I/a (similar to the lap winding). The wave winding requires only two brushes. Wave windings are used in high voltage low current applications.

32.5.4  Paths in Lap-Wound and Wave-Wound Armatures The paths for a four-pole lap-wound simplex dynamo are shown in Figure 32.18 (a). There are four poles and four paths, each path carrying one-fourth of the total current and generating a voltage per path of ep. The total power generated by the dynamo operating as a generator is epl. Figure 32.18 (c) shows the equivalent circuit of this armature operating as a motor at the same speed and flux density. The c.e.m.f. generated per path is ep and the resistance of the winding in each path is rw. In the motor mode, the applied line voltage VL exceeds ep by the voltage drop across the resistance of the windings, IrwJ4 (assuming no brush voltage drop).

Figure 32.18 Armature Paths and Equivalent Circuit for a Four Pole Lap-wound Simplex Dynamo (a,b) Paths in a Lap-wound Dynamo (Generator Mode) (c) Equivalent Circuit of a Lap-wound Dynamo (Motor Mode) The paths for the four-pole wave-wound simplex dynamo are shown in Figure 32.19. It should be noted that there are four poles and two paths since the paths are independent of the number of poles in a wave winding. For conductors of the same currentcarrying capacity as the lap winding above, i.e., 1/4, the current per path in the wave winding is 1/4. The total current—since there

616  Electrical Technology

Figure 32.19 Equivalent Capacity Wave-wound Dynamo Showing Paths and Equivalent Circuit for a Fourpole Simplex Wave Winding (a) Wave-wound Armature (b) Paths (Generator Mode) (c) Equivalent Circuit of a Wave-wound Dynamo (Motor Mode) are but two paths—is 1/2. But now that there are only two paths, the total number of conductors per path is doubled, and the e.m.f. per path is 2ep. The total power generated by the dynamo operating as in the generator mode is 2epI/2, or still epl. The wave-wound dynamo in the motor mode is shown in Figure 32.19 (c). The resistance per path and the voltage per path are twice that of the equivalent lap-wound armature because there are twice as many conductors in series producing a higher c.e.m.f. and higher resistance in opposition to the applied voltage.

32.6 COMMUTATION The rectangular-shaped voltage wave generated within a d.c. armature coil is changed to a unidirectional voltage in the load circuit by means of a mechanical rectifier called a commutator, mounted on the armature shaft. This is illustrated in Figure 32.20 for an elementary two pole machine with one armature coil and a two-bar commutator. Connections to the external terminals are made via small stationary blocks of graphite called brushes that are pressed against the commutator by springs. The generated voltage within the armature coil changes direction every 180 electrical degrees (as shown in Figure 32.20) of rotation, but the voltage in the external circuit remains in the same ­direction. The rotating commutator and stationary brushes constitute a rotary switch that provides a switching action called commutation that switches the internal alternating voltage and current to direct current in the external circuit.

Figure 32.20 Sketches Showing the Commutation Process with a One-coil Armature. As One Coil Moves into the Neutral Plane Another Moves Out, Producing an Essentially Constant Voltage

D.c. Generators  617

When the coil is rotating through the neutral plane, as shown in Figures [32.20 (a) and (c)], it is shorted by the brushes. Since the coil sides are not cutting flux, no armature voltage is generated and no short circuit current occurs. A practical machine has many coils distributed around the armature and the coils pass through the neutral plane one at a time.

32.7  ARMATURE REACTION The form and shape of the magnetic field caused by the main fields, (Figure 32.21) becomes distorted when current is present in the armature windings. This is because the armature windings themselves are wound on a magnetic structure. The magnetic field thus produced in the armature is 90 electrical degrees from that produced by the fields. These two fields become vectorially combined in a distorted result. Figure 32.22 shows a typical undistorted field pole magnetic field that is present with little or no current in the armature. Figure 32.22 shows a visualization of the armature-produced field, which is present to a degree approximately proportional to the armature current.

N

N

S

S

Figure 32.21 Magnetic Flux Distribution due to Field Poles Only Figure 32.23 shows the result of the combination of the two fields. The magnetic field on one side of the field pole is in effect swept aside and reduced. The mid-pole field is about the same as when no current is present in the armature. But the other side of the field pole now has a substantially larger field than before. The voltages generated in the windings are unfortunately generated in direct proportion to the actual field that is present. Followed to its ultimate conclusion, this armature reaction results in the neutral voltage point being appreciably moved in relation to the brush position. As a result, the commutator and brush switching function is no longer spark free and the resulting brush and commutator life is drastically reduced.

32.7.1  Correcting Armature Reaction Effect There are four main ways of combating the armature reaction problem. 1. The first method is to rotate the brush hanger mechanism to find the correct but distorted neutral point. This is very effective at a fixed current load. On the other hand, the brushes must be moved each time the load is changed. This was the early answer, but it has become obsolete. It has the merit that it worked at one time and is easily understood.

Figure 32.22 Magnetic Flux Distribution due to Armature Excitation Only

N

S

Figure 32.23 Combined Magnetic Flux Distribution due to Armature and Field

618  Electrical Technology 2. The second method is to shape or otherwise modify the ends of the field pole shoes so that high flux cannot exist on the ends because of the high path reluctance. This is customarily done on almost all the machinery built today. It reduces but does not eliminate the problem. 3. The third method is to add interpoles or commutating poles to the field structure. These units resemble small main field poles and are installed midway between the main field poles. They function by locally modifying the resultant of the main field pole and armature-caused distorted magnetic field. This distorted field has the local effect around the brushes, causing some voltage generation to take place in the armature coil that is undergoing switching by the commutator and brush. The result of the combined field caused by the main field, the armature reaction field, and the commutating field is that locally around the brushes there is no effective field. Therefore, there is no unwanted voltage generation to spoil the required commutation and brush switching process. Since the effect of armature reaction is related to armature current, its counter action by commutating fields is also needed in relation to armature current. This is readily accomplished by connecting the commutating field windings in series with the armature, so that the variable requirement is automatically met in service. Commutating fields are so effective that they are invariably used in medium and large d.c. machines.   Once experienced, it is immediately obvious when commutating fields are incorrectly adjusted or wrongly connected. The commutator sparks viciously, there is a sound much like frying bacon and a sharp smell of ozone. In such a situation, immediate correction is necessary, because commutator and brush life may be reduced by more than 1000 to 1.

4. Very large machines or machines with very severe duty cycles such as in large machine tools or especially in rolling mill

drives, require the last feature, which is compensating windings (see Figure 32.24). These windings are laid directly into the pole shoes of the main fields. They are parallel to the armature shaft and carry current in the opposite direction to the armature windings immediately adjacent to these. The result is that the main field flux symmetry is no longer distorted, since the armature reaction magnetic flux is equally and oppositely opposed by the compensating winding flux. The function of the commutating fields is still required but to a reduced degree. Compensating windings are expensive and sometimes cumbersome, but their end results in reliable, spark free commutation under conditions that would be impossible without them. This constitutes the fourth method.

w

Figure 32.24  Techniques of Overcoming Armature Reaction

32.8  E.M.F. EQUATION Let

p = Number of pairs of poles f = Flux per pole in Wb Z = total number of conductors on armature = Number of slots × Number of conductors per slot N = speed in revolutions per minute (r.p.m.) e.m.f. generated by armature = e.m.f. generated by one of the parallel paths dφ V e.m.f. generated = Flux cut per second, i.e., e = e = dt

Flux cut by one conductor in making one revolution = 2 pf Flux cut by one conductor per second = 2 pφ ×     \ e.m.f. generated in one conductor =

N 60

( 2 pφ N ) 60

(32.1)

D.c. Generators  619

1. For a wave-wound armature No. of conductors in series per path =        \ e.m.f. generated = 2. For a lap-wound armature Number of conductors in series per path =

\ e.m.f. generated =

Z 2 pφ NZ V (32.2) 60 Z 2p Z 2 pφ N φ NZ × = 2p 60 60

If a is the number of parallel paths through the armature, then 2 p φ NZ     E = × V (32.3) a 60 For a given machine, E varies directly as fN. Example 32.1 An eight-pole lap-wound armature has 960 conductors and a flux per pole of 20 mWb. Calculate the e.m.f. generated when running at 500 r.p.m. Solution: φ NZ E = V 60 =

20 × 10−3 × 500 × 960 60

= 160 V Example 32.2 If the armature in the above example were wave-wound, what would be the e.m.f. generated? Solution:

φ NZ e.m.f. generated = p × 60 V 20 × 10−3 × 500 × 960 60 = 640 V = 4×

Example 32.3 A four-pole 1200 r.p.m. generator with a lap-wound armature has 65 slots and 12 conductors per slot. The flux per pole is 0.02 Wb. Determine the e.m.f. induced in the armature. Solution: e.m.f. induced in the armature =

φ NZ V 60

0.02 × 65′ × 12′ × 1200 60 = 312 V =

620  Electrical Technology Example 32.4 A lap-wound d.c. shunt generator having 80 slots with 10 conductors per slot generates at no load an e.m.f. of 400 V, when running at 1,000 r.p.m. Find out the flux per pole. If this generator is required to generate a voltage of 220 V on an open circuit, at what speed should it be rotated? Solution: φ NZ V e.m.f. generated = 60

400 = φ ×

φ =

1000 × 80 × 10 60

400 × 60 1000 × 80

= 0.03 Wb For a given machine Eα N φ or Since,

E αN φ

f is the same in both the cases = E α N To generate 400 V r.p.m. = 1000 To generate 220 Vp r.p.m. =

1000 220 = 550 400

Example 32.5 (1) A triplex lap-wound armature is used in a 14-pole machine with 14 brush sets, each spanning three commutator bars. Calculate the number of paths in the armature. (2) Repeat part (1) for a triplex wave-wound armature having 2 such brush sets and 14 poles. Solution: 1. mp = 3 × 14 = 42 paths 2.

a =2m = 2 × 3 = 6 paths

Example 32.6 Calculate the generated e.m.f. in each part of Example 32.5 if the flux per pole is 4.2 × 106 times, the generator speed is 60 r.p.m. and there are 420 coils in the armature, each coil having 20 turns. Solution: 1.    Z = 420 coils × 20 turns/coil × 2 conductors/turn E=

4.2 × 106 × 60 × 14 × 10−8 60 × 42

= 235.2 V 2.

E=

4.2 × 106 × 16800 × 60 × 14 × 10−8 60 × 6 × 10

= 1646.4 V Note: 1. Simplex lap windings have as many parallel paths as main field poles a = P. 2. Simplex wave windings have two parallel paths regardless of the number of poles: a = 2. The amount or degree of multiplicity or plex determines the number of parallel paths in the following manner. 3. A lap winding has pole times the degree of plex parallel paths: a = P× plex. 4. A wave winding has two times the degree of plex parallel paths a = 2 × plex. where, P = poles and a = parallel paths.

D.c. Generators  621

32.9  SEPARATELY EXCITED GENERATOR There are three basic types of generators with subtypes and they have distinctly different characteristics. To see how each type and subtype gains its characteristics, it is logical to see how the generator is affected by variations of field strength. Then the variations of field strength that result from various field-to-armature connections can be better understood. It is in the various ways of connecting the field circuit that the generator types develop their individual characteristics as shown in Figure 32.25. If the main fields are supplied from an external source, their ampere turns of excitation can be ­independently controlled. In a practical generator, the output voltage is controlled by many factors. The parameters of number of conductors Z, number of field poles P, and number of parallel paths in the armature windings are designed and built in the features of the particular circuit. I Ia

E

V

Ra

Figure 32.25  Separately Excited d.c. Generator (V=E-IaRa = E-IRa  ) Once it is built, one way to control the generated voltage of a given generator is to vary its rotative speed. The other way is to vary its field flux per pole. The rotative speed is controlled by the characteristics of the prime mover that is coupled to the generator and any gearing or belt drive that may be interposed between the prime mover and the generator. The field flux is determined by the overall magnetic path characteristics which are designed and built in feature, much as the armature windings and number of poles. The field coils are also designed and built with a particular number of winding turns of a particular wire. The ampere turns that are present in the coil are determined by the physical number of turns and the current that flows as a result of the applied field voltage. The applied field voltage can, of course, be of almost any value within reason when applied separately. A separately excited d.c. generator is illustrated in Figure [32.26 (a) and (b)]. The potentiometer in Figure 32.26 (a) permits zero adjustment of the shunt field current as a minimum, whereas the rheostat in Figure 32.26 (b) permits minimum current adjustment but not zero. The separately excited generator of Figure 32.26 (b) combines self-excitation of the series field and separate excitation of the shunt field, providing the advantages of compound operation with the advantages of separate field excitation. The armature current relations of this generator are the same as those for the series generator.

Figure 32.26 Separately Excited Generators (a) Separate Excitation, Shunt Field Using Potentiometer (b) Separate Excitation, Shunt Field Compound Operation

32.10  BASIC GENERATOR TYPES The generated voltage depends on the field magnetic flux. The different ways in which this field flux can be supplied are as follows. 1. The field connections can be tapped directly from the armature. In this way, the voltage that drives the current through the field coil is the full armature circuit voltage. The current can be decreased by the use of a series resistance in the field circuit if desired. This connection, called a self-excited shunt generator or simply a shunt generator, has been illustrated in Figure 32.27 (a).

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2. The field connections can be connected in series with the armature circuit. In this way, the field ampere turns will be

primarily controlled by the resistance of the connected load. There will be no field excitation if no current flows to the load. Known as a series generator, it has been presented in Figure 32.27 (b). Its outputs and characteristics will be seen to be quite different from those of the shunt generator. A series-connected field coil is composed of relatively few turns of heavy wire, whereas a shunt-connected field is composed of many turns of finer wire. The same field ampere turns can be produced (120 turns × 10 amp = 1200 At and 1200 turns × 1 amp = 1200 At). The coils are very different but the ampere turns are the same.

3. The third type, as shown in Figure 32.27 (c) combines a shunt and a series field and is called a compound generator. This type combines the features of the shunt and series generators. There is a further subdivision according to whether the shunt field is connected across the armature alone, called short-shunt compound, or connected across both the armature and the series field, called long-shunt compound. These different varieties of compound generators have slightly different characteristics but normally perform the same tasks. All of these connections are shown in Figure 32.27.

Figure 32.27  Three Connections of the Field Coil (a) Shunt (b) Series (C) Compound Note: Ordinarily, if a generator is separately excited, it is labelled as such: but if it is self-excited, the term may be only implied.

32.11  SCHEMATIC DIAGRAM AND EQUIVALENT CIRCUIT The differences among the three basic types of d.c. generators emerge from the manner in which the d.c. stator field winding’s excitation is produced. The purpose of the generator is to produce a d.c. voltage by conversion of mechanical energy to electrical energy and a portion of this d.c. voltage is used to excite the stationary magnetic field winding.

32.11.1  Shunt Generator When the excitation is produced by a field winding connected across the complete (or almost complete) line voltage produced between the brushes of the armature, the d.c. dynamo is called a shunt generator, as shown in Figure 32.28 (a). The rotor armature is represented (enclosed in a rectangle) as consisting of a source of e.m.f. (Eg ), a resistance (Rw) of the armature winding, a resistance (Rb) of the carbon brushes and the brush contact resistance made with the rotating armature. The entire armature circuit consists of the armature (enclosed in the rectangle of dashed lines) and two optional windings, the compensating winding (Rc) and the interpole winding (Ri) located on the stator. Thus, the portion of the armature circuit that rotates is shown as enclosed in the rectangle and that portion of the armature circuit that is fixed on the stator is outside the rectangle.

D.c. Generators  623

l

Figure 32.28 Shunt Generator. Schematic and Equivalent Circuit. (a) Complete Schematic Circuit Diagram of Shunt Generator (b) Equivalent Circuit of the Shunt Generator For simplicity, all series resistances in the armature circuit may be added and lumped together under a single resistance (Ra), called the armature circuit resistance. In the equivalent circuit of a shunt generator shown in Figure 32.28 (b), the armature circuit consists of a source of e.m.f. (Eg) and an armature circuit resistance (Ra). The field circuit of a shunt generator is in parallel with the armature circuit and, as shown above, consists of the shunt field winding wound on the stationary field poles and a rheostat. The shunt generator, when loaded, is composed of three parallel circuits: (1) the armature circuit; (2) the field circuit; and (3) the load circuit. Since the basic source of e.m.f. and current is the armature, the equivalent circuit Figure 32.28 (b) yields the following current relation Ia = If + Il 

(32.4)

where, Ia is the armature current produced in the same direction as the generated voltage Eg, If is the field current (Vf    /Rf  ) in the field circuit and Il is the load current (V1/R1). For the three circuits in parallel, the same voltage exists across the armature field, and load circuits, respectively, Va = Vf = Vl

(32.5)

where, Va is the voltage across the armature, i.e., Va = Eg – IaRa Vf is the voltage across the field circuit and V1 is the voltage across the load. Example 32.7 A 150 kW, 250 V shunt generator has a field circuit resistance of 50 Ω and an armature circuit resistance of 0.05 Ω. Calculate (1) The full line current flowing to the load; (2) the field current; (3) The armature current; and (4) the fullload generated voltage. Solution: kW × 1000 150 × 1000 W = = 600 A 1. Il = Vl 250 V 2. If =

Vl 250 V = = 5A Rf 50 Ω

3. Ia= If + Il = 50 + 600 = 605 A 4. Eg = Va + IaRa = 250 + 605 × 0.05 = 320.25 V

32.11.2  Series Generator When the excitation is produced by a field winding connected in series with the armature in such a way that the flux produced by the series-connected field winding is a function of the current in the armature and the load, the d.c. dynamo is called a series generator. The complete schematic diagram of the series generator is shown in Figure 32.29 (a). The series field is excited only when a load is connected to complete the circuit. Since this field must carry the full or rated current of the armature, it is constructed of a few turns of heavy wire.

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Figure 32.29 Series Generator. Schematic and Equivalent Circuit (a) Complete Schematic Circuit (b) Equivalent Circuit The compensating winding (Rc) located on the field poles and the interpole winding (Ri) are included in series with the armature winding (Rw) of the rotating armature, which produces a generated e.m.f. (Eg). The equivalent circuit (under load) of a series generator is shown in Figure 32.29 (b). The current in the series field winding (Is) is controlled by a diverter (Rd), which serves to provide a degree of adjustment of excitation of the series field in much the same way as the rheostat in shunt generator. Unlike the shunt generator whose field excitation is virtually independent of the load, the series field excitation (for comparison) depends primarily on the magnitude of resistance of the load. Thus, the diverter serves to provide only minor adjustment of the series field excitation in a series generator. The current relations in a series generator are Ia = Il= Is+ Id (32.6) The voltage relations of a series generator, as shown in the equivalent circuit of Figure 32.29(b), may be summarized as (32.7)

Va = Vl + IsRs where, Va is the voltage across the armature, or Eg – IaRa V1 is the voltage across the load IS  RS is the voltage drop across the series field

32.11.3  Compound Generator When the field excitation is produced by a combination of the two windings, namely, a series field winding excited by the armature or line current, and a shunt field winding excited by the voltage across the armature, the d.c. dynamo is called a compound generator. The complete schematic diagram of a compound generator is shown in Figure 32.30 (a). The stationary field structure is represented as consisting of a shunt field winding and a series field winding wound over the shunt field winding; in addition to the compensating winding inserted in the pole face of the main field poles.

(a)

(b)

(c)

Figure 32.30 Compound Generator: Schematic and Equivalent Long- and ­Short-shunt Connections. (a) Complete Schematic Long-shunt Connection (b) Equivalent Circuit Long-shunt Compound ­Generator and (c) Equivalent Circuit Short-shunt Compound Generator

D.c. Generators  625

The circuit has been simplified to produce two possible equivalent circuits: a long-shunt connection and a short-shunt connection. Figure 32.30 (b) shows the long-shunt compound generator connection in which the field circuit is in parallel with the combined armature and series field circuits as well as with the load circuit. Figure 32.30 (c) shows a short-shunt connection in which the shunt field circuit is in parallel with the armature circuit, and the series field circuit is in series with the load. The current relations of the long-shunt connection of a d.c. compound generator are Ia = If + Il = Is + Id (32.8) The current relations of the short-shunt connection of a d.c. compound generator are Ia = If + Il and Il = Is + Id(32.9) The essential difference between the two is that in the short-shunt connection, the armature current excites the series field, whereas in the long-shunt connection, the load current excites the series field. Example 32.8 A long-shunt compound generator rated at 100 kW and 500 V d.c. has an armature resistance of 0.03 Ω a shunt field resistance of 125 Ω and a series field resistance of 0.01 Ω The diverter carries 54 A. Calculate (1) The diverter resistance at full load and (2) The generated voltage at full load. Solution: 1.

If =

Vf Rf

=

500 V = 4A 125 Ω

Ia = I f + Il = 4 + 200 = 204 A Is = Ia − Id = 204 − 54 = 150 A Since, diverter and series field are in parallel I d Rd = I s Rs and Rd = ( I s Rs ) / R d 150 × 0. 01 = 0.0278 Ω Rd = 54 2. Eg = V1 + I a Ra + I s Rs

= 500 + (204 x 0.03) + (150 x 0.01) = 507.62 V

32.12  CHARACTERISTICS OF D.C. GENERATORS For each form of generator, there are two groups of characteristics to be considered. These are: (1) Open-circuit characteristics; and (2) Load-characteristics.

32.12.1  Separately Excited Generators Under open circuit conditions, the armature winding carries no current, yet it has an e.m.f. induced in it due to its rotation in the magnetic field which is energized from a separate source. The e.m.f. is given by Eαf N, and we may consider this relation for a variation of speed when the field flux is maintained constant and for variation of the field flux when the speed is maintained constant. In the first case, the field flux is constant, provided the field current If is constant, thus EαN. Initially, let the field current be If. Thus, the e.m.f./speed characteristic takes the form shown in Figure 32.31 (a). Higher values of field current If2 and If3 increase the field flux and, therefore, give rise to proportionately greater e.m.f.s for any given speed, but in each case the characteristics are essentially linear. In the second case, let the speed be constant and the field current stay varied. Since the speed N is constant, Eαf. The flux f is varied by the field current If and the relation takes the form defined by the B/H characteristic of the magnetic circuit. It follows that the E/If characteristic also takes a similar form as illustrated in Figure 32.31 (b). By increasing the rotor speed, the e.m.f. characteristic is also increased, thus a family of characteristics can be created with a separate characteristic for each particular speed. None of the characteristics passes through the origin of the graph because even when If is zero, there is still a small e.m.f. induced by the residual magnetism.

626  Electrical Technology I

I

Figure 32.31  Open-circuit Characteristics of a Separately Excited d.c. Generator The function of a generator is to supply current to a load, and generally this is undertaken at a specific supply voltage. Let us suppose that the generator is excited to give the required terminal voltage and that a load is introduced, thus increasing the armature current from zero. Provided that the field current and the rotor speed are maintained as constant, the terminal voltage/load current characteristic takes the form shown in Figure 32.32. This characteristic is called the external V E characteristic of the generator. This characteristic indicates that the terminal voltage Volt drop due to IaRa and armature reaction drops with increase in the load current. There are two reasons for this drop: 1 The armature reaction decreases the field flux and, hence, reduces the induced e.m.f. 2. There is the IaRa voltage drop in the armature. In practice, the drop in output voltage at full load compared with the no-load terminal voltage is small, although appreciable. If necessary, the terminal voltage may be maintained constant, by increasing the field current and/ or the rotor speed.

Full load current

I

Figure 32.32 External or Load Characteristic of a Separately Excited d.c. Generator

32.12.2  Shunt Generator When a shunt generator is operated under no-load conditions, it is the output terminals that are open-circuited, as shown in Figure 32.33. The armature still requires supplying current to the field winding. However, to obtain the open-circuit characteristic, it is a simple test to operate the generator at a constant speed with the field separately excited, thus obtaining the characteristic shown in Figure 32.33 (b). Normally because the field current If is comparatively small, as is the armature resistance Ra, it follows that this characteristic is almost the open-circuit characteristic of the shunt generator and it is not necessary to allow for the very small IaRa voltage drop.

If

E

V

V Operating voltage

Critical field characteristic

Open-circuit characteristic E/Ia

Field winding characteristic V/If

Ra

(a)

(b)

Ia

Figure 32.33  d.c. Shunt Generator Under No-load Conditions The V/If characteristic of the field windings is a straight line, its gradient depending on the resistance of the windings. The generator starts by inducing an e.m.f. due to the residual magnetism. This e.m.f. causes a current to flow in the field windings which enlarges the magnetic field which, in turn, increases the induced e.m.f. This process continues until the

D.c. Generators  627

e.m.f. causes just sufficient current to flow to produce a sufficient field to exactly induce the e.m.f. This condition occurs when the open-circuit characteristic meets the V/I/ characteristic of the field windings. Once the steady state condition has been achieved, there are several possible changes of condition that may be applied to the generator. For instance, if the resistance of the field winding circuit is increased, then the slope of the field characteristic increases and the characteristic intersects at a lower value of E. However, if the line of the field resistance characteristic becomes tangential to the opencircuit characteristic, then a critical point is reached when the generator is unable to sustain the induced e.m.f., which suddenly falls to a very low value. This limitation means that the variation of field resistance is not a suitable method of obtaining control of the output voltage over a large range. Nevertheless, it is often quite sufficient to control the output voltage over a small range near the normal operating voltage. If a wider range of output voltage variation is required, it is necessary to vary the speed of the rotor. If the field-circuit resistance is too high, the shunt generator will fail to excite itself. Thus, a shunt generator can fail to excite because it has no residual magnetism, because the field windings are reverse connected or because the resistance of the field circuit is too high. Other reasons for failure to excite originate from malfunction of components, e.g., dirty brushes on the commutator or a winding being open-or short-circuited. When the shunt generated is operated under load, the load characteristic shown in Figure 32.34 is similar to that of the separately excited generator. In this instance, however, the characteristic is drawn on the basis that the speed N is constant, the field current If now depending on the terminal voltage of the generator. Since this voltage tends to decline with increase of load current I, it follows that If also declines, thus, reducing the field flux f and, hence, the induced e.m.f. E. This, in turn, further reduces the terminal voltage; thus, the drop of the characteristic is slightly more pronounced in the shunt generator than in the separately excited generator. The decline in the terminal voltage is, therefore, caused by 1. The effect of armature reaction. 2. The IaRa voltage drop in the armature winding. 3. The reduction in the field current If mainly due to the other two causes. Over a normal working range, i.e., from no load to full load, the drop in output voltage of a shunt-excited generator is generally less than 5 per cent, which is much smaller than is suggested by the characteristic shown in Figure 32.34. The characteristic in that diagram was exaggerated in order to emphasize that there is a voltage drop with increase of load. If a shunt generator is operated alone, it cannot be excessively overloaded. Instead, when the current reaches a certain value and then tries to increase, the terminal voltage collapses;, thus, the generator protects itself. The voltage collapses because it is unable to maintain sufficient current in the field winding to create sufficient field for the necessary e.m.f.

V E

Volt drop due to IaRa armature reaction and weakened field

Full load current

Figure 32.34 External or Load Characteristic of a d.c. Shunt Generator

Example 32.9 An open-circuit test was carried out on a d.c. shunt generator driven at 1000 r.p.m., the field being separately excited. Terminal voltage V (volts) Field current If (amperes)

312 0.8

357 1.2

390 1.6

414 2.0

435 2.4

The resistance of the field circuit is 200 Ω. Find the open-circuit voltage of the generator when: (1) it is driven at 1000 r.p.m. (2) it is driven at 800 r.p.m. Solution: The results of the test are illustrated in Figure 32.35. The V/lf characteristic is constructed by considering a voltage, say, 400 V and, hence, determining the field current as 400/200 = 2.0 A. By plotting this condition on the graph and joining the point to the origin, as shown, the V/If characteristic can be constructed. The intersection of the open-circuit characteristic of the armature when operating at 1000 r.p.m. and the V/If characteristic occurs at a point corresponding to the terminal voltage of 418 V. EαfN For any given value of field current, the flux remains the same and EαN

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Figure 32.35  For Example 32.9 Thus, in each case the induced e.m.f. is proportional to the rotor speed. If the rotor speed is reduced to 800 r.p.m. from 1000 r.p.m., then the e.m.f. is also reduced. Consider the first set of results in which the terminal voltage was 312 V and let this be E1, the e.m.f. being induced when the rotor rotated at 1000 r.p.m. and the field current was 0.8 A. Given that the speed is reduced to 800 r.p.m. and the field current remains at 0.8 A, then the new e.m.f. E2 is given by E1/E2 = N1/N2

Hence,

By repeating this procedure, we are able to obtain the open-circuit characteristic of the generator operating at 800 r.p.m. as being Terminal voltage V (volts) Field current If (amperes)

250 326 312 331 348 0.8 1.2 1.6 2.0 2.4

The characteristic is drawn and its intersection with the V/lf characteristic indicates that the open-circuit voltage of the generator at 800 r.p.m. is 309 V. Example 32.10 A shunt generator supplies 50 kW to 500 V constant-voltage d.c. system. The armature rotates at 650 r.p.m. and has a resistance of 0.2 Ω. The field current of the generator is 3 A, and this field current remains constant when the output of the generator is reduced to 35 kw. Assuming that the effect of armature reaction is negligible, find the rotor speed required for this smaller output. Solution: I1 = P1/V = 50000/500 = 100 A E1 = V + IalRa = 500 + (100 + 3)0.2 = 520.6 V For 35 kW output, the supply current is given by I2 = P 2 /V = 35000/500 = 70 A and E2 = V + IalRa = 500 + (70 + 3)0.2 = 514.6 V

D.c. Generators  629

The field current is constant and since the armature reaction is negligible E1 N = 1 E2 N2 and N 2 = 650 ×

514.6 = 642.5 r.p.m. 520.6

32.12.3  Series Generator When the series generator is operated with no load, there can be no armature current and, therefore, no current in the field windings to induce an e.m.f. However, a small e.m.f. will be generated due to residual magnetism. When a load is applied to the generator, current can flow in the field windings and an e.m.f. is ­induced. The external or load characteristic of the series generator as shown in Figure 32.36 is similar to that of the open-circuit characteristic which can be obtained by separately exciting the machine. The Critical load difference between the characteristics is caused by a characteristic V voltage drop due to: (a) armature reaction, (b) the IaRa Open-circuit characteristic voltage drop, (c) the IaRf voltage drop The output voltage, therefore, depends on the External characteristic Internal resistance of the load and may be determined by drawing V/Ia voltage drop the voltage/current characteristic of the load as shown to meet the external characteristic. If the load resistance is too large, the slope of the characteristic becomes too steep Load resistance characteristic and exceeds that of the critical load resistance, in which V/I case the generator fails to excite. On the other hand, if the load resistance is too small, the machine protects Full-load current Ia itself by letting the output voltage fall, as illustrated. Figure 32.36  Characteristics of a d.c. Series Generator

32.13 GENERATOR LOSSES

A generator is a machine for converting mechanical energy into electrical energy. During such conversions, losses take place. These losses are dissipated in the form of heat. The principal losses are: 1. Copper loss due to I2R heat losses in the armature and field windings 2. Iron (or core) loss due to hysteresis and eddy-current losses in the armature. This loss can be r­ educed by constructing the armature of silicon steel laminations having a high resistivity and low hysteresis loss. At constant speed, the iron loss is assumed to be constant. 3. Friction and windage losses as a result of bearing and brush contact friction and losses due to air resistance against moving parts (called windage). At constant speed, these losses are assumed to be constant. 4. Brush contact loss between the brushes and the commutator. This loss is approximately proportional to the load current. The total losses in a machine can be quite significant and operating efficiency of 80 per cent and 90 per cent are common.

32.14  POLARITY OF BRUSHES The output terminal of a generator, as with other d.c. power units, possess electrical polarity. In the case of generators, the brush polarity is used to distinguish between the electrical polarity of the brushes and the magnetic polarity of the field poles. Brush polarity markings are often omitted. It can be determined by connecting a voltmeter across the output leads of the generator. Many automotive and aircraft generators are constructed with either the positive or negative brushes grounded to the frame of the generator. When the armature is driven in either direction, an electrical polarity is established at the generator output terminals and at the brushes. If the machine is stopped and then driven in the opposite direction, the field flux is cut in the opposite direction and the brush polarity changes. This has been illustrated in Figure 32.37. As a result, the brush polarity in a separately excited generator can be changed by reversing either the direction of rotation of the armature or the direction of the field current. However, if both the armature direction and field current direction change, the brush polarity will remain unchanged.

630  Electrical Technology Field

A

G

+

Field

B

A

–

G

–

Armature

Field

B

A

+

+

Armature

G

Field

B

A –

– Armature

(a)

G

B +

Armature (b)

Figure 32.37 Factors Affecting Brush Polarity (a) Reversing Armature Rotation Reverses Brush Polarity (b) Reversing Field Current Reverses Brush Polarity A separately excited generator develops voltage for either direction of rotation. This is not true, however, for selfexcited units; they develop voltage in one direction only. The standard direction of rotation of d.c. generators is clockwise when looking at the end of the generator opposite the drive-shaft (usually the commutator end).

32.15  VOLTAGE REGULATION

(d)

The terms voltage regulation and voltage control are often confused. Voltage control refers to intentional changes in the terminal voltage made by manual or automatic regulating equipment such as a field rheostat. Voltage regulation refers to automatic changes in the terminal voltage due to reaction within the generator with changes in the load current. For example, it is inherent in the design of a shunt generator for the output voltage to fall off as the load increases. If the drop is severe, the generator is said to have poor regulation. The voltage regulation of a generator is one of its i­mportant characteristics. Different types of generators have different voltage regulation characteristics. Figure 32.38 i­llustrates the action of the voltage at the terminals of a generator for different values of load current. The drop in terminal voltage is caused by the loss in voltages:

(b) Output voltage

(a)

(c)

Load current

Figure 32.38  Voltage Regulation

1. Across the internal resistance of the armature circuit, including the brush contacts. 2. Due to armature reaction. The curve at (a) is the normal curve for a shunt generator. An ideal curve is shown at (b). Curve (c) illustrates a generator with very poor regulation in the sense that the load voltage drops off considerably as the load current increases. A rising characteristic curve is obtained by using a cumulative compound wound generator curve (d). The term voltage regulation is a measure of the extent to which VL changes as IL varies. It is expressed as a percentage and is calculated from V − VFL Percentage regulation = NL × 100 VFL where, VNL = Terminal voltage under no-load conditions VFL = Terminal voltage under full-load conditions An ideal voltage source would have zero percentage regulation, in which case the regulation curve would be a horizontal line.

32.16  PARALLELING OF GENERATORS There are many reasons for paralleling generators. 1. Power sources such as generators are frequently primary safety items and are, therefore, duplicated or paralleled for reliability. 2. Many major types of machinery, such as generators, run most efficiently when loaded to their design rating. Electric power costs less per kilowatt-hour when the generator producing it is efficiently loaded. Therefore, when the load is reduced, one or more generators can be shut down and the remaining units kept efficiently loaded.

D.c. Generators  631

3. Breakdown or routine maintenance frequently requires that the device being worked on be isolated from its work and shut down. Therefore, if power sources are paralleled, then routine or emergency operations can be performed without disturbing the load conditions. This affects both safety and economy. 4. In the modern world of expanding population, goods and services, the use of electricity is constantly increasing. When added capacity is required, the new equipment can be simply paralleled with the old. This frequently means a more flexible operation, since it increases the choices available for reasons 2 and 3. 5. In many situations not confined to generators, the equipment available to do a particular task may not be available in a sufficiently large capacity or size in a single unit. Here, paralleling must be a design feature just to meet the original load requirement.

32.16.1  Parallel d.c. Generator Requirements The principal types of situations where paralleling of d.c. generators is required are parallel shunt generators of the same or varying sizes as well as parallel compound generators of the same or varying sizes. In all different situations, there are certain requirements that must be met for successful electrical paralleling. A parallel circuit is one in which the same voltage exists across each unit at the paralleling point. This is in accordance with KVL. If the generated voltages of the individual generators are not all the same and they are paralleled, then, three different conditions may be met.

1.  When a generator is producing an internally generated voltage Eg that is appreciably above the voltage at the paralleling point, generator action is taking place and the unit is delivering current to the load. 2. If the generator is developing the same voltage as that existing at the paralleling point, no effective generator action is taking place and no current is flowing to the load. The generator is said to be floating on the line. It is neither contributing nor drawing current and is still being rotated by its own prime mover. 3. If the generator is set so as to develop less internal Eg than the voltage at the paralleling terminals, it will draw current from the paralleling point and will be operating as a motor. In this case, though, the unit is drawing power from— rather than delivering power to—the paralleling junction point.

These three situations are in entire agreement with KCL, as any parallel circuit must be. Furthermore, the following criteria are equally important. 1. The polarities of the generators must be the same or the connections must be interchanged until they are the same. 2. The voltages should be nearly if not actually identical so that each machine will contribute towards the output. 3. The change of voltage with change of load should be of the same character. A positive regulating machine cannot usefully combine with a negative regulation machine, as circulating currents would then dominate the situation. 4. The prime movers that drive the generators should have similar and stable rotational speed characteristics. Example 32.11 Three identical 5 kW shunt dynamos are driven by a single prime mover. Each machine has a field resistance of 60 Ω and an armature circuit resistance of 0.18 Ω. The three machines are set to no-load voltages of 124 V, 120 V and 115 V. They are then connected to a bus that is maintained at 120 V by other machines. Determine for each of the three machines: (1) The line current sent into or drawn from the bus; (2) The armature current; (3) The power drawn from or delivered to the bus; and (4) The power generated. Solution: E g − V1 124 − 120 1. I a = = = 22.0 A acting as a generator 0.18 Ra 120 − 120 = 0 A floating on the line 0.18 115 − 120 = = − 27.8 A acting as a motor 0.18 =

2. Since each machine is connected to the line, each field is seeing 120 V and, thus, drawing current that the armature is presumed to supply, Vf 120 = If = = 2A 60 Rf

632  Electrical Technology Then the armature A = 22.2 + 2.0 = 24.2 A Armature B = 0 + 2.0 = 2.0 A Armature C = –27.8 + 2.0 = –25.5 A The machines are unbalanced to a great extent and yet each machine has a current within the safe range, since from basic power law P 5 × 1000 =I = = 41.7 A rated E 120 3. P = IE,

For machine A 22.2 × 120 = 2664 W For machine B 0 × 120 = 0 W For machine C –27.8 × 120 = –3336 W

Machine C is drawing more power from the bus than machine A is delivering, while machine B is floating. The bus is not aided by the three improperly adjusted machines. Each machine delivers power to its field. Therefore, P = IE –2 × 120 = 240 W to each field 2664 + 240 = 2904 W generated, A 0 + 240 = 240 W generated, B –3336 + 240 = –3096 W motor power, C Machine C will deliver 3096 W motor power back to its prime mover if we decide to neglect all losses but the power to the field. (This is only an approximation.) If each machine had been set to at least the voltage of machine A, there would be a reasonable purpose to the parallel combination. Full-rated load will require that each machine deliver 41.7 A, which with the same procedure as (A) would require E= 120 + IaRa = 120 + [(41.7 + 2) 0.18] = 127.9 V This is 7.9 V greater than the bus voltage. The machine performance is not linear.

S UM M A RY 1. A dynamo can operate either in the motor mode or generator mode. 2. Rotary motion is supplied by a prime mover. 3. The rotor is the part of the dynamo that rotates. 4. The stator is the part of the dynamo that is stationary. 5.  The armature structure serves as a support for the winding conductors. 6. The armature magnetic field sees cyclic reversals. 7.  The armature winding coils are placed in slots in various arrangements. 8. The windings terminate in the rotary switching unit known as the commutator. 9. The basic purpose of the commutator is to change a.c. to d.c. 10. The switching function is shared by the commutator and brushes. 11. The working part of a machine is the armature winding where the voltage is generated or motor action is developed. 

12. A brush is required for each pole of a lap-wound armature. 13.  The wave winding is designed for high-voltage applications. 14.  Rotating the brush hanger mechanism to overcome armature reaction reduces the armature reaction but does not eliminate the problem. 15. Commutating fields are so effective that they are invariably used in medium and large d.c. machines. 16. The use of compensating windings results in spark-free commutation. 17. There are three basic types of generators with subtypes. 18. For each form of generator, there are two groups of characteristics—open circuit characteristics and load characteristics. 19. The efficiency of a generator is a maximum when the variable loss is equal to the constant loss.

D.c. Generators  633

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. The direction of induced voltage in a conductor can be changed by

(a) (b) (c) (d)

Increasing the field strength Reversing the field direction Increasing conductor length Decreasing conductor size

2. Direct current can be supplied to a load by a loop of wire rotating through a field with the use of

11. The winding on an interpole is (a) Made of many turns of fine wire (b) Wound in a direction opposite to that of the armature winding (c) Connected in series with the armature load (d) Connected across the generator terminals

12. To raise generator voltage

3. Induced voltage can be increased in magnitude by









(a) Slip rings (c) A commutator (a) (b) (c) (d)

(b) Brushes (d) A conductor

Using a commutator Using slip rings Decreasing conductor length cut per second Increasing the number of lines of force

4. The induced voltage in a single loop reverses

(a) (b) (c) (d)

Once each revolution Once each half revolution Twice each half revolution Twice each revolution

5. Maximum voltage is induced in a single loop when the sides of the loop are passing

(a) (b) (c) (d)

Perpendicular to the lines of force Parallel to the lines of force At an angle of 45° to the lines of force At an angle of 60° to the lines of force

6. When a commutator is used on a single loop, the voltage at the brushes has a

(a) (b) (c) (d)

Very large magnitude Changing polarity Constant polarity Constant magnitude

(a) (b) (c) (d)

Rotational direction Field direction Current direction Magnetic field strength

(a) (b) (c) (d)

Changes electrical energy to mechanical energy Changes mechanical energy to electrical energy Is always self excited Is always separately excited

9. One of the following is not essential in generating a d.c. voltage

(a) A magnetic field (c) Slip rings

(b) A commutator (d) A conductor

10. Commutating poles are

(a) (b) (c) (d)

13.  Generator output voltage control is usually accomplished by (a) Varying the speed (c) Increasing the flux

(b) A rheostat in the field (d) Decreasing the flux

14. A separately excited generator has the field connected

(a) (b) (c) (d)

Across the armature In series with the armature To an external circuit None of these

15. The voltage of a separately-excited d.c. generator may be increased by

(a) (b) (c) (d)

Increasing the speed of rotation of the armature Decreasing the magnetic flux Both (a) and (b) Neither (a) nor (b)

16. The function of the brushes on a generator is to

(a) Carry the current to the external circuit (b) Prevent sparking (c) Keep the commutator clean (d) Reverse the connections to the armature to provide d.c.



(a) (b) (c) (d)

Reversing the rotation of the armature Reversing the direction of the field current Either (a) or (b) Neither (a) nor (b)

18. Most d.c. generators are

8. A generator

Field current should be increased Field current should be decreased Speed should be decreased Brushes should be shifted forward

17. Electrical polarity at the brushes may be changed by

7. The left head generator rule is usually used to determine

(a) (b) (c) (d)

Fastened to the centre of the commutator Used to regulate the voltage at the armature Secondary poles induced by cross magnetizing the armature Located midway between the main poles



(a) (b) (c) (d)

Self excited Excited by storage batteries Excited separately None of these

19. The field coils of a shunt generator are always connected (a) In parallel with a rheostat (b) In parallel with each other (c) In series with each other (d) Across the armature 20. The voltage of a shunt generator is built up by (a) Permanent magnetism (b) Proper operation of the field rheostat (c) Residual magnetism (d) Increasing the speed

634  Electrical Technology 21. The field windings of a shunt generator must have

26. The series winding must be large enough to carry





(a) (b) (c) (d)

Full line current applied Comparatively low resistance One ohm resistance per volt Comparatively high resistance

22. Cutting resistance out of a shunt field circuit

(a) (b) (c) (d)

Cuts down the magnetic flux Decreases the terminal voltage Increases the load Increases the terminal output voltage

23. Failure of a d.c. generator to build up to its rated voltage can be due to (a) Loss of residual magnetism (b) Resistance greater than the critical field resistance (c) Rotation of the armature opposite to that known to cause a voltage build up (d) Brush contact resistance effectively increasing the field circuit resistance above the critical point (e) Improper connection of the field circuit leads at the brushes (f) All of these

24. Voltage control refers to a change that takes place

(a) Due to the operation of auxiliary regulating equipment (b) When the speed is regulated (c) When the terminal voltage is increased (d) Automatically when the load is changed

25. When the load is raised from minimum to maximum there is

(a) (b) (c) (d)

No change in terminal voltage An increase in terminal voltage A decrease in terminal voltage Less change than in other generators

ANSWERS (MCQ) 1. (b)  2. (c)  3. (d)  4. (b)  5. (b)  6. (c)  7. (c)  8. (b)  9. (c)  10. (b)  11. (c)  12. (a)  13. (b)  14. (c)  15. (a)  16. (a)  17. (b)  18. (d) 

(a) (b) (c) (d)

The total magnetic flux A 300 per cent overload Full line current Full line voltage

27. Select the type of generator that may be used for loads quite distant from the generator

(a) (b) (c) (d)

Over compounded Flat compounded Under compounded Differential compounded

28.  The normal voltage of a compound generator is changed by adjusting the

(a) Series field shunt (b) Brush setting (c) Shunt field rheostat (d) Equalizer

29. The resistance of a series field diverter should be

(a) (b) (c) (d)

Comparatively high Equal to the resistance of the series field A variable resistance Comparatively low

30. To achieve maximum compounding effect the diverter rheostat should be

(a) (b) (c) (d)

Set at its minimum value Set at a high value Set at a midway value Removed from the series field circuit

19. (b)  20. (c)  21. (d)  22. (d)  23. (b)  24. (a)  25. (c)  26. (c)  27. (a)  28. (a)  29. (a)  30. (b).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. On a d.c. machine why is a heavy mainframe construction required? 2. Is an odd number of field poles possible? Why? 3. Name two main functions for the cylindrical laminated steel structure of the main portion of a d.c. armature. 4. Where are armature winding coils placed in a d.c. machine? 5. To what are the ends of the armature winding coils connected? 6. What is the usual material for the commutator bars or segments? 7. What is the function of the commutator?

8. What determines the quantity and spacing of the brushes? 9. Name three functions of the armature shaft bearings? 10. What is the purpose of the end bells? 11. Where are the main field coils located? 12. How are the field coils mounted or attached? 13. What is the purpose of commutating fields? 14. Where are the commutating fields mounted? 15. What is the purpose of compensating windings? 16. Where are compensating windings placed? 17. What is armature reaction and what is its most serious effect?

D.c. Generators  635

18. How do commutating field windings cope with the effect of armature reaction? 19. Why do the brushes normally cover more than two commutator segments? 20. Describe the flow of magnetic flux in the closed magnetic circuit in a d.c. machine. 21. What is meant by a separately excited generator? 22. What is meant by a self-excited generator? 23.  Name three basic d.c. generator types. Draw their schematics. 24. Draw the equivalent circuits of the three basic d.c. generator types. 25. What determines the output voltage polarity of a d.c. generator? 26. What is the difference between the internal and external characteristics of a d.c. generator? 27. Name two major causes of internal voltage loss in a d.c. generator under load conditions. 28.  Does a series generator output voltage continue to increase even with a very serious current overload? 29. What two different connections are recognized for the shunt field in a compound generator? 30. Is residual magnetic field polarity an important consideration in a compound generator? 31.  Name the electrical requirements for successful paralleling of d.c. generators. 32. Why are identical shunt generators comparatively easy to parallel? 33.  What happens if an incoming generator is set substantially above the bus voltage to which it will be paralleled? 34. What happens if an incoming generator is set to deliver exactly the same voltage as the bus to which it will be paralleled? 35. What happens if an incoming generator is set to a voltage below the bus voltage to which it is paralleled? 36. The output of a shunt generator is 500 A at a terminal voltage of 225 V. Armature resistance is 0.03 Ω and shunt field resistance is 50 Ω. What is the e.m.f. generated? 37. A short shunt compound generator delivers 100 A at a terminal voltage of 200 V. Armature resistance is 0.2 Ω series field resistance 0.04 Ω and shunt field resistance 51 Ω.... . Find the total current output of the armature and the e.m.f. generated. 38.  A long shunt compound generator has full load output of l00 kW at 250 V. Armature resistance is 0.05 Ω series field resistance is 0.03 Ω and shunt field resistance is 55 Ω. Find armature current and e.m.f. generated.

39. A four-pole wave-wound armature has 123 conductors and runs at 1200 r.p.m. If the flux per pole is 25 mWb, find the e.m.f. generated. 40. A four-pole lap-wound armature having 320 conductors is run at 900 r.p.m. If the flux per pole is 50 mWb, find the e.m.f. generated. 41. A four-pole lap-wound armature has 120 slots and four conductors per slot. The flux per pole is 50 mW and it generates 240 V. Find the speed. 42. A shunt generator supplies 195 A at 220 V. Armature resistance is 0.02 A and shunt field resistance is 44 Ω. If the iron and friction losses amount to 1600 W, find the e.m.f. generated, copper losses and commercial and electrical efficiencies. 43. A shunt field generator has a field current of 1.13 A and a full load current of 16 A. What is the armature current? 44. The same generator as in CQ 42 has a load voltage of 125 V. What is the armature circuit voltage and the field circuit voltage? 45. If the same generator as in CQ 42 has an armature circuit resistance of 0.693 Q, what is the armature circuit voltage drop at full load using armature current from CQ 42? 46. Under no load conditions, what would be the terminal voltage of the generator in CQ 42 through 44, if no voltage drop other than armature circuit resistance is considered? 47. Calculate the e.m.f. generated by a four-pole wavewound armature, having 45 slots with 18 conductors per slot when driven at 1200 r.p.m. The flux per pole is 0.016 Wb. 48. Derive the e.m.f. equation of a d.c. generator. 49. Describe the various parts of a d.c. generator. 5 0. Explain the difference between lap and wave windings. 51. Describe the conditions essential to build up voltage in a d.c. generator. 52. A d.c. generator develops an e.m.f. of 200 V when driven at 1000 r.p.m. with a flux per pole of 0.02 Wb, it is desired that this e.m.f. be increased to 210 V at 1100 r.p.m. What should be the value of flux per pole under the new circumstances? 53. A voltmeter connected across the terminals of a d.c. shunt generator, rotated at rated speed, shows no appreciable reading. What corrective measures will you take? 54. What do you understand by the term armature reaction in a d.c. generator? 55. Explain why compensating winding is used in large d.c. machines. 56. What are interpoles? Why are they used?

636  Electrical Technology 57. A 7.5 kW d.c. generator is to be paralleled with a 250 V bus. If its voltage is set to 265 V at no load, and it has an armature circuit resistance of 0.523 Ω. (a) What current will it deliver? (b)  Is the current within the machines’ rating? ANSWERS (CQ) 36. 235.1 V  37. 104 A, 224.8   38. 404.5 A, 322. 3 V 39. 122 V  40. 240 V  41. 600 r.p.m.  42. 224 V; 1900 W; 94.5 per cent; 95.8 per cent 

58. A 250 kW d.c. generator has been carrying its full rated load when attached to a 600 V bus. The machine has an armature circuit resistance of 0.083 Ω. If the bus breaker is suddenly opened, what no load voltage will be found to be reproduced by the generator if change of field current is neglected? 43. 17.1 A  44. (a) 125 V (b) 125 V  45. 11.9 V  46. 137 V 47. 518.4 V  52. 0.019 Wb  57. (a) 32.7 A (b) rating 30.0 A 58. 635 V.

33

D.c. Motors OBJECTIVES In this chapter you will learn about:    The development of counter e.m.f.   The development and measurement of torque   Classification of motors   Generator action and motor action and their relationship  Power rating of motors   Construction of motors  Characteristics of d.c. motors  d.c. motor starting problems    d.c. starting switch  d.c. motor reversing   Solution of simple problems on the above topics

Motor and generator action

33.1 INTRODUCTION The d.c. motor and the d.c. generator are essentially the same device. In the d.c. generator, the shaft is rotated by means of mechanical power and direct current is the output from the machine. In the same device—when it is used as a motor—direct current is applied to the machines and the resultant interaction of the magnetic fields causes the motor to produce mechanical energy. A generator can be used as a motor or a motor can be used as a generator. There is a similarity between motors and generators. In a motor as opposed to a generator, there is still a relative motion between conductors and a magnetic field. The rotation of the armature coil will produce a cutting of the magnetic lines of the flux; hence, there will be a voltage induced in the armature. According to the laws of magnetism, the induced voltage will oppose the applied voltage used to drive the motor. The magnitude of this force depends on the field strength the ­direction of rotation, and the speed of rotation. The faster the speed and the stronger the field, the larger will be the induced counter e.m.f. (c.e.m.f.). As far as construction is concerned, there is no difference between a d.c. generator and a d.c. ­motor. The only difference is that in a generator, the generated e.m.f. is greater than the terminal voltage, whereas the generated e.m.f. is less than the terminal e.m.f. in a d.c. motor see Figure [33.1 (a) and (b)]. Since the generated e.m.f. in a motor is in opposition to the applied voltage, it is referred to as the back e.m.f. or counter e.m.f. (c.e.m.f.). If

Ia

IL

If

Rh Shunt field

V

A

(a)

RL

Ia

Is +

Rh Shunt field

V

A

–

(b)

Figure 33.1 (a) A d.c. Shunt Generator, the Generated e.m.f. E = V + IaRa (b) A d.c. Shunt Motor, the Generated e.m.f. E = V – IaRa

638  Electrical Technology Example 33.1 The armature of a d.c. machine has a resistance of 0.1 W. Calculate the generated e.m.f. when it is running as (1) generator giving 80 A at a terminal voltage of 230 V and (2) a motor taking 60 A with applied voltage of 2 30 V. Solution: In both cases the field current is assumed as negligible. 1. Generated e.m.f.  2. Back e.m.f.

E = V + laRa = 230 + 80 × 0.1 = 238 V E = V - IaRa = 230 – 60 × 0.1 = 224 V

33.2  DEVELOPMENT AND MEASUREMENT OF TORQUE Since an electric motor, or any rotary-power producing device is a producer of continuously rotating torque, and since we can equate torque and angular velocity to get power, we need to know how torque is developed and how it is measured. The force acting on a conductor immersed in a magnetic field is given by: F = BI l newtons(33.1) These relations can be related to the torque developed in a motor by multiplying in the following manner. 1. By the number of conductors Z that are in the armature. 2. By the decimal equivalent of the percent of effective pole arc coverage to find the number of the conductors that are in the magnetic field and functioning. 3. By the effective radius of the conductors to convert the force to torque. 4. By convening the current Iin amperes to Ia, the total armature current, and dividing by a the number of parallel paths in the winding which yields the actual current per conductor. T =

Bla PZ (per cent cov)d N- m a 

(33.2)

Example 33.2 A 500 V, d.c. motor is loaded to operate at 500 r.p.m. while developing a torque of 125 N-m. The supply current to the motor is 2.54 A. Find the efficiency of the motor. Solution: 2Nπ T v= = 0.89 60 VI Example 33.3 A lift of mass 250 kg is raised with a velocity of 5.0 m/s. The rope by which the lift is suspended is wound onto a drum of diameter 50 cm and the efficiency of the winding mechanism is 0.85. Determine (1) the rotational speed of the winding drum, (2) the torque applied to the drum by the rope, and (3) the driving torque applied to the shaft of the drum by the winding motor. (4) Given that the winding mechanism is driven by 500 V d.c. motor of efficiency 0.91, determine the supply current to the motor. Solution: 2π Nr 60u ;N = 1. u = ω r = 60 2π r N =

60 × 5 = 159 rev/min 2 p × 0.3

2. T = Fr = Mgr = 250 × 9.81 × 0.3 = 736 N-m P0 =

2π NT 2 pl × 159 × 736 = 12, 260 W = 60 60

= Pi

P0 12, 260 = = 14, 420 W 0.85 n

D.c. Motors  639

Let the shaft torque be Ts. Hence, the shaft power is the input power to the winding mechanism and Ts =

3.

60 Pi 60 × 14420 = = 866 N-m 2π N 2π × 159

The shaft power is the output power of the motor which has an efficiency of 0.91; hence, the electrical power is given by 14420 = Pe = 15850 W 4. 0.91 I = Pe /V = VI = I

15850 = 31.7 A 500

Example 33.4 A d.c. motor takes a total line current of 135 A while rotating at 2550 r.p.m. (267.0 rad/sec). The field pole flux density is 0.8106 Wb/m2 and the field pole coverage is 72 per cent. There are four parallel paths and 96 total conductors, which are effectively in the magnetic structure for 152.4 mm. The effective radius of action is assumed to be the outside of the armature, which has a diameter of 187.3 mm. Determine (1) the gross developed torque and (2) the power that would result if all this torque were available on the rotating shaft. Solution: BI a lZ (per cent cov )d 1. T = a 0.8106(135)(0.1524)(96)(0.72)(0.0936) 4 = 26.97 N-m

=

2. Tw × 10–3 = kW = 26.97 × 267.0 × 10–3 = 7.207 kW

33.3  PRONY BRAKE In its usual form, a prony brake is a screw-actuated clamp brake band, which is wrapped around a rotating water-cooled brake drum. The tendency for the brake mechanism to travel around with the drum is due to the friction-transmitted torque force moving mechanism. The friction between the drum and the brake shoe is related to the band clamping pressure. This force which is attempting to rotate the brake with the moving drum is resisted with a brake arm, which is connected to a fixed anchorage through a calibrated spring scale. With the device under measurement rotating, as illustrated in Figure 33.2, the brake clamp is adjusted until the desired r.p.m. is held. Under this condition, all the torque developed by the prime mover is then absorbed by the brake and shown by the scale reading twice the arm length. In large sizes, the brake requires constant cooling, since all the prime-mover output power is connected to friction caused heat at the brake surface. The prony brake may vibrate or oscillate badly unless carefully prepared but is simple and useful.

F

Figure 33.2 Typical Prony Brake for Torque Measurement

640  Electrical Technology Example 33.5 A d.c. motor is tested with a prony brake loading device and the following data are taken: brake load arm radius from centre 0.6096 m; zero scale reading testing (tare reading) was 13.92 m; operating brake scale reading during test is 90.3 N. The rotating speed is 855 r.p.m. (89.53 rad/sec). (1) What is the torque developed? (2) What power does this represent? Solution: 1. T = F × d = (90.3 – 13.92) (0.6096)

Scale tension setting to set friction Fixed scale support Scale B low force with rotation shown

= 45.53 N-m torque

F = FA – FB in pounds or ounces

2. Tw × 10–3 = kW

f = fA – fB in newtons, kilograms, or grams

= 45.56 × 89.53 × 10–3 = 41.7 kW Note: Tare scale reading cannot be ignored if it is present.

Scale A high force with rotation shown

Friction surface is cord, typically nylon line

Force radius is pulley radius (or motor shaft radius in small units) Rotation with force difference shown

33.4  TWO-SCALE PRONY BRAKE

In small sizes, a two-scale prony brake may be used; as shown in Figure 33.3 Typical Two-scale Prony Brake for Motor Torque Measurement Figure 33.3. In this device, the effective radius arm is the radius of the grooved pulley or, in very small sizes, the radius of the motor shaft. The friction is varied by moving both spring scales and, thus, tightening or loosening the nylon cord brake band. The net torque reading is the difference between the scale readings. The usual scale readings are in milligrams or grams.

33.5 DYNAMOMETER The most versatile and expensive torque measuring device is the dynamometer. A dynamometer may be used to either absorb or produce torques; so it is more flexible in use than a prony brake, which belongs to the class of absorption dynamometers as shown in Figure 33.4. Scale support A dynamometer is usually a d.c. shunt field maForce scale chine of appropriate size and r.p.m. capability. The Rotation key to the operation of the dynamometer is the fact that in any loaded rotating electrical machine, motor action and generator action take place simultaneously. When absorbing power, a dynamometer serves as a generator and the generated voltage is connected to a resistive load bank. The ohmic adjustment of the Tare adjustment load bank—together with the voltage that the dynamometer is adjusted to produce—determines the current that flows. The restraining torque that resists the Force rad. Tachometer rotation exists in the armature windings. Any torque D in feet (optional) force that exists in the armature is there by action d in metres Dynamometer with the magnetic flux. Therefore, when a dynamomsupport Force scale eter is absorbing torque, its field tends to be pulled F in pounds or f in newtons trunnion around equally by the motor action that exists. Motor Dynamometer generator

to be tested

Unit support base

Figure 33.4 Typical Electric Dynamometer for Motor or Generator Torque

  T = f × d converted to newton metres.

(33.3)

The stator and surrounding frame structure of a dynamometer are mounted on low friction ball bearings, which are concentric or coaxial with its armature rotor bearings. The whole of the field and frame of stator assembly are accurately balanced by appropriate weights to remove any gravity torque effects.

D.c. Motors  641

During operation, the dynamometer stator tends to pull its support bearings but is restricted from ­doing so by a torque arm and appropriate scale, much as in the prony brake. The length from the bearing centreline of the stator assembly to the scale attachment point is the torque moment arm. It is usually marked on a conspicuous label attached somewhere on the unit. T = f × d newton metres. (33.4) It is not necessary to measure the electrical quantities in the dynamometer to measure the torque or power. Only the scale arm length, the scale force, and the rotative speed need to be recorded. The speed is usually conveniently read by a directly attached tachometer. Any brush, magnetic and bearing drags also are shown on the spring scale, so the dynamometer is an accurate device. Its great versatility lies in its ability to also function as a motor and to drive any rotating device within its capability and, at the same time, measure the power produced at the shaft coupling. The bulk of energy produced by a dynamometer is normally dissipated in a resistive load bank, which is remotely ­located and maybe appropriately cooled. Example 33.6 It is desired to test a 1865 kw mill motor on a dynamometer at both its low and high rated speeds of 225 and 450 r.p.m. (23.5 and 47.124 rad/sec). If a dynamometer rig is available that has a 1.524 in radius arm, what force scale range must be available in newtons? Solution: f1 =

1865 × 103 kW × 103 = = 51940 N at 23.56 rad/sec cw 1 .524 × 2356

f2 =

1865 × 103 = 25970 N at 47.12 rad/sec 1.524 × 47.12

33.6  BACK ELECTROMOTIVE FORCE IN A MOTOR

Motion

When a motor armature is rotating as a result of the torque that is produced by motor action, it is also acting as a generator at the same time. Since there must be a magnetic field from the field poles in order that motor Field action can take place, that same field then generates voltage in any passing armature conductor. By Lenz’s law and the relation between Fleming’s right-hand rule of generator action and the left-hand rule of motor action, it can be seen that the generated voltage opposes the current produced by the applied voltage that causes the motor action as shown in Figure 33.5. This generated voltage that opposes the applied line voltage is known as back e.m.f. An e.m.f. is induced whenever there is an interchange of energy. Since the back e.m.f. follows the rules and formulas of generators entirely, Current its magnitude is a linear function of rotating speed if the flux is held at a constant. As a motor comes up to speed, its back e.m.f. increases until it is a sub- Figure 33.5 Fleming’s Right-hand Rule stantial part of the imposed line voltage. This back e.m.f. is the very necesof Generator Action sary and beneficial effect that Motion regulates the armature current drawn from the lines. In relation to the current Ia in the armature winding we can, therefore, say that the rate of energy interchange is given by Field P = EIa In a motor, this power is the rate at which electrical energy is changed into mechanical energy (assuming that there are no losses in the transfer). In a generator, this power is the rate at which mechanical energy is changed into electrical energy (with the same assumption). This has been shown in Figure 33.7. V = E ± I a Ra (33.5) Ea fN Current Ta f Ia  Figure 33.6 Fleming’s Left-hand Rule If the machine is operated as a motor, the source must supply both the power of Motor Action that is converted and also the I2R loss.

642  Electrical Technology

Figure 33.7  Power Flow Diagram for a d.c. Motor and a d.c. Generator Example 33.7 A 250 V, d.c. motor is loaded to operate at 1250 r.p.m. and the armature current is 5 A. Given that the resistance of the armature is 2.0 W, determine the output torque of the motor. Solution: I 5.0 A V = E + la Ra V E = V − Ia Ra 250 V E = 250 − (5 × 2) Ra 2.0 Ω M = 240 V It follows that the converted power is P = EIa = 240 × 5 = 1200 W 1200 =

2π NT 2π × 1250 = 60 60

T = 92 N-m Note: 1. In the relation E = Blu the length l is already fixed E ∝ Bu 2. The velocity depends on the rotational speed. E ∝ fN 3. Let K be a suitable constant of proportionality, E = KfN and E1 = Kf 1N1 and E2 = Kf 2N2 Thus, E1 = φ1 N1  E2 = φ 2 N 2   a useful relation for predicting the operation of a d.c. machine.

Figure 33.8  For Example 33.7

(33.6)

Example 33.8 A 500 V d.c. motor has an armature resistance of 0.2 W and is loaded to operate at 1200 r.p.m., the armature current for that load being 40 A. If the load on the motor is increased so that the armature current is 60 A, determine the rotational speed of the rotor, given that 1. The field flux remains unchanged. 2. The field flux is increased by 10 per cent.

D.c. Motors  643

Solution:

E1 = V − I a Ra = 500 − ( 40 × 0.2 ) = 492 V E2 = V − I a 2 Ra = 500 − ( 60 × 0.2 ) = 488 V

1.



(E1/E2) = (f 1 N1/f 2 N2) where, f 1 = f 2 N 2 = ( E2 N1 / E1 ) =

2.

488 × 1200 = 1082 r.p.m. 1.1 × 492

f2 = 1.10 f1 N2 =

E2φ1 N1 1.1φ1 × E1

=

488 × 1200 = 1082 r.p.m. 1.1 × 492

Example 33.9 A d.c. motor, while carrying the normal load, draws 22.5 A from the line at 125 V. It has an armature circuit resistance of 0.45 W. Determine 1. The back e.m.f. being developed. 2. The gross developed mechanical power. Solution:   1. Va - Ec = IaRa, Ec=Va + Ia Ra  2.



Ec = 125 – (22.5) 0.45 = 114.9 V back e.m.f. Ea I a = Pd 114.9 × 22.5 = 2585 W

2585 = 3.465 hp 746 The motor is developing 3.465 hp or 2.585 kW gross mechanical power. Pd = 2.585 kW =

Note: 1. The net useful power is a bit less. 2. The rotating speed of a d.c. motor is the equilibrium result of back e.m.f. due to the speed, allowing just enough current to pass to meet the gross torque requirements. 3. The gross power developed is the equilibrium result of the gross armature circuit wattage from the lines minus the armature I 2R copper loss. 4. Any necessary shunt field wattage is also a loss. 5. The back e.m.f. developed is a major factor in determining motor performance. 6. The various types of motors are fundamentally dependent on the manner in which their field winding circuits are ­arranged.

33.7  CLASSIFICATION OF MOTORS Electric motors can be broadly classified by the type of power source needed to operate them. The three major categories of motors, as shown in Table 33.1 are the d.c. motors, the a.c. motor and the universal motor. Table 33.1 also lists many of the types of motors available within the three major categories of motors. Some of the types of motors listed in the Table 33.1 can be further subdivided. Motors can also be classified by their intended use or special characteristics, e.g., gear motors, ­synchronous motors, multispeed motors, torque motors and so on. Motors also can be grouped into one of the three broad power ratings as: integrated horse power (ihp), fractional horse power (  f hp) and sub-fractional horse power (sf hp). Motors rated at less than 1/20 hp are classified as sfhp motors, and the power is usually expressed in millihorse power (mhp) rather than in fractional horse power. Fractional horse power motors include those motors rated from 1/20 to Ihp. Any motor rated above Ihp is an ihp motor.

644  Electrical Technology Table 33.1  Major Categories of Motors Direct Current

Series Shunt Compound Permanent-magnet Brushless Stepper

Alternating Current

Universal

Split-phase Capacitor-start Permanent-split capacitor Two-value-capacitor Shaded-pole Reluctance Hysteresis Repulsion Repulsion-start Repulsion-induction Induction Consequent pole Polyphase

Non-compensated Compensated

Universal motor is designed to operate from either a.c. or d.c. power. D.c. motors are used for many applications where the control of the speed of a motor is important. There are three types of d.c. motors: series, shunt and the combination of series-shunt or compound. In the series type, the field winding is in series with the armature. In the shunt type, the field winding is arranged in parallel with the armature. In the compound type of motor, there are two sets of field windings—one set is in parallel with the armature while the other set is in series with the armature.

33.8 CONSTRUCTION All d.c. motors—regardless of size—have a stationary field member (usually called a frame or yoke) and a rotating armature member. The frame, which is made of cast or fabricated steel, serves as a means of support for the motor and forms a part of the magnetic circuit connecting the field poles and commutating poles. The field poles upon which the field coils are wound are made of cast steel, forged steel or steel laminations. When cast or forged steel is used, the core is usually made with a circular cross section. Laminated poles, as illustrated in Figure 33.9, are most commonly used (except for very small motors), have a rectangular cross section, and are fastened to the frame by bolts.

Figure 33.9  Laminated Field-pole Piece

Figure 33.10 (a) Isometric View of Armature (b) Preformed Armature Coil

D.c. Motors  645

The armature is made of machine-wound coils embedded in the parallel slots on the surface of the armature core. The core is made of thin wrought-iron or mild, of from 18 to 25 mils in thickness. The brushes are constructed as a means of carrying the current from the external to the internal ­circuit. They are usually made of carbon and are carried in brush holders (Figure 33.11) that are m ­ ounted on brush holder studs or brackets. The commutator is built up of segments of hard-drawn copper insulated from supporting rings by built up mica to form a cylinder. These segments are tightly clamped together by means of a heavy external ring. In order to improve commutation, modern motors are equipped with auxiliary poles (Figure 33.12). The variously termed interpoles or commutating poles are small auxiliary poles placed between the regular field poles. Their purpose is to assist commutation and prevent sparking at the brushes for different loads.

Spring

Brush

Figure 33.11  Brush and Holder for a d.c. Motor

Figure 33.12 Interpoles

33.9  CHARACTERISTICS OF D.c. MOTORS The two principle characteristics are the torque/armature current and speed/armature current relationships. From these Field relations, the torque/speed relationship can be derived.

33.9.1  Shunt Motors A shunt motor is connected exactly as the shunt generator is connected as shown in Figure 33.13. In the usual sense, the line voltage is constant R or nearly so. This means that the field flux f will be a constant value. A The field may or may not use a field rheostat to change the field current and thus the flux value. When the field flux change serves to change the generated voltage in a shunt generator, it affects the rotating speed in a shunt motor. This is true because the change of flux changes the back e.m.f. volts to rotating speed relation. DC Weakening the field requires a higher rotative speed to ­produce the same Figure 33.13  The Shunt Motor required back e.m.f. and, thus, increases the speed. Strengthening the field reduces the rotative speed ­required to produce the required back e.m.f. which, consequently, reduces the effective speed of the motor. The process has a limited effective range, since with a very weak field the motor will tend to be unstable with a high load. There is also a top limit to the field due to saturation. Any one motor may have a 2-to-1 up to 4-to 1 effective range of speed by field control. Wider speed ranges require compensating windings and elaborate controls.

33.9.2  Shunt Motor Speed Characteristics When once adjusted for a particular speed and when holding the same field adjustment, the shunt motor is a relatively constant speed motor over its full normal load range. This follows from ω =

Va − I a R a kφ

where, the field flux f is held

646  Electrical Technology

Speed

Shunt

Compound

nearly constant. The only variable then is Ia, the armature circuit current. If the current increases due to an increasing load, the IaRa terms increases linearly. The effect is that w will drop a small amount over the normal load range (see Figure 33.14).

33.9.3  Shunt Motor Torque Characteristics

Torque

The d.c. shunt motor has the field connected across the line. The field is independent of the armature current and any changes in the load. Any change in armature current changes the torque of the motor. When the load on the Series motor increases the motor slows down and the counter e.m.f. is reduced, allowing a greater armature current to flow and, thus, producing a larger O Input current torque. This larger torque is needed because of the increased load on the Figure 33.14  Speed Characteristics motor. When the load on the ­motor is decreased, there is a corresponding ­increase in motor speed and the counter e.m.f.; the armature current ­decreases and so does the torque. A starting resistance is needed for a shunt motor as a means of limiting the armature current. This limitation of armature current is needed TFL until the required counter e.m.f. is built up by the speed of the motor. In this motor, the starting current is small, and the starting torque is small Compound because of the added resistance in series with the armature. The torque Shunt characteristic curve is, therefore, represented by a straight line as shown Series in Figure 33.15.

33.9.4  Series Motors A series motor is internally connected with the main field coils in series IFL with the armature circuit. This means that all armature circuit current passes 0 Input current through the fields. The field coils are then designed and built with relatively few turns of a relatively large cross-section. The few field coil turns supply Figure 33.15  Torque Characteristic the needed ampere turns with the heavy armature circuit current. The large cross-section results in low resistance and, therefore, low wattage field losses even with high current. In a series motor, as illustrated in Figure 33.16, the field flux is entirely dependent on the armature current. When operating below the knee of the saturation curve, the field magnetic flux is then directly proportional to the d.c. armature current. Figure 33.l6  The Series Motor

33.9.5  Series Motor Speed C ­ haracteristics In a series motor the speed versus back e.m.f. relations are modified



(33.7)

In the series case, the armature circuit has the added resistance of the series fields, which increases the voltage drop due to resistance and reduces the back e.m.f. by that amount. The field flux is now proportional to the fixed number of turns in the field and the particular saturation curve, represented as k , and the field excitation current, which is now Ia. k is then not constant over the full range of operation. The speed of the motor is still directly proportional to the counter e.m.f. although the counter e.m.f. n­ umerator term, Ia(Ra + Rsc), is slightly different. The speed is still inversely proportional to the ­denominator term, but now the denominator has an Ia factor. Armature current la is a direct multiplier to the ­entire denominator and, thus, the la effect is far larger in the denominator. Since the flux f is now directly related to Ia, an increase in Ia must inversely reduce S. The speed curve closely resembles a hyperbola, as represented in Figure 33.16. The high speed at small values of current indicates that this type of motor must not be run on very light loads and, invariably, such motors are permanently coupled to their loads.

D.c. Motors  647

33.9.6  Series Motor Torque Characteristics Torque T varies as Ia (T ∝ Ia). Since the armature and field currents are the same current, I, in a series machine then T ∝ fI over a limited range, before magnetic saturation of the magnetic circuit of the motor is reached (i.e., the linear portion of the B-H curve for the yoke, pole, air gap, brushes, and armature in series). Thus, f ∝ I and T ∝ I 2 After magnetic saturation, f almost becomes a constant and T ∝ I. This is shown in Figure 33.15. The torque increases nearly parabolically and then blends into a straight linear increase.

33.9.7  Compound Motors When a d.c. motor has both a shunt and a series field, it is known as a compound motor, as shown in Figure 33.17. If the motor is connected in such a way that the series field aids the shunt field, it is known as a cumulative compound motor. When the series field is connected to oppose the shunt field, it is then a differential compound motor. The terms are identical to those used in identifying d.c. generators. The compound motor acts with a combination of the characteristics of shunt and series motors. The larger the ­effect of the series field, the more the characteristics r­esemble those of a series motor. Although any range of performance characteristics is available, only certain regions of the potential range are normally used. The cumulative compound motor develops a high torque to match an increase in the torque load as does a series motor. However, the cumulative compound motor has a definite and controllable no-load speed, so that there is Figure 33.17  The Compound Motor no run away problem. This makes the type particularly adaptable to uses requiring sudden application of heavy load, such as in rolling mill drives. A particular advantage under sudden but short duration heavy load is that when the motor drops in rotative speed as it is loaded, it gives up a portion of its kinetic energy to drive the load. If the speed were to be held more closely, the transient would have to be met by high peak currents from the supply line. Cranes, hoists, and elevators use cumulative compound motors since they can smoothly start a heavy load and yet not over speed when unloaded. The series field is frequently cut out of the circuit automatically when the hoist is up to speed. The steady state situation is then handled as if by a shunt motor. A further advantage is the ability to use the motor as an adjustable brake by using it as a generator with a descending load, since it has the shunt field available. The differential compound motor is not too widely used, but it does have some special characteristics that lend it to some special services. When the series field opposition effect is adjusted so that the loss of flux effect just cancels the loss of speed of a shunt motor when loaded, a substantially constant speed results. There are two major performance problems, however, in a differential compound motor that severely limit its use: (1) when set for a flat or even rising speed with load characteristic, the motor will tend to run away or severely increase its speed with a high load; and (2) this same effect of the series field tending to take over will cause a differential compound motor to start in the opposite direction of that desired unless the starting current is held down. Usually, the series field is cut out by special switching during starting. The differential compound motor is a special purpose machine. The speed characteristics of cumulative compound motors can be developed in the following way Va − I a ( R a + R sc ) S = (33.8) kφsh + kk ′I n  Here, the numerator is the same as in the series motor, and the denominator is the sum of the series and shunt cases. Va − I a ( R a + R sc ) For the differential compound motor S =  (33.9) kφsh + kk ′I a These equations differ only in the sign (+ or –) in the denominator. It can be seen that if Ia becomes large enough, the second term in the denominator will overcome the first term. In this case, there is no effective field and full short-circuit current will flow. As the kk´1a term increases, the total effective flux becomes less and less and the motor accelerates dangerously. This has been illustrated in Figure 33.18.

Motor speed (r.p.m. or rad /s)

33.9.8  Compound Motor Speed Characteristics Differential compound

Shunt Cumulative compound Series Rated load current Armature current Ia (A)

Figure 33.18 Speed Characteristic of Compound Generator

648  Electrical Technology

Regardless of compounding, the current in the shunt field circuit and the field flux during starting and running is essentially constant. The current in the series field is a function of the load current drawn by the armature. Starting with a flux equal to the shunt field flux at no load and one that increases with armature current, the cumulative compound motor produces a torque curve that is always higher than that of the shunt motor for the same armature current, as shown in Figure 33.19. For the differential compound motor starting with a flux equal to the shunt field flux at no load, any value of armature current will produce a series field m.m.f. that reduces the total air gap flux and hence the torque. Thus, the differential compound motor produces a torque curve that is less than that of the shunt motor.

N-

33.9.9  Compound Motor Torque Characteristics

Figure 33.19 Comparison of Torque-load Characteristics for d.c. Motors

33.10  RELATION BETWEEN TORQUE AND SPEED OF A MOTOR Since torque is defined as a force tending to produce rotation, increasing the field flux would tend to increase the torque and (possibly) the speed (T = kfIa N-m). On the other hand, increasing the flux would reduce the speed S =

Va − ( I a R a + Brush Drop) Kφ

There is no inconsistency and this is explained. What happens when the field flux is reduced?

Speed (r.p.m.)

1. The field flux of a shunt motor is reduced by decreasing the field current. 2. The counter e.m.f. drops instantly. 3. The speed remains constant as a result of the inertia of large and heavy armature. 4. The decrease in Ec causes an increase in armature current. 5. A small reduction in field flux produces a large increase in the a­ rmature current. 6. The small decrease in flux is more than counterbalanced by a large increase in the armature current. 7. The torque increases more than the flux was reduced. 8. This increase in torque produces a corresponding increase in the speed. To summarize, decreasing field current (and field flux) results in an increase in the speed.

Torque

Shunt Compound

Light

Heavy Series

Figure 33.20  Torque Versus Speed

33.10.1  Torque Versus Speed The torque-speed characteristics of these motors are illustrated in Figure 33.20. The speed of the shunt motors does not change to a great extent as the torque increases. However, the speed of the series motor drops ­considerably as the amount of the required torque is increased. The compound motor has characteristics that lie between those of the series motor and the shunt motor. A heavily compounded motor that acts like a series motor is one that has more series turns than it has shunt turns.

D.c. Motors  649

Example 33.10 A motor develops a torque of 150 N-m and is subjected to 10 per cent reduction in field flux, which produces a 50 per cent increase in the armature current. Find the new torque produced as a result of this change in field flux. Solution: f Ia T Original condition 1.0 1.0 150 N-m New condition 0.9 1.5 ? T = kf Ia Using the ratio method, the new torque is the product of two new ratio changes. 0.9 f 1.5I a T = 150 = 202.5 N-m 1.0 f 1.5I a Example 33.11 A d.c. shunt motor having an armature resistance of 0.25 W and a brush contact voltage drop (BD) of 3 V receives an applied voltage across its armature terminals of 120 V. Calculate the armature current when 1. The speed produces a counter e.m.f. of 110 V at a given load 2. The speed drops (due to application of additional load) and the counter e.m.f. is 105 V. 3. Compute the percentage of change in counter e.m.f. and in armature current. Solution: 1. I a =

V − ( E c + Brush Drop)

=

Ra

120 − (110 + 3) 0.25

= 28 A 2. At increased load,

3.



120 − (105 + 3) 0.25 = 48 A

Ia =

dEc =

110 − 105 , 100 = 4.54 per cent 110

dI a =

28 − 48 , 100 = 71.4 per cent 28

Note: 1. A small increase in counter e.m.f. (4.54 per cent) has resulted in a much larger increase in armature current (71.4 per cent). 2. Small changes in motor speed and counter e.m.f. are accompanied by correspondingly large changes in the ­motor current. 3. In some types of servomotor transducer devices, the motor current is used as an indication of motor load and motor speed. Example 33.12 The shaft torque of a d.c. motor driving a 100 V d.c. shunt-wound generator is 2.5 N-m. The armature current of the generator is 1.6 A at this value of torque. If the shunt field regulator is adjusted so that the flux is reduced by 15 per cent, the torque increases to 35 N-m. Determine the armature current at this new value of torque. Solution:

T1 = k f1Ia T2 = k f2Ia

1

Hence,

2

φ1 I a1 T1 = T2 φ 2 I a 2 =

and

Ia 2 =

16 × 35 = 26.35 A 0.85 × 2.5

φ1 I a1 T2 φ 2T1

650  Electrical Technology Example 33.13 The armature of a d.c. machine has a resistance of 0.25 W and is connected to a 300 V supply. Calculate the e.m.f. generated when it is running 1. As a generator giving 100 A. 2. As a motor taking 80 A. Solution: 1. As a generator



E = V + Ia Ra = 300 + (100) (0.25) = 325 V

2. As a motor

E = V – Ia Ra = 300 – (80) (0.25) = 280 V

Example 33.14 An 8-pole d.c. motor has a wave wound armature with 900 conductors. The useful flux per pole is 25 mWb. Determine the torque exerted when a current of 30 A flows in each armature conductor. Solution: pφ ZI a N-m T = πc –3

p = 4, c = 2, f = 2.5 × 10 Wb, Z = 900, Ia = 30

T =

(4)(25 × 10 −3 )(900)(30) π (2)

= 42.91 N-m

Example 33.15 A 200 V d.c. shunt-wound motor has an armature resistance of 0.4 W and at a certain load has an armature current of 30 A and runs at 1350 r.p.m. If the load on the shaft of the motor is increased so that the armature current increases to 45 A, determine the speed of the motor assuming that the flux remains constant. Solution: E∝f n applies to both generators and motors. For a motor E = V – IaRa Hence, E1 = 200 – (30 × 0.4) = 188 V and E2 = 200 – (45 × 0.4) = 182 V

Since the flux is constant, f1 = f2 1350 188 60 Hence, = and n2 182

φ1n1 E1 = E2 φ 2 n2

n2 =

22.5 × 182 = 21.78 rev/sec 188

The speed of the motor = (21.78 × 60) = 1307 r.p.m. Example 33.16 A 4-pole 440 V d.c. motor takes an armature current of 50 A. The resistance of the armature circuit is 0.3 W. The armature has a wave winding with 850 conductors and the useful flux per pole is 0.025 Wb. Calculate the speed of the motor. If the machine is connected across 220 V supply calculate the new approximate speed. It is to be assumed that the new flux is 0.02 Wb.

D.c. Motors  651

Solution: The generated e.m.f. for a motor

E = V − I a Ra E = 440 − ( 50 × 0.3 ) = 425 V E =

2 Znp φ  850  = 2  × n × 2 × 0.025  a 2 

n = 10 r.p.s. = 600 r.p.m. V φ k 440 × 0.02 n1 V1φ 2 = = n2 V2φ1 220 × 0.025

n=

n1 = 1.6 n2 10 n1 = = 6.25 r.p.ss . = 375 r.p.m. n2 1.6   Example 33.17 A 6-pole 480 V d.c. motor takes an armature current of 110 A. The lap-wound armature has 864 conductors. Calculate (1) the speed and (2) the gross torque developed in the armature. Assume flux per pole to be 0.05 Wb and armature resistance as 0.2 W. Solution: 1. E = 480 – 110 × 0.2 = 458 V 458 = (2 × 864/6) × n × 3 × 0.05 n → r.p.s. n = 10.6 r.p.s. = 636 r.p.m. 2. Mechanical power developed If T is gross torque in N-m

Pm = 458 × 110 = 50380 Watts 2p nT = 50380 50380 T = = 756 Nm N-m 2π n

33.11  DIRECT-CURRENT, MOTOR STARTING PROBLEMS It is desirable to keep the armature circuit resistance of a d.c. motor as low as it reasonably can be. Under normal operating conditions, this low resistance is entirely beneficial. However, during starting and acceleration, excessive current would flow if the full line voltage were placed across the armature circuit. In a complete circuit, current equals voltage divided by resistance I = E/R. In a normal operation, the current is adjusted by the presence of the back e.m.f. Thus, the normal current is limited by the fact that the effective voltage across the low armature resistance is a self-adjusting result of I = (Vp – Vc)/Ra. Since the Ec factor is not available at the start, the Ra must be increased by added resistance or the Vl must be reduced in some fashion. In the usual industrial situation, the line voltage is fixed or nearly so. Such fluctuations as may take place are due to the source characteristics and the total line voltage drops between the source and the motor. It is usually desirable to limit the maximum starting current to 125 to 150 per cent of the normal running current. Occasionally, a value of 200 per cent or more may be used under special conditions. The desired conditions can be equated by the following modifications. Ir × M =

V1 − E c

(33.10)  The full-load rated motor current, Ir times an agreed upon multiplying factor for the allowed starting current M equals the line voltage Vl minus the counter voltage (if there is one) divided by the total armature circuit resistance Ra plus the needed Ra + Rs

652  Electrical Technology starting resistance Rs. The total armature circuit resistance will include all series windings—such as a series field—unless it is specially shorted out, as would be done in a differential compound motor. Vl − E c Rs = Ra (33.11) Ir M 

33.12  D.c. STARTING SWITCH Usually the motor is allowed to accelerate until the high starting current drops down to the normally rated current by the effect of a partial counter voltage. Then the resistance is reduced until the desired maximum current is again reached, whereupon further acceleration takes place. Figure 33.21 shows a simplified schematic diagram of the current-limiting parts of a typical starting switch. The actuating mechanisms are not shown. When the current has dropped from Ir × M to a value Tapped current to or near Ir, which may be called Il, there is some counter limiting resistance voltage. This is shown as Ec=Vl – Il(Ra+Rs) (33.12) The intermediate speeds that are reached when the current decays to the Il arbitrary lower limit is approximated as Series field Ec Control S = S rated × (33.13) or arm Armature El − ( I l × R a )  d.c. supply Ec ω = ω rated × (33.14) El − ( I l × R a ) Figure 33.21  Simplified Starting Switch  These formulas are seen to be the rated angular velocity multiplied by the ratio of the counter or back e.m.f., divided by a term that is the normal operating counter voltage. This approximation is invalid for currents that are greatly different from the rated Il motor with a series field. Example 33.18 A d.c. motor is rated at line conditions of 230 V and 27.5 A at normal full load. It turns at 1750 r.p.m. (183.3 rad/sec) under rated conditions. It has a total armature circuit resistance of 0.803 W, and it is desired to hold its maximum starting current to 150 per cent of the normal full-load current. Determine 1. The current that would flow if there were no added starting resistance. 2. A total starting resistance to meet specified conditions. 3. The rotational speed that may be expected, when the motor has accelerated sufficiently to reduce the line current to the normal rated value. 4. Intermediate values of starting resistances that will allow 150 per cent of rated current to flow for further acceleration and the speeds reached when the current decays to 100 per cent. Solution: 1. There is no added resistance and no back e.m.f.: I= E/R = 230/0.803 =286.4 A This current is obviously excessive and high currents such as this are the reason why starting resistances are used. It is (286.4/27.5) × 100 = 1041 per cent of the specified current or 1041/150 = 6.94 times the desired starting current, which is in itself an overload. V − Ec Rs = l − Ra Ir M 2. =

230 − 0 − 0.803 = 4.77 Ω = R s total 27.5 × 1.5

In a real sense high accuracy is not needed and the resistance would be satisfactory if between 4.5 and 5.0 W.

D.c. Motors  653

3. First determine counter voltage at rated current with Rs tot in the armature circuit E c = Vl − I l ( Ra + R s tot

)

230 – 27.5(0.803 + 4.77) = 76.7V = Ecl Now, S = 5rated 1750

Ec El − ( I l × R a )

76.7 = 645.5 r.p.m. = S1 230 − (27.5 × 0.803)

4. Since a back e.m.f. and a partial speed now exist, the starting resistance is now reduced so that 150 per cent of rated current again flows and the process is repeated. 230 − 767 − 0.803 = 2.91 Ω = R s 2 2751.5 The starting resistance is then reduced from 4.77 W to 2.91 W at the second point on the starter. The second r.p.m. is now determined . 230 − 27.5 ( 0.803 + 2.91 ) = 127.9 V = Ec 2 and

1750 ×

127.9 = 1076 r.p.m. = S2 230 − (27 .5 × 0.803)

The motor increases from 645.5 up to 1076 r.p.m. Repeating the process as many times as necessary, we find steps 3, 4 and 5. Rs3 = 1.67 W S3 = 1363 W Rs4 = 0.845 W

S4 = 1554 W

Rs5 = 0.335 W S5 = 1682 W An Rs6, resistance would be –0.0708 W. The minus value indicates that no additional resistive steps are needed to avoid a current of over 150 per cent of the rated value once a velocity of 1682 r.p.m. is reached. The sixth stage of the starting sequence puts the motor directly across the line with safety. Note: EC5 = 199.8 V,

}

230 − 198 = 37.6 A 0.803

27. 5′ 1. 5 = 41. 25 A The sixth step is not required 37.6 A < 41.25 A Example 33.19 A 120 V d.c. shunt motor has an armature resistance of 0.2 W and a brush volt drop of 2 V. The rated full load armature current is 75 A. Calculate the current at the instant of starting and the percentage of load current. Solution: V − BD 120 − 2 I st = a = Ra 0.2 = 590 A (counter e.m.f. is zero) Percentage of full load =

590 × 100 = 786 per cent 75

Note: The starting current, in this case, is approximately 8 times as great as the rated full-load armature current, due to lack of counter e.m.f. at the instant of starting. Example 33.20 Calculate the various steps (taps) of starting resistance to limit the current in the motor in Example 33.18 to 1. 150 per cent rated load at the instant of starting. 2. A counter e.m.f. that is 25 per cent of the armature voltage Va at 150 per cent rated load.

654  Electrical Technology 3. A counter e.m.f. that is 50 per cent of the armature voltage at 150 per cent rated load. 4. Find the e.m.f. at full load without starting resistance. Solution: 1. At starting, Ec is zero 120 − 2 V − Brush Drop Rs = a 0.2 = 1.05 − 0.2 Ra = 1.5′ 75 Ia = 0.85 Ω 2. R = 120 − ( E c + Brush Drop) Ra = 20 − 30 − 2 0.2 s Ia 1.5′ 75

  

= 0.582 Ω 120 − (60 + 2) 3. R s = 0.2 = 0.516 − 0.2 = 0.316 Ω 1.5′ 75 4. Ec = Va − ( I a Ra + BD ) = 120 −  ( 75 × 0.2 ) + 2  = 103 V A progressively decreasing value of motor starting resistance is required as the motor develops an increased c.e.m.f. owing to acceleration. This is the principle of the armature resistance motor starter. The manner in which a starter is used in conjunction with the three basic types of d.c. dynamos, used as motors, is shown in Figure 33.22. The techniques shown here for starting a motor are schematic diagrams only. Commercial forms of manual and automatic starters and controllers differ somewhat from these. The shunt and compound motors are started with full excitation (i.e., the full line voltage is impressed across the field circuit) in order to develop maximum torque (T = kf Ia). In all the three types, the armature starting current is limited by high-power series connected variable starting resistor. In commercial practice, the initial inrush of armature current is generally limited to a higher value than the full-load current to develop greater starting torque, particularly in the case of large motors that have great inertia and that come up to speed slowly. With the starting arm at position 1 in Figure 33.22 (a), the maximum series resistance will limit the armature current on starting to about 150 per cent of its rated value. As the motor slowly picks up speed, the armature develops c.e.m.f. and the armature current drops to approximately full load. If the starting arm were left at position 1, the armature current would drop somewhat and the speed would stabilize at a value well below the rated speed. In order to a­ ccelerate the motor armature once more, it is necessary to move the arm to position 2. Again, there is an inrush of armature current and the motor rises in speed. This process is continued until the motor armature attains its rated speed, without the need for a series armature resistance and where the c.e.m.f. at that speed is sufficient to limit the armature current. All three types—if started with a mechanical load coupled to the armature—will accelerate more slowly than if started without load. The ­series motor, particularly, should never be started without load coupled to its armature. The shunt and compound motors, on the other hand, may be started with or without mechanical load. Manual starters require some experience in moving the contact arm Figure 33.22 Starter Connections for Shunt, through the various steps of resistance to accelerate the motor to rated Series, and Compound Motors speed without producing excessive armature current. Automatic starters in Schematic Form (a) Shunt are designed electrically to a­ ccelerate the motor to each resistance step, Motor Starter (Schematic Form) regardless of the degree of motor loading, without damage to the motor. (b) Series Motor Starter (ScheNote: Except in the very small sizes, a d.c. motor always needs some matic Form), Compund Motor sort of starting device and frequently needs a speed control. Starter (Schematic Form)

D.c. Motors  655

33.13  D.c. MOTOR REVERSING The direction of rotation of any d.c. motor depends upon the magnetic polarity of its main fields and the direction of the conventional current that flows in the armature windings that are immersed in the fields. The resultant direction is then in accordance with the left-hand rule of motor action. The left hand rule relationship is met under all the field poles in any motor. The methods of winding the armature all accomplish the same thing in a motor: the current is d­ irected oppositely under both north and south poles; so each region pulls in the same direction. When the flux—as shown by the index finger—flows inward to the armature, as under a north pole, the current must go parallel to the assumed shaft direction. If an adjacent and, therefore, south pole is simulated, the flux points out from the shaft. Then, if the hand is further rotated so that the current is also opposite to its original relationship, the thumb (which represents motion) is pointed in the original direction. All this means that rotational direction is determined by four factors: 1. The direction of the field coil winding, which is built in. 2. The connection polarity of the whole field, which may be changed or switched. 3. The direction of the armature coil winding, which is also built in. 4. The connection polarity of the brush group, which is the fixed access point to the armature winding and which may be changed or switched. Factors (2) and (4) are accessible modifications and may be manipulated. Conversely, factors (1) and (3) are fixed by the construction of the unit and cannot realistically be changed. If the overall terminal polarities of a d.c. motor are changed, the current direction of both the armature and the field are changed. Since this, in effect, does the same thing as moving from one pole to another locally, the rotating torque direction is not changed. The exception to this is a permanent magnet field motor where only the armature has coil windings. In this case, reversal can be produced only by reversing the line connections. A reversal is accomplished by changing the polarity of either the armature or the field, but not by changing both (see Figure 33.23).

Figure 33.23  Direct Current Motor Reversing (a) Shunt Motor Reversing (b) Series Motor Reversing (c) Compund Motor Reversing

33.13.1  D.c., Motor Reversing Circuit Connections In a shunt motor, the field normally has much less current than the armature, and so it would seem that the field is the logical place to switch. This is not usually the best practice since the field is a highly inductive circuit. Thus, if any switching is performed before the field current has fully decayed the switch points will arc viciously or even dangerously in large sizes. Furthermore, since any switching is usually the most unreliable part of an electric circuit, it is unwise to switch where the basic action desired contributes to further unreliability. Finally, upon failure of a field reverse switch contact, the next start will take place with little or no field, which causes high-speed run away. For these reasons, the armature circuit is usually the reversed part.

656  Electrical Technology In a series motor as illustrated in Figure 33.23 (b), it does not make much difference since the series coil is much less inductive with its fewer winding turns. The field also has the same current as the ­armature. A compound motor shown in Figure 33.23 (c) must have both fields changed if field reversing is used, so armature switching is less complex. If only one field were reversed, a compound motor would be changed from cumulative to differential, which changes the whole character of the motor.

S UM M A RY 1. The d.c. generator and the d.c. motor are essentially the same device. 2. Since the generated e.m.f. in a motor is in opposition to the applied voltage, it is referred to as the back e.m.f. or counter e.m.f. 3. Any rotary power producing device is a producer of continually rotating torque. 4. T = BIa IZ ( per cent covered) d N-m. 5. Since all the prime mover power is converted to friction caused by heat, in large sizes constant cooling is required. 6. In small sizes, a two-scale prony brake may be used. 7. When a dynamometer is absorbing torque, its field tends to be pulled around equally by the motor action that exists. 8. It is not necessary to measure the electrical quantities in the dynamometer in order to measure the torque or power. 9. The bulk of energy produced by a dynamometer is dissipated in a resistive load bank which is remotely located. 10. The magnitude of back e.m.f. is a linear function of rotating speed. 11. Electric motors can be broadly classified by the power source needed to operate them.

1 2. Motors can also be grouped into three power ratings. 13. There are three types of motors: series, shunt, and ­compound. 14. The two principle characteristics of a motor are the torque/armature current and the speed/armature c­ urrent. 15. A shunt motor is a constant speed motor over its full normal load range. 16. The torque characteristic of a shunt motor is represented by a straight line. 17. The series motor must not be run on very light loads and invariably such motors are coupled to their loads. 18. The torque of series motors increases nearly parabolically and then blends into a straight linear increase. 19. The compound motor acts with a combination of the characteristics of series and shunt motors. 20. A progressively decreasing value of the motor starting resistance is required as the motor develops an increased c.e.m.f. conforming to acceleration. 21. In all three types of motor, the armature starting current is limited by a high-powered series-connected variable starting resistance.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. Torque on a single loop of wire in a magnetic field is (a) Constant (b) The same as the field flux (c) The turning force (d) Never at a maximum value

2. If another loop of wire is added to make a double-loop armature

(a) (b) (c) (d)

The torque becomes steadier The torque becomes less smooth The magnetic field decreases in value Commutator segments must be reduced

3. D.C. motors are rated in

(a) (b) (c) (d)

Voltage, current, frequency and speed Voltage, current and horse power Voltage, current, speed and torque Voltage, current, speed and horse power

4. The generator effect in a motor produces a

(a) (b) (c) (d)

High power factor Counter electromotive force High resistance Reduced line voltage

5. A d.c. motor draws more current with a m ­ echanical load coupled to its shaft because

(a) (b) (c) (d)

Counter e.m.f. is reduced with speed Voltage differential decreases Applied voltage decreases Torque depends on the magnetic strength

6. The speed of a d.c. motor may be reduced below its rated speed without losing torque by reducing the voltage at the (a) Motor (c) Armature

(b)  Series field (d)  Armature and field

7. Advantages of d.c. motors are

(a) (b) (c) (d)

Simplicity in construction Speed control above and below base speed Excellent torque and speed Horse power for size

8. The direction of rotation of a compound interpole motor may be reversed by reversing the direction of current flow through the (a) Armature (b) Armature and field circuit (c) Armature, interpole and series field (d) Shunt field

D.c. Motors  657

9. For a series motor

15. The speed of a d.c. shunt motor

(a) The field is operated below saturation (b) An increase in both the armature current and the load current because of an increase in load (c) The reduction in speed due to an increase in load is greater than that in the shunt motor (d) All of these



10. Brushes of a d.c. motor ride on



(a) Commutator (c) Shaft

(b)  Armature (d)  Commutating pole

11. The twisting effect of a motor is called (a) Turning power (c) Torque

(b)  Horse power (d)  Brake horsepower

12. The twisting effect of a d.c. motor is produced prima­rily by

(a) The armature (b) A current carrying conductor in a magnetic field (c) The rotor (d) Torque in the field coils.

13. A d.c. motor is required to maintain the same speed at full load as at no load. This type of operation can be obtained by using a

(a) (b) (c) (d)

Series motor Cumulative compound motor Shunt motor Differential compound motor

(a) Increases with an increase in load (b) Decreases with an increase in applied voltage (c) Decreases if the field strength is increased (d) Decreases less than in a series motor of the same hp for the same increase in load

16. As load is applied to a d.c. motor the (a) (b) (c) (d)

Field current decreases Field voltage increases Armature current decreases Armature voltage increases

17. The load requirements of a particular d.c. ­motor installation require extremely high starting torque. If speed regulation is not i­mportant, use a

(a) (b) (c) (d)

Series motor Differential compound motor Shunt motor Cumulative compound motor

18. As a load is applied to a cumulative compound-wound d.c. motor its

(a)

Speed decreases (b) Counter e.m.f. decreases (c) Torque decreases (d) Series field current decreases

14. As a load is applied to a d.c. shunt motor the

(a) (b) (c) (d)

Field current increases Counter e.m.f. increases Armature current increases Torque developed decreases

ANSWERS (MCQ) l. (c)  2. (a)  3. (d)  4. (d)  5. (b)  6. (d)  7. (b) 8. (d)  9. (d)  10. (a)  ll. (c)  12. (a)  13. (c)

14. (c)  15. (d)  16. (d)  17. (a)  18. (a).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. Explain the different parts of a d.c. motor. 2. How does a d.c. motor work? 3. Write short notes on the following: (a) Back e.m.f. (b) Gross torque (c) Shaft torque 4. Explain the methods employed to achieve sparkless commutation in an a.c. motor. 5. What are the salient features and fields of application of different types of d.c. motors? 6. Why should a series motor never be run on no load? 7. A motor is running at 950 r.p.m. and the torque exerted at the pulley is 150 N-m. What is the horse power being transmitted? 8. What is torque? 9. What is the source of torque force in a d.c. motor?

1 0. How does a prony brake measure torque? 11. What is the mechanism of torque measurement with a dynamometer? 12. What is the back e.m.f. or counter voltage? 13. How is back e.m.f. beneficial? 14. What is the effective voltage across a d.c. motor armature? 15. What is the meaning of armature power? 16. Is electrical power related to mechanical power? 17. What is meant by equilibrium motor speed? 18. What is meant by the term shunt motor? 19. What is the dominant speed characteristic of a shunt motor? 20. What relationship exists between speed and load torque in a series motor?

658  Electrical Technology 2 1. What is meant by a compound motor? 22. What conditions require the use of a compound motor? 23. What is one major problem in d.c. motor starting? 24. How is a d.c. motor reversed? 25. If the power transmitted by the shaft of a motor is 50 hp, the speed being 480 r.p.m., what is the torque? 26. A 100 V shunt motor is taking a current of 220 A. The armature resistance is 0.015 W and shunt field resistance is 20 W. Calculate the back e.m.f. and power spent in turning the armature. 27. A series motor takes 40 A at 220 V and runs at 800 r.p.m. The armature resistance is 0.2 W, series field resistance 0.1 W, iron and friction losses l/2 kW. Calculate (a) total torque, (b) shaft torque and (c) bhp. 28. Derive the standard torque equation of a d.c. motor from first principles. 29. Draw the speed load characteristics of series, shunt and cumulatively compounded motors. ANSWERS (CQ) 7. 153.8 hp 25. 740 N-m 26. 96.8 V; 20.807 W 27. 99.3 N-m; 93.2 N-m; 10.48 hp;

30. A d.c. shunt motor is rotating at 267.0 rad/sec and developing 57.2 N-m of torque. How many Watts of mechanical power are developed? 31. A 125 V d.c. motor for an electric automobile has an armature circuit resistance of 0.042 W. It is operating at a steady speed and drawing 135A armature current. What is the back e.m.f.? 32. A shunt motor develops a torque of 250 N-m at rated load. When it is subjected to a 15 per cent decrease in field flux, the armature current increases by 40 per cent. Calculate the new torque produced as a result of the change in field flux. 33. A 220 V shunt motor develops a torque 54 N-m at an armature current of 10 A. Find the torque when (a) the armature current is 15 A (b) the armature current is 20 A (c) the armature current is 5 A

30. 15.3 kW 31. 119 V 32. 297.5 N-m 33. 81 N-m; 108 N-m; 27 N-m

Efficiency of Direct ­Current Machinery

34

OBJECTIVES In this chapter you will learn about:  The fixed and variable losses in electric machines (d.c.)   Relationships between losses   Efficiency of electric machines (d.c.)  Simple problems on the above Efficiency

34.1 INTRODUCTION The efficiency of any device or process is simply a ratio of its useful output to its gross input. The output and input must be measured in the same units. Since the operation of any device requires energy in order to produce its desired output and furthermore, since the amount of energy lost in a process is inversely related to its efficiency, the shortage (worldwide) of energy makes this subject extremely important. Forces must move through distances within the desired times in order to perform work at the desired rates. However, when work is performed, energy must be consumed since work and energy are synonymous. Force times distance is work or energy. Power brings in a time relation. The rate of performing work determines the rate of the use of energy. The rate of doing work depends upon the rate of consumption of energy times the efficiency of the process. If a process is 50 per cent efficient, it then takes twice as much energy to perform the process than if it were 100 per cent efficient. The difference is lost and mostly unrecoverable heat and a part, at least, is difficult-to-recover heat. A motor or a generator has a physical size that is directly related to the amount of heat that it has to dissipate in operation. This heat dissipation, of course, depends on the losses within the machine. Since the size of a machine of any sort is a major factor in its weight and cost, the smallest machine that will perform a given task usually has an advantage in the cost. The smaller machine may be more efficiently loaded and, if so, it will use less energy.

34.2  BASIC EFFICIENCY RELATIONSHIPS The efficiency of any machine, such as a motor or generator, is then simply its output power divided by its input power when they are in the same units. The output is the input minus the various accumulated losses. Similarly, the input is the output plus the same losses. For d.c. motors and generators, these losses turn out to be regular and predictable. Some losses can be nearly eliminated. By careful design, all of them can be reduced. Efficiency formulas are output power Efficiency = × 100 per cent = η per cent (34.1) input power  (η is the lower case Greek letter eta) output = input − Σ losses  (34.2) input = ouput + Σ losses  (∑ is the Greek upper case sigma, which is used to mean the summation of) input − Σ losses Efficiency = × 100 per cent = η per cent input 

(34.3)

(34.4)

660  Electrical Technology Efficiency =



output × 100 per cent = η per cent output + Σ losses



(34.5)

Example 34.1 A 14.92 kW motor operating on 125 V takes 144 A when operated at rated conditions. Determine (1) the losses involved and (2) efficiency. Solution: 1.

input power = 125 × 144 = 18000 W or 18.0 kW Input = output + ∑ losses or Out put − input = ∑ losses 18.0 − 14.92 = 3.08 kW losses

2. Efficiency may be determined in a number of ways output 14.92 × 100 per cent = × 100 = 82.9 per cent η = input 18.00

η =

input − Σ losses 18.00 − 3.08 × 100 per cent = × 100 = 82.9 per cent input 18.00

η =

output 14.92 × 100 = 82.9 per cent × 100 per cent = output + Σ losses 14.92 + 3.08

34.3  TYPES OF LOSSES IN D.C. MACHINES With a generator, the difference between mechanical power input and electrical power output is composed of a family of losses. The mechanical power input is normally measured and stated in horse power and is converted to Watts or kilowatts. TS/5252.1 = horse power and the equivalent of horse power × 0.74570 = kilowatts. The most widely used equivalent is 0.746 = kw.

34.3.1  Rotational Losses Following Figure 34.1 on a step by step basis, the mechanical input power is first reduced by the rotational losses. In other words, any power required to turn the armature is a direct subtraction from the input power. Therefore, the rotational power is not available for the development of electrical power. Rotational losses are a summary of bearing friction, brush mechanical friction, cooling fan power, windage loss of armature as a whole and magnetic circuit drag loss. Since the armature rotates within the magnetic field, the magnetic flux path in the armature laminations is required to change continually. Any one portion of the armature must charge and discharge cyclically. The resultant hysteresis losses show in the form of consumption of mechanical rotation power. There is yet another magnetic-related loss called eddy current loss. This loss occurs because any conductor that moves in a magnetic field has a voltage generated within it. Since the laminations themselves are conductive, they have internal voltages that circulate in whatever path that is available. These losses would be very substantial without a laminated structure and would cause much heating in the core. Even with laminations, they are not entirely negligible, but they are small. All these losses can be isolated and measured, but ordinarily they are taken as a whole, where Prot = IaVa (34.6) The voltage and current are taken when running at no load. Note: If there is no field of a particular type, its loss type is not applied; if no series field then no series field loss, etc.

34.3.2.  Winding Resistance Losses The electrical power losses are mostly due to the various resistances that are present in the various parts of the windings. Since the armature windings are undergoing reversals of polarity and current, they have some inductive effect. The various resistances may be summarized as Ra. Since all these windings are subject to the same current, the losses are

P a = I a 2 ×R a 

( 34.7 )

Efficiency of Direct ­Current Machinery  661

Figure 34.1  Generator Losses, Direct Current These losses vary as the load current squared and are thus known as variable losses. I a 2 R a is the largest single loss in the whole machine. After the rotational losses and all the armature circuit losses are subtracted, the series circuit output power remains.

34.3.3  Shunt Field Loss The final generator loss is the winding loss in the shunt field. This is normally considered as a fixed loss, since for any particular excitation adjustment of the shunt field rheostat, it is fixed if the line voltage is fixed. When the shunt field circuit resistance is known, Psh = Ish2 Rsh (34.8) If the field circuit voltage is known, the loss is Psh = Vsh × Ish (34.9) In a long-shunt compound generator, Vsh = Vl. In a short shunt situation, the Vsh voltage is subject to a series field voltage drop Vsh = Vl – Il Rsc (34.10)

662  Electrical Technology Example 34.2 A 75 kW d.c. generator is operated at 230 V. Tests have shown that the rotational loss is 1810 W and the shunt field circuit draws 5.35 A. The armature circuit has a resistance of 0.035Ω and the brush drop is 2.2 V. Calculate (1) the rated current delivered, (2) the total losses, (3) the input power required, and (4) the efficiency at the rated load. Solution: P 75000 1. I = = = 326 .1 A I 230 2.    P = IE = 5.35′ 230 = Psh = 1230 W The rotational loss is given as 1810 W. The variable losses are found from the armature current, which is 326.1 + 5.35 = 331.45 A. Therefore, the copper loss in the armature circuit is Ia 2 × R a = (331.45) 2 (0.035) = Pa = 3846 W. The brush power is (331.45)(2.2) = 729.3 W ≅ PB . The total losses are then 1234 + 1810 + 3846 + 729 = P∑ losses = 7615 W 3. Input power = output power + P∑ losses = 75000 + 7615 = 82615 W = 82.615 kW 4.

η =

75000 × 100 per cent = 90.8 per cent 82165

Note: This is a generator, and the input power must be larger than the output power. Direct current motor losses are shown in Figure 34.2. They are the same types of losses, but they are subtracted in the opposite sequence. In a generator, the input mechanical power must be larger than the output electrical power in order to compensate for the losses. The motor situation is opposite in that the input electrical power must be larger than the output mechanical power to compensate for the losses. Again, if a particular field winding is not present, obviously, its type of loss is not involved.

34.3.4  Relationships Between Losses Motor and generator losses are of two types: fixed losses and variable losses. The fixed losses are not really fixed since they vary with adjustments but they do not differ to a major degree with load current. These are the rotational losses and the shunt field loss. At a single field setting, a shunt machine may vary only a few per cent in rotational speed. A shunt field, rheostat adjustment range may vary the field current and thus the loss by a factor of 3 or 4. Variable losses are the armature circuit resistance losses, since their power loss varies as the square of the current and the current varies with the load. In general, the series field, the commutating field, and the compensating windings (if they are present), are variable I 2R losses. The machine reaches its maximum efficiency when the variable losses equal the fixed losses. P + Vl I sh ≅ I a2 R a at η max rot  (34.11) Eq (34.11) is a general case for any motor or generator. However, some of the terms may not be present in a specific case. For example, in a series motor, the only fixed loss would be the mechanical parts of the rotational loss. Even the magnetic part of the rotational loss would then be a variable. In addition, if a particular type of series winding is not present in a machine, it is not a part of the variable losses. If the fixed and variable losses are equal at maximum efficiency the following is also true. output output ηmax ≅ ≅  (34.12) output + 2 ( variable losses ) output + 2 ( fixed losses )   

ηmax ≅ and   

input − 2 ( variable losses ) input

≅

input − 2 ( fixed losses ) input



(34.13)

Efficiency of Direct ­Current Machinery  663

Figure 34.2  Motor Losses Direct Current where, η is in decimal form.

ηmax ≅ ηmax ≅



Pout  Pout + 2 I a2 R a Pin − 2 I a2 R a Pin



(34.14) (34.15)

Example 34.3 A 10 kW shunt generator having an armature circuit resistance of 0.75 Ω and a field resistance of 125 Ω generates a terminal voltage of 250 V at full load. Determine the efficiency of the generator at full load assuming that the iron, friction, and windage losses amount to 600 W.

664  Electrical Technology Solution: Output power = 10000 W = VI Load current

= I

Field current     = If

10000 = 40 A 250 V 250 = = 2A Rf 125

Armature current = If + I = 2 + 40 = 42 A 10000 VI Efficiency  η = × 100 per cent = 2 2 (VI + I a R a + I f V + C) 10000 + (42 × 0.75) + (2 × 250) + 600 =   

10000 = 80.50 per cent 12423 × 100 per cent

Example 34.4 A 60 hp (44.74 kW) rated, 234 V shunt motor has an armature resistance, including brush resistance, of 0.052 Ω. The field resistance is 48.7 Ω. The motor is stated to have a maximum efficiency of 90.9 per cent. Calculate (1) the line current carried at the maximum efficiency and (2) the rotational loss. Solution: 1. If we assumeI a I a≅ ≅I l I l by neglecting the smaller field current P ηmax≅ ≅ Poutout 2 (The output power is specified by the motor power rating.) ηmax + P Poutout+ 2 I2l2IRl aR a   2. If we modify this by using I × V instead of P then I lVl Vl ηmax ≅ = I lVl + 2 I l 2 R a Vl + 2 I l R a ≅ 90.9 ≅ and

230 230 = 230 + 2 I l (0.052) 230 + 0.104 I l

I l = 221 A

The total fixed losses will match the total variable losses at peak efficiency. Prot = I l 2 Ra − Psh = ( 216.3 ) Note:

2

( 0.052 ) − 1086 = 1347 W

 203  2 − P = ( 216.3 )2 ( 0.052 ) − 1086 = 1347 W a I a − I l − I sh = 221 − Prot = I l =R216 .3shA  48.7  203   2 221 −  I a − I2l −203 I =  = 216.3 A El ( sh )  48 Psh = = 1086 W. 7  2 Rsh 48.7 E 2 ( 203 ) Psh = l = 1086 W Rsh 48.7

34.4  MOTOR ENCLOSURES The motor types begin with the simplest and least expensive and work on up to the most elaborate. More elaborate enclosures usually require larger frame sizes for a given power, since ventilation is restricted.

1. Open Enclosure: An opened frame structure permits maximum air circulation for ventilation. This construction usually is designed in such a manner as to prevent dropped objects from coming into c­ ontact with electrically live or moving parts.

2. Drip-proof Enclosure: An enclosure so constructed that liquid or solid particles that fall at not greater than 15° from vertical will not enter the enclosure either directly or while running off the surface.

Efficiency of Direct ­Current Machinery  665

3. Splash-proof Enclosure: Carries the drip-proof situation further so that particles arriving at up to 100° from the vertical will not penetrate inside. It should be noted that each increase in protection usually reduces ventilation.

4. Guarded Enclosure: This enclosure is so arranged that no accidental or intentional object can ­penetrate. Specifically, a ½ inch diameter rod must not be able to penetrate screens or guards.

5. Weather-proof Enclosure: A variation of the drip-proof and splash proof design that prevents blowing rain, snow, or dust contacting electrical parts.

6. Totally Enclosed Enclosure: Either closed and/or covered but not necessarily air-tight enclosure. 7. Explosion-proof Enclosure: Designed to contain an inside explosion and/or to prevent ignition of specified gases or vapours surrounding the motor. This may be accomplished by specified screens and/or flame traps.

8. Dust Ignition-proof Enclosure: Totally enclosed and constructed to exclude entrance of ignitable dusts or dust that would build up and affect performance.

9. Waterproof Enclosure: Designed in such a way that its total enclosure may be sprayed by the steam from a hose without

detrimental effect. Shaft leakage is allowed if it is drained away in a specified fashion. This enclosure is used in dairy and other food processing machinery where daily cleaning or even sterilization takes place.

34.5  MAINTENANCE AND ACCESSIBILITY A motor or generator must always take into account the need for easy access to inspect, maintain, repair, or replace the unit. With DC machines, the commutator and brushes should be regularly checked for any mechanical wear. The wear products that are primarily finely powdered brush particles, should be periodically cleaned out for maximum life. This dust, when mixed with excess shaft bearing oil or grease, will cling tenaciously to any surface. It is conductive so that it may lead to flash over and damage if not removed. Modern bearings are very long lived but need periodic checking. A normal well-designed ball ­bearing may go from 5000 to 10000 hours without service. However, it will eventually need regreasing. If properly done, another 5000 to 10000 trouble-free hours may be expected. However, if not attended to, the bearings will eventually fail and perhaps spoil the motor windings, owing to progressive o­ verloading. Bearing life is greatly affected by misalignment during installation or by too tight belt tension. Winding temperature and, therefore, winding life expectancy are greatly affected by accumulated dust and grime, since this prevents effective cooling. These problems are interrelated, since a major cause of electrical failure is over lubrication of the bearings. The machine operator should be aware of abnormal sounds, vibrations, smells and excess surface temperature. Each abnormality can usually be traced to improper maintenance and, if detected soon enough, can be cleared up in a few minutes. Ignored symptoms have a way of resulting in serious and expensive break down.

34.6  COOLING AND VENTILATION The life of an electrical machine is largely dependent upon the durability of insulating materials. Insulating materials start deteriorating at relatively low temperatures. Thus, the temperature rise of an electrical machine is limited by the quality of insulating materials used in it. The temperature rise at rated load must be within the prescribed limits. If the ambient temperature under actual working conditions is higher than the standard ambient temperature, then the load on the machine must be reduced. Various cooling methods commonly used for rotating electrical machines are: (1) Radial ventilation; (2) Axial ventilation; (3) Combination of (1) and (2); and (4) Forced ventilation. The method of cooling to be used for a specific machine depends on the size and type of machine.

SUMMARY 1. The efficiency of any device or process is the ratio of its useful output to its gross input. 2. Efficiency is expressed as a percentage and it has no units. 3. Rotational losses are a summary of bearing friction, brush mechanical friction, cooling fan power, windage loss of armature as a whole, and magnetic circuit drag loss. 4. Hysteresis and eddy current losses show as consumption of mechanical rotation power.

5. Prot = IaVa. 6. The electrical power losses are mostly due to the resistance present in various parts of the windings. 7. Ia2Ra is the largest single loss in the entire machine. 8. The shunt field loss is considered as a fixed loss. 9. Variable losses are the armature circuit resistance losses. 10. The machine reaches its maximum efficiency when the variable losses equal the fixed losses.

666  Electrical Technology 11. More elaborate enclosures usually require larger frame sizes for a given power since ventilation is r­ estricted.

12. A motor or generator must always take into account the need for easy access to inspect maintain, repair or replace the unit.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. In a d.c. machine, the constant losses are

(a) (b) (c) (d)

The armature copper losses Commutator losses Iron losses and mechanical loss Friction and windage losses

2. At maximum efficiency there are

(a) (b) (c) (d)

Constant losses = variable losses Constant losses > fixed losses Constant losses < fixed losses None of these

3. Iron losses in a d.c. machine under no load and loaded operation are

(a) The same (b) Different (c) Much larger under loaded operation

4. Mechanical losses in d.c. machines occur as

(a) Bearing friction loss (b) Brush friction loss (c) Bearing friction, brush friction, and air friction loss (d) Air friction and bearing friction loss

ANSWERS (MCQ) l. (c)  2. (a)  3. (a)  4. (c).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. Define efficiency. 2. Must the input of a device be smaller or larger than the output? Why? 3. Name the major categories of losses in a d.c. machine. 4. What are some components of the rotational losses? 5. What are some winding losses in a d.c. machine? 6. Why is the shunt field loss considered separately? 7. What is the loss relation at peak efficiency of a machine? 8. A 5 hp motor draws 34.64 A at 125 V under rated conditions. What is its efficiency? 9. A 7.50 kW shunt motor draws 33.8 A at 250 V under rated conditions. What is its efficiency?

10. A 20 hpdc motor has 89.3 per cent efficiency at rated power. What are its total losses? 11. A 3.5 kW motor is 87.2 per cent efficient at rated power. What is it input power? 12. A 10 hp motor has an input of 8.425 kW, while its losses are 925 W. What is its efficiency? 13. A d.c. motor draws 33.3 A and its shunt field uses 1.35 A. If its armature circuit resistance is 0.385 Ω. What is its armature circuit power loss? 14. A 2.24 kW rated motor has 630 W of total loss. What is its efficiency?

ANSWERS (CQ) 8. 86.2 per cent  9. 88.8 per cent  10. 1787 W 

11. 4014 W  12. 89.0 per cent  13. 407 W  14. 78.0 per cent.

35

D.c. Motor Control OBJECTIVES In this chapter you will learn about:  Motor speed control  Control devices: hand-operated mechanical switches, face plate and drum  Starter switches, two-point, three-point, and four-point  Pilot control devices and their symbols   Relay types and symbols  Comparison of manual and automatic starters  Reversing the direction of shunt, series, and compound motors  Manual reverse control  Various interlocking devices  Retardation and stopping  Dynamic, regenerative and electric brakes  Jogging  Calculation of speed and simple problems on the above issues  Ward Leonard system

Off No-volt trip Overload trip

Arm.

L

F

A

D.c. supply

Shunt field

D.c. Motor control

35.1 INTRODUCTION Any substantial-sized d.c. motor needs current-limiting resistance in the armature circuit for reasonable control of armature current during starting. To reverse a d.c. motor, the polarity of the armature must be changed in relation to the polarity of the field see Figure [35.1 (a) and (b)]. The relatively simple control of rotational speed by changing field flux, armature voltage, load, or a combination introduces another control function. The controlled reduction of rotational speed may require special design attention depending upon the load character or safety requirements. Some of the considerations that should be studied in order to determine a motor control will help put the problem of control selection in its proper perspective.

35.2  CONTROL DEVICES Primary control devices include, but are not limited to the following types of devices: 1. Hand-operated Mechanical Switches: This may be the knife switch in which a blade-like element moves between or is clamped by fixed spring jaws. 2. Rotary Switches: One or more metal segments are rotated into or away from contact with spring-loaded, fixed contacts. These are frequently called drum switches because of their roughly cylindrical construction. Rotary switches may have their fixed contacts arranged like the numbers on a clock face. These contacts are connected by a moving contact arm. The result is called a face plate rotary switch to differentiate it from the drum rotary. These are shown in Figures 35.2 and 35.3. 3. Magnetic Contactors: Magnetic contactors are electromagnetically actuated spring-loaded contacts, usually comprising of at least two circuits, and frequently as many as four or five separate circuits. A contactor requires some form of

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668  Electrical Technology

Figure 35.1 Reversing a d.c. Motor by (a) and (b) Reversing the Armature Current (c) and (d) Reversing the Polarity of the Field

Figure 35.2  Face Plate Rotary Switch Symbols

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Figure 35.3  Drum Rotary Switch Symbol

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pilot device for its actuation. This is really a special power-handling form of a relay. A relay is ­defined as an electromagnetically operated switch. In the ­motor control field, if the device is used for opening or closing the main power lines, it is a contactor. Conversely, if it is a pilot device used to control the ­operation of a contactor, it is a relay. The two different functions may be performed by the same device: the fields of application overlap. 4. Starter Switch: A starter switch may be manually or magnetically operated. In either case, a starter switch is a combination of a contactor and appropriately matched thermal overload devices packaged in a coordinated and standardized assembly. Any one of the overload (OL) heater elements can serve to break the normally closed auxiliary contact that is associated with it. Since all the OL contacts are in series, the opening of any one of them causes the contactor coil to relax and the contactor to drop out. The OL contacts do not carry motor current. The heater elements do carry motor current, as can Typical three line magnetic starter switch for a.c. or d.c. service, L3 −T3 would not be used for d.c. or single phase a.c. Note that if any line be seen in Figure 35.4. overheats due to high current its associated normally closed contacts 5. Current-limiting Resistors: Current-limiting resistors will open and break the coil-holding current. The external pushbutton are primary devices for handling the main a­rmature actuation is not shown. This is really a magnetic contactor with matching associated overload protection. A manual starter will have circuit current. They must be of appropriate resistance mechanical rather than electromagnetic actuation and mechanical and wattage for the task. The resistors may be inserted overload release. or removed from the circuit either by contactors or by a Figure 35.4  Magnetic Starter Switch Symbols face plate rotary switch. 6. Protection Devices: Protection devices such as fuses or circuit breakers are used to break the current flow to the motor if a trouble develops that results in excessive current flow. Larger motors usually r­ equire both types of protection, since each has different characteristics.

35.3  PILOT CONTROL DEVICES Pilot control devices include, but are not limited to, the following types of devices. 1. Push Button Switches: The first link in the chain of control elements is frequently a push button that can be jabbed with a finger. This device simply closes or opens a contact against the force of a light spring. As soon as the force of the finger is removed, the internal spring returns the movable element to its original position. This is said to be momentary action. If the button is mechanically linked to another—as in a start-stop combination—so that when pressed it stays in the on position, it is said to be maintained.   Most frequently, a push button is single and independent. Usually, the push button has dual contacts so that it can be used as either normally open (NO), or normally closed (NC). It is very rugged and long lived. 2. Float Switches: Float switches are operated by a fluid level in a tank or process channel. The actual switch can be NO or NC, or have both types of contacts. The actuation may be either with rising fluid level or dropping level or even a combination of the two. 3. Pressure Switches: Pressure switches may be actuated by increasing or decreasing the pressure, which may be set at low or high. 4. Temperature Switches: Temperature switches may be actuated by increasing or decreasing the temperature. There are many ranges and applications. Either float, temperature, or pressure switches may be NO or NC, or have both types of contacts. All these various types of switches are available in sealed enclosures for hazardous conditions. 5. Flow Switches: Flow switches may be actuated by increasing or decreasing the flow in relation to a set flow point. The actual phenomenon being measured is frequently the difference between two pressures measured in specially shaped portions of the tubing or ducting, such as a venturi section.

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670  Electrical Technology 6. Location, Proximity, or Position Generally Detected by Limit Switches: It is sometimes necessary to actuate a switch within a travel location as about 0.025 mm. In other situations, such as in crane hook travel, a location of ±1 mm is sufficient. It is sometimes necessary to use a switch in the opposite of its normal sense. For example, the circuit may require that the switch be opened at the end of the desired travel, but a NC switch may not allow sufficient over travel. In this case, a NC switch may be held closed with a spring and be forced to open against the spring at the desired limit point. 7. Foot-operated Switches: Foot-operated switches are a special form that is used in industrial control when both the hands are occupied. The switch can be related to a push button or a limit switch, but a special enclosure is required for reasonable life. 8. Relays: Relays are magnetically operated switches. There exist many types of relays as far as size, shape and ­appearance are concerned. Almost any conceivable contact a­ rrangement can be used. Many varieties of relays are available that have adjustable and reproducible ­parameters. Time delay in operation or release or both, or pull in or drop out at particular desired current levels or voltage levels are obtainable. Many control functions can be accomplished by the judicious use of relays. Contacts on a relay can serve to interlock functions so that the prescribed sequences of operation are automatically obtained. Undesired or unsafe operation sequences are locked out until the necessary conditions and sequences are in existence. Figure 35.5  Push Button Circuit Symbols  9. Sensors: Sensors or pickups or transducers may also be used as pilot devices (Figures 35.6 and 35.7). In general, these terms mean a device that converts one phenomenon into ­another. One pilot device can operate another. For example, a fluid level may be sensed by a capacitor device; the fluid becomes the dielectric of the capacitor as it displaces the air or gaseous vapor in the tank. The resulting change in capacitances can be used in a tuned circuit to change a current flow and operate a relay that is sensitively adjusted. 10. Starting Requirements: It is necessary that the motor and its power supply have the same or nearly the same voltage rating. The power supply must make sufficient current available for steady state and peak current requirements. If an operator is used, is a skilled person or a fool proof control required? Is the control manually operated or automatic? If manually operated, is a push button control required or will a rotary lever that requires some judgement be used? If an automatic starting control is to be used, will the beginning of the start cycle be based upon standard pilot devices? Will it be necessary to synchronize the start with another operation or avoid some interfering position condition some form of i­nterlocking is

Figure 35.6  Pilot Device Symbols

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Figure 35.7  More Pilot Device Symbols

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required? What is the starting duty cycle and will it be frequent or infrequent? Are there significant starting torque requirements or does the load have high inertia (either of these conditions would prolong the starting cycle). When starting, is the machine always loaded or always unloaded? Does the motor require small increments of motion or jogging?

35.4  MANUAL D.C. MOTOR STARTERS There are a variety of control circuits. Their differences are in whether the reduction in resistance is accomplished manually or automatically and in the types of protections that are built in. The simplest d.c. starters to understand are the various varieties of face plate switch starters that are known as two-point, threepoint and four-point starters. The names evolve from the number of external connections on the starter control box. These have been illustrated in Figures 35.9 through 35.11. A degree of similarity is apparent among these starters. All have a face plate rotary switch with a connected group of current limiting resistors. The differences lie in the form of protection they contain. These protections are what determine the utility of the specific starter types.

D

Figure 35.9  Two-point Motor Starter

Figure 35.8  Relay Symbols

35.4.1  Two-point Starter A two-point starter is specifically designed for a series motor that is possibly subject to an over speed problem due to loss of load. Here, the spring-loaded control arm is held in the full on position by an electromagnet. This magnet is series wound and its ampere turns are a function of the load current. If the motor loses its load and over speeds, the current will drop down to a low value and the spring-loaded arm will be released. This type of release will also drop out if the line voltage decreases too far. The control will then require a restart by moving the arm.

35.4.2  Three-point Starter A three-point starter is suitable for a shunt or compound motor. The electromagnetic holding coil that is used here is in series with the shunt field coil. The protection gained is against the case where the field circuit opens, owing to internal failure or field rheostat failure. Protection is needed because the shunt motor will over speed in the same manner as the series motor if it loses its field excitation. A three-point starter, however, is sensitive to the field current adjustment and it may drop out when not required by over speed if the line voltage also fluctuates.

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Figure 35.10  Three-point Motor Starter

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35.4.3  Four-point Starter The four-point starter overcomes the last disadvantage by placing the holding coil across the line so that a large range of field adjustments is possible as shown in Figure 35.11.

35.4.4  Drum Rotary Switch Starter

D

Figure 35.11  Four-point Starter

The drum switch provides the same type of control that can be achieved with a face plate starter, as illustrated in Figure 35.12, but it has the advantages of being able to handle greater current and, therefore, larger motor sizes. This is because the physical contact shapes and spacing lend themselves to magnetic arc blow-out features. Furthermore, since the actual acceleration current limiting resistors are not physically part of the drum switch, or even in its housing, they may be made as large as desired. As a result, the drum switch is adaptable to running the motor continuously at intermediate or even the lowest speed. This is because with large enough resistors, the circuit can be used for armature voltage control. Drum switch controllers are the basis of large crane controls.

Figure 35.12  Drum Controller Starter (d.c.)

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The drum switch version of a motor control shown in Figure 35.12 is strictly comparable to the face plate rotary switch control as shown in Figure 35.13 with a few exceptions. The drum control has fewer accelerating steps and does not provide a means for a loss of field current cause an automatic stop. The operation is as follows.   1. The bus line safety switch is usually left on.   2. The circuit breaker is manually or electrically set when the operation is determined.   3. The maintaining relay will close when the circuit breaker is closed. Conversely, it will drop out if voltage is lost during operation and then require that the controller be returned to the off position in order to restart. This relay then is necessary to prevent an automatic restart of the motor if the controller is in run position. It serves the same purpose as the holding coil in Figure 35.13. The auxiliary contacts MA serve to hold the contactor on while voltage exists.   4. The drum controller is then advanced to position 1, which supplies current to the armature through the full bank of current limiting acceleration resistors. This breaks the connection at M.1 and M.2 but the MA contacts hold the armature coil.   5. Deliberate motion to step 2 and then to step 3 progressively cuts out part of the acceleration resistors.   6. When the motor is in step 4, the motor is fully on the line and operating.   7. Speed control of a shunt or compound motor is then available with the shunt field rheostat, which may be left in a preset position.   8. lf the motor has only a series field, the speed adjustment is obtained by leaving the control handle in an intermediate position.   9. The normal means of shutdown is by opening the circuit at the circuit breaker. 10. Breakers are provided for protection against overload currents and fuses for short circuits anywhere in the motor or controller. Note: There are many variations of the face plate and drum controllers. Reversing maybe added to either type. The over speed danger must be covered by a separate sensor and pilot relay if it is considered necessary, because loss of field current will not cause a drop out.

35.4.5  Direct-current Three-point Starter Circuit The complete circuit shown in Figure 35.13 shows a number of normal places where the circuit may be opened either manually or automatically. In a factory installation, normally there is a manual safety switch where the power lines leave the overhead bus supply. The same switch box has the fuses for short-circuit protection, so that a winding failure is also protected. In the machine, there is a manual or magnetic starter switch that makes the machine live or dead as far as electric power goes. This switch does not actually start the machine. It is, however, the usual location for the overload breaker system. The particular symbol shown in Figure 35.13 is that of a magnetic trip breaker. Next in the circuit is the actual three-point starter box. The handle is moved slowly and steadily clockwise until the acceleration cycle is completed when the handle hits a stop and is magnetically held on. Any speed adjustment is controlled by the separate field rheostat with its circuit shown in the Figure 35.13. The motor is shown as a long-shunt compound, but a short-shunt compound or a plain motor may be ­equally applicable.

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D

Figure 35.13 Direct-current Three-point Motor Starter Circuit

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674  Electrical Technology

35.5  AUTOMATIC DIRECT-­CURRENT MOTOR STARTERS Any starting sequence that is initiated either by pushing a button or by the closure of some other form of pilot device must have certain automatic features. The principal feature is that the starting sequence must begin with the currentlimiting acceleration resistors in the armature circuit. Then, some form of automatic control must progressively remove these resistances until the full-line voltage is available to the armature circuit. Even the simplest control must automatically reset either when the stop button is pushed or when another pilot device calls for shut down. This controlled removal of resistance is the heart of the problem. Safety features such as overload release, shortcircuit protection, and low-voltage sequence resetting are usually the same on a manual or automatic motor starting control.

35.6  COMPARISON OF MANUAL ­VERSUS AUTOMATIC STARTER A comparison of manual versus automatic control, their advantages and disadvantages is given in Table 35.1. All automatic control devices are either open or closed loop in function. The distinction between open and closed is determined by whether operation is independent of the performance of the motor for open loop or at least partially dependent upon the performance of the motor for closed loop. In a closed loop, some performance parameter is detected, measured and used to influence the input. Table 35.1  Manual Versus Automatic Control: A Comparison Manual

1. Sequence time may be varied at will, depending upon operator skill and motor response. An unskilled operator may hurry to the point of overload or lag unnecessarily. 2. Simple in construction and maintenance, it is easy to understand and service. 3. R  elatively smaller space and less weight with lower cost. 4. Operator is near major power functions. 5. Can be completely enclosed, waterproof, etc. Less susceptible to moisture or dust damage. Wiping contacts are self cleaning. 6. Size is limited since operator effort increases as contact sizes increase. Locomotive size drum controllers require added contactors to relieve effort and hazard. 7. Ionized gas buildup due to arcing contacts is serious if controller is enclosed or sealed. May even be explosive, as an extreme.

Automatic

1. Sequence time is either fixed or the time is dependent upon internally measured parameters. No skill is ­required. 2. Relatively more complicated and expensive. Some skill and knowledge are needed. 3. Relatively larger space and greater weight and cost. 4. Operator may be remote and at various stations. 5. Certain relay forms are very vulnerable to moisture and dust since pneumatic time delays may be used. Relatively less or no wiping action on contacts. Therefore, less self cleaning. 6. No limit to size and no increase in operator fatigue since push buttons need not grow with the size of the task. 7. Ionized gas buildup is minimized owing to quick make and break of electromagnetic contactors and relays.

A purely timed switching sequence that removes the current limiting resistances is open loop. When the switching of resistances is made dependent on the actual armature circuit, the control is then a closed loop. Open-loop controls are simpler and, thus, easier to understand. Conversely, closed-loop controls are a bit more complicated to understand and to service, but they adjust to circumstances.

35.7  REVERSING CONTROL OF ­DIRECT CURRENT MOTORS Various means of reversing the armature circuits are used in both manual and/or automatic d.c. starters as shown in Figure 35.14. In any case, the whole armature circuit must be reversed as a connected group including commutating fields and compensating windings if they are present. The series field, if it is present, is the only part of the armature circuit that remains unreversed. Series Motor: The connection of either the field winding or of the armature must be reversed. Shunt Motor: Figure [35.15 (a) to (c)] shows the two alternative methods of reversing the direction of rotation of a shunt motor.

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Figure 35.14 Reversing Direction of Rotation of a d.c. Series Motor (a) Diagram of a Series Motor with the Direction of Motion Assumed to be Clockwise (b) Connections to the Field Winding Interchanged, Giving Reversed (Anticlockwise) Rotation (c) Alternative Method when the Armature Connections at the Brushes are Interchanged

Figure 35.15 Reversing the Direction of rotation of a d.c. Shunt Motor

Compound Motor: The easiest way to reverse the direction of rotation of the compound wound motor is to reverse the connections to the armature, as illustrated in Figure [35.16 (a) and (b)]. The reversal of the field connections involves both shunt and series windings, as shown in Figure 35.16 (c).

35.7.1  Manual Reverse Control A typical two-, three-, or four-point starter using a face plate rotary switch may be reversed by interchanging A-1 and A-2 connections of the motor. This is accomplished with a double-pole double-throw switch by interchanging the connections after the motor has stopped. Various interlocking devices are provided to keep this from happening inadvertently. A small drum rotary switch can be used for the reverse in conjunction with the main starter switch, as shown in Figure 35.17. The large and small switch will be so mechanically interlocked in such a way that the reversing switch can only be moved when the main controller is at the off position. Note: 1. Off position is optional and may not be needed. Detents are used to hold chosen position. 2. In hardware or household practice, the toggle double-pole double-throw switch is called a four-way switch because only four terminals are provided.

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676  Electrical Technology 3. Standard six contact rotary switches have the same connections as above, plus two added contacts which are connected together for either forwarder or reverse. The fundamental use of reversing contacts on switches is shown in Figure 35.18. The symmetrical rectangular array of two forward or F contacts or two reverse or R contacts surrounding the armature circuit is frequently used in control diagrams. Again, if commutating fields or compensating windings are present, they must be between the A-1 and A-2 contacts along with the armature itself.

35.7.2  Reversing Requirement ­Considerations 1. Is the motor reversing required regularly, occasionally, or ­never? 2. Might reversing be required in an emergency even though not a normal sequence? 3.  Might plugging be required? 4. Is the normal reverse cycle of operation essentially similar to the forward motion or are these specific and unsymmetrical requirements to meet in one side, and not the other? Note: 1. When a process that is being performed by a motor requires repeated and controlled reversal of direction, a plugging reverse control is used. 2. Incomplete circuits are shown. The safety devices and acceleration circuits previously shown are required in a c­ omplete circuit.

35.8  RETARDATION AND STOPPING There are many situations when it is desired to bring a d.c. motor and its driven load to a controlled reduction of speed. The obvious method is to apply brakes and dissipate the energy in the form of heat. The problem resolves to controllably reverse the torque on the motor. If the motor is converted so that it operates as a generator, it will convert shaft energy to electrical energy. The electrical energy can be dissipated remotely in a number of ways. The motor can absorb as much or more mechanical energy input as a generator than it could deliver as a motor. Figure 35.16 Reversing Direction of Rotation The problems of electric retardation are solved in two principal of a Compound-wound Motor ways. Dynamic Braking: The motor is converted to a generator by circuit arrangement, and the energy it produces is dissipated in a resistive element as heat. Dynamic braking is simple and does not require voltage adjustment in the usual case. Regenerative Breaking: The motor also acts as a generator but, by careful arrangement, the generated electrical energy is fed back into the bus lines to become useful and recoverable electrical power. Regenerative braking involves careful adjustment but is much more efficient since energy is returned.

35.8.1  Electric Brakes The final means of retardation and stopping of a motor is through mechanical brakes. Electrically operated mechanical brakes are usually electromagnetically released and applied by an adjustable spring mechanism. There are frequent situations: the load inertia is small, or the electromagnetic brake is entirely adequate to perform the complete retardation and holding job.

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-

-

-

Figure 35.17 Double-pole Double-throw Switches

Electromagnetically released brakes are very safe devices. They do not release inadvertently and allow a mechanism to drift or run away. On the other hand, they will immediately grab and bring the mechanism to an emergency stop if the power fails or a breaker goes out.

35.8.2 Jogging It is defined as the quickly repeated closing of a circuit in order to start a motor from the rest for the purpose of accomplishing small movements of the driver machines. There are many circumstances where it is desired to produce a

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678  Electrical Technology

Figure 35.18  Fundamental Reversing Control

small motion on a motor-driven device, such as to complete a traversing operation to inch a hoist or a crane hook into a reserved position. Jogging requires a button control that will imitate a slow speed motion and not the whole starting cycle. The circuit problem then comes down to a. push button operation that will close the M contacts without actuating the holding contacts (Figure 35.19). When the button is released, all should stop, as nearly instantaneously possible. Figure 35.19 shows a portion of a reversible jogging circuit. Five push buttons are provided: stop, jog, run, forward, and reverse. The jog-run switch is a maintained pair in such a way that if the jog is depressed it stays and forces run back out. On the other hand, if run is depressed, it stays and forces the jog back out. When run is selected, all F contacts are closed, thus, selecting the armature

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circuit. However, when jog is selected the following steps occur. R

1. Pressing either forward or reverse will start the operation at the lowest speed. The motor does not finish its acceleration cycle but continues at a stabilized low speed. 2. When the forward button is released, operation ceases immediately. 3. Reverse operation is equal and opposite in this control Jogging, however, may be applied to a unidirectional control if desired.

F

F

R F

Example 35.1 A 240 Y d.c. shunt motor has an armature resistance of 0.27 W and field winding resistance of 160 W. On no load, the motor takes a current of 2.9 A and its speed is 1 250 r.p.m. On full load, the motor takes a current of 40.0 A and, although the field current remains the same, the flux per pole is reduced by 5 per cent due to armature reaction. Determine the full load speed of the motor. Solution: Let the e.m.f. under no-load conditions be El and the armature current be Ia1 as shown in Figure 35.20 (a). I a1 = I1 − I f = I1 −

R

Figure 35.19  Partial Reversible Jogging Circuit

V 240 = 2.9 − = 1.4 A Rf 160

E1 = V − I a1 Ra = 240 − (1.4 × 0.27) = 239.6 V Let the e.m.f. under full-load conditions, as seen in Figure 35.20 (b) be I a2 = I 2 − I1 = 40.0 − 1.5 = 38.5 A E2 = V − I a2 R a = 240 − (38.5 × 0.27) = 229.6 V However,

φ1 N1 E1 = E2 φ 2 N 2

and N 2 =

100 229.6 × = 1261 r.p.m. 95 239.6

Figure 35.20  For Example 35.1 Example 35.2 A 500 V d.c. shunt motor has an armature resistance of 1 W and a field winding resistance of 500 W. When loaded to-develop a total torque of 100 Nm, the motor takes a current of 21 A from the supply. Determine the speed of the rotor.

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680  Electrical Technology With the field current remaining unchanged, the motor is further loaded until the torque is 120 Nm. Assuming that the effect of armature reaction is negligible, determine the speed of the motor, Solution: When the total torque is 100 Nm, let the circuit quantities be as shown in Figure 35.21(a) If =

V 500 = = I A; I a1 = I1 − I f = 21 − 1 = 20 A Rf 500 E1 = V − I a1 R a = 500 − (20′ 1) = 480 V

E1 I a1 = I1ω1 =

60 × E1 I a1 2π N1T1 60 × 480 × 20 ; N1 = = = 917 r.p..m. 60 2π T1 2π × 100

When the torque is 120 Nm Let the original quantities be as represented in Figure 3l.21 (b) Ia T1 = 1 T2 I a2 I a2 =

120 × 20 = 24 A 100

E2 = V − I a2 R a = 500 − (24 × 1) = 476 V E2 I a2 = T2ω2 = Figure 35.21  For Example 35.2

N2 =

2π N 2T2 60

60 × 476 × 24 = 909 r.p.m. 2π × 120

Example 35.3 A series motor has a resistance 0.5 W between the terminals. It runs at a speed of 1500 r.p.m. when taking 10 A from 200 V supply. Find the speed at which it will run when taking 20 A from the same supply. Assume no saturation. Solution: Motor resistance R m = 0.5 Q; I a1 = 10 A; I a2 = 20 A Eb 1 = V − I a1 R m = 200 − (10 × 0.5) = 195 V Eb 2 = V − I a2 R m = 200 − (20 × 0.5) = 190 V As there is no saturation:



N1 195 × 20 = N 2 190 × 10 N2 =

1500 × 190 × 10 = 730 r.p.m. 195 × 20

Example 35.4 A 220 V d.c. motor runs at 500 r.p.m. when the armature current is 50 A. Calculate the speed if the torque is doubled, given the armature resistances at 0.2 W. Solution: Since N1 is constant, T ∝ f Ia Since f is constant, T ∝ Ia and Ta1 ∝ Ia, Ta2 ∝ Ia2

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Ia 2

2=

or I a 2 = 100 A

50

E1 = 220 − (50 × 0 .2) = 210 V E2 = 220 − (100 × 0 .2) = 200 V N1 E = 1 N2 E2

and N 2 =

200 × 500 = 476 r.p.m. 210

Example 35.5 A 230 V d.c. shunt motor runs at 1000 r.p.m. when the armature current is 35 A. The resistance of the armature circuit is 0.3 W. Calculate the additional resistance required in the armature circuit to reduce the speed of the motor to 750 r.p.m., assuming the armature current is then 25 A. Solution: Voltage applied to the motor = 230 V Initial speed N1 = 1000 r.p.m. I a at 1000 r.p.m. = 35 A; R a = 0.3 Ω Eb 1 = 230 − (35 × 0.3) = 219.5 V Let the additional resistance be R Total resistance in the armature circuit = (0.3 + R) Ω Total resistance in the armature circuit = (0.3 + R) Ω Armature current in the new situation = 25 A Armature current in the new situation = 25 A Eb 2 = 230 − 25(0.3 + R) Eb 2 = 230 − 25(0.3 + R) As shun t field current is the same, φ1 = φ2 and Eb = kφ N where, k is a constant As shun t field current is the same, φ1 = φ2 and Eb = kφ N where, k is a constant Eb 1 = kφ1 N1 and Eb 1 = kφ2 N 2 Eb 1 = kφ1 N1 and Eb 1 = kφ2 N 2 Eb 1 Eb 2 or

=

φ1 φ2

×

N1 N2

750 222.5 − 25 R φ1 = × φ 2 1000 219.5 R

222500 − 25000 R = 750 × 219.5 R = 2.315 Ω Example 35.6 A 240 V shunt motor runs at 1450 r.p.m. on full load with an armature current of 11 A. The total resistance of the armature and brushes is 0.6 W. If the speed of the motor is to be to 1000 r.p.m. with the same armature current, calculate the amount of resistance to be connected in series with the armature and power lost. Solution: I a2 = I a1 − 11 A = a RaA = 240 − (11 × 0.6) = 233.4 V = VI a1−−I11 = 1450 r.p.m. b 1 = V − I a Ra = 240 − (11 × 0.6) = 233.4 V Ebin = V armature − I a 2 (R circuit + R ) = 240 − 11(0.6 + R) = 233.4 − 11 R With additional resistance, R W N21 the = 1450 r.p.m. a = 1000 ENb 22 = V − I ar.p.m. (R + R a ) = 240 − 11(0.6 + R) = 233.4 − 11 R 2 EI b 1 a2 N E1

N 2 = 1000 r.p.m.

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682  Electrical Technology Eb 2 Eb 1

=

N2 N1

or

233.4 − 11 R 1000 = 233.4 1450

R = 6.585 Ω Power lost = I 2 ( R + R2 ) = 112 × (6.585 + 0.6) = 869.4 Watts Example 35.7 A motor runs at 900 r.p.m. off a 440 V supply. Calculate the approximate speed when the motor is connected across a 200 V supply. Assume the zero flux to be 0.7 of the original flux. V Nr ∝ Solution: V φ If f is the original flux, then N r ∝ φ 460 and kφ = 0.511 900 = 460 kφ 900 = and kφ = 0.511 kφ new voltage new speed = new voltage k × original flux × 0.7 new speed = k × original flux × 0.7 200 Nr = = 559 r.p.m. 200 0.511 × 0.7 Nr = = 559 r.p.m. 0.511 × 0.7 Example 35.8 A d.c. motor takes an armature current of 110 A at 480 V. The resistance of the armature circuit is 0.2 W. The machine has six poles and the armature is lap-connected with 864 conductors. The flux per pole is 0.05 Wb. Calculate: (1) the speed and (2) the gross torque developed by the armature. Solution: 1.

Eg = 480 − (110 × 0.2) = 458 V 458 = 2 ×

846 N r × 3 × × 0.05 6 60

N r = 636 r.p.m. 2. Mechanical power developed by the armature is 110 ×× 458 458 = = 58380 110 58380 W W 636 = 50380 60 Mto=develop 756 r.p.m. (M is the torque in Newton metres exerted on the armature the mechanical power) 2π M ×

M = 756 r.p.m. Nm Example 35.9 A shunt motor, supplied at 250 V, runs at 900 r.p.m. when the armature current is 30 A. The resistance of the armature circuit is 0.4 W. Calculate the resistance required in series with the armature to reduce the speed to 600 r.p.m., assuming that the armature current is then 20 A. Solution: Initial e.m.f. generated = 250 – (30 × 0.4) = 238 V Since the excitation remains constant, the generated e.m.f. is proportional to speed. 600 e.m.f.generated generatedatat600 600r.p.m. r.p.m. = 238 × = 158.7 V 900 Hence, voltage drop due to the total resistance of the armature circuit is 250 – 158.7 = 91.3 V

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and the total of resistance of the armature circuit is 91.3 = 4.567 Ω 20 Therefore, additional resistance required in armature circuit is 4.567 – 0.4 = 4.167 W

35.9  WARD-LEONARD SYSTEM To realize all the virtues of a d.c. motor, an adjustable voltage power supply is required. One possibility is to use an a.c. motor driving a d.c. generator and an exciter to supply field excitation current. In Figure 35.22, the motor-generator set converts a.c. power to adjustable voltage d.c. power. With control rheostat at RCM set for the rated field current, d.c. motor voltage V is controlled by the generator field current through RCG. Motor speed can be controlled smoothly from stand still to rated speed in either direction, reversing the direction of generator field current reverses the direction of motor rotation. Speeds above the rated speed can be obtained by reducing the motor field current. Regenerative braking of the load is possible by reducing the generator field current, thereby reducing V below the motor e.m.f. E = knf and reversing the power flow; the generator acts as a motor, driving the a.c. machine as a generator feeding the rotational energy of the load back into the a.c. line. Regenerative braking saves energy while providing smooth control.

Iaccel

RCX

RCG

A.c. S motor

D.c. motor V

D.c. exciter

Ibrake RCM

+

D.c. generator

Figure 35.22  Ward-Leonard System Adjustable Voltage Supply

S UM M A RY 1. For reasonable control of armature current, d.c. motors need current-limiting resistors in the armature circuit. 2. Primary control devices may be hand-operated mechanical switches, rotary switches, magnetic contactors, starter switches (two-point, three-point or four-point). etc. 3. A starter switch may be either manually or magnetically operated. 4. A starter switch is a combination of a contactor and ­appropriately matched thermal overload devices. 5. Current limiting resistors must be of appropriate resistance and wattage for the task. 6. Protection devices, such as fuses or circuit breakers, are used to break the current flow to the armature, if trouble ­develops that results in excessive current flow. 7. Relays are magnetically operated switches. 8. Drum rotary switch starter has the advantage of being able to handle greater currents and, therefore, larger motor sizes.

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9. The names for face plate switch starters evolve from the number of external connections on the starter control box. 10. The drum control has fewer accelerating steps. 11. There are many variations of the face plate and drum controllers. Reversing may be added to either type. 12. In any type of controller, whatever the type, controlled removal of resistance is the heart of the problem. 13. Manual starters are simple in construction and easy to understand and service. 14. Automatic starters are relatively more complicated and expensive. Some skill and knowledge is needed to operate them. 15. A typical starter may be reversed by interchanging A1 and A2 connections of the motor. 16. The switches are so mechanically interlocked so that the reversing switch can only be moved when the main controller is at the opposition. 17. The problem of controllable reduction of speed resolves to controllably reverse the torque on the motor.

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684  Electrical Technology 18. If the motor is converted to operate as a generator, it will convert the shaft energy to electrical energy. 19. The motor can absorb as much or more mechanical energy input as a generator than it could deliver as a motor. 20. In dynamic braking the motor is converted to a generator by circuit arrangement and the energy it produces is dissipated in a resistive element as heat. 21. Regenerative braking involves careful adjustment but is much more efficient since energy is returned. 22. Electrically operated mechanical brakes are usually electromechanically released and applied by adjustable spring mechanism.

23. Jogging requires a button control that will initiate a slowspeed motion and not the whole starting cycle. 24. To realize all the virtues of a d.c. motor, an adjustable voltage power supply is required. 25. A motor starter accelerates a motor to its normal speed in one direction of rotation. 26. A motor starter limits the current in the armature circuit to a safe value during the starting or acceleration period. 27. The drum controller is used to start, stop, reverse, and vary the speed of a motor.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. For a given d.c. motor, the speed depends upon

5. The speed of a d.c. motor is proportional to

2. It is desired to reverse the speed of a motor. This can be done by

6. What will happen if the field of a d.c. shunt motor is opened?



(a) Flux only (c) Counter e.m.f. only



(b) Applied voltage only (d) (a) and (c)

(a) Reversing the armature and field connections (b) Reversing the armature connections only (c) Reversing the supply connections

3. A motor starter



(a) Accelerates a motor to its normal speed in one direction only (b) Accelerates and reverses a motor (c) Is used for jogging (d) Limits the current in the armature circuit to a safe value

4. Which of the methods of speed control is used to reduce the speed of a shunt motor below its rated speed?





(a) Counter e.m.f. (c) Eb f

(b) 1/f (d) Eb /f

(a) The speed of the motor will be reduced (b) It will continue to run at its normal speed (c) The speed of the motor will be enormously high and may damage it (d) The current in the armature will decrease

7. In a d.c. motor the starting resistance is used

(a) (b) (c) (d)

Across the motor In series with the field winding In series with the armature In series with the motor as a whole

(a) Armature control (b) Field control (c) Ward-Leonard control (d) Armature and field control

ANSWERS (MCQ) l. (d)  2. (b)  3. (a) and (d)  4. (a)  5. (d)  6. (c)  7. (c).

CON V E NTI O NA L Q UE S TI O NS (C Q ) 1. What is the major operating difference between a manual and an automatic d.c. motor starter? 2. What is the distinction between a pilot device and a primary control device? 3. What is meant by the duty cycle of a motor? 4. What is the function of a limit switch? 5. What does the term plugging mean? 6. What is low-voltage protection? 7. What electrical functions are performed by a manual starting switch? 8. Explain the difference between a two-point, ­three-point, and four-point starting switch?

M35_AUTH_ISBN_C35.indd 684

9. What features of a d.c. motor necessitate current limiting resistors in a starter circuit? 10. What is the difference between a normally open (NO) and a normally closed (NC) contact on a control element? 11. When reversing a d.c. motor, what element of the motor circuit usually has its circuit sequence changed? 12. What is meant by the term jogging? 13. What simple circuit provisions are necessary for a ­motor control to have a jogging feature? 14. What limits the range of the d.c. motor speed control when it is performed by varying shunt field current? 15. What is the economic disadvantage of speed control by means of series armature resistance?

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16. Why is the line voltage control of a d.c. motor more efficient than series armature resistance control? 17. What is the disadvantage of combined series and shunt armature resistance speed control? 18. A d.c. series motor of resistance 1 W between terminals runs at 800 r.p.m. at 200 V with a current of 15 A. Find the speed at which it will run when connected in series with a 4 W resistance and taking the same supply voltage. 19. Briefly explain the different methods of speed control of d.c. motors citing their relative merits and demerits. 20. What is the necessity of a starter for a d.c. motor? With the help of a diagram, describe the working of a three point starter for a d.c. shunt motor. 21. Explain how the speed of a d.c. shunt motor may be varied both above and below the speed with which it runs with full field current. 22. Discuss the Ward-Leonard method of speed control for d.c. motors and explain its merits over other methods. 23. A 220 V shunt motor runs at 950 r.p.m. when the armature current is 50 V. The armature circuit resistance is 0.5 W. Calculate the value of resistance to be connected in series with the armature to reduce the speed to 760 r.p.m. when the armature current of the motor is 40 A. ANSWERS (CQ) 18. 540.5 r.p.m. 23. l.l W 24. 3.591 W

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24. A 220 V shunt motor running at 1000 r.p.m. has an armature resistance of 0.3 W and an armature current of 15 A at a certain load. What resistance should be placed in series with the armature to reduce the speed of the motor to 750 r.p.m.? 25. A shunt motor takes an armature current of 40 A at a certain load. The armature circuit resistance is 0.6 W. Find the resistance required in series with the armature circuit to reduce the speed of the motor by 60 per cent. The load torque remains constant. 26. A 10 hp, 240 V series motor has a line current of 38 A and a rated speed of 600 r.p.m.. The armature circuit and series field resistances are, respectively, 0.4 Q and 0.2 O. The brush volt drop is 5 V. Assume that the motor is operating on the linear ­portion of its saturation curve below rated armature current. Calculate (a) Speed when the load current drops to 20 A at half rated load. (b)  The no-load speed when the line current is 1 A. (c) The speed at 150 per cent rated load when the line current is 60 A and the series field flux is 12.5 per cent of full-load flux due to saturation.

25. 1.96 W 26. (a) 1198 r.p.m. (b) 25185 r.p.m. (c) 450.1 r.p.m.

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36

Transformers — Single Phase OBJECTIVES Primary

In this chapter you will learn about:   Mutual inductance and methods of controlling mutual inductance  Dot convention and its significance   Transformer types and construction of trans­formers (single-phase only)   Current and voltage relations in an ideal transf­ormer   General transformer equation and simple prob­lems involving it  Practical transformer and circuit relations  Transformer ratings and connected problems   Impedance transformation in transformers–transformation ratio  Step-up and step-down autotransformers    Transformer equivalent circuit—referred to pri­mary and secondary    Transformer regulation; open circuit test and shortcircuit test   Pulse transformers, power transformers and instrumentation transformers  Parts of a transformer and transformer cooling  Transformer connections

Secondary

180° 270° 360° 0°

90°

(a)

Flux lines build upward in primary Primary

Secondary

180° 270° 360° 0°

90°

(b)

Flux lines collapse downward in secondary Primary

Secondary

180° 270° 360° 0°

90°

(c)

Flux lines build downward in primary Primary

Secondary

180° 270° 360° 0° (d)

90° Flux lines build collapse upward in secondary

Transformer action

36.1 INTRODUCTION When electric energy is transmitted, losses occur which depend on the magnitude of the current. In order to minimize these losses, higher voltages are used and, thus, lower currents are needed for the same energy. On the other hand, it is necessary to return to lower voltages for the sake of distribution at the place where power is being consumed and, furthermore, to adjust these distribution voltages to the various requirements. The reason only alternating current is given priority over direct current is due to the fact that the former can be more easily transformed. By using very high voltage in power distribution lines, the required current can be kept low. This, is turn, means lower losses in the lines. The problem with d.c. power is that it can only be generated at a relatively few thousand volts owing to the limitations of the generator’s commutator and brushes. Once generated, the voltage cannot be raised to a higher level by convenient means as far as high power level is concerned. This was a real limit in the past. Alternating current can be and is generated at 15000 volts and higher. Then, by the use of transformers, it can be raised to levels of many hundreds of thousand of volts; thus, enormous power can be ­carried over long distance at low current levels (see Figure 36.1).

Transformers — Single Phase  687

Figure 36.1 Typical Higher Voltage Transformer and Associated Switching Structure

36.2  MUTUAL COUPLING The property of mutual inductance is associated with the flux of one coil linking with the turns of a second coil. The alternating current i1 in the coil illustrated in Figure 36.2 creates an alternating magnetic flux, part of which only links coil 1 (this is referred to as the linkage flux) and the remainder links coil 2 causing an induced voltage, V2, whose magnitude, in part, depends on i1 and the value of the mutual inductance between the coils. Mutual inductance, M, like self inductance, is measured in henrys. The mutual inductance is 1 Henry if—when the current i is instantaneously changing at the rate of one ampere per second—the induced voltage v2, is 1 volt. The factors that determine the value of mutual inductance are; the number of turns for N1 and N2, the cross-sectional area of the coils, the orientation of their axes and the nature of their cores. Induced voltage, V2 = M × rate of change in i1. In terms of sine wave values V2r.m.s = 2pf Mi1r.m.s  (36.1) where, f is the frequency of the current i1. Provided the two coils are wound in the same sense, V2 lags i1 by 90°, but if the coils are in the ­opposite sense V2 leads i1 by 90°. The property of mutual inductance is reversible, so that if the same rate of change of current, i1 is flowing in coil 2, then the voltage induced in coil 1 is (M × rate of change in i1).

Linkage flux

i1

Coil 1 N1 turns

Linkage flux

Figure 36.2  Mutual Coupling

Coil 2 N2 turns

V2

688  Electrical Technology If the coils are tightly wound, one on top of the other, with a common soft-iron core, the leakage flux is negligible. Assuming perfect flux linkage between the coils, the mutual inductance is M =

µ0 µ1 N1 N 2 A Henrys l 

(36.2)

where, A is the cross-sectional area of each of the coils in square metres m r is the relative permeability of soft-iron. l is the length of the coils in metre µ0 = 4π × 10−7

µ0 µ1 N12 A Henrys (36.3) l µ µ N2A and that of L 2 = 0 1 2 Henrys (36.4) l 2 Then = M L= L1 L 2 (36.5) 1 L2 and M The self - inductance The ofself - inductance of L1 =

In case leakage flux cannot be neglected, then only a fraction, k, of the total flux links the two coils. The fraction k which cannot exceed 1 is called the coefficient of coupling or coupling factor. Equation (36.5) can now be written as M = k L1 L 2 (36.6) Also M = k L1 ×

N2 N = kl2 × 1 (36.7) N1 N2

If L1 = L2 = L then    k = M/L  and  M = kL (36.8) Note: If a steady direct current flows through L1, the linkage flux is constant in magnitude and direction and the voltage induced in L2 is zero.

36.3  SERIES CONNECTION OF MUTUALLY-COUPLED COILS The individual fluxes of two mutually coupled coils, connected in series, can aid or oppose depending on the sense in which the coils are wound (see Figure 36.3(a)).

A

B M k L1

L1 + L2 + 2M

L2 Coil #1

Coil #2

(a)

(b)

L1 + L2 + 2k × L1L2

Figure 36.3  Coils in Series Aiding If the coils are in series aiding, as shown in Figure 36.3(a), then, If the fluxes are opposing, then

LT+ = L1 + L 2 + 2 M = L1 + L 2 + 2k L1 L 2

LT− = L1 + L 2 − 2 M = L1 + L 2 − 2k L1 L 2

and M =

LT+ − LT− 4



(36.9)



(36.10)

(36.11)

Transformers — Single Phase  689

The (+) and (–) signs are for aiding and opposing, respectively. If L1 = L2 = L, then + L1 = 2( L + M ) = 2 L(l + k ) 

(36.12)

and L1− = 2( L − M ) = 2 L(1 − k ) (36.13) LT+ + LT−

and L = M = and k =

(36.14)

4 LT+ − LT− 4

LT+ − LT− LT+ + LT−

(36.15)

(36.16)

36.4  PARALLEL CONNECTION OF MUTUALLY COUPLED COILS Figure 36.4 represents two mutually coupled coils in parallel. If the sense of the winding is such that the coils are in parallel aiding, then A L1 L 2 − M 2 LT+ = (36.17) L1 + L 2 − 2 M However, if the sense of the winding is such that the coils are in parallel opposing, the sign of M is reversed and L1 L 2 − M 2

and

LT−



1 1 1 (36.19) = + LT L1 ± M L 2 ± M

=

L1 + L 2 + 2 M

M L1

L2

(36.18)

Note: If M = 0, Eqs. 36.17 through 36.19 reduce to the familiar product over sum (L1L2 / (L1+L2)) formula, If L1=L2= L L+M 2  L−M and LT− = 2  LT+ =

B

Figure 36.4 Parallel Coils with Mutual Inductance

(36.20) (36.21)

Example 36.1 Two coils whose self inductances are 75 mH and 125 mH have a mutual inductance of 155 mH. What is the coupling factor? Calculate the equivalent inductance if the coils are connected in (1) series–aiding (2) series opposing. Solution: 15.5 k = = 0.155 Coupling factor, 75 × 125 1. LT = L1 + L2 +2M  = 75+ 125+ 2 × 15.5 = 231 mH 2. LT = 75+125 – 2 × 15.5= 169 mH Example 36.2 Two coils, whose self-inductances are 65 mH and 85 mH, are connected in parallel-aiding, with a coupling factor of 0.35. What is the total equivalent inductance of the parallel combination? If one of the coils is now reversed without changing the coupling factor, what is the total equivalent inductance?

690  Electrical Technology Solution:

M = k L1 L 2 = 0.35 65 × 85 = 26 mH

parallel -- aiding, parallel aiding, LT =

L1 L2 − M 2 L1 + L2 − 2M

65 × 85 − 262 = = 49.5 mH 65 + 85 − 52   L1 L 2 − M 2 parallel− -opposing, opposing LT = parallel L1 + L 2 + 2 M 65 × 85 − 262 65 + 85 + 52

=

= 24.0 mH

36.5  THE DOT CONVENTION In Figure 36.5, two coils are shown with a common core which channels the magnetic flux f. This arrangement results in close coupling. To determine the proper sign on the voltages of mutual induction, apply the right-hand rule to each coil. f

f1

R1 v1

+ –

f2

L1

i1

I

R2

L2

i2

M

+ –

Thumb v2

(a)

(b)

Figure 36.5  (a) Close Coupling (b) Right-hand Rule If the fingers wrap around in the direction of the assumed current, then the thumb points in the direction of the flux. Resulting positive directions for f1 and f2 are shown in the Figure. If fluxes f1 and f2 aid one another, then the signs on the voltages of mutual inductance are the same as the signs on the voltages of self-inductance. In Figure 36.6, source v1 drives a current i1, with a corresponding flux f1 as shown. Lenz’s law ­implies that the polarity of the induced voltage in the second circuit is such that if the circuit is completed, a current will pass through the second coil in such a direction as to create a flux opposing the main flux established by i1. When the switch is closed, as illustrated in Figure 36.6(b) flux f2 will have the direction shown. The right-hand rule, with the thumb pointing in the direction of f2 provides the direction of the natural current i2. The induced voltage is the driving voltage for the second circuit, as suggested in Figure 36.6(b). This voltage is present whether or not the circuit is closed. When this switch is closed, current i2 is established with a positive direction as shown. R2

R1 v1

+ –

i1

φ2

φ1

L2

L1 M

(a)

R2 i2

L2 di1 M dt

i2

– + (b)

Figure 36.6 (a) A Passive Second Loop (Switch Open) (b) Passive Second Loop (Switch Closed) The sign on a voltage of mutual inductance can be determined if the winding sense is shown on the circuit diagram as in Figures 36.5 and 36.6. To simplify matters, the coils are marked with dots at the terminals which are instantaneously of the same polarity.

Transformers — Single Phase  691

To assign the dots to a pair of coupled coils, select a current direction in one coil and place a dot at the terminal where the current enters the winding. Determine the corresponding flux by application of the right-hand rule, as illustrated in Figure 36.7. φ

φ

i i

Figure 36.7  Establishing the Sign on Voltage of Mutual Inductance The flux of the other winding, according to Lenz’s law, opposes the first flux. Now let us place a dot at the terminal of the second winding where the natural current leaves the winding. This terminal is simultaneously positive with the terminal of the first coil when the initial current had entered. With the instantaneous polarity of the coupled coils given by the dots, the pictorial representation of the core with its winding sense is no longer needed. The following dot rule may now be used: 1. When both the assumed currents enter or both leave a pair of coupled coils by the dotted terminals (Figure 36.8), the signs on the M-terms will be the same as the signs on the L-terms, but. 2. If one current enters by dotted terminal, while the other leaves by a dotted terminal, the signs on the M-terms will be opposite to the signs on the L-terms. + i1 v

+ i1

i2

+ –

v

i2

+ –

– (a)

– (b)

Figure 36.8  The Dot Rule

36.6  PREVENTING MUTUAL INDUCANCE Mutual inductance can be reduced or prevented by methods such as: (1) Axis Orientation; (2) Physical Separation; and (3) Shielding. 1. Axis Orientation: In Figure 36.9(a), the two coils are near each other along a common axis, resulting in magnetic coupling. Now if we suppose that the centre axes of the two coils are at 90° to each other, as shown in Figure 36.9(b); under this condition, very little of the flux from one coil cuts the other coil. This amounts to saying that very little coupling occurs between the two coils (inductors). When the axes of the coils are lined up and close together, as shown in Figure 36.9(a), the result is mutual inductance. Magnetic field

Output

Magnetic field

Axis

Coils near each other along a common axis Generator (a)

(b)

Figure 36.9 Magnetic Coupling (a) High Mutual Inductance (b) Very Little Mutual Inductance

692  Electrical Technology 2. Physical Separation: When inductors are physic­ally separated, then mutual inductance is ­reduced. The further apart the inductors are, the less the mutual inductance (coupling of flux) they have. 3. Shielding: An inductor which is enclosed in a magnetic shield has very little mutual inductance with surrounding inductors (Figure 36.10). The flux from surrounding inductors passes through low–reluctance, highpermeability shield rather than through the shield enclosed inductor.

36.7 TRANSFORMER The transformer is a simple, reliable and efficient device for changing an a.c. voltage from one level to another. The ratio of voltage change, k, can be almost Figure 36.10 A Radio Frequency any reasonable number either to step up or step down the voltage. Coil with Its Shield When two coils of wire are inductively coupled, the magnetic flux that passes through one coil passes through the other to a greater or lesser degree. In other words, the magnetic circuit is common or largely common to both coils. If the magnetic flux is cyclically varying—because the coil that creates it has a cyclically varying current in it—the magnetic flux linkage to the second coil is also cyclically changing. As a result, the varying flux generates a second varying voltage in the second coil. The second voltage is transformer voltage and is said to be created by transformer action (see Figure 36.11). The availability and use of transformer action is one of the major reasons for commercial preference of alternating current for power use. Alternating current is also subject to power losses in transmission by the I 2R effect. This product results in Watts of lost power. If the current can be substantially reduced by raising the voltage, the I 2R loss can be drastically reduced. This is the major reason for long transmission lines using high voltage levels. Very little power is lost in the transformer. Transformers change voltage and inversely change current. They also may serve for electrical isolation and to change impedances. The input a.c. connection to the primary coil has been i­llustrated in Figure 36.11. The primary coil may be the high or low voltage end. If the input is the high-voltage end, the transformer is called a step-down transformer. When operated in the reverse order, with the primary coil using the low voltage, the unit is called a step-up transformer. Transformer symbols are given in Figure 36.12. The figure is self explanatory.

Flux

Primary coil

Alternating current

(b)

(c)

(d)

(e)

(f)

Secondary coil

– + Primary source of power

(a)

+ – Alternating voltage

Transformer Action: The Primary Figure 36.11  Receives the Power, the Secondary Delivers it

Figure 36.12 Transformer Symbols (a) Air Core (b) Iron Core (c) Shielded (d) Variable Autotransformer (e) Multiple Secondaries (f) Center-Tapped Secondary

36.8  TRANSFORMER CONSTRUCTION The primary and secondary coils are connected magnetically by the laminated iron core, as has been shown in Figure 36.13. They are not connected by electrical conductors. If the core forms a simple rectangle in construction, with the primary

Transformers — Single Phase  693

coil surrounding one leg, and the secondary another parallel leg, it is known as a core-type transformer, as illustrated in Figure 36.14. Conversely, if the core forms a rectangular figure eight, with both coils concentrically mounted on the middle leg, it is known as a shell-type transformer as shown in Figure 36.15.

(a)

(b)

        

I laminations E laminations I laminations

I and E joints (c)

Figure 36.13 Stacking of E and I Laminations (a) First Four Layers (b) Second Four Layers

H1 High voltage

X1 Low voltage

H2

X2

Figure 36.14  The Core-type of Transformer

X1

H1

X2 H2

Figure 36.15  Shell Type of Transformer

694  Electrical Technology The two types are different in construction and, as might be expected, they have somewhat different properties. The differences are such that neither type has dominated the field. Both types are available. In general, the core type has a longer mean length of the core and a shorter mean length of coil turn. The core type also has a smaller cross-section of iron, and will need a greater number of turns of wire, since, in general, not as high a flux may be reached in the core. However, the core-type is better adapted for some high-voltage services since there is more room for insulation. The shell type has better provision for mechanically supporting and bracing the coils. This ­allows better resistance to the very high mechanical forces that develop during a high current short circuit. Many satisfactory forms of lamination stacks have been developed using simple rectangular sheared or punched plates of core steel. Higher production units use E-shaped or L-shaped punchings. Finally, the core may be wound of a continuous ribbon of special high permeability steel. With either of the transformer constructions, the coils are carefully wound, mechanically supported in bobbins or coil forms, and impregnated with insulation. The terminals are carefully supported and brought out to a terminal board or insulators. The transformer core is assembled inside the coils, and is mechanically clamped, otherwise supported in position. Clamping is necessary to keep the laminations from moving under cyclic magnetic forces. The electrical and magnetic assembly is then placed in a case except for the very smallest sizes. Very small units are sometimes of open construction with mounting brackets attached to the core. Medium and large units are fully cased. Small and medium transformers are air cooled even if of cased construction. Large sizes are filled with an insulating transformer oil in order to transmit heat from the coils and core to the outside surface where air cooling is available. Still larger sizes have either cooling fans or actual oil-to-air radiators to enable the heat to be transmitted to the surrounding air.

36.9  IDEAL TRANSFORMER If a transformer could be constructed with ideal properties it would need to have the following not quite attainable properties. 1. All magnetic flux created by the primary coil would ideally link with the secondary coil. There would be no leakage flux. This is nearly achieved in a carefully designed iron core transformer. 2. The primary and secondary coils would have zero resistance. Again, this is nearly achieved, but some resistance is present, since the conductor cross-section is limited. A transformer having these ideal properties is shown in Figure 36.16.

φm X1 I1

E1

H1 E2

V1

I2

V2

Load

H2

X2

Figure 36.16  Ideal Transformer The operation of the ideal is then as follows. 1. With the incoming primary coil voltage momentarily being positive, the direction of the primary current is as shown with the arrow I1. This produces the magnetic flux fm in the direction shown. The subscript m signifies mutual flux. In an ideal transformer, that is the only flux present. 2. Since this flux fm is changing, it is inducing a voltage E1, which opposes the applied voltage V1. The dot convention shows that the induced voltage is positive at the top of the coil when applied voltage is positive. This is in accordance with Lenz’s law. 3. At the same time, the magnetic flux is also inducing a voltage E2 in the secondary coil. 4. Again, in accordance with Lenz’s law, this voltage must be of such polarity that any current, I2, that it produces must also oppose the mutual flux.

Transformers — Single Phase  695

If these conditions are true, as they certainly are qualitatively even in a real transformer, then

1. If there is no load, or the secondary circuit is open I2 = 0 A.

2. Since the applied voltage V1 is alternating in polarity, its resultant current I1 is also alternating. I1 produces the flux fm which also alternates at the same frequency. The alternating flux induces voltage E1, which constantly opposes V1, and induces voltage E2. These are also alternating voltages whose instantaneous polarities follow the dot convention. 3. There is a small component of the I1 current that remains, because I1 is not completely cancelled. Thus, E1 does not quite equal V1. This small component is called Im or the magnetizing current. This is the current necessary to overcome the reluctance of the magnetic circuit. The reluctance is low in a magnetic circuit, but it is not zero. The magnetizing current is the only current during no load. 4. Figure 36.l7(a) shows the relations of these various phasors under no-load condition. Im lags the primary voltage by 90° because the coil is assumed to be a pure inductance (no resistance). The flux fm is in phase with the current. 5. The varying flux leads the voltage that it induces by 90°. Yet another way of saying this is that induced voltages E1 and E2 are lagging the flux by 90°. This brings E1 180° out of phase with V1; or E1 opposes V1. This is entirely in ­accordance with an ideal inductance at this stage since the secondary coil is an open circuit and does not yet have any effect. 6. Let us assume that an inductance or lagging load is connected to the secondary terminals, as illustrated in Figure 36.17(b). A lagging load is by far the most usual situation for a power transformer and this is most realistic. The lagging current I2 can be seen to lag the secondary voltage E2 by the power factor angle q2. Where θ ′1 = θ 2

I2 E1 (also E2)

V1

E2

V1

θ2 θ ′1

Im

I1 = Im

φm

φm (a)

I ′1

(b) I2 E2

Where θ ′1 = θ 2

θ2

V1

θ ′1

Im

θ1

Φm

I ′1 I1

(c)

Figure 36.17 Ideal Transformer Phase Relations (a) Primary Coil Relations, No Load (b) Secondary Coil Relation, Under Load (c) Primary Coil Relations, Under Load 7. Even though this is an assumed ideal transformer it, nonetheless, has real properties. Its secondary current I2 and secondary coil turns N2 together produce a demagnetizing flux that is proportional to I2N2 and which opposes fm. This effect, if not compensated, would tend to reduce both E2 and E1 voltages. 8. Thus, the assumed direction of E1 voltage causes the primary component of the load current I'1 flow in the primary, I' N1 = I2N2, so that the same number of ampere turns of magnetization that was lost in step 7 is restored. The sinusoidal flux level in a transformer then remains nearly constant. The symmetry between I2 and q2 on the one hand and I'1, and q´1 on the other should be kept in mind. 9. With reference to Figure 36.17(c) it can be seen that now there are two components of the primary coil current Im and I´1, The phasor sum of these currents is then I1, which may be seen to have a different power factor angle q1. Thus, in loading an ideal transformer, its primary current will assume a lower power factor than its original q = 90° when unloaded. The power factor angle q1 reflects the load power factor angle as one of its component parts. Therefore, the power factor angle q1 of the primary current is not the same as q2, of the load current and as a result, angle q1 > q2.

696  Electrical Technology

36.10  TRANSFORMATION RATIO The effect of loading is roughly analogous to the loading of a d.c. shunt motor. The back e.m.f. of generator action decreases along with the speed until enough additional current is drawn for the load to be supported. Here, the presence of I2 phasor forces I´1 to grow. In turn, the phasor sum of I´1 and Im grows until sufficient current is drawn to support the load and maintain the magnetizing ampere turns. The equality between the secondary demagnetizing ampere turns and the primary magnetomotive force ampere turns can be equated. Thus I´1 N1 = I2N2(36.22) The above equation can also be written as I2 ′

=

N1 = k (36.23) N2

I1 Here k is a transformation ratio, or ratio of primary coil turns to secondary coil turns. This is popularly known as the turns ratio. The turns ratio is a fixed quantity, depending on the actual number of turns in the winding coils as the transformer is wound and connected. It is not a constant in a fundamental sense, but is rather a built-in fixed ratio. Example 36.3

The high-voltage coil of a transformer is wound with 700 turns of wire, and the low-voltage coil is wound with 292 turns. When used as a step-up transformer, the load current is 10.5 A. Find: (1) The transformation ratio k, and (2) The I´1 current, which is the load component of the primary current. Solution: 1.      N1/N2 = k 292/700 = k = 0.417 2. I2/I´1 = k  or  I´1 = I2/k = 10.5/0.417= 25.2 A Note: 1. In a real situation, the actual current is a few per cent higher than I´1 This is to allow for the component Im. 2. There will also be added currents necessary to account for the leakage of magnetic flux. 3. The high voltage side always has the smaller current and vice versa. Example 36.4 Using the same transformer as in Example 36.3, calculate its transformation ratio when used as a step-down transformer. Solution: Now, the high-voltage winding is the primary. N1/N2 = 700/292 k = 2.40 Note: 1. Equation 36.22 works both ways. 2. Subscript 1 means primary. Subscript 2 means secondary. 3. The transformer can be used either for step-up or for step-down. 4. k depends on how the transformer is used. 5. The two numbers are reciprocals. 6. = k

N1 = N2

E1 = E2

I2 V = 1 I1 V2

7. E1 > V1 and E2 > V2 (by a small amount in a practical transformer). 8. V1I1 >V2I2.

Transformers — Single Phase  697

In an ideal transformer, the volt-amperes in the primary circuit equals the volt amperes in the secondary circuit if we neglect Im. If I´1 is much larger than Im, we may say that E1I1 = E2I2. Finally, for an ideal transformer with no flux losses, no I 2R losses and negligible magnetizing current, V1I1 = V2 I2. In a real transformer of very large size, the difference between the output power and its input power is around 1 per cent. Even in small to medium sizes, less than 10 per cent loss is not usual. Transformers are rated in volt-amperes or kilovolt-amperes rather than Watts. The volts per turn is constant since each turn is cut by the same flux. Example 36.5 A 4600 to 230 V, 60 Hz, 7.5 kVA transformer is rated at 2.6 volts per turn of its winding coils. Assuming it to be an ideal transformer, calculate the (1) Step-up ratio, (2) step-down ratio, (3) total turns of the high voltage coil, (4) total low-voltage turns, (5) Primary current as a step-down, and (6) secondary current as a step-up. Solution: 1. V1/V2 = k = 0.05 1 1 = = 20 2. k 0.05 3. 4.



4600 Vt = 1769 turns 2.6 V 230 Vt = 8.5 turns 2.6 V

5.  As a step-down transformer, 4600 V coil is the primary 7.5 kVA(1000 VA) = 9.63 A (4600 V) kVA 6. 230 V is the secondary rating 7500 VA = 32.6 A 230 V

36.11  GENERAL TRANSFORMER EQUATION To develop the basic voltage relationship in a transformer, on a basis that does not involve the turns ratio, we start with Eq. 36.24. φ volts E av = t turns (36.24) Equation 36.24 relates a steady value of the magnetic flux f, with a time to link that flux. As a result, the voltage produced is given as Eav, since nothing is assumed about the uniformity or the lack of uniformity of the velocity required to traverse the flux in time t. If, we now realize that the winding turn does not move, but that flux changes sinusoidally and cyclically from maximum through zero to an opposite polarity maximum, we can then write:

E av = 4φm f N



(36.25)

where, fm refers to the peak mutual flux f = the cyclic frequency of flux variation N = the number of turns required (Primary or Secondary) Since the voltage term wanted is the root mean squared rather than the average, the exact relation needed is. Eav=4(1.1l)fm  f N (36.26) This is the general transformer equation. The peak flux is a limitation of the magnetic material chosen, and then related to the physical size of the unit in question. The higher the total flux, the less turns are needed to induce a given voltage, as a transformer is enlarged (it has room for a larger cross-section core), The peak magnetic flux chosen is normally not far around

698  Electrical Technology

Flux density – β , webers per square metre

the knee of the curve—as can be seen in Figure 36.18—since a non-linear change in flux with the primary coil current will introduce a non-linear relation between applied voltage and induced voltage. This will result in appreciable harmonics in the transformer output.

Annealed sheet silicon steel

1.8 1.6 1.4

Cast steel

1.2 1.0 0.8 0.6 0.4

Cast iron

0.2 0

0

1000

2000

3000

4000

5000

6000

7000

H ampere turns per metre

Figure 36.18  Magnetization Curves A transformer is also limited in its usable frequency. Since the maximum permissible flux density may not be exceeded, a reduction in frequency must be accompanied by a reduction in the applied voltage. Example 36.6 A 4600 to 230 V, 60 Hz, step-down transformer has core dimensions of 76.2 mm by 111.8 mm. A maximum flux density of 0.930 Wb/m2 is to be used. Calculate the following assuming 9 per cent loss of area due to stacking factor of laminations, (1) primary turns required, (2) turns per volt, (3) secondary turns required and (4 ) transformation ratio. Give your comments. Solution: Although the mechanical dimensions are normally given in millimetres, the calculations are performed in metres. 0.0762 m (0.118 m)(0.91) = 0.007752 m 2 0.00752 m 2

0.93000 Wb = φ = 0.0072 Wb m2 N =

E 4.4428 f φ pm

4600 Np = = 2397 turns 1.    4.4428(60)(0.0072) 2. 2397t/4600 V = 0.521 t/V 3. 230 V × 0.521 t/V = 120 turns 4. k = 4600/230 = 20 Comments: 1. The transformer specified is a fairly large one owing to its core size. Had its core area been less, say, 0.1 times the area specified, the required turns would have been 10 times as much; because 0.1 times the area would have meant 0.1 times the peak mutual flux. This, in turn is reasonable, since fm (maximum flux density or flux per unit area) is an inherent quality of the magnetic steel.

Transformers — Single Phase  699

2. Since the frequency f is also a linear component of the basic equations, if the frequency is changed, the voltage must change in a given transformer. A transformer designed for a low frequency can be used at a higher frequency. If the voltage is not changed the peak magnetic flux will be less. The other way around, if the frequency is reduced the voltage must also decrease because the magnetic flux cannot substantially increase. Example 36.7 A 120 V to 27.5 V, 400 Hz, step-down transformer is to be operated at 60 Hz. Find (1) the highest safe input voltage and (2) the transformation ratio in both frequency situations. Give your comments. Solution: 1. Since the peak mutual flux cannot increase, the allowed voltage must decrease in proportion ­because the general transformer equation must fit in either situation. 120 (60/400) =18 V 2. The 27.5 V secondary must also decrease by the same ratio since the volts per turn are the same for both primary and secondary. Therefore, 120 V1 = = 4.36 27.5 V2

k =

The general transformer equation does not specify the current lm. However, the magnetization of the core requires ampere turns through the coil. This current is related to the lamination steel used and how far beyond the true linear portion of the magnetization curve—as shown in Figure 36.18—the designer cares to go. Single-phase transformers will develop severe harmonics in the input voltage and current if the operating point of fm is too far up the curve. Example 36.8 The transformer in Example 36.6 is operated with no load or an open secondary. The mean magnetic path length is 1219 mm. Find the current required to maintain the core magnetization. Solution: The specified magnetic flux densities are entered into Figure 36.l8 to find the required A-t/m of the core length. For b = 0.930 Wb/m2, m = 420 A-t/m. (using the curves for annealed sheet silicon steel). The required ampere turns is (404 × 1.219) = 494 turns. It is not possible to interpret the curve to sufficient accuracy. The peak magnetization takes place during the peak current (for a.c. currents are rated in r.m.s v­ alues) 495 AT/2397 T = 0.207 peak amperes = 0.146 A=Im. This will still be under ideal transformer condition, and there will be some loss. A transformer of this size would have a rating in the 15 to 20 kVA range, so that its rated Ip current would be in the 2-3 A range.

36.12  PRACTICAL TRANSFORMER The ideal is not really achieved. One basic problem is that the inductive coupling between the coil windings is not perfect. In spite of all precautions in the design of the magnetic circuit, there is some small magnetic flux leakage from each coil. Each coil has some resistance which produces I 2 R losses. Each coil has inductance by virtue of its construction so that the currents passing through the coil see IZ voltage drop through the impedance. A loaded practical transformer then corresponds to Figure 36.19. Z1 =



( R1 ) 2 + ( X L1 ) 2

Z1 = R1 + jXL1



E1 = V1 − IZ



V1 = E1 + I1Z



k =

N1 = N2





(36.27) (36.28) (36.29)

 

(36.30)

E1 E2 

(36.31)

700  Electrical Technology Z 2 = ( R 2 ) 2 + ( X L2)2    

(36.32)

Z 2 = R 2 + jX L 2    

(36.33)

Ê 2 = V2 + I 2 Z 2     

(36.34)

V2 = Ê 2 − jI 2 Z 2    

(36.35)

Where, f1 is the primary leakage flux



f2 is the secondary leakage flux



R1 is the primary coil resistance



XL1 is the primary coil inductive leakage reactance due to leakage flux

R2 is the secondary coil resistance and XL2 is the secondary coil inductive leakage reactance due to leakage flux. These are simply the real-world circuit values that must exist since it is not possible to build coils around a magnetic core without some resistance and substantial inductance. The R and X components are really combined in the same wire length of the coils. These values are minimized by careful construction, as are the unavoidable leakage fluxes. The shell construction will nearly eliminate the leakage fluxes. However, the coils will then be longer (have a longer mean length) and the resistance will be inevitably greater. The actual physical design is a compromise of many factors. Improvements in the properties of lamination steel directly improve the transformer design. Higher attainable flux density would allow less winding turns for the same-sized core owing to a larger total flux in a smaller physical size. This effect would reduce the winding size and resistance. From Figure 36.19, as well as the ordinary a.c. circuit theory, it can be seen that the following instantaneous relations are true if the correct phasor relations are used: Primary leakage flux = Φ 1 Mutual flux

I1Z1 R1

Φm

XL1 I1

V1

Secondary leakage flux = Φ 2

E1

I2Z2 XL2

E2

Φ1

Φ2

N1 turns

R2 I2

V2

ZL

N2 turns

Figure 36.19  Practical Transformer Circuit Relations E1′ = V1′ − I1′ Z1 or V1′ = E1′ + I1′ Z1    and

E 2′ = V2′ + I 2′ Z 2



(36.36) (36.37)

′ ′ ′ 2 2 Further E1 = V1 − I1 ( R 1 ) + ( XL1 ) 

(36.38)

E ′ = V2′ + I 22 ( R 2 ) 2 + ( X L 2 ) 2 2  For a loaded practical transformer with lagging PF, V1 > E1  and  V2 > E2

(36.39)

The simple algebraic addition or subtraction of the IZ terms corresponds to the worst possible power factor, but it is a useful first approximation.

Transformers — Single Phase  701

36.13  TRANSFORMER RATINGS The manufacturer of an electrical machine usually indicates on the nameplate the normal operating conditions. A typical nameplate might read ‘4600 : 230 V, 60 Hz, 10 kVA’. The design voltages of the two windings are 4600 V and 230 V r.m.s and the turns ratio is 1:20; either side may be the primary. At a frequency of 60 Hz, the design voltages bring the operation near the knee of the magnetization curve and the core losses are not excessive. The actual primary voltage will be that necessary to provide rated secondary voltage under rated load. On large transformers, taps on the windings allow small adjustments in the turns ratio. Using either side as a secondary, the rated output of 10 kVA (full load) can be maintained continuously without excessive heating, and the consequent deterioration of the winding insulation. Since the heating is dependent on the square of the current, the output is rated in apparent power (kVA) rather than in power (kW). Supplying a zero power-factor load, a transformer can be operating at rated output while delivering zero power. The cross-section of the iron core is determined by the operating voltage and frequency. The cross section of the copper conductor is determined by the operating current. Knowing the effect of these factors, an engineer can operate a device under changed operating conditions. When electrical isolation between primary and secondary is not required, an auto transformer offers a low cost highefficiency alternative. In the auto-transformer, all the turns link the same flux, but some of the turns carry both primary and secondary currents. Example 36.9 A 2300 V to 230 V step-down transformer is rated at 750 kVA and 60 Hz. Its windings have the following resistances and inductances: R1 = 0.093 W, XL1 = 0.280 W, R2 = 0.00093 W and XL2 = 0.00280 W. The transformer is operating at rated load. Calculate the following: 1. Primary and secondary currents. 2. Primary and secondary winding impedances. 3. Primary and secondary winding voltage drops. 4. Primary and secondary induced voltages. 5. The transformation ratio. 6. Ratio of terminal voltages. Solution: kVA 75000 1. I1 = = V1 2300   = 326.1 A

750000 = 3216 A 230 750000 = )3216 2 2) 2= = in(the I1 when determined in this fashion transformer. Z1 (assumes R1 ) 2 + ( no X LI1losses 0.093 + (0A.280) 2 = 0.295 Ω 230 I2 =

2.

Z1 ( R1 ) 2 + ( X L1 ) 2 = Z 2 ( R 2 )2 + ( X L 2 )2 =

(0.093) 2 + (0.280) 2 = 0.295 Ω (0.00093) 2 + (0.00280) 2 = 0.00295 Ω

= I1Z1 326 = .1(0.295) 96.2 V 3.     I 2 Z 2 = 326.1(0.00295) = 0.962 V 4.     E1 = V1 − I1Z1 = 2300 − 96.2 = 2204 V E2 = V2 − I 2 Z 2 = 230 + 9.62 = 239.6 V 2204 k = 5.    = 9.198 239.4 V1 2300 = = 10 6.    V2 130

702  Electrical Technology Note: 1. A transformer actually has a turns ratio that is numerically less than the terminal voltage ratio. 2. A transformer can be used either way within its ratings. Example 36.10 Using the same transformer, as in Example 36.8, as a step-up unit with the rated kVA at the low- voltage winding, determine the probable secondary voltage. Solution: E1 = V1–I1Z1 = 230 – 0.92 = 220.4 V = k E= 2

1 = 0.1087 9.193

E1 220.4 = = 2028 V k 0.1087

E 2 = V2 + I 2 Z 2 or V2 = E2 − I 2 Z 2 (2028 − 96.2)V = V2 V2 = 1932 V Comparable ratios of performance hold for any transformer under reversed conditions. As a result, many transformers have tapped windings or a choice of many turns ratios. Example 36.11 The high voltage side of a transformer has 500 turns and the low-voltage side has 100 turns. When connected as a step-down transformer, the load current I2 is 12 A. Calculate: (1) transformation ratio, (2) load component of primary current (I1) and (3) transformation ratio if the transformer is used as step-up transformer. Solution: 500 N1 1. k = = =5 100 N 2 2. 3.

I1 =

I 2 12 = = 2.4 A 5 k

k =

N1 100 = = 0.2 500 N2



Example 36.12 A 2300/115 V, 60 Hz, 4.6 kVA step-down transformer is designed to have an induced e.m.f. of 2.5V/turn. Assuming an ideal transformer, calculate (1) Number of high-side and low-side turns, (2) Rated primary and secondary current, (3) Stepdown and step-up ratio, using answers to part (1), and (4) Step-down and step-up ratios using answers to part (2). Solution: 2300 = 920t = N1 1. N h = 2.5 115 = 46t = N 2 Nl = 2.5 4600 = 2A 2300 4600 = 40 A Il = 115

2. I h =

Transformers — Single Phase  703

920 = 20 46 46 = 0.05 kl = 920

3. kh =

40 = 20 2 2 kl = = 0.05 40

4. kh =

Example 36.13 A 1 kVA, 220/110 V, 400 Hz transformer is desired to be used at a frequency of 60 Hz. Calculate: (1) the maximum r.m.s voltage that may be applied to the high voltage side and maximum voltage output of the low-voltage side, (2) The kVA rating of the transformer under conditions of reduced frequency. Solution: 1. To maintain the same permissible flux density, both voltages of the high and low sides must change in the same proportion as the frequency. 60 Eh = 220 × = 33 V and 400 Eh 33.0 = = 16.5 V El = k 2 2. The original current rating of the transformer is unchanged, since the conductors still have the same current-carrying capacity. Thus, 1 × 103 Ih = = 4.545 A and the new VA rating is 220 Vh I h = Vl I l = 33 × 4.545 − 150 VA = 0.15 kVA

36.14  TRANSFORMER EQUIVALENT CIRCUITS The ability to lump together the equivalent combined circuit of a primary and secondary depends on the concept of reflected impedance.

36.14.1  Reflected Impedance Referring to Figure 36.19 and using the nomenclature shown, we may develop an impedance reflection scheme as follows: 1. If the load impedance ZL is removed, the secondary is open circuit, I2 = 0 and ZL = ∞. 2. Looking back into the transformed secondary terminals, the impedance is Z2 = V2 /I2. 3. Looking into the primary terminals, the impedance is Z1 = V1/I´1 4. A change in the load current I2 is reflected by a change in the primary current. Since the primary and secondary are related, we can use a single equivalent circuit with the secondary reflected to the primary. Thus V1 = kV2 and I1′ = I 2 /k2 kV2 V = Z1 = k 2 = k 2 Z 2 or k 2 = Z1/Z 2. (36.40) I / k I2 2     The fact that the ratio of the input to output impedance is equal to the transformation ratio squared is of vital importance. Primary and secondary impedance are related to Ohm’s law, so that 1. The high-voltage side has a relatively lower current and, thus, higher impedance. 2. The low-voltage side has a relatively higher current and thus lower impedance. 3. When reflected to the high-voltage side, impedance is higher, and vice versa.

704  Electrical Technology Example 36.14 Find the turns ratio of the transformer to provide impedance matching in the circuit in Figure 36.20. Solution: 10,000 Ω If the connection is made direct, as in Figure 36.20 very little power develops in the 4 W load. A transformer, as represented in Figure 36.21 is, therefore, inserted as k2 50 V will have to equal 4/10000.

4Ω

k 2 = 2500 = k

(a)

= 2500 50

Figure 36.20  For Example 36.14 I1

k = N1/N2

Z1

V2

I2

50:1 V1 10,000 Ω

ZL

4Ω

10,000 Ω 50 V

Z12 = k 2Z2 (a)

(b)

Figure 36.21  (a) Impedance-Matching (b) Impedance-Matching Transformer Example 36.15 An audio output transformer connected between audio amplifier and its speaker has 500 primary turns and 25 secondary turns. If the speaker impedance is 8 W, calculate: 1. Impedance reflected to the transformer primary at the output of the amplifier. 2. Matching transformer primary current if the output of the amplifier is 10 V. Solution: 500 N = 20 1. k = 1 = 25 N2 Z1 = k 2 Z L = (20) 2 × 800 = 3200 Ω = 3.2 kΩ 2. = I1

10 V1 = = 3.124 A Z1 3200

36.14.2  Impedance Matching The transformation ratios are V1/V2 = N1/N2 = k and, corr­espondingly, I1/I2 = N2/N1 = 1/k. If we divide the voltage ratios by the current ratios, we get the respective impedance ratios. Z V1/V2 k = = 1 = k2 (36.41) 1/ k I1/ I 2 Z2 Z1 = k 2Z L     Z1 = k 2  ZL The above relation establishes the transformer as an impedancematching device. It states that the ‘total impedance on the secondary of a transformer, ZL, is reflected back to the primary as a primary impedance which is ZL(k2), where ZL is the matching transformer secondary load (where ZL > > Z2) and k is the turns ratio N1/N2.

36.l4.3  The Three Winding Transformer as an Impedance Matching Device

Z1 = Z 2′ II Z ′3

N2 N1

VP

Z1

V2

N3 V3

Z2 x y Z3

Figure 36.22 shows an impedance-matching trans­former having two Figure 36.22 Impedance-matching Transseparate secondary windings, N2 and N3 each connected to two separate former with Two Sep­ arate loads, Z2 and Z3, respectively. Treating each impedance separately Secondary Loa­ds

Transformers — Single Phase  705

(36.42) ( NN )  N = Z′ = Z ( ) (36.43) N

Impedance reflected by

Z 2 = Z 2′ = Z 2

1

Impedance reflected by

Z3

1

3

3

2

2

2

2

The two reflected impedances, Z´2 and Z´3, may be considered as two unequal parallel impedances reflected as the primary impedance, where Zt = Z´2||Z´3 or Z 2′ Z 3′ Zt = Ω (36.44) ′ ′ Z Z + 2 3  N1 2 N1 2 N N = Z2, Z 3′ = Z 2′ = = Z 3 , V2 = V p 2 and V3 = 3 N2 N2 N1 N1

( )

( )

36.14.4  Tapped Matching Transformers Frequently, tapped-matching transformers are used with their loads connected, as shown in Figure 36.23. In this case, load Z3 sees its impedance reflected to the primary as Z´3= [N1/(N2 + N3)2] Z3, since Z3 is connected across the entire secondary. A relatively common type of tapped impedance-matching transformer used to match the output of a transistorized audio amplifier to a 4 W, or 8 W speaker load is shown in Figure 36.24. The load should present the same primary impedance, Zp, to the audio amplifier in order to match the output impedance of the transistorized audio amplifier for maximum power transfer. A N1 VP

N3

NP

Z1 N2 G

B

ZP

Z3

8Ω 4Ω

N2 N1

Z2

G

Figure 36.23 Tapped-secondary Impedancematching Transformer

Figure 36.24 Impedance-matching Transformer with Two Taps for Single Secondary Load

Example 36.16 The output impedance of a (single channel 100 W) transistorized power amplifier is 3.2 kW. A tapped impedance-matching transformer, as shown in Figure 36.24 having 1500 primary turns is used to match the amplifier output to either 4 W, or an 8 W speaker. Calculate: (1) total number of secondary turns, N2, to match an 8 W impedance speaker, (2) number of turns, N1, to match a 4 W impedance speaker, (3) impedance that must be connected between the 4 W and 8 W terminals to reflect a primary impedance of 3.2 kW. Solution: Zp 3200 1. k = = = 20 ZL 8 N2 = 2. k = N1 =

NP 1500t = = 75t k 20 ZP 3200 = = 28.284 ZL 4 Np k

=

1500t = 53t 28.284

3. N − N = 75t − 53t = 22t 2 1 Zp 3200 ZL = 2 = = 0.69 Ω (1500/ 22 ) 2 k 1. The 4 W tap is not placed at the centre of the secondary winding N2 but at some point approximately 0.7 of the total winding with respect to terminal G. 2. As shown by part (3) of the solution, improperly connecting a 4 W or 8 W speaker between the 8 W and 4 W taps produces a severe ‘mismatch’. The proper load impedance between these connections (for correct impedance match) is 0.69 W. Such a mismatch reduces the power (and volume) to the speaker severely.

706  Electrical Technology

36.14.5  Equivalent Circuits It can be seen from Figure 36.25 that the secondary winding resistance and inductive reactance and the load have been reflected back to the primary. The resulting circuit has the primary, secondary and magnetization circuits shown in series parallel. The primary current is composed of the magnetization current, Im and the load component I´1. I1′k2Z2

I1Z1

I1

k 2XL2

k2 R 2

XL1

R1

I1′

Im

Vj XLm

Rm

kV2

k 2ZL

Figure 36.25  Equivalent Circuit of Loaded Power Transformer Rm represents the in-phase component of the magnetization current; XLm represents the inductive reactance component of the transformer with an open secondary. The total current Im is almost 90º lagging with respect to the voltage V1. If the transformer is unloaded I´1 is zero and the right branch does not affect the circuit. In Figure 36.26, the Rm and XLm block has been shifted to the input voltage V1 side of the R1 and XL1 components. This would involve the magnetizing current being fed from the full V1 voltage without the small reduction due to I1Z1­ drop. Since the Im current is already very small in relation to I1 and at a very substantial phasor angle to it, this does not appreciably affect the I1 current. However, it does allow the primary resistance and reactance components and the reflected secondary resistance and reactance to be lumped as shown in Figure 36.26. I1′Ze1 R 1+ k 2R2 I1′

I1 Vj

XL1 + k 2XL2

Im kV2 Rm

k 2ZL

XLm

Figure 36.26 Equivalent Circuit Approximation with Combined Primary and Reflected Secondary Impedances Then, since the Im current component is considered negligible, when the transformer is loaded, the lm branch can be discarded leaving Figure 36.27 as the final simplification. The following combinations have been made. I I1Ze1 I1

V1

Re1

Xe1

kV2

k 2ZL

Figure 36.27 Simplified Equivalent Circuit with Magnetizing Current Im Assumed as Negligible

Transformers — Single Phase  707

Rel =R1+ k R2 Xel=XL1+k2XL2

(36.45) (36.46)

2

Z = ( R e1 ) 2 + ( X e1 ) 2 or(36.47) e1 Z e1 = R e1 + jX e1  (36.48)    As a result of the assumption that Im is minimal I1 ≅ I´1 We can now combine the resistive and reactive components of the load V1 (36.49) I1 =  ( R e1 + k 2 R L ) 2 + ( X e1 ± k 2 X L ) 2 or in complex form I1 =

R e1

V1  + k R L + j ( X e1 ± k 2 X L )

(36.50)

2

The ± sign allows for load power factor, the plus sign for lagging load power factor, the minus sign for leading load power factor, and the k2XL term drops out for unity power factor.

36.15  SECONDARY VOLTAGE PHASOR RELATIONS Transformer impedances can be reflected either all to the primary or all to the secondary. To determine the secondary power factor and voltage regulation, it is desirable to reflect impedances to the secondary. The impedance relationships become equivalent secondary relations. R1 R e2 = R 2 + 2 (36.51) k  X X e2 = X L 2 + L21 (36.52) k 

Z e2 =

( Re 2 )2 + ( X e 2 )2

Z e 2 = R e 2 + jX e 2

(36.53)



(36.54)



The relation between these impedances is shown in Figure 36.26.

When the phasors are combined in a diagram form as illustrated in Figure 36.28, voltage regulation of a power transformer can be calculated. With a unity power factor load, the output current I2 is drawn in line with the output terminal voltage V2. Thus, the voltage phasor I2 Re2 is also in line with the output voltage V2. Since the reflected primary and secondary impedances must be accounted for, the I2Ze2 phasor is drawn as shown. Ze3 is the combined impedance of both windings. The relationship between V1 and V2 is V1 = (V2 + I 2 R e2 ) 2 + ( I 2 X eL ) 2 (36.55) k 

I2 Z

Figure 36.28  Unity Power Factor

I2

Z e2 I2 X

I2 = I1k

e2

V1 k

q2 2

co

sq

q2

V

V2

V

e2

2

sin

2

I2 R

When the load power factor is lagging, the load current lags behind the load voltage V2 by the power factor angle q, as shown in Figure 36.29. Since the current in the secondary winding is in phase with the load current, the voltage phasor I2Re2 is drawn parallel to the load current I2, establishing the impedance triangle relationship to the V2 phasor. The voltage phasor V2 has been drawn horizontally; as is the conventional diagram attitude (the trigonometric visualization is simplified).

I2 Re2

V2

I2 = I1k

This is very similar to the voltage regulation of a synchronous alternator under the unity power factor.

36.15.2  Lagging Power Factor Voltage Relations

e2

V1 k

I2 Xe2

36.15.1  Unity Power Factor Voltage Relations

Figure 36.29  Lagging Power Factor

708  Electrical Technology

V1 = k

(V2 cos θ 2 + I 2 R e 2 ) 2 + (V2 sin θ 2 + I 2 R e 2 ) 2 (36.56a)

36.15.3  Leading Power Factor When the load power factor is leading, the voltage phasor diagram assumes the relationship shown is Figure 36.30. Since the load current leads the load voltage, I2 is shown leading V2, and V2 is still shown horizontally in a reference position.

V1 = k

(V2 cos θ 2 + I 2 R e 2 ) 2 + (V2 sin θ 2 − I 2 R e 2 ) 2 (36.56b)

XL2

Z

e2

Xe2

k2

XL 1

The minus sign occurs in the last term since the V2 sin q2 phasor sense is opposite to the I2 Xe2 phasor. Note: The relations in all three power factor situations involve the same reflected equivalent impedance triangle as can be seen in Figure 36.31.

I2 Xe2

s q2

V2 V1 k I2 = I1k

V2 sin q2

co

R1

q2 V2

k2

I2 Re2

Figure 36.30  Leading Power Factor

R2 Re2

Figure 36.3l Reflected Equivalent Impedance Relations

36.16  TRANSFORMER VOLTAGE R ­ EGULATION The fall in output terminal voltage is known as regul­ation and increases as the load increases. It is usually expressed as a percentage of the secondary no load voltage. Per cent regulation =

No load secondary voltage − full load secondary voltage full load secondary voltage

In Eq. 36.56, the last term is plus (+) for unity and lagging power factor and minus (–) for leading power factor. Again, if the power factor is unity, the cos q2 term is unity and the sin q2 term is zero. Equation 36.56 becomes Eq 36.55. The output voltage of a transformer is not intended to match the input voltage, but rather the input voltage times the transformation ratio (V1/k). The voltage regulation then becomes V1 / k − V2 100 = per cent regulation V2 

(36.57)

36.17  MAXMIMUM POWER TRANSFER Figure 36.32 shows a practical a.c. source, V, whose internal impedance, ZS, is essentially resistive. The terminal voltage of the source is V1. Since the transformer is ideal, the primary induced voltage E1 = V1. The load impedance ZL is resistive, RL, and is coupled to the source via an ideal matching transformer. The load impedance reflected to the primary is Zp, where, Zp = k2ZL = k2RL.

Transformers — Single Phase  709 I1

Zs = Rs

Zp

Vs

+

V1 = E1

V

k=

N1 N2

I2

N2

N1

E2 = V2

–

ZL = RL

V2

Ideal matching transformer

Source

Figure 36.32 Using Matching Transformer for Maximum Power Transfer from Source (V) to Load (RL) For maximum power transfer in a.c. circuits, the following equalities occur. 1. The reflected impedance to the primary must be the conjugate of the source impedance. For the resistive case shown in Figure 36.32. R= R= k 2 R L , and = k P S

ZP = ZL

RS RL

2. The terminal voltage V1 is equal in magnitude to the internal voltage drop across the source impedance or half the source voltage V: V V V= = 1 S 2 3. Then the secondary terminal voltage V2, using the transformation ratio, is

V2 =

V1 V = k 2k

4. The secondary current, I2 and primary current I1 are, respectively, I2 =

V2 V V V = 1 = = 2kZ1 2kR L ZL kZ L

I1 =

I2 V V V = = = 2 ZP 2 R S k Z1 + Z P

5. Then the power transferred to the load is PL = I 22 R L 6. The power supplied by the source or the total system power, PT is PT = VI1 cos θ = PL + PS = I 22 R L + I12 RS 7. And the power transfer efficiency, η is

η =

PL PT

50 per cent

Example 36.17 For the circuit shown in Figure 36.32 the supply voltage of the source is 10∟0◦ V, the resistance of the source is 1 kW, and the load resistance, is 10 W. Calculate: (1) required transformation ratio of the matching transformer for MPT; (2) terminal voltage of the source at MPT; (3) terminal voltage across the load at MPT; (4) secondary load current I2, by at least two independent methods; (5) primary load current drawn from the source, I1 by at least two methods; (6) maximum power dissipated in the load; (7) power dissipated internally within the source (8) power transfer efficiency.

710  Electrical Technology Solution: 1. = k

RS = RL

1000 = 10 10

ο 2. V1 = V = 10 0 V = 5 V 2 L V1 5 V = = 0.5 V 3. V= 2 k 10

0.5 V 10 V V2 V 4. I 2 = Z = 10 W = 50 mA, I 2 = 2kR = 10′10 W = 50 mA 1 L 5. I1 =

10 V 50 mA I2 V = = 5 mA; I1 = = = 5 mA 2RS 21 kW 10 k

6. PL = I 22 RL = (50 mA) 2 10 Ω = 25 mW 7. PT = VI1 cos θ = (10 V)(5 mA)(1) = 50 mW; PT + PS = 25 mW = 50 mW 8. η =

PL Pr

=

25 mW = 50 per cent 50 mW

36.18  LOSSES IN TRANSFORMERS A transformer heats up when in operation. The basic purpose of a transformer is not to provide heat but to transfer energy from the primary to the secondary. Any heat produced by a transformer represents a power loss or inefficiency. The efficiency η of a transformer is PS percentage efficiency, η = × 100 (36.58) PP  The power consumed by the transformer is referred to as a power loss. This is caused by hysteresis loss, eddy current loss, and copper or I 2 R loss. 1. Hysteresis Loss: This has been dealt with in detail in Chapter 10. 2. Eddy Current Loss: The changing magnetic flux induces a voltage in the coil conductors as well as the core. The voltage induced in the core causes the current to circulate in the core. This current is called eddy current. The eddy current, in flowing through the resistance of the core, produces heat. An eddy current is defined as a local current induced in a conducting, body by a varying or relatively moving magnetic flux. The production of eddy currents is a serious disadvantage firstly because energy is expended in producing them, and secondly because the eddy currents produce heat and may cause an undesirable temperature rise. In a.c. circuits, the eddy currents are continuously induced; consequently, their presence is a much more serious matter here than in a simple d.c. circuit It is not possible to prevent the induction of eddy currents but their magnitude can be restricted by providing a high resistance path. For this reason, high resistivity is a desirable feature in soft magnetic materials when carrying alternating fluxes. The cores of instruments that are subjected to ­alternating fluxes are always subdivided by constructing them from thin laminated sheets (Figure 36.33) or stampings, each one insulated from its neighbour by a film of ­insulating varnish. This lamination Oxide on has very little effect on the magnetic path but offers maximum reeach surface sistance to eddy currents. The path of eddy currents includes these of the insulating gaps in series, and the gaps further afford a considerable laminations reduction in the magnitude of eddy currents. For the higher frequency currents, dust cores or ferrites are employed. The use of a special high permeability iron, such as mumetal, for the stampings reduces the loss by reducing the dimensions Figure 36.33  Laminated Core required for the iron circuit. Refer Figure 36.34.

Transformers — Single Phase  711

3. Copper Loss: Copper loss refers to the power dissipated in the windings of the transformer. It is also called the I 2R loss, where, R is the d.c. resistance of the turns in the winding. Copper loss is minimized by using as large a ­conductor as possible in the windings. However, the conductor size is limited by the area of the windows (openings) in the core into which the winding must fit.

(a)

(b)

36.19  SHORT CIRCUIT TEST

(c)

The short circuit test of a transformer, as illustrated in Figure 36.35, is performed by shorting out the low voltage winding, and applying voltage to the high voltage winding until the rated current flows in the high voltage winding, at which time the rated current should also flow in the low voltage winding.

(d)

HV

Power meter

Isc VSource

Figure 36.34 Production of ­Transformer Core Stampings LV

Vsc

VSource adjusted until rated current flows in windings. Values referred to measured windings. zeq =

Vsc Isc

Req =

P Isc2

Xeq = Zeq2 – Req2

Figure 36.35  Transformer Test Connections—Short Circuit Test The voltage and power is recorded. Because core losses are proportional to V 2, and the voltage ­required to cause rated current to flow with a shorted winding is low, the core losses are very low during the short circuit test. The core losses are low enough in the short circuit tests to be ignored for all but the most accurate calculations. All the input power drawn during the short circuit test is used in overcoming the total reflected primary copper loss. Little or no power is consumed by the magnetic losses during the short c­ ircuit test.

36.20  OPEN CIRCUIT TEST Figure 36.36 shows the connections for the open circuit test. The low voltage winding is supplied with the rated voltage which should result in the rated voltage on the high voltage side. The current drawn should be low enough so that the copper losses are very low and the power measured is almost all from the core losses. The current drawn during the open circuit test will be the excitation current. Power meter VSource At rated low voltage

Isc

LV

HV

Voc

Ioc = Iex LV = jIm + Ih + e = jIex sin θ + Iex cos θ Poc =

Voc2 RcLV

cos θ =

Poc Voc Ioc

Figure 36.36  Transformer Test Connections—Open Circuit Test

712  Electrical Technology During the open circuit test, there are virtually no copper losses in the primary winding and none in the secondary winding, since I is very small compared to the rated I. As a result, the power consumed in the open circuit test is all chargeable to the magnetic circuit losses. These losses include the hysteresis and eddy current losses and the magnetization power losses. When both the losses have been found, the overall transformer efficiency can be calculated.

36.21  TRANSFORMER EFFICIENCY The efficiency of any device is its output power divided by its input power. When the efficiency is desired in per cent the answer is multiplied by 100. output × 100 = η per cent Efficiency = (36.59) input  The short circuit test will enable determination of the winding copper losses in a transformer. The open circuit test will ­enable the determination of the combined magnetic circuit losses in a transformer. Since these are all of the losses involved, Efficiency =

input − ∑ losses



input

× 100 per cent 

(36.60)

or output × 100 per cent output + ∑ losses

(36.61)  In rating the efficiency of distribution transformers used to supply a load district or a commercial area, a term called the allday efficiency is used. This is simply the total energy delivered by the transformer in 24 hours divided by the total energy input. A determination of this efficiency shows the advantage of the fact that the best efficiency is placed somewhere near the mid-load. The all-day efficiency takes into account periods of no load when the input is solely the Pcore requirement. Distribution transformers and voltage-dropping transformers for machine tools are never switched off. There is an advantage in keeping the windings mildly warmed by the magnetizing current. This helps prevent insulation breakdown in a humid climate since the windings tend to stay dry when they are warm. Transformers are the most efficient devices ever invented. A transformer that is properly designed, built and cooled can be more than 99 per cent efficient. Maximum efficiency always occurs at the load point where the fixed losses equal variable losses. For maximum efficiency in a transformer, variable copper losses must equal fixed core losses, or

I 22 R e2 = Pc



Pc R e2  Where, I2 is the load current, at which maximum efficiency occurs. I2 =

(36.62) (36.63)

Example 36.18 A 20 kW lighting transformer of ordinary efficiency, 95 per cent is on full load for six hours a day. Find the all day efficiency if the full-load losses are equally divided between copper and iron. Solution: Total losses on full load = 5 per cent of 20 kW = 1 kW Iron losses = 0.5 kW and full load copper losses = 0.5 kW Output of 20 kW for 6 hours = 120 kWh Copper losses for 6 hours/day = 3kW Iron losses for 24 hrs/day = 12 kWh Energy intake during 24 hours = 135 kWh All day efficiency = 120/135 = 88.9 per cent Example 36.19 The total copper loss of a transformer determined by the short circuit test at 20 °C is 630 W, and the copper loss determined by the true ohmic resistance at the same temperature is 504 W. What is the load loss at a working temperature of 75 °C?

Transformers — Single Phase  713

Solution: Eddy current loss at 20 °C = 630 – 504 = 126 W True copper loss at 75 °C = (504) (75 + 234.5)/(20 + 234.5) = 613 W Eddy current loss at 75 °C = (126) (20 + 234.5)/(75 + 234.5) = 104 W Load loss at 75 °C = 613 + 104 = 717 W Example 36.20 A 50 kV A, 2200/200 V, 60 Hz transformer has a core loss, determined by the open circuit test, of 3500 W and a copper loss at rated current of 630 W, determined by the short circuit test. Find the efficiency: (1) at full load, unity power factor; (2) at three fourth load, unity power factor; (3) at full kVA rating, 80 per cent power factor; and (4) at three fourth of rated kVA, 80 per cent power factor. Solution: 50000 = 98.1 per cent 1. 50000 + 350 + 630 2. 3. 4.

3 /4 (50000) = 98.22 per cent 3 / 4 (50000) + 350 + (3 / 4) 2 630 50000(0.80) = 97.6 per cent 50000(0.80) + 350 + 630 3/ 4 (50000)(0.80) = 97.7 per cent 3 / 4 (50000)(0. 80) + (3 / 4)2 (130)

36.22 AUTOTRANSFORMER In an autotransformer, part of the energy transfer is by conduction. In a normal transformer, all of the energy transfer is by induction. The special field of utility of an autotransformer is where the transformation ratio is relatively small or near unity. The three principle types of autotransformers are shown in Figure 36.37. It is composed of two separate windings that are electrically joined at a common point. An autotransformer is more efficient than a normal transformer. In a step-down autotransformer, V1 > V2, so that I1 < I2. In a step-up autotransformer, the situation is the other way round so that I2 I1

I1

a b

V1

I2

V2 V1

c

Load

V2

Ic

Load

Ic

(b)

(a) I2 I1 Movable brush V1

Ic

Prepared winding surface

V2

Load

(c)

Figure 36.37 Autotransformer Circuits (a) Step-down Autotransformer (b) Step-up Autotransformer (c) Variable Autotransformer

714  Electrical Technology V1 < V2  and  I1 > I2. The real difference in an autotransformer is that in a step-down autotransformer I2 = I1 + Ic(36.64) And in a step-up autotransformer I1 = I2 + Ic(36.65) The term Ic means the common current. The common current is always the arithmetic difference between I1 and I2. The normal transformer has two windings for the I 2R losses. The autotransformer has a sharply reduced loss in the common winding portion and a normal loss in the extended winding leg. The reason for the greatly reduced current in the common winding portion is that I1 and I2 currents pass through the common portion and they are always in the opposite sense. The current caused by the induced voltage flows opposite to the input current. In an autotransformer, the secondary current is this induced current.

36.22.1  VARIABLE AUTOTRANSFORMERS An autotransformer is made variable in the voltage adjustment in a similar fashion to the output of a potentiometer. These are built by winding a single coil on a toroidal magnetic core. The outer surface of the winding is specially plated and prepared for use as a commutator-like switching surface. The movable secondary connection is usually a special carbon composition brush. It is arranged in such a way that the connection is never opened, but rather a continuous variation of secondary turns is selected turn by turn. The secondary voltage is then adjustable in very small steps from about 0 to 1.2 or more times the primary voltage. The result is an efficient source of variable output a.c. voltage. Autotransformer circuits are shown in Figure 36.37.

36.22.2  Autotransformer Power Division The power divided between the conducted power, Pcon, and the transformed power, Ptr, is such that the output power equals the sum of the two parts. P2 = Pam + Ptr(36.66) This assumes that the power factor is 100 per cent and the load is purely resistive. In Figure 36.37(a) if we label the upper part of the winding, that sees the input current, as a, b and the winding that sees the combined current as bc, the total length of the winding is then a.c. and winding portion ab considered to be the primary of the transformer, winding portion bc is consiered to be the secondary of the transformer. The ampereturns NabI1 equal the ampere turns Nbc(I2–I1). This is in accordance with the relation N1I1 @ N2I2. In an autotransformer, I´ is more nearly equal to I1 than in a normal transformer, because there is less inductive work to be performed, so there is less Im current. P = (V1 − V2 ) / I1 = V2 ( I 2 − I1 ) (36.67) tr 

Ptr = P1 (1 − 1/ k ) Pcond = P2 − p1





(36.68) (36.69)

Example 36.21 An autotransformer is adjusted for an output voltage of 86.3 V when operated from a 117 V line. The variable power load draws 3.63 kW at unity power factor at this setting. Determine the transformed power and the conduced power at this setting. Solution: 117/85.3 = k = 1.37 If we assume 100 per cent efficiency, the power input is 3.63 kW Ptr = 3630(1 – 1/1.37) = 980 W Pcon = 3630 – 980 = 2650 W Note: In this situation 1. 73 per cent of the total output is directly conducted to the load. 2. 27 per cent of the power is transformed.

Transformers — Single Phase  715

3. The ratio of transformed power to conducted power becomes small as k approaches unity. 4. The power handling ability of an autotransformer is misused if the k required is large. Example 36.22 A normal 5 kVA, 2300/208 V step-down distribution transformation is connected as an autotransformer to step the line voltage down from 2508 to 2300 V. The transformer is connected with the 208 V winding as ab and the 2300 V winding as bc (Figure 36.37(a)). Assume that the transformer is working to its full 5 kVA rating. Assume unity power factor and calculate the total power input. Solution: k = 2508/2300 = 1.090 P1 =

5000 Ptr = = 60555 W 1 − 1/ k 1/1.090

Note: 1. With a turns ratio not far from one, when connected for voltage adjustment, this transformer can handle a load of over 60 kW. 2. A distribution transformer with tapped windings, and with double-primary and double-secondary windings, can be used to handle a wide variety of power-line voltage adjustments. 3. Very large levels of power can be handled with moderate-sized units and with very high efficiency. 4. An autotransformer is not suitable for large percentage voltage reductions as a distribution transformer.

36.23  POWER TRANSFORMERS Power transformers are normally oil-immersed transformers used for substations and connection to large commercial and industrial customers. Power transformer capacities vary over a wide range. Power transformers are designed for very high efficiency and are routinely maintained. Power transformers are well protected both for safety and economy. Almost all large power transformers are three phase as opposed to three single phase units. A power transformer one-line drawing with protection is shown in Figure 36.38. Power transformer Primary Lightning arrester

Circuit breaker

Fuse CB Current transformer

Current transformer and potential transformer for protective relays

Lightning arrester

Figure 36.38  Power Transformer One-line Drawing with Protection

36.23.1  Parts of a Transformer The design, type, and accessories of a power transformer, depend on its size, application and location. The principal components and accessories are shown in Figure 36.39. The steel tank is used for housing the core and windings and for mounting various accessories ­required for the operation of the transformer. The core is constructed of insulated iron sheet laminations in order to keep iron losses low. The laminations are interleaved at the ends to form the frame of the required shape and size. There is one primary and one secondary winding in a single-phase transformer and three primary and three secondary windings in a three-phase transformer that are suitably interconnected. The low-voltage winding is placed next to the core and the high-voltage winding is placed concentrically over it. The tap changing switch (not shown) is used for changing the transformation ratio of the windings to allow for variations on the primary side. The conservator is a small cylindrical tank mounted on top of oil-filled transformers to provide space

716  Electrical Technology Conservator tank

Bushing Steet tank Insulating oil Cooling tubes

Silica gel breather

Winding

Core Wheels

Figure 36.39  Parts of a Transformer for expansion and contraction of oil. The conservator is equipped with a sight glass at one end to indicate the level of oil. The breather is the device through which all movement of air from and into the transformer takes place. The breather is equipped with a dehydrator containing silica gel which absorbs moisture from the air passing through it. It can be reactivated. In all transformers above 50 kVA rating, a thermometer is fitted to indicate the temperature of hot oil. In very large units, temperature indicating devices are fitted in the winding itself to record temperature. A ­temperature sensitive alarm is also incorporated to sound the alarm in the event of an abnormal temperature rise. Pressure relief, fitted on top of the transformer, provides protection against pressure build-up inside a transformer. The transformer can explode in the absence of pressure relief. A number of valves for the purpose of filling, drawing, filtering, sampling etc., are provided on each transformer. Different methods of cooling of transformers are illustrated in Figure 36.40. Water

(a)

Core windings

(b)

(c)

Figure 36.40 Methods of Cooling Transformers (a) Tank with External Tube (Natural Cooling) (b) Water Cooling (c) Air Blast Cooling

36.24  INSTRUMENT TRANSFORMERS Transformers used in distribution include: power transformers, autotransformers, distribution transformers and instrument transformers. Distribution transformers provide the final link with the customer. A typical high quality pole top transformer has a life expectancy of up to 40 years. They are protected with a minimum of a primary fuse and a lightning arrester at or near the transformer primary. In direct current circuits, the useful range of meters may be extended beyond their self-contained capacities by the use of calibrated resistors in the form of shunts in the current circuits and multipliers in the voltage circuits. These devices in general, are unsuited to alternating current measurements, especially because they do not provide insulation of the meters from the relatively high voltages so extensively used in alternating current circuits. Such circuits necessitate the use of auxiliary apparatus to permit greater convenience in measurement to insulate the meters from the high-voltage circuits as a safeguard to life. Instrument transformers are employed on all alternating current circuits which exceed predetermined voltage and current limits for safe and economic operation of self-contained meters.

Transformers — Single Phase  717

There are two distinct classes of instrument transformers: voltage or potential transformers and current transformers. These two classes of transformers differ only as to the method of use. HV Bushings

36.24.1  Potential Transformers

The capacity of a potential transformer, as illustrated in Figure 36.41 is small compared to that of power transformers. The load on potential Oil transformers is relatively light and it is not necessary to have potential transformers with a capacity greater than 100 to 500 VA. Core The high-voltage primary winding of a potential trans­former has the same voltage rating as the primary circuit. Let us assume that it is Winding required to measure the voltage of a 4600 V, single-phase line and the ratio between the primary and secondary windings is 20:1. A voltmeter connected across the secondary of the potential transformer indi- Winding cates 230 V. To indicate the actual reading on the high-voltage circuit, the instrument reading (in this example), must be multiplied by 20. In some cases, the voltmeter is calibrated to read the actual value of the Figure 36.41  Potential Transformer voltage on the primary side; thus, the possibility of errors is reduced. All instrument transformers now manufactured have subtractive polarity as shown in Figure 36.42. One of the secondary leads of the transPrimary H1 H2 former is grounded to eliminate high-voltage hazards. Potential transformers have highly accurate ratios; generally, the error is less than 0.5 per cent.

36.24.2  Current Transformer Secondary x1

x2

Meter

Figure 36.42 Principle of Operation of Instrument Transformers

Current transformers, as shown in Figure 36.43 have very accurate ratios between the primary and secondary current values. The error of most modern current transformers is less than 0.5 per cent. The secondary circuit should never be opened when there is current in the primary winding. When the primary winding has a large current rating, it may consist of a straight conductor passing through the centre of a straight hollow core as shown in Figure 36.43(b). Secondary winding (N 2 turns)

50 A Connected in series with Very low voltage circuit

Insulated primary (I turn)

5A Higher voltage

Toroidal ‘clock spring’ core

Primary Secondary (a)

(b)

Figure 36.43 (a) Connections for Current Transformer (b) Current Transformer with a Bar Primary The secondary winding is wound on the core. This assembly is called a bar-type current t­ransformer. All standard current transformers with ratings of 1000 amperes or more are bar-type transformers.

36.25  PULSE TRANSFORMERS A pulse transformer is a special type of wide-band transformer designed to transmit voltage or current pulses with specific requirements on wave-shape fidelity. A pulse transformer is more like a transmission line than a conventional power transformer. In general, the turns ratio of a pulse transformer must be low in order to provide good wave shape fidelity.

718  Electrical Technology A thyristor-trigger transformer is a pulse transformer designed especially for gate-triggering applications. The triggering and the impedance characteristics of thyristors and the inter-winding dielectric strength are the key factors in the design. Most applications call for a step-down transformer, since the voltage required for effective triggering is relatively low. Several secondaries may be needed for simultaneous triggering of thyristors. Since trigger transformers usually interface between low-power sensitive control circuits and high-voltage high power circuits, they need a high interwinding dielectric strength. An increase in dielectric strength causes an increase in transformer size which causes an increase in pulse rise time.

36.26  TRANSFORMER CONNECTIONS The multiple-winding transformer, as seen in Figure 36.44 is really two complete sets of windings wound on one core. There are two 2300 V primary coils and two 230 V secondary coils. Each coil can transfer the energy involved in 25 kVA at any reasonable power factor. Normally, all coils are used in any one of the four possible connections. These combinations are accomplished by series and parallel connections. Only coils having identical voltage may be paralleled. H1

X1

H3

X3

4600 V

2300 V

H1

X1

H3

X3

460 V H2

460 V

X2

H2

X4

H4

H1

X1

H1

X1

H3

X3

H3

X3

H2

X2

H2

X2

X4

H4

H4

(a)

230 V

2300 V

X2

(b)

X4

230 V

4600 V

H4

(c)

(d)

X4

Figure 36.44 Typical Distribution Transformer Showing Voltage Connections (a) 4600/460 V Series-Series Connection (b) 2300/460 V Parallel-Series Connection (c) 4600/230 V Series-Parallel Connection (d) 2300/230 V Parallel-Parallel Connection Conventionally, the odd-number terminals have the same instantaneous polarity. This means that the coils, when properly interconnected, will oppose each other so that there will be no circulating current. Therefore, H1 may be connected to H3, and H2 to H4. The same situation holds for the secondary connections. In order that transformers may be successfully and economically paralleled, some vital conditions must be met. 1. The voltage ratings of the primaries and secondaries and, therefore, the turns ratios should be i­dentical. 2. The proper polarities must be observed at interconnection. 3. With different kVA rating transformers, the equivalent impedances should be inversely proportional to kVA ratings if circulating currents are to be avoided. 4. The equivalent resistances and equivalent reactances should have the same ratio to avoid ­circulating currents and operation at different power factors. If the equivalent impedances are nearly inversely proportional to their kVA ratings, it is usually sufficient. If one unit were to be of core construction and the other of shell construction, the probability of a successful match is much less, an actual test would be needed to obtain the required data. The circulating current, Ic, is seen to be the difference between secondary voltages divided by the sum of reflected secondary impedances: (V − V2b ) I c = 2a V2 a + V2b (36.70)

Transformers — Single Phase  719

where, Ic = circulating current V2a and V2b = secondary voltages of transformers a and b Ze2a and Ze2b = reflected equivalent secondary impedances of transformers a and b When the transformation ratio is exactly similar and the secondary impedances are nearly inversely proportional to the transformer load ratings, there is an acceptable load matching. Example 36.23 A small 10 VA, 115 V primary transformer has two secondary windings, 6.3 V and 5.0 V, respectively, with impedances of 0.2 W and 0.15 W, respectively. Calculate 1. Rated secondary current when the L-V windings are connected in series aiding 2. Circulating current when the L-V windings are paralleled, and per cent overload produced Assume that both impedances are at the same phase angle. Solution: 1. Both coils must be series-connected and used to account for the full VA rating of the transformer. Hence, the rated current is I2 =

100 VA 10 VA = = 0.885 A (6.3 + 5) V 11.3 V

2. When the windings are paralleled, the net circulating current is due to the net voltage applied across the total internal impedance of the windings, or Ic =

(63.5) V 1.3 V = = 3.71 A (0.2 + 0.15) W 0.35 V

The percent overload is

3.17 A = 420 per cent 0.835 A

Example 36.24 High side short circuit test data for the 20 kVA transformer shown in Figure 36.45 is 115 V, 87 A, 250 W. Calculate (1) equivalent impedance referred to the high side when the coils are series-connected; (2) equivalent-impedance referred to the low side when the coils are series connected; (3) rated secondary current when the coils are series-connected; and (4) secondary current when the coils are short circuited and rated voltage applied to the high-voltage side, and the per cent overload produced. Solution: 4.5 V Z eh = = 0.0517 Ω 1.        87 A Z eh = 0.0517 2.   I 2 rated = 3.    

Vf Vr 115 V a.c.

Figure 36.45  For Example 36.24

20 = 3.91 × 10−4 Ω 230

20 kVA × 1000 = 1000 A = 1 × 103 A 20 V

20 V I 2 sc = = 51 × 103 −4 3 . 91 × 10  

720  Electrical Technology Example 36.25 A 15 kVA, 4600/208 V transformer with a secondary impedance of 0.0100 W is to be paralleled with a 10 kVA, 4600/208 V transformer with a Ze2b = 0.0148 W. When the two transformers are paralleled and carrying a combined load of 21.5 kVA, calculate; (1) the individual load currents, (2) the per cent load capacity used with each transformer. Solution: 1.

The total load current for 21.5= kVA

The individual transformers being matched in secondary impedance I 2b = I 2a 93.5 =

21500 = 93.5 A 230

voltage will carry a load current in inverse proportion to their reflected 0.0100 0.0100 = I 2b ; I 2a 0.0148 0.0148

0.0100 I 2a + I 2a 0 .0143

 I tot = I 2 a + I 2b

= I 2 a 55 = .8 A, I 2b 37.7 A 15000 10000 I a (= .2 A I brated 65 = = 43.5 A rated ) = 230 230 2.

Per cent rated current =

55.8 × 100 = 85.6 per cent ( transformer a ) 65.2

Percent rated current =

37.7 × 100 = 86.6 per cent ( transformer b) 43.5

S UM M A RY 1. Alternating current can be more easily transformed. 2. The property of mutual inductance is associated with the flux of one coil linking with the turns of a second coil. 3. M = µ0 µ r N1 N 2 Henrys. 4. M =

L1 L2 .

5. If the coils are wound tightly—one on top of the other—on a common soft iron core, the leakage flux is negligible. 6. M = k1 L1 L2 . 7. LT = L1 + L2 ± 2 M . 8. 1/LT = 1/(L1 ± M) + 1/(L2 ± M). 9. Mutual inductance can be prevented by axis orientation, physical separation and shielding. 10. Transformers change voltage and current levels. 11. The transformer primary and secondary are connected magnetically. 12. The transformer construction is either core type or shell type. 13. The magnetizing current is the only current during no load. 14. I1′ N1 = I 2 N 2 .

15. = k

N1 = N2

E1 = E2

I1 V = 1 . I 2 V2

16. Eav = 4φm f N . 17. There are losses (flux leakage, I 2R, hysteresis and eddy current) in a practical transformer. 18. The secondary impedance can be reflected to the primary and vice versa. 19. Transformers are used for isolation and impedance matching. 20. An eddy current is a local current induced in a conducting body by a varying magnetic field. 21. The magnitude of eddy currents can be reduced by providing a high-resistance path. 22. Transformers are the most efficient devices ever invented. 23. Autotransformers are used where the transformation ratio is near unity. 24. Power transformers are designed for very high efficiency. 25. The tap-changing switch is used for changing the transformation ratio. 26. Distribution transformers provide the final link with the customer. 27. Potential transformers have a highly accurate ratio.

Transformers — Single Phase  721

28. In current transformers when the primary winding has a large current rating it may consist of a straight conductor passing through the centre of a straight hollow core. 29. A pulse transformer is a wide-band transformer designed to transmit voltage or current pulses.

30. Conventionally, the odd-number terminals have the same instantaneous polarity. 31. In order that transformers may be successfully connected in parallel, some vital conditions must be met.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. When the primary winding has more turns than the secondary, the voltage in the secondary is

(a) Increased (c) Doubled

(b) Decreased (d) Halved

2. In the coils of a transformer the motion of the flux is caused by

(a) Direct current (c) Rotating primary

(b) Moving secondary (d) Alternating current

3. Energy is transferred from the primary to the secondary coils, without a change of

(a) Frequency (c) Voltage

(b) Current (d) Ampere-turns

4. Transformer efficiency averages

(a) 70 per cent (c) 50 per cent

(b) 97 per cent (d) 100 per cent

5. A transformer has a primary winding rated at 150 V, and a secondary winding rated at 300 V. The primary winding has 500 turns. How many turns does the secondary have?

(a) 250 (c) 2500

(b) 1000 (d) 10000

6. A control transformer is a step-down transformer. Compared to the secondary winding, the primary winding is

(a) Larger in wire size (b) Smaller in wire size (c) The same size as the secondary (d) Connected to the load

7. The current in the secondary winding is

(a) Higher than that in the primary (b) Is lower than that in the primary (c) Controls the current in the primary

8. Double-wound transformers contain

(a) (b) (c) (d)

One main winding A primary and a secondary winding One main winding with two coils A primary and double wound secondary

9. The autotransformer may be used as a

(a) (b) (c) (d)

Power transformer Current transformer Potential transformer Compensator motor starter

10. A transformer in which the secondary is part of the primary is

(a) (b) (c) (d)

A series parallel connection An autotransformer A double-wound transformer An isolating transformer

11. Parallel operation of single phase ­transformers can be accomplished when the

(a) Voltage and percentage impedance ratings are identical (b) Voltage and current ratings are equal (c) Cooling methods are identical (d) Primary and secondary voltage ratings are equal

12. Primary taps are designed to

(a) (b) (c) (d)

Raise the voltage of the secondary Drain the oil Lower the voltage of the secondary Raise or lower the voltage of the secondary

13. A slight voltage drop at the secondary t­erminals from no load to full load is called

(a) Reactance (b) Regulation (c) Transformation (d) Taps

14. A transformer has subtractive polarity when the

(a) Two primary coil voltages oppose each other (b) Two secondary coils have opposite polarities

15. A transformer has an additive polarity when

(a) Two primary coils are in series (b) Two secondary coils have aiding polarities

ANSWERS (MCQ) 1. b 2. d 3. a 4. b 5. b 6. b 7. c 8. c 9. a 10. b 11. a 12. d 13. b 14. b 15. b.

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. Derive an expression for the induced e.m.f. in a transformer. 2. How are the voltage and current transformation ratios in a transformer related?

3. How does the leakage reactance affect the equivalent circuit of a transformer? 4. Define regulation of a transformer. Deduce the expression for percentage regulation. Under what conditions is it maximum?

722  Electrical Technology 5. Discuss in detail the losses in a transformer. On what factors do they depend? Explain the method of measuring these losses 6. How will you perform open-circuit and short-circuit tests on a transformer? What is the ­significance of these tests? 7. Write short notes on: (a) Rating of a transformer (b) Cooling of transformers (b) Efficiency of a transformer (d) Impedance matching 8. Draw the equivalent circuit of a transformer referred to primary. 9. Explain the construction of a transformer with the help of a suitable sketch. 10. What is an autotransformer? How can it be used as (a) a step-up transformer (b) step-down transformer? What are its normal advantages and disadvantages over a conventional transformer? 11. Explain transformer action. 12. What advantageous use is made of transformers in long-distance electric power transmission? 13. What is meant by mutual flux? 14. Why is volts per turn a constant? 15. What is meant by reflected impedance? 16. Why is transformer efficiency normally at its highest at about one-half full load? 17. Why are transformers frequently built with two identical primary and two identical secondary windings? 18. In using multiple-winding transformers, why must only windings with identical voltages be paralleled with each other? 19. Why is the efficiency of an autotransformer normally higher than that of a standard transformer of similar voltage ratio and kVA rating? 20. How can a normal two-winding transformer be used as an autotransformer? 21. Why is an autotransformer potentially unsafe for large voltage ratios? 22. What is a potential transformer and how is it normally used? 23. What is a current transformer and how is it normally used? 24. A transformer has 120 primary turns and 720 secondary turns. If its load current is 0.833A, what is its primary current load component? ANSWERS (CQ) 24. 5.0 A  25. (a) 10 (b) 1830 turns (c) 183 turns  26. (a) 115 V (b) 13.22 W (c) 4.35 A (d) 52.9 W  27. 5.7 A 28. 0.919 A   29. 11 kVA  30. 38.33 to 1 

25. A 2300 to 230 V, 60 Hz, 2 kVA transformer is rated at 1.257 V/turn of its winding coils. Assuming an ideal transformer, calculate (a) step-down transformation ratio (b) total turns of the high-voltage coil (c) total turns of the low voltage coil. 26. A transformer has a turns ratio of k = 2. If its input voltage is 230 V and its output current is 8.70 A, what is (a) secondary voltage (b) load impedance (c) primary current (d) primary input impedance? 27. What is the current in the common winding portion of an autotransformer if its primary current is 22.3 A and its secondary current is 28.0 A? 28. An autotransformer is used as a step-up unit; its input voltage is 2.08 V while its output is 230 V. If its load is 2 kV, what is the current in the common winding portion? 29. A 1000 VA transformer is connected as an autotransformer to step 2530 V down to 2300 V. Its normal 230 V secondary is connected to its normal 2300 V primary. Under this situation, how much load in volt amperes can be handled? 30. A potential transformer is desired to allow safe reading on a 4600 V line. What voltage ratio transformer should be specified? 31. A current transformer is desired to handle 2000 A line from a high powered alternator. What nominal current ratio is desired? 3 2. Paralleling is proposed between a 15 kVA, 4600/230 V step-down transformer and a 10 kVA, 4600/208 V stepdown transformer. The reflected equivalent secondary impedance of the first is Z e2 = 0.0100 W, and that of the second is Ze2b, = 0.0122 W. Determine their circulating secondary current at no load. 33. Two mutually coupled coils are joined in a series-aiding arrangement if L1 = 4 H, L2 = 8 H, and the coupling factor is 0.6 (a) what is the total inductance (b) what is the total inductance if one of the coils is reversed? 34. The primary of an ideal transformer contains 500 turns, and the secondary contains 20,000 turns. If the supply voltage is 15 mV, 2 kHz and the load resistance is 8 W, calculate (a) turns ratio (b) secondary voltage (c) secondary current (d) primary current (e) secondary power (f) primary power (g) reflected resistance. 35. A transformer with coupling coefficient 0.9 has 10 turns in the primary and 400 turns in the secondary. If the supply voltage is 120 V and the supply current is 2 A, calculate (a) turns ratio (b) output power.

31. 2000–to–5 or 400–to–l  32. .001 A  33. (a) 18.8 H (b) 2 H 34. (a) 0.025 (b) 0.6 V (c) 75 mA (d) 3 A (e) 45 mW (f) 45 mW (g) 5 m W  35. (a) 0.025 (b) 2.16 W.

37

Three-Phase Transformers OBJECTIVES In this chapter you will learn about:

Primary phase 1

  Understand the unique advantages that polyphase circuits possess   Understand balanced polyphase voltages    Understand the difference between a bank of three identical single-phase tra­nsformers and polyphase transformers having semi windings on a connective   Learn to solve simple problems on the per unit system   Learn to raise or lower the voltage with a bank of single-phase transformers conn­ected in three-phase arrangements   Understand the techniques used in paralleling three-phase transformers   Learn to solve simple problems on the above topics

Primary phase 2

Primary phase 3

1

5

9

2

6

10

3 4

7 8

11 12

Secondary phase 2

Secondary phase 3

Secondary phase 1

Three-phase transformers

37.1 INTRODUCTION

(a)

(b)

The generation and transmission of electrical power is more efficient in poly-phase systems employing a combination of two, three or more sinusoidal voltages. In addition, poly-phase circuits and mechines posses some unique advantages, for example, power in a three-phase circuit is constant rather than pulsating as in a single-phase circuit. Also, three-phase motors start and run much better than single-phase motors. The most common form of poly-phase system employs three balanced voltages, equal in magnitude and differing in phase by 360°/3 = 120° (see Figure 37.1). To transfer a three-phase voltage a bank of three identical single phase transformers are required or single-phase transformer having six windings on a common core is used. This is shown in Figure 37.2.

37.2  THREE-PHASE CONNECTIONS OF SINGLE-PHASE TRANSFORMERS

(c)

Figure 37.1  Balanced Three-phase Voltages

Most a.c. power is generated and distributed as threephase. The voltage is raised or lowered with a bank of single-phase transformers connected in three-phase arrangements as shown in Figure 37.3. The current and voltage relationships between phase and line values for a wye connection as illustrated in Figure 37.3(a) are = V line

= 3 V phase and I line

I phase 

(37.1)

724  Electrical Technology

(i)

(ii)

(iii)

(a)

(b)

Figure 37.2 (a) Three Identical Single-phase Transformers (i, ii and iii) Showing Polarity Markings and Rated at 1370/230 V, 10 kVA each (b) Three-phase Core-type Transformers

Figure 37.3 Three-phase Connections of Single-phase Transformers

Three-Phase Transformers  725

The current and voltage relationships between phase and line values for a delta connection are = I = and Vline V phase  (37.2) 3 I phase line The delta-delta bank shown in Figure 37.3(b) and in Figure 37.4(a) has the advantage of being able to operate continuously with one of the three transformers disconnected from the circuit. This open-delta connection, also called a V-V connection, provides a connection means for inspection, maintenance, testing and replacing of transformers one at a time, with only a brief power interruption. The open-delta connection is also used to provide three-phase service in applications where a possible future increase in load is expected. This increase may be accommodated by adding the third transformer to the bank at a later date. Transformers selected for a delta-delta or open-delta connection must have the same turns ratio and the same per cent impedance in order to share the load equally. A phasor-diagram illustrating the current and voltage relationships for a delta-connected secondary is given in Figure 37.4(b). The phase currents, also called coil currents are Iaa, Ibb, and Icc. The three line currents, determined by applying kCL to the secondary junctions in Figure 37.4(a) are I1 = I aa ′ , + I b ′ b    

(37.3)

I 2 = I bb ′ , + I c ′c    

(37.4)

I 3 = I cc ′, + I a ′a    

(37.5)

Performing the indicated operations in Figure 37.4(b), the magnitudes of the three line currents, as determined by geometry, are shown to be equal to √3 or 1.73 times the phase currents. Disconnecting one transformer as shown in Figure 37.4(c) does not change the secondary line voltages, V1–2 and V2–3 are the same as before and V3–1=Vcc+Vbb as determined from phasor addition is     V3–1=Vc ′c+Vb ′b

(37.6)

Performing the indicated phasor additions in Figure 37.4(d) shows V3–1 to be the same whether connected delta-delta or open-delta. Since the three secondary line voltages are the same whether operating delta-delta or open-delta, and the load impedance has not changed, line currents I1, I2 and I3 must also be the same when operating delta-delta or open-delta. As evidenced in Figure 37.4(c), however, the coil currents in the two remaining transformers must increase to equal the line current that is Ib'b →I1 and Icc'→I3. Figure 37.4(d) shows the coil current Icc' increasing in magnitude and shifting its phase 30° to coincide with that of the line current I3, and the coil current Ib'b, increasing in magnitude and shifting its phase 30° to coincide with that of line current I1. Thus, if a delta-delta bank is operating at the rated load, and one transformer is removed, the current in the two remaining transformer coils will increase to 1.73 times its normal rating. To prevent overheating and possible roasting of the windings when operating open-delta, the bank current and, hence, the bank apparent power must be rerated to reflect the lower kVA capacity. Thus I ∆ − ∆ , rated = 0.577 × I ∆ − ∆, rated  IV −V , rated = (37.7) 3 Connecting the transformer bank open-delta did not change the three line voltages. Hence, the bank rating when connected open-delta is S ∆ − ∆, rated = 0.577 × S ∆ − ∆, rated  SV −V , rated = (37.8) 3 Example 37.1 Three 25 kVA, 480-120 V, single-phase, 60 Hz transformers are connected Δ–Δ. The total load on the bank is 50 kVA. A fault in one transformer requires its removal, and the bank is operating in open-delta. Determine the maximum allowable power that the open-delta bank can handle without overheating. Solution: The capacity of the delta-delta bank = 25 × 3 = 75 kVA The capacity when operating open-delta = 75 × 0.577 kVA = 43.3 kVA

726  Electrical Technology

C

C

A

Primary

Primary

B

B 1 a′ A

2

b′

b

Secondary c′ a

1

c

b

2

b′

Secondary C

c c′

C

3 I3

I1

I2

I1

I3

I2

3 Balanced 3-phase unity PF load

Balanced 3-phase unity PF load

(a)

(c)

Vcc′ = V2—3 Vcc′ = V2—3 Ib′b

Icc′

I3

I1 I3

30

30 Iaa′

30

Va′a = V3—1

Vaa′ = V3—1

30

30

Ic′c

Ibb′

Vbb′ = V1—2

I1

Icc′ Ib′b

30 Ia′a

Vb′b

Vbb′ = V1—2

I2 (b)

I2

Vc′c

(d)

Figure 37.4  (a) Delta-delta Bank (b) Phasor Diagram for (c)

Example 37.2 It is desired to use two transformers in open-delta to supply a balanced three-phase load that draws 50 kW at 120 V and 0.9 power factor lagging. The input voltage to the transformer bank is 450 V and 60 Hz. Determine the minimum power rating required for each transformers. Solution: P= 50, 000 =

3 Eline I line PF 3 × 120 × I line × 0.9

I line = 267.2918 A

Three-Phase Transformers  727

When operating open-delta, the transformer phase current equals the line current. Thus, the minimum apparent power rating of each transformer is. 120 × 267.2918 = 32.1 kVA 1000 Example 37.3 A three-phase transformer has 420 turns on the primary winding and 36 turns on the secondary winding. The supply voltage is 3700 V. Find the secondary line voltage on no load when the windings are connected as (1) star/delta (2) delta/star. Solution: 1. Primary phase voltage = 3700 /1.73 = 1908 V Secondary phase voltagge =

1908 × 36 = secondary line voltage = 163.54 V 420

2. Primary phase voltage = 3700 V 3300 × 36 = 283 V 420 Secondary line voltage = 283 × 1.732 = 490 V Secondary phase voltage

37.3  THREE-PHASE TRANSFORMERS Three-phase transformers have all three-phases wound on a single magnetic core, as shown in Figure 37.5(a) for shell-type construction and in Figure 37.5(b) for core-type construction. The core-type transformer is simple in construction, and limits third-harmonic fluxes and, hence, third-harmonic voltages have a relatively small value. Three-phase transformers use much less material than three single-phase transformers for the same three-phase power and voltage ratings. Hence, they weigh less and cost much less to produce. Furthermore, since all the three-phases are in one tank, the wye or delta connections can be made internally, reducing the number of external high-voltage connections from six to three. The principal disadvantage of a three-phase transformer compared with its three-transformer counterpart is that the failure of one phase puts the entire transformer out of service. A decision whether to use a three-phase transformer or three single-phase transformers, however, depends on many factors, including initial cost, cost of operation, cost of spares, cost of repairs, cost of down time, space requirement, and need for continued operation in case one phase is disabled.

(a)

(b)

Figure 37.5  Basic Construction of Three-phase Transformer: (a) Shell-type and (b) Core-type

728  Electrical Technology

37.4  PARALLELING THREE-PHASE TRANSFORMER BANKS There is an angular displacement called phase shift between the corresponding primary and secondary line voltages in the Y-Δ bank and with Δ-Y bank, as shown in Figure 37.6(b), with the low voltage lagging the high voltage by 30°. There is no angular displacement between corresponding primary and secondary line voltages in a Y-Y bank, Δ-Δ bank, or V-V bank. Because of the phase shift inherent in the Y-Δ and Δ-Y banks, they must not be paralleled with Y-Y, Δ-Δ, or V-V banks; to do so would cause large circulating currents and severe over-heating of the windings, even in no load conditions. Only banks with the same phase shift should be operated in parallel. The bank ratio (ratio of line voltages) for Y-Y, Δ-Δ V-V banks is equal to the respective turns ratios. This may be deduced from Figure [37.6 (a), (b) and (c)].

Figure 37.6 Paralleled Three-phase Transformer Banks Whose Corresponding Output Voltages are 30° Out of Phase (a) Circuit Diagram (b) Phasor Diagram Showing One Set of Corresponding Output Voltages (c) Simplified Circuit

37.5  HARMONIC SUPPRESSION IN THREE-PHASE CONNECTIONS The magnetizing current that produces a sinusoidal flux and, hence, a sinusoidal output voltage is itself non-sinusoidal, containing many odd harmonic components. Suppressing any one of the harmonic components will result in a nonsinusoidal flux and, hence, a non-sinusoidal secondary voltage. Figure 37.7(a) shows a wye-connected generator supplying a wye-wye transformer bank, with the neutral of the transformer bank connected to the neutral of the generator. The fundamental and third harmonic components of the magnetizing currents for phases A, B, and C are shown in Figure 37.7(b). The three phases are separated vertically for easier viewing and are placed on the fundamental time axis. The corresponding phasor diagrams are shown in Figure 37.7(c). The waves representing the respective fundamentals are 120° apart, but the corresponding third harmonics are in phase with each other. The third harmonic currents and their multiples, called triplen harmonics, have zero phase sequence and they are not three-phase quantities. The three-third harmonics (one from each transformer) are in phase with each other, the three-ninth

Three-Phase Transformers  729

harmonics (one from each transformer) are in phase with each other. Non-triplen harmonics, however, such as the second, fourth, fifth and seventh, are three-phase quantities and must be treated as such. Since the respective triplen harmonic currents of each phase of a wye-connected transformer bank are all in phase—all going in or all going out—they require a neutral line to the wye-connected source, as has been shown in Figure 37.7(a), only the third-harmonic currents are indicated. If the neutral is not connected, the third-harmonic currents will be suppressed, the flux will not be sinusoidal. The secondary output will have an appreciable third-harmonic voltage that may result in a resonance rise in voltage and overcurrent due to partial series resonance between the capacitive reactance of the lines and leakage reactance of the transformer at the third-harmonic frequency. For this reason, a wye-wye bank without a line connecting the neutral of the wye primary to the neutral of a wye source is not desirable for distribution systems.

(a)

(b)

(c)

Figure 37.7  Harmonic Suppression in Three-phase Connections: (a) Y-Y Bank with Neutral Connection to Primary (b) Waves of Fundamental and Third Harmonic (c) Phasors of Fundamental and Third Harmonic

730  Electrical Technology

S UM M A RY 1. The generation and transmission of electrical power is more efficient in a poly-phase system. 2. To transfer a poly-phase voltage a bank of three identical single-phase transformers or a single poly-phase transformer having six windings on a common core is required. 3. The per-unit system yields several advantages over other types of power calculations. 4. All per unit functions are written since they are ratios of the same units. 5. One of the advantages of the use of per unit quantities is that three-phase circuits are treated identically as d.c. or single-phase systems. 6. The open-delta (V-V) connection provides a convenient means for inspection, maintenance, testing and replacement of transformers, one at a time, with only a brief power interruption.

7. Transformers selected for a delta-delta or open-delta connection must have the same turns ratio and the same per unit impedance in order to share the load equally. 8. The core-type three-phase transformer is simpler in construction and limits third harmonic fluxes to a relatively very small value. 9. Three-phase transformers weigh less and cost much less to produce. 10. Failure of one phase puts the entire three-phase transformer out of service. 11. Only banks with the same phase shift should be operated in parallel. 12. A wye-wye bank without a line connecting the neutral of the wye-primary to the neutral of a wye source is not desirable for distribution systems.

M U LT IP LE C H O I C E Q UE S TI O NS ( M C Q ) 1. The voltage transformation ratio of a transformer depends on



2. The rating of a transformer is expressed in





(c) Transformer 1, wye/delta, transformer 2, delta/wye (d) none-of the above

(a) ES /EP (c) NS /NP

(b) EP /ES (d) NP /NS

8. With peaked e.m.f. with transformer windings the hysteresis loss is

(a) W (c) kW

(b) kVA (d) None of the above

9. The oil used in transformer is

3. The power factor of a transformer is

(a) Reduced (c) Constant

(b) Increased (d) None of the above

(a) Natural oil (b) Vegetable oil (c) Animal oil (d) None of the above (a) Always leading (b) Very close to unity and leading (c) Very close to unity and lagging 10. Oil used in transformers helps in (d) Dictated by the power factor of the load (a) Dissipating heat (b) Improving dielectric strength of the insulation 4. For their protection, transformers are normally equipped (c) Neither (a) nor (b) (d) Both (a) and (b) with (a) Breather (b) Conservator 11. The magnetizing current in a transformer is (c) Bocholt’s relay (d) All of the above (a) Third harmonic (b) Fifth harmonic 5. Wye-delta connection for three-phase transformers is (c) Seventh harmonic very common for (d) Ninth harmonic (a) Low voltage, large rating transformers 12. Two transformers of identical voltage but of (b) Power supply transformers different capacities are operating in parallel. For (c) Low voltage, large rating transformers (d) Distribution transformers satisfactory loads having (a) Impedance must be equal 6. The most suitable connection for three-phase distribu- (b) Per-unit impedance must be equal tion transformers is (c) Per-unit impedance and (X/R) ratio must be equal (a) delta/wye (b) delta/delta (d) Impedance and (X/R) ratio most be equal (c) wye/delta (d) wye/wye

7. Which of the following three phase transformers cannot be operated in parallel?

(a) Transformer 1, wye/wye; transformers 2, delta/delta (b) Transformer 1, wye/delta, transformer 2, delta/delta

ANSWERS (MCQ) 1. (a)  2. (b)  3. (d)  4. (d)  5. (b)  6. (a)  7. (b)  8. (a)

9. (a)  10. (d)  11. (d)  12. (c).

Three-Phase Transformers  731

CON V E N TI O NA L Q UE S TI O NS ( C Q ) 1. When paralleling transformers, why must turns ratios be identical, or very nearly so? 2. Why must the correct polarities be strictly observed while-paralleling transformers. 3. What feature will allow transformers with different kilovolt-amperes ratings but the same turns ratios to be paralleled? 4. What features are necessary for matching parallel transformers? 5. Name some features that must be identical, or nearly so in a three-phase transformer bank. 6. What might determine whether three separate transformers or three sets of windings in a single threephase transformer might be chosen for a three-phase application? 7. What type of three-phase transformer connection is necessary in order to have a neutral connection available? 8. Why is the neutral point of a wye transformer connection usually grounded? 9. Name two advantages of a Δ-Δ transformer connection. 10. What is the principle type of use for a Y-Δ transformer connection? 11. What other type of the three-phase connection may be paralleled with a Y-Δ connection? 12. Name two advantages of the V-V or open delta threephase connection?

ANSWERS (CQ) 13. 991 A 14. I2a=17.76 A, I2b=12.24 A

13. Paralleling is proposed between a 15 kVA, 4,600/230 V step-down transformer and a 10 kVA, 4,600/208 V step-down transformer. The reflected equivalent secondary impedance of the first is Ze2 or = 0.0100 Ω, and that of the second is Ze2b = 0.0122 Ω. Determine their circulating secondary current at no load. 14. A 4 kVA, 2300/20 BV transformer with a secondary impedance Ze2a = 0.0310 Ω is to be paralleled with a 3 kVA, 2300/208 V transformer with Ze2b = 0.0450 Ω. When a combined load of 6.25 kVA is carried, calculate the individual load currents. 15. In an open-delta transformer bank the load power factor is 0.803, what are the individual transformer power factors? 16. Industrial plant draws 1004 at 0.7 PF lagging from the secondary of a 2300/230 V, 60 kVA, Y-Δ distribution transformer bank. Calculate (a) Power connected by the plant in kW and apparent power in kVA (b) Rated secondary phase and line currents of the transformer bank (c) Percent load on each transformer (d) Primary phase and line current drawn by each transformer (e) The kVA rating of each transformer

1 5. PF1=0.993 PF2=0.397 16. (a) 27.9 kW, 39.8 kVA (b) 87 A, 150.6 A, (c) 66.4 per cent, (d) 104, (e) 20 kVA.

Synchronous Generators — Alternators

38

OBJECTIVES In this chapter you will learn about:   The difference between d.c. generators and alternators    The physical construction of a.c. machines   Slot structure in a.c. dynamos   Coil and pole group connections   Stator windings   Pitch factor and distribution factor   Alternator performance at various load power factors   Voltage regulation of an alternator at various power factors   Winding resistance and synchronous impedance   Equivalent circuit of a synchronous alternator   Basic voltage generation formula   Requirements for parallel operation of alternators   How to meet these requirements   Different methods of synchronization

Slot structure in alternators

38.1 INTRODUCTION Alternating current dynamos are visually different from d.c. machines of comparable sizes. The difference in appearance has a fundamental basis because the machines are inside out in relation to d.c. machines. Most a.c. machines have the armature in the field or stator position and the field in the moving or rotor position. This arrangement is the natural order of things for a few very good reasons (see Figure 38.1). 1. The high voltage, high current and, therefore, high powerhandling element is the armature on any a.c. or d.c. rotating electrical machine. Armatures coils are, therefore, larger than field coils. 2. Since no alternate switching of coil polarities is needed on an a.c. machine, no commutator function is needed. Thus, the high-power windings may be made stationary for direct connection. The universal motor is an exception to this condition. 3. The field structure and coils are not ordinarily required to handle more than a fraction of the total power. Thus, their

Figure 38.1  A Cylindrical Rotor

Synchronous Generators — Alternators  733

rotating electrical connection may be made smaller. Since no polarity switching is required, collect rings are usually used. 4. The armature and field coils are both placed in slots in the punched magnetic structure, but the stationary armature structures can be conveniently made with deeper slots to handle the required larger coils. 5. It is easier to cool the stator than the rotor, which is an advantage of the normal a.c. construction.

38.2  PHYSICAL CONSTRUCTION OF D.C. MACHINES Except for specialized types, such as the universal a.c. - d.c. motor, which appears much like a d.c. series motor, almost all a.c. motors and generators are built to take advantage of the natural relation of having the armature fixed and surrounding the field and the field moving and inside of the armature.

38.2.1  Fixed Armature or Stator The fixed and outside armature has a complete ring of teeth and slots on its inner face. In the usual machine, all the slots are filled with similar and symmetrical coils. As a result, it may not be at all obvious how many poles or phases are present in the winding. In the field rotor, the constructions may be much like that of a d.c. armature with a complete circular magnetic structure having a continuous group of slots and teeth on the outer surface. Again, these slots are filled with similar and symmetrical coils and it is not readily apparent how many poles or phases are present in the machine. With salient pole field construction, the number of poles is visible, as in a d.c. machine. A salient pole machine is illustrated in Figure 38.2. If the winding slot sides are parallel to each other in a single slot, which is a frequent construction, it may be seen from Figure 38.3 that Figure 38.2  A Salient Pole Machine the stator tooth structure becomes stronger as it grows deeper. On the other hand, Figure 38.3 shows that a rotor tooth becomes weaker as it grows deeper. This tooth structure advantage for the stator is used in the a.c. stator.

Figure 38.3  Typical Magnetic Lamination Slot Structure in an a.c. Dynamo

734  Electrical Technology Parallel side slots are not necessary in small sizes. However, on large sizes, where the coils are wound with large crosssectional wire and where the insulation must be most carefully distributed, the parallel side slot is required. Since large coils are preformed, bound with insulation and impregnated with varnish and baked, they cannot readily change shape after installation in the magnetic coil. Smaller a.c. machines are wound with loose coils of round wire, which may be slipped down into the slot turn by turn during winding or installation. In this fashion, almost any slot shape may be used. Full use of the slot cross-section secures to require parallel side slots in larger sizes. In any shaped slot, some provision must be made to capture and hold the windings in place. As a result, the slot will have some provision for a covering wedge, even if parallel sided. In the a.c. machine stator, the current is continually varying at the frequency repetition rate. The resulting magnetic flux then varies cyclically, and there are hysteresis and eddy current losses in the magnetic structure. Minimizing these losses requires the use of a laminated magnetic structure. The structure is built up of thin plates of silicon steel alloy that are readily punched to shape in press dies built for the task. The punched stator laminations usually cover full circle in small- and-medium sized machines. Since punch dies are expensive, only a few different numbers of slots and teeth are provided for a basic size. The larger-sized machines are built up with laminations in segments of reasonable sizes. The size depends on the available stock width and press die size. Lamination stock thickness is dictated by eddy current loss considerations. Thin lamination has less eddy current loss but becomes difficult to handle and the teeth will bend too easily. A stock thickness of about 0.35 mm has long been used for 60 Hz a.c. machines. The numbers of slots are standardized around 36, 48, 60 and 72 slots, and so on, for good mechanical reasons. These will become more apparent when actual windings are discussed.

38.2.2  Rotating Field Structure The mechanical construction of the rest of an a.c. motor or generator follows closely that for d.c. machines except for the lack of a commutator with synchronous alternators and synchronous motors, slip rings—which are used to carry d.c. power into and out of rotating field—are used in a similar location to a commutator. A slip ring as illustrated in Figure 38.4 is a copper alloy ring that is insulated from the rotor shaft and connected to the rotor windings. A carbon brush is supported in a brush carrier rigging to complete the connection. Since there is no requirement for a particular internal resistance to aid commutation, the slip ring brush is harder and denser than a commutator brush. It has a lower voltage drop and is therefore responsible for less power loss than its d.c. counterpart.

(a)

(b)

Figure 38.4 In an a.c. Generator the Rotating Conductor is Connected to the Load Through the Slip Rings and Brushes (a) Electrical Parts of an a.c. Generator (b) Simplified Sketch of the Conductor Loop When the rotor windings carry three-phase a.c. power, three slip rings are used. In some cases regarding larger synchronous motors, multiple windings are used and five or mover slip rings may be present. The high-power armature windings are placed on the stator structure, which has relatively larger winding space. Overall, an a.c. machine can usually be smaller than its d.c. counterpart with the same power rating. The lack of a commutator also contributes to size reduction. An a.c. machine averages about 50 per cent more power within the same frame size.

Synchronous Generators — Alternators  735

38.3  ALTERNATOR WINDINGS The types of windings used in a.c. machinery are closely related to d.c. windings. Both lap and wave windings are used, but lap winding is much more common owing to shorter coil connections. In single phase machines, a winding form called concentric coils is used not because of any circuit advantage, but because it lends itself to prepared coil structure that can be put in place rapidly. Here the economics of construction dominate. Since three-phase machines require three identical groups of windings spaced 120 electrical degrees apart, and since poles must exist in pairs, certain rules affect windings and magnetic structure slot spacing. Electrical degrees refer to the cycle angle of the repeating sine wave, where one full cycle is 360 electrical degrees. Since opposite magnetic poles produce opposite voltages in a moving coil to pole situation, the maximum voltage difference along a sine wave is found 180 electrical degrees apart. This 180 electrical degrees shift takes place between each successive field pole. The total electrical degrees in a 360 mechanical degree (Figure 38.5) rotation is then simply 180 times the number of poles or the total electrical degrees in one revolution = 180 P (38.1) where, P is the number of poles. For example, take a three-phase four-pole machine. Let us assume that 36 slots are available in the stator lamination stock. There are 36/4 = 9 slots per pole position and 9/3 = 3 slots per phase per pole. These 36 slots would allow twice as many or six, per phase per pole for a two-pole machine, and similarly, two slots per pole per phase for a six-pole machine.

(a)

(b)

Figure 38.5 Electrical and Mechanical Degrees (a) Four-pole Magnetic Field (b) 720 Electrical Degrees Per Revolution

38.3.1  Chording of Windings

Figure 38.6 Four-pole a.c. Stator with Full Pitch Coils

One design factor is the chording of the pole windings. If, on the 36 slot, four-pole machine an individual coil enters slot 1 and comes back from slot 9 (see Figure 38.6); it will have spanned 90 mechanical degrees of the stator circular structure. Since there are four poles by definition, in this case, 90 mechanical degrees is 180 electrical degrees (Eq. 38.1). Thus, the two sides of the coil are in the same relative position on the adjacent north and south pole positions. This is a full-pitch coil construction (see Figure 38.6). The more usual a.c. machine coil will cover less of the periphery of the machine and is then said to be fractional pitch. A typical coil situation might have a coil enter slot 1 and leave slot 7. This then covers six out of a possible nine slot pitches and is a 6/9 or 66.7 per cent pitch. The majority of a.c. machine coils are of fractional pitch type, for which there are a few important advantages.

736  Electrical Technology 1. The ends of the coil are shorter, which means less copper loss due to less total length. 2. The ends of the coil can be formed more compactly. The end bells will need less winding space, resulting in a shorter unit. 3. There is a distinct reduction in machine harmonics due to cancellation of higher harmonics. Since all a.c. equipment is designed to operate on a pure sine wave, the generation of harmonics is to be avoided.

38.3.2  Coil Group Connections Figure 38.7 illustrates the manner in which the coils are laid into the slots. A vast majority of lap or wave-wound machines use this double-layer winding arrangement. This is very similar to the manner of winding a d.c. armature. This interconnection of the coils will result, in this 36-coil situation, in 12 groups of the coils per group. Each group is then involved with one phase and one pole. Since there are four poles in this simple but real situation, there are four coil groups in each phase. This is the usual situation, even when more slots or coils are used.

Figure 38.7  Double-layer Coils in a.c. Stator A 72-slot, six-pole machine, when wound for three phase, would have 72/6 = 12 slots per pole and 12/3 = 4 slots per phase per pole. Here the coils would be connected in groups of four, and there would be six of these four-coil groups per phase. Many varieties of coil group connections are possible, but only a relatively few are used today. In a three-phase machine, the coil groups per phase are connected for all the poles, and this larger grouping is usually divided into two parts. On the 36-slot machine, 2 three-coil groups are permanently connected. There are then two of these six coil connections per phase. If they are series connected, then the motor or generator will be set for operation on the higher of its two rated voltages. If parallel connected, the lower of the two rated voltages may be accommodated. In this way, a motor or generator may be operated on either 110 V or 220 V or perhaps 220 V or 440 V and so on. Great installation flexibility is inherently obtained. As these coil groups are gathered together, the direction of winding or connection on opposite poles must be opposite. Thus, each adjacent series connected per phase per pole group must be reversed for proper polarity, as shown in Figure 38.8 for typical coil interconnection. A three-phase machine, when gathered into phase coil groups, is then connected either in wye or in delta, and also in series or parallel as shown for wye connections in Figure 38.8. Three-phase motor stator coil group ends are normally numbered from 1 to 9 as shown, and the points 10,11,12, are normally buried unless specially needed.

38.3.3  Winding Distribution Since coils are usually laid in the manner shown in Figure 38.6, they are seen to be spaced uniformly around the periphery of the machine stator. Returnning to the 36-coil, four-pole situation, it can be seen that in this specific case any one-pole has three phase groups of three series connected coils per pole. The voltages generated in coils of single-phase groups of three coils are not simply additive. Since each coil is not swept or cut by the same intensity of magnetic flux at the same time, they are not in the same time phase relation even though they are a part of the same phase winding. The individual coil voltages must be combined as phasors. All these factors, which make up a multi phase, two level, chorded or whole-pitch distributed winding, are applicable for a variety of a.c. generators and motors, both large and small.

Synchronous Generators — Alternators  737

Figure 38.8  Typical Coil and Pole-group Connections

38.4  SYNCHRONOUS ALTERNATOR The synchronous alternator is the basic a.c. generator. It is called synchronous because its generated frequency is directly related to its number of armature and field poles and to its rotative speed. An individual coil of winding generates a full cycle of a.c. voltage each time it is swept by a pair of magnetic poles. The generated frequency is converted from cycles per pole pair to a machine basis by the following relation. f =

PS 120 

where, f = frequency in hertz P = number of poles    S = speed in rpm This relationship is of fundamental importance and can be easily followed if constructed from its basics.

(38.2)

738  Electrical Technology cycle 1. One full cycle of alternating current is developed for each pair of magnetic poles swept by a winding: 2 poles poles 2. There are a fixed number of poles in a full circle of construction or one revolution: rev Note: There must be an even-integer number of poles, as in a d.c. machine. 3. The rotative speed is measured in revolutions per minute: rev min 4. There is 1 minute for each 60 seconds: min 60sec Gathering and cancelling, we have cycle poles rev min PS cycle × → f = × ×P ×S (38.3) 2 poles rev 120 sec min 60 sec    P ω cycles (38.4) f = × 4π sec   Note: cycle/sec = Hertz If the rotative speed is given in radians per second or w, then ω rad (38.5) S rpm = sec  There are only a few recognized and used a.c. power frequencies. These are 25, 50, 60 and 400 Hz, with 50 Hz and 60 Hz by far the most common. 400 Hz is used almost exclusively for aircraft a.c. power because it allows small high speed machines which also require less magnetic structure size and weight. Any prime mover inherent rotative speed can be matched by a combination of pole and frequency combination. Example 38.1 A large hydroelectric power plant is under consideration. Its hydraulic head or water level difference above and below the dam and its power requirement dictate its water turbine or runner must turn at from 137.00 rpm (14.387 rad/sec) to 140.00 rpm (14.661 rad/sec) to reach peak efficiency. Power required is 60 Hz. 1. How many poles must a direct-connected alternator have? 2. What rotational speed must be used? Solution: 1. P =

or

PS 4 π f = ω 120

P=

4π 60 = 52.5 poles 14.347

P=

4π 60 = 51.4 poles 14.661

Poles can only exist in even integer numbers. So we must have P  =  52 poles 2. Using 52 poles, we have Checking with rpm

f =

PS 120 f 120(60) = or S = = 138.46 rpm 120 P 52 rpm = 138.46 × 0.10472 = 14.500 rad/sec.

Synchronous Generators — Alternators  739

38.5  STATOR WINDINGS In a synchronous machine, the stator winding is the armature winding in which the operating e.m.f. is induced. Two types of windings namely: single-layer winding and double layer winding are considered.

38.5.1.  Single-layer Winding The main difficulty with single-layer windings is to arrange the end connections so that they do not obstruct one another. Figure 38.9 illustrates one of the most common methods of arranging these end connections for a four-pole, three-phase synchronous machine having two slots per pole per phase, i.e., six slots per pole or a total of 24 slots.

Figure 38.9  End Connections of a Three-phase Single-layer Winding In Figure 38.9 all the end connections are shown bent outward for clearness. In actual practice, the end connections are usually shaped in the manner shown in Figure 38.10 and in section in Figure 38.11. This method has the advantage that it requires only two shapes of end connections, namely, those marked (in Figure 38.11 which

Figure 38.10 End Connections of a Three-phase Single-layer Winding

Figure 38.11  Sectional View of End Connections

740  Electrical Technology are brought straight out of the slots and bent so as to lie on a cylindrical plane, and those marked D. The later, after being brought out of the slots, are bent roughly at right angles, before being bent again to form an arch alongside the core. The connections of the various coils are more easily indicated by means of the developed diagram of Figure 38.12.

Figure 38.12  Three-phase Single Layer Winding The solid lines represent the Red phase, the dot/dash lines the Y Vp When the synchronous motor is deliberately adjusted this way, it draws a leading Ia current from the line by contributing an internal Ia sin θ reactive current component. At comparable leading and lagging power factors, the Ia stator current is the same. The benefit is that the created Ia sin θ component is opposite in phase to whatever Ia sin θ components may exist with the rest of the installation that the synchronous motor serves. Substantial system power factor improvement is achieved commercially by the over-excited large synchronous motors while they work. When a synchronous motor is used solely to produce a large leading Ia sin θ current component, it is called a synchronous capacitor. This name comes from effect, which is the same as if a giant capacitor were placed across the line. When the motor carries a normal mechanical load, it is also normally used as a synchronous power factor corrector when it is deliberately over excited.

39.7  SYNCHRONOUS MOTOR V CURVE The power factor response of a synchronous motor under various d.c. field excitation currents while holding a constant power is shown by the V-curve test. The curves are so named because of their distinctive shape when plotted. A synchronous motor that is to be tested is connected to instrumentation and a variable load, as shown in Figure 39.7. The two Wattmeter method is shown, but any of the acceptable Wattmeter circuits are applicable, bearing in mind that the three-phase loads are balanced.

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Figure 39.7  Circuit Connection for V Curves of Synchronous Motor Three voltmeters are shown but only one reading is needed if the balance is shown. The motor is tested by applying a load and varying the d.c. field excitation in logical steps. Data for all meters are recorded at each step so that volt-amperes and power can be determined. This allows the power factor of the motor to be determined for each field current setting for each load. Figure 39.8 shows a typical family of V curves. The curves may be taken over more closely spaced increments to truly define the curve shapes, depending on the ease of load control and the period of time available. In Figure 39.8 (a) the no-load curve drops to a minimum but not zero. The minimum current can be related to a minimum power necessary to overcome the fixed internal losses, such as the rotational losses that are always present. The shape of the curves clearly shows that for each load, there is a distinct minimum armature phase current Ia at a specific d.c. field current If. This specific field current is known as normal excitation. Unless otherwise specified, the labelled d.c. field current for a synchronous motor will be the current that produces minimum armature winding current in the region of 80 per cent to full load. Figure 39.8 (b) shows the same data plotted as load power factor versus d.c. field current. These curves show that a synchronous motor can be over excited and carry a substantial leading power factor. This process is limited by the maximum current rating of the stator windings. Even though the increasing d.c. field current brings higher and higher leading power factors, the main stator current is increasing at the same time. The current handling capability of the motor is fairly well taxed for full-load currents at 100 per cent PF. If it is desired to carry a strong leading PF for load power factor improvement of a factory, and at the same time, to power the factory air compressors, conveyor systems, etc., the synchronous may well need to be a larger size. This is because a synchronous motor may be rated at unity PF or pole at 80 per cent PF at a given load. The name plate usually states the load and power factor conditions. If a more leading PF is desired, it can be met by a motor of one or more frame sizes larger than the basic power requirement ordinarily needed. The V-curve intersection with the normal excitation line in Figure 39.8 (a) illustrates the phasor relation in Figure 39.6 (a). The V-curve intersection with 0.8 PF lagging dotted line in Figure 39.8 (a) illustrates a phasor relation like that shown in Figure 39.6 (b). The intersection of a load curve with 0.8 PF leading line in Figure 39.8 (a) is illustrated as a phasor relation in Figure 39.6 (c). Example 39.1 A factory has an average total electrical load of 41300 kW at 0.810 PF lagging. Part of the load is incurred by a large three-phase induction of 5073 kW, which operates at 0.730 PF lagging and at 92 per cent efficiency. The motor is in need of rewinding and requires extensive mechanical rebuilding so that replacement is scheduled. Two different synchronous motors are investigated, one to carry the same 6800 hp (5073 kW) load at unity PF and the same efficiency. The other one

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770  Electrical Technology

50

pe

rc en

tl

oa

d

is a larger frame unit and carries the same load at the same efficiency and at 0.780 PF leading. Calculate the following: 1. Overall system power factor using unity PF motor. 2. Overall system power factor using the 0.780 leading PF motor. 3. The difference in required kilovolt-ampere rating of the two motors. Solution: 1. The original system kilowatt power will be expected to remain through these alternatives since the motor load and efficiency remain the same. In that case the kilovolt-amperes and kilovolt-amperes reactive are

41300 kW = 50988 kVA originally 0.810 PF cosθ = 0.810, θ = 39.9º, sinθ = 0.586 (50988) (0.586) =29 900 kVAr originally The original motor kVA and kVAr are

6800 hp(0.746 kW ) = 5514 kW motor input 0.92 eff (hp) cos θ = 0.730 (motor), θ = 43.1°, sin θ = 0.683

5

t

en

rc

e 0p

5514 = 7553 kVA motor input 0.730 50

per

cen

t

Figure 39.8 Synchronous Motor V Curves (a) Armature Current vs Field Current (b) Power Factor vs Field Current

(7953) (0.683) = 5159 kVAr motor The original and still factor load less than motor is 41300 kW – 5514 kW = 39790 kW The kVAr component is 2990 kVAr = 5159 kVAr = 24740 kVAr The unity PF motor will then create a total factory kW the same as the original or 41300 kW, but no more kVAr than the factory without the motor or 24740 kVAr. The total factory power factor will then be arc tan

24740 = 0.599 41300

which corresponds to θ = 30.92º; thus cos θ = 0.858 or 85.8 per cent PF

2. A synchronous motor with the same horse power (or kW) and a 0.780 PF leading will have the same kW as the other motors, but will contribute a leading kVAr component

5514 kW = 7069 kVA 0.780 PF cos θ = 0.780, θ = 38.74°, sin θ = 0.626 (7069) (0.626) = 4424 kVAr leading The total plant kW is the same 41300 kW, but the kVAr is less: 24740 kVAr – 4424 kVAr = 20316 kVAr

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Synchronous Motors  771

The total plant power factor is then arc tan

20316 = θ = 26.19° 41300

cos 26.19° = 9.897 or 89.7 per cent PF 3. Step (1) requires a kVA rating the same as its kW rating owing to unity PF: 5514 kVA. The step (2) situation requires 7069 kVA; 7069 – 5514 = 1555 kVA The motor that produces the improvement in plant PF from 85.8 per cent to 89.7 per cent requires an additional kVA rating of over 1500 kVA. This means a substantially large size is needed to decrease the added heat, since 100-92 or 8 per cent of this difference is lost heat in the stator windings.

39.8  SYNCHRONOUS CAPACITORS A number of specialized synchronous motors are deliberately manufactured without any shaft extensions at all. They are intended to be used solely for power factory correction. Also, they are incapable of driving a mechanical load. Any over-excited synchronous motor that is not used to drive a load may be classed as a synchronous capacitor. Although there is no mechanical load on a synchronous capacitor to contribute to the armature current, the Figure 39.9  Synchronous Capacitor Phase Relationships V-curves of Figure 39.8 show that when over-excited, even at no load, the stator armature current is high. This is not a disadvantage, considering that the synchronous capacitor armature current may be raised to its rated value at an extremely low leading PF for use in power factor correction. As shown in Figure 39.9 when a synchronous motor is over-excited without load, the resultant phase impedance Er is quite high, despite the very small torque angle a, producing a relatively large leading armature current Ia, which is practically at 90° with respect to the bus phase voltage. Synchronous capacitors are preferred for power factor correction over commercial capacitors. The former can be constructed much less expensively in extremely high kVA (and even MVA) ratings as well as in high voltages (100 kV to 800 kV), as compared to fixed commercial capacitor of the same voltage and kVA rating. Such synchronous capacitors, driving no mechanical load whatsoever, merely float on the transmission lines of a power system for purposes of power factor improvement.

39.8.1 Power Factor Correction Advantages Loads having moderate to low lagging or leading power factors (below 0.65) result in a severe loss of electrical power to the utility supplying power to a given industrial or commercial occupancy. Lower power factors require the utility to increase their capacity or apparent power in order to supply a higher current for the lower power factor loads. This added capacity (and higher current) is needed all along the line, from the generating station, through the transformers and the transmission lines, to the load. The cost of this additional added capacity is kept to a maximum by means power factor correction. Many other advantages also emerge from power factor correction. These are: 1. Since the power capacity and line current are both lower, the power losses (I 2R) in the lines are reduced. 2. Similarly, the line volt drop across the line impedance is reduced, making the voltage regulation task easier in maintaining rated voltage to occupancies supplied by the utility. 3. Transmission efficiency from source to load is increased. 4. Utility costs are decreased, reflecting a savings (theoretically) to the consumer. Almost all commercial, industrial, and residential loads tend to have lagging power factors (i.e., current lags voltage) due to inductive reactive loads (motors, fluorescent lights, etc.). Consequently, power factor correction consists of adding capacitive loads in parallel with existing inductive loads to raise the power factor. It is customary not to attempt any correction of the power factor of a system all the way to unity power factor. The economic reason placing a limit on maximum power factor correction can be inferred from the data in Table 39.1 for a 10000 kVA system. A 10000 kVA system operating at 0.6 PF is capable of delivering only 6000 kW; whereas, at unity PF

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772  Electrical Technology it could deliver 10000 kW at the same current and at the same line drop. Any increase in output, however, is at the expense of reactive kilovolt-ampere. In improving the PF from 0.65 to 0.70, for example, there is an increase in output of 500 kW at a correction cost of 460 kVAr. In improving the PF from 0.80 to 0.85, the increase in 500 kW is made at a higher correction cost of 730 kVAr. At each successively higher power factor level, the kVAr cost is greater for a further improvement of 0.05 in the power factor. In improving the power factor from 0.95 to unity, the 500 kW increase in output entails a correction cost of 3120 kVAr. Thus, it is economically prohibitive generally to range the power factor beyond 0.85 lagging. Table 39.1  Total Reactive Kilovolt-amperes of Correction Required at Various Power Factors System PF

0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

Output in kW

Kilovolts Available

Kilovolts to Correct From Next Lower PF

Cumulative Total Kilovolts Required in Correction

6000 6500 7000 7500 8000 8500 9000 9500 10 000

8000 7600 7140 6610 6000 5270 4360 3120 0

— 400 460 530 610 730 910 1240 3210

— 400 860 1390 2000 2730 3640 4880 8000

Example 39.2 A 3-phase synchronous motor of 8000 Watt 1100 V has synchronous reactance of 8 Ω per phase. Find the minimum current and the corresponding induced e.m.f. for full load condition. The efficiency of the machine is 0.8. Neglect armature resistance. Solution: The current in the motor is minimum when the power factor is unity, That is, when cos θ = 1. Motor input = (motor output) /efficiency Pi= 8000/0.8=10000 W=10 kW Pi = 3 VL I L cos φ IL =

Pi 3VL cos θ

=

10 × 103 3 × 1100 × 1

= 5.249 A

2

 1100  2 2 V 2 + ( Ia X s ) =   + ( 5.249 ×8 )  3  per phase f = 636 E =E 636.49 V.49 perVphase

For unity factor

f

Example 39.3 A 3-phase, 400 V synchronous motor takes 52.5 A at a power factor 0.8 leading. Determine the induced e.m.f. and the power supplied. The motor impedance per phase is (0.25+j 3.2) Ω. Solution: For leading power factor E 2f = (V cos φ − I a Ra ) 2 + V (sin φ + I a X s ) 2 2

  400   400 = × 0.8 − 52.5 × 0.25  +  × 0.6 + 52.5 × 3.2    3   3 = (171.6 ) + ( 306 + 0.57 ) 2

2

2

E f = 391.3 V

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Line e.m.f. = 3 × 351.3 = 605.8 V Power supplied = 3 VL I a cos φ = 3 × 400 × 52.5 × 0.8 = 29098 W Example 39.4 Find the three highest speeds at which synchronous motor generator sets could run to link up with a 25 Hz and a 60 Hz system. Solution: f = (RPS) × pole pair f 60 f RPM = × 60 = pole pair pole pair

=

60 × 25 for 25 Hz and pole pair

60 × 60 for 60 Hz pole pair

Equating the two, we have P25 25 25 60 1 = = or = P25 P60 P65 60 2.4 That is, the pole pairs of 60 Hz = 2.4 pole pairs of 60 Hz; both must be integers. So, the first three values of P25 will be 5, 60 × 25 60 × 25 60 × 25 10 and 15 and the corresponding RPM will be = 300 RPM, = 150 RPM and = 100 RPM. 5 10 15

S UM M A RY 1. A synchronous motor is a machine that converts a.c. electric power, to mechanical power at a constant speed called synchronous speed. 2. A synchronous motor is a doubly excited machine. 3. The synchronous motor is an alternator (a.c. generator) used as a motor. 4. Synchronous motors are constant speed machines. 5. Synchronous machines designed specifically for power factor correction have no external shafts and are called synchronous condensers. 6. The float on the bus supply reactive power to the system. 7. The direction of the reactive power is adjusted by changing the field excitation of the machine. 8. When discussing individual motors, it is assumed that the machine is connected to an infinite bus. 9. The exciter (a d.c. shunt generator) is placed on the same shaft as the motor and a small portion of the motor torque is required to generate the d.c. required for its field excitation.

1 0. A synchronous motor is not inherently self starting. 11. The stator has a single phase or poly-phase winding that is identical to that of the alternator. 12. The rotor is generally a salient pole rotor, except in types of exceedingly high speed. 13. In order to eliminate hunting and develop the necessary starting torque when an a.c. voltage is applied to the stator, the rotor poles contain pole-face conductors that are short circuited at their ends. 14. The synchronous motor will not start by itself without a damper winding. 15. The name plate usually states the load and power factor condition. 16. Any over-excited synchronous motor that is not used to drive a load may be classed as a synchronous capacitor. 17. Synchronous capacitors are preferred for power factor correction over commercial capacitors. 18. Power factor correction consists of adding capacitive loads in parallel with existing inductive loads to raise the power factor.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. An alternator operates on the principle of

(a) Electromagnetic induction (b) Self induction (c) Mutual induction (d) (c) or (b)

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2. The starter core of a synchronous machine is built up of

(a) (b) (c) (d)

Stainless steel laminations Silicon steel laminations Cast iron laminations Cast steel laminations

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774  Electrical Technology 3. In a salient pole field structure, the pole shoes cover about

(a) (b) (c) (d)

One-third of pole pitch One-half of pole pitch Two-third of pole pitch Whole of the pole pitch



4. For a two-layer winding, the number of stator slots is equal to the number of

(a) Poles (c) Coil sides

(b) Conductors (d) Coils

5. The rating of a universal machine is usually governed by the

(a) Speed (c) Weight

(b) Temperature rise (d) None of these

6. The synchronous reactance of the machine is the

(a) Reaction due to armature reaction of the machine (b) Reactance due to leakage flux (c) Combined reactance due leakage flux and armature reaction (d) Reactance due to armature reaction or leakage flux

7. A poly-phase field is

(a) Pulsating and stationary (b) Pulsating and rotating (c) Constant amplitude and rotating at synchronous speed (d) Constant in amplitude and stationary in space

8. When does a synchronous motor operate with leading power factor current?

(a) (b) (c) (d)

While it is under excited While it is critically excited While it is over excited While it is heavily loaded

9. A salient pole synchronous motor is running on no-load. If it’s excitation is cut off, it will

(a) Continue running at synchronous speed (b) Continue running at a speed slightly less than synchronous speed (c) Stop (d) None of these

10. The speed of a synchronous motor can be varied by varying its

(a) Excitation (c) Supply frequency

(b) Supply voltage (d) Load

11. If the field of a synchronous motor is under excited, the power factor will be

(a) Lagging (c) Unity

(a) Load angle is equal to internal angle θ (b) Input power factor is unity (c) Load angle is 45° (d) Load angle is 0°

13. Synchronous motors when operated at power factor ranging from lagging through unity to leading for voltage control are called

(a) (b) (c) (d)

Voltage boosters Synchronous reactors Mechanical synchronizers None of these

14. Synchronous condenser means

(a) A synchronous motor with capacitor connected across terminals to improve PF (b) A synchronous motor operating at full load with leading PF (c) An over-excited synchronous motor partially supplying mechanical load, and also improving the PF of the system to which it is connected (d) An over-excited synchronous motor operating at no load with leading PF used in large power station for improvement of PF

15. A synchronous motor may fail to pull in synchronism owing to

(a) Excessive load (c) High friction

(b) Low excitation (d) Any of these

16. Which of the following used for synchronizing threephase generator is considered the best one?

(a) Three dark lamp method (b) Two bright and one dark lamp method (c) Synchroscope (d) None of these

17. An infinite bus bar has

(a) Constant voltage (b) Constant frequency (c) Infinite voltage

(d) Both (a) and (b)

18. Which of the following synchronous motors is cost comparable to that of an induction motor?

(a) (b) (c) (d)

High kW output high speed High kW output low speed Low kW output low speed Low kW output high speed

(b) Leading (d) More than unity

ANSWERS (MCQ) 1. (a)  2. (b)  3. (c)  4. (d)  5. (b)  6. (c)  7. (c)  8. (c) 9. (b)  10. (c)  11. (a)  12. (a)  13. (b)  14. (d)  15. (d)

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12. A synchronous motor will deliver maximum power when

16. (c)  17. (d)  18. (b).

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CON V E NTI O NA L Q UE S TI O NS (C Q ) 1. Name two factors that will cause an alternator to ‘motorize’. 2. Give the equation that determines the average speed of a synchronous motor. 3. Explain why a synchronous motor is not inherently self starting. 4. Explain why a synchronous motor will run at synchronous speed or not at all. 5. Give four methods used for starting synchronous motors. Which of the four methods is the most commonly used and why? 6. How can the speed of a synchronous motor be adjusted? 7. How does an amortisseur winding reduce hunting caused by pulsating loads? 8. State how a synchronous motor can be started, stopped and reversed 9. What are the two components of synchronous motor torque? What are they due to? 10. Give one inherent advantage of synchronous motor over an induction motor as a source of mechanical power 11. What is a synchronous capacitor and how can it be distinguished from a synchronous motor? ANSWERS (CQ) 2. S = (120f)/P 16. 150 RPM

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12. Is there an economic limit to improvement of PF? If so, what is it? 13. In addition to correction of PF and a source of mechanical power, give an additional application of the synchronous motor 14. What is meant by the torque angle of a synchronous motor? What factors affect the magnitude of this angle? 15. Differentiate between pull-in torque, pull-out torque and locked, rotor torque. 16. Determine the speed of a 40-pole synchronous motor operating from a 3-phase, 50 Hz, 4600 V system. 17. A 3-phase 50 hp, 2300 V, 60 Hz synchronous motor is operating at 90 r.p.m. Determine the number of poles in the rotor. 18. Calculate (a) the frequency of the voltage that must be applied to the stator of a 10 pole, three-phase, 40 V synchronous motor, required to operate at 1200 r.p.m. (b) the number of pole required for a 220 V, three-phase synchronous motor to operate at a speed of 500 r.p.m. when 50 Hz is applied to the stator (c) The full load speed of 36 pole, 60 Hz, 220 V synchronous motor in r.p.m. and rad/second.

17. 80 poles 18. (a) 100 Hz (b) 12 poles (c) 16 poles, 20.9 rad/s.

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Induction Motors (Three Phase)

40

OBJECTIVES In this chapter you will learn about:    Rotating magnetic fields   Direction of rotation   Slip and its effect   Construction of a three-phase induction motor  WRIM and SCIM   Losses and efficiency  Maximum power  Induction motor characteristics    Starting techniques for induction motors  Induction motor parameters   Simple problems on the above topics  Induction motor with squirrel cage motor

S

N Rotor sΩs

Rotor

Slip sΩs

Stator sΩs

Stator

40.1 INTRODUCTION The three-phase induction motor has a simple yet exceedingly robust construction which is comparatively cheap to manufacture. It also has good operating characteristics that make it a suitable drive for many production machines such as lathes or fans. These features are reflected by the fact that the induction motor is the most commonly used type of a.c. motor. The induction motor is chosen for its simplicity, reliability and low cost. These features are combined with good efficiency, good overload capacity and minimal or no service requirement. When the facts of very wide availability and simple installation by relatively little trained personnel are added, the choice of an induction motor seems well founded. The motor is supplied from a three-phase source and, therefore, it requires three supply conductors that are connected to three windings attached to the stator. The rotor also has a conductor system but this is not connected to its supply. Instead, the current in the rotor conductors is induced by a transformer action from the stator windings. As a result of this transformer action, the stator windings are sometimes termed the primary windings and the rotor windings may be termed the secondary windings. Thus, the three-phase induction motor is a sort of transformer with a rotating secondary winding. There are two principal methods of connecting the rotor conductors together and this gives rise to two classes of induction motor. These are: the cage-rotor induction motor and the wound-rotor inductor motor. In either case, the principle of operation is essentially the same and this depends on the action of the stator windings which gives rise to a rotating magnetic field.

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40.2  THE ROTATING MAGNETIC FIELD The most simple form of three-phase stator winding is embedded in the inner surface of the cylindrical stator. This form of construction is illustrated is Figure 40.1.

Figure 40.1  Simple Three–phase Stator Construction The windings are spaced at 120° intervals around the stator surface. One may be tempted to think that there is only 60° between the windings but when describing the angle, we consider the difference to be between the start of each winding the start being the point at which the supply is connected. We can see that there is 120° between the external connection of the R-phase and that of the Y-phase, whilst there is a further 120° to the connection of the B-phase. A further 120° remains between the B-phase and R-phase, thus, completing a revolution of the stator surface. The windings are inserted into the stator surface in a manner similar to that already described with the case of armature winding in the d.c. machine. For convenience, each phase winding is shown to be distinctly separate from the winding supplied by the next phase. This makes it easy to observe the different windings, but in practice, the windings would be spread out to cover the entire surface of the stator. Each winding has a distinct beginning and an end. Such windings are called phase windings, and are not to be confused with the commutator windings which form closed loops tapped by means of commutators. The start terminals of the phase windings are connected to the supply lines. In the case shown in Figure 40.2, the finish terminals are connected together. Thus, the windings are star connected but there is no reason why the windings should not have been delta connected instead. Each winding sets up a flux which acts along the axis of the winding. This action is shown in Figure 40.2, showing the field arrangement for only one winding. By considering different instants during a cycle of ­alternating current flowing in the coil, we can see that the flux always acts in the same axis but with varying magnitude and with alternating direction. In the three-phase system there are three windings, each producing a flux and each acting in a ­different direction within the stator. If we combine the three fluxes, we find that a similar magnetic field is set up within the stator except that the magnitude of the total field is bigger than that associated with any one winding. Actually, it is 50 per cent larger and for the arrangement shown in Figure 40.1, we can see in Figure 40.3 the manner in which the three fluxes add together to give the larger total flux. This diagram has been drawn for the instant at which the R-phase current is at its maximum value. It follows that the currents in other two phases at the same instant are half the maximum value and flowing in opposite directions. By applying our observations of Figure 40.3, the R-phase winding produces its maximum flux and the axis of the field is shown by an arrow which indicates the direction of the field. The fields of the other two windings act in relatively opposite directions to that of the first winding because the currents in these other windings are flowing in ‘negative’ directions. Effectively, the

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Figure 40.2  The Pulsating Flux Due to Alternating Current in a Single Coil

Figure 40.3  Field Distribution in a Three-phase Stator with Three-phase Windings

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Induction Motors (Three Phase)  779

three separate fields are similar, although not in identical directions and their combined effect is to produce a field greater than that of the R-phase winding and acting in the direction of its field. We can use the Right Hand Grip Rule to confirm the direction of the separate phase fields and, hence, the direction of the combined field. With one winding we could obtain a certain field. With three windings, we have only increased the resultant field by 50 per cent, which is not a particularly substantial increase. The importance of this arrangement only becomes apparent when we consider subsequent instants. In Figure 40.4, for example, we can once again look at the flux arrangement as was considered in Figure 40.3, but at an instant 30° later in supply cycle. At this instant, the current in the R-phase has fallen to 0.87 of its maximum value, while that in the Y-phase is zero. The current in the B-phase has risen in value to 0.87 of its maximum value and it is again flowing in the negative direction. Figure 40.4 shows the separate phase windings fields and the resultant combined field.

Figure 40.4 Field Distribution in a Three-phase Stator with Three-phase Windings at an Instant Later than that of Figure 40.3 In this case, we again produce a somewhat bigger resultant field than is produced by one winding but the important observation is that the axis of the resultant field has shifted from that seen in Figure 40.3, the field axis has been shifted by 30°. We have seen that a field set up by three windings fixed in space produces a magnetic field whose axis has shifted relative to the windings. It should be sufficient to draw the resultant field only for a selection of instants throughout the supply cycle. This has been illustrated in Figure 40.5 and, by taking instants throughout the complete cycle, it will be seen that the resultant field rotates one complete rotation. We conclude that a system of windings fixed in space and excited from a three-phase a.c. supply produces a rotating field. Note: It has been assumed that the construction of the three-phase machine is symmetrical, i.e., the windings are identical and displaced at equal angles from one another. It has also been assumed that the supply is symmetrical and, subsequently, the phase currents have all been of equal maximum instantaneous values. This symmetry is implicit to all rotating machine studies.

40.3  SPEED OF THE ROTATING MAGNETIC FIELD Spaces have been left between the conductors of one phase winding and the conductors of the next phase winding. This failure to use all the surface space of the stator is wasteful and in practice, conductors cover the entire stator surface area, as indicated in Figure 40.6(a). However, drawing the individual conductors is time consuming and it is usual to indicate the winding groups of conductors by shaded areas as shown in Figure 40.6(b).

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780  Electrical Technology

Figure 40.5  Rotating Magnetic Field

Figure 40.6  Three-phase, Two Pole Winding Arrangement Each phase winding has been shown to set up its own field, which acts from one side of the ­stator across to the other. This has been shown in Figures 40.7 and 40.8. Seen from the rotor, it would seem that the field emanates from a N-pole at one side and terminates in a S-pole at the other. The same ­observations may be made of the resultant field, and for this reason, the winding arrangement is d­ escribed as a two-pole winding. By using six windings instead of three, it is possible to have a four-pole machine, while nine windings give a sixpole machine, and so on; although unusual, it is quite practical to have a machine with about a 100 poles. The winding arrangements of four-pole and six-pole machines are shown in Figure 40.8.

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Induction Motors (Three Phase)  781

Figure 40.7 (a) Elementary Three-phase Induction Motor (b) Three-phase Flux Waves and (c) Instantaneous Direction of Resultant Stator Flux

Figure 40.8  Four- and Six-pole, Three-phase Stator Windings One cycle of the supply causes the field to rotate through one revolution. The field axis starts half way between the R-phase conductors and after half a cycle rotates past one set of R-phase conductors, i.e., one side of R-phase windings to reach a position again half way between the R-phase conductors yet acting in the opposite direction. A further half cycle completes the movement to reach the original relationship in space between the conductor position and the field axis. In the four-pole machine, the movement of the field past one set of conductors is only sufficient to rotate the field through 90° and the completion of the cycle only rotates the field through 180°. At this instance, the field system appears again to be the same as the field system when it started to move and this is because there are two possible situations which are identical, i.e., when the field system is in its initial position and when it is upside down.

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782  Electrical Technology Developing this argument, the four-pole machine requires a further cycle of the supply to rotate the field back to its original position. Thus, a four-pole machine requires two cycles of the supply in order that the field rotates through one revolution. The six-pole machine can, similarly, be shown to require three cycles of the supply for the field to rotate through one revolution. We can observe that the number of cycles of the supply required for one revolution of the magnetic field is always half the number of poles. However, poles in such machines always come in multiples of two, i.e., in pairs of poles. It follows that the number of pole pairs is equal to the number of cycles of the supply required for one complete revolution of the magnetic field. If the number of pole pairs is p and the supply frequency is ƒ hertz, then the number of revolutions of the magnetic field per second is N = ƒ/P

(40.1)

where, ‘n’ is the rotational speed in revolution per second. since ‘n’ revolutions per second is equivalent to N revolutions per minutes. N = 60 ƒ/P (40.2) Example 40.1 A six-pole, three-phase, 50 Hz induction motor sets up a rotating field. At what speed does it rotate? Solution: n = ƒ/P = 50/3 = 16.7 rev/s N = 60 ƒ/P = (60 × 50)/3 = 1000 rev/min The speed at which the rotating magnetic field revolves is termed the synchronous speed. In most machines, the supply frequency and the number of poles are fixed and it follows that the synchronous speed is constant for any given machine. However, it is possible to produce variable frequency supplies, and even to change the number of poles, in either of which cases the synchronous speed can be changed, but these cases are relatively unusual. The relationship between the speed of the rotating flux and the number of stator poles may be visualized by comparing the mechanical degrees of circular arc travelled by the flux in motors with different numbers of stator poles. This is shown in Figure 40.9, where the circular arc travelled by the rotating field of a four-pole motor is twice that of an eight-pole motor, assuming the same frequency and the same time period; the centre line of flux rotates 60° (phase A to phase C) for a fourpole winding and 30° (A to C) for an eight-pole winding, as illustrated in Figures [40.9(a) and (b)], respectively. The synchronous speed of an induction motor operating from a fixed frequency system can be changed by changing the number of poles in the stator, and by using a frequency converter to change the frequency or both.

Figure 40.9  Coil Spans for the Four-pole Winding (a) and Eight-pole Winding (b)

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Induction Motors (Three Phase)  783

Example 40.2 Determine the synchronous speed of a six-pole, 460 V, 60 Hz induction motor if the frequency is reduced to 85 per cent of its rated value. Solution: 120 ( 60 × 0.85 ) 120 × f s ns = = P 6 = 1020 rev/min

40.4  DIRECTION OF ROTATION If the supply connections to any two windings are reversed, then the phase sequence of the stator current is reversed. The result of this change is to cause the field to rotate in the opposite direction, as indicated in Figure 40.10.

Figure 40.10  Direction of Rotation of the Magnetic Field As a common application of the three-phase induction machine is to serve as a motor and since the rotor turns in the same direction as the rotating flux, it follows that the direction of rotation of a motor can be reversed simply by interchanging any two of the three supply conductors.

40.5  SLIP AND ITS EFFECT ON ROTOR FREQUENCY AND VOLTAGE The difference between the speeds of the rotating flux and the speed of the rotor is called slip speed and the ratio of slip speed to synchronous speed is called slip. Expressed mathematically N = ns – nr (40.3) s=

ns − nr ns

(40.4) 

n = slip speed (rev/min) ns = synchronous speed (rev/min) nr = rotor speed (rev/min) s = slip (pu) The slip, as expressed in Eq. 40.4, is called per-unit slip. The slip depends on the mechanical load connected to the rotor shaft (assuming a constant supply voltage and a constant supply frequency). Increasing the shaft load decreases the rotor speed, thus, increasing the slip. If the rotor is blocked to prevent turning nr = 0, and Eq. (40.4) reduces to Where,

s=

ns − 0 =1 ns

(40.5)  Releasing the brake allows the rotor to accelerate. The slip decreases with acceleration and approaches zero when all mechanical load is removed. If operating with the no shaft load and if the windage and friction are sufficiently small, the very low relative motion between the rotor and the rotating flux of the stator may cause the rotor to become magnetized along an axis of minimum reluctance. If this occurs, the rotor will lock in synchronization with the rotating flux of the stator; the slip will be zero, no induction motor torque will be developed and the motor will act as a reluctance-synchronous motor. However, the application of a small shaft load will cause it to pull out of synchronism, and the induction-motor action will again occur. Solving Eq. (40.4) for nr expresses the rotor speed in terms of slip: nr = ns (1 – s) (40.6)

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784  Electrical Technology

40.5.1  Effect of Slip on Rotor Frequency The frequency of the voltage induced in a rotor loop by a magnetic field is given by P×n fr = 120 

(40.7)

where, ƒr = rotor frequency (Hz) P = number of stator poles n = slip speed (rev/min)

substituting Eq. (40.3) into Eq. (40.7) fr =

P(ns − n r ) 120



From Eq. (40.4) ns– n­r = sns Substituting in Eq. (40.7) sPns fr = 120  If the rotor is blocked so that it cannot turn, s = 1, and Eq. (40.9) becomes Pn s f BR = 120  where, ƒBR = frequency of voltage generated in the blocked rotor (BR) Substituting Eq. (40.6) into Eq. (40.9) results in the general expression for rotor frequency. Thus, ƒr = sƒBR At blocked rotor, also called locked rotor, there is no relative motion between rotor and stator, the slip is 1.0, frequency of the voltage generated in the rotor is identical to the frequency of the applied stator voltage. That is ƒBR = ƒstator

(40.8)

(40.9)

(40.10)

(40.11) and the (40.12)

40.5.2  Effect of Slip on Rotor Voltage The voltage generated in a rotor loop, as illustrated in Figure 40.11 (formed by two rotor bars and end connections), as it is swept by the rotating stator flux is given by Er= 4.44 Nƒr φmax (40.13)

Figure 40.11 (a) Rotating Field Sweeping a Rotor Bar (b) Direction of Flux Generated Around Rotor Bar and (c) Direction of Rotor-bar Current Substituting Eq. (40.11) into Eq. (40.13) Er = 4.44 NsƒBR φmax  At blocked rotor, s = 1, Eq. (40.14) becomes EBR­= 4.44 NƒBR φmax Substituting Eq. (40.15) into Eq. (40.13) Er = sEBR

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(40.14) (40.15) (40.16)

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Induction Motors (Three Phase)  785

Equation (40.16) is the general expression for voltage induced in a rotor loop at any rotor speed in terms of blocked-rotor voltage and slip. Example 40.3 The frequency and induced voltage in the rotor of a certain six-pole wound-rotor induction motor, whose shaft is blocked, are 60 Hz and 100 Hz, respectively. Determine the corresponding values when the rotor is running at 1100 rev/min Solution: 120 f s 120 × 60 ns = = = 1200 rev/min P 6 ns − n r 1200 − 1100 = 0.0833 s= = 1200 ns f r = s f BR = 0.0833 × 60 = 5.0 Hz Er = sEBR = 0.0833 × 100 = 8.33 V Example 40.4 A four-pole induction motor operating at a frequency of 60 Hz has a full-load rotor slip of 5 per cent. Calculate the rotor frequency: (1) At the instant starting; and (2) At full load. Solution: 1. At the instant of starting

ns – nr s = n s where, nr is the rotor speed. Since the rotor speed ns at that instant is zero, s =1, or unity slip. The rotor frequency is ƒr = sƒBR = 1.0 × 60 = 60 Hz 2. At full load the slip is 5 per cent (given) and, therefore, s = 0.05  and  ƒr = s × ƒ = 0.05 × 60

Rotor frequency = 3 Hz

Example 40.5 A three phase two pole induction motor is connected to a 50 Hz supply. Determine the synchronous speed of the motor in rev/min. Solution: ns= 50/l rev/s

= 50 × 60 rev/min = 3000 rev/min

Example 40.6 A stator winding supplied from a three-phase 60 Hz system is required to produce a magnetic flux rotating at 900 rev/min. Determine the number of poles. Solution: 900 = 15 rev / s 60 ns = f / p then p = f /ns = 60/15 = 4

Synchronous speed = Since, Example 40.7

A three-phase two pole motor is to have a synchronous speed of 6000 rev/min. Calculate the frequency of the supply voltage. Solution: ns = f / p, then f = ns p since, 6000 2 = × = 100 Hz 60 2

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786  Electrical Technology

40.6  CONSTRUCTION OF A THREE-PHASE INDUCTION MOTOR Many a.c. motors are classified as induction motors. An induction motor is a motor that has no electric connections between the power sources and the rotor, yet the rotor has conductors which carry current. The current in the rotor conductors is an induced current. It is induced by the magnetic field created by the stator windings. A wound-rotor induction motor, as shown in Figure 40.12, has a regular three-phase winding similar to that of the stator and is wound with the same number of poles. The phases are usually wye-connected and terminate at the slip rings. A wyeconnected rheostat with a common lever is used to adjust the ­resistance of the rotor circuit. The rheostat provides speed control, torque adjustment at locked rotor and current limiting during starting and acceleration.

Figure 40.12  Cutaway View of Wound-rotor Induction Motor The stator core is an assembly of thin laminations stamped from silicon-alloy sheet steel; the use of silicon steel for the magnetic material minimizes hysteresis losses. The laminations are coated with oxide or varnish in order to minimize eddy-current losses. Insulated coils are set in slots within the stator core and the overlapping coils are connected in series or parallel arrangements in order to form phase groups and the groups are connected in wye or delta. The connections wye or delta, series or parallel, are dictated by voltage and current requirements. The rotors are of two basic types (squirrel cage and wound rotor). Small squirrel-cage rotors use a slotted core of laminated steel into which the molten aluminium is cast to make the conductors, end rings, and fan blades. Large squirrel rotors, as shown in Figure 40.13, use brass bars and brass end rings that are brazed together to form the squirrel cage. There is no insulation between the iron core and the conductors, and none is needed; the current induced in the rotor is contained within the circuit formed by the conductors and end rings also called end connections. Skewing the rotor slots helps avoid ­crawling and reduces vibration. A wound-rotor induction motor uses insulated coils that are set in slots and connected in wye arrangement. The rotor circuit is completed through a set of slip rings, carbon brushes and a wye-connected rheostat. The three-phase rheostat is composed of three rheostats connected in a wye; a common lever is used to simultaneously adjust all three rheostat arms. The transfer of energy occurs in a manner similar to that in a transformer. The stator is often referred to as the primary and the rotor as the secondary. The air gap is made quite small so as to offer minimum reluctance. Each coil of an induction motor stator spans a portion of the stator circumference equal to or slightly less than the pole pitch. The pole pitch is equal to the stator circumference divided by the number of stator poles, it may be expressed in terms of stator slots or arc. If the span is less than a pole pitch, it is called a fractional pitch winding.

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Induction Motors (Three Phase)  787

Figure 40.13  Conductor Arrangement in a Squirrel-cage Rotor

40.7  ROTOR IMPEDANCE AND CURRENT When an induction motor is stationary, the stator and rotor windings form the equivalent of a transformer, as illustrated in Figure 40.14(a), which represent one phase of an inductor-motor rotor. The power and torque developed by a three-phase motor is three times that developed by one of its phases. It will be assumed that the stator is an ideal stator in that it produces a rotating magnetic field of constant amplitude and constant speed, and that it has no core losses, no copper losses and no voltage drops. The rotor is represented by an electrically isolated closed circuit containing resistance and reactance acted on by an induced rotor voltage Er. The rotor voltage is generated at a frequency ƒr by the rotating flux of the stator. The model shows one phase of a wound rotor, or one phase of an equivalent wound rotor, if the rotor is squirrel cage. The rotor resistance is dependent on the length, cross-sectional area, resistivity and the skin effect of the rotor conductors as well as external rheostat resistance if it is a wound rotor like that shown in Figure 40.12. The inductive reactance Xr of the rotor, called leakage reactance, is caused by leakage flux and is ­dependent on the shape of the rotor conductors, its depth in the iron core, the frequency of the rotor voltage and the length of the air gap between the rotor and the stator iron. The leakage reactance of the rotor expressed in terms of rotor frequency and rotor inductance is Xr= 2pƒrLr (40.17) Substituting Eq. (40.11) into Eq. (40.17) and simplifying Xr = 2p (sfBR) Lr = s (2p fBRLr) = sXBR (40.18) Replacing Xr, Er and ƒr in Figure 40.14(a) with their equivalent values in terms of slip results in Figure 40.14(b): the rotor impedance as determined from the associated impedance diagram in Figure 40.14(c) is Zr = Rr + jsXBR (40.19) Applying Ohm’s law to the rotor circuit in Figure 40.14(b) Ir =

sEBR sEBR = Zr R r + jsX BR

(40.20) 

Dividing both numerator and denominator by s, E EBR I r = BR = Zr s Rr /s + jX BR

(40.21)

 A modified equivalent circuit and associated impedance diagram corresponding to Eq. (40.19) are shown in Figures [40.14(d) and (e)], respectively. The constant blocked-rotor voltage in Figure 40.14(d), combined with an equi­ valent rotor resistance that varies with slip provides a convenient tool for analysis of induction-motor behaviour. Expressing the rotor current in terms of magnitude and phase angle, Ir =

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EBR 0° Z r /s θ r

=

EBR −θ r Z r /s

(40.22)

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788  Electrical Technology

Figure 40.14 Equivalent Circuits and Corresponding Impedance Diagrams for an Induction Motor with an Ideal Stator and a Real Rotor

The magnitude of the rotor current is EBR Z r /s  Expressing Zr /s and θr in terms of their components, as shown in Figure 40.14(e) Ir =

Ir =

EBR ( Rr / s) 2 + X BR 2

θr = tan −1

X BR R r /s



(40.23)

(40.24)  (40.25)

Example 40.7 The frequency of the supply of an eight-pole induction motor is 50 Hz and the rotor frequency is 3 Hz. Determine (1) the slip, and (2) the rotor speed.

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Induction Motors (Three Phase)  789

Solution: 1. ƒr= sƒ Hence, 3 = s × 50 = 0.06 or 6 per cent 2. Synchronous speed, ns = ƒ/p = 50/4 = 12.5 rev/s 12.5 × 60 = 75 rev/min ns – nr 12.5 – n s = ns = 0.06 =  12.5 r (0.06) (12.5) = 12.5 – nr Rotor speed nr = 12.5 – 0.75 = 11.75 rev/s or 705 rev/min Example 40.8 The rotor of a certain 25 hp, six-pole, 60 Hz induction motor has equivalent resistance and equivalent reactance per phase of 0.10 Ω and 0.54 Ω, respectively. The blocked rotor voltage/phase (EBR) is 150 V. If the rotor is running at 1164 rev/min, determine (1) synchronous speed, (2) slip, (3) rotor impedance, (4) rotor current, (5) rotor impedance if changing the shaft load resulted in 1.24 per cent slip, (6) speed for the conditions in (5). Solution: 120 f 120 × 60 1. ns = = = 1200 rev/min 6 P 2. s =

ns − nr 1200 − 1164 = = 0.030 ns 1200

Rr 0.010   3. Z r = + JxBR = + j 0.54 = 3.3768 9.20 = 3.38 9.20 Ω s 0 . 0 3 0 

EBR 150 0   = 4. I r =  = 44. 421 − 9. 2 = 44 . 4 − 9 . 2 A ZR . . 3 3768 9 2 0 Rr 0.10   5. Z r = + JxBR = + j 0.54 = 8.09257 3.83 = 8.093.83 Ω s 0.0124 6. nr = ns (1 − s ) = 1200(1 − 0.0124) = 1185 rev/min

40.8  LOCUS OF THE CURRENT The changes that take place in rotor impedance angle qr and rotor-current magnitude Ir, as an unloaded induction motor accelerates from standstill (blocked rotor) to synchronous speed are illustrated in Figure 40.15(a). The curves are plots of Eqs. (40.23) and (40.24), respectively. It can be seen that the rotor current and the rotor impedance angle have their greatest values at blocked rotor both decrease in values as the rotor ­accelerates and both approach zero as the rotor approaches synchronous speed. For low values of slip (s < 0.05), the rotor current is proportional to the slip. A phasor diagram representing the magnitude and phase angle of the rotor for values of slip from s = 1 to s = 0 is shown in Figure 40.15(b). The current phasor changes in both magnitude and phase angle as the machine accelerates from blocked rotor (s = 1) to synchronous speed (s = 0). The locus of the current phasor is a semicircle. Proof of its semicircle character is obtained by expressing Zr /s in terms of sin qr and then substituting into Eq. (40.23). Thus, from Figure 40.14(e), Zr X = BR s sinq r  Substituting Eq. (40.26) into Eq. (40.23) and simplifying

(40.26)

EBR sinθ r (40.27) X BR  Equation (40.27) is the polar equation for a circle that is tangent to the horizontal axis at the origin and whose diameter is EBR /XBR . Ir =

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790  Electrical Technology

Figure 40.15 (a) Rotor Current and Rotor Impedance Angle Versus Speed for Representative Induction Motor (b) Locus of Rotor-current Phasor

40.9  LOSSES AND EFFICIENCY Calculations involving overall motor efficiency must take into account the losses that occur in both the stator and the rotor. The stator losses include all hysteresis losses and eddy-current losses in stator and rotor (called core losses), I 2R losses in the stator winding (called stator copper losses). Given the input power to the stator, and the stator losses, the net power crossing the air gap is Pgap= Pin – Pcore – Pscl (40.28) where, Pin= total three-phase power input to stator Pcore = core loss P scl = stator conductor loss or stator copper loss Figure 40.16 shows the flow of power from stator input to shaft output and accounts for the loss in both stator and rotor. The power flow diagram is a useful adjunct to problem solving. It suggests a convenient method of solution.

Figure 40.16 Shows the Flow of Power from Stator Input to Shaft Output and Accounts for the Losses in Both Stator and Rotor Power

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Induction Motors (Three Phase)  791

Power P = 2p nT where T is the torque in Newton metres; hence, torque T = P/(2pN). If P2 is the power input to the rotor from the rotating field, and Pm is the mechanical power (including friction losses), then = T From which Hence,

P2 Pm = 2 pns 2 pn r

(40.29)



P2 Pm Pm nr = = or ns nr P2 ns 1 − Pm n =1− r P ns 2 P2 − Pm ns − n r = =s P2 ns 

(40.30)

But P2 – Pm is the electrical or copper loss in the rotor, i.e., P2 – Pm = Ir2R2 Hence slip

s=

rotor copper loss I r 2 R2 = P2 rotor output

Or power input to the rotor P2 =

I r 2 R2 s 

(40.31)

See Figures 40.16 and 40.17 As has been shown in Figure 40.17(a), when an induction motor is stationary, the stator and rotor windings form the N  equivalent of a transformers and E2 =  2  E1 (40.32a)  N1   As can be seen in Figure 40.17(b), when running, the induced emf is proportional to the slip, s. Hence, when running,  NRr  the rotor emf per phase = Er = sE2 = s  (40.32b) E1  E1  

(a)

(b)

Figure 40.17  Rotor e.m.f., Frequency, Impedance and Current (a) When Stationary (b) When Running

Motor Efficiency, Motor Efficiency, η =

output power  Pm  =  × 100 per cent  input power  P1 

(40.33)

Example 40.9 The power supplied to a three-phase induction motor is 32 kw and the stator losses are 1200 w. If the slip is 5 per cent, determine (1) the rotor copper loss, (2) the total mechanical power developed by the rotor, (3) the output power of the motor if friction and windage losses are 150 w and (4) the efficiency of the motor neglecting rotor iron loss.

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792  Electrical Technology Solution: 1. input power to rotor = stator input power-stator losses     32 kw – 1.2 kw = 30.8 kw rotor copper loss rotor input 5 rotor copper loss = 30.88 100 Slip =

Rotor copper loss = (0.05) (30.8) = 1.54 kw 2. Mechanical power developed by the rotor = rotor input power – rotor losses = 30.8 – 1.54 = 29.26 kw 3. Output power of motor = power developed by the rotor – friction and windage losses = 29.26 – 0.75 = 28.51 kw  output power  4. Efficiency of inductor motor =   × 100 per cent  input power   28.51  =  × 100 per cent = 8 9.10 per cent  32  Example 40.10 The speed of induction motor in Example 40.9 is reduced to 35 per cent of its synchronous speed by using external rotor resistance. If the torque and rotor losses are unchanged, determine (1) the rotor copper loss, and (2) the efficiency of motor. Solution: 1. Slip  n − nr  s= s × 100 per cent = ( ns − 0.35ns ) ×100 per cent  ns  = ( 0.65 ) (100 ) = 65 per cent Input power to rotor = 30.8 kw Since, rotor copper loss ; rotor copper loss = ( s ) ( rotor input ) rotor input  65  Rotor copper loss =  ( 30.8 ) = 20.02 kw  100  s=

2. Power developed by rotor = input power to rotor – rotor copper loss = 30.8 – 20.02 = 10.78 kw Output power of motor = power developed by rotor-friction and windage losses = 10.78 – 0.75 = 10.03 kw Efficiency, Efficiency, η =

output power  10.03  × 100 per cent =   × 100 per cent = 31.34 per cent input power  32 

40.10  AIR GAP POWER The power transferred electromagnetically across the air gap between the stator and rotor is called air-gap power or gappower. Sgap= EBR Ir cos qr + j EBR Ir sin q r 

M40_AUTH_ISBN_C40.indd 792

(40.34)

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Induction Motors (Three Phase)  793

The active and reactive components of gap power are (Active power) Pgap = EBR Ir cos qr (Reactive power) Qgap = EBR Ir sin qr where,

(40.35) (40.36)

EBR = blocked rotor voltage Ir = magnitude of rotor current

qr = rotor impedance angle cos qr = power factor of rotor Active component Pgap: Supplies the shaft power output as well as friction, windage, as well as heat losses in the output. Reactive component Qgap supplies the reactive power for the alternating magnetic field about the rotor current. Component Qgap is not dissipated. It follows a sinusoidal pattern as it see-saws across the gap between the stator and the rotor. Components Pgap and Qgap may be represented in a power diagram as the two sides of a right triangle whose diagonal is Sgap as illustrated in Figure 40.18 Sgap= Pgap+ j Qgap

(40.37)

Figure 40.18  Power Diagram for Air Gap Power Example 40.11 For the motor operating at 1164 rev/min in Example 40.8, determine the total three-phase apparent power crossing the air gap its active and reactive components, and the rotor power factor. Solution: S gap = 3 × 190 0 × 44.421 +9.2 = 19.989 9.2 VA Converting to rectangular form S gap = (19732 + j 3197) VA Pgap = 19732 W, Qgap = 3196 Var PF = cos 9.2 = 0.99

40.11  MAXIMUM TORQUE Torque T =

 1   I r2 R2  Pα = 2πns  2πns   s 

from Eq. (40.31) P2 =

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I r2 R2 s

and

Ir =

 s ( N 2/ N1 )E1 R 2 2 + ( sX 2 )

(40.38)

2



8/28/2012 6:51:38 PM

794  Electrical Technology 2  1   s 2 ( N 2 /N1 ) E12   R2  Hence, torque per phase T =      2πns   R 2 2 + ( sX 2 )2   s 

Or

2  1   s ( N 2 /N1 ) E12 R2  T =    2πns   R 2 2 + ( sX 2 )2 

 (40.39)



2  m   s ( N 2 /N1 ) E12 R2  If there are m phases, then torque T =      2πns   R2 2 + ( sX 2 )2 

 m ( N 2 /N1 )2    sE12 R2 T =  2 2 2πns    R 2 + ( sX 2 ) 

(40.40)

(40.41) 

  sE 21 R2 T = K 2  where, K is a constant for a particular machine.  R 2 2 + ( sX 2 )    sE 21 R2 T∝ 2  R 2 2 + ( sX 2 ) 

i.e., torque

(40.42) (40.43)



Under normal conditions, the supply voltage is usually, constant; hence, Eq. (40.43) becomes

 sR2 T ∝ 2  R2 + sX 2 2

(

)

   R2  ∝ 2    R s + ( sX )2  2 2 

(40.44)

The torque will be a maximum, when the denominator is a minimum and this occurs when R22 / s = sX 22 i.e., when s = R2 / X 2

or

R2 = sX 2 = X r

(40.45)



Thus, the maximum torque occurs when rotor resistance and rotor reactance are equal i.e., R2 = X2

(40.46)

Example 40.12 A 415 V, three-phase, 50 Hz, four pole star connected induction motor runs at 24 rev/s on full load. The rotor resistance and reactance per phase are 0.35 Ω and 3.5 Ω, respectively, and the effective rotor-stator turns ratio is 0.85:1. Calculate: (1) the synchronous speed; (2) the slip; (3) the full load torque; (4) the power output if mechanical losses amount to 770 W; (5) the maximum torque; (6) the speed at which maximum torque occurs; and (7) the starting torque. Solution: 1. Synchronous speed, n= s 2. Slip, s =

f 50 = = 25 rev/s or 1500 rev / min p 2

ns − nr 25 − 24 = = 0.04 or 4 per cent ns 25

3. Phase voltage  m ( N 2 / N1 )2 E1 =  2πns 



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 sE 21  2   R 2 2 + ( sX 2 )

  3 ( 0.85 )2 =   2π ( 25 )

  0.04 ( 239.6 )2 ( 0.35 )  2   0.352 + ( 0.04 × 3.5 )

(

)

   

 803.71  = ( 0.01380 )  = 78.05 V  0.1421 

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Induction Motors (Three Phase)  795

4. Output power, including friction losses, Pm= 2p nrT (2p) (24) (78.05) = 1170 W Hence, power output = Pm – mechanical losses 1170 – 770 = 11000 W = 11 kW 5. Maximum torque occurs when R2 = Xt = 0.35 Ω Slip,

Slip, s =

Hence, maximum torque, T

m

R2 0.35 = = 0.1 X2 3.5

 sE 21 R 2 = ( 0.01380 )  2  R 2 2 + ( sX 2 )

  

 ( 0.1 ) ( 239.6 )2 ( 0.35 )   = (0.01380)   ( 0.35 )2 + 0.352   

(

)

 2009.29  = ( 0.0 1380 )   = 113.18 Nm  0.245  6. For maximum torque, slip s = 0.1 n − nr 25 − nr s= s i.e., 0.1 = ns 25 7.

Hence, (0.1) (25) = 25 – nr  and  nr­= 25 – (0.1) (25) The speed at which maximum torque occurs nr = 22.5 rev/s = 1350 rev/min At the start (at stand still), slip s = 1 Hence, starting torque  m ( N 2 / N1 )2 = 2π ns 

  E2 R  ( 239.6 )2 ( 0.35 )    2 1 2 2  = ( 0.01380 )    R 2 + X 2   ( 0.35 )2 + ( 3.5 )2

  ( 2009.86 )   = ( 0.01380 )     ( 12.3725 ) 

Starting torque = 22.41 Nm

40.12  INDUCTION MOTOR TORQUE-SPEED CHARACTERISTICS The normal starting torque may be less than the full load torque. The speed at which maximum torque occurs is determined by the value of rotor resistance. At synchronous speed, slip s = 0 and torque is zero. From these observations, the torquespeed and torque-slip characteristics of an induction motor can be drawn (Figure 40.19). The rotor resistance of an induction motor is usually small compared with its reactance, so that the maximum torque occurs at a high speed, typically about 80 per cent of the synchronous speed. Curve P in Figure 40.19(a) is a typical characteristic of an induction motor. The curve P cuts the full-load torque line at X, showing that at full load the slip is 4–5 per cent. The normal operating conditions are between O and X. It can be seen that for normal operation the speed variation with load is quite small the induction motor is an almost constant speed machine. Redrawing the speed torque characteristic between O and X gives the characteristic shown in Figure 40.19(b), which is similar to that of the d.c. shunt motor. If maximum torque is required at starting, then a high resistance rotor is necessary, which gives ­characteristic Q in Figure 40.19(a), however, the motor has a full load slip of over 30 per cent which results in a drop in efficiency. Also, such a motor has a large speed variation with variation of load curves R and S, as seen in Figure 40.19(a), are characteristic for values of rotor resistance between those of P and Q. Better starting torque than for curve P is obtained but with lower efficiency and with speed variations under operating conditions. A squirrel-cage induction motor (SCIM) would normally follow characteristic P. This kind of ­machine is highly efficient and about constant speed under normal running conditions. However, it has a poor starting torque and must be started off load or very highly loaded. Also, on starting, the current can be four or five times the normal full load current due to the motor acting like a transformer with secondary short circuited. A wound-rotor induction motor (WRIM) would follow characteristic P when the slip rings are short circuited, which is the normal running condition. However, the slip-rings allow for the addition of resistance to the rotor circuit externally and

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796  Electrical Technology

(a)

(b)

Figure 40.19  (a) Torque-speed Characteristics (b) Figure 40.19(a) Redrawn as a result for starting, the motor can have a characteristic similar to curve Q in Figure 40.19(a). The high starting current experienced by the squirrel cage induction motor can be overcome. In general, for three-phase induction motors, the power factor is usually between about 0.8 and 0.9 lagging and the fullload efficiency is usually about 80–90 per cent. As torque is proportional to the square of the supply voltage, any voltage variations, therefore, would seriously affect the induction motor performance.

40.13  WRIM AND SCIM: A COMPARISON The advantages of the WRIM compared with the SCIM are as follows. 1. They have a much higher starting toque. 2. They have a much lower starting current. 3. They have a means of varying speed by use of external rotor resistance. The advantages of the SCIM compared with WRIM are as follows. 1. They are cheaper and more robust. 2. They have slightly higher efficiency and power factor. 3. They are explosion proof, since the risk of sparking is eliminated by the absence of slip rings and brushes.

40.14  STARTING TECHNIQUES FOR INDUCTION MOTORS Induction motors of almost any horse power may be started by connecting them across full voltages, as shown in Figure 40.20, and most are started that way. In many cases, however, the high in rush current associated with full-voltage

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Induction Motors (Three Phase)  797

Figure 40.20  Starting by Connecting Induction Motors Across Line starting can cause large voltage dips in the distribution system. Lights may dip or flicker, and unprotected control systems may drop out due to low voltage. The impact torque that occurs when starting at full voltage can damage the driven equipment. The methods commonly used for reducing in-rush current are reduced voltage starting using auto transformers current limiting through wye-delta connections of stator windings, part winding connections, series impedance and solid-state control. Auto Transformer Method: Auto transformer reduces the stator voltage and, thus, the starting current. However the the starting torque is seriously reduced, so the voltage is reduced only sufficiently to give the desired reduction of the starting current. A typical arrangement is shown in Figure 40.21. A double-throw switch connects the autotransformers in the circuit for starting and when the motor is up to speed the switch is thrown to the run position which connects the supply directly to the motor.

Figure 40.21  Auto Transformer: Starting of Induction Motors Wye-delta Starting Method: This method is used for starting the connections to the stator that are wye-connected, so that the voltage across each phase winding is 1/ 3 (i.e., 0.577) of the line voltage for running the windings are switched to delta connections. A typical arrangement is shown in Figure 40.22. This method of starting is less expensive than that by autotransformer. Part Winding Method: The part-winding method uses a stator with two identical three-phase windings, each capable of supplying one-half of the rated power. The power circuit for starting a part-winding motor is shown in Figure 40.23. Contacts 1 are closed first, energizing one winding. After a brief time delay, contacts 2 are closed energizing both windings. The part winding starter is one of the least expensive starters but is limited to dual-voltage motors that are operated on the low-voltage connections.

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798  Electrical Technology

Figure 40.22  Wye-delta Starting of Induction Motors

Figure 40.23  Part-winding Starter Series Impedance Starter: The series impedance starter shown in Figure 40.24(a) uses a resistor and inductor in series with each phase of the stator windings to limit the current during start-up. The running contacts (R) are open when starting, to limit the in-rush current and are closed to short out the impedance when the motor is near rated speed. The Ohmic values of the resistor or reactor are selected to provide approximately 70 per cent rated voltage at the motor terminals when starting. The series impedance stator provides smooth acceleration and is the simplest method of starting induction motors. At each step, the motor operation will transfer from one characteristic to the next so that the overall starting characteristic will be as shown by the bold line in Figure 40.24(b). Solid-state Starters: A solid-state starter, shown in Figure 40.25, uses back-to-back thyristors to limit the current. The control circuit (not shown) allows a gradual build up of current. The smooth build up permits a soft start with no impact loading and no significant voltages dips. Solid-state starters can be designed to incorporate many special features such as speed control, power factor control, protection against overload and single phasing.

40.15  DETERMINATION OF INDUCTION MOTOR PARAMETERS In those cases where induction motor parameters are not readily available from the manufacturers, they can be approximated from a d.c. test, a no-load test and a blocked-rotor test.

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(a)

(b)

Figure 40.24 (a) Series Impedance Start (b) At Each Step the Motor Characteristic Transfers from One to the Other

Figure 40.25  A Solid-state Starter

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800  Electrical Technology

40.15.1  D.c. Test The purpose of the d.c. test is to determine R1. This is accomplished by connecting any two stator leads to a variable voltage d.c. source as shown in Figure 40.26(a). The d.c. source is adjusted to provide approximately rated stator current and the resistance between the two stator leads is determined from voltmeter and ammeter readings. Thus, from Figure 40.26(a)

Figure 40.26  Circuit for d.c. Test to Determine R1 Rd.c. =

Vd.c. I d.c.

(40.47)

 If the stator is wye-connected, as shown in Figure 40.26(b),

(40.48a)

Rd.c.= 2R1wye and R1 wye = Rd.c. /2 If the stator is delta connected, as shown in Figure 40.26(c) Rd.c. =

R1, ∆ .2R1.∆ R1, ∆ + 2R1, ∆

=

2 R1, ∆ 3

and

(40.48b)

R1, ∆ = 1.5 Rd.c. 

40.15.2  No-load Test The no-load test is used to determine the magnetizing reactance XM and the combined core, windage, and friction losses. These losses are essentially constant for all load conditions. The connections for the no-load test are shown in Figure 40.27. The rotor is unblocked and allowed to run unloaded at both rated voltage and rated frequency. At no-load, the operating speed is very close to synchronous speed and the slip = 0, causing the current in R2 /s branch to be very small. For this ­reason, the R2 /s branch is drawn with dotted lines as shown in Figure 40.27, and omitted from the ­no-load current calculations. Since IM >> Ife , I0 ≈ IM; thus, the Rfe branch is also drawn with dotted lines and omitted from the no-load current calculations. Referring to the approximate equivalent circuit shown in Figure 40.28 for the no-load test the apparent power input per phase is (40.49) SNL = VNL.INL The reactive power per phase is determined from SNL= P2NL + Q2NL  (40.50)

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Induction Motors (Three Phase)  801

Figure 40.27  Basic Circuit for No-load Test and Blocked-rotor Test Solving for QNL, QNL = S 2NL– P 2NL

(40.51)

Expressing the reactive power in terms of current and reactance, and solving for the equivalent reactance at no-load QNL= I 2NL.XNL   and   XNL = QNL /I 2NL

(40.52)

where as indicated in Figure 40.28 XNL = X1­ + XM (40.53) Substituting X1 as determined from blocked rotor test, into Eq. (40.53) permits the determination of XM. The input power per phase at no load includes the core loss, stator copper loss, windage loss and friction loss (all per phase). That is PNL= I 2NLR1 + Pcore +PW, f (40.54) Separation of friction and windage losses from the no-load loss may be accomplished by plotting the no-load power versus voltage squared for low values of voltage and extrapolating to zero voltage.

Figure 40.28  Equivalent Circuit Per Phase for No-load Test

40.15.3  Blocked-Rotor Test The blocked-rotor test is used to determine X1 and X2. When combined with data from the d.c. test, it also determines R2. The test is performed by blocking the rotor so that it cannot turn, and measuring the line voltage, line current, and three-phase power input to the stator. Connections for the test are shown in Figure 40.29. An adjustable voltage a.c. supply (not shown) is used to adjust the blocked-rotor current to approximately rated current. If instrument transformers and single-phase Watt meters are used, the effect of transformer ratios and the direction of Watt meter readings (whether a positive or negative) must be considered.

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802  Electrical Technology Since the exciting current (I0) at blocked rotor is considerably less than the rotor current (I2) the exciting current may be neglected, enabling a simplification of the equivalent circuit shown in Figure 40.29, where Xm and Rfe are drawn with dotted lines and omitted when making blocked-rotor calculations. IBR.15

R1

jX1

R2/s

jX2

I0

PBR.15

ZBR.15 VBR.15

Rfe

jXM

Figure 40.29  Equivalent Circuit Per Phase for Blocked-rotor Test The IEEE test code recommends that the blocked-rotor test be made using 25 per cent rated frequency with the test voltage adjusted to obtain approximately rated current. Thus, a 60 HZ motor would use 15 Hz test voltage. The total reactance calculated from the 15 Hz test is then corrected to 60 Hz by multiplying by 60/15. The total resistance calculated from the 15 Hz test is essentially correct, however, and must not be adjusted Referring to Figure 40.29, when all values are per phase R1 + R2­ = RBR.15 Z BR.15 =

VBR.15 I BR.15

(40.55) R BR.15 =

and

PBR.15 I 2 BR.15

(40.56) 

Resistance R2 is obtained from RBR.15 by substituting R1 from the d.c. test into Eq. 40.55 Thus R2= RBR.15 – R1 From Figure 40.29 2 2 Z BR.15 = R BR.15 + X BR.15 

(40.57) (40.58a)

And

X BR.15 = Converting XBR.15 to 60 HZ

Z 2 BR.15 − R 2 BR.15  X BR.60 =

60 15X BR.15

where, XBR.60 = X1 + X2



(40.58b)

(40.59a) (40.59b)

S UM M A RY 1. The induction motor is the most commonly used type of a.c. motor. 2. The current in the rotor conductors is induced by transformer action. 3. The number of cycles of the supply required for one revolution of the magnetic field is always half the number of poles. 4. The speed at which the rotating magnetic field revolves is termed the synchronous speed. 5. The synchronous speed is constant for any given machine. 6. The direction of rotation of a motor can be reversed by interchanging any two of the three supply conductors.

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7. The difference between the speed of the rotating flux and the speed of the rotor is called slip speed. 8. The ratio of slip speed to synchronous speed is called slip. 9. The slip decreases with acceleration and approaches zero when all mechanical load is removed. 10. At blocked rotor the slip is 1.0. 11. The connections wye or delta, series or parallel are dictated by voltage and current requirements. 12. The rotors are of two basic types: squirrel cage and wound rotor. 13. The stator is often referred to as the primary and the rotor as the secondary.

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14. The rotor current and the rotor impedance have their greatest values at blocked rotor; both decrease in value as the rotor accelerates and both approach zero as the rotor approaches synchronous speed. 15. The stator losses include all hysteresis losses and eddy current losses in the stator and rotor (called core losses).

16. The power transferred electromagnetically across the air gap between the stator and rotor is called the air-gap power or gap power. 17. The maximum torque occurs when rotor resistance and rotor reactance are equal. 18. The induction motor parameters are determined by d.c. test, no-load test and blocked rotor test.

M U LT IP LE C H O I C E Q UE S TI O NS ( M C Q ) 1. The speed at which the rotating magnetic field, produced by the stator current rotates is

6. The no-load current of a three-phase induction motor in terms of its full-load current is of the order of



(a) 10 per cent (c) 50 per cent

(a) (b) (c) (d)

Rotor speed Synchronous speed Greater than synchronous speed Less than synchronous speed

2. The no-load slip of a three-phase induction motor is of the order of

(a) (b) (c) (d)

1 per cent 2 per cent 6 per cent 4 per cent

(b)  ns (d)  sn

4. The stator of a three-phase induction motor is lami­ nated to

(a) (b) (c) (d)

Reduce eddy current losses Reduce copper losses in the stator winding Reduce hysteresis losses All of the above

5. In a three-phase induction motor iron losses occur in

(a) (b) (c) (d)

7. The efficiency of a three-phase induction motor as compared to that of a transformer is (a) Lesser (c) Higher (e) Comparable

(b)  Much less (d)  Much higher

8. The nature of the PF of a three-phase induction motor is

3. The rotating field of the rotor rotates relative to the ­stator core at a speed equal to (a) sns (c) n

(b)  20 per cent (d)  25 per cent

Stator winding Rotor winding Stator core and teeth Rotor core and teeth

ANSWERS (MCQ) 1. (b)  2. (b)  3. (b)  4. (a)  5. (c)  6. (d)  7. (a)  8. (e)



(a) (b) (c) (d) (e) (f)

Leading Unity Lagging May be leading or lagging Always lagging Always leading

9. The slip of a three-phase induction motor under blocked rotor test is (a) 1.0 (c) Zero

(b)  0.5 (d)  0.2

10. The starting torque of three-phase squirrel cage induction motor is

(a) (b) (c) (d)

High Zero Low Equal to full-load torque

9. (a)  10. (c).

CON V E N TI O NA L Q UE S TI O NS ( C Q ) 1. What is meant by a doubly excited motor? 2. How does a uniform strength of a rotating magnetic field induce voltage in a.c. induction motor rotor? 3. How do voltages that are induced in rotor produce a rotating magnetic field? 4. What is slip in an induction motor? 5. Why must some slip be present for motor action? 6. Why does running an induction motor unloaded enable the rotational losses to be determined?

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7. What is the utility of the blocked rotor test? 8. Why is it desirable to know the resistance of the stator winding? 9. Why is induction motor rotor current related to slip? 10. What is the difference between gross developed torque and net output torque? 11. Why is maximum torque called breakdown torque? 12. What is the synchronous speed of an induction motor with six poles operating on 60 HZ?

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804  Electrical Technology 13. What is the synchronous speed of an induction motor with four poles operating on 400 HZ? 14. An induction motor operates at 4.45 per cent slip and has four poles. What is its r.p.m. on 60 HZ? 15. A three-phase induction motor draws 4.5 A from its lines at 230 V line-to-line at a power factor of 0.153 while running at no load. Its d.c. resistances line to line between two phases of the stator is 1.863 Ω. What is its rotational loss? ANSWERS (CQ) 12. 1200 r.p.m.  13. 12000 r.p.m.  14. 1720 r.p.m.  15. 204 W

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16. An induction motor is tested in the blocked – rotor test. Its rated line current of 8.5 A is drawn when the line voltage is 16.6 V and the total voltage is 48.8 W. Under these conditions what is (a) The equivalent resistance reflected to the stator per phase? (b) The equivalent impedance per phase? (c) The equivalent inductive reactance per phase?

16. (a) 0.255 Ω (b) 1.13 Ω (c) 1.11 Ω.

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Induction Motors (Single Phase) OBJECTIVES

41 Shading coils

Squirrel-cage rotor

Shading coils

In this chapter you will learn about:  Phase-splitting circuits  Pole–speed relationship  Single-phase induction motor  The shaded-pole motor  The universal motor  The synchronous motor  Getting the motor started  Centrifugal switch  The capacitor start split-phase motor  Resistance start split-phase motor  Quadrature windings: running and starting  Torque-speed characteristic  Two-value capacitor motors  Permanent-split capacitor motors  Reversing single-phase induction motors  Dual-voltage operation

Magnetic core Magnetic core Field coil

Field coil

Main coils

To line

Squirrel-cage rotor

Shading coills

Single-phase induction motors

41.1 INTRODUCTION Single-phase induction motors are used extensively in industrial, commercial, and domestic applications. They are used in clocks, fans, blowers, pumps, washing machines, and machine tools, as well as range in size from a fraction of a horse power (HP) to about 15 HP. Large single-phase induction motors are split-phase machines that have two separate windings phy­sically displaced by ninety electrical degrees and phase-splitting circuits are those that cause the current and associated flux of one winding to lag or lead the current and associated flux of the other winding (Figure 41.1). The net effect is the production of a rotating magnetic field that sweeps a squirrel-cage rotor developing induction motor action. Smaller single-phase induction motors use a much simpler device called a shading coil to provide the phase-splitting effect.

Figure 41.1 Single-Phase Induction Motor (a) The Squirrel-cage Rotor (b) Rotor and Starting Switch of a Split-phase Motor

806  Electrical Technology Phase-splitting circuits are also used to operate three-phase induction motors from a single-phase source. This enables larger motors to be operated in isolated areas where three-phase sources are not available. Note: Generally, the term small motor means a motor of less than 1 HP, i.e. fractional HP motor.

41.2  CLASSES OF INDUCTION MOTORS There are four basic classes of single-phase motors that are used roughly in equal quantities, which are as follows: 1. Single-phase induction motors are used for personal and small business tasks; furnace oil burner pumps, hot water circulators, or hot air circulators are some of their typical uses. Refrigerator compressors and power tools, such as lathes and bench-mounted circular saws are also powered with induction motors. Their power outputs usually range from about 1/6 HP (0.125 kW) up to ¾ HP (0.560 kW), although higher ranges are also used. Four-pole motors are usually used, with lesser quantities of two and six poles. The pole–speed relationship is similar to that of two threephase motors, because the operation of the motor is essentially the same. S = 120 f / p r.p.m.  or  ω = 4π f / p rad/sec (41.1) 2. Shaded-pole motors, as shown in Figure 41.2, are used in smaller sizes for quiet, low-cost applications. Typical uses are in small fans and blowers with less power ratings than those described in (1). There is, of course, an overlap in use. The shaded-pole motor is simple and reliable but has a low starting torque and efficiency relative to that of the induction motor.

Figure 41.2  Shaded-pole Motor (the Copper Strap is Necessary for Starting the Motor) 3. The universal motor closely resembles a d.c. series motor and, as its name implies, will operate on any a.c. frequency or on direct current. They are widely used because they can develop very high speed while loaded and very high power for their size. Any service that requires a speed beyond that possible with a two-pole induction motor, where S = 3600 r.p.m., is a normal use for a universal motor.

SInduction Motors (Single Phase)  807

4. The last basic type of single-phase motor is the synchronous motor. The single-phase synchronous motor, as shown in Figure 41.3, is one of the various forms of magnetized but unwound motors. Consequently, they do not have the ability to control power factor. The exact speed relation of a synchronous motor is used in electric clocks and various cycle-timing devices. These motors are built in the smallest of sizes, where the power output is only a few Watts.

Figure 41.3  Synchronous Motor The relative quantities of these four basic types of single-phase motors are almost evenly distributed on a one-for-one basis in modern homes. The single-phase induction motor usually has a very high power rating compared with the other types mentioned here; they collectively develop as much power as the aggregate of the other three put together. The split-phase motors employ two separate windings having different reactance resistance ratios. The current reaches its maximum with high-reactance winding at a later time and the rotor experiences a shift in magnetic field, providing the necessary starting torque. When the motor is nearly up to speed, the high-resistance winding is disconnected by a centrifugal switch. The capacitor motor shown in Figure 41.4 employs a capacitor in series with an auxiliary winding to provide the necessary phase shift. For improved performance, two capacitors are used. The larger capacitor provides good starting torque and it is then switched out by a centrifugal switch. The smaller one remains in the circuit to provide better operating efficiency and power factor.

Figure 41.4  Single-phase Induction Motors (a) Shaded Pole (b) Split Phase (c) Capcitor 41.3  GETTING THE ROTOR STARTED The problem in single-phase induction motor design is to get the rotor started. There are several ingenious methods of doing this. In the shaded-pole motor, Figure 41.4(a), a heavy copper coil is wound around one half of each salient stator pole. Induced circuits in the shorted turn delay the build-up of magnetic flux in that region of the pole. The magnetic flux vector appears to shift as a function of time and the rotor experiences the effect of a partially rotating field.

808  Electrical Technology The running winding consists of insulated copper wire. It is placed at the bottom of the stator slots. The wire size in the starter winding is smaller than that of the running winding. These coils are placed on top of the running winding coils in the stator slots closest to the rotor. Both the starting and running windings are connected in parallel to the single-phase line when the motor is started. After the motor accelerates to speed (approximately two-thirds to three quarter of the rated speed), the starting winding is disconnected automatically from the line by means of a centrifugal switch. The rotor for the split-phase motor is similar to that of the squirrel-cage induction motor, that is, the rotor consists of a cylindrical core, assembled from steel laminations. Copper bars are mounted near the surface of the rotor. The bars are brazed or welded to two copper endings. In some motors, the rotor is a one-piece cast aluminium unit. The rotor fans are a part of the squirrel-cage rotor assembly. These rotor fans maintain air circulation through the motor to prevent large increase in the temperature of the rotor windings. The centrifugal switch is mounted inside the motor as shown in Figure 41.5, and disconnects the starting winding after the rotor reaches a predetermined speed. The switch consists of a stationary part and rotating part. The stationary part is mounted on one of the end shields and has two contacts Figure 41.5 Operation of a Centri­ that act like a single-pole single-throw switch. The rotating part of the cenfugal Switch trifugal switch is mounted on the rotor. A simple diagram illustrating the operation of the centrifugal switch is given in Figure 41.5. When the rotor is at a standstill, the pressure of the spring on the fibre ring of End-bell the rotating part keeps the contacts closed. When the rotor reaches approximately three-quarters of the rated speed, the Switch centrifugal action of the rotor causes the rotor to release its pressure on the fibre ring and the contacts open. As a result, Centrifugal the starting winding is disconnected from the line. mechanism

41.4 CONSTRUCTION The split-phase induction motor consists of a stator, a rotor, a centrifugal switch located inside the motor (Figure 41.6), two end shields housing the bearings that support the rotor shaft and a cast steel frame into which the stator core is pressed. The two end shields are bolted to the cast steel frame. The bearings housed in the end shields keep the rotor centred within the stator so that it rotates with a minimum of friction and without striking Figure 41.6  Cut-away View of a Split-Phase Motor or rubbing the stator core. The stator for a split-phase motor consists of two windings held in place in the slots of a laminated steel core. The two windings consist of insulated coils distributed and connected to make up two windings spaced 90o apart. One winding is the running winding and the second one is the starting winding.

41.5  PHASE SPLITTING Initiating a rotating magnetic field from a single-phase source, without resorting to mechanical means, requires the use of two stator windings and a phase splitting circuit. The physical layout of the windings for an elementary two-pole split-phase motor is shown in Figure 41.7(a), and the corresponding equivalent circuit diagram of the motor is shown in Figure 41.7(b). The main winding supplies the direct-axis flux (φd) and an auxiliary winding displaced at ninety electrical degrees from the main winding supplies the quadrature flux (φq).The auxiliary winding is also called the starting winding. The phase splitter is connected in series with the auxiliary winding, causing the current in the auxiliary winding to be out of phase with the current in the main winding. As the magnetic field due to a current is in phase with the current that produces it, the quadrature field and the main field will be out of phase, resulting in a rotating flux and induction motor action.

SInduction Motors (Single Phase)  809

Figure 41.7 (a) Elementary Two-pole Single-phase Motor with Phase S ­ plitter (b) Equivalent Circuit Diagram Phase splitting may be accomplished through the use of capacitance or resistance. If accomplished through the use of capacitance, the motor is called a capacitance-start split-phase motor; if it is a­ ccomplished through the use of resistance, it is called a resistance-start split-phase motor. Regardless of the means used to start the rotor turning (be it phasesplitting or mechanical action), once it starts turning, self-excitation will maintain the quadrature field, and the auxiliary winding with its phase splitter may be disconnected.

41.6  LOCKED-ROTOR TORQUE The locked-rotor torque of a split-phase motor is proportional to the magnitudes of the locked-rotor current in each winding times the sum of the angle of phase displacement between the two currents. Expressed mathematically,

Tlr = k sp ⋅ I mw ⋅ I aw sin α

α = θi , mw − θi , aw



where, ksp = machine constant, split-phase motor

Iaw = current in auxiliary winding (A)



Imw = current in main winding (A)



qi, aw = phase angle of current in auxiliary winding



qi, mw = phase angle of current in main winding



α = phase displacement between Iaw and Imw.



(41.2) (41.3)

810  Electrical Technology Example 41.1 The main and auxiliary windings of a hypothetical 120 V, 60 Hz, split-phase motor have the following locked-rotor parameters: Rmw = 2.00 Ω  Xmw = 3.50 Ω Raw = 9.15 Ω  Xaw = 8.40 Ω The motor is connected to a 120 V, 60 Hz system. Determine the (1) locked-rotor current in each winding; (2) phase displacement angle between the two currents; (3) locked-rotor torque in terms of the machine constant; (4) external resistance required with the auxiliary winding to obtain a 30º phase displacement between the two currents; (5) lockedrotor torque for the conditions in (3); and (6) per cent increase in torque due to the addition of external resistance. Solution: The circuit for the original conditions is shown in Figure 41.8(a)   Zmw = 2.00 + j 3.50 = 4.03311 − 60.2551° Ω    Zaw = 9.15 + j 8.40 = 12.4211 − 42.5530° Ω

1.    

I mw =

120 0° = 29.7688 4.0311 60.2551°

− 60.2551° A

I aw =

120 0° = 9.6110 12.4211 42.5530°

− 42.6° A

α = θ i , mw − θ i , aw = −60.2551 − (−42.5530) = 17.7021° 2.   = 17.7° 3. Tlr = k sp I mw I aw sin α = k × 29.7688 × 9.6610 × sin 17.7021 = 87.45k sp 4. The circuit for the new condition, with a resistor in series with the auxiliary winding, is shown in Figure 41.8(b). A phasor diagram showing the respective currents for the old condition and the desired location of the new, auxiliary winding current I ′aw is shown in Figure 41.8(c). The required phase angle for I ′aw is as follows:

θi, aw = −60.2551° + 30° = −30.2551° Applying Ohm’s law to the auxiliary branch in Figure 41.8(b), we obtain I ′aw =

VT Z ′aw

= I ′aw

− 30.2551° =

VT

0°

Z ′aw θ ′z , aw

θ ′z , aw = 30.2551° From the impedance diagram for the new auxiliary circuit branch in Figure 41.8(d), we obtain X aw ′ = tan θ aw R aw + R x Rx = =

X aw − R aw ′ ) tan (θ aw 8.40 − 9.155 = 5.25 Ω tan ( 30.2551° )

V ′ = T= I aw 5.   Z′

12.0 0 ° = 7.1979° − 30.2551° A 9 15 + 5.2508 + j8.40 . aw Tlr = k sp I mw I aw sin α = k sp × 29.7668 × 7.1979 × sin 30° = 107.1 k sp

SInduction Motors (Single Phase)  811

Figure 41.8 For Example 41.1(a) Original Circuit (b) Modified Circuit (c) Phase Diagram for Determining the Required Phase Angle of Auxiliary Current for New ­Conditions (d) Impedance Diagram for the New Auxiliary-Circuit Branch

6.

107.1− 87.45 × 100 = 22.5% increase 87.45

Note: The added resistance in the auxiliary winding circuit decreased the auxiliary winding current, but increased the locked-rotor torque. Example 41.2 Carry out a graphical analysis for Example 41.1 Solution: Because only the auxiliary winding has series-connected elements to provide phase splitting, the current in the main winding may be assumed to be constant, permitting Equation 41.2 to be written as (41.4) Tlr α I aw sin α  Graphs of Іaw, µ1 and (Іaw sin α) as a total resistance of the auxiliary winding circuit in Example 41.4, as Rx is increased from 0 Ω to 20 Ω, are shown in Figure 41.9. Note the following: 1. The current in the auxiliary winding decreases with increasing resistance. 2. Angle α increases with increasing resistance. 3. The locked-rotor torque Iaw sin α1 reaches a peak value with an auxiliary circuit resistance of ­appr­oximately 14.2 Ω and decreases with increase in resistance. Note: For every split-phase motor there is an optimum value of auxiliary circuit resistance that will maximize the locked rotor torque. The phase displacement for this optimum value of resistance is generally between 25º and 30º.

41.7  RESISTANCE-START SPLIT-PHASE MOTORS The circuit diagram for a general-purpose resistance-start split-phase motor is shown in Figure 41.10(a). The auxiliary winding is wound with a smaller diameter than the main winding, causing the auxiliary winding to have a higher ratio of

812  Electrical Technology

Figure 41.9 Graphs of Auxiliary Winding Current, Phase-displacement Angle a, and Locked-rotor Torque Represented by Iaw Sin a, for the Split-phase Motor in Example 41.1 resistance to reactance than the main winding. The switch with auxiliary circuit is a magnetic relay, a solid-state switch, or a centrifugally operated switch. The centrifugally operated switch, as shown in Figure 41.10(a), is closed when the motor is at rest and open when the rotor is at 75–80 per cent synchronous speed. A solid-state switch, called a triac, is shown with broken lines in Figure 41.10(a); the switch closes when starting and is set to open at approximately 75 per cent synchronous speed. A magnetic relay (not shown) is closed by a high motor-staring current, and springs open when the acceleration of the motor reduces the current to approximately 80 per cent of the locked-rotor current. A representative phasor diagram for the motor (when starting) is shown in Figure 41.10(b).

SInduction Motors (Single Phase)  813

A typical torque-speed characteristic for a resistance-start split-phase motor is shown Figure 41.10(c). This motor is adaptable to loads such as centrifugal pumps, oil burners, blowers and other loads of ­similar characteristics that require moderate torques and constant speed. This motor offers no means for speed control from a fixed frequency source other than that obtained by recounting for different pole arrangements.

Figure 41.10 Resistance-start Split-phase Motor (a) Circuit Diagram (b) Phase Diagram (c) Torque-speed Characteristic

41.8  CAPACITOR-START SPLIT-PHASE MOTORS A capacitor-start split-phase motor develops a much larger (Іaw, sin α), and hence a much larger locked-rotor torque, than does the resistance-start split-phase motor. The value of capacitance that produces the greatest locked-rotor torque in a capacitor-start split-phase motor causes a phase-displacement angle α of between 75° and 88° compared with the 25° and 33° phase-displacement angle of the resistance-start split-phase motor. The circuit diagram and phase relationships for the capacitor-start split-phase motor are shown in Figure [41.11(a, b)], respectively. A typical torque-speed characteristic for the motor is shown in Figure 41.11(c). The starting curve shows the motor characteristics, with both the auxiliary and main windings energized. The running curve shows the characteristic behaviour after the auxiliary winding is disconnected. A comparison of the characteristic with that of the resistance start split-phase motor in Figure 41.10(c) shows that the running characteristics of both machines are essentially the same. The significant difference between the two machines is the starting torque; about 130 per cent rated for the resistance-starts split-phase motor and 300 per cent rated for the capacitor-start split-phase motor. The high starting torque and good speed regulation of the capacitor-start motor make it well suited for applications in stokers, compressors, reciprocating pumps and other loads of similar characteristics. This motor offers no means for speed control from a fixed frequency source other than that obtainable by reconnecting for a different number of poles.

814  Electrical Technology

Figure 41.11 Capacitor Motors (a) Circuit for Capacitor-start Motor (b) Phase Diagram ­Corresponding to (a) (c) Torque-speed Characteristic for Motor in (a) (d) Permanent-split Capacitor Motor (e) Two-value Capacitor Motor Neither the resistance-start split-phase motor nor the capacitor-start split-phase motor can attain synchronous speed. The rotating flux depends on current in the rotor to produce the quadrature field. As the rotor approaches synchronous speed, the speed-voltage is induced in the rotor, the associated current in the rotor and the quadrature flux approaches zero. Hence, the accelerating torque will become zero at slightly below synchronous speeds. However, permanent-split capacitor motors and two-value capacitor motors are in effect two-phase motors, and at no load could attain synchronous speed.

41.8.1  Permanent-split Capacitor Motors A permanent-split capacitor motor utilizes a permanently connected auxiliary circuit containing a cap­a­citor. There is no switch in the auxiliary circuit, and its operation is smoother and quieter than a capacitor-start or resistance-start motor of the same power rating. The values of capacitance for this type of motor are smaller than the one used in the capacitor-start

SInduction Motors (Single Phase)  815

motor and is a compromise between the best starting and best running performances. The primary field of application for a permanent-split capacitor motor is for shaft-mounted fans used in heaters and for ventilating fans. Its speed may be varied by a tapped or slide-wire autotransformer in the main line, as shown in Figure 41.11(d); by using an external resistor or reactor in series with the main winding or in series with both windings or by adjusting the number of turns in the main winding through the use of taps and a selector switch or by solid-state control.

41.8.2  Two-value Capacitor Motor A two-value capacitor motor, shown in Figure 41.11(e), provides a greater amount of capacitance for starting than for running. This provides a greater locked-rotor torque than is obtainable with the permanent-split capacitor motor, and a reduced capacitance when running results in improved power factor, improved efficiency and higher breakdown torque.

41.9  REVERSING SINGLE-PHASE INDUCTION MOTORS The reversing of the direction of rotation of the motor is accomplished by stopping the machine, interchanging the leads to the auxiliary circuit and then restarting. This reverses the quadrature-axis flux, causing flux rotation to be in the opposite direction. This is illustrated in Figure 41.12.

Figure 41.12  Reversing the Direction of Rotation of a Split-Phase Induction Motor

41.10  DUAL-VOLTAGE OPERATION Single-phase motors often have dual-voltage ratings. To obtain these ratings, the running winding consists of two sections. Each section of the winding is rated. With reference to Figure 41.13, the motor is rated at 115/230 V. One section of the running winding is marked T1 and T2 and the other section is marked T3 and T4. If the motor is to be operated on 230 V, the two 115 V windings are connected in series across the 230 V line. If the motor is to be operated on 115 V, then two windings are connected in parallel across the 115 V line. This is illustrated in Figure 41.13. The starting winding, however, consists of only one 115 V winding. The leads of the starting winding are generally marked T5 and T6. If the motor is to be operated on 115 V, both sections of running winding are connected in parallel with the starting winding, as shown in Figure 41.13(a). For 230 V operation, the connection jumpers are changed with the terminal box so that the two 115 V sections of the running winding are connected in series across the 230 V line. This is shown in Figure 41.13(b). The 115 V starting winding is connected in parallel with one section of the running winding. If the voltage drop across this section of the running winding is 115 V, then the voltage across the starting winding is also 115 V. Some dual-voltage split-phase motors, as shown in Figure 41.14, have a starting winding with two sections and a running winding with two sections. The running winding sections are marked T1 and T2 for one section and T3 and T4 for the other section. One section of the starting winding is marked T5 and T6 and the other section of the winding is marked T7 and T8. Figure 41.13 shows the winding arrangement for a dual-voltage motor with one starting winding and two running windings. The correct connections for 115 V operation and for 230 V operation are given in the table shown in Figure 41.14.

816  Electrical Technology

Figure 41.13 Dual-voltage Operation of Single-phase Motors: (a) Motor Connected for 115 V (b) Motor Connected for 230 V

Figure 41.14  Winding Arrangement with Two Starting and Two Running Windings Example 41.3 Using the given data for the split-phase motor windings in Example 41.1, determine (1) the capacitor required in series with the auxiliary winding to obtain a 900 phase displacement between the current in the main winding and the current with auxiliary winding at locked rotor; (2) locked-rotor torque in terms of the machine constant. Solution: 1. The winding impedances in Example 41.1 are Zmw = 2.00 + j 3.50 = 4.03311 − 60.2551° Ω Zaw = 9.15 + j 8.40 = 12.4211 − 42.5530° Ω The circuit for original conditions is shown in Figure 41.15(a) 120 0° I mw = = 29.1688 4.0311 60.2551°

− 60 2551° A

120 0° = 9.6610 − 42.5530° A I aw = 2 11 42.5530° 12 4 .   The circuit diagram for the new condition (with a capacitor in series with the auxiliary winding) is shown in Figure 41.15(b), and a phasor diagram showing the respective currents for the original condition and the desired location of the new auxiliarywinding current is shown in Figure 41.15(c). The required phase angle for I ′aw is

θi,′ aw = 90° − 60.26° = 29.74° Applying Ohm’s law to the auxiliary branch in Figure 41.15(b), we get VT 0° Z ′aw = Z ′aw θ z′, aw = ′ 29.74° I aw

SInduction Motors (Single Phase)  817

Figure 41.15 For Example 37.3 (a) Original Circuit (b) Modified Circuit (c) Phasor D ­ iagram for Determining Required Phase Angle of Auxiliary Current for New Conditions, (d) Impedance Diagram for New Auxiliary-circuit Branch ′ = Thus θ z , aw −29.74° From the impedance diagram shown in Figure 41.15(d) for the new auxiliary circuit branch, we get

(

)

tan θ z′, aw =

X aw − X c ; Raw

(

X c = X aw − Raw tan θ z′, aw

)

X c = 8.40 − 9.15 × tan ( −29.74° ) = 13.628 Ω Xc =

1 1 = = 194.6 µ F 2 π f c 2 π × 60 × 13.628

′ = I aw 2.   

120 ∠0° = 11.387 29.74° 9.15 + j 8.40 − j 13.628

Tlr = k sp I mw I aw , siin α = k sp × 29.7688 × 11.387 × sin 90° Tlr = 338.9 k sp Note: The per cent increase in locked-rotor torque obtained by capacitor start in Example 41.3 with respect to the lockedrotor torque obtained by resistor start in Example 41.1 is 338.9 − 107.1 × 100 = 216 per cent 107.1 Example 41.4 Carry out a graphical analysis of Example 41.3.

818  Electrical Technology Graphs of Іaw, α and (Іaw sin α), plotted against Xc for Example 41.3, are shown in Figure 41.16. The current in the auxiliary winding increases and then decreases with increasing capacitive reactance (resonance phenomena); angle α increases with increasing capacitive reactance and the locked-rotor torque, represented by Іaw sin α, increases to some peak value and then decreases with increasing capacitive reactance. Note that for the given winding parameters, the optimum value of capacitive reactance that resulted in a phase displacement angle of approximately 75° produced the greatest locked-rotor torque. A comparison with the Іaw sin α curve for the ­capacitor motor in Figure 41.16(c) with that of the splitphase motor in Figure 41.9(c) shows that with the same windings, phase shifting with capacitance can produces ­significantly greater locked-rotor torques than can phase shifting with resistance.

41.11  SHADED-POLE MOTORS The starting methods employed so far are generally based on the principle of producing a rotating magnetic field to initiate rotor rotation. Split-phase motors employ stators with uniform air gaps with respect to their rotor and stator windings, which are uniformly distributed around the periphery of the stator. The shaded-pole motor, illustrated in Fig­ure 41.17(a), utilizes a short-circuited coil or copper ring, called a shading coil, to provide the starting torque. The shading coil is wound around a part of the pole face and acts as the short-circuited secondary of a transformer. Figure 41.17(a) shows the general construction of a salient two-pole shaded-pole motor. The special pole pieces are made up of laminations, and a short-circuited Figure 41.16 For Example 41.4 Graphs of Auxiliary shading coil is wound around the ­smallest segment of Winding Current, Phase Displacement the pole piece. The shading coil, separated from the Angle α, Locked-Rotor Torque Representmain a.c. field winding, serves to provide a phase-splited by Iaw sin α, for the Capacitor Start ting of the main field flux by delaying the change of flux Motor in Example 41.3 with the smaller segment. As shown in Figure 41.17(b), when the flux with field poles tends to increase, a short-circuit current is induced in the shading coils, which by Lenz’s law opposes the force and flux producing it. Thus, as the flux increases in each field pole, there is a concentration of flux in the main segment of each pole, while the shaded segment opposes the main field flux. At point C, as shown in Figure 41.17(e), the rate of change of flux and current is zero, and no voltage is induced in the shaded coil. Consequently, the flux is uniformly distributed across the poles. When the flux decreases, the current reverses in the shaded coil to maintain the flux in the same direction. The result is that the flux now crowds in the shaded segment of the pole. An examination of Figs. 41.17(b, c, d) will reveal that at intervals b, c, and d, the net effect of the flux distribution in the pole has been to produce a sweeping motion of flux across the pole face representing a clockwise rotation. The flux with the shaded-pole segment is always lagging the flux in the main segment in time as well as in physical space (although a true 90° relation does not exist between them). The result is that a rotating magnetic field is produced, sufficient to cause unbalance in rotor torques (double-revolving field theory) such that the clockwise torque exceeds the counter-clockwise torque (or vice versa) and the rotor always turns in the direction of the rotating field.

SInduction Motors (Single Phase)  819

Figure 41.17  Shaded-pole Motor Construction, Operation and Characteristics The electrical characteristics of shaded-pole motor are shown in the torque-slip curve of Figure 41.17( f ). The staring torques are very small and nominally about 25 per cent of the full-load torque. Rated torque, depending on the horse power occurs nominally at about 10–25 per cent slip. Maximum breakdown torque is slightly higher than rated and occurs at slips between 30 and 40 per cent. Efficiencies vary from 5 to 35 per cent. Split-phase induction motors are manufactured in both fractional and integral HP motor sizes. The shaded-pole motor is usually a small fractional HP motor not exceeding 1/10 HP, but motors up to ¼ HP have been produced. The great advantage of this motor lies in its utter simplicity – a single-phase rotor winding, a cast squirrel-cage rotor and special pole pieces. No centrifugal switches, capacitors, special starting windings or commutators are used.

41.11.1 Reversing Shaded-pole Motors To reverse the direction of rotor rotation, it would be necessary to unbolt the pole structure and reverse it physically. To eliminate such a slow and complicated process, newer techniques have been devised for producing reversible shaded-pole motors. The first of these techniques is to connect the shading coils in series on corresponding shading segments and shortcircuit them through a switch. As shown in Figure 41.18(a), the shading coils on trailing salient pole tips on one side are short-circuited for CW rotation and those on trailing pole tips on the opposite side of the pole are short-circuited for CCW rotation. At no time, however, are both sets of trailing-pole tips short-circuited. The second method is generally used with non-salient pole stators. Two separate distributed w ­ indings, 90o in space with respect to the short-circuited shaded poles, are shown as windings A and B in Figure 41.18(b). When winding A, not shown as distributed, is energized, the flux pattern is in a CW order: winding A, shaded pole A', winding A (in the location of B) and shaded pole B'. When winding B is energized, the flux pattern is CCW. Winding B, shaded coil A', winding B distributed at A and shaded coil B'. The third method, as shown in Figure 41.18(c), also employs a single continuous distributed winding with appropriate taps at the 90° points. When the taps of one set are energized by the double-pole double-throw, switch, the rotor rotates clockwise. When the taps of a second set, displaced by 90° with respect to the shading coils, are energized the motor rotates in the counter-clockwise direction. Note: While the shaded-pole motor using the methods described in Figure 41.18 is a reversible motor, it is not a reversing motor. Once started in a specified direction of rotation, it must be brought to a standstill before the motor direction is reversed.

820  Electrical Technology

Two pairs of shading coils alternately short circuited by switch (main winding not shown)

1 φ a.c. supply Two individual distributed main fields at 90°

Tapped distributed winding method

Figure 41.18  Methods of Reversing Shaded-pole Motors

S UM M A RY 1. The stator of single-phase induction motors has a starting winding and a running winding. 2. The rotor of a single-phase induction motor is similar in construction to the rotor of a two-phase squirrel-cage motor. 3. The centrifugal switch disconnects the starting winding after the rotor reaches a predetermined speed, usually two-third or three-quarter of the rated speed. 4. The motor is called a split-phase motor because it starts like a two-phase motor from a single-phase line. 5. The motor must have both the starting and running winding energized at the instant the motor c­ircuit is closed to create the necessary starting torque. 6. If the mechanical load is too great when a split-phase motor is started, or if the terminal voltage is too low, then the motor may fail to reach the speed required to operate the centrifugal switch. 7. To reverse the direction of rotation of the motor, simply interchange the leads of the starting winding. 8. The speed regulation of a split-phase induction motor is very good. 9. The starting torque of the split-phase motor is comparatively poor. 10. In a capacitor-start induction-run motor, the capacitor provides a higher starting torque than is o­ btainable with a standard split-phase motor.

11. The capacitor limits the starting surge current to a lower value than is developed by the standard splitphase motor. 12. The capacitor is used to improve the starting torque and does not improve the power factor of the motor. 13. The capacitor-start induction-run motor is used in those applications where there are relatively few starts in a short period of time. 14. In a capacitor-start capacitor-run motor, the starting winding and the capacitor are connected in the circuit at all times. 15. The capacitor-start capacitor-run motor has a very good starting torque. 16. The problem in single-phase induction motors is to get the rotor started. 17. The auxiliary winding has a higher ratio of resistance to reactance. 18. For dual-voltage operation, the running winding consists of two sections. 19. The shaded-pole motor utilizes a short-circulated coil or copper ring called a shading coil. 20. The shading coil acts as the short-circuited secondary of a transformer. 21. While a shaded-pole motor is a reversible motor, it is not a reversing motor.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. The starting torque in a single-phase induction motor is

a) High c) Low e) Zero

b) Very high d) Very low

2. The efficiency of a single-phase induction motor as compared to that of a three-phase induction motor of the same power rating is



a) Higher c) Lower

b) Much higher d) Much lower

3. The no-load current of a single-phase induction motor is of the order of

a) 30 per cent c) 60 per cent

b) 45 per cent d) 75 per cent

SInduction Motors (Single Phase)  821

4. The centrifugal switch closes when the rotor reaches approximately

a) b) c) d)

One-fourth of the rated speed Two-third of the rated speed Three-fourth of the rated speed Rated speed

5. Initiating a rotating magnetic field from a single-phase source requires



c) Two stator windings and a phase splitting circuit d) None of the above

6. Shaded-pole motor is a

a) Reversible motor b) Reversing motor c) None of the above

a) Two stator windings b) A phase-splitting circuit

ANSWERS (MCQ) 1. (e)  2. (c)  3. (b)  4. (c)  5. (c)  6. (a).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. What prevents a single-phase induction motor from being self-starting unless it has special starting circuit provisions? 2. Describe the basis of the double-revolving field theory. 3. How does the creation of a second artificial phase enable a single-phase-motor to develop starting torque? 4. How is the required phase shift accomplished in a resistance split-phase motor? 5. What is the function of the centrifugal switch in a single-phase motor? 6. What happens when the centrifugal switch fails to open? 7. How is the required phase shift accomplished in a ­capacitor-start induction motor?

8. What circuit change enables a resistance split-phase or capacitor-start induction motor to be reversed? 9. What advantage does a capacitor-start capacitor-run motor have over a capacitor-start motor? 10. How does an auto transformer enable a capacitor to perform as two different values of capacitor in a capacitor-start capacitor-run motor? 11. What future limits the utility of a permanent-split ­capacitor motor? 1 2. Why is a single-phase induction motor less efficient than a comparable power three-phase i­nduction motor?

42

Specialized Motors OBJECTIVES Rotor

In this chapter you will learn about:  The reluctance principle  Reluctance-start induction motors  Hysteresis motors  Reluctance torque and hysteresis torque   Stepper motors for precise positioning of mechanical systems  Types of stepper motors and their applications  Step angle per input pulse  Permanent magnet stepper motors  Variable-reluctance stepper motors  Hybrid stepper motors  Linear induction motors   Unrolling conventional squirrel-cage induction motor to produce a linear induction motor  Universal motors  Simple problems on the above

Stator cup A

Coil A

Coil B

Stator cup B Output shaft

Permanent magnet stepper motor

42.1 INTRODUCTION A.c. and D D .c. machines and combinations of these machines are used, in general, for the conversion of mechanical energy to electrical energy and vice versa. There are, however, other kinds of dynamos and combinations of dynamos that perform similar energy conversion and are more specialized in nature and in application. Machines, such as reluctance motors and hysteresis motors, are used for timing devices, tape recorders, tachometers, and other such devices with constant speed requirements. They are used extensively in process industries, such as the manmade fibre industry, where many components of the process line must operate in sync. Stepper motors are used in conjunction with pulse-driving circuits for precise positioning of mech­anical systems. They are essential components of disk drives, printers, plotters, and other applications that require step-by-step positioning. Linear induction motors (LIMs) are used to apply mechanical forces and to cause movement in a straight or curved line. They are used in conveyer systems, door openers, aircraft launchers, electromagnetic guns, liquid metal pumps for nuclear reactors, high-speed rail transportation, etc. Universal motors have applications in low-power apparatus, such as vacuum cleaners, small-power tools, and kitchen appliances.

42.2  RELUCTANCE-START INDUCTION MOTOR The reluctance motor is an induction motor with a modified squirrel-cage rotor such as that shown in Figure 42.1. The notches, flats, or barrier slots provide equally spaced areas of high reluctance. The sections of rotor periphery between the high-reluctance areas are called salient poles; the number of salient poles must match the number of slots. The stator winding may be three phase or single phase.

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Figure 42.1  Types of Rotor Laminations Used in Reluctance Motors According to the reluctance principle, mechanical force is exerted on a sample of magnetic material located in a magnetic field. The force tends to act on the material in such a way as to bring the material into the portion of the magnetic field that has the greatest density. If the sample is irregularly shaped, it will tend to be aligned in such a way as to produce minimum magnetic reluctance, and, consequently, maximum flux density. Thus, particles of iron filings are aligned in the presence of a magnetic field parallel to the field direction. The reluctance-start induction motor whose starting is initiated by the reluctance principle is not the same as a nonexcited synchronous motor. The reluctance principle states that where the air gap is small, the self-inductance of the field winding is high, causing the current in the field winding to lag the flux that produced it; conversely, where the air gap is high, the self-inductance is reduced and the current is more exactly in phase with the flux. The mutual air gap flux is delayed, therefore, in the vicinity of air gap, producing a sweeping effect similar to that produced in the shaded-pole motor. Since the fluxes are displaced somewhat in time and also in space, a rotating magnetic field is produced at all field poles at instants t1, t2, and t3 successively, as shown in Figure 42.2.

Figure 42.2  Reluctance-start Induction Motor and Development of Rotating Field The running torque characteristics of the salient-pole reluctance-start induction motors are not as good as those of the non-salient-pole shaded-pole motor. This is evident because, in order for the speed e.m.f. to develop a rotating magnetic field once rotation has been initiated, the air gap must be fairly uniform. Furthermore, similar to the shaded-pole motor, the starting torque of the reluctance-start motor is also poor. Other than reversing the poles on the stator, there is no way of changing the direction of rotation of the reluctance-start induction motor. Operation is always in the direction from high to low air gap, i.e. to maximize the field. The shaded-pole motor is generally preferred over the reluctance-start induction motor since it is less expensive to manufacture, has higher efficiency and better running torque characteristics and it is reversible. Speed control is the same for both the motors.

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824  Electrical Technology

42.3  HYSTERESIS MOTORS Single-phase cylindrical (non-salient-pole) synchronous induction or shaded-pole motors are classified as hysteresis motors. The difference between this motor and the reluctance motor is in (1) the shape of the rotor and (2) the nature of torque produced. The reluctance motor is pulled into synchronism and runs on reluctance torque, whereas the hysteresis motor pulls into synchronism and runs on hysteresis torque. Hysteresis-type laminations are made of hardened high-retentivity steel rather than commercial low retentivity dynamo steel. The stator of the hysteresis motor is the same as that for an induction motor. The rotor, however, consists of a smooth cylinder made of very hard permanent-magnet alloy material and a nonmagnetic support as shown in Figure 42.3(a). Rotating magnetic field produced by phase splitting or a shaded-pole stator induces eddy currents in the steel of the rotor and travel across the two bar paths (Figure 42.3(b)). A high-retentivity steel produces a high hysteresis loss, and an appreciable amount of energy is consumed from the rotating field in reversing the direction of the rotor. At the same time, the rotor magnetic field set up by eddy currents causes the rotor to rotate. A high starting torque is produced as a result of the high rotor resistance (proportional to the hysteresis loss). As the rotor approaches synchronous speed, the frequency of current reversal in the crossbars decreases and the rotor becomes permanently magnetized in one direction as a result of the high retentiveness of the steel rotor. With two field poles, the rotor shown in Figure 42.3(b) develops a speed of 3600 r.p.m. at 60 Hz. The motor runs as a hysteresis motor on hysteresis torque because the rotor is permanently magnetized.

(a)

(b)

Figure 42.3 (a) Rotator for Hysteresis Motor (b) Hysteresis-type Laminations of Hardened High-retentivity Steel

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The principle of hysteresis motor action is further explained using the elementary hysteresis motor shown in Figure 42.4. The magnets represent the stator flux, which serves to induce opposite magnetic polarity in the hardened alloy rotor. With the magnets stationary, as shown in Figure 42.4(a), the magnetic axis of the rotor poles is coincident with the magnetic axis of the stator. Spinning the stator magnets, with the rotor blocked as shown in Figures 42.4(b, c), provides a rotating magnetic field that exerts a torque on the induced magnetic poles of the rotor. As the stator poles rotate, the induced magnetic poles in the rotor constantly reform in new positions, following the rotating flux. Because of hysteresis, the rotor poles always lag the stator poles by angle δh. The constant lag angle results in a constant force of attraction, and hence a constant accelerating torque. Releasing the rotor, and assuming no overload, the constant torque will accelerate the rotor to synchronous speed.

Figure 42.4 Hysteresis Motor Behaviour: (a) Magnets and Rotor Stationary (b) and (c) ­Rotor Blocked and Magnets Rotating The amount of torque produced as a result of this magnetization is not as high as that of a reluctance torque. However, hysteresis torque is extremely steady in both amplitude and phase despite fluctuations in supply voltage; hence, it is widely used in high- quality cassette players, compact disk players, record players, and tape recorders. As reluctance torque can be produced more cheaply than hysteresis torque for the same fractional horse power, high-torque hysteresis motors are more expensive than reluctance synchronous motor of the same rating. Because of their low inertia, smaller single-phase hysteresis motors accelerate to their synchronous speed in a few cycles of input. These motors find great application in timing and clock mechanisms (Figure 42.5), where the synchronous speed (for two poles) is 3600 r.p.m. This speed lends itself quite well to high-torque gear reductions, i.e., 1 r.p.m. for the second hand and/or 1 r.p.m. for the minute hand. Yet another important application of the polyphase hysteresis motor is found in inertial guidance and gyroscope rotors, which require absolutely constant speed as a function of line frequency. The following are some unique features of the hysteresis motor: 1. The constant-hysteresis torque (Figure 42.6) from locked rotor to synchronous speed permits the hysteresis motor to synchronize any load that it can accelerate; no other motor can perform in this manner.

Figure 42.5  Timing Clock

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Figure 42.6 Torque–speed Characteristic of the Hysteresis Motor

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826  Electrical Technology 2. The smooth rotor provides quiet operation. It does not suffer from magnetic pulsations caused by slots and/or salient poles that are present in the rotors of other motors. 3. The relatively high resistance and high reactance of the hysteresis rotor limit the starting current to approximately 150 per cent rated current. This contrasts significantly with the reluctance rotor, whose low reactance and low resistance result in a locked-rotor current of approximately 600 per cent rated current.

42.4  STEPPER MOTORS Stepper motors, also called stepping motors are highly accurate pulse-driven motors that change their angular position in steps, in response to input pulses from digitally controlled systems. Stepper motors are used for precise positioning of mechanical systems and may be used without feedback. Examples of their applications include head positioning in computer disk drives, positioning of carriage, ribbon, point head and paper feed in typewriters, as well as printers, robots, etc. The step angle per input pulse depends on the construction of the stepper motor and the control system used. Stepper motors with a 45° step angle provide a resolution of 360/45 equal to eight steps per revolution; stepper motors with a 1.8 step angle provide a resolution of 360/1.8 equal to 200 steps revolution etc. The total angle travelled by the rotor is equal to the step angle times the number of steps. It can be expressed mathematically as follows: Resolution = steps/rev =

   

360° β 

θ = β×steps

(42.1) (42.2)

where, β = step angle (deg/pulse)    θ = total angle travelled by rotor (degree) The speed of a stepper motor is a function of the step angle and stepping frequency (called the pulse rate). Thus,







n=



b× fp 360 

(42.3)

where, n = shaft speed (rps)   

ƒp = stepping frequency (pulses/s)

Example 42.1 A stepper motor has a 2.0° step angle. Determine (1) resolution, (2) number of steps required for the rotor to make 20.6 revolutions, (3) shaft speed if the stepping frequency is 1800 pulses/s. Solution: 1.

Resolution = steps/rev

 

= 360/2.0=180

2.   θ = β × step  

= 20.6 × 360 = 2.0 × steps

Steps =     n=

3.     

20.6 × 360 = 3708 2.0

β × fp

=

360 = 10 rev /s

2.0 × 1800 360

42.4.1  Types of Stepper Motors In all types of stepper motors, rotation is produced by switching suitably connected windings in some predetermined sequence to produce angular discrete rotation steps that are essentially uniform in magnitude. The three most popular types of stepper motors are as follows:

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1. Variable-reluctance (VR) type, also known as reactive-rotor type. 2. Permanent magnet (PM) type, sometimes called active-rotor type. 3. Hybrid type, a combination of PM and VR.

42.4.2  Variable-reluctance Stepper Motors The toothed stator and toothed rotor of a variable-reluctance stepper motor, as shown in Figure 42.7, are constructed from soft steel that retains very little residual magnetism. Coils wound around the stator teeth provide the magnetic attraction that establishes the rotor position. The reluctance of the magnetic circuit formed by the rotor and stator teeth varies with the angular position of the rotor. Energizing one or more stator coils causes the rotor to step forward, or to step backward, to a position that forms a path of least reluctance with the magnetized stator’s teeth.

Figure 42.7 Variable-reluctance Stepper Motor Showing Different Step Positions ­Corresponding to the Switching Sequence in (f) A simple circuit arrangement for sequencing current to the stator coils is shown in Figure 42.7(f). The eight stator coils are connected in two-coil groups to form four separate circuits called phases. Each phase has its own independent switch. Although shown as mechanical switches in Figure 42.7, in actual practice, switching of phases is accomplished with solidstate control. Figure 42.7(a) illustrates the position of the rotor with SW1 closed, energizing phase A; the r­otor is in a position of minimum reluctance with rotor teeth 1 and 4 aligning with stator teeth 1 and 5, respectively. Closing switch SW2 and opening switch SW1 energize phase B, causing rotor teeth 3 and 6 to align with stator teeth 4 and 8, respectively, as shown in Figure 42.7(b), for an angular step of 15º. Closing switch SW3 and opening switch SW2 energize phase C, causing rotor teeth 2 and 5 to align with stator teeth 3 and 7, respectively, as shown in Figure 42.7(c). As each switch is closed and the preceding one opened, the rotor moves an additional step angle of 15º. The stepping sequence, as shown in Figure 42.7(a, c), follows the sequence of switches repeating 1 through 4, over and over, until the ­desired number of revolutions or a fraction of a revolution is achieved.

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828  Electrical Technology The direction of rotation for 1–4 switching sequence (shown in Figure 42.7) results in clockwise (CW) stepping of the rotor. Reversing the sequence of pulses by closing the switches in the order 4.3.2.1 will cause counter-clockwise stepping. The relationship between step angle and the number of teeth in the rotor and the number of teeth in the stator is

β=

Ns − Nr Ns Nr

× 360



(42.4)

where, β = step angle in space degrees    Ns = number of teeth in stator core   

Nr = number of teeth in rotor core

42.4.3  Permanent-magnet Stepper Motors A simplified diagram of a permanent-magnet stepper motor is shown in Figure 42.8. The rotor shown in Figure 42.8(b) has two toothed sections separated by a permanent magnet. The two sections are offset from each other by one-half of a tooth pitch. The magnet provides opposite polarity to each section, developing north poles in one section and south poles in the other section. Figures 42.8(a, c) show the two end views of the combined rotor and stator. All north poles are on one end and all south poles are on the other end. The stator coils shown in Figures 42.8(a, c) span both rotor sections. An axial view of the assembled stepper motor is shown in Figure 42.8(d).

Figure 42.8 Permanent Magnet Stepper Motor: (a) Stator and South Section of Rotor (b) Rotor (c) Stator and North Section of Rotor (d) Axial View of Assembled Motor

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Each rotor section contributes to the development of torque. In effect, the sections are in parallel. The net effect is that of a five-tooth rotor with a four-tooth stator (in this illustration). The step angle for the stepper in Figure 42.8 is β =

Ns − Nr Ns Nr

× 360 =

4−5 4×5

× 360 = 18° 

(42.5)

The principle of operation of a permanent magnet stepper motor is developed using the circuit diagram, switching table, and the corresponding rotor positions in Figure 42.9. For simplicity, only the south section of the rotor is shown. The rotor positions are keyed to the switching sequence for clockwise rotation; phase A is energized by SWI and phase B is energized by SW2.

Figure 42.9 Circuit Diagram of a Permanent Magnet Stepper Motor with Rotor Positions Keyed to Switching Sequence for Clockwise Rotation

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830  Electrical Technology

42.4.4  Hybrid Stepping Motors The hybrid stepper motor is a combination of the PM and the VR types. Typically, most hybrid stators have eight poles and each pole has between two and four teeth. Two phases are wound on the eight poles, i.e., there are four poles per phase. The rotor always has a permanent magnet along with soft-iron pole structure containing an even number of teeth (typically 18). The PM or stepper motors, along with the stepping angle, is independent of the number of phases and is purely a function of the number of rotor teeth. For each change of stator excitation, the stepping angle is

β=

90° degrees  P

(42.6)

where, P is the number of rotor teeth. Example 42.2 A hybrid stepper motor has fifty variable-reluctance teeth. Calculate the stepping angle in degrees. Solution:

β=

90° 90° = = 1.8° P 50

42.4.5  Comparison of Stepper Motor Types The major advantage of hybrid motors is their small stepping angle. This is important whenever high-resolution angular positioning is required. The torque produced by hybrids is greater than that for VR and/or PM types for a given motor volume. Consequently, whenever high torque and small stepping angles are required and space is limited, the hybrid stepping motor is used. VR stepping motors are chosen for the following two major applications: 1. The VR motor is used whenever the load must be moved by a considerable distance requiring several revolutions of the motor. As the stepping angle is greater, fewer steps and correspondingly fewer excitation changes are needed to reach the required distance. This results in less time to produce the change. 2. The inertia of the V/R motor is lower because it does not carry a permanent magnet. This reduced inertia enables the VR motors to accelerate the load faster and reduces the possibility of overshoot or oscillation at the end of a step. PM stepper motors exhibit the highest inertia and the highest rotational speed because they usually rotate at higher stepping angles. Because of their inherently higher speed, the torque for a given hp rating is lower. Production of PM stepper motors is, therefore, limited to the smallest power ratings.

42.5 LIM The major difference between conventional induction motor (producing rotary motion) and a linear induction motor (producing linear motion) is the difference in their respective air gaps. The rotating induction motor has a closed air gap, whereas the linear induction motor has an open air gap with an entry end and exit end. Figure 42.10(a) shows the cross-section of a conventional squirrel-cage induction motor. Primary conductors are embedded in the stator core and secondary conductors are embedded in the rotor core. The air gap is closed upon itself. If we imagine that the conventional squirrel-cage induction motor in Figure 42.10(a) is unrolled to the left and right, as shown in Figure 42.10(b), we obtain the LIM. In effect, the magnetic rotor core may now be considered as a magnetic strip and the s­ econdary rotor conductors as a conductive strip. The primary conductors embedded in a magnetic flat-slotted bed still continue to produce a moving (gliding) flux as a result of their polyphase currents. If we assume that the primary winding in Figure 42.10(b) is stationary and produces a gliding flux continuously from left to right, the secondary conductive and magnetic strip will also move from left to right, but not at the same speed as the flux. There must be some slip theoretically in order to develop force on the secondary.

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Figure 42.10 Unrolling Conventional Squirrel-cage Induction Motor to Produce a Linear Induction Motor Figure 42.11(a) shows one common form of LIM with a short primary and relatively long secondary magnetic sheet and conductive sheet. In this short primary single-sided linear induction motor (SLIM), the secondary is stationary and the primary is capable of motion. In this mode, SLIM is used for long operating distances, because it would be too expensive to design a full-length primary winding. As the secondary is stationary and fixed, the induced secondary currents produce flux to propel the primary along the conductive strip. This design is typically used in cranes, where the three-phase power is available in the crane cab, and the secondary is a steel I-beam. Figure 42.11(b) shows the short-secondary single-sided SLIM, in which the secondary conductors are embedded in a flat-slotted core. This type of LIM is suitable for limited distances but develops relatively high thrust forces. Figure 42.11(c) shows the double-sided primary LIM with a coreless secondary. The double-primary construction provides a more definite magnetic circuit. It is essentially the design used with a railway car LIM. The secondary sheet shown in Figure 42.11(c) may be either a magnetic or a non-magnetic material.

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832  Electrical Technology

Figure 42.11  Flat Three-phase Linear Induction Motors

42.6  UNIVERSAL MOTORS Universal motor is a name given to a type of motor that can operate either on a.c. or d.c. with about the same speed characteristics. This is a series-connected motor, in which the armature and the field coils are connected in series. An elementary universal motor (Figure 42.12(b, c)) has its rotating part, called the armature, connected in series with the series-field winding. The rotating commutator and stationary brushes constitute a rotary switch that reverses the direction of the current in the armature coil as the coil rotates. The equivalent circuit diagram is shown in Figure 42.12(d). The direction of developed torque, and hence the direction of armature rotation, is independent of polarity of the a.c. source. This is shown in Figure 42.12(b, c) for the respective alternations of the a.c. source. The direction of the mechanical force exerted on each conductor is determined by the flux bushing rule. The torque developed by the universal motor is proportional to the flux density of the series field and the current in the armature conductors. That is,   TDα Bp Ia (42.7) where, TD = developed torque   

Bp = flux density due to current in series-field winding

   Ia = armature current As indicated in Figure 42.12(d), however, the current in the series-field is the armature current. Hence, neglecting magnetic saturation effects, Bp α Ia (42.8) Substituting into Equation (42.7) TD α Ia2

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(42.9)

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Specialized Motors  833 Series-field winding Rotation

F

Conductor B

B A Brush

Commutator bar I

To a.c. source (a) Series-field winding Rotation

N

B A

F

Conductor B

S

Brush

Commutator bar

I + To a.c. source (b) –

Series field

To a.c. source

Armature

(c)

Figure 42.12  Universal Motor and its Equivalent Circuit Hence, the torque developed by a universal motor is approximately proportional to the square of the armature current. Universal motors can develop higher torques, can accelerate to higher speeds, and have a higher power-to-weight ratio than induction motors of the same power rating. Reversing the direction of rotation of a universal motor is achieved by reversing the direction of current in series field or in the armature, but not in both. Speed adjustment is accomplished by using an auto transformer or solid-state control to reduce the voltage applied to the motor; reducing the ­applied voltage reduces the armature current, which reduces the developed torque, and hence reduces the speed. The torque and speed characteristics of the universal motor are essentially the same, whether operating on a.c. or d.c. Furthermore, because of its relatively small dimensions, no load speeds in excess of 12,000 r.p.m. are achieved without damage. Universal motors have applications in vacuum cleaners, portable power tools, and kitchen appliances. Series motors, operating from a 25 Hz single-phase system, are used for traction purposes on some electrified railroads.

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834  Electrical Technology

S UM M A RY 1. The reluctance motor is an induction motor with a modified squirrel cage. 2. Where the air gap is small, the self-inductance of the field winding is high. 3. Where the air gap is high, the self-inductance is reduced. 4. The mutual air gap flux is delayed in the vicinity of the air gap. 5. Operation is always in the direction from high to low air gap. 6. The reluctance motor runs on reluctance torque. 7. The hysteresis motor runs on hysteresis torque. 8. With the magnets stationary, the magnetic axis of the rotor poles is coincident with the magnetic axis of the stator.

9. Spinning the stator magnets with the rotor blocked provides a rotating magnetic field. 10. Hysteresis torque is steady in both amplitude and phase. 11. Stepper motors change their angular position in steps. 12. The total angle travelled by a stepper motor is equal to the step angle multiplied by the number of steps. 13. Hybrid stepper motors have a small stepping angle. 14. VR stepper motors have a greater stepping angle. 15. PM stepper motors have the highest inertia and the highest rotational speed. 16. LIMs produce linear motion. 17. Universal motors can operate either on a.c. or on d.c. 18. The torque developed by a universal motor is approximately proportional to the square of the armature current.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. The synchronous speed of a linear induction motor does not depend on

a) b) c) d)

Supply frequency Width of pole pitch Number of poles Any of the above

2. The secondary of a linear motor normally consists of

a) b) c) d)

Distributed single-phase winding Solid conducting plate Distributed three-phase winding Concentrated single-phase winding

3. Which of the following motors can be run on both a.c. and d.c. supply?

a) b) c) d)

Repulsion motor Reluctance motor Universal motor Synchronous motor

4. Which motor has a rotor with no teeth or winding?

a) b) c) d)

Hysteresis motor Universal motor Split-phase motor Reluctance winding

5. Which of the following applications use a universal motor?

a) b) c) d)

Oil expellers Portable tools Lathe machines Floor-polishing machines

ANSWERS (MCQ) 1. (c) 2. (b) 3. (c) 4. (a) 5. (b) 6. (c) 7. (b) 8. (c) 9. (c)

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6. Which of the following motors do not have a winding on it?

a) Repulsion motor c) Hysteresis motor

b) Reluctance motor d) Universal motor

7. Which motor will make least noise?

a) Shaded-pole motor c) Universal motor

b) Hysteresis motor d) Reluctance motor

8. Which of the following motors is used in mixies?

a) b) c) d)

Hysteresis motor Reluctance motor Universal motor Repulsion motor

9. The direction of rotation of a universal motor can be reversed by reversing the flow of current in

a) Armature winding c) Either (a) or (b)

b) Field winding d) Neither (a) nor (b)

10. For which of these applications is reluctance motor preferred?

a) Electronic shavers b) Lifts and hoists c) Refrigerators d) Signalling and timing devices

11. Which stepper motor has the least stepping angle?

a) VR c) Hybrid

b) PM

12. Which stepping motor has the greatest stepping angle?

a) VR c) Hybrid

b) PM

10. (d) 11. (c) 12. (b).

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CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. Explain the principle of reluctance-motor operation: how does it start, accelerate, and synchronize? 2. Explain the principle of hysteresis motor operation: how does it start, accelerate, and synchronize? 3. Will the overall accuracy of a stepper motor be greater at 100 steps than at 10 steps? Explain. 4. Name and briefly explain the three types of stepper motors.

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5. Explain why any rotary motor principle may have a linear counterpart? 6. Explain the principle of linear-induction-motor operation, and state how a LIM may be reversed? 7. Explain why the torque developed by a universal motor varies as the square of the armature current. 8. How may the speed of a universal motor be adjusted?

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43

Servos and Synchros OBJECTIVES In this chapter you will learn about:  Important characteristics of servos   Angular displacement   Operational amplifier characteristics   Transfer function   Operational amplifier specifications   Different types of d.c. servomotors   A.c. servomotors   Magnetic damping   Types of synchros   Self-synchronous system   Advantages of self synchronous units

Antenna

Gearing

Elevation axis

Transducer

Elevation angle sensing (tracking system)

43.1 INTRODUCTION Servos are closed-loop control systems used to determine the position, velocity, or acceleration of mechanical loads. There are four important characteristics of servos, which are as follows: 1. A servo is actuated by an error; this error is the difference between the desired output and the actual output. 2. A servo’s output power is larger than that available from the input information; for instance, a potentiometer control knob takes only a finger touch for control, but an antenna weighing many kilos may be controlled by this potentiometer as a part of a servo. 3. The power applied to the load is proportional to a combination of the error signal its derivatives and its integrals. 4. Practical servos are stable. Two basic areas of servo performance characteristics are servo accuracy as represented by a number of errors and the time it takes for transients to settle down after a change in command has taken place. A special type of small motor or generator is the synchro. These synchros are motors or generators that use alternating current synchros, which are basically used for the transmission of rotational or angular-position information to remote points. For example, in the rotation of a device, such as a rudder aboard ship or a rotating antenna, there are two synchros: one is used at the rotating device and the other is remotely located where the operator is. Every time the rudder or antenna moves (in this example), the synchro that is attached or geared, also moves. This angular displacement is transmitted by means of wires to the receiving synchro; the receiving synchro indicates the exact angular position of the original device.

43.2  OPERATIONAL AMPLIFIERS With remarkable advances in electronics and integrated circuits (ICs), the requirement to implement design from discrete components has given way to easier and more reliable methods of signal conditioning. Many special circuits and generalpurpose amplifiers are now contained in IC packages, producing a quick solution to signal conditioning problems together with small size, low power consumption, and low cost. In general, the application of ICs requires familiarity with an available line of such devices and their specifications and limitations before they can be applied to a specific problem. Apart from these specialized ICs, there is also a type of amplifier

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that finds wide application as the building block of control system. This device, called an operational amplifier (op amp) has been in existence for many years. It was first constructed from tubes, then from discrete transistors, and now as integrated circuits. Although many lines of op amps with diverse specifications exist from many manufacturers, they all have common characteristics of operation that can be employed in basic designs relating to any general op amp.

43.2.1  Op Amp Characteristics An op amp is a circuit composed of resistors, transistors, diodes, and capacitors. It typically requires connection of bipolar power supplies one, that is both +Vs and –Vs with respect to ground. When considered as a functional element of some larger circuit, however, we need to know only its input and output signals. Hence, the op amps are usually used in a larger circuit using their own schematic symbol, as shown in Figure 43.1. The power supply connections are not shown; only two input terminals and an output terminal is depicted in the figure.

(a)

(b)

Figure 43.1  The Schematic Symbol and Response of an Operational Amplifier One input is labelled with a minus (-) sign and is called the inverting input. The other input is labelled with a plus (+) sign and is called the non-inverting input. The sign labels are part of the symbol and must always be included.

43.2.2  Transfer Function Figure 43.1(b) shows the relationship between the two input terminals and the resulting output voltage. The output voltage, Vout, is plotted against the difference between the two input voltages, (V2 − V1). This input is called the differential input voltage. Figure 43.1(a) shows that there is no ground connection directly in the op amp. However, the voltages shown are with respect to ground. We have applied V1 to the non-inverting input and V2 to the inverting input terminal. When V2 is much large than V1, so that (V2 − V1) is positive, the output is saturated at some negative voltage −Vsat; conversely, when V1 is much larger than V2, the output is saturated at some positive voltage +Vsat. Hence, the terminals are called inverting and non-inverting, respectively. When the voltage on the (–) terminal is more positive, the output voltage is negative (i.e., the sign is inverted). There is a narrow range of differential input voltage, labelled ΔV in Figure 43.1(b), within which the output changes from +saturation to –saturation. For most op amp, this input voltage range is less than (a milli volt) where the saturation voltages are typically on the order of 10 V. Thus, the slope of the transition between saturation levels is very large, typically exceeding 100,000 V/V. Other characteristics of the op amp are: (a) the input impedance is very high (typically exceeding 1 MΩ); (b) the output impedance is very low (typically less than 100 Ω). The device is always used with feedback; such feedback permits implementation of many special relationships between input and output voltage.

43.2.3  Ideal Inverting Amplifier Consider the circuit in Figure 43.2. Resistor, R2, is used to feedback the output to the non-inverting input of the op amp and R1 connects the input voltage Vin to the same point. The common connection is called the summing point.

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Figure 43.2  Inverting Op Amp

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838  Electrical Technology

43.2.4  The Op Amp Inverting Amplifier With no feedback in the (+) grounded, Vin > 0 saturates the output negative and Vin < 0 saturates the output positive. With feedback, the output adjusts to a voltage such that: (a) the summing point voltage is equal to the (+) op amp input level, zero in this case and (b) no current flows through the op amp input terminals because of the assumed infinite input impedance. In this case, the sum of the currents at the summing point must be zero.











І1 + I2 = 0

(43.1)

Vin Vout + =0 R1 R2

(43.2)

where, І1 is the current through R1 І2 is the current through R2 By Ohm’s law we have

From the Eq. 43.2, we can write the circuit response as R2 Vin R1

(43.3)  The circuit in Figure 43.2 is an inverting amplifier with gain R2/R1 that is shifted 180° in phase (inverted) from the input. This device is also an attennator by virtue of making R2 < R1. This example suggests two rules that can be applied to analyze the ideal operation of any op amp circuit. In most cases, such an analysis will provide the circuit transfer function. The design rules are: (a) assume that no current flows through the op amp input terminals – that is the inverting and noninverting terminals and (b) assume that there is no voltage difference between the op amp input terminals – that is V + = V-.

Vout =

43.2.5  Non-ideal Effects Analysis of op amp circuits with non-ideal response is performed by considering the following parameters: 1. Finite open-loop gain: The design rules presented here assume the op amp has infinite open-loop gain. Of course, a real op amp does not have an infinite open loop gain as shown in Figure 43.1(a) and again in Figure 43.3(b). The gain is defined as the slope of the voltage-transfer function: A

∆Vout 2Vsat  (43.4) ∆ (V2 − V1 ) ∆V

For a typical op amp Vsat 10 V and ΔV 100 μv, so AΔ 200,000. 2. Finite input impedance: A real op amp has finite input impedance and consequently a finite voltage across and current through its input terminals. 3. Nonzero output impedance: A real op amp has a non-zero output impedance, although this low-output impedance is typically only a few ohms.

(a)

(b)

Figure 43.3 Non-ideal Input/Output Characteristics of an Op Amp Include Finite Gain, Finite Impedance, and Offsets In most modern applications, these non-ideal effects can be ignored in designing op amp circuits. For example, consider the circuit in Figure 43.3(b) where the finite impedances and gain of the op amp have been included. We can employ standard circuit analysis to find the relationship between input and output voltages for this circuit.

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Servos and Synchros  839

Summing the currents at the summing point, gives I1 + I2 + I3 = 0 Then, each current can be identified in terms of circuit parameters to give

(43.5)

Vin − Vs Vo − Vs Vs − = 0 (43.6)   + Ri R2 Z in Finally, V0 can be related to the op amp gain as  V − Vs  V0 = AVs −  0  Z0 ,  R2  Combining the equations    V0 = −

R2  1  −  Vin (43.7) R1  1 − µ 

Where,

µ=

 Z 0   R2 R2   1 + R  1 + R + Z  in 1 1

  

 Z0   A + R  2

(43.8) 

If we assume that µ is very small compared with unity, then Eq. (43.7) reduces to the ideal case given by Eq. (43.3). Indeed, if typical values for an IC op amp are chosen for a case when R2/R1 = 100, we can show that  Qin or rise if Qout>1, the denominator of Eq. (44.5) becomes essentially G; therefore, the closed-loop gain is 1; this indicates that when open-loop gain is high, the closed-loop gain is unity. For an open-loop gain G 0 is (a) Linear (c) Dynamic

(b)  Non-linear (d)  Time varying

3. Linear systems obey

(a) (b) (c) (d)

Reciprocity principle Principle of super-position Principle of maximum power transfer All of the above

4. If the initial conditions for a system are in erentlyzero, what does it signify?

(a) The system stores energy (b) The system does not store energy (c) The system is working with zero reference input (d) None of the above

5. In a control system, the use of feedback (a) Increases the reliability (b) Eliminates the chances of instability

(c) Reduces the effects of disturbance and noise signals in the forward path (d) Increases the influence of component parameters on system performance

6. Which one of the following effects is not caused by negative feedback?

(a) (b) (c) (d)

Reduction in gain Increase in bandwidth Increase in distortion Reduction in output impedance

7. Damping in a control system is a function of (a) Gain (c) √gain

(b)  1/(gain) (d) 1/V gain

8. In the time domain specification, the time delay is the time required for the response to reach

(a) (b) (c) (d)

75 per cent of the final value 50 per cent of the final value 33 per cent of the final value 25 per cent of the final value

9. How can a steady-state error in a system be reduced

(a) (b) (c) (d)

By increasing the system gain By decreasing the system gain By decreasing the static error constant By increasing the input

ANSWERS (MCQ) 1. (b)  2.(a)  3. (d)  4. (c)  5. (c)  6. (c)  7. (d)  8. (b)  9. (b).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1.  Find the closed-loop gain G’(  jω) and the error gain ∈( jω)/R(  jω) for the following values of G(  jω): (a) 100∠0º (b) 10∠0º (c) 1∠0º (d) 0.5∠0º (e) 100∠–90º (f) 10∠–90º (g) 1∠–90º (h) 0.5∠–90º (i) 1∠–135º (j) 1∠–175º (k) 1∠–180º 2. Find the time integrals of the following functions and state the units of the answers: (a) 7 V from 0 to 35 (b) 8t V, where, t = time from 0 to 2.5 (c) 4e–5t V from 0 to 0.2 S 3. Explain the concept of a closed-loop system. Describe the manner in which it differs from an open-loop system.

4. Describe the essential components of a closed-loop system with which you are familiar. What are the advantages and disadvantages associated with the use of feedback in such a control system? 5. Describe the settling time in a servo system. 6.  Explain the terms under-damped, over-damped and critically damped. 7. Explain the manner in which a transducer can produce a signal proportional to angular displacement. 8. Why don’t power inputs appear in a control system block diagram? 9. What information on system performance is provided by step response? 10. Explain in system terminology how body temperature is regulated? 11. Define transform, transform function and step response.

ANSWERS (CQ) 1. (a) 0.99∠0° 0.0099∠0° (b) 0.91 ∠0°, 0.091∠0° (c) 0.5 ∠0°, 0.5 ∠0° (d) 0.33∠0°, 0.67∠0° (e) 1∠–.57°, 0.01∠+89.43° (f) I∠+84.3° (g) 0.707∠–45°, 0.707∠+45°

(h) 0.45∠–63.4°, 0.9∠+26.6° (i) 1.3∠–67.5°, 1.3∠+67.5 (j) 11.5∠+87.5° (k) –∞, +∞ (2.(a)) 21 V.s, (b) 16 V.s (c) 0.506 V.s.

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45

Inverters and Converters OBJECTIVES In this chapter you will learn about:  The difference between converters and inverters   Various types of converters   Rectification-half-wave, full-wave and bridge   Rectifiers, single-phase and threephase   Uninterrupted power supply (UPS)   Single-phase inverters   Three-phase inverters   Six-step inverters   Different types of converting machinery

Three-phase a.c. input

A.c. / d.c. converter

D.c. / a.c. inverter

A.c. motor

D.c. intermediate circuit Lead washer Insulating washer Spring washer

Metal washer

Insulating washer Copper disc Copper oxide film Bass connection Tin alloy Selenium Nickel plating Steel

Metal spacing washers

Inverters and converters

45.1 INTRODUCTION Despite the increasing use of alternating current (a.c.) for all industrial purposes, there are nevertheless many cases in which the use of direct current (d.c.) is either necessary or advantageous to a degree sufficient to make it preferable. For electric traction on suburban railways, for electrolysis, battery charging, and chemical processes; and for conditions where extensive and economical control of speed control of motors are desired, d.c. is employed. Since economy and the need for standardization have led to the use of a.c. for generation and transmission, some form of converting plant is necessary when bulk supplies of d.c. are to be obtained from an a.c. network. Since about 1920, conversion has become increasingly effected by rectification, and currently the importance of new rotating converter plants is small. However, since large numbers of converters have been in service for a long time, they are still of interest, and in any case can compete with rectifiers for low-voltage outputs. One of the most widely used electronic circuits that is used to convert voltage is the rectifier circuit. Since the rectifier circuit uses diodes to convert a.c. voltage to d.c., it is also called a converter circuit. Power that is supplied to a modern factory is a.c.; hence, it is important to have circuits that convert the a.c. power to d.c. power to operate. Some industrial applications such as a.c. variable-frequency motor drivers and welders will have a section where a.c. voltage is converted to d.c. and yet another section where d.c. voltage must be converted back to a.c. voltage. The circuit that changes d.c. voltage back to a.c. voltage is called an inverter. The inverter circuit is required to change the d.c. voltage back to a.c. voltage because the frequency of commercial a.c. mains is fixed at 50 Hz and the output section of the drive needs to be able to provide the frequency of the output voltage in the range of 0–100 Hz.

45.2  CONVERTING MACHINES There are various designs of converting machines designed as per conditions and they fall into three classes—motor generators, synchronous (or rotary) converters and motor converters.

Inverters and Converters  865

Motor generators consist of an a.c. motor (usually of the synchronous, but sometimes of the induction, type) and a directcoupled d.c. generator. This is the simplest and most obvious converting unit. When a synchronous machine and d.c. machine are combined, the resulting machine is termed a synchronous converter, a type that has been widely developed. Motor converters are a combination of the induction motor-generator and the synchronous converter, where the conversion of energy takes place partly mechanically through the shaft and partly electrically by the interconnection of the motor and converter windings. A comparison between the three machines is provided in Table 45.1. The choice of a converting machine for a given case will be influenced primarily by economic considerations and secondly by technical matters, although the two are, of course, mutually dependent to a great extent. Table 45.1  Comparison of Converting Machines Machine

Advantages

Disadvantages

Induction motor-generator

Self-starting Any frequency Operates directly on moderately high voltages Reliable and simple Wide regulation of d.c. voltage possible Free from reversal of polarity and flashover No limit to d.c. voltage

Low efficiency Low power factor unless  compensated Not readily reversible   (i.e. a.c. to d.c. only)

Synchronous motor-generator

Any frequency Operates directly on moderately high voltages High power factor Wide d.c. voltage regulation Operation reversible Free from reversal of polarity and flashover No limit to d.c. voltage

Low efficiency Requires special starting gear Liable to fall out of step

Synchronous converter

High efficiency High power factor Cheap

Requires step-down transformer D.c. voltage regulation limited Liable to flashover and reversal   of polarity D.c. voltage (1200–1500 V)

Motor converter

Self-starting Any frequency Operates directly on moderately   high voltages Reliable and simple High power factor Wide regulation of d.c. voltage possible Relative freedom from reversal of   polarity and flashover

Low speed, therefore expensive D.c. voltage (1700–2000 V)

45.3  RECTIFIERS: A.C. TO D.C. CONVERSION One of the most widely used electronic circuits to convert voltage is the rectifier circuit. Since the rectifier circuit uses diodes to convert a.c. voltage to d.c., it is called a converter circuit. Power that is supplied to a modern factory is a.c., so it is important to have circuits that can convert the a.c. power to d.c. power, since most solid-state devices require d.c. power to operate.

45.3.1  Single-phase Rectifier Circuits Single-phase rectifier circuits have been used since the advent of vacuum-tube diodes. When vacuum tubes were first introduced to control voltage and current, they required various d.c. power supplies. Since d.c. power supplies originate from a.c. voltage, vacuum-tube diode rectifiers were used to convert a.c. voltage to d.c. voltage. After solid-state devices were developed, one of their first uses was to provide rectification of a.c. voltages to provide the necessary d.c. voltages.

45.3.2  VI Characteristics of Rectifiers Rectifiers can be of copper oxide, selenium, silicon or germanium. The maximum peak inverse voltage (PIV) rating of copper oxide rectifiers is about 2 V and that of selenium rectifiers about 10 V. Copper oxide and selenium rectifiers can

866  Electrical Technology handle only limited amounts of current. The PIV rating of germanium rectifiers is about 300 V and these rectifiers can handle currents of up to 100 mA. The PIV rating of silicon rectifiers is 1000 V and these rectifiers can safely handle currents of up to 500 mA. Silicon and germanium rectifiers are invariably used in modern electronic equipments. The VI characteristic of rectifiers is non-linear as shown in Figure 45.1. Ideal rectifiers should have zero resistance in the forward direction and an infinite resistance in the reverse direction. The forward resistance of practical rectifiers is not zero, it is minimum. Similarly, the reverse resistance of practical rectifiers is not infinite, though it is maximum. To deliver high current, these rectifiers should have a very low forward resistance. The current through the rectifier in the reverse direction, often referred to as leakage current, should be extremely low. Rectifiers and their characteristics, both in the forward and reverse direction, are shown in Figure 45.1.

Figure 45.1  VI Characteristic of Rectifiers

45.3.3  Rectifier Operation Figure 45.2 shows a circuit that uses a single solid-state diode rectifier. This type of power supply uses a transformer to increase or decrease the voltage from a 220-V a.c. supply voltage. The secondary supplies the requisite voltage to the rectifier. A resistor is shown to indicate a typical load. The diagram also shows the typical waveforms for input voltage and rectified voltage. Since only one diode is used, the half-wave in the output waveform occurs when the a.c. input voltage is positive between 0° and 180°. No output voltage will be supplied during the point from 180° to 360°, where the a.c. voltage is negative.

Figure 45.2 Rectifier Power Supply. The Waveforms Show a.c. Voltage Supply and Half-wave d.c. at the Load Resistor. (a) Diode Circuit Converts its a.c. Input to Fluctuating d.c. Output (b) A Single-diode (Half-wave)

Inverters and Converters  867

The average d.c. voltage read with a d.c. voltmeter is called the d.c. voltage for the rectifier. For a single-diode (half-wave) rectifier, it is as follows:

Vd.c.av = where,

VP

π

= 0.318 VP

VP = r.m.s. volts × 1.414

(45.1)

The diode in this circuit will drop approximately by 0.7 V . In larger-power diodes, the drop may be as much as 2 V. Vd.c. av =

VP − 0.7

π



(45.2)

Example 45.1 Calculate Vd.c.av voltage for a single-diode half-wave rectifier that has an input voltage of 220 V a.c. r.m.s. (including the 0.7 V d.c. drop for the diode). Solution: VP = Vr.m.s. × 1.414 = 220 × 1.414 = 310.08 V

Vd.c.av =

VP − 0.7

π

=

310.08 − 0.7

π

=

310.62

π

= 98.83 V

A drawback of the single-diode half-wave rectifier is that it produces only a half-wave d.c. output. If a second diode is added to this circuit and a centre-tapped transformer is used, the output waveform will be two positive half-waves. The first diode provides an output half-wave when the supply voltage is between 0° and 180°, and the second half-wave output is provided by the second diode when the supply voltage is between 180° and 360°. Figure 45.3 shows the electrical diagram of the two-diode full-wave bridge circuit. This diagram also shows the sine wave for the single-phase input voltage and the waveform of the two positive half-waves for the output.

Figure 45.3  (a) Single-phase Rectifiers (b) Full-wave Rectification

868  Electrical Technology Another circuit that provides a full-wave output uses four diodes and a regular transformer without the centre tap. This circuit uses two diodes at a time to rectify each half of the sine wave. Figure 45.4 shows an example of this type of circuit. The input sine wave and the output full wave (two half-cycles) are also depicted.

Figure 45.4 The Current Path of the Positive and Negative Half-cycles of the Sine Wave as it is Rectified Through the Bridge A.c. voltage from the bottom terminal of the transformer is applied to the bridge where the cathode of the diode 1 and the anode of diode 4 are connected and from the top terminal of the transformer where the cathode of diode 2 and the anode of diode 3 are connected. This indicates that the a.c. is connected where the anode of one diode is connected to the cathode of the second diode. The output for the bridge circuit will have its positive d.c. voltage terminal at the point where the cathodes of diode 3 and diode 4 are connected. The negative point of the circuit will be where the anodes of diode 1 and diode 2 are connected. This point is also grounded. When a.c. voltage is applied to the four-diode full-wave bridge rectifier, the positive half of the sine wave is rectified by diodes 1 and 3 and the negative half by diodes 2 and 4, as illustrated in Figure 45.4. The path the electrons would travel through the bridge is also shown. Electron flow is always against the arrows of the diode. In some industrial power supplies, the four-diode full-wave bridge rectifier is drawn slightly differently even though it operates exactly as in the previous circuit. Figure 45.5 shows an example of the full-wave bridge drawn with the diode bridge turned on its side so that it looks like a square rather than a diamond. The bridge is illustrated in this way because the six-diode three-phase bridge rectifier uses a similar pattern. The equation for calculating the d.c. average voltage by the two-diode full-wave bridge rectifier is derived in two steps after the peak secondary voltage is determined. Since this is a centre-tapped transformer, the Vout (peak) can be calculated in either of the two ways: using X1−X2 as the full secondary voltage divided by 2 or using X1-CT as the amount of voltage for half the transformer. It should however be ensured that the voltage drop for each diode is subtracted from the secondary voltage.  Vsecondary (x1 − x2 )  Vout (peak) = 1.414 ×   − 0.7 Vsecondary (x1 − x2 )  (45.3a) 2  Or

Vout (peak) = 1.414 × ( Vsecondary (x1 − CT ) ) − 0.7 V  (45.3b)

Next, the d.c. average voltage is calculated. Since now there are two diodes in the circuit, the output voltage will include voltage from both half-waves, which can be represented as follows: 2V p Vd.c.av =  (45.4)

π

where, Vp is Vout (peak) from either of the previous equations.

Figure 45.5 Electrical Diagram of the Four-diode Bridge where the Diodes are Shown in a Box Formation Rather than as Diamonds. The Circuit Function is Exactly Similar to the Bridge Shown in Figure 45.4

Inverters and Converters  869

Example 45.2 Calculate the Vd.c.av voltage for a two-diode full-wave rectifier that uses a centre-tapped transformer. The secondary voltage is 110 V a.c. r.m.s., as measured between X1 and X2 exactly similar to the single-diode rectifier circuit, and the voltage from X1 to the CT is 55 V. Solution: The secondary peak voltage Vp = 155.54 (X1−X2) needs to be divided by 2 to obtain Vp for the value of voltage from one line to the centre tap (X1-CT), Vp = 77.77 V and this value can be used directly in the equation. The answer is ~49.06 V. The slight difference in voltage is because of the voltage drop of the second diode. Vd.c.av =

2 V p ( X 1 − CT )  − 0.7 2 ( 77.77VD − 0.77 ) = = 49.06 Vd.c. π π

The equation for calculating the d.c. average voltage at the output of the four-diode bridge rectifier is similar to the twodiode full-wave bridge except that the equations for the four-diode bridge must account for a 0.7 V drop in each of the two diodes used to rectify the positive and negative half-cycles. The equation to calculate the Vd.c.av must be calculated in two steps. The first step will use an equation to determine the peak voltage ( Controlling torque (b) = Controlling torque (c) < Controlling torque

(a) Moving coil (c) Hot wire (e) Electrostatic

(b) Moving iron (d) Thermocouple

3. The two control springs are

7. Friction is practically eliminated in

(a) Wound in opposite directions (b) Wound in the same direction (c) Not wound

(a) Moving-coil instruments (b) Taut band suspension instruments (c) Moving-iron instruments

4. The control spring should have a

8. Eddy current damping cannot be employed in

(a) Large number of turns (b) Small number of turns

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(a) Moving-coil instruments (b) Moving-iron instruments   (c) Hot-wire instruments

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936  Electrical Technology 9. No-return springs are used in (a) Moving-coil instruments (b) Moving-iron instruments (c) Electrodynamometer instruments

10. A crossed-coil movement is used in (a) Electrodynamometer instruments (b) Moving-coil instruments (c) Moving-iron instruments

14. In a permanent moving-coil instrument, the deflecting torque is proportional to (a) I2 (b) 1/I (c) I (d) 1/I2 15. The resistance of a voltmeter compared with that of an ammeter is (a) Very low (c) Very high

(b) Equal (d) Equal to twice

11. Which instrument is slow to respond to changes?

16. Ammeters and voltmeters come under the category of

(a) Moving coil (c) Hot wire



(b) Thermocouple (d) Electrostatic

12. Which instrument cannot be adapted to measure current or resistance? (a) Moving coil (b) Electrostatic (c) Moving iron

17.  Dynamometer-type moving-coil instruments can be used to measure power in

13. Damping used in moving-iron instruments is (a) Air friction (c) Fluid friction

(a) Indicating instruments (b) Recording instruments (c) Integrating instruments (d) Standard instruments

(b) Eddy current (d) Any one of these

ANSWERS (MCQ) 1. (b)  2. (b)  3. (a)  4. (a)  5. (c)  6. (a)  7. (b)  8. (b)  9. (c)  10. (a)  11. (b)  12. (b)  13. (a) 

(a) a.c circuits only (c) Both a.c and d.c circuits

(b) d.c circuits only (d) None of these

18. In instruments provided with spring control (a) Td α I   (b) Tc α I   (c) Tc α θ   (d) Td α θ

14. (c)  15. (c)  16. (d)  17. (c)  18. (c).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. What are the different torques in indicating instruments? How are they produced? 2. Derive an expression for the deflecting torque of a moving-coil instrument. 3. What is the difference between an ammeter and a voltmeter? 4. Differentiate between moving-coil and moving-iron instruments. 5. With the help of a suitable sketch, explain in detail the working of a moving-iron repulsion type metre. 6. Explain why some instruments have uniform (linear) scales while others have cramped (nonlinear scales). 7. Compare the moving-coil movement with the movingiron movement.

8. How does a hot-wire instrument work? What are its limitations? 9. How is a taut band suspension movement superior to a moving-coil movement? 10. Compare the merits and demerits of moving-iron instruments and dynamometer-type instruments. Which one is superior and why? 11. A permanent magnet moving-coil instrument has a fullscale deflection of 90º for a current of 2A. Find the current required for a deflection of 30º if the instrument is (1) spring controlled and (2) gravity controlled. 12. A weight of 5 g is used as the controlling weight in a gravity-controlled instrument; find its distance from the spindle if the deflecting torque corresponding to a deflection of 60º is 1.13×10-3 N.m.

ANSWERS (CQ) 11. (1) 1.18 A (2) 1.414 A  12. 26.6 mm

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Ammeters, Voltmeters and Ohmmeters

50

OBJECTIVES In this chapter you will learn about:  Difference between ammeters and voltmeters and how to connect them   Accuracy of meters in terms of full-scale deflection  Requirement of a shunt to bypass excess current   Multi-range ammeters  Calculation of the value of shunt for different currents  Different methods of connecting shunts in multi-range ammeters   Universal shunt and its significance  Calculation of the value of universal shunt  Ammeter loading and how to minimize its effect  Requirement of multiplier to drop excess voltage

  Multi-range voltmeters  Calculation of the value of multipliers for different ranges  Different methods of connecting multipliers in multirange voltmeters  Voltmeter loading and how to minimize its effect   Voltmeter sensitivity in Ω/v   Series-type and shunt-type ohmmeters for the measurement of resistance   Basic requirements of ohmmeters   Multi-range ohmmeters   Measurement of insulation resistance   Electric circuit and movement of the Megger   Ohmmeter scales

Measuring electric parameters

938  Electrical Technology

50.1 INTRODUCTION There is only one path for the flow of current in a series circuit. If this current is to be measured, the ammeter is required to be connected in series with the circuit. Shunts of suitable ohmic resistance are connected across the ammeters to bypass the current excess of the meter’s full-scale deflection (FSD) current and thus to enhance the ammeters’ current-measuring capability. The ohmic value of the shunt is inversely related to the internal resistance of the ammeter. Current measurements are made much less frequently than either voltage or resistance measurements. This is because the circuit has to be physically interrupted to insert the meter. This is shown in Figure 50.1 and the meter can be connected anywhere in the circuit, as shown in Figure 50.2.

(a)

(b)

Figure 50.1 Current Being Measured (Current Must Flow Through Both the Meter and the Load) (a) Pictorial Presentation (b) Schematic Diagram Voltage measurements are the easiest and the most common electrical measurements. They are made with power connected to the circuit. Figure 50.3 shows the circuit connections for measuring voltage. The switch is in the closed position. A load has voltage across it only when the current is flowing through it. When resistance is measured, the ohms-adjust control is rotated until the meter indicates zero on the ohms scale and the test lead tips are touching each other. The ohms-adjust control must be adjusted for each range of the ohms function and whenever the range is changed. The ohmmeter function uses a cell, battery or power supply inside the meter housing. It has its own source of energy. Therefore, any other energy source must be disconnected from any circuit in which the resistance is to be measured. Never measure the resistance of a load when the power is connected to the circuit (see Figure 50.4) as this will damage the ohmmeter.

50.2  SPECIAL FEATURES The special features of meters measuring electrical quantities are: (a) Internal resistance, (b) sensitivity and (c) accuracy. 1. Internal resistance: The internal resistance of meters depends on the thickness of the wire and the number of turns (length) of the wire. The thicker the wire the lesser the resistance and vice versa. The longer the wire or the more the number of turns, the higher the resistance. Internal resistance of the moving-coil meters is far less than that of the moving-iron meters. 2. Sensitivity: The maximum amount of current that a meter can safely withstand is called its FSD current. It can also be defined as the amount of current required to move the pointer from the position of rest (zero) to the maximum graduation mark (full scale) on the scale. Moving-coil meters are more sensitive than moving-iron meters. 3. Accuracy: Amount of current read by a meter and the actual current are two different quantities. If both the readings are identical, the accuracy is 100%. Practically, some deviations from accuracy are bound to be there. The basis for accuracy is FSD. As such, accuracy is always expressed in terms of FSD. Percentage error at FSD is minimum and for current less than FSD it increases. Suppose the accuracy of a meter is ± 2 per cent of FSD, it will be ±4 per cent at half FSD, ±8 per cent at one-fourth FSD and so on (see Figure 50.5).

Ammeters, Voltmeters and Ohmmeters  939

Figure 50.2 Ammeter Location: Location of the Ammeter with Respect to the Lamp or the Switch Does Not Change the Amount of Current

(a)

(b)

Figure 50.3 Lamp Voltage Being Measured: Switch Must be Closed (a) Pictorial Presentation (b) Schematic Diagram

940  Electrical Technology

(a)

(b)

Figure 50.4 Measuring Resistance: Power Source is Disconnected from the Load by the Open Switch (a) Pictorial Presentation (b) Schematic Diagram

50.3 AMMETERS

Figure 50.5  Accuracy of Meters

An ammeter is inserted in series with the circuit under test so that the current being measured passes through the instrument. The ammeter resistance must be low compared with the circuit resistance so that the current will not be appreciably reduced and the voltage drop and power loss introduced by the insertion of the meter will not be excessive. Care should be taken neither to overload the meter nor to exceed the FSD. A shunt, as shown in Figure 50.6, is a resistance of very low value connected in parallel with the basic meter movement. They are usually made from materials with very low temperature coefficients. They are generally precision, low-tolerance (±2% or less) resistors.

50.3.1 Ammeter Shunts A shunt diverts most of the current around the meter movement. For example, a 100 µA movement, as shown in Figure 50.6, is converted to a 1 mA with a meter by shunting of 900 µA around the movement. The 1 mA splits at the junction of the meter movement and the shunt, 100 µA passes through the movement and causes FSD of the pointer while the other 900 µA passes through the shunt. Figure 50.6 Ammeter with Shunt: Shunts Extend the Range of the Basic Meter Movement

Figure 50.7  Ammeter Shunts

If an ammeter is required to read more than one range of current, different shunts for each range will have to be connected. These shunts can be connected across the meter by using a switch or by using different marked sockets for each range. If a switch is to be used, one end of all the shunts is strapped together and connected to one end of the meter (see Figure 50.7). The other end of the meter is connected to the pole of the switch. The number of ways of using the switch will depend on the number of ranges to be read by

Ammeters, Voltmeters and Ohmmeters  941

the meter. Shunt resistors, for low ranges of current are connected inside the meter. For extremely higher ranges of currents, shunts are required to be connected externally. Such a shunt resistor is shown in Figure 50.8. Connecting them externally protects the meter from damage and also helps them to dissipate the heat in the atmosphere. In some meters a tapped resistor is used as a shunt. For the lowest range of current, the whole of the tapped resistor is used as a shunt. For higher ranges of current, part of the shunt resistor is not used, whereas the remaining part is effective in bypassing current in excess of its full-scale deflection. For the highest range of current, tapping with the two lowest ohmic values is used as a shunt. Tapped shunts are made of resistance wire with zero temperature coefficient of resistance. For extremely high ranges of current, shunt resistors have almost negligible ohmic value. If a switch is used to connect them in circuit, the contact resistance of the switch will be almost equal to the ohmic value of the shunt resistor. This will introduce errors in meter readings. The only practical solution of the above problem is the use of a universal shunt. For different ranges of current, part of the universal shunt comes in series with it and the remaining part is effective as a shunt for that particular range. This swamps the contact resistance of the switch.

Figure 50.8  Tapped Shunt Resistor

Figure 50.9  Two Types of Shunts for Measurement of Very High Currents

50.3.2  Calculating the Value of Shunts Referring to Figure 50.10, if the meter current is i amperes and the total current is I amperes, the balance of current (I − i) MS flows in the shunt. The joint resistance R Ω of the meter (MΩ) shunted by S Ω is R = . The p.d. across the meter (M + S ) is ν = IR = iM, from which the current ratio is obtained.

942  Electrical Technology

i

i

Figure 50.10  Shunted Ammeter

I M M (M + S ) (M + S ) = = = =n (50.1) R MS S i , where n is the multiplying power of the shunt, i, i.e, it is the number by which the current reading (i) on the meter is multiplied to give the current value of the main circuit (I = ni). From the ratio

(M + S ) =n, S

M + S = nS  and  M = S(n−1) M (50.2)  (n −1) In other words, to reduce the meter current to 1/n of the main current, a shunt having a resistance equal to 1/(n−1) times the meter resistance must be applied. The addition of a shunt reduces the meterr circuit resistance from M to MS/(M+S). This reduction in circuit resistance is equal to or S =

MS M 2 .(50.3) = (M + S ) (M + S ) A compensating resistance of this value must be added in series with the main circuit if it is undesirable to disturb the total circuit resistance. M −

50.3.3  Universal Shunt A universal shunt is used with highly sensitive galvanometers to prevent damage caused by the passage of heavy currents through the movement. This shunt has a number of resistors of different values joined in series. Any resistance value can be selected by a movable contact arm, thus making it possible to vary the multiplying power according to the requirements (see Figure 50.11). The arrangement of a universal shunt is shown in Figure 50.11, which also indicates the values of the various resistors and the multiplying power at each adjustment. The shunt has three terminals. The galvanometer is joined across the pairs (G1 G2) embracing the whole of the resistors, and the external circuit under test is connected between the common terminal (G1 T1) and the movable contact switch (T2). With the switch in the infinity position so that no current flows in the galvanometer. The multiplying power

n=

Meter resistance + shunt resistance (50.4). shunt resistance

In the universal shunt, the sum (M + S) is constant and the multiplying power of the shunt is inversely proportional to the resistance of the shunt.

Ammeters, Voltmeters and Ohmmeters  943

Figure 50.11  Universal Shunt

50.3.4  Calculating the Value of Universal Shunts Let us suppose that currents I1, I2 and I3 are to be measured. When the switch is in position 1, the current flowing through the meter will be Im. This corresponds to the FSD current of the meter. The remaining current (I1 - Im) will flow through RS1. If the internal resistance of the meter is Rm, then I1 I2 I3

Figure 50.12  Universal Shunt Calculating Procedure at Position 1 of the Switch

I m Rm = (I1 − I m ) Rs1 , let

I1 = n1 then Im

I m Rm = I1 Rs1 − I m Rs1 I m Rm + I m Rs1 = I1 Rs1 Rm +Rs1 R I1 = 1+ m I = n1 = R Rs1 m s1



(50.5)

n1 Rs1 = Rs1 +Rm ; n1 Rs1 − Rs1 = Rm ; Rs1 (n1 − 1) = Rm Rs1 =

Rm n1 − 1

n1 times the FSD current can be read by the meter in position 1 of the switch. When the switch is in position 2, Rs1−Rs2 will come in series with the meter, whereas Rs2 will operate as the shunt for this range. The remaining part of I2, i.e. (I2 - Im), will flow through the meter. Rs1 I m (Rs1 − Rs 2 + Rm ) = (I 2 − I m )Rs 2 I m Rs1 − I m Rs 2 + I m Rm = I 2 Rs 2 − I m Rs 2 I m (Rs1 + Rm ) = I 2 Rs2 I 2 Rs1 + Rm I = , let 2 = n2 Im Rs2 Im n2 =

Rs1 + Rm R + Rm ; Rs 2 = s1 n1 Rs 2

Rs1 I m (Rs1 − Rs 2 + Rm ) = (I 2 − I m )Rs 2 I m Rs1 − I m Rs 2 + I m Rm = I 2 Rs 2 − I m Rs 2 944  Electrical Technology

I m (Rs1 + Rm ) = I 2 Rs2 I I 2 Rs1 + Rm = , let 2 = n2 Im Rs2 Im n2 =

Rs1 + Rm R + Rm ; Rs 2 = s1 n1 Rs 2

n2 times the FSD current can be read by the meter in position 2 of the switch.

Figure 50.13  Calculating Procedure at Position 2 of the Switch When the switch is in position 3, Rs1−Rs3 will come in series with the meter, whereas Rs3 will work as the shunt. The remaining I3–Im current will flow through the shunt. I m (Rs1 − Rs 3 + Rm ) = (I 3 − I m )Rs 3 I m Rs1 − I m Rs 3 +I m Rm =I 3 Rs 3 − I m Rs 3 I m Rs1 + I m Rm = I 3 Rs 3 I m (Rs1 + Rm ) = I 3 Rs 3 n3 =

let

I3 = n3 Im

Rs1 + Rm R + Rm ; Rs 3 = s1 Rs 3 n3

n3 times the FSD current can be read by the meter in position 3 of the switch.

Figure 50.14  Calculating Procedure at Position 3 of the Switch Example 50.1 An ammeter has an FSD of 1mA and an internal resistance of 27 Ω. Find the value of shunt resistance required for measuring 10 mA with this meter. Solution: The range of the ammeter has to be multiplied ten times in it, from 1 mA to 10 mA. If 10 mA of current is divided into ten equal parts, only one part is required for FSD as shown in Figure 50.15, while the remaining nine parts have to be shunted. As resistance is inversely proportional to the amount of current, the ohmic value of this shunt will be one-ninth the meterr resistance so that it can divert nine times the FSD current. Figure 50.15  For Example 50.5

Rs =

27 = 3Ω 10 − 1

Ammeters, Voltmeters and Ohmmeters  945

Example 50.2 The FSD current of an ammeter is 50 µA and the meter resistance is 100 Ω. Find the value of shunt resistance required to adopt this meter for measuring 100 mA. Solution: Multiplying factor =

100 × 10 50 × 10

−3

−6

=

100000 50

= 2000

Out of these 2000 parts, only one part (FSD current) will flow through the meter, whereas the remaining 1999 parts have to be shunted as shown in Figure 50.16. 100 Rs = = 0.05 Ω approximately. 1999 Figure 50.16  For Example 50.2

Example 50.3

The FSD current of an ammeter is 1 mA and its internal resistance is 30 Ω. This meter has to be adapted to measure 5 mA, 50 mA and 100 mA. Find the values of the shunts required and in how many ways can these shunt resistors be connected? Solution: 1. If 5 mA is divided into five equal parts, each of 1 mA, one part will flow through the meter and the remaining four parts have to be by passed through the shunt. The ohmic value of this shunt will obviously be one-fourth that of the meter resistance. Rs1 =

30 = 7.5 Ω 5 −1

2. Reasoning on the same grounds Rs2 =

30 = 0.612 Ω 50 − 1

3. By the same logic Rs3 =

30 = 0.303 Ω (100 − 1)

Instead of three discrete shunts as shown in Figure 50.17(a), a single shunt with tappings as shown in Figure 50.17(b) can also be used. Rs1 + Rs 2 + Rs 3 = 7.5 Ω Rs1 + Rs 2 = 0.612 Ω.

Rs 3 = ( 7.5 − 0.612) Ω = 6.888 Ω. Rs1 + Rs 2 = 0.612 Ω.

Rs 2 = ( 0.612 − 0.303) Ω = 0.309 Ω. Figure 50.17(b)

Figure 50.17  Different Ways of Connecting Shunts (For Example 50.3)

946  Electrical Technology Example 50.4 Compute the values of the shunt resistor for the different ranges in a multi-range ammeter shown in Figure 50.18. Solution: Consider the input current applied to the tap point ‘a’ of Figure 50.18. In this position, the FSD of the movement should correspond to an applied current of 300 µA that should pass through the meter movement and 200 µA that should pass through the shunt resistor. Thus, to provide this current division R1 + R 2 + R3 + R 4 = 0.5 R m = 1000 Ω. Considering now the input tapping point ‘b’ for which FSD of the meter movement should correspond to an input of 1 mA Figure 50.18 Universal Current Shunt to Obtain 900 (R2+ R3+ R4 ) =100 (R1 + Rm), Direct Current Ranges in a Multiwhere,    R1 = 700 Ω. meter (for Example 50.4) And by similar calculation    R2 = 200 Ω; R3 = 70 Ω; R4 = 30 Ω. In a practical multi-meter, as many as seven or eight direct ranges are provided by this technique. Example 50.5 A multimeter with a resistance of 10 Ω is connected with a shunt of 0.01 Ω. What will be the current passing through the instrument if it is connected to a circuit in which a current of 1 A is flowing? Solution: I=1A Rm =10 Ω R Rsh = m n −1 10 0.01 = ; n =1001 n −1 1 I I n = ; Im = = Rex = 0.01 Ω Im n 1001 Figure 50.19  For Example 50.5 I m = 0.000999 Α Example 50.6 A moving-coil milliammeter has a resistance of 5 Ω and an FSD of 15 mA. Determine the value of shunt resistance to be used so that the instrument could measure current up to 600 mA at 20 °C. What is the percentage error in the instrument while operating at 40 °C, given that the temperature coefficient of copper is 0.004 per °C at 20 °C. Solution: Rm= 5 Ω;    Im=15 mA. The voltage across the instrument is 5 × 15 × 10-3 = 75 mV. Total current is 600 mA. Current passing through the shunt is (600 – 15) mA = 585 mA. 75 mV Rsh = = 128.205 mΩ 585 mA R m 40 = [1+0.004(40 − 20) ] = 5.4 Ω R sh 40 = 0.128205 [1 + 0.004(40 − 20) ] = 0.1384614 Ω 5.4 × 0.1384614 = 0.135 Ω →  Rm || Rsh. 5.4 + 0.1384614 Voltage (V) = 600×10-3×0.135 = 0.081 V.

Total resistance =

= Im

0.081 = 0.015 A = 15 mA 5.4

Percentage error =

15 − 15 × 100 = 0 15

Ammeters, Voltmeters and Ohmmeters  947

Example 50.7 A moving-coil instrument gives a reading of 25 mA when the potential difference across its terminals is 75 mV. Calculate the shunt resistance for FSD corresponding to 50 A. Solution: 75 mV Meter resistance     Rm = = 3Ω 25 mV Main current is I = 50 A. FSD current is 25 mA = 0.025 A. 50 n= = 2000 0.025 Shunt resistance (Rsh) is

 

Rm 3 3 = = Ω n − 1 2000 − 1 1999

Example 50.8 A moving-coil instrument has a resistance of 2 Ω and it reads up to 250 V when a resistance of 5000 Ω is connected in series with it. Find the current range of the instrument when it is used as an ammeter with the coil connected across a shunt resistance of 2 mΩ. Solution: 250 = 0.04998 A = 49.98 mA Rm= 2 Ω: Im for FSD is 2 + 5000 Rsh= 2×10-3Ω, Ish =

Im Rm R sh

=

49.98 × 10−3 × 2 2 × 10−3

= 49.98 mA

Current range of instrument is Im+ Ish = 0.04998 + 49.98 = 50 A.

50.3.4  Ammeter Loading Ammeters are connected in series and have some inherent internal resistance. Their insertion or removal should not alter the working of the circuit. With reference to Figure 50.20, the current flowing in the circuit before connecting it is 10 A, whereas the current is reduced to 5 A after the ammeter is connected in the circuit. This is an example of extreme loading. There is nothing wrong with the meter but it has not been used properly. If an ammeter of 1 Ω internal resistance is inserted in the previous circuit, current will be reduced to 9.1 A. Now the loading effect has been considerably reduced even though it is still there. Per cent Loading =

Current difference with and without ammeter ×100 Current without ammeter

Figure 50.20  Ammeter Loading If an ammeter with an internal resistance of, say 10 Ω, is to be used, it should be used in a circuit with an ohmic resistance extremely higher than 10 Ω. Suppose this meter is to be used in a circuit having 10 kΩ resistance, 100 = 0.01 A. Current before ammeter is inserted = 10000 = Current after ammeter is inserted

100 = 0.0099 A 10010

948  Electrical Technology Per cent Loading=

0.01 − 0.0099 × 100 = 1 per cent. 0.01

We conclude that 1. Insertion of an ammeter in a circuit alters the circuit conditions to some extent. 2. Ammeter with low internal resistance introduces minimum loading. 3. Ammeter with high internal resistance introduces considerable loading.

50.4 VOLTMETERS The terminals of a voltmeter are connected across the points whose potential difference is to be measured. To avoid drawing excessive current from the supply terminals and lowering the potential at the points to be measured, it is necessary for a voltmeter to possess a high value of resistance compared with that of the circuit being measured. 2 This will also ensure that the power (V  /R) absorbed by the meter is kept to a minimum. While measuring the p.d. across a resistor X (Figure 50.21), the shunting effect of the voltmeter Figure 50.21 Method of Connecting a Voltmeter to Load reduces the resistance across the points A and B, the Provided that the Meter Resistance is High circuit current increases, the potential drop across Compared with that of the Load Resistance X; R increases and the p.d. to be measured across the Shunting effect and the Error will be Small the points A and B falls due to the presence of the voltmeter. The degree of inaccuracy of the measurements depends on the relationship between the voltmeter resistance and the value of X. Voltmeter resistances are expressed in ohms per volt for FSD. For example, a voltmeter whose terminals present a resistance of 50,000 Ω and has FSD corresponding to a reading of 50 V shows a resistance of 10,000 Ω per V. Most voltmeters show a deflection that is proportional to the current (or to the square of the current) flowing in the moving system (the electrostatic type is an exception). If an FSD requires a current of I amperes, as the resistance R Ω of the moving system is constant, it follows that a p.d. of V = IR across the moving system will always produce an FSD. If the ratio of the p.d.-to-current remains constant, the deflection will be proportional to the p.d. (or to the square of the p.d.) and the scale can be calibrated in volts.

50.4.1  Voltmeter Multipliers The range of a voltmeter may be extended by connecting a fixed resistor in series with it. For example, if a voltmeter of 5000 Ω resistance is connected in series with a 5000 Ω resistor and the combination connected across a source of p.d, only one-half of this p.d will appear across the meter terminals and the scale reading will require a multiplication by two to give the correct value. The problem of calculating the series resistance necessary to extend the range of a voltmeter may be regarded from the viewpoint of dropping the excess volts across the series resistance, or alternatively, as a question of adding sufficient series resistance to limit the FSD current to its correct value where the increased p.d appears. Referring to Figure 50.22, suppose that I amperes is the current required to produce an FSD of V volts on a given instrument and the resistance of the meter is R Ω and R = V/1. Now suppose that it is desired to extend the voltmeters range to read n volts (n > 1). To limit the p.d across the meter to V volts, it is necessary to drop the excess voltage, n V − v = V (n − 1) volts, across the resistor X Ω. At a current of I amperes the value of the series resistance is given by X =

V ( n − 1) I

Ω.

Alternatively, to maintain the current at I amperes when the total p.d is n volts, the total resistance will need to be n V/s Ω.

Figure 50.22  Extending the Range of a Voltmeter

Ammeters, Voltmeters and Ohmmeters  949

The voltmeter resistance is equal to V/I so that the value for the added resistance is =

nV V V ( n −1) Ω. − = I I I

Figure 50.23 The Voltmeter Multiplier Extends the Voltage Range of the Basic Meter Movements Example 50.9 What value multiplier is needed to make an 5V voltmeter from a 1 mA, 100 Ω meter movement? Solution: I m = I m A1.Rm = 100 Ω. Vm = 0.001 A × 100 Ω = 0.1 V. Vmt = 5 V - 0.1 V = 4.9 V. Rmt =

4.9 V = 4900 Ω. 0.001 A

Example 50.10 The FSD current of an ammeter is 10 mA and its internal resistance is 1000 Ω. Calculate the value of the multiplier required for making this meter to measure 10 V. Solution: Total voltage to be measured is 10 V. FSD current is 10 mA. 10 Total resistance is Rm (meter) + Rmt (multiplier) = = 1000 Ω . 10 × 10−3 Multiplier resistance = (1000 – 1000) = 0 Ω. In this particular case, no multiplier resistance is required as the meter can directly read 10 V. Example 50.11 The FSD current of an ammeter is 1 mA and its internal resistance is 1000 Ω. If it is to be used as a voltmeter for measuring 1 V, 10 V, 100 V and 1000 V, calculate the value of multiplier resistances. Solution: For 1 V, 1 = 1000 Ω. Total resistance is 1× 10−3 Rm1= (1000 - 1000) = 0 Ω. No external resistance is required. For 10 V, 10 = 10 kΩ. Total resistance is 1× 10−3 Rm2= (10 kΩ - 1 kΩ) = 9 kΩ. For 100 V, Total resistance is 100 kΩ Rm3 = (100 kΩ - 1 kΩ) = 99 kΩ.

950  Electrical Technology For 1000 V, Total resistance is

1000 1× 10−3

= 1000 kΩ Rm4 = (1000 kΩ - 1 kΩ) = 999 kΩ.

Example 50.12 In how many different ways will you connect the multipliers given in Example 50.11? Solution: Instead of connecting different multipliers for different ranges, each one being a high ohms (high voltage) resistance, there is yet another method of connecting them. In this method, as shown in Figure 50.24, the multipliers for different ranges are all connected in series. Rm1 = 0 Ω, Rm2 = 9 kΩ, Rm3 = 99 kΩ - 9 kΩ = 90 kΩ and Rm4 = 999 kΩ - (90 kΩ + 90 kΩ) = 900 kΩ. This is illustrated in Figure 50.25. Example 50.13

Figure 50.24  Ways of Connecting Multipliers

A moving-coil instrument gives an FSD of 10 mA when a potential difference of 10 mV is applied across its terminals. How will you use the instrument to measure (1) currents up to 100 A and (2) voltages up to 500 V?

Ammeters, Voltmeters and Ohmmeters  951

Solution: 1. Let Rsh be the value of the shunt resistance required to extend the range of the meter to 100 A, as shown in Figure 50.25(a). The current flowing through the shunt is 100-(10×10-3)A. As the voltage across the instrument will be the voltage across the shunt, we have Rsh =

10 × 10−3 −3

100 − (10 × 10 )

=

0.01 = 0.0001 Ω 99.99

2. Total voltage across the circuit is 500 V, see Figure 50.25(b). Total resistance of the circuit is

500 10 × 10−3

Resistance of the instrument (Rm) =

= 50 kΩ

10 × 10−3 10 × 10−3

= 1 Ω.

Value of the multiplier resistance is 50 kΩ − 1 Ω = 49,999 Ω.

Figure 50.25  For Example 50.13 Example 50.14 It is required to measure the voltage across a 100 Ω resistor in the circuit as shown in Figure 50.26. Two voltmeters are available for measurement. Voltmeter A has a sensitivity of 90 Ω/V and voltmeter B has a sensitivity of 900 Ω/V. Calculate (1) the reading of each voltmeter and (2) the percentage error in each case. Solution: Refer to Figure 50.26. 150 = 0.5 A. 100 + 200 Voltage drop across 100 Ω = 100 × 0.5 = 50 V. 1. Current in 100 Ω resistance is

Voltmeter A: Its resistance is 50 × 90 = 4500 Ω (Figure 50.27(a)). RT = 200 + Current in the circuit =

4500 × 100 = 297.826 Ω 4500 + 100

150 = 0.504 A 297.826

Reading of voltmeter A is

4500 × 100 × 0.504 = 49.301 V. 4500 + 100

Voltmeter B: Its resistance is 50×900 = 45,000 Ω. (Figure 50.27(b)). RT= 200 + Current in the circuit is

4500 × 100 = 297.778 Ω 4500 + 100

150 = 0.5004 A 297.778

Reading of voltmeter B is

4500 × 100 × 0.5004 = 49.926 V 4500 + 100

Figure 50.26  For Example 50.14

952  Electrical Technology 49.301 − 50 × 100 = 1.397 per cent. 50 49.926 − 50 × 100 = −0.148 per cent. Percentage error in the second case is 50

2. Percentage error in the first case is

Figure 50.27  Solution for Example 50.14

50.4.2  Voltmeter Loading A voltmeter should have an infinite resistance so that its insertion or removal from the circuit does not alter circuit conditions. In Figure 50.28(a), before connecting the voltmeter the voltage drop across each resistance is 50 V, i.e., half of the total voltage. If a voltmeter of 100 kΩ is connected across one of these resistors, the circuit conditions are no longer the same. Now the total resistance becomes 150 kΩ. The voltmeter will now read with the voltmeter but it has not been used in the right place.

150 kW ×100 = 33.3 V. There is nothing wrong 150 kW

Instead of a 100 kΩ resistance voltmeter, if a 2 MΩ resistance voltmeter is used as shown in Figure 50.28(b), the change in voltage with the insertion of the voltmeter will be almost negligible. The voltmeter will now read

95.23 kW × 100 = 48.8 V  195.23 kW

see Figure 50.29(b).

Figure 50.28 Voltmeter Loading Because the Total Resistance in the Circuit has Changed. This is Further Elaborated in Figure 50.29 The change in voltage brought about by the insertion of the voltmeter is called voltmeter loading, because it is basically due to the voltmeter. To conclude

Ammeters, Voltmeters and Ohmmeters  953

1. A change in voltage should not occur by insertion of a voltmeter into the circuit. 2. Voltmeter with a higher Ω/V rating introduces less loading of the circuit. 3. Voltmeter with a low Ω/V rating introduces more loading of the circuit. Voltmeter loading is much more common than ammeter loading. It can occur in series and series-parallel circuits. It does not occur in parallel circuits because the voltmeter is connected across the power source, ie, it is in parallel with all other parts of the circuit. In general, voltmeter loading occurs in high-resistance circuits such as those found in many electronic devices. Manufacturers of electrical and electronic equipment often specify the voltages at various points in their circuits. Usually they specify either the sensitivity or the input resistance of the meter used for measuring these voltages. When using a meter different from the one specified, one must be aware of the possible meter loading. The voltmeter’s input resistance should be 20 times greater than the resistance across which the voltage is to be measured. Under these conditions, the loading will change the resistance of the circuit by less than 5 per cent.

Figure 50.29  Figure 50.29 Further Elaborated Example 50.15 Assume the voltage across R2 in Figure 50.30 was measured with a 2000 Ω/V meter on the 5 V range. How much voltage would the meter indicate? Solution: Input resistance = sensitivity × range

= ( 2000 × 5) = 10 kΩ. 5k × 10k = 3.33 kΩ. 5k + 10k 6 V × 3.33 kΩ = 2.4 V = 5 kΩ + 3.33 kΩ

R2 = VR 2

The measured voltage is 2.4 V.

50.5 OHMMETERS To utilize the moving-coil movement for the measurement of resistance requires the provision of a source. In most multi-meters (multirange, multifunction instruments) this takes the form of one or more cells depending on the magnitude of resistances that can be measured and the number of ranges provided. Two methods are provided for making this type of measurements. In the series arrangement, as shown in Figure 50.31, the unknown resistance is connected in series with the meter movement. FSD current corresponds to the unknown resistance of 0 Ω and it is on the right-hand side of the scale.

Figure 50.30  For Example 50.15

954  Electrical Technology

Figure 50.31 Series Type Ohmmeter Circuit R1 is Adjusted to Provide FSD when the Terminals are Shorted Together. In the Shunt Arrangement Shown in Figure 50.33, the Unknown Resistance is Connected in Parallel with the Meter Movement. Short Circuiting the Meter Terminals Gives 0 Ω on the Left-Hand Side of the Scale and an Open Circuit or Infinite Resistance Gives FSD

Figure 50.32  Shunt-Type Ohmmeter Circuit Example 50.16 Discuss the adoption of a 100 µA, 2000 Ω meter movement for measurement of resistances assuming the source to be a 1.5 V cell. Solution: For the series-type arrangement, the adjustment resistor that must be connected in series, as shown in Figure 50.33, should have a value such that when the meter terminals are short-circuited, the current through the meter will be 100 µA. 1.5 Ra + Rm = = 15, 000 Ω 100 × 10−6

Figure 50.33  For Example 50.16

Ammeters, Voltmeters and Ohmmeters  955

Note: 1. In practice Ra must be adjustable to allow for the changes in the terminal voltage of the battery voltage as it ages. 2. The adjustment resistor could comprise a fixed 10 kΩ resistor and a variable 5 kΩ resistor in series. 3. For both the series and shunt arrangements, the adjustment resistor has the same value for a given meter movement sensitivity and source voltage.

50.6  MEASUREMENT OF INSULATION RESISTANCE In the type of instrument employed for measuring insulation resistance, as shown in Figure 50.34, the moving system comprising both a pressure coil and a current coil is deflected under the action of mutually opposing electromagnetic forces. The force acting upon the pressure coil depends on the current in that coil, which is practically proportional to the p.d. applied to the instrument terminals. The force acting upon the current coil is proportional to the current that flows in it. This is also the current in the external circuit. These two forces urge the moving system in opposite directions and the position taken by the pointer Figure 50.34 ‘MEG’ Insulation Testers (Ohmmeter) depends on the ratio p.d./current in the resistance of the external circuit. The ohmmeter consists of two parts, the indicating system and a small hand-driven permanent magnet-type d.c. generator. The same permanent magnet system is utilized both for the moving system of the meter and for the generator field. The general principle of the meter is illustrated in Figure 50.35.

Figure 50.35  Principle of the Ohmmeter P is the pressure coil and A is the current coil. The generator maintains a constant p.d. across the pressure coil and also across the unknown resistance X connected to the ohmmeter terminals L and E. The same current flows through A and X. The magnetic and electric circuits of the ohmmeter are shown in Figure 50.36. Two powerful bar magnets fitted with pole pieces provide the field for the generator at one end and for the moving system at the other. An e.m.f. is generated in the armature when it is rotated from the external handle through the intermediary of a suitable gearing and a clutch. This clutch, of centrifugal type, is designed to slip at a critical speed of rotation—about 100 r.p.m.—in order that the armature speed and e.m.f. of the generator remain constant. The generator e.m.f. is 500 V in most instruments of this type. The generator commutator consists of four rings, each of two segments with each segment connected to the armature. Four pairs of carbon brushes insulated from one another are used. Constancy of terminal voltage is not in itself essential because the ratio VI is being measured. It is however particularly important when testing circuits containing inductance or capacitance (underground lines, for example) a variable voltage would produce surge currents and produce unsteady readings.

956  Electrical Technology

Figure 50.36  Ohm Meter—Magnetic and Electrical Circuits The moving system illustrated in Figure 50.37 comprises three coils moving about a fixed cylindrical core that is split along its length. The whole system carrying the pointer is supported on jewelled bearings. There is no mechanical controlling torque. Connections to the coils are made by fine phosphor–bronze strip. The current coil A carries the current that flows in the external circuit connected to the meter terminals. Its function is in every way similar to that of the moving coil of a milliammeter. Its current and torques are inversely proportional to the resistance under test. A protective resistor R2 is inserted in series to limit the maximum current value. The pressure coil P with the compensating coil C is arranged approximately at right angles to the current coil. Connected in series, they are energized directly from the generator and provide the controlling torque. A limiting resistor R1 is connected in series with these coils. The compensating coil is fixed to the outer edge of the pressure coils. It is wound in opposite to the pressure coil, the two coils together forming an astatic pair. The combination is largely independent of stray magnetic fields from sources external to the instrument. A specially shaped S pole piece allows the compensating coil to move over it and so to oppose the force on the pressure coil. The two external terminals are marked line (L) and earth (E). A third terminal, the guard terminal or guardning (G), is provided to present a passage through the circuit coil of leakage currents flowing over the surface of the instrument between terminals L and E. The guard terminal is connected Figure 50.37  Movement of Bridge Megger to the negative terminal of the generator. While measuring the insulation resistance of a table, this guard terminal is connected to the insulating material, between conductors and sheath of the cable, to nullify the effect of any leakage current that passes over the open end of the cable. When the instrument is idle, there is no current and no controlling torque on the moving system. The pointer is entirely free and will remain at any position on the scale. With the generator operating, if no connection is made between terminals L and E, i.e., if the external resistance is infinitely high, no current will flow in the current coil. The pressure coil is polarized by current from the generator and will set itself in the position shown in Figure 50.36, i.e., at right angles to the air-gap flux. In this position, the pointer is opposite to the infinity mark on the scale. When a resistance is applied across L and E terminals, a current (inversely proportional to the sum of the external and internal resistances) will flow in the current coil. This coil will tend to set itself at right angles to the flux lines but it is, however, subjected to the opposing torque of the pressure coil. The current coil deflects the static system into a gradually increasing magnetic field and the moving system is held in equilibrium due to the balanced torques. The corresponding resistance value is indicated directly by the pointer against the scale.

Ammeters, Voltmeters and Ohmmeters  957 ms

ms

goh

goh

Me

Me

Megohms ms Thousand oh

Me

Tho

goh

usa

ms

nd

ohm

s

Figure 50.38  Ohmmeter Scales

S UM M A RY 1. Ammeters have a very low resistance and are connected in series. 2. Voltmeters have an extremely high resistance and are connected in parallel. 3. The circuit has to be physically broken for insertion of the ammeter. 4. Shunts enhance the current-measuring capacity of ammeter by bypassing the excess current around the meter movement.

5. In a multi-range ammeter, the number of shunts depends on the number of ranges. 6. Shunts in multi-range ammeters can be inserted into the circuit by different methods. 7. A universal shunt is used with highly sensitive galvanometers. 8. Multipliers enhance the voltage-measuring capacity of voltmeters by dropping excess voltage.

958  Electrical Technology 9. In a multi-range voltmeter, the number of multipliers depends on the number of ranges. 10. Multipliers in multi-range voltmeters can be introduced into the circuit by different methods. 11. Ammeters with high internal resistance introduce considerable loading. 12. Voltmeters with a high Ω/V introduce the least loading. 13. Ohmmeters use one or more cells as a source of energy, depending on the magnitude of resistances to be measured and the number of ranges provided.

14. There are two types of ohmmeters—series and shunt— depending on whether the resistance to be measured is connected in series with or across the meter movement. 15. Ohmmeters comprise an adjustable resistance to compensate for the ageing of the source of power or of components. 16 The insulation tester consists of two parts: the indicating system and a small hand-driven permanent magnet d.c. generator.

M U LT IP LE C H O I C E Q UE S TI O NS ( M C Q ) 1. The internal resistance of the voltmeter is

6. The value of shunt for higher ranges of current is

(a) Zero (c) High

(a) Extremely high (b) Extremely low

(b) Very small (d) Extremely high

2. The shunt used in the milliammeter

7. How many coils are required in the megger



(a) One (c) Three

(a) Will extend the range and increase the meter resistance (b) Will extend the range and decrease the meter resistance (c) Will decrease the range and meter resistance (d) Will decrease the range and increase the meter resistance

3. The range of an ammeter can be extended by using suitable (a) Shunts (b) Multipliers (c) Both (a) and (b)

(b) Two (d) Four

8. In a multi-range voltmeter, the ohmic value of the multiplier (a) Increases with range (b) Decreases with range (c) Is not affected by range

9. In a multi-range ammeter, the ohmic value of the shunt

4. Measurement of resistance requires a suitable (a) Shunt (b) Source (c) Multiplier

(a) Increases with range (b) Decreases with range (c) Remains the same

10. The magnitude of the strength of the source in an ohmmeter

5. Measurement of alternating current/voltage requires a suitable (a) Shunt (b) Rectifier (c) Multiplier

(a) Depends on the magnitude of the unknown resistance (b) Depends on the magnitude of the unknown resistance and range provided (c) Depends on the ranges provided

ANSWERS (MCQ) 1. (d)  2. (b)  3. (a)  4. (b)  5. (b)  6. (b)  7. (b)  8. (a)  9. (b)  10. (b)

CON V E N TI O NA L Q UE S TI O NS ( C Q ) 1. How will you convert a single-range ammeter into a multi-range ammeter? 2. How will you convert a single-range voltmeter into a multi-range voltmeter? 3. What are the factors that decide the ohmic values of shunts and multipliers? 4. What are the different methods of connecting shunts and multipliers in a multi-range instrument.Explain with the help of suitable illustrations?

5. Discuss the limitations of switching arrangement in multi-range multi-function measuring instruments? 6. Explain briefly the following: (a) Shunt, (b) Tapped shunt, (c) Multipliers, (d) Tapped multiplier, (e) Universal shunt 7. Explain the difference between a series-type ohmmeter and a shunt-type ohmmeter? 8. How is the accuracy of the following specified? (a) Ammeter, (b) Voltmeter, (c) Ohmmeters

Ammeters, Voltmeters and Ohmmeters  959

9. What are the advantages and drawbacks of shunt and multipliers? 10. A moving-coil instrument gives FSD when its current coil is connected across 50 µV supply. Calculate the current in the coil of instrument if its resistance is 5 Ω? 11. Find out the value of the resistance of the shunt and power dissipation to extend the range of 1 mA, 80 Ω resistance meter to measure a 1 A current? 12. What is the value of internal resistance of the meter coil that is rated at 50 µA, 200 mV? 13. Determine the value of the multiplier resistor to be connected to make 10 V voltmeter with a coil of instrument that has 1 mA current passing through the coil and a resistance of 100 Ω. 14. A moving-coil instrument gives FSD when connected across 50 mV supply. What modifications are required to measure the following: (1) 100 V potential (2) 10 A current. ANSWERS (CQ) 10. 10 mA 11. 0.08 Ω, 0.07992 W 12. 4000 Ω 13. 9900 Ω

15. A moving-coil ammeter has a resistance of 5 Ω and FSD of 20 mA. Determine the resistance of the shunt required so that the instrument could measure currents up to 500 mA at 20 °C. What is the percentage error with the instrument operating at a temperature of 40 °C? The temperature coefficient of copper is 0.0039 per °C. 16. A moving-coil meter can read up to 1 mA and has a resistance of 0.02 Ω. How could this instrument be adopted to read (i) voltage up to 300 V and (ii) current up to 100 A? 17. A moving-coil voltmeter reading up to 20 mV has a resistance of 2 Ω. How can this instrument be adopted to read voltages up to 300 V?

14. Rm = 19.99 kΩ, Rsh = 0.005 Ω 15. Error = 0.07 16. Rm = 299.98 Ω, Rsh = 0.000202 Ω 17. Rm = 29998 Ω (29. 998 kΩ)

Watt Meters and Energy Meters

51

OBJECTIVES In this chapter you will learn about:  The difference between ampere meters, watt meters, ampere-hour meters and watt-hour meters    Electrodynamic (dynamometer) and inductiontype watt meters   The construction and principle of operation of watt meters   Energy meters—both commutator type and mercury type    The essential parts of induction watt-hour meters   The significance of the different parts   Production of fluxes and driving torque   Reading watt-hour meter   Poly-phase induction watt-hour meters   Errors in energy meters   Different methods of measuring power in three-phase circuits   Simple problems on the above

Watthour meter

51.1 INTRODUCTION The measurement of electricity distributed for commercial purposes is based on the following units: (1) Ampere (the unit of current), (2) volt (the unit of e.m.f.), and (3) the watt (unit of electric power). Ampere hour and watt hour are derived from (1) and (3) , which means, respectively, an ampere of current and a watt of power each maintained for one hour. Since the latter units combine the elements of both rate and time, they are adapted for the measurement of the supply of electricity, which is electrical work or energy. The watt hour is the more useful of the two units because it is a direct measure of electrical work and its multiple, the kilowatt hour (1000 Watt hours) is the unit generally employed for commercial purposes. Ampere-hour meters and watt-hour meters derive their names from the units in which they register. An ampere-hour meter integrates, or adds together, products of current and time without regard to the voltage at which current is supplied. Consequently, an ampere-hour meter is useful only when it is desired to know merely the quantity of electricity, as in storage battery charging or in electroplating. A watt-hour meter integrates, or adds together, units of electrical energy (watts) with respect to time and since watt includes the resultant effect of both voltage and current, watt-hour meters are used universally in the commercial measurement of electrical energy. Electricity can be measured through the different effects it produces. The chemical effect, such as the decomposition of an electrolyte, has been utilized in some ampere-hour meters; the magnetic effects are the basis of all the motor type of meters now in general use, and the heating effects are the basis of hot-wire type of instruments and thermal demand meters.

51.2  WATT METERS A watt meter is a combination of an ammeter and a voltmeter and therefore consists of two coils (Figure 51.1) known as current coil and pressure coil. The operating torque is produced due to the interaction of fluxes because of current in the

Watt Meters and Energy Meters  961

current and pressure coils. The current coil is inserted in series with the line carrying the current to be measured and the pressure coil, in series with a high non-inductive resistance R, is connected across the load or supply terminals.

(a)

(b)

Figure 51.1 (a) Watt Meter Connections: (b) For Clockwise Torque, the Instantaneous Currents with Current Coils and with Voltage Coil Must be in the Same Direction The watt meter gives a reading that is proportional to the current flowing (1) through its current coil (2) p.d. across potential coil and (3) cosine of the phase angle between voltage and current. The watt meter measures the power lost in CC or PC in addition to load power. Normally, the power lost in CC or PC is very small compared with that measured and therefore can be neglected (Figure 51.2). There are two methods of connecting watt meters in the circuit for measurement of power, as shown in Figure 51.3. The circuit in Figure 51.3(a) is used for circuits carrying small currents, whereas the circuit shown in Figure 51.3(b) is used for measurement of power in circuits carrying large currents. There are four types of watt meters namely: (1) Dynamometer-type watt meters, (2) Induction-type watt meters, (3) Electrostatic-type watt meters and (4) Thermal-type watt meters. Of these, the dynamometer and induction type are the most commonly employed.

Voltage coil

Disc

Current coils

Figure 51.2 Induction-type Watt Meter

51.3  DYNAMOMETER-TYPE WATT METER When used as a watt meter, the fixed coil, which is divided into two equal portions to provide a uniform field, is used as a current coil and the moving coil is used as a pressure coil, that is, the fixed coil carries the current flowing through the circuit and the moving coil carries a current proportional to the voltage across the circuit. A high non-inductive resistance is connected in series with the moving coil to limit the current in it. The magnetic fields of the fixed and moving coils react with one another, causing the moving coil to turn about its axis (Figure 51.4). The movement is controlled by hair springs, which also lead the current into and out of the moving element. Damping is provided by light aluminium vanes moving in an air dash pot. Eddy current damping cannot be used, as the introduction of a permanent magnet required for damping will

962  Electrical Technology

Figure 51.3 Watt Meter Connections: (a) Circuit with Small Currents (b) Circuit with Large Currents greatly distort the weak working magnetic field. The knife-edge pointer is fixed to the moving coil spindle and moves over a suitably calibrated mirror-type scale. Let v be the supply voltage, i the load current and R the total resistance of the moving coil circuit (Figure 51.5).

Figure 51.5 Connection Diagram of Watt Meter; Measurement of Power in a.c. Circuits Current through the fixed coil is   If = i Current through the moving coil is Vm = u/R

Figure 51.4 The Dynamometer-type Watt Meter

Deflecting torque,

Td α I f Vm α

iv R

In a d.c. circuit, power is given by the product of voltage and current; hence, the deflecting torque is directly proportional to power. In an a.c. circuit, the instantaneous torque is proportional to instantaneous power. instantaneous α ν = (51.1) where, k is a constant. Tav α

1 2π

2π

∫ Vmax sin θ × I max sin(θ − φ) d θ 0

taking ν = Vmax sin θ and i = I max sin (θ − φ) V I V I Tav α max max × 2 π cos φ α max ⋅ max cos φ α VI cos φ 4π 2 2 where, V and I are the r.m.s. values. Thus, Td α V I cos θ α true power.  (51.2) Owing to the large time constant of the moving system it cannot follow the rapid variation of the torque having double the frequency of the voltage and the instrument takes up a position at which the average deflecting torque is balanced by the controlling torque. Thus, an electrodynamic instrument, when connected as shown in Figure 51.6, indicates the average power irrespective of the fact whether it is connected in an a.c. or d.c. circuit.

Watt Meters and Energy Meters  963

Fixed coil

Figure 51.6  An Electrodynamic Instrument Indicates the Average Power The scale of the dynamometer-type watt meter is more or less uniform because its deflection is proportional to the average power and for spring control torque it is proportional to the deflection. Therefore, θ∝ power.

(51.3)

The watt meter has four external terminals ±V and ± I. It is necessary to connect the ± I terminal and the ±V terminal to the same wire of the incoming supply line. In this way, the fixed coils and the moving coil will be at about the same potential because most of the voltage across the voltage branch is dropped by the high-value series resistor. An electric field would arise between the potential and the current coils if they were at different potentials. The force of attraction due to the field could slightly restrict the movement of the moving coil and give an erroneous reading. The meter will always read upscale when the instrument is correctly connected in the circuit in which the power is to be measured. If, for any reason, the meter reads backward, the current coil connections and not the potential coil connections should be reversed. The overall errors in commercially manufactured dynamometer instruments lie between ± 0.1 per cent and ± 0.5 per cent when operated between their specified frequencies. These high-accuracy instruments are used as laboratory standards of power. The watt meter is rated in terms of its maximum current voltage and power. Each of these ratings must be observed to prevent damage to the instrument. In low-power factor circuits, either of these limits could be exceeded. Ranges: 1. Current 0.25–200 A without employing current transformers. 2. Potential 5 to 750 V without employing potential transformers.

51.4  COMPENSATING COIL Suppose the watt meter W (Figure 51.7) is measuring the power in circuit AB. Then it is usual to connect the pressure coil across AB. If i is instantaneous current in AB and is the instantaneous current in the pressure coil, then (i+i1) is the instantaneous current in the current coil. Therefore, torque at any instant, which is proportional to the product of the instantaneous currents in the pressure, and current coils is proportional to i1(i+i1). Now, the current in the pressure coil is equal at any instant to e/R. Instantaneous torque ∝ (ei/R + ei1/R) ∝ (ei + ei1) ∴ Reading, which ∝ average torque ∝ average of (ei) + average of (ei1)    ∝ average power in AB + average power in pressure coil (51.4) A correction for the power used in the pressure coil, therefore, has to be applied. In standard watt meters, this correction is applied automatically by means of a compensating coil. This is a small coil placed with its axis along the axis of the current coil and having the same number of turns as the current coil, but connected in series with the pressure coil. It is so arranged that its ampere-turns neutralize the extra ampere-turns with the current coil due to the current i1. When the current in the circuit AB is very small, the pressure coil of the watt meter is sometimes connected to points A and C as shown in Figure 51.7(a). In this case, the watt meter reading-power in AB+ power in current coil.

964  Electrical Technology

Figure 51.7  (a) Watt Meter Connections (b) Connections of Compensating Coil

51.5  INDUCTION-TYPE WATT METERS The operating principle of the instrument is discussed in detail in the section Watt Meters. These instruments can only be used in a.c. systems, whereas dynamometer-type watt meters can be used on either a.c. or d.c. systems. Induction-type instruments are useful only when the frequency and supply voltage are approximately constant. Figure 51.8 shows the arrangement of the circuits in the Lipman type of induction watt meter, which has a low VA consumption in its windings. The voltage winding VV is wound on the main magnet core, which also carries coils FF for power factor compensation, the resistor R being adjustable for calibration purposes to bring the operating fluxes, Fp and Fc, 90° out of phase with one another. Within the moving element, or rotor, D—an aluminium drum—is a fixed core, of cruciform shape, completing the magnetic circuit.

Figure 51.8  Induction Watt Meter—Lipman Type

Watt Meters and Energy Meters  965

The current coil ML is wound on one limb of this cruciform core. The voltage coil flux passes horizontally through the rotor, whereas the flux due to the current coils acts on the motor or in a vertical direction (Table 51.1).

51.5.1  Dynamometer-type, Watt Meter and Induction-type Watt Meter a Comparison Table 51.1  A Comparison of Dynamometer-type Watt Meters and Induction-type Watt Meters Sl. Dynamometer-type No. Watt Meters

1 2 3 4 5 6

The instrument can be used on both d.c. and a.c. systems The instrument can have a high degree of accuracy if carefully designed Power consumption is comparatively low Weight of moving system in comparatively low The instrument has uniform scale The instrument has comparatively weaker working torque

Induction-type Watt Meters

The instrument cannot be used on d.c. system The instrument is less accurate. It is accurate only at stated frequency and temperature Power consumption is comparatively high Weight of moving system is comparatively high The instrument has uniform long scale The instrument has comparatively stronger working torque

51.6  ENERGY METERS The amount of electricity used by a consumer must be metered; thus, energy meters are perhaps one of the most familiar electrical instruments. The development of methods of measuring electricity has been interesting. The first measurements were made with meters that were nothing more than clocks measuring the time the load was applied. The first type was started and stopped by the flow of direct current. Knowing the number of lamps, the current taken per lamp and the duration of current flow, the ampere hours could be computed. The first commercially successful ampere-hour meter was the Edison chemical meter developed between 1878 and 1881. This meter involved an electrotype cell containing a metallic salt and two zinc electrodes. Ampere hours were calculated by weighing the plates at intervals. The early 80s also brought about the development of the first commutation-type meters. These meters, similar to the chemical meter, were ampere-hour meters, as were also the mercury-motor meters, developed in England in 1888 by Hookhamand Ferranti. The well-known Thomson recording watt meter was the first really successful watt meter. This meter that is of the commutator type and the mercury motor watt-hour meter remain in principle, though in an improved form, as a means of measuring kilowatt hours on direct current circuits. The Thomson meter also served as a means of measuring energy on the early alternating current circuits; for this purpose, however, it was superseded by the induction meter, developed between 1886 and 1890. The early induction meters were ampere-hour meters, followed in 1886 by the watt-hour meter (Figure 51.9). Many improvements were made in the commutator, mercury motor and induction type and these types of meters with many refinements are the current accurate meters in use. Direct current distribution systems have been largely superseded by alternating current networks. Hence, the use of direct current watt-hour meter is now practically confined to special applications. Figure 51.9  Watt-hour Meter

51.7  GENERAL CLASSIFICATION Watt-hour meters are divided broadly into direct and alternating current types. Those in present use are all of the motor type and comprise three essential elements: a motor causing rotation; a means for providing the necessary load or drag; and a registering mechanism that will integrate, or sum up, the instantaneous values of electrical energy.

966  Electrical Technology Motor-type watt-hour meters may be classified according to the type of motor used, into the commutator type, the mercury type and the induction type. The commutator and mercury types are used on direct currents and the induction type is universally used in alternating current circuits.

51.8  DIRECT CURRENT TYPES 51.8.1  Commutator-type Meters As shown in Figure 51.10, the commutator-type watt-hour meter usually has connected in the circuit to be measured two open-wound field, or current coils b. Between them rotates an armature a, the windings of which are connected across the circuit. In series with the armature is placed a high value resistance r, so that the current in the armature is very small. As the armature revolves at a slow speed, practically no electromotive force is generated, and the current in the armature is proportional to the potential of the circuit. Since, the armature circuit is permanently closed when the meter is in series, the armature current flows continuously. The current in the field coils, however, depends on the use of energy in the circuit to which the meter is connected. Current in the field coils causes a magnetic flux, and the reaction between this flux and that of the current in the armature conductors causes the armature to rotate. The commutator c, against which bear the stationary brushes e, connects the coils successively in circuit as they come into active positions, m, and ensures the proper direction of current flow within their coils. The torque exerted on the moving element of the watt-hour meter is proportional to the product of the values of currents in its field coils and armature. The field coils carry the main current or a fixed part of it, and the current in the armature is proportional to the voltage of the main current. Consequently, the torque of the watt-hour meter is proportional to the rate at which energy is expended in the circuit.

Figure 51.10  Commutator Type Meter

Watt Meters and Energy Meters  967

To make the speed of the meter proportional to the torque, a load must be provided that will vary directly as the speed. This is accomplished by introducing a generator consisting of a disc d (Figure 51.10) of copper or aluminium mounted on the axis of armature and revolving between the jaws of the permanent magnet m. The disk forms a closed conductor of practically constant resistance in which eddy currents are generated directly proportional to the number of magnetic lines of force cut in a given time or to the speed of the meter. The relative directions of eddy currents with respect to the poles of one of the permanent magnets are as indicated in Figure 51.11, which shows a top view of the disc rotating counter clockwise between the pole pieces of the magnet m (Figure 51.11(b)). The polarity of the magnet m is indicated by the letters N and S (Figure 51.11(c)). The eddy currents produce a flux with polarity indicated by the letters N′ and S′ (Figure 51.11(b)). The pole N will repel N′ and attract S, and similarly the pole S will repel S′ and attract N the poles shown below the disc. All these repulsions and attractions are in opposition to the motion of the disc. As the opposing flux is proportional to the eddy currents, it is proportional also to the speed of the disc; hence, the drag of the generator element of the meter on the motor element is directly proportional to the speed.

Figure 51.11 The Drag on the Generator Element of the Meter on the Motor Element is Directly Proportional to Speed The revolutions of the moving element of the watt-hour meter are recorded by a suitable registering mechanism geared to the shaft, part of which is represented, in Figure 51.10, by a gear wheel driven by a worm on the shaft of the moving element of the meter. Commutator-type meters on alternating current are not as accurate as the induction-type meter, which is also lower in cost and maintenance. Therefore, in modern practice, commutator-type meters are confined to use on direct current circuits. A form of the Thomson watt meter with its cover removed is shown in Figure 51.12.

51.8.2  Mercury-type Meters For their operation, mercury motor meters depend on the principle that a metallic disc will rotate when current is passed between its axis and periphery if the path of the current across the disc is through a magnetic field. The disc in these meters is immersed in mercury, and the current is conducted to and away from it by the mercury. The flux acting on the disc may be produced by a permanent magnet or by an electromagnet energized by a potential circuit. The retarding, or generator, element is practically the same as that used in commutator-type meters; in some meters, the drag magnets act directly on the disc constituting the motor element, and in others on a separate disc. The high-load adjustment is usually obtained with a thermocouple compensating device consisting of two strips of dissimilar metals joined together and surrounded by a heating coil or resistance wire in series with the potential magnet coils. The purpose of the thermocouple is to pass a small low-potential current through the armature to give a slight torque necessary to overcome bearing friction. To obtain proper compensating values, the current from the thermocouple is regulated by means of a suitable variable resistance. Light-load compensation may also be obtained by passing some of the potentialcoil current through the armature disc, thus giving an initial or starting torque that may be regulated by an adjustable shunt resistance in the potential coil circuit.

968  Electrical Technology The construction of mercury-type watt-hour meters is exemplified clearly by the Sangamo watt-hour meters, the general inside appearance of which is shown in Figure 51.13(a), the case and the register being removed. The electrical circuits of these meters are shown diagrammatically in Figure 51.13(b). The lettering in Figure 51.13(a) corresponds to that in Figure 51.13(b), as far as the parts referred to are visible in it. The service binding posts are represented by a1 and a2 and the binding posts for the load by b1 and b2. The copper disc c is immersed in the mercury chamber d constituting the motor element, and the contact ears, which conduct the current to and from the mercury, are shown on each side of the mercury chamber at e. A laminated shunt magnet f is energized by the potential coil g. The current circuit is from the binding post a, through one of the compounding turns around the magnet, the function of which is to build up the speed of the disc on heavy loads; then to the left contact ear e and by means of the mercury contact to the disc c. The current then crosses the copper disc c, and as the resistance is much less than that of the mercury, leaves by right hand contact ear e and passes through another compounding turn around the potential magnet to the terminal post b. The potential circuit of the meter is from post a1 through thermocouple coil I and the shunt magnet winding g and back to binding post a2 on the other side of the circuit.

Figure 51.12 A Typical Thomson Watthour Meter with its Cover Removed

(a)    (b)

Figure 51.13 (a) Sangamo Mercury-type Watt-hour Meter (b) The Electrical Circuits of the Meter Shown in Figure 51.13(a)

Watt Meters and Energy Meters  969

The line current between i is subject to the influence of the flux through disc c between the poles of the electromagnet; an iron plate is imbedded with insulation just above the disc so that the magnetic flux passes twice through the disc. There is, therefore, a magnetic flux proportional to the voltage acting on the current that passes through the disc, so that a continuous torque is obtained, which is proportional to the power. A hard wood float k attached to the armature is so proportioned as to give buoyancy or a slight lift to the entire moving system when armature and float are immersed in the mercury. There is, therefore, no weight on the lower bearing but a slight upward thrust on the upper one. The alignment of the moving system is maintained by two ring bearings, one each on either end of the shaft. The heating coil i in the potential circuit surrounds two strips of dissimilar metal fastened together and the heating of this junction causes a flow of current. The free ends are connected by a slotted terminal to the posts l1 and l2. Connecting the ends in the same relative position to the terminals l2 and l3 will reverse the direction of the thermocouple current through the armature circuit. Therefore, its direction can be adjusted to correspond to that of the main current in the armature irrespective of how the meter is connected in the circuit. The current in the thermocouple circuit is adjusted by means of the clamp m between the lower wire, which is of copper, and the upper wire, made of special resistance metal. When the clamp m is set over the joint just to the right part of the conductor marked n, the compensation for light load becomes zero as all the thermocouple current will flow directly through the clamp. When the clamp is at the position as shown Figure 51.13, the current will flow from terminal l1 through the armature circuit back along the upper resistance wire of the shunt through clamp m and the lower copper wire of the shunt to l2. As the adjustment clamp is moved to the right, less resistance will be included in the thermocouple circuit and therefore more current will flow around the armature circuit, giving increased torque. When the clamp is set to position n, a slight reversed current will be set up in the armature circuit. Any tendency to creep because of the surrounding conditions can thereby be eliminated. The full-load or main speed adjustment of the meter is made not by shifting the drag magnets with respect to the disc, but by shunting more or less of the flux between the upper poles of the two drag magnets by means of a soft iron disc o (Figure 51.13(a)). The iron disc is mounted on a vertical screen so that it can be easily raised or lowered. The lower the position of the disc, the more the magnetic flux shunted across the two upper poles and the less will be the drag of the magnets on the disc g, so that the faster will be the speed of the meter for a given driving torque. Sangamo watt-hour meters for capacities exceeding 10 amperes direct current are used with current shunts and are adjusted for use with the shunt by means of a low-resistance wire p (Figure 51.13(b)), connected through clamp q, in series with the armature. By moving the clamp, the voltage drop through the armature circuit may be adjusted to the correct value for the shunt. The resistance of the shunt is such that at full load a current of 10 amperes will pass through the meter. A proper ratio is used in the recording train, corresponding to the ratio between the total current and that shunted through the meter, so that the dials read correctly in kilowatt hours.

51.9  ALTERNATING CURRENT TYPES (INDUCTION WATT-HOUR METERS) The induction watt-hour meter is surprisingly simple in its structural assembly. This type of meter is remarkably accurate over a wide range of loads, power factors, voltages, frequencies and temperatures. The induction watt-hour meter depends on the principles of induction for both the motivation and control of its moving element. Accordingly, it can be used only on alternating current circuits. The action of the motor element of the induction watt-hour meter is analogous to that of the poly-phase induction motor. The poly-phase induction motor is a transformer with a secondary, which is capable of rotating with respect to the primary. Each phase winding of the primary is spaced from every other phase winding, by the same number of electrical degrees as the corresponding voltages are displaced in time phase, in the poly-phase supply system. The result is a primary field that rotates progressively around the rotor axis. In sweeping through the conducting material of the rotor, the rotating primary field induces voltages that circulate currents in the closed-circuit paths of the rotor. The reaction between these rotor currents and the primary fluxes creating them produces the driving torque of the motor. The essential parts of the induction watt-hour meter and its circuit’s connections are clearly shown in their relative positions in Figure 51.14. 1. A motor element to create driving torque, consisting of an electromagnet e, current coil c and potential coil p. 2. A generator element formed by permanent magnets f to create control of torque. 3. A moving element, disc d, rotating on shaft s, serving as a rotor to both motor and generator elements. 4. The register element r, with its register gear train g and dials h, which records rotations of the moving element in energy turns. 5. The necessary bearings for the moving element and the base in which the assembled parts are mounted. The top bearing and guide are indicated at b, a worm or pinion at u, the pivot and lower jewel bearing for the shaft s at l, and anticreep holes are shown at a.

970  Electrical Technology A chart outlining briefly the effect produced by the different induction watt-hour meter elements and their relation to each other is shown in Figure 51.15. The rotor is usually a thin aluminium disc mounted on a vertical spindle and arranged to rotate through the air gaps of both the composite electromagnet comprising the motor element and the permanent magnets comprising the generator element. Revolutions of the disc, over any period of time, are recorded by the positions of the pointers, or hands that are moved over the register dial through the aid of the gear train, which in turn is meshed with the worm (or a pinion) on the upper end of the rotating disc shaft. As shown in Figure 51.14, two bearings are required for the vertical shaft on which the disc revolves: a top or guide bearing and a lower or main bearing. The top bearing, whose main function is merely to hold the moving element centred, consists of a stationary pivot that projects downward into a hole in the top of the disc shaft. The area of contact in the bearing is exceedingly small, but even with the lightest of moving elements the unit pressure is enormous and can be a source of friction in the meter. The motor and generator actions take place in only the relatively small sections of the disc in the immediate vicinity Figure 51.14  Induction Watt-hour Meter of the respective air gaps. The permanent magnets provide the damping or braking function in the generator element of the induction watt-hour meter, similar to that in the corresponding element of the commutator-type watt-hour meter. The composite electromagnet of the motor element is made up of two units: a potential (voltage) electromagnet and a current electromagnet. These correspond in function to the primary windings of the induction motor. To minimize power losses and resultant heating, potential electromagnet coil is wound with many turns of small wire; this winding offers a relatively high impedance to current flow when line voltage is applied. To avoid an objectionable voltage drop as well as the power losses and heating, the current electromagnet coil is constructed of relatively few turns of heavy wire. The coil thus introduces negligible resistance to the flow of the load current through the line with which the coil is connected in series. The cores of these units are made integral parts of a common magnetic frame, a typical form of which is illustrated in Figure 51.16. The poles of the potential and current electromagnets are so located as to apply their magnetic fluxes to the disc in a displaced space relationship, and the major portion of the potential flux is circulated locally, with only a small leakage flux crossing the disc air gap from p to c, which is designated as the potential unit is indicated at a. Sometimes the two coils on the legs of the current electromagnet core may have an unequal number of turns, as shown in Figure 51.16(b).

51.9.1 Production of Fluxes Driving Torques Of the two electromagnets comprising the motor element of the induction watt-hour meter shown in Figure 51.14, the flux of one must be proportional to line voltage and the flux of the other must be proportional to load current if the motor element is to develop a rotating field whose strength is proportional to the load (volts x amperes) at unity power factor. Accordingly, the potential electromagnet is connected across the line, which establishes a useful flux proportional to voltage, and the current electromagnet is connected in series with the line so that the load current flowing through it establishes a flux proportional to current. If the meter disc is to develop a speed that is strictly proportional to the watts load at any definite magnitude of line voltages and load current, it must result from a meter torque that is also proportional to the power factor of the load. Maximum net meter torque must be developed with the load at unity power factor and zero net meter torque must be developed with the load at zero power factor. Furthermore, the induction watt-hour meter must be capable of operating from a single-phase source, which in itself cannot produce a rotating electromagnetic field. The time-phase displacement of fluxes necessary for the production of such a field must accordingly be obtained by artificial means. The rotating field is produced artificially by making the potential unit of the composite electromagnet as purely inductive as possible, and the current unit as purely non-inductive as possible. This construction secures almost a 90-degree quadrature relationship of the total electromagnetic fluxes produced by the two units, and it is known as the split phase method of obtaining the effects of two-phase currents from a single phase. The required 90-degree or quadrature relationship of potential and current fluxes may be obtained exactly, however, by interposing an air gap in the magnetic structure of the potential unit as at “a” in Figure 51.16(a), which results in leakage flux being shunted through the meter disc between p and c. A lagging device, consisting of a closed-circuit loop or coil, sometimes called a shading coil, is placed on the potential electromagnet pole tip in such a position that it embraces the leakage flux passing through the disc gap. This loop or coil

Watt Meters and Energy Meters  971

Figure 51.15  Chart Showing the Effect Produced and Their Relation to Each Other may be in the form of either several turns of small wire, or a single turn or plate punched from a sheet of conducting material. The wire coil is placed in a fixed position in the leakage 3 flux path and is closed through an adjustable resistance. The relation of the meter potential coil and the lagging device is comparable with that of the primary and secondary windings, respectively, of a transformer with large leakage reactance. The potential flux in the meter establishes a current in the closedcircuit lagging device by simple transformer action. The flux set up by this small current, in turn, reacts with and causes the already almost quadrature leakage flux, making it lag even further behind the applied voltage. The resistance of the lag coil may accordingly be varied so as to produce the exact quadrature relationship between the voltage and current fluxes acting on the disc. When the power factor of the load decreases, the lag of the load current behind the line voltage increases; at the same time it approaches an in-phase relationship to the useful potential flux cutting the disc in the meter. This flux had been brought into quadrature with the line voltage by the lagging device. If the line voltage and current are constant in magnitude, the rotating field strength is proportional to the sine of the time-phase angle between the current and the lagged potential flux. This is mathematically the same as the cosine of the angle between the load current and the line voltage. Therefore, the torque decreases in step with a decreasing power factor.

972  Electrical Technology

(a)

(b)

Figure 51.16  The Composite Electromagnet of the Motor Element (a) Potential Flux (b) Current Flux The disc driving torque is produced by a combination of transformer action and motor action. The alternating fluxes from both electromagnets sweep through the disc metal and induce small voltages in the disc near the pole tips. The disc metal acts like the short-circuited secondary of a transformer and currents flow through the local sections of the disc in which the electromagnetic fluxes are effective. Two sets of these currents are produced in the disc, with one set being caused by the potential electromagnet and the other set by the current electromagnet. Because of the physical displacement of the poles of the composite electromagnet, the currents thus generated in the disc by one electromagnet flow through part of the space occupied also by the inducing from the other electromagnet. The conductor of the induced current, which in this case is the disc, tends to move out of the field by simple motor action. The torque driving the disc is thus made of two similar components. One may be accounted for by the motor action developed between the alternating flux from the electromagnet and the currents induced in the disc, through transformer action by the alternating flux from the second electromagnet. The other component is conversely produced. As the respective electromagnetic fluxes are proportional to the current producing them and since the current that flows in the coil of the potential electromagnet varies as the applied voltage, the driving torque is proportional to both voltage and current, and therefore to the watts at unity power factor.

51.9.2  Reading Watt-hour Meters The motion of the rotating element of the watt-hour meter is transmitted through a system, or train of gears, to a series of hands, or pointers, that revolve over dials. A series of dials is imprinted on a dial face, and the positions of the hands with respect to the dials indicate the reading of the meter. The gears and dials are all part of the register. To simplify, the gearing used is sometimes made of a register constant, the number by which the register reading must be multiplied to determine the amount of electrical energy measured by the watt-hour meter. The register constants of meters of recent types are 10 or 100, increasing with the capacity of the register to 100 or 100 times its maximum direct reading. The registers of most of the meters have four dials and they record electrical energy in kilowatt hours, as shown in Figures 51.17 to 51.21. The right-hand dial indicates units. The next tens, the next hundreds and the next thousands. For this reason, the right hand dial is sometimes called the units dial, the next tens dials and so on. Each dial is divided into 10 divisions numbered from 1 to 0. A complete revolution of the units hand carries it over ten divisions and causes the hand on the tens division to move one division. A complete revolution of the tens hand carries the hundred hand over one division and so on. To distinguish the dials, some manufacturers mark them with 1 s, 10 s, 100 s (units, tens, hundreds) and so on as shown in Figure 51.18. In other makes of the instruments, as in Figure 51.17, the dials are not marked; the positions of the dials indicate the value of their reading. In still others, as in Figure 51.21, the value of a complete revolution of the hand is marked above the dial. The highest amount that a direct-reading four-dial register can record is 9999, but register constants may be employed to increase this amount. The register constant is always fixed by the manufacture and marked plainly on the dial face, the usual method of use being simply to multiply by 10 or multiply by 100 as the case may be. This is shown in Figure 51.20. Registers reading in watt hours usually have four dials (Figure 51.22), the lowest dial registering 1000 Watt hours of one revolution for the hand or 100 Watt hours for each division of the dial, the number above or below a dial indicating the value of one revolution of the corresponding hand (note: motor meters can be used on both d.c. and a.c. circuits).

Watt Meters and Energy Meters  973

Figure 51.17 The Register by Taking into Account the Revolutions of the Moving Element with Proper Gear Ratio Records the Energy in Kilowatt Hours that has Passed Through the Meter

Figure 51.18  Dial Systems and Markings

Figure 51.20  Dial Systems and Markings

Figure 51.19  Dial Systems and Markings

Figure 51.21  Register Constant

Figure 51.22 Registers Reading in Watt Hours Usually have Four Dials

974  Electrical Technology In principle, the motor meter is a small motor of the d.c. or a.c. type whose instantaneous speed of rotation is proportional to the circuit current in case of an ampere-hour meter and to the power of the circuit in case of a watt-hour meter.

51.10  POLY-PHASE INDUCTION WATT-HOUR METERS The most common type of poly-phase induction watt-hour meter consists of two or more single-phase meters combined with the disc mounted on a common shaft and arranged to record on a single register. The performance of poly-phase meters is comparable to that of single-phase meters. Owing to the fact that two or more elements comprising the poly-phase meter are mechanically coupled by acting upon a common moving element, full-load adjustment for individual elements loses its significance. The adjustments must include some additional features by which the torque of the individual element may be brought to equality. This equalizing adjustment is called balancing, which is sometimes secured by changing the air gap between voltage and current electromagnets of either element. Decreasing the air gap increases the torque contributed by the element being adjusted. In other types of adjustments, balance is secured by moving an open loop of magnetic material in or out of the potential electromagnet air gap; or by a tilting movement of both potential and current electromagnets to alter their position relative to the discs (Figure 51.23).

   (a)       (b)

Figure 51.23 (a) A General Electric Company Type Two—Element Poly Phase Meter (b) The Motive Elements of the Meters

51.11  ERRORS IN ENERGY METERS The various errors that may occur in energy meters are given in Table 51.2, along with the remedy for each.

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Table 51.2  Errors in Energy Meters Sl. No.

Error

Causes

Effects

Remedy

1

Phase error, i.e., angle between V and Ip is not 90°

This is because of some Energy meter registers resistance of the coil energy even though the and iron losses actual energy passing through the meter is zero. This is because torque is not zero at zero power factor

This error is eliminated by adjusting the position of the shading band provided on the central limb of the upper magnet

2

Speed error

This is due to improper position of the brake magnet

3

Frictional error

This error is due to fric- Energy meter registers tion at bearing, etc. less energy consumed by the load

For eliminating the faster error brake magnet is moved towards the centre of the disc and vice versa for slower error

4

Creeping

The slow but continuous rotation of the disc, even though there is no load, is termed as ‘creeping’. This is due to (1) excessive friction compensation, (2) excessive supply voltage, (3) Stray magnetic field Using the instrument for other than the designed frequency

Energy meter will record even though no load is connected to it

5

Frequency error

This error is introduced due to change in resistance of the coil as temperature changes

This changes the impedance of the coil; thus, error is introduced Energy meter registers slightly less energy

6

Temperature error

The disc moves either faster or slower

This error is eliminated by placing two short-circuited bands on the outer limbs of the shunt magnet Two holes are drilled in the disc, on the opposite sides of the spindle at the same distance. This causes distortion of the field, thus preventing the rotation of the disc under no load condition

Instrument should be used only for the designed frequency This error is very small; hence, no means are provided for eliminating this error

51.12  MEASUREMENT OF POWER IN THREE-PHASE CIRCUITS It is possible to measure the power in a circuit without a watt meter by using either three ammeters or three voltmeters in conjunction with one inductive resistor. These methods are not, however, of much practical importance.

976  Electrical Technology

51.12.1  Three-voltmeter Method The connections are as shown in Figure 51.24, in which V1, V2 and V3 are three voltmeters and R is a non-inductive resistor connected in series with the load. From the vector diagram of Figure 51.24(b), we have V 21 = V 22 + V 23 + 2 V2V3 cos φ (51.5)

Figure 51.24  Three-voltmeter Method of Measuring Single-phase Power Now IV3 cos φ is the power in the load, so that

Power in load = I V3 cos φ =

V 21 − V 22 − V 23 2R

The power factor is given by cos φ =

V 21 − V 22 − V 23 2 V2V3

(51.6)

(51.7)

The assumptions are made that the current in the resistor R is the same as the load current, and that this resistor is entirely non-inductive.

51.12.2  Three-ammeter Method This method is somewhat similar to the three-voltmeter method. The necessary connections are shown in Figure 51.25. The current measured by ammeter A1 is the vector sum of the load current and that taken by the non-inductive R (this latter being in phase with the voltage E). From the vector diagram, I 21 = I 22 + I 23 + 2 I 2 I 3 cos φ (51.8) But

I 2 = E /R

\ I 21 = I 22 + I 23 + 2

E .I 2 cos φ R

Figure 51.25  Three-ammeter Method of Measuring Single-phase Power Hence, the power EI2 cos φ is given by EI 2 cos φ =

( I 21 − I 22 − I 23 ) R (51.9) 2

Watt Meters and Energy Meters  977

and cos φ =

( I 21 − I 22 − I 23 ) (51.10) 2 I 2 I3

51.13  MEASUREMENT OF THREE-PHASE POWER Three-phase power can be measured by (1) three-watt meter method, (2) two-watt meter method and (3) one-watt meter method.

51.13.1 Three-watt Meter Method of Measuring Three-phase Power The connection for this method is shown in Figure 51.26, in which the load is star-connected. W1, W2 and W3 are the three watt meters, connected as shown. The arrows denote the direction of current and voltage, which are conventionally considered positive. If the symbols representing currents and voltages denote instantaneous values, then Instantaneous power in the load = e1i1 +e2i2+e3i3(51.11) Let v be the potential difference between the star point of the load and the star point O of the watt meter voltage coils. Then we have i1

φ

e1

e´1

e2 connection

e3

e´3 i3

W3 e´2

i2

W2

Figure 51.26  Three-watt Meter Method of Measuring Three-phase Power e′1 + v = e1 e′2 + v = e2 e′3 + v = e3 Therefore, total instantaneous power, by substitution is

( e′1 + v ) i1 + ( e′2 + v ) i2 + ( e′3 + v ) i3 = e′1i1 + e′2i2 + e′3i3 + v ( i1 + i2 + i3 ) (51.12) = e′1 i1 + e′2 i2 + e′3 i3

(i1+i2+i3) = 0 in any three-phase, three-wire system, whether balanced or not. Now e′1v1 + e′2 v2 + e′3 v3 is the total instantaneous power measured by the three watt meters, and thus the sum of the readings of the watt meters will give the main value of the total power.

51.13.2 Two-watt Meter Method of Measuring Three-phase Power This is the most common method of measuring three-phase power. It is particularly useful when the load is unbalanced. The connections for the measurement of power in the case of starconnected three-phase load are shown in Figure 51.27. The current coils of the watt meters are connected in lines (1) and (3) and their voltage coils between lines (1) and (2) and (3) and (2), respectively. Figure 51.28 gives the vector diagram for the load circuit, assuming a balanced load, that is, the load currents and power factors are the same for all three phases. E10, E20 and E30 are the vectors representing the phase voltages, and are supposed to be equal,

Figure 51.27 Two-watt Meter Method of Measuring Three-phase Power

978  Electrical Technology whereas I1, I2 and I3 are vectors representing the line currents. The voltages applied to the voltage-coil circuits of the watt meters are E12 and E32, which are the vector sums of the phase voltages as shown. Then, the instantaneous power in the load    = e1i1 + e2 i2 + e3i3 (51.13) where, e1, e2 and e3 are the instantaneous phase voltages and i1, i2 and i3 are the instantaneous line currents. Since, i1 + i2 + i3 = 0, i2 = −i1 − i3 Therefore, total instantaneous power = e1 i1 + e2 ( − i1 − i3 ) + e3 i3 = i1 (e1 − e2 ) + i3 (e3 − e2 ) Now, i1 (e1–e2) is the instantaneous power deflecting watt meter W1 and i3 (e3 – e2) is the instantaneous power deflecting watt meter W2. These watt meters measure I1 E12 cos α and I3 E32 cos β, respectively, where α and β are the phase angles between I1 and E12 and between I3 and E32. The sum of the watt meter readings thus gives the mean value of the total power in the load. Now α = 30º + ϕ and β = 30º – φ Also E12 = E32 = √3E, where, E is the phase voltage. Figure 51.28 Vector Diagram—Two-watt Meter Method

Therefore, the sum of the watt meter readings is, W = If

(

)

(

3 IE cos 30o + φ + 3 IE cos 30o − φ

)

I1 = I 2 = I 3 = I W =

3 IE cos  (30ο + φ ) + cos(30ο − φ )  =

3 IE (2 cos 30ο cos φ )

= 3 IE cos φ

which is, of course, the total power in the load. Note: 1. If one of the voltages (such as E12) is more than 90º out of phase with the current associated with this voltage in the watt meter, the voltage-coil connections must be reversed in order that the instrument may give a forward reading. Under these circumstances, the watt meter reading must be reckoned as negative, and the algebraic sum of the readings of the two instruments gives the mean value of the total power. 2. If the power factor of the load is 0.5—so that I1 lags 60º behind E10 (cos 60º being 0.5). Thus, the phase angle between E12 and I1 is 90º and watt meter W1 should read zero. Power factor = If W1 and W2 are the two watt meter readings, (W1 + W2) gives the total power W1 − W2 = =

(

)

(

)

3 IE  cos 30o + φ − cos 30o − φ  3 IE −2 sin 30o sin φ = − 3 IE si n φ

(

)

W1 − W2 − 3 IE sin φ − tan φ = = W1 + W2 3 IE cos φ 3 or tan φ =

3 ( W2 − W1 )

( W1 + W2 )

from which φ and power factor, cos φ, of the load may be found. Example 51.1 A single-phase energy meter has a constant of 1500 rev/kWh. If 8 lamps of 100 W, 6 fans of 60 W and 2 heaters of 1000 W operate for one hour, the disc makes 4500 revolutions. Find out whether the meter reads correctly. If not, find the percentage error.

Watt Meters and Energy Meters  979

Solution: Power supplied = (8 × 100) + (6 × 60) + (2 × 1000) = 3160 Watts = 3.16 kW Energy supplied = 3.16 × 1 = 3.16 kWh Number of revolutions to be made = 3.16 × 1500 = 4740 Actual revolutions made = 4500 Hence, the energy meter is slow and records less energy than consumed. 4500 Energy recorded = = 3 kWh 1500 Per cent errors =

3 − 3.16 × 100 = −5.0663 per cent (records less). 3.16

Example 51.2 A single-phase energy meter has a constant speed of 1300 revolutions/kWh. The disc revolves at a rate of 4.2 revolutions/ minute when a load of 150W is connected to it. If the load is on for 11 hours, how many units are recorded as error? What is the percentage error? 150 Actual energy consumed in 11 hours = × 11 = 1.65 kWh 1000 Revolutions made by the disc in 11 hours = 3.5 × 11 × 60 = 2310 Energ y consumption recorded by the meter

2.310 = 1.777 kWh 1300

R ecording error = 1.777 − 1.63 = 0.127 kWh (excess) Percentage error =

1.777 − 1.65 × 100 1.65

= 7.69 per cent

51.13.3 One-watt Meter Method of Measuring Three-phase Power This method can be used only when the load is balanced. The connections for a star-connected system are shown in Figure 51.29. The current coil of the watt meter is connected in one of the lines, and one end of the voltage coil is connected to the same line, the other being connected alternately to the first one and then the other of the remaining two lines by means of the switch S. The vector diagram for this method measurement is given in Figure 51.30. E01, E02 and E03 represent the three phase voltages and I1, I2 and I3 the three line currents. In a balanced system, these three voltages are each equal to E1 and the three currents are each equal to I. The phase angles are also each equal to φ. The vector E12 is the vector difference between E01 and E02, and is the voltage applied to the voltage coil when the switch S is on ‘contact (a)’. Similarly, E13 is the vector difference between E01 and E03 and is applied to the watt meter voltage coil when the switch is on ‘contact (b)’.

Figure 51.29  One-watt Meter Method of Measuring Three-phase Power

980  Electrical Technology E12 = E13 = √3E

Then,

Watt meter reading when switch S is on ‘contact α’ =

3 EI cos (30o + φ )

The sum of these two readings is 3 EI cos φ, as shown in the analysis of two-watt meter method, and this is the total power in the circuit. In the same way, the angle φ is given by tan φ =

3 ( W2 − W1 ) W1 + W2

and the power factor is cos φ or Figure 51.30 Vector Diagram, One Watt Meter Method

 cos  tan −1  

3 ( W2 − W1 )    W1 + W2 

S UM M A RY 1. Ampere hour is an ampere of current maintained for one hour. 2. Watt hour is a watt of power maintained for one hour. 3. Ampere-hour meter integrates products of current and time without considering voltage at which current is supplied. 4. Watt-hour meter integrates units of electrical energy with respect to time. 5. Electricity can be measured through the effects it produces. 6. A watt meter is a combination of an ammeter and a voltmeter and therefore consists of a current coil and a pressure coil. 7. There are four types of watt meters: dynamometer type, induction type, electrostatic type, and thermal type. 8. When used as a watt meter (dynamometer type) the fixed coil is employed as a current coil and the moving coil is used as a pressure coil. 9. The magnetic fields of the fixed and moving coils react with one another causing the moving coil to turn about its axis. 10. Damping is provided by a light aluminium vane moving in an air dash pot. 11. The movement is controlled by hair springs. 12. The deflecting torque is directly proportional to power. 13. An electrodynamic instrument indicates the average power. 14. Induction-type instruments can only be used on a.c. systems. 15. Energy meters are one of the most familiar electrical instruments.

16. Motor-type meters comprise three essential elements: a motor to cause rotation, a means for providing the necessary load or drag, and a registering mechanism. 17. To make the speed of the motor proportional to the torque, a load that will vary directly as the speed must be provided. 18. The light-load adjustment is usually obtained with a thermocouple-compensating device. 19.  The motor and generator actions take place in the relatively small sections of the disc in the immediate vicinity of their respective air gaps. 20. The rotating field is produced artificially by making the potential unit of the composite electromagnet as purely inductive as possible and the current unit as purely non-inductive as possible. 21. The torque decreases in step with a decreasing power factor. 22. The register, by taking into account the revolutions of the moving element with proper gear ratio, records the energy in kilowatt hours that has passed through the meter. 23. Use is sometimes made of a register constant, the number by which the register reading must be multiplied to measure the amount of energy measured by the watthour meter. 24. The registers of most of the motors have four dials and they record energy in kilo watthours. 25. Power in single-phase circuits can be measured by either the three-voltmeter method or the three-ammeter method. 26. Power in three phase circuits can be measured by the three-watt meter method, the two-watt meter method or the one-watt meter method. 27. The two-watt meter method is the most common method of measuring three-phase power.

Watt Meters and Energy Meters  981

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. The watt-hour meter can be classified as a

(a) Deflecting instrument (b) Digital instrument (c) Indicating instrument (d) Recording instrument

2. The moving system of an indicating type of electrical instrument is subjected to a (a) Deflecting torque (c) Damping torque

(b) Controlling torque (d) All of the above

10.  Dynamometer-type moving-coil instrument can be used to measure power in

(a) a.c. circuits only (b) d.c. circuits only (c) Both a.c. and d.c. circuits (d) None of these

11. A single-phase domestic electric meter indicates

3. The damping force acts on the moving system of an indicating instrument only when it is



(a) Stationary (b) Moving (c) Just started to move

12. The instruments that indicate the magnitude of the electrical quantity being measured instantaneously are called

4. The most efficient type of damping employed in electrical measuring instruments is

(a) Integrating instruments (b) Recording instruments (c) Indicating instruments (d) All of these

(a) Fluid friction (c) Eddy current

13. Energy meter is

(b) Air friction

5. Induction type instruments find extensive application as (a) Ammeters (c) Voltmeters

(b) Watt-hour meters (d) Frequency meters

6. Induction watt-hour meters are free from (a) Phase errors (c) Frequency errors

(b) Temperature errors

7. Which of the following instruments can be used to measure only d.c.?

(a) Moving-iron instruments (b) Moving-coil instruments (c) Induction-type instruments (d) None of these

8. In a permanent magnet moving-coil instrument, the deflecting torque is proportional to (a) I2    (b) 1/I (c) I

(d) 1/I2

9. In a moving-iron instrument the deflecting torque is proportional to (a) I2

(b) 1/I

(c) I   (d) 1/I2



(a) Energy in joules (b) Power in kilowatt (c) Energy in kilowatt hours (d) Energy in watt hours

(a) An integrating instrument (b) An indicating instrument (c) A recording instrument (d) An absolute instrument

14. The Watt meter

(a) Has three connections, two of which are used at a time (b) Can measure d.c. power but not 50Hz a.c. power (c) Has voltage and current coils to measure true power (d) Only measures apparent power

15. The steady speed of the disc in an energy meter is achieved when

(a) Braking torque is zero (b) Braking torque is half of the operating torque (c) Braking torque is more than operating torque (d) Operating torque is equal to braking torque

16. The induction type energy meter is

(a) An ampere-hour meter (b) True watt-hour meter (c) Watt meter (d) Reactive volt-ampere meter

ANSWERS (MCQ) 1. (c)  2. (d)  3. (b)  4. (c)  5. (b)  6. (c)  7. (b)  8. (c)  9. (a)  10. (c)  11. (c)  12. (a)  13. (a)  14. (c)  15. (d)  16. (b).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. Differentiate between recording and integrating types of instruments. Give two examples of each. 2. Describe the following in case of measuring instruments: a) Deflecting torque b) Controlling torque c) Damping torque

3. Describe the construction and working principle of a shaded-pole-type induction instrument. How are errors due to change of frequency and temperature minimized in such an instrument? 4. Explain the working principle of an induction-type watt meter with the help of a diagram.

982  Electrical Technology 5. Describe the construction and working principle of an energy meter. 6. Explain the sources of error in a single-phase energy meter. How are they eliminated? 7. Write short notes on the following: a) Watt meter b) Energy meter 8. Explain how the following adjustments are made in a single-phase induction type energy meter? a) Adjustment for friction compensation b) Overload 9. Explain the difference between an ampere-hour meter and a watt-hour meter. 10. What are the essential requirements of a watt-hour meter? 11. Briefly explain the working of a commutator-type meter. 12. With the help of a diagram explain the electrical circuit of a mercury-type meter. 13. What are the essential parts of an induction watt-hour meter? ANSWERS (CQ) 18. 20per cent 19. +2.08per cent (fast) 20. 0.893 (lagging).

14. The disc driving torque is produced by a combination of transformer action and motor action. Explain. 15. Explain the significance of register constant. 16. State the working principle of a dynamometer-type watt meter and show its connections. 17. Draw the sketch of a single-phase induction-type energy meter and name its parts. 18. A single-phase energy meter has a constant of 1000 rev/kWh. When a total load of 5 kW is used for 2 hours, the disc makes 12,000 revolutions. Find out whether the meter is reading correctly. If not, find the percentage error. 19. A 40 amperes, 230 volts energy meter on full-load test makes 60 revolutions in 46 seconds. If the normal disc speed is 500 revolutions per kWh, find the percentage error with proper sign by assuming the load to be purely resistive. 20. In the two-watt meter method of power measurement in three phases, the readings of the watt meter are 1000 W and 550 W. What is the power factor of the load?

Multimeters – V.O.Ms.

52

OBJECTIVES In this chapter you will learn about:  Difference between analog and digital displays   Deflecting torques, control torque and damping torque   Direct current, voltage and resistance measurements by analog techniques   Current transformer, potential transformer and clamp meter specifications and their significance   Circuit details of Simpson Model 2.60 Multimeters seven-segment displays: three-and-a-half digit and eight-and-a-half digit readouts   The principle of measurement of current, voltage and resistance by digital techniques simple problems on the above

Multimeters analog and digital

52.1 INTRODUCTION Analog measurements are those involved in continuously monitoring the magnitude of the signal or quantity to be measured. The use of analog instrumentation is very extensive; although digital instruments are ever increasing in number, versatility and applications, it is likely that analog devices will remain in use for many years and for some applications it seems unlikely to be replaced by digital services. For example, it is possible for an operator to assimilate a far greater amount of information from a multi-analog display than from a multi-digital display. A large number of analog instruments are electro-mechanical in nature, making use of the fact that when an electric current flows along a conductor, the conductor becomes surrounded by a magnetic field. This property is used in electrical and mechanical instruments to obtain the deflection of a pointer (1) by the interaction of the magnetic field around a coil with a permanent magnet, (2) between the ferromagnetic vanes in the coil’s magnetic field or (3) through the interaction of the magnetic fields produced by a number of coils. This is illustrated in Figure 52.1. Constraining these forces to form a turning movement produces a deflecting torque, which is a function of the current in the instruments coil and the geometry and type of the coil system. To obtain a stable display, it is necessary to equalize the deflecting torque with an opposing or control torque. The magnitude of this control torque must increase with the angular deflection of the pointer and this is obtained by using spiral springs or a ribbon suspension so that the control torque = Cθ Nm, where θ is the angular deflection in radians and C is the control constant in Newton meters per radian and will depend on the material and geometry of the control device. The moving parts of the instrument will have a moment of inertia and when a change in the magnitude of deflection takes place, it will produce an acceleration torque. As the movable parts are attached to a control spring, they combine to form a mass-spring system. To prevent excessive oscillations when the magnitude of the electrical input is changed, a damping torque must be provided, which will only act when the movable parts are in motion. Most digital instruments display the measurand in discrete numerals, thereby eliminating the parallax error and reducing the human errors associated with analog pointer instruments. In general, digital instruments have superior accuracy to analog pointer instruments, and many incorporate automatic polarity and range indication, which reduce operator training, measurement error and possible instrument damage through overload. In addition to these features, many digital instruments have an output facility enabling permanent records of measurements to be made automatically.

984  Electrical Technology

   (a) (b)



Figure 52.1 Electromechanical Instruments (a) d’ Arsonval Movement Principle and (b) Iron-vane Meter Movement

Digital instruments are, however, usually more expensive than analog instruments. They are also sampling devices, that is the displayed quantity is a discrete measurement made, either at one instant in time or over an interval of time by using digital electronic techniques. Analog and digital displays are illustrated in Figure 52.2.

   (a)     (b)   (c)

Figure 52.2  Displays: (a) Analog (b) Digital 35 Dots and (c) Digital 14 Bars

52.2 DIRECT CURRENT RANGES (Analog) If the ammeter in to be used in multi-range instruments, different shunts (one for each range) will have to be connected. These shunts can be connected across the meter by using a switching arrangement, or by using separate sockets for each range, or by using a tapped resistor each tapping providing a range. All three arrangements are illustrated in Figure 52.3.

Multimeters— V.O.Ms.  985

(a) (b)

(c)

Figure 52.3 Connecting Shunts in Multi-range Ammeters (a) by Using a Switch (b) by Providing Separate External Sockets for the Different Ranges and (c) by Using a Tapped Resistor

52.3 DIRECT VOLTAGE RANGES (Analog) Full-scale deflection current of an ammeter multiplied by its internal resistance is called its full-scale deflection voltages (VFSD). It is the maximum voltage that can be safely applied directly across the meter movement. VFSD = I FSD ´ R m 

(52.1)

If a voltage in excess of VFSD is to be measured, the excess voltage has to be dropped across a series resistance, of appropriate Ohmic value and wattage, and connected in series with the meter movement. This resistance is called the multiplier resistance for that range. For more than one range, more than one multipliers and provision for their insertion into the measuring system have to be provided. Similar to the case of multi-range ammeters, multipliers can also be introduced into the circuit as illustrated in Figure 52.4. Rn

100 V 30 V 10 V 3V 1V

Re Rd Rc Rb

Movement

Ra

Swamp resistor

    Rn

(a) Rn

Rn

Rn

Rn

Rn

100 V

30 V

10 V

3V

1V Movement Swamp resistor

(b)

Figure 52.4  Multi-range Voltmeters. Switching to Large Multipliers Increases the Range

986  Electrical Technology

52.4 MULTI-RANGE OHMMETERS A single Ohmmeter range is not practical when it is necessary to measure very small or very large resistance values. Commercially available Ohmmeters are designed to provide multiple Ohmmeter ranges as well as compensation for a change in the battery voltage. Commercial multi-meters provide for resistance measurement from less than 1 Ω up to many megohm’s in several ranges. The range switch in Figure 52.5 shows the multiplying factors for the Ohms scale. A multiplying factor is given for each Ohms range as Ohmmeter ranges always multiply the scale reading by ‘R × factor’. On voltage ranges, you may have to multiply or divide the scale reading to match the full-scale voltage with the value on the range switch. In the specification relating to resistance range of a multi-meter, the accuracy of the resistance measurement is normally noted as a percentage of the mid-scale value of the resistance range and not the full-scale value as is common in other ranges, as shown in Figure 52.5.

Range switch

R × 10 R × 100 R × 1000

Figure 52.5 A Multiple-Range Ohmmeter

Figure 52.6  Shunt Arrangement for Multi-range Ohmmeters Note: When the zero Ohm adjustment cannot deflect the pointer all the way to zero at the right edge, it usually is an indication that the battery voltage is too low and must be replaced.

52.5 ALTERNATING CURRENT RANGES The deflection of the moving coil meter depends on the instantaneous value of the applied current, and if the frequency of the applied signal is greater than a value determined by the mechanical properties of the movement, it will remain stationary. Thus, to facilitate the measurement of alternating quantities it is necessary to convert them to a direct current so that the signal applied to the meter movement is unidirectional. A copper oxide or silicon rectifier bridge circuit (Figure 52.7) is commonly used to give full-wave rectification of the applied waveform. However, the resulting instrument is an average or mean sensing device, and that the scale will almost always be calibrated in r.m.s values on the assumption that the measurand is a single frequency sine wave, ie, Vr.m.s.= 1.11 V mean. Hence, readings made of signals with distorted waveforms will be subject to considerable error. The characteristics of the rectifier used for this conversion are important because the resistance of the rectifier changes with the current passing through it (Figure 52.8). Therefore, a multi-range ammeter cannot be produced using shunts if the same linear scale is to be used for both a.c. and d.c. ranges. To overcome the problem, an arrangement based on the circuit in Figure 52.9 is used where the current transformer enables the various current ranges to be scaled to an appropriate magnitude for the rectifier characteristic.

Multimeters— V.O.Ms.  987

During these half-cycles current flows through circuit in this direction

No

During these half-cycles current flows through circuit in this direction

rre

nt

No

nt

re

cu

No

r cu

nt

cu

rre

rre

nt

No

cu

Note that instrument current is always in the same direction

Figure 52.7  Working Principle of a Rectifier-type Meter

Figure 52.8  Working Principle of a Rectifier-type Meter: Rectifier Resistance Characteristic

988  Electrical Technology

52.6 ALTERNATING VOLTAGE RANGES While providing alternating voltage ranges the nonlinearity of the rectifier characteristic again creates a certain amount of difficulty. This may be overcome by resorting to the use of an instrument transformer. In this a step-up voltage transformer must be used if ranges that have a full-scale value of less than 10 V or so are required; the resulting arrangement is as shown in Figure 52.9. Another type of instrument that is frequently used in conjunction with a rectifier is the clamp ammeter. It, however, is not a precision instrument but is used for rough approximation of current in conductors and in motor or transformer leads. Principally, the instrument consists of a current transformer of the inserted-primary type, and a rectifier connected to a permanent magnet, a moving-coil type instrument. In Figure 52.9(d) the conductor forms the transformer primary winding, which induces a current in the transformer secondary.



  

  

   (a) (b)

(c)

Transformer secondary winding

Conductor whose current it is desired to measure

Measuring instrument

Rectifier

10

25

50

100

250

500

Shunt taps (d)

Figure 52.9 (a) Current Transformer Principle. The Conductor in which the Current is Being Measured Becomes a Single-Turn Primary, (b) The Use of a Current Transformer to Obtain Alternating Current Ranges, (c) Clamp Type Meter and (d) A Clamp Type Meter Circuit Used with a Rectifier-type Meter

Multimeters— V.O.Ms.  989

Note: The use of instrument transformer enhances the voltage and current-measuring capabilities of measuring instruments. These transformers are not so costly, and are used as attachments for measuring instruments. These transformers can also be in-built. Basically, instrument transformers reduce voltage and current required to be measured to the level that can be safely handled by the meter movement. Meter movement, in both the cases, is connected across the secondary of the transformers. Potential transformers (PT) are voltage step-down transformers whereas current transformers (CT) are current step-down transformers.

Figure 52.10  Multi-meter Circuit for a Low-voltage Range In some multi-meters the current and voltage transformers are combined into a single unit and operated with a resistor network to provide all the scaling for the alternating voltage ranges. This is shown in Figure 52.11. Clamp-type meters, as shown in Figure 52.9(d), can also be used for measuring voltage. Two connections coming out of clamp-type meters are marked probes. Multiplier resistors are connected internally for measuring different voltage ranges. A switch provided in these meters has two positions: I for measuring current and V for measuring voltage.

52.7 V.O.M. SPECIFICATIONS Multi-range, multi-function meters are called multi-meters. They are also known as V.O.Ms (Volt-Ohm-Milli ammeters). The properties of the moving coil movement, in particular its linear scale and good sensitivity (torque-to-weight ratio), have resulted in its uses in the display of multi-range instruments designed either for a single function or for multi-function applications. One of the most difficult things to understand about V.O.Ms. is what is meant by the various manufacturer’s specifications. There is a wide variety of V.O.Ms. being marketed commercially (AVO METER, AVOMINOR TRI PLETT, PHILIPS, SIMPSON, SANWA, BPL, etc—to name just a few). Each one of them is priced according to the measurement facilities provided by the manufacturer and the accuracy and ease of handling of the instrument.

Figure 52.11  A Circuit Arrangement that Uses a Combined Current and Voltage Transformer

990  Electrical Technology All measure voltage (both d.c. and a.c.) and current (both d.c. and a.c.) and resistance. The question that confronts the technician is which meter to choose for a specific application. The answer to this question lies in proper interpretation of the specifications. These are discussed as follows: 1. Accuracy: Probably the most straightforward specification is the V.O.M’s. accuracy. Accuracy is conventionally expressed in terms of error. The accuracy of a V.O.M. on its voltage and current scales is given as a percentage of the full-scale value of the range in use. If a meter with a rated accuracy of ± 3 per cent is being used on its 10 V range, the deviation error is ± 3 per cent of 10 V or ± 0.3 V. Thus, if the meter indication was 4 V, the true value of the voltage being measured could be any value between 3.7 V and 4.3 V. Although the 0.3 V error is 3 per cent of the full-scale value of 10 V, it is a much larger percentage of the measured voltage of 4 V. In 30 this case, the error is a percentage of indication ( ×100=7.5 per cent) and 4 not 3 per cent. Figure 52.12 shows the error as a percentage of indication Figure 52.12 Error as a Percentage at various points on the scale of a meter that has a specified accuracy of of Indication 3 per cent of its full-scale range. The error as a percentage of indication becomes greater at the lower end of the scale. For this reason, measurements should be made on a range that will allow the indication to fall in the upper part of the scale. Manufacturers specify the accuracy of voltmeters and milliammeters as a percentage of full-scale because this allows the accuracy to be expressed as a single figure. This error is not a constant percentage of the actual indication. V.O.Ms. are available with accuracies of 2–4 per cent. Usually the accuracy on a.c. ranges is less than that on d.c. ranges because of the extra components needed for a.c. measurements. For example, a V.O.M. with a rated accuracy of 2 per cent on d.c. scales might be rated 3 per cent for a.c. measurements. Figure 52.13 illustrates the same voltage scale but with an Ohmmeter scale added. In general, the error amounts to a constant number of degrees of pointer deflection, as with the voltmeter, but the Ohmmeter scale is highly nonlinear, that is, the numbers are spread apart at low-resistance values and are crowded together at high-resistance values. Thus, although the error in a voltmeter amounts to a constant number of volts, the error in an Ohmmeter does not amount to a constant number of Ohms. The number of Ohms is least at the low-resistance end of the scale, simply because the scale tends to spread out there. The number of Ohms tends to be greatest at the high-resistance end of the scale simply because the numbers are crowded there. The error of an Ohmmeter as a percentage of the indication is smallest at the mid scale point and becomes greater Figure 52.13 Voltage Scales in Figure 52.11 with an Ohmmeter at each end of the scale. This is shown in Figure 52.13, which gives error Scale Added and Error as a in percentage at various points on the scale that has an error of 3 per Percentage of Indication cent at its midscale indications. It can be seen that as far as accuracy is concerned, the Ohmmeter is simply not of the same class as voltmeters or ammeters. Even in the highest priced V.O.Ms., the Ohmmeter accuracy over most of the scale is nowhere near as good as that of the voltage and current scales. This is not a serious problem because in most circuits resistance values are not very critical. Many resistors have tolerances as high as ± 20% of their rated value. However, it is worth learning the limitations of the instrument so that it is not misled by its indications. 2. Voltmeter sensitivity: The name sensitivity is misleading because what it really specifies is the influence that the meter will have on the voltage being measured. As there is no source of power in a voltmeter, the current required to deflect the pointer must come from the circuit being tested. The amount of voltage in the circuit that will change as a result of this power borrowing depends on both resistance values in the circuit and the sensitivity of the voltmeter. Voltmeter sensitivity is usually specified in terms of Ω/V, which is the resistance of the meter on its various ranges. For example, a voltmeter with a sensitivity of 20,000 Ω/V will have a 20 kΩ resistance on its IV range, 200 kΩ on its 10 V range and 20 MΩ on its 100 V range. The way in which a low-sensitivity meter will change the voltage in the circuit being tested is illustrated with the help of Figure 52.14. With the circuit open, as shown in Figure 52.14(a), the voltage across the arrows would be 150 V. When the circuit is closed and the voltmeter connected, as shown in Figure 52.14(b), the circuit is actually equivalent to that of Figure 52.14(c), where 100 kΩ is the total resistance of the meter. Now there will be a voltage drop across the 50 kΩ resistance. The amount of current and voltage will depend on the resistance of the meter, which in this example is

Multimeters— V.O.Ms.  991

1000 Ω/V or 100 kΩ in total; for the 100 V range. Now there are, in effect, two 50 kΩ resistors in series across the 150 V source. The voltage will then drop evenly across, that is, 75 V across the top resistor and 75 V across the meter. In this case, we have a loading error of 25 per cent as a result of connecting the meter. Now let us consider how the circuit of Figure 52.14 will be influenced by a meter having a sensitivity of 50 kΩ/V. This value is typical of some of the better V.O.Ms. available today. The situation is shown in Figure 52.15(a). It can be seen that the voltage between the points to be monitored was 100 V before connecting the meter. As the meter has a sensitivity of 50 kΩ/volt and is set to 100 V range, it will have a resistance of 5 MΩ. Connecting a 5 MΩ resistor across a 100 kΩ resistor, will have a negligible effect on the total circuit voltage.

Figure 52.14 Power Borrowing Depends on Both Resistance Value in the Circuit and the Sensitivity of the Voltmeter

Figure 52.15  Reducing Loading Effect by Using an Enhanced Sensitivity V.O.M.

992  Electrical Technology The Ohmic values of multiplier resistors for different voltage ranges can be calculated from the sensitivity of the voltmeter as follows: Sensitivity=1/IFSD Ω/V(52.2) If the meter is set to a voltage range of say 100 V, for example, the total resistance on 100 V range is 100×Sensitivity. Denoting meter resistance by Rm and that of the multipliers for 100 V range by RM (100), we get RM(100)=(100×Sensitivity)–Rm (52.3) Sensitivity in terms of Ohms/volt applies mainly to indicating instruments that do not include tubes and transistors. For instruments that contain electronic circuits, the sensitivity ratings are simply given in Ohms. For example, 11 MΩ is a typical sensitivity rating for a VTVM. Sometimes, however, the 11 MΩ rating is referred to as input resistance or input impedance of the VTVM. Meter movements that are designed to give full-scale deflection at low currents are more sensitive, more expensive and easily damaged than those that deflect full scale with higher current values. This higher cost of the meter movement is the main reason for the higher cost of a high-sensitivity meter. Another reason for the higher cost is the fact that in a sensitive meter, movement must be well protected from excessive current, physical shock, or vibration. 3. Ranges: In most practical instruments used for testing and troubleshooting, convenience and speed in the ability to change the range and function for the instrument is important. Instrument designers provide easy methods for changing the range and function. As the indication of a voltmeter is the most accurate on the upper two-thirds of the scale, it is important that the V.O.M. has enough voltage ranges to make such indications possible. In contrast, there should not be too many scales such that it becomes difficult to read Figure 52.16 Basic Scales of a Typical any one of them. The same scale can be used for different ranges, V.O.M. but the scale markings must be multiplied by a suitable factor at different positions of the range switch; common voltage scales for V.O.Ms. are 1, 2.5, 5 and 10 V maximum. Ranges are ×1 ×10, × 100 and sometimes ×1000, of these basic scales as shown in Figure 52.16. 4. Frequency response: Although practically all V.O.Ms. have a.c. voltage ranges, the frequency range over which they will accurately indicate voltages is definitely limited. Older V.O.Ms. were not often accurate much above the power line frequency of 50 Hz. This is not a limitation when 50-Hz voltages are the only a.c. voltages that will be measured. In most of the measurements, it is very useful to be able to measure audio frequency signals. Most modern V.O.Ms. will measure voltages through the audio range of about 50 Hz to 15 kHz. Such instruments are valuable in signal tracing, in audio circuits, and in measuring frequency response of audio amplifiers. 5. Overload protection: The moving coil of a meter mechanism is wound with very fine wire and it can easily be burned out if too much current is passed through it. For this reason, many V.O.Ms. are equipped with some sort of protection circuit. Normally, any one or more of the following safety devices are incorporated in these meters: i. Mechanical protection (used in Avo Meter Model 8X) ii. Fuse (used in the majority of low-priced V.O.Ms.) iii. Protection diodes across the meter movement (used in Simpson Multi-meter Model 260) iv. Transistorized voltage-sensitive circuit (used in some Philips Multi-meter Models) The most common meter protection circuits consist simply of two diodes connected back-to-back across the meter mechanism as shown in Figure 52.17. The diodes do not allow current to flow through them in the forward direction until the voltage across them exceeds some fraction of a volt. This voltage is greater than the voltage across the meter mechanism when the full-scale current is flowing. So normally no current flows through either diode. When the meter is overloaded, the voltage across tends to become greater than the breakdown voltage of the diodes. Thus, one or the other diode will start to conduct, depending on the polarity of the voltage applied to the meter. This action limits the current through the meter to a safe value.

Voltmeter

Figure 52.17  Protection Diodes

Multimeters— V.O.Ms.  993

Although this arrangement protects the meter from damage, it provides no protection for the shunt and multiplier resistors. If a meter having this type of protection is subject to a severe overload, the mechanism will be protected, but one of the shunt or multiplier resistors may be destroyed. The meter will not operate on at least one of its ranges; but it will be much less expensive to repair them if the mechanism were not protected. Some complex protection circuits protect not only the meter mechanism but also other components of the circuit. The heart of such an arrangement is a transistorized voltage that will trip on either positive or negative voltages. This circuit is connected across the terminals of the meter mechanism. When the voltage drop across the mechanism is three or four times the normal fall-scale voltage, the sensing circuit will be activated. This circuit may drive a relay that will open the circuit from the common lead of the V.O.M. The relay will latch and keep the circuit open until it is reset by pressing a button. As the relay contacts are in series with all of the V.O.M. circuits, regardless of what range is in use, it protects all of the components of the V.O.M. The transistorized circuit draws a nominal current.

52.8 THE SIMPSON MODEL 260 MULTI-METER A representative example of a commercial multi-meter is the Simpson Model 260. The circuit diagram of the meter is given in Figure 52.18. The basic movement of the meter has a full-scale current of 50 µA and an internal resistance of 2000 Ω. The meter is a combination of a d.c. milliammeter, a d.c. voltmeter, an a.c. voltmeter, a multi-range Ohmmeter and an output meter.

a.c.

Figure 52.18  The Simpson Model 260 Multi-meter

52.8.1 D.C. Current The following ranges of d.c. current are available (see Figure 52.19): (1) 10 mA, (2) 100 mA, (3) 500 mA and (4) 5 A. The equivalent circuit for measurement of 10 mA d.c. current is shown in Figure 52.20(a). A resistance of 3000 Ω is connected with the meter movement and the whole of the combination is shunted by a 25 Ω universal shunt. The equivalent circuit for measurement of 100 mA d.c. current is illustrated in Figure 52.20(b). The resistance in series with the meter is now increased to 3022.5 Ω and the whole of the combination is shunted by a 2.5 Ω tapping on the universal shunt. For measurement of a d.c. currents up to 500 mA, the resistance in series with the meter is increased to 3024.5 Ω and the whole of combination is shunted by a 0.5 Ω tapping on the universal shunt. The equivalent circuit is illustrated in Figure 52.20(c). For measurement of d.c. currents up to 10 A, the resistance in series with the meter movement is further increased to 3,024978 Ω and the whole of the combination is shunted by a 0.025 Ω tapping on the universal shunt. The equivalent circuit is illustrated in Figure 52.20(d). The value of the shunt required for the measurement of d.c. current to up to 10 A is extremely low (0.025 Ω). That is why the range selector switch excludes the 10 A d.c. range and a separate socket marked +10 A is brought out. The contact resistance of the switch, if used, would pose problems and introduce considerable errors.

994  Electrical Technology

Figure 52.19  Jacks and Switch Position for Direct Current

 (a) (b)

 



(c) (d) Figure 52.20 Equivalent Circuit for Measurement of (a)10 mA d.c. Current (b) 100 mA d.c. Current (c) 500 mA d.c. Current and (d) d.c. Current up to 10 A

Multimeters— V.O.Ms.  995

The complete circuit for d.c. current measurement is shown in Figure 52.21. A 100 µA d.c. current range is also provided for measurement of base current of transistors. In this range, equal amounts of current flow through the meter movement and shunt because resistance in both branches is equal.

52.8.2 D.C. Voltage The following d.c. voltage ranges are available (see Figure 52.22): (1) 2.5 V, (2) 10 V, (3) 50 V, (4) 250 V, (5) 1000 V and (6) 5000 V.

Figure 52.21 D.c. Current Measuring Section

Figure 52.22 Jacks and Switch Positions for Direct Voltage

The full-scale deflection current of the meter is 50 mA. The internal resistance of the meter is 2000 Ω. This corresponds to a full-scale deflection voltage of 100 µV. This is the maximum voltage that can be directly applied across the meter. The equivalent circuit for the measurement of 2.5 V d.c. is given in Figure 52.23(a). A multiplier resistance of 48 kΩ is connected in series with the meter movement to drop the excess voltage. The equivalent circuit for the measurement of 10 V d.c. is given in Figure 52.23(b). Yet another multiplier resistance of 150 kΩ is connected in series with the multiplier resistance for the 2.5 V d.c. range (48 kΩ). The equivalent circuit for the measurement of 50 V d.c. is given in Figure 52.23(c). Another multiplier resistance of 800 kΩ is connected in series with it.

M

M

Figure 52.23 Equivalent Circuit for the Measurement of (a) 2.5 V d.c., (b) 10 V d.c., (c) 50 V d.c., (d) 250 V d.c., (e) 1000 V d.c. and (f) 5000 V d.c.

996  Electrical Technology The multipliers for 2.5 V d.c. and 10 V d.c. ranges are 48 kΩ+150 kΩ. The equivalent circuit for the measurement of 250 V d.c. is given in Figure 52.23(d). Another multiplier resistance of 4 MΩ is connected in series with the multipliers for 2.5 V, 10 V and 50 V ranges (48 kΩ + 150 kΩ + 800 kΩ). The equivalent circuit for the measurement of 1000 V d.c. is given in Figure 52.23(e). Another multiplier resistance of 15 MΩ is connected in series with the multipliers for 2.5 V, 10 V and 250 V d.c. ranges (48 kΩ + 150 kΩ + 800 kΩ + 4 MΩ). The equivalent circuit for the measurement of 5000 V d.c. is given in Figure 52.23(f). Yet another multiplier resistance of 80 Figure 52.24  D.c. Voltage Measurement MΩ is connected in series with the multipliers for 2.5 V, 10 V, 50 Section V and 1000 V d.c. ranges (48 kΩ + 150 kΩ + 800 kΩ + 4 MΩ + 15 MΩ). The insulation resistance of the switch is far less than the combined value of multipliers on the 5000 V range. That is why the switch is excluded for the measurement of d.c. voltages on the 5000 V range. A separate socket marked 5000 V is brought out. The complete circuit for the measurement of d.c. voltages is given in Figure 52.24.

52.8.3 A.C. Voltage The following a.c. voltage ranges are available (see Figure 52.25): (1) 2.5 V, (2) 10 V, (3) 50 V, (4) 250 V and (5) 1000 V. Suitable multiplier resistors are used for different ranges. The complete circuit diagram for a.c. voltage measurement is given in Figure 52.27. A half-wave rectifier is used in series with the meter movement to measure a.c. voltage. Another diode is used to protect the meter from voltages in the reverse direction. A shunt is connected across the meter movement. Due to heavy current flow through this combination, rectification takes place on the linear portion of the diode voltage current characteristic ensuring linearity of scales on a.c. voltage measurement. This is illustrated in Figure 52.26. The complete a.c. voltage measurement section is shown in Figure 52.27.

Figure 52.26 VI Characteristic of Rectifier Diode

Figure 52.25 Jacks and Switch Position for a.c. Volts

Figure 52.27 Complete a.c. Voltage Measurement Section

52.8.4 Resistance The following resistance measurement ranges are available (see Figure 52.28): (1) R × 1, (2) R × 100 and (3) R × 10 k.

Multimeters— V.O.Ms.  997

Figure 52.28  Jacks and Switch Positions for Resistance Equivalent circuits for the different resistance measurements are given in Figure 52.29. A 1.5 V cell is used as the source of power for R × 1 and R × 100 ranges of resistance measurement. A total of five 1.5 V cells are used as the source of power for R × 10 k range of resistance measurement. The current flowing through the meter movement decreases for increasing values of resistance to be measured. To force a current equal to full-scale defection current, a higher source of power is required, in particular on the R × 10 k resistance measurement range.

Figure 52.29 Equivalent Circuits for Different Resistance Measurement Ranges (a) R×10, (b) R×100 and (c) R×10 k

52.9  DIGITAL MULTI-METERS Digital instruments sample the measurand perform evaluation using digital electronics and normally display the measure in discrete numerals. In general, contemporary instruments use either L.E.D or liquid crystal seven-segment displays as shown in Figure 52.30. The major advantage of a digital display is that it eliminates parallax errors and reduces human errors associated with interpreting the position of a pointer on an analog scale. Most digital instruments display superior accuracy and input characteristics to analyze instruments. They may also incorporate automatic polarity indication, range selection and provide a digitally coded output, properties that reduce operator training, which is the possibility of instrument damage through overload, and improve measurement reliability.

998  Electrical Technology

(a) E 1 D 2 Anodes 3 C 4 Decimal 5

E 1

10 G 9 F 8 Anodes 7 A 6 B

D 2 Cathodes 3 C 4 Decimal 5

10 G 9 F 8 Cathodes 7 A 6 B

(b)    (c)

  Figure 52.30 Seven-Segment Displays (a) LED Segment Arrangement and Typical Display (b) Common Anode (c) Common Cathode

52.9.1 Digital Panel Meter The digital panel meter shown in Figure 52.31 eliminates the need to decide which mark is closest to the pointer. There is no guess work in trying to decide if the meter reading is 1.999 or 1.998. Digital meters are usually specified by the number of digits in their readout. When the most significant (left-most) digit can only be a 0 or 1, it is only counted as a half-digit. The 2-volt DPM as shown in Figure 52.31 is only a three-and-a-half digit meter. Even though it is called a 2-voltmeter, it can measure a maximum of 1.999 V. Figure 52.31  Digital Panel Meter

  52.9.2 Digital Voltmeters (Voltage D.C.)

Figure 52.32 Alternating Voltage Ranges in a Mean Sensing Multi-meter

The measuring circuits of a multi-function digital voltmeter have a high input resistance and are connected to the output of voltage-divider networks as shown in Figure 52.32. The input resistance is thus approximately constant on the direct voltage ranges. A common value being 10 MΩ, although an increasing number of models have higher input resistance (100 MΩ and even 1 GΩ), the upper limit of voltage is usually 1000 V and the smallest measurable value (the resolution on the most sensitive range) is 10 μV. However, developments are reducing this to 1 μV and even to 0.1 μV. Digital voltmeters can be classified according to the form of analog to digital converter (ADC) they use, the most common being successive approximation, ramp, dual ramp and pulse width. The successive approximation and ramp, form are the examples of sampling meters in which they provide digital values equivalent to the voltage at a particular time instant. The dual ramp and pulse width form are examples of integrating meters in which the average value of the voltage is given over a fixed measurement time to carry out a measurement but have better noise rejection. With the successive approximation form, a sample of the input voltage is compared with a voltage that is increased in

Multimeters— V.O.Ms.  999

increments until its total value equals the input voltage. Sampling times are typically of the order of 1000 times or more per second. The successive approximation form of digital voltmeter is one of the faster responding voltmeters. For very fast response, instead of comparing the input voltage with steadily mounting voltage in circuits and building up to the required voltage, the comparison can be made simultaneously with a large range of voltages, with each one being linked to a digital code and the matching voltage found rapidly. Such a form of voltmeter is said to employ a flash converter and has conversion times of the order of 10s. The ramp form is the simplest and cheapest form of digital voltmeter, the input voltage being compared with steadily increasing ramp voltage, where the time between the two voltages is equal and the end of the ramp voltage is a measure of the input voltage. Owing to nonlinearities in the shape of the ramp waveform and its lack of noise rejection, accuracy is limited to about ±0.05 per cent. Sample rates can be up to about 1000 times per second. The dual ramp form involves a capacitor being charged during a time equal to 1 cycle of the line frequency. The resulting potential difference is then compared with a steadily increasing ramp voltage and the time taken for the two to become equal is a measure of the input voltage. It has the advantage of noise and line frequency rejection, but since it integrates the signal over 1 cycle of the main frequency, it has a conversion time of only the reciprocal of the main frequency. Accuracy is about ±0.005 per cent. Pulse width form produces pulses whose width, that is, duration, is proportional to the input voltage. The duration of the pulse is then measured by a clock. By integrating over 1/50s rejection of the line frequency occurs and a high resolution is possible. Digital meters provide a numerical read-out that eliminates parallax as well as interpolation errors. The resolution of such an instrument corresponds to the voltage that gives a change in the least significant bit of the meter display. Displays are generally between three and a half and eight-and-a-half digits, the half being because the most significant bit can only take the value of 0 or 1. A three-and-a-half digit display has a resolution of 1 in 2000 and a eight-and-a-half display has 1 in 2×108. Typically, such instruments have an input resistance of 10 MΩ or higher, capacitances of 40 pF and good stability. Voltage ranges differ from 100 mV to about 1000 V, with the limit of resolution being about 1 μV.

52.9.3 Voltage A.C. The methods used to convert a.c. to d.c. are similar to those used with permanent magnet moving coil instruments. Rectification methods give average values and as instruments are generally scaled to read r.m.s. values the result needs correction for non-sinusoidal waveforms. True r.m.s. readings can be obtained by using a thermocouple to monitor the temperature of the resistor across which the input voltage is applied. Typical accuracy varies from about ±1 per cent of the reading plus three digits with a three-and-a-half digit display to ±0.05 per cent of the reading plus 0.03 per cent of the full-scale reading for an eight-and-a-half digit display. The frequency range varies from about 45 Hz to 10 kHz for a three-and-a-half digit display to 10 Hz to 100 kHz for a eight-and-a-half digit display. The input impedance is about 10 MΩ with 100 PF. Voltage ranges vary from full-scale readings of about 100 mV to 1000 V r.m.s.

52.9.4 Current Both d.c. and a.c. currents are determined by the digital voltmeter being used to measure the potential across a standard resistor as shown in Figure 52.33. Typically, the accuracy is about ±0.2 per cent of the reading plus two digits for d.c. and ±1 per cent of the reading plus two digits for a.c.. For both d.c. and a.c. the ranges are from about 200 µA to 2 A and the voltage drop less than 0.3 V. The frequency range is about 45 Hz to 1 kHz.

Figure 52.33  Current Functions of a Digital Multi-meter

52.9.5 Resistance Resistance can be measured using a digital voltmeter by passing a known current through the resistance and determining the resulting potential difference across it as shown in Figure 52.34.

1000  Electrical Technology

Figure 52.34  Resistance Function of a Digital Multi-meter Higher accuracy is, however, obtained by passing the same current through a standard resistor and the unknown resistor and comparing the potential differences across the two. Accuracy varies from about ±0.1 per cent of the reading plus 1 digit for three-and-a-half digit meter to ±0.0002 per cent of the reading to ±0.0004 per cent of the full-scale reading for an eight and a half digit display. The resistance ranges are from 200 Ω to 1000 MΩ.

S UM M A RY 1. Analog measurements continuously monitor the magnitude of a signal. 2. A large number of analog instruments are electromechanical in nature. 3. Deflecting torque is a function of the current within the instrument’s coil and its shape and geometry. 4. For a stable display, it is necessary to equate the deflecting torque with an opposing control torque. 5. A damping torque must be provided, which will only act when the movable parts are in motion. 6. Digital instruments display the measure in discrete numerals. 7. Digital instruments incorporate automatic polarity and range indication. 8. Shunts enhance the current-measuring capacity of an ammeter. 9.  Full-scale deflection voltage is the minimum voltage that can be safely applied directly across the meter movement. 10. Shunts are connected across the meter movement to bypass the excess current. 11. Multipliers are connected in series with the meter movement to drop the excess voltage. 12. Zero Ohms adjustment is provided in Ohmmeter to compensate for the changes in battery voltage. 13. The resistance of the rectifier changes with the current passing through it. 14. The current transformer enables the various current ranges to be scaled to an appropriate magnitude for the rectifier characteristic. 15. In some instruments, the current and voltage transformers are combined in a single unit. 16. Commercial multi-meters are priced according to the measurement facilities provided by the manufacturer.

17. Accuracy is conventionally expressed in terms of error. 18. Accuracy of voltmeter and ammeter scales is given as a percentage of the full-scale value of the range in use. 19. Because of the spread out of the Ohmmeter scale in the lower end and crowding at the upper end of the scale, the accuracy of Ohmmeter scales is expressed as a percentage of the middle-scale value. 20. A clamp ammeter is not a precision instrument. 21. The higher the Ω/V of a voltmeter, the lower the loading it will introduce when inserted into a circuit. 22. Sensitivity in terms of Ω/V is for indicating instruments. 23. The sensitivity of instruments, which includes active devices, is expressed in Ohms. 24. Instrument designers provide easy methods for changing range and function. 25. The most common meter protection circuits consist of two diodes connected back-to-back across the meter movement. 26. Contemporary digital instruments use either L.E.D or liquid crystal seven-segment displays. 27. The ramp form is the simplest and one of the cheapest forms of digital voltmeters. 28. Digital meters provide a numerical readout that eliminates parallax and interpolating errors. 29. Both the a.c. and d.c. currents are determined by the digital voltmeter being used to measure the potential across a standard resistor. 30. Resistance can be measured by passing a known current through a resistor and by determining the resultant potential difference across it.

Multimeters— V.O.Ms.  1001

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. Measurement of resistance requires a suitable

11. V.O.Ms. are

(a) Shunt

(a) Single range multi-function instruments (b) Multi-range multi-function instruments (c) Multi-range single function instruments

(b) Source

(c) Multiplier

2. The zero of the Ohms scale on the right-hand side of the scales means (a) Series arrangement (b) Shunt arrangement (c) Both a and b

3. Measurement of alternating current/voltage requires a suitable (a) Shunt (c) Multiplier

(b) Rectifier

(b) In-built

5. The value of the shunt for higher ranges of current is (a) Extremely high

(b) Extremely low

6. Shunts (a) Provide electrical isolation to the meter (b) Do not provide electrical isolation to the meter

7. The current flowing in the secondary of instrument transformer is standardized as (a) 5 A (b) 10 A (c) 15 A 8. The voltage across the secondary of instrument transformers is standardized at (a) 50 to 100 V (b) 100 to 150 V (c) 150 to 200 V

15. The error of an Ohmmeter as a percentage of indication is smallest at the (a) Mid-scale point (b) Each end of scale

16. Voltmeters with a higher sensitivity have (a) Less loading effect (b) Higher loading effect

17. Sensitivity in terms of Ω/V applies mainly to (a) Indicating instruments (b) Instrument that contains electronic circuits

18. Sensitivity in case of instruments that contain electronic circuits is expressed in (b) Ω/V

19. Most of the measuring instruments are

(b) Standardized

10. The current flowing in the secondary of instrument transformers is independent of load conditions in the case of (a) Current transformers

(a) Lesser than that on d.c. ranges (b) Greater than that on d.c. ranges

(a) Ohms

9. Measuring instruments used with instrument transformers are (a) Not standardized

(a) Lower at the lower end of the scale (b) Greater at the lower end of the scale

14. The accuracy on a.c. ranges is

4. Instrument transformers are (a) Used as attachments (c) Both (a) and (b)

1 2. Accuracy of V.O.Ms. varies from (a) 1 to 2 per cent (b) 2 to 3 per cent (c) 2 to 4 per cent 13. The error as a percentage of indication becomes

(b) Potential transformers

(a) Provided with suitable protection (b) Not provided with protection

20. The most common protection circuit is in the form of (a) Fuse (c) Protection diode

(b) Mechanical

21. Protection diodes provide protection to the (a) Meter movement only (b) Meter movement and external components

ANSWERS (MCQ) 1. (b) 2. (a) 3. (b) 4. (c) 5. (b) 6. (b) 7. (a) 8. (b) 9. (b)     17. (a) 18. (a) 19. (a) 20. (c) 21. (a). 10. (a) 11. (b) 12. (c) 13. (b) 14. (a) 15. (a) 16. (a)

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. How will you convert a single-range ammeter into a multi-range ammeter? 2. How will you convert a single-range voltmeter into a multi-range voltmeter? 3. What are the factors that decide the Ohmic value of shunts and multipliers?

4. What are the different methods of connecting shunts and multipliers? Explain with the help of suitable illustrations. 5. Discuss the limitations of switching arrangements in multifunction multi-range instruments. 6. Briefly explain: (a) Shunt (b) Tapped shunts (c) Univeral shunt (d) Multiplier (e) Tapped multiplier

1002  Electrical Technology 7. Explain the difference between series-type Ohmmeter and shunt-type Ohmmeter. 8. How will you measure alternating currents and voltages with the help of a moving coil movement? 9. Discuss the influence of instrument rectifiers on the linearity of scales. 10. What are the drawbacks of shunts and multipliers? 11. Explain the difference between current transformers and potential transformed. 12. Briefly explain the working of a clamp-type meter. 13. What are the different V.O.M. specifications? Discuss their importance. 14. How is the accuracy of the following specified? (a) Ammeters, (b) Voltmeters, (c) Ohmmeters

15. How will you specify voltmeter sensitivity? Given voltmeter sensitivity and meter resistance, calculate the value of multipliers for different voltage ranges. 16. Meter movement needs to be protected from overloads. What are the different methods of providing protection to the meter movement and external components? Which of these methods are commonly used? 17. What are the factors that you will consider while selecting a V.O.M. 18. Assume the voltage across R2 in Figure 52.35 is measured with a 2000 Ω/V voltmeter on the 5 V range. How much voltage will the meter indicate?

Figure 52.35  For CQ18

The Oscilloscope

53

OBJECTIVES In this chapter you will learn about:    The various parts of a cathode ray tube (CRT) and their significance  The formation of the electron beam   Various controls associated with a CRT   Fluorescence and phosphorescence   The fermentative electron and its detection and focusing  Internal and external graticules  The significance of graticule calibration   The block diagram of a CRT and the function of each block  Time base and its importance

53.1 INTRODUCTION

General purpose oscilloscope

If electronic instruments were chess pieces the oscilloscope would be the queen as it can duplicate all the measurements of voltage: current, resistance, frequency, phase and so on, performed by other instruments; in addition, it allows one to evaluate rapidly changing phenomena including those events that happen only once and last for a few nanoseconds. The peculiar power of the oscilloscope lies in its ability to produce an electronic picture to show the variation of a quantity with regard to another; these quantities are electrical signals as found in a circuit or obtained from a transducer. The use of transducer makes it possible for the oscilloscope to measure the pulse rate, to analyze engine performance, or to portray many other non-electrical quantities. However, oscilloscopes are mostly used to display variations in voltage amplitude over a period of time. In this respect, the oscilloscope performs a function similar to that of a barograph or seismograph, in which a pen tracer adds an ink line on a chart that moves forward at a steady rate in the case of electrical signals; however, they mostly repeat the same amplitude variation continuously at a final rate. For instance, a train of pulses may have a repetition rate of 1 kHZ. The period of time on the leading edge of one pulse to the same point on the next pulse is therefore one millisecond. If we set our oscilloscope pen to travel across its chart in 1 ms, whereas at the same time it rises and falls in accordance with the amplitude variations of the pulse it will draw a picture of the pulse. This is Figure 53.1 Making Voltage Waveform Visible shown in Figures 53.1 and 53.2. As it lasts only for a 1000th of a second, we cannot see this picture unless we freeze it, as in a photograph. However, if we make the pen repeat its sweep across the chart every millisecond, in synchronization with the pulse train, successive pulses will be superimposed continuously to form a visible image. It will serve as if the signal is standing still on the screen (see Figure 53.3), as a spinning wheel seems stationary as under a synchronous strobe light.

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1004  Electrical Technology

Figure 53.2 Oscilloscope Display of Square Wave Response of an Amplifier

Figure 53.3  Freezing the Display

A mechanical pen is limited by its mass to a minimum frequency response of around 150 Hz. But the electron-beam pen in the oscilloscope CRT is practically weightless and is capable of responding to frequencies beyond a GHz.

53.2 CRT A simple CRT is shown in Figure 53.4. It is the heart of the oscilloscope. It consists of the base neck (in which an electron gun in enclosed), a bulb, and a faceplate (screen). The face, bulb and neck are made of glass, although in some CRTs the bulb is metal. Usually, the base plugs into either a standard octal or twelve-pin socket from which connections to the circuit that operates the tube are made.

Figure 53.4  Construction of a Conventional Cathode Ray Tube The electron gun, as shown in Figure 53.5, consists of a cathode, which, when heated, emits electrons, a control grid, a cylindrical anode, which, when a high positive voltage is applied, attracts the electrons in a stream through it, and two sets of deflecting plates. The set of deflection plates that moves the beam of electrons in a horizontal direction is called the horizontal deflection plates. The other set, called the vertical deflection plates, moves the beam in a vertical direction as shown in Figure 53.6. The beam starting from the cathode passes through the control grid and the anode between the two sets of deflection plates and strikes the face of the CRT. The internal surface of the face is covered with a fluorescent material or phosphor,

Figure 53.5  The Electron Gun

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Figure 53.6 The Horizontal and Vertical Deflection Plates

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The Oscilloscope  1005

which illuminates at the spot at which the beam is directed. In modern CRTs (see Figure 53.7), the anode consists of two anodes. The first anode is the focusing anode and is located closer to the cathode than to the second anode, which is called the accelerating anode.

Figure 53.7  The Focusing and Accelerating Anodes By proper adjustment of the voltage rates between the two anodes, the best focus (sharpest spot) is obtained. In a practical oscilloscope, the voltage on the control grid can be varied to change the brightness or intensity of the spot on the front part of the CRT. The internal surface of the bulb is coated with aquadag, a carbon-conductive coating. The aquadag is connected electrically to the first or focusing anode. The purpose of the aquadag is to collect secondary electrons that are dislodged by the electron beam from the florescent screen. It they are not collected, these secondary electrons would settle back in a random distribution on the screen and produce light. This extra light will reduce the contrast and sharpness of the spot. A modern CRT is illustrated in Figure 53.8.

Figure 53.8  Modern Cathode Ray Tube

53.3 INTENSITY The intensity of the spot on the florescent screen will depend on the energy contained in the electron beam, that is on the number and velocity of electrons bombarding the screen at any instant. To obtain a sufficiently powerful electron beam an anode potential, of the order of 1000 V is required in the normal hot cathode or low-voltage tube. The potential of the control grid will affect the electrons to a certain extent, as in the same way the grid potential affects the anode current in a thermionic valve. Figure 53.9 illustrates how the control grid potential affects the anode current. It can be seen that the more negative the control grid becomes, the smaller the number of electrons drawn toward the anode, but due to the concentrating effect of the control grid, the greater the percentage of these electrons a­ rriving on the fluorescent screen.

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Figure 53.9  Characteristics of a Cathode Ray Tube These two factors have a conflicting effect on the number of electrons actually arriving at the screen and at some point, on the curve, the screen will receive the maximum number of electrons and the light spot will have maximum brilliance. The cathode with its small emitting surface is surrounded by the control grid, which is maintained at a negative potential with respect to the cathode. This potential is adjusted by an intensity control, which varies the brightness of the pattern being displayed. The greater the negative potential applied to the control grid, the greater the repelling effect on the electrons leaving the control grid and hence the reduction in the brightness of the pattern.

53.4  FLUORESCENT SCREEN Most substances are fluorescent, that is, they have the property that they emit light when subjected to electron bombardment. This property is, however, possessed by different substances to different degrees; for instance, one substance may emit a greater intensity of light than another for a given rate of bombardment. The frequency, that is the colour of light emitted, varies from substance to substance. All fluorescent materials continue to emit light for some time even after the electron bombardment has ceased. This is called afterglow and its duration varies with different substances from a few milliseconds to many seconds. In some applications, it is an advantage for a tube to have a long afterglow. In general, fluorescent materials consist of a crystal metallic salt containing a minute trace of impurity. This impurity, known as the activator, is essential, as the base substance in its pure state frequently has no fluorescent properties. The short representation list of fluorescent materials used for the screen of the cathode ray tube is given in Table 53.1. Table 53.1  Fluorescent Screen Materials Base

Zinc silicate Zinc sulphide Zinc sulphide Zinc sulphide

Activation

Colour of Trace

Manganese Manganese Silver Copper

Blue green Orange Blue Green

Note: Zinc silicate occurs as willemite and is the material most commonly used for CRT screen. A screen that emits a high intensity of light for a given rate of electron bombardment is desirable in practically all cases as otherwise high anode voltage would be required to produce the required brilliance of the image. It is also essential that the substance can be applied to the end of the tube in such a way that it produces a uniform screen.

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The Oscilloscope  1007

If the tube is to be used for visual examination of waveforms, as is usually the case, the tracer must be of a colour that produces minimum fatigue and eye strain whether viewed in daylight or artificial light. If the waveform under examination is recurrent and the spot of light may be made to trace the same path again and again, an afterglow of 10–20 microseconds will be sufficient with the natural persistence of vision to give an impression of a stationary trace at all but the lowest frequencies. For visual examination of very low-frequency waveforms and in particular for transients, i.e. non-recurring waveforms, a longer afterglow is desirable and it may be of the order of several seconds. If the tube is intended only for photographic work, a blue trace is desirable; the blue light being more active than the green, which has a greater effect on the light-sensitive material of the photographic film for a given exposure. In a generalpurpose tube intended for both visual and photographic examinations, a screen that gives a blue green trace is used. The phosphor screen can be designed for fluorescence in a certain colour with a short medium or long persistence. Table 53.2 gives details about the type of phosphor, its colour and persistence. Table 53.2  Type of Phosphor, its Colour and Persistence Persistence

Type Flworescence Phosphorescence

Medium

Long

Medium

Long

Short

P1 Green -

P2 Green -

P4 White -

P7 Blue yellow

P11 Blue -

The P1 phosphor is used in the general-purpose oscilloscope. The P4 phosphor is used in television picture tubes. The P7 phosphor is a two-layered phosphor screen, one of short persistence (blue) and the other of long persistence (yellow). This type of screen may be used with colour filters to provide a dual characteristic, which flashes blue and persists in yellow for observation of slow process. The P11 phosphor is used for high-speed photography of transients. Short-persistence screens fluoresce for 1/1000 s, medium persistence for 1–2 s, and long persistence for longer than 2 s.

53.5 FOCUSING For oscillograph work, a very small sharply defined spot is required and although a certain amount of focusing can be obtained by adjustment of the control grid potential, this in itself is not sufficient and it is necessary to adopt some additional focusing device. There are two principal methods by which this may be achieved. The first method is electrostatic focusing in which the electron beam is passed through an electrostatic field so shaped as to cause the electrons to converge on the screen. This type of focusing has the advantage that it can be controlled easily. The second method, electromagnetic focusing, is rarely used, except in tubes intended for television work. The effect of the electrostatic field between the cathode and the control grid causes the beam to taper sharply and cross over (see Figure 53.10). The beam is then accelerated towards the fluorescent screen by the high-anode potentials of the

Figure 53.10  An Elementary Electron Gun with Electric Field Focusing (a) Electron Gun (b) Electron Lens

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1008  Electrical Technology anode system. However, after the fist crossover point, the beam is subjected to dispersion and if allowed to persist it would produce a blurred luminous spot. Therefore, the beam must be focused to produce a second crossover point in the vicinity of the screen. To accomplish this, the beam is tapered and brought into sharp focus on the screen by adjusting the electrostatic field existing in the anode system. The accelerating anode is maintained at a constant positive potential of about 1,000 V, whereas the positive potential of the focus anode is varied by means of a focus control. The electric field set up by these potentials through which the beam must pass is the principle of electron optics, which is closely analogous to optics concerning light rays. The direction of electric lines in the field is such that the cathode rays tend to converge and run parallel to the central axis of the electron gun. However, the convergence is graded due to the accelerating speed of the electrons causing this second crossover point to occur at the fluorescent screen, thus resulting in a sharp and well-focused luminous dot.

53.6 DEFLECTION It is now necessary to investigate the means by which the electron beam may be moved across the screen under the influence of externally applied waveforms. There are two methods known as electrostatic and electromagnetic deflection. In the electrostatic deflection method, two plates are arranged one on each side of the beam as illustrated in Figure 53.11. If a voltage is applied across the deflector plates, the beam will be attracted towards the positive plate and repelled from the negative plate so that the spot of light changes its position. Consider two parallel plates, P1 and P2, with a voltage Vd applied between them as illustrated in Figure 53.11.

Figure 53.11 Attraction Between Unlike Electric Charges and Repulsion Between Like Electric Charges Form the Basis of Electrostatic Deflection These plates produce a uniform electric field in the Y direction as a result of which electrons entering the field will experience a force. Under the influence of this force, these electrons can be accelerated in the Y direction. As there is no force acting on these electrons in any other direction (X or Z), these electrons will not experience acceleration in any other direction. (53.1) The loss of potential energy PE = Ev, 

Where, e is the charge of the electron (coulomb) and Va is the anode potential (volts) The gain in kinetic energy KE = ½ ne Ve where, Ve is the velocity of electron (m/s) entering the field of the deflecting plate and the mass of electrons (kg). The gain in kinetic energy is equal to the loss of potential energy. 1 eVa = mVe2 2 

(53.2)

2eVa (53.3) m  The velocity in the X direction remains the same throughout the passage of electrons through the deflection plates as there is no force acting in that direction. The deflection corresponding to a particular value of voltage (or current) will be directly proportional to the strength of the electrostatic field, to the length of the electron path lying in the field, and to the distance of the fluorescent screen from the deflecting system. For high-deflection sensitivity, all of these constant factors are made as large as possible subject Ve =

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The Oscilloscope  1009

to limitations of space. The deflection also depends on the anode voltage to which it is inversely proportional, as a higher anode voltage gives an increased velocity and hence a smaller deflection. In Figure 53.12, the deflection plates are held very close together at the end nearer the cathode to give maximum sensitivity but diverge toward the fluorescent screen in order that a wider deflection of the beam may be possible. If the voltage across the deflector plates (or the current through the deflecting coils) is alternating, the spot will follow the alteraFigure 53.12 Electrostatic Deflection in tions in voltage (or current) exactly and without appreciable time a Cathode Ray Tube lag. As the spot of light is moved back and forth under the influence of an alternating potential applied to the deflection plates, it will trace out vertical (Figure 53.13) or horizontal (Figure 53.14) straight lines. This is due partly to the persistence of vision and partly to the afterglow properties of the fluorescent material and will appear as a continuous line unless the frequency is very low, in which case the actual motion of the spot may be followed by the eye.

Figure 53.13  Vertical Deflection

Figure 53.14  Horizontal Deflection

In electromagnetic deflection (see Figure 53.15), a pair of deflection coils is arranged around the neck of the tube to produce a magnetic field at right angles to the electron beam. The direction of deflection is given by Fleming’s left-hand rule, the electron beam being subject to a deflecting force in just the same way as in a current-carrying conductor. It is important that the magnetic field produced by the coils be uniform and systematic. The deflection is proportional to the magnetic field that is proportional to the current passing through the deflection coil. Such a coil has inductive impedance and may therefore disturb any circuit under test.

53.7  TIME BASE

Figure 53.15 Electromagnetic Deflection in a Cathode Ray Tube

If a second pair of deflection plates is so fitted as to produce an independent deflection at right angles to the first then the spot of light may be moved to any position on the screen instead of merely in a straight line. The plates that cause a horizontal deflection are called the X plates and those causing vertical deflection are called the Y plates. If the voltage under examination is applied across the Y plates and at the same time a voltage is applied to the X plates, which will cause the spot to travel at a uniform rate across the screen (Figure 53.16(a)), then clearly the trace on the screen will be an accurate graph showing the instantaneous voltage.

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1010  Electrical Technology

Figure 53.16 (a) Formation of an Oscilloscope Trace with a Linear Time Base (b) Controlling the Defecting Voltage on the X Plates by Synchronizing the Time Base with the Voltage under Examination

53.7.1  Plotted Against Time If the voltage applied to the Y plates is a transient, that is, it occurs once only, it may be impossible to examine the trace by the eye and one must therefore photograph the trace as it is produced or use a fluorescent screen that will retain the trace for a reasonable period after its formation. If, however, the voltage under examination is recurrent, then voltage on the X plates may be controlled so that it moves the spot uniformly across the screen from the left to right and having reached the limit of its sweep to the right returns the spot very rapidly to the left, where it begins the sweep again. It can be arranged that the second and subsequent traces lie exactly on top of the first, then the eye will obtain the impression of a stationary trace on the screen. The means whereby the deflecting voltage on the X plates is controlled is called the time base and to ensure that a stationary trace appears on the screen, it is necessary to synchronize the time base with the voltage under examination, that is, the time occupied by a whole number of cycles of the recurrent voltage on the Y plates.

53.8 GRATICULES

Figure 53.17 Transparent Graph Screen or Graticule (Internal)

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CRT screens have calibrated vertical and horizontal marks, Figure 53.17, to facilitate the use of the oscilloscope; the accuracy of these marks depends on how close the graticule marks can be placed to the actual phosphor to eliminate parallels. Early oscilloscope tubes used an external graticule to provide the necessary marks, but the distance between the marks on the graticule and the actual phosphor coating could be nearly 1 cm, which causes measurement errors if not used carefully. In internal graticules, lines are etched on the surface of the front glass of CRT; the distance separating the phosphor and the graticule is nearly zero and parallax errors are nearly nonexistent. Internal graticules cause two problems. First, as the graticule cannot be aligned once, the tube has been assembled; any misalignment between the deflection plates and the internal graticule must be corrected by electronic means. This is usually done by supplying a magnetic field by wrapping the CRT with a wire carrying current. The magnetic field rotates the electron beam and effectively rotates the CRT trace. Second, it is somewhat more difficult to illuminate the internal graticule lines for

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The Oscilloscope  1011

photographic purposes and therefore some CRTs have special electrode guns that flood the entire phosphor screen to enhance the internal graticule lines.

53.9  BLOCK DIAGRAM The block diagram of a typical oscilloscope is given in Figure 53.18. To enable the CRT to function modern oscilloscopes use the circuits shown in block form.

Figure 53.18  Block Diagram of a Typical Oscilloscope The direct application of the vertical or Y signal to the deflection plates would severely limit the versatility and sensitivity of the oscilloscope. To overcome both of these difficulties a range of attenuator and amplifier arrangement is inserted between the incoming signal and the deflector plates. The steps of sensitivity are usually given in V or V/div of deflection and arranged in a 1-2-5 sequence. The range of sensitivities varies between oscilloscopes, but in a general-purpose instrument it will probably be from 5 mV/div to 20 V/ div: a division is commonly 8 mm. You feed the waveform you wish to observe to the vertical amplifier. This has sensitivity control (vertical attenuator control). It reduces the amplitudes of the input signals so that after amplification, it has a size that will fit on the screen; the dial of this control is marked in volts and millivolts per vertical scale division. This means that if you set it at volt per division, for example, one volt input will cause the electron beam to be deflected one graticule division up or down, depending on whether the voltage is positive or negative. Most vertical amplifiers are differential amplifiers. These are used in test equipment because of their ability to reject common mode signals. Identical signals will pass identical parts to cancel each other out. If they are not identical, the difference will be amplified and appear at the output. Vertical amplifiers may also be considered as d.c.-coupled video amplifiers because of their bandwidth. These amplifiers use negative feedback to obtain a flat response over as wide a range as ­possible, with peaking coils to provide high-frequency compensation. The gain of these amplifiers can be adjusted by vertical gain control. The vertical amplifier has a push–pull output, so that one vertical deflection plate pulls, whereas the other pushes the electron beam as it passes between them in the absence of an input signal. With the horizontal deflection plates disconnected the electron beam strikes the beam dead centre producing a stationary glowing spot. If a very low frequency signal is now applied to the vertical amplifier input, the resultant alternating potentials on the deflection plates will cause the spot to move up and down. At higher frequencies, the spot will move too fast for your eyes to follow so that it blurs into a vertical straight line. The vertical position control (not shown) allows you to move this up or down on the CRT screen by changing the bias on the push–pull stage, thus forcing one plate to a higher potential than the other. In practical oscilloscopes, the time base will be adjustable so that signals have a wide range of frequencies that can be displayed on a convenient time scale. A typical range of horizontal deflection sweeps is from 25/cm to 200 ns/cm in 1, 2, and 5 unit steps. To synchronize the time base and the Y deflecting signal, a triggering circuit is used. This is the circuit that is sensitive to the level of voltage applied to it, so that when a predetermined level of voltage is reached a pulse is passed from the trigger circuit to initiate one sweep of the time base. The trigger circuit of an oscilloscope is adjustable so that a particular point on either the positive or the negative half-cycle may be selected and used to trigger the time base.

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1012  Electrical Technology The horizontal amplifier converts the single-ended saw tooth output from the saw tooth (sweep) generator with a push–pull signal suitable for the horizontal plates. In this, it is similar to many d.c.-coupled audio amplifiers except for the addition of controls peculiar to oscilloscopes. The gain of the amplifier can be adjusted with the help of the horizontal gain control. The horizontal position control (not shown) is a potentiometer that changes the d.c. level of the amplifier so as to alter the voltage balance at the output, thereby shifting the display horizontally.

S UM M A RY 1. The oscilloscope produces an electronic picture to show the variation of one quantity with regard to another. 2. Electrical signals are mostly the same amplitude variations repeated continuously at a fixed rate. 3. The electron beam pen in the oscilloscope CRT is practically weightless and is capable of responding to frequencies beyond a gigahertz. 4. The CRT is the heart of the oscilloscope. 5. The beam from the cathode of the CRT strikes the inner surface of the face covered with a phosphor, which illuminates at the spot at which the beam is directed. 6. The intensity of the spot depends on the number and velocity of electrons bombarding the screen at any instant. 7. Fluorescent screen emits light when subjected to electron bombardment. 8. A screen that emits a high intensity of light for a given rate of electron bombardment is desirable in practically all cases. 9. The trace must be of a colour that produces minimum fatigue and eye strain whether viewed in daylight or artificial light. 10. The beam must be focused to provide a second crossover point in the vicinity of the screen. 11. The principle of electron optics is closely analogous to optics concerning light rays.

12. The means by which the deflecting voltage on the X plates is controlled is called the time base. 13. It is necessary to synchronize the time base with the voltage under examination. 14. CRT screens have calibrated vertical and horizontal marks to facilitate the use of the oscilloscope. 15. Vertical amplifiers have a sensitivity control. 16. Vertical amplifiers have a push–pull output. 17. Most vertical amplifiers are differential amplifiers. 18. The vertical position control allows the CRT gun to move the display up or down on the CRT screen. 19. Horizontal amplifiers have a horizontal gain control. 20. The horizontal position control shifts the display horizontally. 21. Most oscilloscopes have fairly high input impedance. 22. An oscilloscope can measure not only the amplitude of a signal but also the frequency and phase while displaying the waveform. 23. Zinc silicate occurs as willemite and is the most common material used for CRT screen. 24. The phosphor screen can be designed for fluorescence of a certain colour with a short, medium or long persistence.

M U LT IP LE C H O I C E Q UE S TI O NS (M C Q ) 1. The oscillograph can display

5. Short persistence screen persists for about

(a) Electrical quantities (b) Nonelectrical quantities

(a) 2 seconds (c) 1/1000 seconds

2. The use of transducers makes it possible for the oscillograph to display

6. The method of focusing used in a CRT is

(a) Electrical quantities (b) Non-electrical quantities

7 Deflection depends on the anode voltage to which it is

3. The CRT has (a) One set of deflecting plates (b) Two sets of deflecting plates

4. The intensity of the spot on the fluorescent screen depends on (a) The number of electron (b) The velocity of electron

(c) Both a and b

M53_AUTH_ISBN_C53.indd 1012

(a) Electromagnetic

(b)  1–2 seconds

(b)  Electrostatic

(a) Inversely proportional (b) Directly proportional

8. Graticules can be (a) Internal (b) External (c) a or b

9. Parallax error is almost nonexistent in CRTs with (a) Internal graticules (b) External graticules

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The Oscilloscope  1013

10. Vertical sensitivity controls

12. In practice the time base is

(a) Reduce the amplitude of the incoming signal (b) Increase the amplitude of the incoming signal

(a) Adjustable (b)  Not adjustable

11. Direct application of the vertical signal to the deflection plates is

13. Most oscilloscopes have a

(a) Desirable

(b)  Not desirable

(a) Low-input impedance (b) High-input impedance

ANSWERS (MCQ) 1. (a)  2. (b)  3. (b)  4. (c)  5. (c)  6. (b)  7. (a)  8. (c)  9. (a)  10. (a)  11. (b)  12. (a)  13. (b).

CON V E N TI O NA L Q UE S TI O NS (C Q ) 1. Explain the basic components comprising the gun in a CRT. 2. In which direction is the electron beam moved by (a) the horizontal deflection plates (b) the vertical deflection plates? 3. Name the two anodes found in modern CRTs and describe their relative locations and their purpose. 4. What is meant by retrace; illustrate and explain? 5. What is the purpose of horizontal or sweep oscillator in an oscilloscope? 6. How does the adjustment of the brightness or intensity control affect the display on the CRT? 7. What are the major blocks of a CRO and what does each do? 8. What are the major components of a CRT?

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9. How is the electron beam focused to a fine spot on the face of the CRT? 10. How is the vertical axis of an oscilloscope deflected? How does it differ from the horizontal axis? 11. Why is an attenuator probe used? 12. What are the advantages and disadvantages of oscilloscopes? 13. Write short notes on (a) Afterglow (e)  Synchronizing (b) Fluorescent screen (f)  Focusing (c) Graticule (g)  Deflection (d) Time base (h) First and second crossovers 14. What are the factors that control the movement of a spot on CRT screen?

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54

Oscilloscope Techniques OBJECTIVES In this chapter you will learn about:

Electron gun Screen

 Preliminary checks on the oscilloscope   Screen patterns obtained with deflection voltages   Lissajous figures in the range from 0o to 360o   Frequency ratios  Voltage and current measurement

Bulb Base 1

2

3

1. Base 2. Stem 3. Heater-cathode Control-grid assembly

4

5

6

7

4. Anodes 5. Focusing electrode 6. Deflection plates

8

7. Inter plate shield 8. Glass-support bead 9. Getter

The cathode ray tube

54.1  INTRODUCTION The most glamorous and important electrical/electronic test and measuring instrument is the cathode ray oscilloscope. The scope is widely used for the visual observation of electrical work forces. In addition, the oscilloscope is finding diversified applications in many nonelectronic industrial and scientific uses where physical effects and phenomena are converted into electrical signals. Oscilloscopes range from general purpose to elaborate special purpose types. A modern cathodes ray tube (CRT) is shown in Figure 54.1. The measurement capabilities of the oscilloscope are limited only by the skill of the operator. The oscilloscope must also be in good working condition. Otherwise, a defect in the electrical system may cause a misleading pattern. To avoid this, periodic checks should be conducted on: intensity and focus, positioning, synchronizing, deflection, deflection polarity, equalizing X and Y deflection, voltage calibration, and deflection sensitivity (see Figures 54.2 through 54.11).

Figure 54.1  Simplified Diagram of a Modern Cathode Ray Tube

54.2  PRELIMINARY CHECKS The CRT beam should be controlled to produce a small luminous dot by adjusting simultaneously the intensity and focus controls. A point (dot) is said to have position, but no magnitude (area). Therefore, the first thing in checking the operating condition of an oscilloscope is to adjust the cathode ray beam for correct intensity and to focus it to produce a very fine luminous spot (dot; see Figure 54.2). If this is not possible then trouble may exist in the electron gun or power supply.

Oscilloscope Techniques  1015 Anode

Vertical-deflection plates Electron beam

Spot

Bulb Base Cathode

Neck

Control grid

Horizontal-deflection plates

Screen

Figure 54.2 Adjust the Cathode Ray Beam for Correct Intensity and Focus it to Produce a Very Fine Luminous Spot For adjusting the intensity and focus, power must be applied to the oscilloscope and the intensity and focus controls must be turned fully clockwise. Both the horizontal and vertical gain controls must be turned fully counter-clockwise (no deflection). The positioning controls must be adjusted so that the fluorescent dot is in the centre of the screen (See Figure 54.3). The intensity and focus control must be adjusted simultaneously to obtain a very fine dot of light. It should be possible to reduce the dot to a point still visible. This will allow one to check the intensity and focus controls individually. The intensity control must be rotated through its entire range. When the controls are turned fully counter-clockwise the beam should be cut off and when turned fully clockwise the spot will offer a very bright brilliance. The spot should be cut off for one-third of the rotation of the control, after a fine dot of light should appear. The focus control, as shown in Figure 54.4, causes the dot to increase in size when rotated on each side of its fine dot position. About one-third of its rotation from the fully counter-clockwise position should produce the correct focus or the smallest dot area. However, with the correct intensity and focus adjustment, there should be at least one-third in rotation on the control in either direction. In different oscilloscopes, these fine dot control positions will vary but the adjustments should come well within the range of intensity and focus controls. Focus

Intensity

a.c. Off Vert position

Horiz position

Fine frequency

Horiz Deflection Horiz Amp

Vert gain

100 Kc

Horiz

10 Kc gain 10 1001Kc

Sync amplitude Vert

GND

Coarse frequency

Ext sync

Int Ext line

Sync selector

Test Intensity mod signal

Horiz GND

Figure 54.3  A Typical Cathode Ray Oscilloscope

Figure 54.4  Beam Focus

When the spot is motionless, the screen is subjected to a concentrated electron beam bombardment, causing the fluorescent material to become permanently desensitized in that area. In view of this, it is necessary to make a rapid observation. In addition, high brightness patterns when stationary for long periods might burn themselves into the screen materials; therefore, it is a good practice to reduce brightness (intensity) to a usable minimum level.

1016  Electrical Technology With the vertical and horizontal controls turned ­fully clockwise, the vertical positioning control must be ­rotated through its entire range and the displacement of the spot on the Y axis observed. Erratic movement of the spot during this test will indicate a defective control or component in the positioning circuit. The test should be repeated using the horizontal positioning control and the displacement of the spot on the X axis noted (see Figure 54.5). The oscilloscope must be switched on, the sync selector turned to the initial sync position and the sync amplitude turned control fully counter-clockwise. The coarse frequency control is turned to a frequency range that includes 50 Hz. A 50-Hz test signal is applied to the vertical input terminals and vertical gain control is turned up for normal viewing. Fine frequency (vernier) control should be adjusted (see Figure 54.6) until one complete cycle appears and is almost stationary. The sync amplitude control is turned slowly clockwise until the pattern becomes stationary. This adjustment is important as too much sync voltage will distort the pattern. Fine frequency control is readjusted to obtain two complete cycles; the adjustment is continued to obtain five cycles.

Figure 54.5  Vertical and Horizontal Positioning

Figure 54.6  Sweep Frequencies

One cycle will appear on the screen if the sweep frequency is equal to the frequency of the test signal (50 Hz). The horizontal time base excursions will then be 1/50th of a second. To check the deflection linearity, a sine wave signal is applied to the vertical input and the time base is synchronized to produce one cycle. Although the sine wave of the power frequency is fixed at 50 Hz, it provides a good standard signal as a starter. To check the horizontal linearity, the oscilloscope is switched on, the horizontal time base adjusted to produce one sine wave, and the normal amount of sync is applied to lock in the pattern. Horizontal gain control is turned fully counterclockwise. Vertical gain control is adjusted to a produce an ~75-mm vertical line. The horizontal gain control is turned up gradually and the horizontal expansion of the sine wave pattern is noted. There should be an even expansion of the sine waveform on each side of the centre as shown in Figure 54.7(a). To check the vertical linearity, the vertical gain control is turned fully counter clockwise. The horizontal gain control is adjusted to produce an ~75-mm horizontal line. The vertical gain control is turned up gradually and the peak-to-peak expansion of the sine wave pattern noted. There should be an even expansion of the sine wave on each side of the baseline as shown in Figure 54.7(b).

(a)

Figure 54.7  Horizontal and Vertical Linearity

(b)

Oscilloscope Techniques  1017

Terms such as positive going and negative going must not be confused with positive or negative half sine waves. As shown in Figure 54.8, sine wave contains both positive going and negative going cycles. A half cosine wave is either positive going or negative going depending on the direction of wave motion as shown in Figure 54.9.

(a)

(b)

Figure 54.8 Positive and Negative Swings: (a) Positive and Negative Peaks of Sine Wave (b) Positive and Negative Swings in a Half Sine Wave

(a)

(b)

Figure 54.9 Positive and Negative Swings: (a) Positive and Negative Peaks of Cosine Wave (b) Positive or Negative Peaks of a Cosine Wave To check the deflection polarity, the oscilloscope is switched on and the necessary control is ­adjusted to produce an image. Horizontal gain control is turned fully counter-clockwise and the vertical input attenuator, if used, is adjusted to X1. This will provide maximum deflection sensitivity. A 0.25-µF capacitor is connected across the test leads or input binding posts. This short circuits the random noise pulse and holds the spot steady. The vertical positioning control is adjusted until the spots are set at the bottom of the screen. The test leads are connected across a 1-V cell observing polarity, positive to vertical input lead and negative to ground lead. On contact with the cell, the spot should be deflected up and return. The spot movement is only momentary, but sufficiently long enough to observe the direction. The capacitor is discharged and the test is repeated. If the oscilloscope is provided with a polarity reversal switch, it should be in the normal position. Deflection polarity is illustrated in Figure 54.10. When measuring the phase shift or making other tests requiring equal X and Y traces, it is necessary to equalize both deflection traces. This is important as the vertical deflection sensitivity is slightly greater than the horizontal deflection sensitivity and X and Y amplifiers will show unequal traces for equal gain of a standard input signal connected to the X and Y amplifiers. When the vertical and horizontal forces are equal, their combined force is represented by a 45o diagonal trace (see Figure 54.11). Figure 54.10  Deflection Polarity

io n es ul ta nt de fle ct

3"

R

Vertical deflection

1018  Electrical Technology

45º

3" Horiz deflection

Figure 54.11 Equal Vertical and ­Horizontal Forces

A 50-Hz test signal is connected to both vertical and horizontal input terminals and sweep control is switched to horizontal amplifier and oscilloscope turned on. Vertical gain control is turned fully counter-clockwise and horizontal gain control adjusted to provide a 75-mm trace. Horizontal input lead is removed and vertical gain control is adjusted to provide a 75-mm trace. Horizontal gain control setting should not be disturbed. Now, the horizontal input lead is reconnected. A diagonal trace should appear that is 45o off horizontal as shown in Figure 54.11. This indicates that the X and Y traces are equal and the resultant trace represents the vector sum. If an elliptical pattern appears on the screen, 50-Hz phase shift between the vertical and horizontal amplifiers should be corrected.

54.3  SCREEN PATTERN OBTAINED WITH DEFLECTION VOLTAGES

With no external voltage applied to either plate, the spot rests on the centre of the screen (see Figure 54.12(a)). If we apply an a.c. voltage between the vertical input to vertical and ground, a vertical line is displayed, as shown in Figure 54.12(b). If the signal voltage is applied to the horizontal input terminal and earth, a horizontal line will be displayed on the screen (see Figure 54.12(c)).

Figure 54.12 Screen Patterns: (a) Spot at the Centre of the Screen (b) Vertical Line (c) Horizontal Line A peak-to-peak voltage is equivalent to a d.c. voltage of the same value but a d.c. source does not provide the same sustained up and down motion of the beam unless the d.c. voltage is switched on and off repeatedly. Response of the beam to d.c. voltages is illustrated in Figure 54.13.

Figure 54.13 Response of CRT to d.c. Voltages: (a) Zero Voltage Applied (b) 15 V Positive (c) 15-V Negative

54.3.1  Lissajous Figures A sine wave voltage is applied to both the plates of the CRT. A diagonal straight line is displayed on the screen (Figure 54.14). Note: The vertical and horizontal deflection voltages are equal in amplitude and pass through zero at the same instant that are in phase. This is the requirement for displaying a straight set line at 45o angle. If the sine waves applied to the two sets of deflection plates have the same amplitude and the same frequency but are 90o different in phase, a circular pattern is displayed on the screen as illustrated in Figure 54.15.

Oscilloscope Techniques  1019

Figure 54.14 In-phase Equal Amplitude Sine Waves Applied to Both Pairs of Deflecting Plates, the Resulting Pattern is a Straight Line

Figure 54.15 Development of a Circular ­Pattern by Two Sine Waves with the Same Frequency and Amplio tude but 90 Different in Phase

Phase differences in the range from 0 to 90o produce elliptical Lissajous figures as exemplified in Figure 54.16. Any phase angle can be measured as shown in Figure 54.17. The ellipse is carefully centred on the screen and the interval M and N are measured. Then the phase angle between the vertical and horizontal voltages is M/N. Figure 54.18 shows the progress of patterns in this situation for a range of 360o in 45o steps.

Figure 54.16 Elliptical Lissajous Figures Produced by Two Sine Waves with the Same o o Frequency and Amplitude, but with 30 and 60 Phase Differences

1020  Electrical Technology

Figure 54.17 Phase Angle Difference of the Deflection Voltages is Equal to Arc Sin M/N

Figure 54.18  Lissajous Figures in the Range from 0 o to 360o

Figure 54.19 Ellipses Produced by Unequal Signal Voltages Having a Phase Difference of 90 : (a) Horizontal Voltage>Vertical Voltage (b) Vertical Voltage>Horizontal Voltage o

Note: The 45o ellipse leans to the right, whereas the 135o ellipse leans to the left. The frequency ratio is given by the ratio of number of tangencies to vertical and horizontal ­boundaries of the pattern as illustrated in Figure 54.20.

Figure 54.20 Lissajous Patterns Produced by Sine Wave Voltages that Have Equal Amplitudes, but that Differ in Frequencies

54.4  VOLTAGE AND CURRENT MEASUREMENTS The parameter of voltage that must easily be determined using an oscilloscope for a sine wave is the peak to peak value. The magnitude is determined using the engraving on the graticule in conjunction with the calibrated ranges of the input amplifier. For example, for the waveform in Figure 54.21, the amplifier sensitivity is set to 20 mV/dv. The peak-to-peak

Oscilloscope Techniques  1021

amplitude is 20×6=120 mV. Should it be the RMS value of voltage that is required, then (assuming a distortion-less sine wave) this peak-to-peak value must be divided by 2 2 that is Vrms =

peak to peak deflection × amplifier sensitivity 2 2



(54.1)

The oscilloscope is a high-input impedance instrument and therefore cannot be directly used for the measurement of current. Current can, of course, be measured as voltage drops across resistors, but care must be taken in connecting the oscilloscope leads to a resistor for this purpose, because unless a differential input amplifier is being used, one side of the voltage dropping resistor will have to be at earth potential.

peak to peak

Figure 54.21  Measurement of Voltage from a CRO Display

S UM M A RY 1. The scope is widely used for the visual observation of electrical waveforms. 2. The measurement capabilities of the oscilloscope are limited only by the skill of the operator. 3. With the correct intensity and focus control adjustment, there should be at least one-third in rotation on the control in either direction. 4. When the spot is motionless, the screen is subjected to a concentrated electron beam bombardment.

5. Too much sync voltage will distort the pattern. 6. It is necessary to equalize both deflection traces. o o 7. The 45 ellipse leans to the right, whereas the 135 ellipse leans to the left. 8. Phase angle difference of the deflection voltages is arc sin M/N. 9. The frequency ratio is given by the ratio of number of tangencies to vertical and horizontal boundaries of the pattern.

Index A a.c. servomotors, 842–843 a.c. voltage, 996, 999 air gap power, 792–793 ranges, 986–987 alternating voltage ranges, 988–989 alternator windings, 735–737 equivalent circuits, 753–754 parallel operation, 754–757 requirements for parallelling synchronous, 757–759 voltage regulation, 745–751 ammeters definition, 940 loading, 947–948 shunts, 940–944 automatic control systems, 849 automatic frequency control, 861–862 autotransformer, 713–714 B Biot-Savart relationship, 595–598 C constant-speed compound generator, 860–861 control system, 847–848, 850 complete, 882 thyratron control, 877 thyristor, 877–881 universal motor control, 881–882 converter circuit, 865 converting machines, 864–865 D damping, 852–854 d.c. current, 993–995 d.c. generator armature reaction, 617–618 armature structure, 610 armature windings, 612–616 brush polarity, 629–630 characteristics, 625–629 commutation, 616–617 commutators and brushes, 611

construction, 609–610 e.m.f. equation, 618–619 excitation of, 621 losses, 629 paralleling of, 630–631 schematic diagram and equivalent circuit, 622–625 types, 621–622 voltage regulation and voltage control, 630 d.c. machine cooling methods for, 665 enclosures, 664–665 losses, 660–663 maintenance and accessibility, 665 physical construction of, 733–734 ventilation for, 665 voltage generation formula, 744 d.c. motor, 592 back electromotive force in, 641–642 characteristics of, 645–648 classification, 643–644 construction of, 644–645 control devices, 667–671 dynamometer, 640–641 prony brake of, 639–640 relation between torque and speed of, 648 retardation and stopping of, 676–679 reversing of, 655–656, 674–676 starters, 671–674 starting problems, 651–652 starting switch, 652 torque measurements, 638 d.c. servomotors, 840–842 d.c. voltage, 995–996 digital multi-meters, 997–1000 digital panel meter, 998 direct voltage ranges, 985 display transfer function, 908 distribution factor, 741–743 doubly excited system, 606 dynamometer, 640–641

E efficiency formulas, 659–660 electrostatic voltmeter, 932–935 energy conversion devices, 589–590 energy meters, 965–966 errors in, 974–975 F Faraday’s law, 590–591 feedback control system, 854–857 Fleming’s rule, 593 G galvanometers, 932 generator, 593–594. see also d.c. generator conversion process in, 601–602 e.m.f. between the brushes of multicoil armature, 595 energy balance, 604–605 linear and rotary motion, 602–603, 606 methods of analysis, 603–604 motor action vs generator action, 599 power flow diagrams, 599–600 H hysteresis motors, 824–826 I impedance, 787–788 as per unit quantity, 885 induction motors, three-phase, 786–787 losses and efficiency, 790–791 instrument transformers, 716–717 measurement of, 955–957 integrator, 857–859 intelligent instruments, 908 inverters, 871–874 L LIM, 830–832 locus of current phasor, 789

1024 Index M maximum power transfer, 708–709 measurements accuracy, definition, 904 analog, 983 calibration, 908–909 electron performance, 905 error, definition, 905 factors affecting accuracy, 905–906 instrument, definition, 904 measurement systems, 906–908 precision, definition, 904–905 resolution, definition, 905 sensitivity, definition, 905 measuring instruments classification of, 917–918 digital instruments, 983–984 dynamometer instrument, 924–926 electrostatic voltmeter, 932–935 Ferraris-type induction instruments, 927 galvanometers, 932 gravity-controlled instruments, 918–919 hot-wire instruments, 930 induction-type instruments, 926–930 induction-type watt meters, 929–930 moving-iron instruments, 919–922 polarized moving-iron instrument, 922–923 shaded-pole type, 928–929 thermocouple instruments, 931–932 metre movements design principles, 911–914 metres, 915 special features of, 938–940 moving-coil instruments, 915–917 mutual coupling, 687–688 dot convention, 690–691 parallel connection for, 689 series connection for, 688–689 mutual inductance, 687, 691–692 O ohmmeters, 953–957 multi-range, 986 open-loop system, 849 operational amplifiers, 836–839 practical issues associated with, 840

oscilloscope amount of focusing, 1007–1008 block diagram of, 1011–1012 CRT, 1004–1005 deflection method, 1008–1009 fluorescent screen, 1006–1007 function of, 1003 graticules, 1010–1011 intensity of the spot, 1005–1006 preliminary checks, 1014–1018 screen pattern obtained with deflection voltages, 1018–1020 time base for plotting, 1009–1010 voltage and current measurements, 1020–1021 P power transformers, 715–716 pulse transformers, 718–719 R rectifiers, 865–867 three-phase full-wave, 869–870 regulators, 851 reluctance-start induction motor, 822–823 resistance, 996–997, 999–1000 as per unit quantity, 885 rotating magnetic field, 777–783 S selsyns, 844–845 servomechanism, 851 servomechanisms, 848–849 shaded-pole motors, 818–820 multi-meter, 993 single-phase induction motors capacitor-start split-phase motor, 813–815 classes, 806–808 construction, 808 dual-voltage operation, 815–816 locked-rotor torque of, 809 phase splitting, 808–809 resistance-start split-phase motor, 811–813 reversing of, 815 slip, 783–785 stator windings, 739–741 stepper motors, 826–830 synchronous alternator, 737–738 synchronous motors construction, 764 general, 763–764 operation of, 764–766

power factor control, 767–768 starting of, 766–767 synchronous capacitors, 771–772 V-curve, 768–769 T three-phase circuits and systems measurement of power in, 975–977 as per unit quantity, 886 three-phase induction motors parameters, 798–802 starting technique, 796–798 torque-speed characteristics, 795–796 WRIM and SCIM, 796 three-phase power, measurement of one-watt meter method, 978–980 three-watt meter method, 977 two-watt meter method, 977–978 transfer function, 850–851, 908 connections, 718–719 construction, 693–694 efficiency, 712 equivalent circuits, 703–707 general equation, 697–698 harmonic suppression in three-phase, 728–729 ideal, 694–695 losses in, 710–711 maximum power transfer, 708–709 open circuit test of, 711–712 paralleling three-phase, 728 as per unit quantity, 887–895 practical, 699–701 ratings, 701 short circuit test of, 711 three-phase, 727 three-phase connections of single-phase, 723–725 transformation ratio, 696 voltage regulation, 708 voltage relations, 707–708 transient periods, 852 U uninterruptible power supplies (UPS), 870–871 unit or identity matrix, 114 universal motor, 832–833 universal shunt, 942–944 V voltmeter digital, 998–999

Index

voltmeters definition, 948 loading, 952–953 multipliers, 948–949 V.O.Ms (volt-Ohm-milli ammeters), 989–993

W Ward-Leonard system, 683 watt-hour meter, 965–966 commutator-type, 966–967 induction, 969–972 mercury-type, 967–969

1025

poly-phase induction, 974 reading, 972–974 watt meter, 960–961 compensating coil, 963 dynamometer-type, 961–963 induction-type, 964–965