Aircraft Engineering Principles


Roll-Royce RB211-524 Engine fitted to a Boeing 747 Aircraft


Aircraft Engineering Principles

Lloyd Dingle Mike Tooley

AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD • PARIS SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO


Elsevier Butterworth Heinemann Linacre House, Jordan Hill, Oxford OX2 8DP 30 Corporate Drive, Burlington, MA 01803 First published 2005 Copyright © 2005, Lloyd Dingle and Mike Tooley. All rights reserved The right of Lloyd Dingle and Mike Tooley to be identified as the authors of this work has been asserted in accordance with the Copyright, Design and Patents Act 1988 No part of this publication may be reproduced in any material form (including photocopying or storing in any medium by electronic means and whether or not transiently or incidentally to some other use of this publication) without the written permission of the copyright holder except in accordance with the provisions of the Copyright, Designs and Patents Act 1988 or under the terms of a licence issued by the Copyright Licensing Agency Ltd, 90 Tottenham Court Road, London, England W1T 4LP. Applications for the copyright holder’s written permission to reproduce any part of this publication should be addressed to the publishers British Library Cataloguing in Publication Data Dingle, Lloyd Aircraft engineering principles 1. aerospace engineering I. Title II. Tooley, Michael H. (MichaelHoward), 1946– 692.1 Library of Congress Cataloguing in Publication Data A catalogue record for this book is available from the Library of Congress ISBN 0 7506 5015 X For information on all Elsevier Butterworth-Heinemann publications visit our website at www.books.elsevier.com Typeset by Charon Tec Pvt. Ltd, Chennai, India www.charontec.com Printed and bound in Great Britain

Contents
Preface Acknowledgements viii x

PART 1 Chapter 1 1.1 1.2 1.3 1.4 1.5

INTRODUCTION Introduction The aircraft engineering industry Differing job roles for aircraft maintenance certifying staff Opportunities for training, education and career progression CAA licence – structure, qualifications, examinations and levels Overview of airworthiness regulation, aircraft maintenance and its safety culture

1 3 3 3 7 15 18

PART 2 Chapter 2

2.1 2.2 2.3 2.4 2.5 Chapter 3 3.1 3.2 3.3 3.4 Chapter 4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10

SCIENTIFIC FUNDAMENTALS Mathematics General introduction Non-calculator mathematics Introduction Arithmetic Algebra Geometry and trigonometry Multiple choice questions Further mathematics Further algebra Further trigonometry Statistical methods Calculus Physics Summary Units of measurement Fundamentals Matter The states of matter Mechanics Statics Dynamics Fluids Thermodynamics

31 33 33 34 34 34 53 73 100 109 109 118 131 144 165 165 165 170 178 182 183 184 207 240 257


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4.11 4.12 PART 3Chapter 5 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 5.20 Chapter 6 6.1 6.2 6.3 6.4 6.5 PART 4 Chapter 7 7.1 7.2 7.3 7.4 7.5 7.6 7.7

Light, waves and sound Multiple choice questions ELECTRICAL AND ELECTRONIC FUNDAMENTALS Electrical fundamentals Introduction Electron theory Static electricity and conduction Electrical terminology Generation of electricity DC sources of electricity DC circuits Resistance and resistors Power Capacitance and capacitors Magnetism Inductance and inductors DC motor/generator theory AC theory Resistive, capacitive and inductive circuits Transformers Filters AC generators AC motors Multiple choice questions Electronic fundamentals Introduction Semiconductors Printed circuit boards Servomechanisms Multiple choice questions FUNDAMENTALS OF AERODYNAMICS Basic aerodynamics Introduction A review of atmospheric physics Elementary aerodynamics Flight forces and aircraft loading Flight stability and dynamics Control and controllability Multiple choice questions

277 297 309 311 311 313 315 319 322 326 333 341 353 355 369 379 386 397 402 414 418 423 429 438 451 451 456 511 515 531 539 541 541 541 545 561 570 579 587 595 597 601

APPENDICES A Engineering licensing examinations B Organizations offering aircraft maintenance engineering training and education


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C D E F Index

The role of the European Aviation Safety Agency Mathematical tables Systeminternational and imperial units Answers to “Test your understanding”

603 605 615 623 637


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Preface
The books in the series have been designed for both independent and tutor assisted studies. For this reason they should prove particularly useful to the “self-starter” and to those wishing to update or upgrade their aircraft maintenance licence. Also, the series should prove a useful source of reference for those taking ab initio training programmes in JAR 147 (now ECAR Part-147) and FAR 147 approved organizations and those on related aeronautical engineering programmes in further and higher education establishments. This book has primarily been written as one in a series of texts, designed to cover the essential knowledge base required by aircraft certifying mechanics, technicians and engineers engaged in engineering maintenance activities on commercial aircraft. In addition, this book should appeal to the members of the armed forces, and students attending training and educational establishments engaged in aircraft engineering maintenance and other related aircraft engineering learning programmes. In this book we cover in detail the underpinning mathematics, physics, electrical and electronic fundamentals, and aerodynamics necessary to understand the function and operation of the complex technology used in modern aircraft. The book is arranged into four major sections: • • • • IntroductionScientific fundamentals Electrical and electronic fundamentals Fundamentals of aerodynamics career progression routes, professional recognition, and the legislative framework and safety culture that is so much a part of our industry. In the section on Scientific fundamentals we start by studying Module 1 of the JAR 66 (now ECAR Part-66) syllabus (see qualifications and levels) covering the elementary mathematics necessary to practice at the category B technician level. It is felt by the authors, that this level of “non-calculator” mathematics is insufficient as a prerequisite to support the study of the physics and the related technology modules, that are to follow. For this reason, and to assist students who wish to pursue other related qualifications, a section has been included on “further mathematics”. The coverage of JAR 66 Module 2 on physics is sufficiently comprehensive and at a depth, necessary for both category B1 and B2 technicians. The section on Electrical and electronic fundamentals comprehensively covers ECAR 66 Module 3 and ECAR Part-66 Module 4 to a knowledge level suitable for category B2 avionic technicians. Module 5 on Digital Techniques and Electronic Instrument Systems will be covered in the fifth book in the series, Avionic Systems. This book concludes with a section on the study of Aerodynamics, which has been written to cover ECAR Part-66 Module 8. In view of the international nature of the civil aviation industry,all aircraft engineering maintenance staff need to be fully conversant with the SI system of units and be able to demonstrate proficiency in manipulating the “English units” of measurement adopted by international aircraft manufacturers, such as the Boeing Aircraft Company. Where considered important, the English units of measure will be emphasized alongside the universally recognized SI system. The chapter on physics (Chapter 4) provides a thorough introduction to SI units, where you will also find mention of the English system,

In the Introductory section you will find information on the nature of the aircraft maintenance industry, the types of job role that you can expect, the current methods used to train and educate you for such roles and information on the examinations system directly related to civil aviation maintenance engineering. In addition, you will find information on typical


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Preface

with conversion tables between each system being provided at the beginning of Chapter 4. To reinforce the subject matter for each major topic, there are numerous worked examples and test your knowledge written questions designed to enhance learning. In addition, at the end of each chapter you will find a selection of multiple-choice questions, that are graded to simulate the depth and breadth of knowledge required by individuals wishing to practice at the mechanic (category A) or technician (category B) level. These multiplechoice question papers should be attempted after you have completed your study of the appropriate chapter. In this way, you will obtain a clearer idea of how well you have grasped the subject matter at the module level. Note also that category B knowledge is required by those wishing to practice at the category C or engineer level. Individuals hoping to pursue this route should make sure that they thoroughly understand the relevant information on routes, pathways and examination levels given later. Further information on matters, such as aerospace operators, aircraft and aircraft component manufacturers, useful web sites, regulatory authorities, training and educational establishments and comprehensive lists of terms, definitions and references, appear as appendices

at the end of the book. References are annotated using superscript numbers at the appropriate point in the text. Lloyd Dingle Mike Tooley

Answers to questions Answers to the “Test your understanding” questions are given in Appendix F. Solutions to the multiple choice questions and general questions can be accessed by adopting tutors and lecturers. To access this material visit https://books.elsevier.com/manuals and follow the instructions on screen.

Postscript At the time of going to press JAR 66 ad JAR 147 are in the process of being superseded by the European Civil Aviation Regulations (ECAR) 66 and 147. Wherever in this volume reference is made to JAR 66 and JAR147, then by implication, these are referring to ECAR Part-66 and ECAR Part-147 (see Appendix C for details).


Acknowledgements
The authors would like to express their gratitude to those who have helped in producing this book. Jeremy Cox and Mike Smith of Britannia Airways, for access to their facilities and advice concerning the administration of civil aircraft maintenance; Peter Collier, chairman of the RAeS non-corporate accreditation committee, for his advice on career progression routes; The Aerospace Engineering lecturing team at Kingston University, in particular, Andrew Self, Steve Barnes, Ian Clark and Steve Wright, for proof reading the script; Jonathan Simpson and all members of the team at Elsevier, for their patience and perseverance. Finally, we would like to say a big ‘thank you’ to Wendy and Yvonne. Again, but for your support and understanding, this book would never have been produced!


P A R T

1

Introduction


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Chapter

Introduction

1
1.1 The aircraft engineering industry The global aircraft industry encompasses a vast network of companies working either as large international conglomerates or as individual national and regional organizations. The two biggest international aircraft manufacturers are the American owned Boeing Aircraft Company and the European conglomerate, European Aeronautic Defence and Space Company (EADS), which incorporates airbusindustries. These, together with the American giant Lockheed-Martin, BAE Systems and aerospace propulsion companies, such as Rolls-Royce and Pratt and Whitney, employ many thousands of people and have annual turnovers totalling billions of pounds. For example, the recently won Lockheed-Martin contract for the American Joint Strike Fighter (JSF) is estimated to be worth 200 billion dollars, over the next 10 years! A substantial part of this contract will involve BAE Systems, Rolls-Royce and other UK companies. The airlines and armed forces of the world who buy-in aircraft and services from aerospace manufacturers are themselves, very often, large organizations. For example British Airways our own national carrier, even after recent downsizing, employs around 50,000 personnel. UK airlines, in the year 2000, employed in total, just over 12,000 aircraft maintenance and overhaul personnel. Even after the events that took place on 11th September 2001, the requirement for maintenance personnel is unlikely to fall. A recent survey by the Boeing Corporation expects to see the demand for aircraft and their associated components and systems rise by 2005, to the level of orders that existed prior to the tragic events of 11th September 2001. Apart from the airlines, individuals with aircraft maintenance skills may be employed in general aviation (GA), third-party overhaul companies, component manufacturers or airframe and avionic repair organizations. GAcompanies and spin-off industries employ large numbers of skilled aircraft fitters. The UK armed forces collectively recruit around 1500 young people annually for training in aircraft and associated equipment maintenance activities. Aircraft maintenance certifying staff are recognized throughout Europe and indeed, throughout many parts of the world, thus opportunities for employment are truly global! In the USA approximately 10,000 airframe and propulsion (A&P) mechanics are trained annually; these are the USA equivalent of our own aircraft maintenance certifying mechanics and technicians. Recent surveys carried out for the UK suggest that due to demographic trends, increasing demand for air travel and the lack of trained aircraft engineers leaving our armed forces, there exists an annual shortfall of around 800 suitably trained aircraft maintenance and overhaul staff. Added to this, the global and diverse nature of the aircraft maintenance industry, it can be seen that aircraft maintenance engineering offers an interesting and rewarding career, full of opportunity.

1.2 Differing job roles for aircraft maintenance certifying staff Individuals may enter the aircraft maintenance industry in a number of ways and perform a variety of maintenance activities on aircraft or on their associated equipments and components. The nature of the job roles and responsibilities for licensed certifying mechanics, technicians and engineers aredetailed below. The routes and pathways to achieve these job roles, the opportunities for career progression, the certification rights and the nature of the


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Aircraft engineering principles

necessary examinations and qualifications are detailed in the sections that follow. 1.2.1 The aircraft maintenance certifying mechanic Since the aircraft maintenance industry is highly regulated, the opportunities to perform complex maintenance activities are dependent on the amount of time that individuals spend on their initial and aircraft-type training, the knowledge they accrue and their length of experience in post. Since the knowledge and experience requirements are limited for the certifying mechanic (see later), the types of maintenance activity that they may perform, are also limited. Nevertheless, these maintenance activities require people with a sound basic education, who are able to demonstrate maturity and the ability to think logically and quickly when acting under time constraints and other operational limitations. The activities of the certifying mechanic include the limited rectification of defects and the capability to perform and certify minor scheduled line maintenance inspections, such as daily checks. These rectification activities might include tasks, such as a wheel change, replacement of a worn brake unit, navigation light replacement or a seat belt change. Scheduled maintenance activities might include:replenishment of essential oils and lubricants, lubrication of components and mechanisms, panel and cowling removal and fit, replacement of panel fasteners, etc., in addition to the inspection of components, control runs, fluid systems and aircraft structures for security of attachment, corrosion, damage, leakage, chaffing, obstruction and general wear. All these maintenance activities require a working knowledge of the systems and structures being rectified or inspected. For example, to replenish the hydraulic oil reservoirs on a modern transport aircraft requires knowledge of the particular system, the type of oil required (Figure 1.1), the replenishment equipment being used, all related safety considerations and knowledge of the correct positioning of the hydraulic services prior to the replenishment.

Figure 1.1 Identification label showing the type of oil contained within the drum.

Figure 1.2 Boeing 767 hydraulic reservoir charging point, showing contents gauge, changeover valve and hydraulic hand pump.

In addition, for this task, the mechanic must be able to recognize the symptoms for internal or external hydraulic oil leakage when carrying out these replenishment activities on a particular hydraulic system reservoir. For example, Figure 1.2 shows the hydraulic reservoir replenishing point for the Boeing 767. The replenishment process requires the changeover valve to be selected and oil sucked into the reservoir, via the replenishmenthose (Figure 1.3) which is placed in the oil container. The certifying mechanic then operates the hand pump (see Figure 1.2) to draw the hydraulic fluid up into the reservoir. When the reservoir is full, as indicated by the contents gauge, the hose is withdrawn from the container, blanked and stowed. The changeover valve is put back into the flight position, the panel is secured and the


Introduction

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Figure 1.3 Hydraulic reservoir replenishment hose, removed from stowage point.

Figure 1.4 Boeing 767 flap drive motor and associated drive mechanism.

appropriate documentation is completed by the certifying mechanic, who will have a company approval to perform this task. For this job role, like all those that follow, there is a statutory requirement for a particular period of training and experience before a maintenance mechanic is issued with limited certifying privileges. Within the armed forces a similar job role exists for those who have undergone training as aircraft mechanics, for flight line operations or similar maintenance activities. 1.2.2 The aircraft maintenance category B certifying technician The role of the category B certifying technician is subdivided into two major sectors: category B1 (mechanical) and category B2 (avionic). B1 maintenance technicians will have an in depth knowledge of airframe, engine and electrical power systems and equipment in addition to a thorough knowledge of aircraft structures andmaterials. While category B2 maintenance technicians will have an in-depth integrated knowledge of aircraft electrical, instrument, autopilot, radio, radar, communication and navigation systems. The knowledge and skills gained from their initial training, together with aircraft-type knowledge and a substantial period of practical experience, will enable category B technicians, once granted approvals, to undertake one or

more of the following maintenance operations: • In-depth scheduled inspection activities. • Complex rectification activities. • Fault diagnosis on aircraft systems, propulsion units, plant and equipments. • Embodiment of modifications and special technical instructions. • Airframe and other aircraft repairs. • Strip-down and aircraft re-build activities. • Major aircraft component removal, fit and replacement tasks. • Use and interrogate built-in test equipment (BITE) and other diagnostic equipments. • Functional tests and checks on aircraft systems, propulsion units and sub-systems. • Trouble-shooting activities on base and away from base. • Aircraft engine ground running activities. • Rack and re-rack avionic equipments and carry out operational tests and checks on avionic systems. • Supervise and certify the work of less experienced technicians and mechanics. As can be seen from the above list of maintenance operations, the category B maintenance technician can be involved in a very wide and interesting range ofpossible activities. For example, Figure 1.4 shows a photograph of the Boeing 767 flap drive motor and associated linkage mechanism. The main source of power is via the hydraulic motor, scheduled servicing may involve the


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Aircraft engineering principles

1.2.3 The base maintenance category C certifying engineer Before detailing the job role of the category C licensed engineer, it is worth clarifying the major differences in the roles performed by line maintenance certifying staff and base maintenance certifying staff. In the case of the former, the inspections, rectification and other associated maintenance activities take place on the aircraft, on the “live side” of an airfield. Thus the depth of maintenance performed by “line maintenance personnel” is restricted to that accomplishable with the limited tools, equipment and test apparatus available on site. It will include “first-line diagnostic maintenance”, as required. Base maintenance, as its name implies, takes place at a designated base away from the live aircraft movement areas. The nature of the work undertaken on base maintenance sites will be more in-depth than that usually associated with line maintenance and may include: in-depth strip-down and inspection, the embodiment of complex modifications, major rectification activities, off-aircraft component overhaul and repairs. These activities, by necessity, require the aircraft to be on the ground for longer periods oftime and will require the maintenance technicians to be conversant with a variety of specialist inspection techniques, appropriate to the aircraft structure, system or components being worked-on. The category C certifier acts primarily in a maintenance management role, controlling the progress of base maintenance inspections and overhauls. While the actual work detailed for the inspection is carried out by category B technicians and to a limited extent, category A base maintenance mechanics, in accordance with the written procedures and work sheets. These individual activities are directly supervised by category B maintenance certifying technicians, who are responsible for ensuring the adequacy of the work being carried out and the issuing of the appropriate certifications for the individual activities. The category C certifier will upon completion of all base maintenance activities sign-off the aircraft as serviceable and fit for flight. This is done using a special form known as a certificate of

Figure 1.5 Technicians working at height considering the alignment of the APU prior to fit.

operation and inspection of this complex system, which in turn requires the certifying technician to not only have the appropriate system knowledge, but also the whole aircraft knowledge to ensure that other systems are not operated inadvertently. Figure 1.5 shows two technicians working at height on highway staging, considering the alignment of theaircraft auxiliary power unit (APU), prior to raising it into position in the aircraft. To perform this kind of maintenance, to the required standards, individuals need to demonstrate maturity, commitment, integrity and an ability to see the job through, often under difficult circumstances. Similar technician roles exist in the armed forces, where the sub-categories are broken down a little more into, mechanical, electrical/ instrument and avionic technicians, as well as aircraft weapons specialists known as armament technicians or weaponeers. In fact, it is planned from January 2004 that the Royal Air Force (RAF) will begin initial training that follows the civil aviation trade categories. That is mechanical technicians, who will undertake airframe/engine training and to a lesser extent electrical training and avionic technicians, who will eventually cover all avionic systems, in a similar manner to their civil counterparts. Cross-training of existing maintenance personnel is also planned to take place over the next 10 years. The armament technician and weaponeer will still remain as a specialist trade group.


Introduction

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1.3 Opportunities for training, education and career progression Those employed in civil aviation as aircraft certifying staff may work for commercial aircraft companies or work in the field of GA. The legislation surrounding the training and education of those employed in GA is somewhat different (butno less stringent) than those employed by passenger and freight carrying commercial airline companies. The opportunities and career progressions routes detailed below are primarily for those who are likely to be employed with commercial carriers. However, they may in the future, quite easily, be employed by GA organizations. Commercial air transport activities are well understood. In that companies are licensed to carry fare paying passengers and freight, across national and international regulated airspace. GA, on the other hand, is often misunderstood for what it is and what place it holds in the total aviation scene. Apart from including flying for personal pleasure, it covers medical flights, traffic surveys, pipeline inspections, business aviation, civil search and rescue and other essential activities, including pilot training! With the advent of a significant increase in demand for business aviation, it is likely that those who have been trained to maintain large commercial transport aircraft will find increasing opportunity for employment in the GA field. In the UK, and indeed in many countries that have adopted our methods for educating and training prospective aircraft maintenance personnel, there have been, historically, a large number of different ways in which these personnel can obtain initial qualifications and improver training. Since the advent of the recent Joint Aviation Requirements (JAR) legislation on personnellicensing, the methods for obtaining initial education and training have become somewhat more unified. Although there still exist opportunities for the “self-starter”, achievement of the basic license may take longer. The schematic diagrams that follow are based on those issued by the Civil Aviation Authority (CAA),1 Safety Regulation Group (SRG). They show the qualification and experience

Figure 1.6 Category C maintenance engineer explaining the complexity of the technical log to the author.

release to service (CRS). Thus the category C certifying engineer has a very responsible job, which requires a sound all-round knowledge of aircraft and their associated systems and major components (Figure 1.6). The CRS is ultimately the sole responsibility of the category C certifying engineer, who confirms by his/her signature that all required inspections, rectification, modifications, component changes, airworthiness directives, special instructions, repairs and aircraft re-build activities have been carried out in accordance with the laid-down procedures and that all documentation have been completed satisfactorily, prior to releasing the aircraft for flight. Thus, the category C certifying engineer will often be the shift maintenance manager, responsible for the technicians and aircraft under his/her control. The requirements for the issuing of an individual category C licence and the education, training and experience necessary before theissue of such a licence are detailed in the sections that follow. The military equivalent of the category C licence holder will be an experienced maintenance technician who holds at least senior non-commissioned officer (SNCO) rank and has a significant period of experience on aircraft type. These individuals are able to sign-off the military equivalent of the CRS, for and on behalf of all trade technicians, who have participated in the particular aircraft servicing activities.


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Aircraft engineering principles

routes/pathways for the various categories of aircraft maintenance certifying staff, mentioned earlier. 1.3.1 Category A certifying mechanics JAR 147 approved training pathway The JAR 147 approved training organization is able to offer ab initio (from the beginning) learning programmes that deliver JAR 66 basic knowledge and initial skills training that satisfy the regulatory authority criteria. In the case of the UK our regulatory authority is the CAA.

Note that a list of CAA JAR 147 approved training organizations, together with other useful education and training institutions, will be found in Appendix B, at the end of this book. Ab initio programmes in approved training organizations often encompass the appropriate CAA examinations. If the examinations have been passed successfully, then an individual requires 1 year of approved maintenance experience before being able to apply for a category A aircraft maintenancelicense (AML). Note also the minimum age criteria of 21 years, for all certifying staff, irrespective of the category of license being issued (Figure 1.7).

Figure 1.7 Category A qualifications and experience pathways.


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Skilled worker pathway The requirement of practical experience for those entering the profession as non-aviation technical tradesmen is 2 years. This will enable aviation-orientated skills and knowledge to be acquired from individuals who will already have the necessary basic fitting skills needed for many of the tasks likely to be encountered by the category A certifying mechanic. Accepted military service pathway Experienced line mechanics and base maintenance mechanics, with suitable military experience on live aircraft and equipments, will have their practical experience requirement reduced to 6 months. This may change in the future when armed forces personnel leave after being cross-trained. Category B2 AML pathway The skills and knowledge required by category A certifying mechanics is a sub-set of those required by B1 mechanical certifying technicians. Much of this knowledge and many of the skills required for category A maintenance tasks are not relevant by the category B2 avionic certifying technician. Therefore, in order that the category B2 person gains the necessary skills and knowledge required for category A certification, 1 year of practical maintenance experience is considerednecessary. Self-starter pathway This route is for individuals who may be taken on by smaller approved maintenance organizations or be employed in GA, where company approvals can be issued on a task-by-task basis, as experience and knowledge are gained. Such individuals may already possess some general aircraft knowledge and basic fitting skills by successfully completing a state funded education programme. For example, the 2-year full-time diploma that leads to an aeronautical engineering qualification (see Section 1.3.4). However, if these individuals have not practiced as a skill fitter in a related engineering discipline, then it will be necessary to complete

the 3 years of practical experience applicable to this mode of entry into the profession. 1.3.2 Category B certifying technicians The qualification and experience pathways for the issue of category B1 and B2 AMLs are shown in Figures 1.8 and 1.9. Having discussed in some detail pathways 1–5 for the category A licence, it will not be necessary to provide the same detail for the category B pathways. Instead you should note the essential differences between the category B1 and B2 pathways as well as the increased experience periods required for both, when compared with the category A license. Holders of the category A AML require a number of years experience based on their background. This is likely to be less for those wishing to transfer to a category B1 AML, rather than to aB2 AML, because of the similarity in maintenance experience and knowledge that exists between category A and B1 license holders. Conversion from category B2 to B1 or from B1 to B2 requires 1 year of practical experience practicing in the new license area. Plus successful completion of the partial JAR 66 examinations, as specified by the CAA and/or JAR 147 approved training organization. 1.3.3 Category C certifying engineers The three primary category C qualification pathways are relatively simple to understand and are set out in Figure 1.10. Qualification is either achieved through practising as a category B1 or B2 certifying technician, for a minimum period of 3 years or entering the profession as an engineering graduate from a recognized degree. Those individuals wishing to gain a category C AML, using the category B route, will already have met the examination criteria in full. However, those entering the profession as engineering graduates will have to take category B1 or B2 knowledge examinations in full or in part, depending on the nature of the degree studied. Examples of


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Aircraft engineering principles

Figure 1.8 Category B1 qualifications and experience pathways.

Figure 1.9 Category B2 qualifications and experience pathways.


Figure 1.10 Category C qualifications and experience pathways.

Figure 1.11 Non-standard qualification and experience pathways.


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Aircraft engineering principles

Figure 1.12Routes to an honours degree and category A, B and C licenses.

non-standard entry methods and graduate entry methods, together with the routes and pathways to professional recognition are given next. 1.3.4 Non-standard qualification and experience pathways Figure 1.11 illustrates in more detail two possible self-starter routes. The first shows a possible progression route for those wishing to gain the appropriate qualifications and experience by initially serving in the armed forces. The second details a possible model for the 18+ school leaver employed in a semi-skilled role, within a relatively small aircraft maintenance company.

In the case of the semi-skilled self-starter, the experience qualifying times would be dependent on individual progress, competence and motivation. Also note that 18+ is considered to be an appropriate age to consider entering the aircraft maintenance profession, irrespective of the type of license envisaged. 1.3.5 The Kingston qualification and experience pathway In this model, provision has been made for qualification and experience progression routes for category A, B and C AML approval and appropriate professional recognition (Figure 1.12).


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Figure 1.13 Fast-track routes to category B and C AML.

Figure 1.13 also shows the various stoppingoff points, for those individuals wishing to practice as either category A, B or C certifiers. Figure 1.13 shows two possible fast-trackroutes for the qualification and award of either a category B or C license. Fast track in this case means that because of the partnership between Kingston University2 and KLM the total programme is recognized by the CAA for ab initio approval, which reduces the qualifying times to a minimum, as shown in Figures 1.8–1.10. The appropriate practical experience being delivered at KLM’s JAR 147 approved training school at Norwich Airport. Kingston University also has a partnership with the City of Bristol College, which is a JAR 147 approved organization. With the expansion of Kingston’s highly successful programme there will be more opportunities for 18+ school leavers, to undertake ab initio training, leading to the CAA examinations and the

award of a foundation or full B.Eng.(Hons) degree. The Royal Aeronautical Society (RAeS) recognizes that full category B JAR 66 AML holders, with appropriate experience and responsibilities, meet the criteria for professional recognition as incorporated engineers and may, subject to a professional review, use the initials I.Eng. after their name. Honours degree holders, who also hold a full category C AML may, with appropriate further learning to masters degree level, apply for recognition as chartered engineers through the RAeS. This is the highest professional accolade for engineers and recognized internationally as the hallmark of engineering ability, competence and professionalism. Figure 1.14shows where the full category A, B and C aircraft maintenance certifiers sit, within the professional engineering qualification framework. Thus the category A mechanic, can


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Figure 1.14 Routes to aerospace engineering professional recognition. Table 1.1 Type of engineering degree Mechanical engineering bias Aeronautical engineering or Air transport engineering bias Electrical or Electronic engineering bias Module exemption Module 1 Mathematics and Module 2 Physics Module 1 Mathematics, Module 2 Physics and Module 8 Basic aerodynamics Module 1 Mathematics, Module 2 Physics, Module 3 Electrical fundamentals and Module 4 Electronic fundamentals Module 1 Mathematics, Module 2 Physics, Module 3 Electrical fundamentals, Module 4 Electronic fundamentals and Module 8 Basic aerodynamics Complete exemption from Modules 1 to 10. Approved as fast-track route to “C” licence

Avionic engineering bias

Kingston University B.Eng.(Hons) aircraft engineering degree (mechanical engineering bias)

with suitable structured training and experience, gain engineering technician status. The full category B technician, again with appropriate structured training and experience, can apply for Incorporated Engineer recognition. The category C engineer, can with an appropriate masters degree or bachelor (Hons) degree and further learning to masters degree level, eventually gain professional recognition as a charteredengineer. Partial exemptions from JAR 66 examinations may be awarded to recognized engineering

degrees, dependent on the type of degree being studied. These limited exemptions, by degree type are detailed in Table 1.1. No other exemptions are allowed and all other modules applicable to the licence category need to be passed by CAA approved JAR 66 examination. Note: The one exception, where a large amount of exemption is given for graduates of the Kingston B.Eng.(Hons) aircraft engineering degree, which is directly aimed at preparing aircraft maintenance engineers, for their licence examinations.


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15

1.4 CAA licence – structure, qualifications, examinations and levels 1.4.1 Qualifications structure The licensing of aircraft maintenance engineers is covered by international standards that are published by the International Civil Aviation Organization (ICAO). In the UK, the Air Navigation Order (ANO) provides the legal framework to support these standards. The purpose of the licence is not to permit the holder to perform maintenance but to enable the issue of certification for maintenance required under the ANO legislation. This is why we refer to licensed maintenance personnel as “certifiers”. At present the CAA issue licences under two different requirements depending on the maximum take-off mass of the aircraft. For aircraft that exceeds 5700 kg, licenses are issued under JAR 66. The JAR 66 license is common toall European countries who are full members of the Joint Aviation Authority (JAA). The ideal being that the issue of a JAR 66 licence by any full member country is then recognized as having equal status in all other member countries throughout Europe. There are currently over 20 countries throughout Europe that go to make-up the JAA. In US, the US Federal Aviation Administration (USFAA) is the equivalent of the JAA. These two organizations have been harmonized to the point where for example, licences issued under JAR 66 are equivalent to those licences issued under FAR 66, in countries that adhere to FAA requirements. Holders of licences issued under JAR 66 requirements are considered to have achieved an appropriate level of knowledge and competence, that will enable them to undertake maintenance activities on commercial aircraft. Licences for light aircraft (less than 5700 kg) and for airships, continue to be issued under the UK National Licensing Requirements laid down in British Civil Airworthiness Requirements (BCAR) Section L. The intention is that within a few years, light aircraft will be included within JAR 66. At present, this has implications for people who wish to work and obtain licences in GA, where many light aircraft are operated.

Much of the knowledge required for the JAR 66 licence, laid down in this series, is also relevant to those wishing to obtain a Section L licence for light aircraft. Although the basicSection L licence is narrower (see Appendix B) and is considered somewhat less demanding than the JAR 66 licence it is, nevertheless, highly regarded as a benchmark of achievement and competence within the light aircraft fraternity. As mentioned earlier, the JAR 66 license is divided into categories A, B and C, and for category B license, there are two major career options, either a mechanical or avionic technician. For fear of bombarding you with too much information, what was not mentioned earlier was the further subdivisions for the mechanical license. These sub-categories are dependent on aircraft type (fixed or rotary wing) and on engine type (turbine or piston). For clarity, all levels and categories of license that may be issued by the CAA/FAA or member National Aviation Authorities (NAA) are listed below. Levels Category A: Line maintenance certifying mechanic Category B1: Line maintenance certifying technician (mechanical) Category B2: Line maintenance certifying technician (avionic) Category C: Base maintenance certifying engineer Note: When introduced, the light AML will be category B3. Sub-category A A1: Aeroplanes turbine A2: Aeroplanes piston A3: Helicopters turbine A4: Helicopters piston Sub-category B1 B1.1: Aeroplanes turbine B1.2: Aeroplanes piston B1.3: Helicopters turbine B1.4: Helicopters piston


16

Aircraft engineering principles Table 1.2 Syllabus modules by subject Module 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1516 17 18 Content Mathematics Physics Electrical fundamentals Electronic fundamentals Digital techniques and electronic instrument systems Materials and hardware Maintenance practices Basic aerodynamics Human factors Aviation legislation Aeroplane aerodynamics, structures and systems Helicopter aerodynamics, structures and systems Aircraft aerodynamic structures and systems Propulsion Gas turbine engine Piston engine Propeller Airship (to be developed)

Note that the experience requirements for all of the above licences are shown in Figures 1.7–1.10. Aircraft-type endorsements 3 Holders of JAR 66 aircraft maintenance licences in category B1, B2 and C may apply for inclusion of an aircraft-type rating subject to meeting the following requirements. 1. The completion of a JAR 147 approved or JAA/NAA approved type training course on the type of aircraft for which approval is being sought and one which covers the subject matter appropriate to the licence category being endorsed. 2. Completion of a minimum period of practical experience on type, prior to application for type rating endorsement. Type training for category C differs from that required for category B1 or B2, therefore category C type training will not qualify for type endorsement in category B1 or B2. However, type courses at category B1 or B2 level may allow the licence holder to qualify for category C level at the same time, providing they hold a category C basic licence.Licence holders seeking type rating endorsements from the CAA must hold a basic JAR 66 licence granted by the UK CAA. 1.4.2 JAR 66 syllabus modules and applicability The JAR 66 syllabus may be taught and examined on a module-by-module basis. The subject matter of individual modules may vary according to the category of licence being studied. The depth of the subject matter may also vary according to the category. Where this is the case, in this series of books, the greatest depth of knowledge required by category will always be covered. In all, there are currently 17 modules in the JAR 66 syllabus. These modules are tabulated in Table 1.2, together with Table 1.3 indicating their applicability to a particular category and mechanical sub-category.

1.4.3 Examinations and levels The JAR 66 examinations are modular and designed to reflect the nature of the JAR 66 syllabus content. These modular examinations may be taken on CAA premises, or on the premises of approved JAR 147 organizations. The number and type of examination conducted by JAR 147 approved organizations will be dependent on the exact nature of their approval. A list of approved organizations and examination venues will be found at the end of this book in Appendix A. For candidates taking the full modular JAR 66 examinations, information on the conduct and procedures for these examinations will be found in Chapter 23 of the JAA Administrative and Guidance Material.4 TheJAR 66 module content may vary in terms of the subjects covered within the module and the level of knowledge required according to whether or not a category A, B1 or B2 license is being sought. Thus, in this book, we will cover in full JAR 66 Modules 1, 2, 3, 4 and 8. Module 1 (Mathematics, Chapter 2 in this book), will


Introduction Table 1.3 Module applicability to category and mechanical sub-category Module A or B1 aeroplanes with: Turbine engine 1 2 3 4 5 6 7 8 9 10 11b 12 13c 14d 15 16 17
a b

17

A or B1 helicopter with: Turbine engine Piston engine

B2 avionic

Piston engine

a

a

a

a

– – – – – – – – – – – – –

– – – – –

– –

– – –

This module is not applicable to category A. Module 11 is applicable only to mechanical certifying staff. c Module 13 is only applicable to B2 avionic certifying technicians. d Module 14 offers a less in depth treatment of propulsion, designed for study by B2 avionic certifying technicians.

be covered to the depth required by the B1 and B2 technician examination. Further mathematics (chapter 3) is also included, which is designed to assist understanding of Module 2, Physics. The further mathematics is not subject to JAR 66 examination but is still considered by the authors to be very useful foundation knowledge. Those studying for the category A licence should concentrate on fully understanding, the non-calculator mathematics given in Chapter 2 of this book.They should also be able to answer all the test questions at the end of this chapter. Module 2 (Physics, Chapter 4 in this book) is covered to a depth suitable for category B technicians, no distinction is made between B1 and B2 levels of understanding,5 the greatest depth being covered for both categories, as appropriate. The Module 2 content not required by category A mechanics, is mentioned in the introduction to the chapter and reflected in the physics test questions given at the end.

Module 3 (Electrical fundamentals, Chapter 5 in this book) is covered at the category B technician level, with clear indications given between the levels of knowledge required for the category A and B license requirements. Module 4 (Electronic fundamentals, Chapter 6 in this book) is not required by category A mechanics but, as before, the treatment of the differing levels of knowledge for category B1 and B2 will be taken to the greater depth required by B2 technicians. The differences in level again being reflected in the test questions given at the end of the chapter. Module 8 (Basic aerodynamics, Chapter 7 in this book) will be covered in full to category B level, with no demarcation being made between category A and B levels. For the sake of completeness, this chapter will also include brief coverage of aircraft flight control taken from Module 11.1. The typical examination questions directly related to Module 8 will be clearly identified at theend of the chapter.


18

Aircraft engineering principles

Full coverage of the specialist aeroplane aerodynamics, high-speed flight and rotor wing aerodynamics, applicable to Modules 11 and 13 will be covered in the third book in the series, Aircraft Aerodynamics, Structural Maintenance and Repair. Examination papers are mainly multiplechoice type but a written paper must also be passed so that the licence may be issued. Candidates may take one or more papers, at a single examination sitting. The pass mark for each multiple-choice paper is 75%! There is no longer any penalty marking for incorrectly answering individual multiple-choice questions. All multiple-choice questions set by the CAA and by approved organizations have exactly the same form. That is, each question will contain a stem (the question being asked), two distracters (incorrect answers) and one correct answer. The multiple-choice questions given at the end of each chapter in this book are laid out in this form. All multiple-choice examination papers are timed, approximately 1 min and 15 s, being allowed for the reading and answering of each question (see Table 1.4). The number of questions asked depends on the module examination being taken and on the category of licence being sought. The structure of the multiple-choice papers for each module together with the structure of the written examination for issue of the license are given in Table 1.4. More detailed andcurrent information on the nature of the license examinations can be found in the appropriate CAA documentation,6 from which the examination structure detailed in Table 1.4 is extracted. Written paper The written paper required for licence issue contains four essay questions. These questions are drawn from the JAR 66 syllabus modules as follows: Module 7 9 10 Paper Maintenance practices Human factors Aviation legislation Question 2 1 1

1.5 Overview of airworthiness regulation, aircraft maintenance and its safety culture 1.5.1 Introduction All forms of public transport require legislation and regulation for their operation, in order to ensure that safe and efficient transport operations are maintained. Even with strict regulation, it is an unfortunate fact that incidents and tragic accidents still occur. Indeed, this is only to self-evident with the recent spate of rail accidents where the Potters Bar accident in 2002, may very likely be attributable to poor maintenance! When accidents occur on any public transport system, whether travelling by sea, rail or air, it is an unfortunate fact, that loss of life or serious injury may involve a substantial number of people. It is also a fact that the accident rate for air travel is extremely low and it is currently one of the safest forms of travel. The regulation of the aircraft industry can only lay down the framework for the safe and efficient management of aircraft operations, inwhich aircraft maintenance plays a significant part. It is ultimately the responsibility of the individuals that work within the industry to ensure that standards are maintained. With respect to aircraft maintenance, the introduction of the new harmonized requirements under JAA and more recently ECAR should ensure that high standards of aircraft maintenance and maintenance engineering training are found not only within the UK, but across Europe and indeed throughout many parts of the world. In order to maintain these high standards, individuals must not only be made aware of the nature of the legislation and regulation surrounding their industry, but also they need to be encouraged to adopt a mature, honest and responsible attitude to all aspects of their job role. Where safety and personal integrity must be placed above all other considerations, when undertaking aircraft maintenance activities. It is for the above reasons, that a knowledge of the legislative and regulatory framework of the industry and the adoption of aircraft


Introduction Table 1.4 Structure of JAR 66 multiple-choice examination papers Module Number of Time allowed questions (min) Module Number of questions Time allowed (min)

19

1 Mathematics Category A 16 20 Category B1 30 40 Category B2 30 40 2 Physics Category A 30 40 Category B1 50 65 Category B2 50 65 3 Electrical fundamentals Category A 20 25 Category B1 50 65 Category B2 50 65 4 Electronic fundamentalsCategory A – – Category B1 20 25 Category B2 40 50 5 Digital techniques/electronic instrument systems Category A 16 20 Category B1 40 50 Category B2 70 90 6 Materials and hardware Category A 50 65 Category B1 70 90 Category B2 60 75 7 Maintenance practices Category A 70 90 Category B1 80 100 Category B2 60 75 8 Basic aerodynamics Category A 20 25 Category B1 20 25 Category B2 20 25 9 Human factors Category A 20 25 Category B1 20 25 Category B2 20 25

10 Aviation Legislation Category A 40 50 Category B1 40 50 Category B2 40 50 11 Aeroplane aerodynamics, structures and systems Category A 100 125 Category B1 130 165 Category B2 – – 12 Helicopter aerodynamics, structures and systems Category A 90 115 Category B1 115 145 Category B2 – – 13 Aircraft aerodynamics, structures and systems Category A – – Category B1 – – Category B2 130 165 14 Propulsion Category A – – Category B1 – – Category B2 25 30 15 Gas turbine engine Category A 60 75 Category B1 90 115 Category B2 – – 16 Piston engine Category A 50 65 Category B1 70 90 Category B2 – – 17 Propeller Category A 20 25 Category B1 30 40 Category B2 – –

Note: The time given for examinations may, from time to time, be subject to change. There is currently a review pending of examinations time based on levels. Latest information may be obtained from the CAA website.

maintenance safety culture, becomes a vital part of the education for all individuals wishing to practice as aircraftmaintenance engineers. Set out in this section is a brief introduction to the regulatory and legislative framework, together with maintenance safety culture and the vagaries of human performance. A much fuller coverage of aircraft maintenance legislation and safety

procedures will be found in, Aircraft Engineering Maintenance Practices, the second book in this series. 1.5.2 The birth of the ICAO The international nature of current aircraft maintenance engineering has already been mentioned. Thus the need for conformity of


20

Aircraft engineering principles

standards to ensure the continued airworthiness of aircraft that fly through international airspace is of prime importance. As long ago as December 1944, a group of forward thinking delegates from 52 countries came together in Chicago, to agree and ratify the convention on international civil aviation. Thus the Provisional International Civil Aviation Organization (PICAO) was established. It ran in this form until March 1947, when final ratification from 26 member countries was received and it became the ICAO. The primary function of the ICAO, which was agreed in principle at the Chicago Convention in 1944, was to develop international air transport in a safe and orderly manner. More formerly, the 52 member countries agreed to undersign: certain principles and arrangements in order that international civil aviation may be developed in a safe and orderly manner and thatinternational air transport services may be established on the basis of equality of opportunity and operated soundly and economically. Thus in a spirit of cooperation, designed to foster good international relationships, between member countries, the 52 member states signed up to the agreement. This was a far-sighted decision, which has remained substantially unchanged up to the present. The ICAO Assembly is the sovereign body of the ICAO responsible for reviewing in detail the work of ICAO, including setting the budget and policy for the following 3 years. The council, elected by the assembly for a 3-year term, is composed of 33 member states. The council is responsible for ensuring that standards and recommended practices are adopted and incorporated as annexes into the convention on international civil aviation. The council is assisted by the Air Navigation Commission to deal with technical matters, the Air Transport Committee to deal with economic matters and the Committee on Joint Support of Air Navigation Services and the Finance Committee. The ICAO also works closely with other members of the United Nations (UN) and other non-governmental organizations such as the

International Air Transport Association (IATA) and the International Federation of Air Line Pilots to name but two. 1.5.3 The UK CAA The CAA was established by an act of parliament in 1972, as an independent specialist aviation regulator and provider of air trafficservices.7 Under the act it is responsible to the government for ensuring that all aspects of aviation regulation are implemented and regulated in accordance with the ANO formulated as a result of the act. Following the separation of National Air Traffic Services (NATS) in 2001, the CAA is now responsible for all civil aviation functions, these are: economic regulation, airspace policy, safety regulation and consumer protection. The Economic Regulation Group (ERG) regulates airports, air traffic services and airlines and provides advice on aviation policy from an economic standpoint. Its aim is to secure the best sustainable outcome for users of air transport services. The Directorate of Airspace Policy (DAP) is responsible for the planning and regulation of all UK airspace including the navigation and communication infrastructure to support safe and efficient operations. Both civilian and military experts staff this group. The Consumer Protection Group (CPG) regulates travel organizations, manages the consumer protection organization, air travel organizers’ licensing (ATOL) and licenses UK airlines, in addition to other functions. The Safety Regulation Group (SRG) ensures that UK civil aviation standards are set and achieved in a cooperative and cost-effective manner. SRG must satisfy itself that aircraft are properly designed, manufactured, operated and maintained. It is also the responsibility of this group to ensure the competence offlight crews, air traffic controllers and aircraft maintenance engineers in the form of personal licensing. All the major functions of this group are shown in Figure 1.15. Note: with the recent introduction of European Aviation Safety Agency (EASA),


Introduction

21

Figure 1.15 CAA-SRG functions and responsibilities.

some of these functions (particularly with the certification of individuals and the approval of organizations, concerned with aircraft maintenance) will gradually be transferred from the CAA-SRG to EASA. 1.5.4 Civil aviation requirements The broad international standards on airworthiness set up by the ICAO were backed up by detailed national standards, overseen in the UK by the National Authority for Airworthiness the CAA. These national standards were known in the UK as BCAR and in the USA as, Federal Airworthiness Regulations (FAR). Many other countries adopted one or the other of these requirements, with their own national variations. As international collaborative ventures became more wide spread, there was increasing pressure to produce a unified set of standards, particularly in Europe. Thus came into being (under

the auspices of the JAA) the European Joint Aviation Requirements or JAR, for short. Then, with increasing collaborative ventures between Europe, the USA and other major economies around the world, there became a need to harmonize these European requirements (JAR), with those of the USA, FAR.This harmonization process is still ongoing and is not without difficulties! It is unnecessary in this brief introduction to go into detail on the exact nature of JAA in overseeing the European JAR airworthiness requirements and design protocols. Suffice to say8 that the Civil Aviation Authorities of certain countries have agreed common comprehensive and detailed aviation requirements (JAR) with a view to minimizing type certification problems on joint aviation ventures, to facilitate the export and import of aviation products, and make it easier for maintenance and operations carried out in


22

Aircraft engineering principles

one country to be accepted by the CAA in another country. One or two of the more important requirements applicable to aircraft maintenance organizations and personnel are detailed below: JAR 25 – Requirements for large aircraft (over 5700 kg) JAR E – Requirements for aircraft engines JAR 21 – Requirements for products and parts for aircraft JAR 66 – Requirements for aircraft engineering certifying staff, including the basic knowledge requirements, upon which all the books in this series are based JAR 145 – Requirements for organizations operating large aircraft JAR 147 – Requirements to be met by organizations seeking approval to conduct approved training/examinations of certifying staff, as specified in JAR 66. 1.5.5 Aircraft maintenance engineering safety culture and human factors If you have managedto plough your way through this introduction, you cannot have failed to notice that aircraft maintenance engineering is a very highly regulated industry, where safety is considered paramount! Every individual working on or around aircraft and/or their associated equipments, has a personal responsibility for their own safety and the safety of others. Thus, you will need to become familiar with your immediate work area and recognize and avoid, the hazards associated with it. You will also need to be familiar with your local emergency: first aid procedures, fire precautions and communication procedures. Thorough coverage of workshop, aircraft hangar and ramp safety procedures and precautions will be found in Aircraft Engineering Maintenance Practices, the second book in the series. Coupled with this knowledge on safety, all prospective maintenance engineers must also foster a responsible, honest, mature and

Figure 1.16 Control column, with base cover plate fitted and throttle box assembly clearly visible.

professional attitude to all aspects of their work. You perhaps, cannot think of any circumstances where you would not adopt such attitudes? However, due to the nature of aircraft maintenance, you may find yourself working under very stressful circumstances where your professional judgement is tested to the limit! For example, consider the following scenario. As an experienced maintenance technician, you have been tasked with fittingthe cover to the base of the flying control column (Figure 1.16), on an aircraft that is going to leave the maintenance hanger on engine ground runs, before the overnight embargo on airfield noise comes into force, in 3 hours time. It is thus important that the aircraft is towed to the ground running area, in time to complete the engine runs before the embargo. This will enable all outstanding maintenance on the aircraft to be carried out over night and so ensure that the aircraft is made ready in good time, for a scheduled flight first thing in the morning. You start the task and when three quarters of the way through fitment of the cover, you drop a securing bolt, as you stand up. You think that you hear it travelling across the flight deck floor. After a substantial search by torchlight, where you look not only across the floor, but also around the base of the control column and into other possible crevices, in the immediate area, you are unable to find the small bolt. Would you: (a) Continue the search for as long as possible and then, if the bolt was not found, complete


Introduction

23

the fit of the cover plate and look for the bolt, when the aircraft returned from ground runs? (b) Continue the search for as long as possible and then, if the bolt was not found, inform the engineer tasked with carrying out the ground runs, to be aware that a bolt is somewhere in the vicinity of the base of the control column on the flight deckfloor. Then continue with the fit of the cover? (c) Raise an entry in the aircraft maintenance log for a “loose article” on the flight deck. Then remove the cover plate, obtain a source of strong light and/or a light probe kit and carry out a thorough search at base of control column and around all other key controls, such as the throttle box. If bolt is not found, allow aircraft to go on ground run and continue search on return? (d) Raise an entry in the aircraft log for a “loose article” on the flight deck. Then immediately seek advice from shift supervisor, as to course of action to be taken? Had you not been an experienced technician, you would immediately inform your supervisor (action (d)) and seek advice as to the most appropriate course of action. As an experienced technician, what should you do? The course of action to be taken, in this particular case, may not then be quite so obvious, it requires judgements to be made. Quite clearly actions (a) and (b) would be wrong, no matter how much experience the technician had. No matter how long the search continued, it would be essential to remove the cover plate and search the base of the control column to ensure that it was not in the vicinity. Any loose article could dislodge during flight and cause possible catastrophic jamming or fouling of the controls. If the engine run is to proceed, actions (a) and (b) are still not adequate. A search of the throttle box area for the bolt wouldalso need to take place, as suggested by action (c). Action (c) seems plausible, with the addition of a good light source and thorough search of all critical areas, before the fit of the cover plate, seems a reasonable course of action to take, especially after the maintenance

log entry has been made, the subsequent search for the bolt, cannot be forgotten, so all is well? However, if you followed action (c) you would be making important decisions, on matters of safety, without consultation. No matter how experienced you may be, you are not necessarily aware of the total picture, whereas your shift supervisor, may well be! The correct course of action, even for the most experienced engineer would be action (d). Suppose action (c) had been taken and on the subsequent engine run the bolt, that had been lodged in the throttle box, caused the throttle to jam in the open position. Then shutting down the engine, without first closing the throttle, could cause serious damage! It might have been the case that if action (d) had been followed, the shift supervisor may have been in a position to prepare another aircraft for the scheduled morning flight, thus avoiding the risk of running the engine, before the loose article search had revealed the missing bolt. In any event, the aircraft would not normally be released for service until the missing bolt had been found, even if this required the use of sophisticated radiographic equipment to findit! The above scenario illustrates some of the pitfalls, that even experienced aircraft maintenance engineers may encounter, if safety is forgotten or assumptions made. For example, because you thought you heard the bolt travel across the flight deck, you may have assumed that it could not possibly have landed at the base of the control column, or in the throttle box. This, of course, is an assumption and one of the golden rules of safety is never assume, check! When the cover was being fitted, did you have adequate lighting for the job? Perhaps with adequate lighting, it might have been possible to track the path of the bolt, as it travelled across the flight deck, thus preventing its loss in the first place. Familiarity with emergency equipment and procedures, as mentioned previously is an essential part of the education of all aircraft maintenance personnel. Reminders concerning the use of emergency equipment will be found in hangars, workshops, repair bays and in many


24

Aircraft engineering principles

Figure 1.17 Typical aircraft hangar first aid station.

Figure 1.18 Typical aircraft hangar fire point.

other areas where aircraft engineering maintenance is practiced. Some typical examples of emergency equipment and warning notices are shown below. Figure 1.17 shows a typical aircraft maintenance hangar first aid station, complete with explanatory notices, first aid box and eye irritation bottles. Figure 1.18 shows anaircraft maintenance hangar fire point, with clearly identifiable emergency procedures in the event of fire and the appropriate fire appliance to use for electrical or other type of fire. Figure 1.19 shows a grinding assembly, with associated local lighting and warning signs, for eye and ear protection. Also shown are the drop-down shields above the grinding wheels to prevent spark burns and other possible injuries to the hands, arms and eyes. Figure 1.20 shows a warning notice concerning work being carried out on open fuel tanks and warning against the use of electrical power. In addition to this warning notice there is also

a no power warning at the aircraft power point (Figure 1.21). You may feel that the module content contained in this book on principles is a long way removed from the working environment illustrated in these photographs. However, consider for a moment the relatively simple task of inflating a ground support trolley wheel (Figure 1.22). Still it is a common practice to measure tyre pressures in pounds per square inch (psi), as well as in bar (Figure 1.23). Imagine the consequences of attempting to inflate such a tyre to 24 bar, instead of 24 psi, because you mis-read the gauge on the tyre inflation equipment! The need to understand units, in this particular case is most important. It cannot happen I hear you say; well unfortunately it can, the above is an account of an actual incident. Fortunately the technician inflatingthe tyre, followed standard safety procedures, in that he stood behind the tyre, rather than along side it, during the inflation process. The tyre separated from


Introduction

25

Figure 1.21 Ground power warning.

Figure 1.19 Grinding wheel assembly, with associated lighting and warning signs. Figure 1.22 Oxygen bottle trolley, showing trolley wheel.

Figure 1.20 Open fuel tanks warning notice. Figure 1.23 Pressure gauges graduated in bar and in psi.

the wheel assembly and shot sideways at high velocity. If the technician had been to the side of the tyre and wheel assembly he would have sustained serious injury! At that time this technician was unaware of the difference in units between the bar and for him, the more familiar imperial units of psi. Thus the need to adopt a

mature attitude to your foundation studies is just as important as adopting the necessary professional attitude to your on-job practical maintenance activities.


26

Aircraft engineering principles

Completing the maintenance documentation When carrying out any form of maintenance activity on aircraft or aircraft equipment, it is vitally important that the appropriate documentation and procedures are consulted and followed. This is particularly important, if the maintenance technician is unfamiliar with the work, or is new to the equipment being worked on. Even those experienced in carrying out a particular activity should regularlyconsult the maintenance manual, in order to familiarize themselves with the procedure and to establish the modification state of the aircraft or equipment being worked on. The modification state of the documentation itself should not only be checked by the scheduling staff, but also by the engineer assigned to the task to ensure currency. When certifying staff sign-up for a particular maintenance activity, there signature implies that the job has been completed to the best of their ability, in accordance with the appropriate schedule and procedures. Any maintenance engineer, who is subsequently found to have produced work that is deemed to be unsatisfactory, as a result of their negligence, during the execution of such work, may be prosecuted. It should always be remembered by all involved in aircraft maintenance engineering that mistakes can cost lives. This is why it is so important that certifying staff always carry out their work to the highest professional standards, strictly adhering to the laid-down safety standards and operational procedures. Human factors The above examples concerning the dropped bolt and the mistakes made when attempting to inflate the ground support trolley tyre illustrate the problems that may occur due to human frailty. Human factors9 impinges on everything an engineer does in the course of their job in one way or another, from communicating effectively with colleagues to ensuring they have adequate lighting tocarry out their tasks. Knowledge of this subject has a significant impact on the safety

standards expected of the aircraft maintenance engineer. The above quote is taken from the CAA publication (CAP 715) which provides an introduction to engineering human factors for aircraft maintenance staff, expanding on the human factors syllabus contain in JAR 66 Module 9. A study of human factors, as mentioned earlier, is now considered to be an essential part of the aircraft maintenance engineers education. It is hoped that by educating engineers and ensuring currency of knowledge and techniques, that this will ultimately lead to a reduction in aircraft incidents and accidents which can be attributed to human error during maintenance. The study of human factors has become so important that for many years the CAA has cosponsored annual international seminars dedicated to the interchange of information and ideas on the management and practice of eliminating aviation accidents, resulting from necessary human intervention. Numerous learned articles and books have been written on human factors, where the motivation for its study has come from the need to ensure high standards of safety in high risk industries, such as nuclear power and of course air transport! Aircraft maintenance engineers thus need to understand, how human performance limitations impact on their daily work. For example, if you are the licensed aircraft engineer (LAE)responsible for a team of technicians. It is important that you are aware of any limitations members of your team may have with respect to obvious physical constraints, like their hearing and vision. As well as more subtle limitations, such as their ability to process and interpret information or their fear of enclosed spaces or heights. It is not a good idea to task a technician with a job inside a fuel tank, if they suffer from claustrophobia! Social factors and other factors that may affect human performance also need to be understood. Issues such as responsibility, motivation, peer pressure, management and supervision need to be addressed. In addition to general fitness, health, domestic and work-related stress, time pressures, nature of the task, repetition, workload and the effects of shift work.


Introduction

27

The nature of the physical environment in which maintenance activities are undertaken needs to be considered. Distracting noise, fumes, illumination, climate, temperature, motion, vibration and working at height and in confined spaces, all need to be taken into account. The importance of good two-way communication needs to be understood and practiced. Communication within and between teams, work logging and recording, keeping up-to-date and the correct and timely dissemination of information must also be understood. The impact of human factors on performance will be emphasized, wherever and whenever it is thoughtappropriate, throughout all the books in this series. There will also be a section in the second book in this series, on Aircraft Engineering Maintenance Practices, devoted to the study of past incidents and occurrences that can be attributed to errors in the maintenance chain. This section is called learning by mistakes. However, it is felt by the authors that human factors as contained in JAR 66 Module 9, is so vast that one section in a textbook, will not do the subject justice. For this reason a list of references are given at the end of this chapter, to which the reader is referred. In particular an excellent introduction to the subject is provided in the CAA publication: CAP 715 – An Introduction to Aircraft Maintenance Engineering Human Factors for JAR 66. We have talked so far about the nature of human factors, but how do human factors impact on the integrity of aircraft maintenance activities? By studying previous aircraft incidents and accidents, it is possible to identify the sequence of events which lead to the incident and so implement procedures to try and avoid such a sequence of events, occurring in the future. 1.5.6 The BAC One-Eleven accident By way of an introduction to this process, we consider an accident that occurred to a BAC One-Eleven, on 10th June 1990 at around 7.30 a.m. At this time the aircraft, which had taken off from Birmingham Airport, had climbed to a height of around 17,300 ft

Figure 1.24 A Boeing767 left front windscreen assembly.

(5273 m) over the town of Didcot in Oxfordshire, when there was a sudden loud bang. The left windscreen, which had been replaced prior to the flight, was blown out under the effects of cabin pressure when it overcame the retention of the securing bolts, 84 of which, out of a total of 90, were smaller than the specified diameter. The commander narrowly escaped death, when he was sucked halfway out of the windscreen aperture and was restrained by cabin crew whilst the co-pilot flew the aircraft to a safe landing at Southampton Airport. For the purposes of illustration, Figure 1.24 shows a typical front left windscreen assembly of a Boeing 767. How could this happen? In short, a task deemed to be safety critical was carried out by one individual, who also carried total responsibility for the quality of the work achieved. The installation of the windscreen was not tested after fit. Only when the aircraft was at 17,300 ft, was there sufficient pressure differential to check the integrity of the work! The shift maintenance manager, who had carried out the work, did not achieve the quality standard during the fitting process, due to inadequate care, poor trade practices, failure to adhere to company standards, use of unsuitable equipment and long-term failure by the maintenance manager to observe the promulgated procedures. The airline’s local management product samples and quality audits, had not detectedthe existence of inadequate standards employed by the shift maintenance manager because they did


28

Aircraft engineering principles

not monitor directly the work practices of shift maintenance managers. Engineering factors There is no room in this brief account of the accident to detail in full all the engineering factors which lead up to the windscreen failure; however, some of the more important factors in the chain of events are detailed below: • Incorrect bolts had been used with the previous installation (A211-7D). • Insufficient stock of the incorrect A211-7D bolts existed in the controlled spare parts carousel dispenser. Although these bolts were incorrect, they had proved through 4 years of use to be adequate. • No reference was made to the spare parts catalogue to check the required bolts’ part number. • The stores system, available to identify the stock level and location of the required bolts was not used. • Physical matching of the bolts was attempted and as a consequence, incorrect bolts (A2118C) were selected from an uncontrolled spareparts carousel, used by the maintenance manager. • An uncontrolled torque limiting screwdriver was set up outside the calibration room. • A bi-hexagonal bit holder was used to wind down the bolts, resulting in the occasional loss of the bit and the covering up of the bolt head. Hence the maintenance manager was unable to see that the countersunk head of the bolts, was furtherrecessed than normal. • The safety platform was incorrectly positioned leading to inadequate access to the job. • The warning from the storekeeper that A2118D bolts were required did not influence the choice of bolts. • The amount of unfilled countersunk left by the small bolt heads was not recognized as excessive. • The windscreen was not designated a “vital task” therefore no duplicate (independent) inspection was required.

Figure 1.25 Simplified schematic cross-section of a typical windscreen requiring external fit.

• The windscreen was not designed so that internal pressure would hold it in place, but was fitted from the outside (Figure 1.25). • The shift maintenance manager was the only person whose work on the night shift was not subject to the review of a maintenance manager. • Poor labelling and segregation of parts in the uncontrolled spare-parts carousel. • The shift maintenance manager did not wear prescribed glasses when carrying out the windscreen change. The impact of human factors The above series of events does not tell the whole story. For example, why was it that the shift maintenance manager was required to perform the windscreen change in the first place? A supervisory aircraft engineer and a further LAE, normally part of the shift, were not available that night. In order to achieve the windscreen change during the night shift and have the aircraft ready for a pre-booked wash


Introduction Table 1.5 Part No.A211-8D A211-8C A211-7D Shank length (in.) 0.8 0.8 0.7 Diameter (in.) 0.1865–0.1895 0.1605–0.1639 0.1865–0.1895 Thread size 10 UNF 8 UNC 10 UNF Comments Correct bolts 84 bolts used Bolts removed

29

early in the morning, the shift maintenance manager decided to carry out the windscreen change by himself. His supervisory aircraft engineer and other airframe engineer were busy rectifying a fault on another BAC One-Eleven aircraft, which needed to be completed before departure of the aircraft the following morning. Also in the early hours of the morning when the windscreen change took place, the bodies’ circadian rhythms are at a low ebb. This, coupled with a high workload, may have lead to tiredness and a reduced ability to concentrate. The highway staging platform was incorrectly positioned for easy access to the job, had this been correctly positioned the maintenance manager may have been better able to notice that the bolt heads were recessed in the countersink, significantly more than usual. The assumption that the bolts removed from the aircraft windscreen were correct was made by the maintenance manager. Thus one of the most important dictums was ignored; never assume, check! The non-availability of the bolts (A211-7D) even though incorrect, in the controlled spare parts carousel, lead the manager to search in a non-controlled carousel, where parts were poorly labelled or incorrectly segregated. This in turn lead the manager toselect the bolts using visual and touch methods. This resulted in the final error, in the chain, being made. The bolts selected were of the correct length but were crucially 0.026 of an inch, too small in diameter. The illustrated parts catalogue (IPC), which should have been consulted before replacing the old bolts, specifies that the attachment bolts should be part number (A211-8D). The specification for these bolts, together with those selected from the carousel (A211-8C) are shown in Table 1.5. The windscreen change on this aircraft was not considered a vital point. The CAA state

that the term “vital point” is not intended to refer to multiple fastened parts of the structure, but applies to a single point, usually in an aircraft control system. In September 1985 BCARs introduced a requirement for duplicate inspections of vital points, which are defined as: any point on an aircraft at which a single malassembly could lead to a catastrophe, resulting in loss of the aircraft or fatalities. Had the windscreen been considered a vital maintenance operation, then a duplicate inspection would have been performed and the excessive recess of the bolt heads may very well have been noticed. Also, there are no CAA requirements for a cabin pressure check to be called up after the work has been carried out on the pressure hull. Such checks are written into the aircraft maintenance manual at the discretion of the aircraft design team, and were notcalled up on the BAC One-Eleven. Had they been necessary, then the sub-standard integrity of the incorrectly fitted windscreen would have been apparent. A full account of this accident, the events leading up to it and the subsequent safety recommendations will be found on the Air Accident Investigation Board website,10 from which some of the above account has been taken. The safety recommendations As a result of the above accident and subsequent inquiry, eight safety recommendations were given. Briefly, these recommendations are as follows: • The CAA should examine the applicability of self-certification to aircraft engineering safety critical tasks following which the components or systems are cleared for service without functional checks. Such a review should include the interpretation of single mal-assembly within the context of vital points.


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Aircraft engineering principles

• British Airways should review their quality assurance system and reporting methods, and encourage their engineers to provide feedback from the shop floor. • British Airways should review the need to introduce job descriptions and terms of reference for engineering grades, including shift maintenance manager and above. • British Airways should provide the mechanism for an independent assessment of standards and conduct an in depth audit into work practices at Birmingham Airport. • The CAA should review the purpose and scope of their supervisoryvisits to airline operators. • The CAA should consider the need for periodic training and testing of engineers to ensure currency and proficiency. • The CAA should recognize the need for corrective glasses, if prescribed, in association with the undertaking of aircraft engineering activities. • The CAA should ensure that, prior to the issue of an air traffic controller (ATC) rating, a candidate undertakes an approved course of training, that includes the theoretical and practical handling of emergency situations. The above recommendations are far reaching and provide an example of human factors involvement, far removed from the direct maintenance activity, but very much impacting on the chain of events leading to an accident or serious incident. It is these complex interactions that may often lead to maintenance errors being made, with subsequent catastrophic consequences. No matter how sophisticated the policies and procedures may be, ultimately due to the influence of human factors, it is the integrity, attitude, education and professionalism of the individual aircraft maintenance engineer, that matters most, in the elimination of maintenance errors.

an insight into the demanding and yet very rewarding work, offered to aircraft maintenance certifying staff. No matter at what point you wish to enter the industry, you will find routes and pathways that enable you to progress to any level, dependent only, on your own ambitions andaspirations. The training and education to reach the top of any profession is often long and arduous and aircraft maintenance engineering is no exception! The subject matter that follows may seem a long way removed from the environment portrayed in this introduction and yet, it forms a vital part of your initial educational development. Therefore, you should approach the subjects presented in Chapters 2 and 3 of this book, with the same amount of enthusiasm and dedication as you will with the practical activities you find yourself engaged in, when qualified to practice your profession. The non-calculator mathematics, you are about to meet, may seem deceptively simple. However, do remember that the pass rate is 75%, as it is for all your JAR 66 examinations. This is likely to be significantly higher than any other examination pass rate, you may have encountered up till now. It is, therefore, very important that you become familiar with all the subject matter contained in the following chapters, if you are to be successful in your future CAA examinations. There are numerous examples, multiple-choice questions and other types of questions provided to assist you in acquiring the necessary standard. References
1. CAA-SRG Engineer Standards, papers 3–6 (May 2001). 2. Kingston University, Rationale for Aerospace Programmes (May 2001). 3. CAA-SRG, JAR-66 Information for New Applicants Leaflet 2 Issue 16 (October 2001). 4. JAA Administration andGuidance Material (1999). 5. JAR-66 Appendix 2 Section 1 Levels (April 2002). 6. CAA-SRG JAR-66 Syllabus and Examinations No. 6 (issued 16/10/01). 7. CAA Corporate Information, page 1–3. (April 2002). 8. JAR-66 Certifying Staff Maintenance, page F1 (April 2002). 9. CAP715 An Introduction to Aircraft Maintenance Human Factors for JAR-66 (January 2002). 10. UK Air accident investigation branch (AAIB). www.dft. gov.uk/stellent/groups/dft_accidentinvest_page.hcsp

1.5.7 Concluding remarks It is hoped that this short introduction into the aircraft maintenance industry has given you


P A R T

2

Scientific fundamentals


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Chapter

Mathematics

2
General introduction This chapter aims to provide you with a sound foundation in mathematical principles, which will enable you to solve mathematical, scientific and associated aircraft engineering problems at the mechanic and technician level. Mathematics is divided into two major parts: Non-calculator mathematics, which covers all of the mathematics laid down in Joint Aviation Requirements (JAR) 66 Module 1, up to the level appropriate for aircraft maintenance category B certifying technicians. The other part of mathematics is Further mathematics (Chapter 3), which in the opinion of the authors, is necessary for a thorough understanding of the physics and electrical principles that follow. A second objective of Further mathematics is to provide themathematical base necessary for further academic and professional progression, particularly for those individuals wishing to become Incorporated Engineers, after successfully obtaining their category B license. We start with some elementary arithmetic. In particular, we review the concepts of number and the laws that need to be followed, when carrying out arithmetic operations, e.g. addition, subtraction, multiplication and division. The important concept of arithmetic estimates and estimation techniques involving various forms of number are also covered. While revising the fundamental principles of number, we consider both explicit numbers and literal numbers (letters), in order to aid our understanding of not only arithmetic operations, but also the algebraic operations that will follow later. Decimal numbers and the powers of 10 are then considered, after which fractional numbers and the manipulation of fractions are covered. The algebraic content of JAR 66 Module 1 is introduced with the study of powers and exponents (indices) of numbers. This, together with your previous knowledge of fractions and fractional numbers, will provide you with the tools necessary to manipulate algebraic expressions and equations. The essential skill of transposition of formulae is also covered. This will be a particularly useful mathematical tool, when you study your physics and electrical principles. We finish our study of algebra by consideringbinary and other number systems and their application to simple logic circuits. In our study of geometry and trigonometry, we start by looking at the methods used for the graphical solution of equations and other functions. This section clearly lays out the idea of graphical axes and scales. We then consider the nature and use of the trigonometric ratios and the solution of right-angled triangles and the circle. The nature and use of rectangular and polar co-ordinate representation systems, for finding bearings and angles of elevation and depression are then considered. We finish our study of noncalculator mathematics with a study of the more important theorems of the circle, together with some geometric constructions, considered particularly useful to solve engineering problems, in particular, as an aid to engineering drawing and marking out. In our Further mathematics (Chapter 3) we build on our initial study of algebra by considering more complex algebraic and logarithmic expressions, functions and formulae. We will use our basic knowledge of graphs to represent more complex algebraic and logarithmic functions and to solve equations and engineering problems, which involve these functions. In addition, we will briefly introduce the concept of complex numbers, which will be found particularly valuable for those wishing to pursue an avionic pathway. Our further study of trigonometry will include the use of trigonometric ratios to solveengineering problems involving measurement.


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Aircraft engineering principles

Next, we introduce and use a variety of statistical methods to gather, manipulate and display scientific and engineering data. We will then consider the ways in which the elementary rules of calculus arithmetic may be used to solve problems involving simple differentiation and integration of algebraic and trigonometric functions. Finally, we use the calculus to solve some elementary engineering problems, which involve rates of change and the summation of areas and volumes. In order to aid your understanding of mathematics, you will find numerous fully worked examples and test your understanding exercises, spread throughout this chapter. In addition, typical example JAR 66 license questions are given at the end of this chapter. Important note: Only very familiar units, such as mass, weight, pressure, length, area and volume are used in this part of the mathematics. The detailed study of units appears in the chapters on physics and electrical principles (Chapters 4 and 5, respectively), where their nature and use is fully explained. Some of the JAR 66 questions, found at the end of this chapter, require the reader to have some understanding of units, which may be gained by studying other sections of the book (in particular, Chapter 4).

2.2 Arithmetic 2.2.1 Numbers and symbols It is generally believed that our present number system began with theuse of the natural numbers, such as 1, 2, 3, 4, . . . . These whole numbers, known as the positive integers, were used primarily for counting. However, as time went on, it became apparent that whole numbers could not be used for defining certain mathematical quantities. For example, a period in time might be between 3 and 4 days or the area of a field might be between 2 and 3 acres (or whatever unit of measure was used at the time). So the positive fractions were introduced, e.g. 1 , 1 and 3 . These two groups of numbers, 2 4 4 the positive integers and the positive fractions, constitute what we call the positive rational numbers. Thus, 711 is an integer or whole number, 1 is a positive fraction and 234 3 is a rational 4 5 number. In fact a rational number is any number that can be expressed as the quotient of two integers, i.e. any number that can be written in the form a/b where a and b represent any integers. Thus 2 , 8 and 1 are all rational numbers. 5 9 The number 1 can be represented by the quotient 1 = 1, in fact any number divided by itself 1 must always be equal to 1. The natural numbers are positive integers, but suppose we wish to subtract a larger natural number from a smaller natural number, e.g. 10 subtracted from 7, we obviously obtain a number which is less than zero, i.e. 7 − 10 = −3. So our idea of numbers must be enlarged to include numbers less than zero called negative numbers. The number zero (0) is unique, it isnot a natural number because all natural numbers represent positive integer values, i.e. numbers above zero and quite clearly from what has been said, it is not a negative number either. It sits uniquely on its own and must be added to our number collection.

NON-CALCULATOR MATHEMATICS
2.1 Introduction As mentioned earlier, this part of the mathematics has been written explicitly to cover all of the syllabus content laid down in JAR 66 Module 1. It can thus be studied independently, by those only wishing to gain the knowledge necessary to pass the Civil Aviation Authority (CAA) examination for this module. However, in order to offer the best chance of success in the JAR 66 physics and electrical and electronic principles modules and as a preparation for further study, the authors, strongly recommend that you should also study the further mathematics contained in Chapter 3.

Key point
The natural numbers are known as positive integers.


Mathematics

35

So to the natural numbers (positive integers) we have added negative integers, the concept of zero, positive rational numbers and negative natural numbers. What about numbers like √ 2? This is not a rational number because it cannot be represented by the quotient of two integers. So yet another class of number needs to be included, the irrational or non-rational numbers. Together all, the above kinds of numbers constitute the broad class of numbers known as real numbers.They include positive and negative terminating and non-terminating decimals (e.g. ± 1 = 9 ±0.1111 . . . , 0.48299999, ±2.5, 1.73205 . . .). The real numbers are so called to distinguish them from others, such as imaginary or complex numbers, the latter may be made up of both real and imaginary number parts. Complex numbers will not be considered during our study of mathematics. Key point
A rational number is any number that can be expressed as the quotient of two integers, i.e. a/b where a and b are any two integers.

the number with its sign changed. For example, the absolute value of +10 is 10 and the absolute value of −10 is also 10. Now the absolute value of any number n is represented by the symbol |n|. Thus |+24| means the absolute value of +24. Which is larger, |+3| or |−14|? I hope you said |−14| because its absolute value is 14, while that of |+3| is 3 and of course 14 is larger than 3. We are now ready to consider the laws of signs. Key point
The absolute value of any number n is always its positive value or modulus and is represented by |n|.

Although we have mentioned negative numbers, we have not considered their arithmetic manipulation. All positive and negative numbers are referred to as signed numbers and they obey the arithmetic laws of sign. Before we consider these laws, let us first consider what we mean by signed numbers. Conventional representation of signed numbers is shown below, with zero at themidpoint. Positive numbers are conventionally shown to the right of zero and negative numbers to the left: · · · −4 −3 −2 −1 0 +1 +2 +3 +4 · · · The number of units a point is from zero, regardless of its direction, is called the absolute value of the number corresponding to the point on the above number system when points are drawn to scale. Thus the absolute value of a positive number, or of zero, is the number itself. While the absolute value of a negative number is

The laws of signs You are probably already familiar with these laws, here they are: First law: To add two numbers with like signs, add their absolute values and prefix their common sign to the result. This law works for ordinary arithmetic numbers and simply defines what we have always done in arithmetic addition. For example, 3 + 4 = 7 or in full (+3) + (+4) = +7. After the introduction of the negative numbers, the unsigned arithmetic numbers became the positive numbers, as illustrated above. So now all numbers may be considered either positive or negative, and the laws of signs apply to them all. Does the above law apply to the addition of two negative numbers? From ordinary arithmetic we know that (−7) + (−5) = −12. This again obeys the first law of signs, because we add their absolute value and prefix their common sign. Second law: To add two signed numbers with unlike signs, subtract the smaller absolute value from the larger and prefix the sign of the number with thelarger absolute value to the results. So following this rule, we get for example: 5+(−3) = 2; and so on. −12+9 = −3; 6+(−11) = −5


36

Aircraft engineering principles

The numbers written without signs are, of course, positive numbers. Notice that brackets have been removed when not necessary. Third law: To subtract one signed number from another, change the sign of the number to be subtracted and follow the rules for addition. For example, if we subtract 5 from −3, we get −3 − (+5) = −3 + (−5) = −8. Now what about the multiplication and division of negative and positive numbers, so as not to labour the point the rules for these operations are combined in our fourth and final law. Fourth law: To multiply (or divide) one signed number by another, multiply (or divide) their absolute values; then, if the numbers have like signs, prefix the plus sign to the result; if they have unlike signs, prefix the minus sign to the result. Therefore, applying this rule to the multiplication of two positive numbers, e.g. 3 × 4 = 12, 12 × 8 = 96 . . . and so on, which of course, is simple arithmetic! Now applying the rule to the multiplication of mixed sign numbers we get e.g. −3 × 4 = −12, 12 × (−8) = −96 . . . and so on. We can show, equally well, that the above rule yields similar results for division. Example 2.1 Apply the fourth law to the following arithmetic problems and determine the arithmetic result: (a) (−4)(−3)(−7) = ? (b) 14/−2 = ? (c)5(−6)(−2) = ? (d) −22/−11 = ? (a) In this example we apply the fourth law twice, (−4)(−3) = 12 (like signs) and so 12(−7) = −84. (b) 14/–2 applying the third law for unlike signs immediately gives −7, the correct result. (c) Again applying the third law twice. 5(−6) = −30 (unlike signs) and (−30)(−2) = 60. (d) −22/−11 applying the third law for like sign gives 2, the correct result. The use of symbols We have introduced earlier the concept of symbols to represent numbers when we defined rational numbers where the letters a and b were used to represent any integer. Look at

the symbols below, do they represent the same number? √ IX; 9; nine; + 81 I hope you answered yes, since each expression is a perfectly valid way of representing the positive integer 9. In algebra we use letters to represent Arabic numerals such numbers are called general numbers or literal numbers, as distinguished from explicit numbers like 1, 2, 3, etc. Thus a literal number is simply a number represented by a letter, instead of a numeral. Literal numbers are used to state algebraic rules, laws and formulae; these statements being made in mathematical sentences called equations. If a is a positive integer and b is 1, what is a/b? I hope you were able to see that a/b = a. Any number divided by 1 is always itself. Thus, a/1 = a, c/1 = c, 45.6/1 = 45.6. Suppose a is again any positive integer, but b is 0. What is the value of a/b? What we are asking is what is thevalue of any positive integer divided by zero? Well the answer is that we really do not know! The value of the quotient a/b, if b = 0, is not defined in mathematics. This is because there is no such quotient that meets the conditions required of quotients. For example, you know that to check the accuracy of a division problem, you can multiply the quotient by the divisor to get the dividend. For example, if 21/7 = 3, then 7 is the divisor, 21 is the dividend and 3 is the quotient and so 3 × 7 = 21, as expected. So, if 17/0 were equal to 17, then 17 × 0 should again equal 17 but it does not! Or, if 17/0 were equal to zero, then 0 × 0 should equal 17 but again it does not. Any number multiplied by zero is always zero. Therefore, division of any number by zero (as well as zero divided by zero) is excluded from mathematics. If b = 0, or if both a and b are zero, then a/b is meaningless. Key point
Division by zero is not defined in mathematics.

When multiplying literal numbers together we try to avoid the multiplication sign (×), this is because it can be easily mistaken for the letter x.


Mathematics

37

Thus, instead of writing a × b for the product of two general numbers, we write a · b (the dot notation for multiplication) or more usually just ab to indicate the product of two general numbers a and b. Example 2.2 If we let the letter n stand for any real number, what does each of the following expressions equal? (a) n/n =? (b) n × 0 = ? (c) n × 1 = ? (d) n + 0 = ? (e) n − 0 = ? (f) n − n = ? (g) n/0 = ? (a) n/n = 1, i.e. any number divided by itself is equal to 1. (b) n × 0 = 0, any number multiplied by zero is itself zero. (c) n × 1 = n, any number multiplied or divided by 1 is itself. (d) n + 0 = n, the addition of zero to any number will not alter that number. (e) n − 0 = n, the subtraction of zero from any number will not alter that number. (f) n − n = 0, subtraction of any number from itself will always equal zero. (g) n/0, division by zero is not defined in mathematics. The commutative, associative and distributive laws We all know that 6 × 5 = 30 and 5 × 6 = 30, so is it true that when multiplying any two numbers together, the result is the same no matter what the order? The answer is yes. The above relationship may be stated as: The product of two real numbers is the same no matter in what order they are multiplied. That is, ab = ba this is known as the commutative law of multiplication. If three or more real numbers are multiplied together, the order in which they are multiplied still makes no difference to the product. For example, 3 × 4 × 5 = 60 and 5 × 3 × 4 = 60. This relationship may be stated formally as: The product of three or more numbers is the same no matter in what manner they are

grouped. That is, a(bc) = (ab)c; this is known as the associative law of multiplication. These laws may seem ridiculously simple, yet they form thebasis of many algebraic techniques, which you will be using later! We also have commutative and associative laws for addition of numbers, which by now will be quite obvious to you, here they are: The sum of two numbers is the same no matter in what order they are added. That is, a + b = b + a. This is known as the commutative law of addition. The sum of three or more numbers is the same no matter in what manner they are grouped. That is, (a + b) + c = a + (b + c). This is known as the associative law of addition. You may be wondering where the laws are for subtraction. Well you have already covered these when we considered the laws of signs. In other words, the above laws are valid no matter whether or not the number is positive or negative. So, for example, −8 + (16 − 5) = 3 and (−8 + 16) − 5 = 3 In order to complete our laws we need to consider the following problem: 4(5 + 6) = ? We may solve this problem in one of two ways, firstly by adding the numbers inside the brackets and then multiplying the result by 4, this gives: 4(11) = 44. Alternatively, we may multiply out the bracket as follows: (4 × 5) + (4 × 6) = 20 + 24 = 44. Thus, whichever method we choose, the arithmetic result is the same. This result is true in all cases, no matter how many numbers are contained within the brackets! So in general, using literal numbers we have: a(b + c) = ab + ac This is the distributive law. In words, it is rather complicated: Thedistributive law states that: the product of a number by the sum of two or more numbers is equal to the sum of the products of the first number by each of the numbers of the sum. Now, perhaps you can see the power of algebra in representing this law, it is a lot easier to remember than the wordy explanation!


38

Aircraft engineering principles

Remember that the distributive law is valid no matter how many numbers are contained in the brackets, and no matter whether the sign connecting them is a plus or minus. As you will see later, this law is one of the most useful and convenient rules for manipulating formulae and solving algebraic expressions and equations. Key point
The commutative, associative and distributive laws of numbers are valid for both positive and negative numbers.

in your head; we will use a particular method of long multiplication to obtain the result. The numbers are first set out, one under the 35 other, like this: where the right-hand integers 24 5 and 4 are the units and the left-hand integers are the tens, i.e. 3 × 10 and 2 × 10. We multiply the tens on the bottom row by the tens and units on the top row. So to start this process, we place a nought in the units column underneath the bottom row, then multiply the 2 by 5 to get 1 × 10, carry the 1 into the tens column and add it to the product 2 × 3; i.e.: 35 24 0 then multiply the 2 × 5 = 10, put in the nought of the ten and carry the one 35 24
1

Example 2.3If a = 4, b = 3 and c = 7, does a(b − c) = ab − ac The above expression is just the distributive law, with the sign of one number within the bracket, changed. This of course is valid since the sign connecting the numbers within the bracket may be a plus or minus. Nevertheless, we will substitute the arithmetic values in order to check the validity of the expression. Then: 4(3 − 7) = 4(3) − 4(7) 4(−4) = 12 − 28 −16 = −16 So, our law works irrespective of whether the sign joining the numbers is positive or negative. Long multiplication It is assumed that the readers of this book will be familiar with long multiplication and long division. However, with the arrival of the calculator these techniques are seldom used and quickly forgotten. CAA license examinations, for category A and B certifying staff, do not allow the use of calculators; so these techniques will need to be revised. One method of long multiplication is given below. Long division will be found in Section 2.3, where the technique is used for both explicit and literal numbers! Suppose we wish to multiply 35 by 24, i.e. 24 × 35. You may be able to work this out

00

now multiply 2 × 3 = 6 (the tens) and add the carried ten to it, to give 7, then 35 24 700 We now multiply the 4 units by 35. That is 4 × 5 = 20 put down the nought carry 2 into the ten column, then multiply the 4 units by the 3 tens or, 4 × 3 = 12 and add to it the 2 we carried to give 140, i.e.: 35 24 700140 All that remains for us to do now is add 700 to 140 to get the result by long multiplication, i.e.: 35 24 700 140 840


Mathematics

39

So 35 × 24 = 840. This may seem a rather longwinded way of finding this product. You should adopt the method you are familiar with. This process can be applied to the multiplication of numbers involving hundreds, thousands and decimal fractions, it works for them all! For example, 3.5 × 2.4 could be set out in the same manner as above, but the columns would be for tenths and units, instead of units and tens. Then we would get: 3.5 2.4 7.0 1.4 8.4 Notice that in this case the decimal place has been shifted two places to the left. If you do not understand why this has occurred you should study carefully the section on decimals and the powers of 10 that follows. Example 2.4 Multiply: (1) 350 × 25 (2) 18.8 × 1.25 In both the cases, the multiplication is set out as shown before. 1. With these figures, hundreds, tens and units are involved. You will find it easier to multiply it by the smallest or the least complex number. 350 Now we multiply by 25 in a similar 25 manner to the previous example. Multiply first by the 2 × 10, which means placing a nought in the units column first. Then multiply 2 × 0, putting down below the line the result, i.e. zero. Then: 2 × 5 = 1 × 10, again put down the nought and carry the single hundred. So we get: 350 25
1

the nought, we first put down). This part of theprocess was the equivalent of multiplying 350 × 20 = 7000. So we get: 350 25 7000 We now multiply the number 350 by 5, where 5 × 0 = 0; put it down below the line; 5 × 5 = 25 put down the 5 and carry the 2. Finally, 5 × 3 = 15, add the 2 you have just carried to give 17. So the total number below the 7000 is 1750 = 350 × 5 and we get: 350 25 7000 1750 Finally we add the rows below the line to give the result, i.e.: 350 25 7000 1750 8750 Then 350 × 25 = 8750. 2. For this example the multiplication is laid out in full, without explanation, just make sure you can follow the steps. 18.8 1.25 18800 3760 940 23.500 Then, 18.8 × 1.25 = 23.5. Note that the decimal point is positioned three places to the left, since there are three integers to the right of the decimal points. You should now attempt the following exercise, without the aid of a calculator!

00

We continue the process by multiplying 2 by the 3 hundreds and adding the single hundred or 2 × 3 + 1 = 7 to give 7000 (remembering


40

Aircraft engineering principles

Test your understanding 2.1
1. 6, 7, 9, 15 are ___________ numbers. 2.
8 1 7 5 , 4 , 64

Key point
Decimal numbers may be rational, irrational or real numbers.

are ___________ numbers.

3. Rewrite the numbers 5, 13, 16 in the form a/b, where b = 6. 4. Express the negative integers −4, −7, −12 in the form a/b, where b is the positive integer 4. √ 5. + 16 can be expressed as a positive ___________. Itis ___________. √ 6. 10 cannot be expressed as a ___________ number; however, it is a ___________. 7. Express as non-terminating decimals: (a) 1 , (b) 1 , (c) 2. 3 7 8. Find the value of: (a) a(b + c − d), where a = 3, b = −4, c = 6 and d = −1 (b) (21 − 6 + 7)3 (c) 6 × 4 + 5 × 3 9. Which of the following has the largest absolute value: −7, 3, 15, −25, −31? 10. −16 + (−4) − (−3) + 28 = ? 11. Find the absolute value of −4 × (14 − 38) + (−82) = ? 12. What is (a)
15 −3

Essentially then, decimal numbers may be expressed in index form, using the powers of ten. For example: 1,000,000 = 1 × 106 100,000 = 1 × 105 10,000 = 1 × 104 1000 = 1 × 103 100 = 1 × 102 10 = 1 × 101 0 =0 1/10 = 0.1 = 1 × 10−1 1/100 = 0.01 = 1 × 10−2 1/1000 = 0.001 = 1 × 10−3 1/10,000 = 0.0001 = 1 × 10−4 1/100,000 = 0.00001 = 1 × 10−5 1/1,000,000 = 0.000001 = 1 × 10−6 I am sure you are familiar with the above shorthand way of representing numbers. For example, we show the number one million (1,000,000) as 1 × 106 , i.e. 1 multiplied by 10, six times. The exponent (index) of 10 is 6, thus the number is in exponent or exponential form, the exp button on your calculator! Note that we multiply all the numbers, represented in this manner by the number 1. This is because we are representing one million, one hundred thousand, one tenth, etc. When representing decimal numbers in index (exponent) form, the multiplier is always a number which is ≥1.0 or XL , respectively. √Note that, when XL > XC , Z = [R2 + (XL − XC )2 ] and φ = arctan(XL − XC )/R, similarly, √ when XC > XL , Z = [R2 + (XC − XL )2 ] and φ = arctan(XC − XL )/R. It is important to note that a special case occurs when XC = XL in which case the two equal but opposite reactances effectively cancel each other out. The result of this is that the circuit behaves as if only resistance, R, is present (in other words, the impedance of the circuit, Z = R). In this condition the circuit is said to be resonant. The frequency at which resonance occurs is given by: XC = XL thus 1 = 2πfL 2πfC from which f2 = and thus 1 4π2 LC 1 √

Figure 5.162 Phasor diagram for the series R–C circuit when X L > X C .

f =

2π LC

Figure 5.164 Impedance triangle for the series R–C circuit when X L > X C .

Figure 5.163 Phasor diagram for the series R–C circuit when X C > X L .

Figure 5.165 Impedance triangle for the series R–C circuit when X C > X L .


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where f is the resonant frequency (in Hz), L is the inductance (in H) and C is the capacitance (in F). Example 5.83 A series circuit comprises an inductor of 80 mH, a resistor of 200 and a capacitor of 22 µF. If a sinusoidal current of 40 mA at 50 Hz flows in this circuit, determine: (a) (b) (c) (d) (e) (f) the voltage developed across the inductor, the voltage dropped across the capacitor, the voltage dropped across the resistor, the impedance of the circuit,the supply voltage, the phase angle.

=√

0.159 4× 10−10

=

0.159 0.159 = 2 × 10−5 2 × 10−5

= 7950 = 7.95 kHz At the resonant frequency the circuit will behave as a pure resistance (recall that the two reactances will be equal but opposite) and thus the supply current at resonance can be determined from: I= V V 20 = = = 0.4 A Z R 50

5.15.9 Parallel and series–parallel AC circuits As we have seen, in a series AC circuit the same current flows through each component and the supply voltage is found from the phasor sum of the voltage that appears across each of the components present. In a parallel AC circuit, by contrast, the same voltage appears across each branch of the circuit and the total current taken from the supply is the phasor sum of the currents in each branch. For this reason we normally use voltage as the reference quantity for a parallel AC circuit rather than current. Rather than simply quote the formulae, we shall illustrate the techniques for solving parallel, and series–parallel AC circuits by taking some simple examples. Example 5.85 A parallel AC circuit comprises a resistor, R, of 30 connected in parallel with a capacitor, C, of 80 µF. If the circuit is connected to a 240 V, 50 Hz supply determine: (a) (b) (c) (d) the current in the resistor, the current in the capacitor, the supply current, the phase angle.

Solution (a) VL = IXL = I × 2πfL = 0.04 × 25.12 =1V (b) VC = IXC = I × 1/(2πfC) = 0.04 ×144.5 = 5.8 V (c) VR = IR = 0.04 × 200 = 8 V (d) Z = R2 + (XC − XL )2 = √2002 + (144.5 − 25.12)2 = 54,252 = 232.9 (e) V = I × Z = 0.04 × 232.9 = 9.32 V (f) φ = arctan(XC − XL )/R = arctan(119.38/ 200) = arctan(0.597) = 30.8a—Š Example 5.84 A series circuit comprises an inductor of 10 mH, a resistor of 50 and a capacitor of 40 nF. Determine the frequency at which this circuit is resonant and the current that will flow in it when it is connected to a 20 V AC supply at the resonant frequency. Solution Using: 2π LC −3 where L = 10 × 10 H and C = 40 × 10−9 F gives: f = 6.28 10 × 10−3 × 40 × 10−9 √ 1 f = 1 √

Solution Figure 5.166 shows the parallel circuit arrangement showing the three currents present; I1 (the current in the resistor), I2 (the current in the capacitor) and IS (the supply current). Figure 5.167 shows the phasor diagram for the parallel


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(d) The phase angle, φ, can be determined from: cos φ = 8 I1 in-phase current = = = 0.8 supply current IS 10 φ = 36a—Š 52 (leading) Example 5.86
Figure 5.166 A parallel AC circuit: see Example 5.85.

from which:

A series–parallel AC circuit is shown in Figure 5.168. If this circuit is connected to a 110 V, 400 Hz AC supply, determine: (a) (b) (c) (d) the current in the resistive branch, the current in the inductive branch, the supply current, the phase angle.

Figure 5.167 Phasor diagram for the circuit shown in Figure 5.166.

circuit. From thissecond diagram it is important to note the following: • the supply voltage, V, is used as the reference phasor; • the capacitor current, I2 , leads the supply voltage, V (and the resistor current, I1 ) by 90a—Š . (a) The current flowing in the resistor can be determined from: 240 V = = 8 A (in-phase with the I1 = R 30 supply voltage) (b) The current flowing in the capacitor can be determined from: I2 = V = XC V
1 2πfC

Figure 5.168 A series-parallel AC circuit.

Solution Figure 5.169 shows the phasor diagram for the parallel circuit. From the phasor diagram it is important to note the following: • the supply voltage, V, is once again used as the reference phasor; • the phase angle between the supply voltage, V, and supply current, IS , is denoted by φ; • the current in the inductive branch, I2 , lags the supply voltage (and the current in the resistive branch) by a phase angle, φ2 . (a) The current flowing in the 22 be determined from: I1 = resistor can

= V × 2πfC

Thus I2 = 240 × 6.28 × 50 × 80 × 10−6 = 6 A (leading the supply voltage by 90a—Š ). (c) Since I1 and I2 are at right angles to one another (see Figure 5.167) we can determine the supply current, IS , from: IS =
2 2 I1 + I 2 =

√ 82 + 62 = 100 = 10 A

110 V = = 5 A (in-phase with the R 22 supply voltage)


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Figure 5.169 Phasor diagram for the circuit shown in Figure 5.168.

Figure 5.170 Phasor diagram showing total in-phaseand total quadrature components in Example 5.86.

(b) The current flowing in the capacitor can be determined from: I2 = V = Z V R2 +
2 XL

The total quadrature current, Iy , is given by: Iy = I2 sin φ2 = 8.14 × 0.93 = 7.57 A The supply current, IS , can now be determined from: √ IS = 8.012 + 7.572 = 64.16 + 57.3 √ = 121.46 = 11.02 A (d) The phase angle, φ, can be determined from: cos φ = in-phase current 8.01 = = 0.73 supply current 11.02 φ = 43.4a—Š (lagging) Key point
We use current as the reference phasor in series AC circuit because the same current flows through each component. Conversely, we use voltage as the reference phasor in a parallel AC circuit because the same voltage appears across each component.

=

V R2 + 2πf L
2

from which: I2 = = 110 52 + 6.28 × 400 × 5 × 10−3 110 110
2

110 =√ = 2 13.52 182.75 52 + (12.56)

= 8.14 A Thus I2 = 8.14 A (lagging the supply voltage by φ2 ). The phase angle for the inductive branch, φ2 , can be determined from: cos φ2 = or XL 12.56 = = 0.93 Z 13.52 from which φ2 = 68.3a—Š Hence the current in the inductive branch is 8.14 A lagging the supply voltage by 68.3a—Š . (c) In order to determine the supply current we need to find the total in-phase current and the total quadrature current (i.e. the total current at 90a—Š ) as shown in Figure 5.170. The total in-phase current, Ix , is given by: sin φ2 = Ix = I1 + I2 cos φ2 = 5 + (8.14 × 0.37) = 5 + 3.01 = 8.01 A R2 5 = = 0.37 Z 13.52from which:

5.15.10 Power factor The power factor in an AC circuit containing resistance and reactance is simply the ratio of true power to apparent power. Hence: Power factor = true power apparent power

The true power in an AC circuit is the power that is actually dissipated as heat in the resistive component. Thus: True power = I 2 R


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where I is r.m.s. current and R is the resistance. True power is measured in watts (W). The apparent power in an AC circuit is the power that is apparently consumed by the circuit and is the product of the supply current and supply voltage (which may not be in phase). Note that, unless the voltage and current are in phase (i.e. φ = 0a—Š ) the apparent power will not be the same as the power which is actually dissipated as heat. Hence: Apparent power = IV where I is r.m.s. current and V is the supply voltage. To distinguish apparent power from true power, apparent power is measured in volt-amperes (VA). Now since V = IZ we can re-arrange the apparent power equation as follows: Apparent power = IV = I × IZ = I Z
2

Example 5.88 A coil having an inductance of 150 mH and resistance of 250 is connected to a 115 V, 400 Hz AC supply. Determine: (a) the power factor of the coil, (b) the current taken from the supply, (c) the power dissipated as heat in the coil. Solution (a) First we must find the reactance of the inductor, XL , and the impedance, Z, of the coilat 400 Hz. XL = 2π × 400 × 150 × 10−3 = 376 and Z=
2 R2 + XL =

2502 + 3762 = 452

Now returning to our original equation: Power factor = = true power I2R = apparent power IV I2R I2R R = 2 = I × IZ I Z Z

We can now determine the power factor from: R 250 Power factor = = = 0.553 Z 452 (b) The current taken from the supply can be determined from: V 115 I= = = 0.254 A Z 452 (c) The power dissipated as heat can be found from: True power = power factor × VI = 0.553 × 115 × 0.254 = 16.15 W Key point
In an AC circuit the power factor is the ratio of true power to apparent power. The power factor also the cosine of the phase angle between the supply current and supply voltage.

From the impedance triangle shown earlier in Figure 5.153, we can infer that: R Power factor = = cos φ Z Example 5.87 An AC load has a power factor of 0.8. Determine the true power dissipated in the load if it consumes a current of 2 A at 110 V. Solution Now since: Power factor = cos φ = true power apparent power

Test your understanding 5.15
1. In a circuit containing pure capacitance the _____ _____ will lead the _________ by an angle of _________. 2. Determine the reactance of a 220 nF capacitor at (a) 400 Hz and (b) 20 kHz.

True power = power factor × apparent power = power factor × VI Thus: True power = 0.8 × 2 × 110 = 176 W


414
3. Determine the reactance of a 60 mH inductor at (a) 20 Hz and (b) 4 kHz. 4. A 0.5 µF capacitor is connected toa 110 V, 400 Hz supply. Determine the current flowing in the capacitor. 5. A resistor of 120 is connected in series with a capacitive reactance of 160 . Determine the impedance of the circuit and the current flowing when the circuit is connected to a 200 V AC supply. 6. A capacitor or 2 µF is connected in series with a 100 resistor across a 24 V, 400 Hz AC supply. Determine the current that will be supplied to the circuit and the voltage that will appear across each component. 7. An 80 mH coil has a resistance of 10 . Calculate the current flowing when the coil is connected to a 250 V, 50 Hz supply. 8. Determine the phase angle and power factor for Question 7 (supra). 9. An AC load has a power factor of 0.6. If the current supplied to the load is 5 A and the supply voltage is 110 V determine the true power dissipated by the load. 10. An AC load comprises a 110 resistor connected in parallel with a 20 µF capacitor. If the load is connected to a 220 V, 50 Hz supply, determine the apparent power supplied to the load and its power factor.

Aircraft engineering principles

Figure 5.171 The principle of the transformer.

due to leakage). A sinusoidal current flowing in the primary winding produces a sinusoidal flux within the transformer core. At any instant the flux, , in the transformer core is given by the equation: =
max

sin(ωt)

5.16 Transformers Syllabus Transformer construction principles and operation; Transformer losses andmethods for overcoming them; Transformer action under load and no-load conditions; Power transfer, efficiency, polarity markings; Primary and secondary current, voltage, turns ratio, power, efficiency; Auto transformers. Knowledge level key
A – B1 2 B2 2

where max is the maximum value of flux (in Wb) and t is the time in seconds. You might like to compare this equation with the one that you met earlier for a sine wave voltage in Section 5.14.5. The r.m.s. value of the primary voltage (VP ) is given by: VP = 4.44fNP
max

Similarly, the r.m.s. value of the secondary voltage (VS ) is given by: VS = 4.44fNS
max

From these two relationships (and since the same magnetic flux appears in both the primary and secondary windings) we can infer that (Figure 5.172): VP NP = VS NS If the transformer is loss-free the primary and secondary powers will be equal. Thus: PP = PS Now PP = IP × VP and PS = IS × VS

5.16.1 Transformer principles The principle of the transformer is illustrated in Figure 5.171. The primary and secondary windings are wound on a common low-reluctance magnetic core consisting of a number of steel laminations. All of the alternating flux generated by the primary winding is therefore coupled into the secondary winding (very little flux escapes


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Example 5.90 A transformer has 1200 primary turns and is designed to operated with a 110 V AC supply. If the transformer is required toproduce an output of 10 V, determine the number of secondary turns required. Solution
Figure 5.172 Transformer turns and voltages.

Since

VP VS

= NP we can conclude that: NS NP V S 1200 × 10 = = 109.1 VP 110

So IP × VP = IS × VS VS NS From which IP = VP and thus IP = NP IS IS Furthermore, assuming that no power is lost in the transformer (i.e. as long as the primary and secondary powers are the same) we can conclude that: NS IP = IS NP The ratio of primary turns to secondary turns (NP /NS ) is known as the turns ratio. Furthermore, since ratio of primary voltage to primary turns is the same as the ratio of secondary turns to secondary voltage, we can conclude that, for a particular transformer: Turns-per-volt (t.p.v.) = VP VS = NP NS

NS =

Example 5.91 A transformer has a t.p.v. rating of 1.2. How many turns are required to produce secondary outputs of (a) 50 V and (b) 350 V? Solution Here we will use NS = t.p.v. × VS (a) In the case of a 50 V secondary winding: NS = 1.5 × 50 = 75 turns (b) In the case of a 350 V secondary winding: NS = 1.5 × 350 = 525 turns Example 5.92 A transformer has 1200 primary turns and 60 secondary turns. Assuming that the transformer is loss-free, determine the primary current when a load current of 20 A is taken from the secondary. Solution Since
IS IP

The t.p.v. rating can be quite useful when it comes to designing transformers with multiple secondary windings. Example 5.89 A transformerhas 2000 primary turns and 120 secondary turns. If the primary is connected to a 220 V AC mains supply, determine the secondary voltage. Solution Since
VP VS

=

NP NS

we can conclude that: IS NS 20 × 60 = = 1A NP 1200

IP = =
NP NS

we can conclude that:

5.16.2 Transformer applications Transformers provide us with a means of coupling AC power from one circuit to another without a direct connection between the two.

VP NS 220 × 120 VS = = = 13.2V NP 2000


416 Table 5.6 Core material Air Typical power rating Typical frequency range Typical efficiency Typical applications Less than 100 mW 10 MHz to 1 GHz See note Radio receivers and transmitters Ferrite Less than 10 W 1 kHz to 10 MHz 95% to 98% Pulse circuits, switched mode power supplies

Aircraft engineering principles

Laminated steel (low volume) 100 mW to 50 W 50 Hz to 20 kHz 95% typical Audio and lowfrequency amplifiers

Laminated steel (high volume) 3 VA to 500 VA 45 Hz to 500 Hz 90% to 98% Power supplies

A further advantage of transformers is that voltage may be stepped-up (secondary voltage greater than primary voltage) or stepped-down (secondary voltage less than primary voltage). Since no increase in power is possible (like resistors, capacitors and inductors, transformers are passive components) an increase in secondary voltage can only be achieved at the expense of a corresponding reduction in secondary current, and vice versa (in fact, thesecondary power will be very slightly less than the primary power due to losses within the transformer). Typical applications for transformers include stepping-up or stepping-down voltages in power supplies, coupling signals in audio frequency amplifiers to achieve impedance matching and to isolate the DC potentials that may be present in certain types of circuit. The electrical characteristics of a transformer are determined by a number of factors including the core material and physical dimensions of the component. The specifications for a transformer usually include the rated primary and secondary voltages and currents the required power rating (i.e. the rated power, usually expressed in VA), which can be continuously delivered by the transformer under a given set of conditions, the frequency range for the component (usually stated as upper and lower working frequency limits) and the per-unit regulation of a transformer. As we shall see, this last specification is a measure of the ability of a transformer to maintain its rated output voltage under load. Table 5.8 summarizes the properties of some common types of transformer (note how the choice of core material is largely responsible

Figure 5.173 Various transformers.

for determining the characteristics of the transformer) (Figure 5.173). 5.16.3 Transformer regulation The output voltage produced at the secondary of a real transformer falls progressively, as the load imposed on thetransformer increases (i.e. as the secondary current increases from its no-load value). The voltage regulation of a transformer is a measure of its ability to keep the secondary output voltage constant over the full range of output load currents (i.e. from no-load to fullload) at the same power factor. This change, when divided by the no-load output voltage, is referred to as the per-unit regulation for the transformer. This can be best illustrated by the use of an example. Example 5.93 A transformer produces an output voltage of 110 V under no-load conditions and an output


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417

voltage of 101 V when a full-load is applied. Determine the per-unit regulation. Solution The per-unit regulation can be determined for: Per-unit regulation = = VS(no-load) − VS(full-load) VS(no-load)

110 − 101 110 = 0.081 (or 8.1%)
Figure 5.174 Hysteresis curves and energy loss.

5.16.4 Transformer efficiency and losses As we saw earlier, most transformers operate with very high values of efficiency. Despite this, in high power applications the losses in a transformer cannot be completely neglected. Transformer losses can be divided into two types of loss: • Losses in the magnetic core (often referred to as iron loss). • Losses due to the resistance of the coil windings (often referred to as copper loss). Iron loss can be further divided into hysteresis loss (energy lost in repeatedly cycling the magnet flux in the corebackwards and forwards) and eddy current loss (energy lost due to current circulating in the steel core). Hysteresis loss can be reduced by using material for the magnetic core that is easily magnetized and has a very high permeability (see Figure 5.174 – note that energy loss is proportional to the area inside the B–H curve). Eddy current loss can be reduced by laminating the core (e.g. using E- and I-laminations) and also ensuring that a small gap is present. These laminations and gaps in the core help to ensure that there is no closed path for current to flow. Copper loss results from the resistance of the coil windings and it can be reduced by using wire of large diameter and low resistivity. It is important to note that, since the flux within a transformer varies only slightly between the no-load and full-load conditions, iron loss is substantially constant regardless of

the load actually imposed on a transformer. On the other hand, copper loss is zero when a transformer is under no-load conditions and rises to a maximum at full-load. The efficiency of a transformer is given by: Efficiency = from which Efficiency = and Efficiency = 1 − losses × 100% input power input power − losses × 100% input power output power × 100% input power

As we have said, the losses present are attributable to iron and copper loss but the copper loss appears in both the primary and the secondary windings. Hence:  iron loss  + primary copper loss  +secondary copper loss Efficiency = 1 − input power × 100% Once again, we shall explain this with the aid of some examples. 


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Example 5.94 A transformer rated at 500 VA has an iron loss of 3 W and a full-load copper loss (primary plus secondary) of 7 W. Calculate the efficiency of the transformer at 0.8 power factor. Solution The input power to the transformer will be given by the product of the apparent power (i.e. the transformer’s VA rating) and the power factor. Hence: Input power = 0.8 × 500 = 400 W Now Efficiency = 1 − (7 + 3) × 100% = 97.5% 400

5.17 Filters Syllabus Operation, application and uses of the following filters: low pass, high pass, band pass and band stop. Knowledge level key
A – B1 1 B2 1

5.17.1 Types of filter Filters provide us with a means of passing or rejecting AC signals within a specified frequency range. Filters are used in a variety of applications including amplifiers, radio transmitters and receivers. They also provide us with a means of reducing noise and unwanted signals that might otherwise be passed along power lines. Filters are usually categorized in terms of the frequency range that they are designed to accept or reject. Simple filters can be based around circuit (or networks) of passive components (i.e. resistors, capacitors and inductors) whilst those used for signal (rather than power) applications can be based on active components (i.e.transistors and integrated circuits). Most filters are networks having four terminals; two of these terminals are used for the input and two are used for the output. Note that, in the case of an unbalanced network, one of the input terminals may be linked directly to one of the output terminals (in which case this connection is referred to as common). This arrangement is shown in Figure 5.175. The following types of filter are available: • low-pass filter, • high-pass filter,

Test your understanding 5.16
1. Sketch a diagram to illustrate the principle of the transformer. Label your diagram. 2. The core of a power transformer is ___________ in order to reduce _______ _______ current loss. 3. Sketch a B–H curve for the core material of a transformer and explain how this relates to the energy loss in the transformer core. 4. A transformer has 480 primary turns and 120 secondary turns. If the primary is connected to a 110 V AC supply determine the secondary voltage. 5. A step-down transformer has a 220 V primary and a 24 V secondary. If the secondary winding has 60 turns, how many turns are there on the primary? 6. A transformer has 440 primary turns and 1800 secondary turns. If the secondary supplies a current of 250 mA, determine the primary current (assume that the transformer is loss-free). IP NS 7. Show that, for a loss-free transformer, = . IS NP 8. Explain how copper loss occurs in a transformer. How can this loss be minimized? 9. Atransformer produces an output voltage of 220 V under no-load conditions and an output voltage of 208 V when full-load is applied. Determine the per-unit regulation. 10. A 1 kVA transformer has an iron loss of 15 W and a full-load copper loss (primary plus secondary) of 20 W. Determine the efficiency of the transformer at 0.9 power factor.

Figure 5.175 A four-terminal network.


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Figure 5.176 Frequency response for a low-pass filter.

Figure 5.178 Frequency response for a high-pass filter.

Figure 5.177 A simple C–R low-pass filter.

Figure 5.179 A simple C–R high-pass filter.

• band-pass filter, • band-stop filter. Key point
Filters are circuits that pass or reject AC signals within a specified frequency range. Simple passive filters are based on networks of resistors, capacitors and inductors.

from which: 1 2πCR where f is the cut-off frequency (in Hz), C is the capacitance (in F) and R is the resistance (in ). f = 5.17.3 High-pass filters High-pass filters exhibit very low attenuation of signals above their specified cut-off frequency. Below the cut-off frequency they exhibit increasing amounts of attenuation, as shown in Figure 5.178. A simple C–R high-pass filter is shown in Figure 5.179. Once again, the cut-off frequency for the filter occurs when the output voltage has fallen to 0.707 of the input value. This occurs when the reactance of the capacitor, XC , is equal to the value of resistance,R. Using this information we can determine the value of cut-off frequency, f , for given values of C and R. Since R = XC or R= and once again: 1 R= 2πfC f = 1 2πCR 1 2πfC

5.17.2 Low-pass filters Low-pass filters exhibit very low attenuation of signals below their specified cut-off frequency. Beyond the cut-off frequency they exhibit increasing amounts of attenuation, as shown in Figure 5.176. A simple C–R low-pass filter is shown in Figure 5.177. The cut-off frequency for the filter occurs when the output voltage has fallen to 0.707 of the input value. This occurs when the reactance of the capacitor, XC , is equal to the value of resistance, R. Using this information we can determine the value of cut-off frequency, f , for given values of C and R. Since R = XC or


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where f is the cut-off frequency (in Hz), C is the capacitance (in F) and R is the resistance (in ). Key point
The cut-off frequency of a filter is the frequency at which the output voltage has fallen to 0.707 of its input value. Figure 5.180 Frequency response for a band-pass filter.

Example 5.95 A simple C–R low-pass filter has C = 100 nF and = 10 k . Determine the cut-off frequency of the filter. Solution Now f = = 1 2πCR

1 6.28 × 100 × 10−9 × 10 × 104 100 = 6.28 = 15.9 Hz Example 5.96 A simple C–R low-pass filter is to have a cut-off frequency of 1 kHz. If the value of capacitance used in the filter is to be 47 nF, determinethe value of resistance. Solution Now 1 2πCR from which 1 1 = R= 2πfC 6.28 × 1 × 103 × 47 × 10−9 f = = 106 = 3.39 k 295.16

Figure 5.181 A simple L–C band-pass filter (or acceptor ).

attenuation outside this range. This type of filter has two cut-off frequencies: a lower cut-off frequency (f1 ) and an upper cut-off frequency (f2 ). The difference between these frequencies (f2 − f1 ) is known as the bandwidth of the filter. The response of a band-pass filter is shown in Figure 5.180. A simple L–C band-pass filter is shown in Figure 5.181. This circuit uses an L–C resonant circuit (see Section 5.15.8) and is referred to as an acceptor circuit. The frequency at which the band-pass filter in Figure 5.181 exhibits minimum attenuation occurs when the circuit is resonant, i.e. when the reactance of the capacitor, XC , is equal to the value of resistance, R. This information allows us to determine the value of frequency at the center of the pass band, f0 : XC = XL thus 1 = 2πf0 L 2πf0 C from which
2 f0 =

5.17.4 Band-pass filters Band-pass filters exhibit very low attenuation of signals within a specified range of frequencies (known as the pass band) and increasing

1 4π2 LC


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421

and thus f0 = 2π LC 1 √

where f0 is the resonant frequency (in Hz), L is the inductance (in H) and C is the capacitance (in F). The bandwidth of the band-pass filter is determined by its Q-factor. This, in turn, is largelydetermined by the loss resistance, R, of the inductor (recall that a practical coil has some resistance as well as inductance). The bandwidth is given by: Bandwidth = f2 − f1 = f0 2πf0 L = Q R

Figure 5.182 Frequency response for a band-stop filter.

where f0 is the resonant frequency (in Hz), L is the inductance (in H) and R is the loss resistance of the inductor (in ). 5.17.5 Band-stop filters Band-stop filters exhibit very high attenuation of signals within a specified range of frequencies (know as the stop-band) and negligible attenuation outside this range. Once again, this type of filter has two cut-off frequencies: a lower cut-off frequency (f1 ) and an upper cut-off frequency (f2 ). The difference between these frequencies (f2 − f1 ) is known as the bandwidth of the filter. The response of a band-stop filter is shown in Figure 5.182. A simple L–C band-stop filter is shown in Figure 5.183. This circuit uses an L–C resonant circuit (see Section 5.15.8) and is referred to as a rejector circuit. The frequency at which the band-stop filter in Figure 5.183 exhibits maximum attenuation occurs when the circuit is resonant, i.e. when the reactance of the capacitor, XC , is equal to the value of resistance, R. This information allows us to determine the value of frequency at the center of the pass band, f0 : XC = XL thus 1 = 2πf0 L 2πf0 C

Figure 5.183 A simple L–C band-stop filter (or rejector ).

from which
2 f0 =

1 4π2 LC

and thus f0= 1 √ 2π LC

where f0 is the resonant frequency (in Hz), L is the inductance (in H) and C is the capacitance (in F). As with the band-pass filter, the bandwidth of the band-pass filter is determined by its Qfactor. This, in turn, is largely determined by the loss resistance, R, of the inductor (recall that a practical coil has some resistance as well as inductance). Once again, the bandwidth is given by: Bandwidth = f2 − f1 = f0 2πf0 L = Q R

where f0 is the resonant frequency (in Hz), L is the inductance (in H) and R is the loss resistance of the inductor (in ).


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Example 5.97 A simple acceptor circuit uses L = 2 mH and C = 1 nF. Determine the frequency at which minimum attenuation will occur. Solution Now f0 = = 2π LC 1 √ = 2π 2 × √ 1 10−3 × 1 × 10−9

106 = 112.6 kHz 8.88
Figure 5.184 Improved T-section and π-section filters.

Example 5.98 A 15 kHz rejector circuit has a Q-factor of 40. Determine the bandwidth of the circuit. Solution Now Bandwidth = 15 × 103 f0 = = 375 Hz Q 40

5.17.6 More complex filters The simple C–R and L–C filters that we have described in earlier sections have far from ideal characteristics. In practice, more complex circuits are used and a selection of these (based on T- and π-section networks) are shown in Figure 5.184. The design equations for these circuits are as follows: Characteristic impedance: Cut-off frequency: Inductance: Capacitance: L C 1 fC =√ 2π LC Z0 L= 2πfC Z0 = C= 1 2πfC Z0

(in F). Note that the characteristic impedance of a network is the impedance seen looking into an infinite series of identical networks. This can be a difficult concept to grasp but, for now, it is sufficient to know that single section networks (like the T- and π-section filters shown in Figure 5.184) are normally terminated in their characteristic impedance at both the source (input) and load (output). Example 5.99 Determine the cut-off frequency and characteristic impedance for the filter network shown in Figure 5.185.

Figure 5.185

Solution Comparing the circuit shown in Figure 5.185 with that shown in Figure 5.184 shows that the filter is a high-pass type with L = 5 mH and C = 20 nF.

where Z0 is the characteristic impedance (in ), fC is the cut-off frequency (in Hz), L is the inductance (in H) and C is the capacitance


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Now fC = = and Z0 = L = C 5 × 10−3 = 20 × 10−9 5 × 103 20 1 √ = 6.28 5 × √ 1 10−3 × 20 × 10−9

5.18 AC generators 2π LC Syllabus Rotation of loop in a magnetic field and waveform produced; Operation and construction of revolving armature and revolving field type AC generators; Single-, two- and three-phase alternators; Three phase star and delta connections advantages and uses; Calculation of line and phase voltages and currents; Calculation of power in a three phase system; Permanent magnet generators (PMG). Knowledge level key
A B1 2B2 2 –

105 = 15.9 kHz 6.28

= 0.5 × 103 = 500

Test your understanding 5.17
1. Sketch the typical circuit for a simple C–R low-pass filter. 2. Sketch the circuit of (a) a simple C–R lowpass filter (b) a simple C–R high-pass filter. 3. A simple C–R high-pass filter has R = 5 k and C = 15 nF. Determine the cut-off frequency of the filter. 4. Signals at 115, 150, 170 and 185 kHz are present at the input of a band-stop filter with a center frequency of 160 kHz and a bandwidth of 30 kHz. Which frequencies will be present at the output? 5. Identify the type of filter shown in Figure 5.186.

5.18.1 AC generators AC generators, or alternators, are based on the principles that relate to the simple AC generator that we met earlier in Section 5.13.2. However, in a practical AC generator the magnetic field is rotated rather than the conductors from which the output is taken. Furthermore, the magnetic field is usually produced by a rotating electromagnet (the rotor) rather than a permanent magnet. There are a number of reasons for this including: (a) The conductors are generally lighter in weight than the magnetic field system and are thus more easily rotated. (b) Thicker insulation can be used for the conductors because there is more space and the conductors are not subject to centrifugal force. (c) Thicker conductors can be used to carry the large output currents. It is important to note that the heat generated in the output windings limitsthe output current that the generator can provide. By having the output windings on the outside of the machine they are much easier to cool! Figure 5.187 shows the simplified construction of a single-phase AC generator. The stator consists of five coils of insulated heavy gauge wire located in slots in the high-permeability laminated core. These coils are connected in series

Figure 5.186 See Question 5 of Test your knowledge 5.17.
6. The cut-off frequency of a filter is the frequency at which the _______ voltage has fallen to _______ of its _______ voltage. 7. The output of a low-pass filter is 2 V at 100 Hz. If the filter has a cut-off frequency of 1 kHz what will the approximate output voltage be at this frequency? 8. An L–C tuned circuit is to be used to reject signals at 15 kHz. If the value of capacitance used is 22 nF determine the required value of inductance. 9. Sketch the frequency response for (a) a simple L–C acceptor circuit and (b) a simple L–C rejector circuit. 10. A T-section filter has L = 10 mH and C = 47 nF. Determine the characteristic impedance of the filter.


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Figure 5.187 Simplified construction of a single-phase AC generator.

By adding more pairs of poles to the arrangement shown in Figure 5.187, it is possible to produce several cycles of output voltage for one single revolution of the rotor. The frequency of the output voltage produced by an AC generator is given by:pN f = 60 where f is the frequency of the induced e.m.f. (in Hz), p is the number of pole pairs and N is the rotational speed (in rpm). Example 5.100
Figure 5.188 Output voltage produced by the single-phase AC generator shown in Figure 5.187.

to make a single stator winding from which the output voltage is derived. The two-pole rotor comprises a field winding that is connected to a DC field supply via a set of slip rings and brushes. As the rotor moves through one complete revolution the output voltage will complete one full cycle of a sine wave, as shown in Figure 5.188.

An alternator is to produce an output at a frequency of 60 Hz. If it uses a four-pole rotor, determine the shaft speed at which it must be driven. Solution Re-arranging f = pN to make N the subject 60 gives: 60f N= p


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Figure 5.190 Output voltage produced by the two-phase AC generator shown in Figure 5.189. Figure 5.189 Simplified construction of a twophase AC generator.

A four-pole machine has two pairs of poles thus p = 2 and: N= Key point
In a practical AC generator, the magnetic field excitation is produced by the moving rotor whilst the conductors from which the output is taken are stationary and form part of the stator.

of time, a multi-phase supply will transmit a more evenly distributed power and this, in turn, results in a higher overall efficiency. Key point
Three-phase AC generators are more efficient andproduce more constant output than comparable single-phase AC generators.

60 × 60 = 1800 rpm 2

5.18.3 Three-phase AC generators The three-phase AC generator has three individual stator windings, as shown in Figure 5.191. The output voltages produced by the threephase AC generator are spaced by 120a—Š as shown in Figure 5.192. Each phase can be used independently to supply a different load or the generator outputs can be used with a threephase distribution system like those described in Section 5.18.4. In a practical three-phase system the three output voltages are identified by the colours red, yellow and blue or by letters, A, B and C, respectively. 5.18.4 Three-phase distribution When three-phase supplies are distributed there are two basic methods of connection: • star (as shown in Figure 5.193); • delta (as shown in Figure 5.194).

5.18.2 Two-phase AC generators By adding a second stator winding to the singlephase AC generator shown in Figure 5.187, we can produce an alternator that produces two separate output voltages which will differ in phase by 90a—Š . This arrangement is known as a two-phase AC generator (Figures 5.189 and 5.190). When compared with a single-phase AC generator of similar size, a two-phase AC generator can produce more power. The reason for this is attributable to the fact that the two-phase AC generator will produce two positive and two negative pulses per cycle whereas the singlephase generator will onlyproduce one positive and one negative pulse. Thus, over a period


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Figure 5.191 Simplified construction of a three-phase AC generator.

Figure 5.192 Output voltage produced by the three-phase AC generator shown in Figure 5.191.

Figure 5.193 Star connection.

A complete star-connected three-phase distribution system is shown in Figure 5.195. This shows a three-phase AC generator connected a three-phase load. Ideally, the load will be balanced in which case all three-load resistances (or impedances) will be identical. The relationship between the line and phase voltages shown in Figure 5.195 can be determined from the phasor diagram shown in Figure 5.196. From this diagram it is important to

Figure 5.194 Delta connection.


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Figure 5.195 A complete star-connected threephase distribution system.

Figure 5.196 Phasor diagram for the three-phase system shown in Figure 5.195.

An alternative, delta-connected three-phase distribution system is shown in Figure 5.197. Once again this shows a three-phase AC generator connected a three-phase load. Here again, the load will ideally be balanced in which case all three-load resistances (or impedances) will be identical. In this arrangement the three line currents are 120a—Š apart and that the line currents lag the phase currents by 30a—Š . We can also show that: √ IL = 3IP It should also be obvious that:

notethat three line voltages are 120 apart and that the line voltages lead the phase voltages by 30a—Š . In order to obtain the relationship between the line voltage, VL , and the phase voltage, VP , we need to resolve any one of the triangles, from which we find that: VL = 2(VP × cos 30a—Š ) Now cos 30a—Š = and hence:
√ 3 2

a—Š

VP = VL Example 5.101 In a star-connected three-phase system the phase voltage is 240 V. Determine the line voltage. Solution

√

VL =

√ √ 3VP = 3 × 240 = 415.68 V

VL = 2 VP × from which: VL = √ 3VP

3 2

Example 5.102 In a delta-connected three-phase system the line current is 6 A. Determine the phase current. Solution IL = √ 3IP

Note also that the phase current is the same as the line current, hence: IP = IL


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Figure 5.197 A complete delta-connected three-phase distribution system.

from which: IL 6 IP = √ = = 3.46 A 1.732 3 5.18.5 Power in a three-phase system In an unbalanced three-phase system the total power will be the sum of the individual phase powers. Hence: P = P1 + P2 + P3 or P = (V1 I1 )cos φ1 + (V2 I2 )cos φ2 + (V3 I3 )cos φ3 However, in the balanced condition the power is simply: P = 3 VP IP cos φ where VP and IP are the phase voltage and phase current, respectively, and φ is the phase angle. Using the relationships that we derived earlier, we can show that, for both the star and deltaconnected systems the total power is given by: √ P =3VL IL cos φ Example 5.103 In a three-phase system the line voltage is 110 V and the line current is 12 A. If the power factor is 0.8 determine the total power supplied.

Solution Here it is important to remember that: Power factor = cos φ and hence: √ P = 3VL IL × power factor √ = 3 × 110 × 12 × 0.8 = 1829 = 1.829 kW

Key point
The total power in a three-phase system is the sum of the power present in each of the three phases.

5.18.6 A practical three-phase AC generator Finally, Figure 5.198 shows a practical AC generator which uses a “brushless” arrangement based on a rotating rectifier and PMG. The generator is driven from the engine at 8000 rpm and the PMG produces an output of 120 V at 800 Hz which is fed to the PMG rectifier unit. The output of the PMG rectifier is fed to the voltage regulator which provides current for the primary exciter field winding. The primary exciter field induces current into a three-phase rotor winding. The output of this winding is fed to three shaft-mounted rectifier diodes which produce a pulsating DC output which is fed to the rotating field winding.


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Figure 5.198 A practical brushless AC generator arrangement.

The main exciter winding is wound so as to form six poles in order to produce an output at 400 Hz. The output voltage from the stator windings is typically 115 V phase, 200 V line at 20 kVA, or more. Finally, it is important to note that theexcitation system is an integral part of the rotor and that there is no direct electrical connection between the stator and rotor. Key point
A three-phase AC generator can be made “brushless” by incorporating an integral excitation system in which the field current is derived from a rotor-mounted rectifier arrangement. In this type of generator the coupling is entirely magnetic and no brushes and slip rings are required.

5. In a star-connected three-phase system the phase voltage is 220 V. Determine the line voltage. 6. In a star-connected three-phase system the line voltage is 120 V. Determine the phase voltage. 7. In a delta-connected three-phase system the line current is 12 A. Determine the phase current. 8. A three-phase system delivers power to a load consisting of three 8 resistors. Determine the total power supplied if a current of 13 A is supplied to each load. 9. In a three-phase system the line voltage is 220 V and the line current is 8 A. If the power factor is 0.75 determine the total power supplied. 10. Explain, with a simple diagram, how a brushless AC generator works.

5.19 AC motors Syllabus Construction, principles of operation and characteristics of AC synchronous and induction motors both single and poly-phase; Methods of speed control and direction of rotation; Methods of producing a rotating field: capacitor, inductor, shaded or split pole. Knowledge level key
A – B1 2 B2 2

Test your understanding 5.18
1.Sketch the arrangement of a simple two-pole singlephase AC generator. 2. An alternator with a four-pole rotor is to produce an output at a frequency of 400 Hz. Determine the shaft speed at which it must be driven. 3. Sketch (a) a star-connected and (b) a delta-connected three-phase load. 4. Explain the advantage of two- and three-phase AC generators compared with single-phase AC generators.


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5.19.1 Principle of AC motors AC motors offer significant advantages over their DC counterparts. AC motors can, in most cases, duplicate the operation of DC motors and they are significantly more reliable. The main reason for this is that the commutator arrangements (i.e. brushes and slip rings) fitted to DC motors are inherently troublesome. As the speed of an AC motor is determined by the frequency of the AC supply that is applied, AC motors are well suited to constant-speed applications. The principle of all AC motors is based on the generation of a rotating magnetic field. It is this rotating field that causes the motor’s rotor to turn. AC motors are generally classified into two types: • synchronous motors, • induction motors. The synchronous motor is effectively an AC generator (i.e. an alternator) operated as a motor. In this machine, AC is applied to the stator and DC is applied to the rotor. The induction motor is different in that no source of AC or DC power is connected to the rotor. Of these twotypes of AC motor, the induction motor is by far the most commonly used. 5.19.2 Producing a rotating magnetic field Before we go any further it is important to understand how a rotating magnetic field is produced. Take a look at Figure 5.199 which shows a three-phase stator to which three-phase AC is applied. The windings are connected in delta configuration, as shown in Figure 5.200. It is important to note that the two windings for each phase (diametrically opposite to one another) are wound in the same direction. At any instant the magnetic field generated by one particular phase depends on the current through that phase. If the current is zero, the magnetic field is zero. If the current is a maximum, the magnetic field is a maximum. Since the currents in the three windings are 120a—Š out of phase, the magnetic fields generated will also be 120a—Š out of phase.

Figure 5.199 Arrangement of the field windings of a three-phase AC motor.

Figure 5.200 AC motor as a delta-connected three-phase load.

The three magnetic fields that exist at any instant will combine to produce one field that acts on the rotor. The magnetic fields inside the motor will combine to produce a moving magnetic field and, at the end of one complete cycle of the applied current, the magnetic field will have shifted through 360a—Š (or one complete revolution). Figure 5.201 shows the three current waveforms applied to the field system. These waveforms are 120a—Š out of phase witheach other. The waveforms can represent either the three


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Figure 5.201 AC waveforms and magnetic field direction.

alternating magnetic fields generated by the three phases, or the currents in the phases. We can consider the direction of the magnetic field at regular intervals over a cycle of the applied current (i.e. every 60a—Š ). To make life simple we take the times at which one of the three current waveforms passes through zero (i.e. the point at which there will be no current and therefore no field produced by one pair of field windings). For the purpose of this exercise we will use the current applied to A and C as our reference waveform (i.e. this will be the waveform that starts at 0a—Š on our graph). At 0a—Š , waveform C-B is positive and waveform B-A is negative. This means that the current flows in opposite directions through phases B and C, and so establishes the magnetic polarity of phases B and C. The polarity is shown on the simplified diagram above. Note that B is a

north pole and B is a south pole, and that C is a north pole and C is a south pole. Since at 0a—Š there is no current flowing through phase A, its magnetic field is zero. The magnetic fields leaving poles B and C will move towards the nearest south poles C and B. Since the magnetic fields of B and C are equal in amplitude, the resultant magnetic field will lie between the two fields, and will have the direction shown. At the nextpoint, 60a—Š later, the current waveforms to phases A and B are equal and opposite, and waveform C is zero. The resultant magnetic field has rotated through 60a—Š . At point 120a—Š , waveform B is zero and the resultant magnetic field has rotated through another 60a—Š . From successive points (corresponding to one cycle of AC), you will note that the resultant magnetic field rotates through one revolution for every cycle of applied current. Hence, by


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applying a three-phase AC to the three windings we have been able to produce a rotating magnetic field. Key point
If three windings are placed round a stator frame, and a three-phase AC is applied to the windings, the magnetic fields generated in each of the three windings will combine into a magnetic field that rotates. At any given instance, these fields combine together in order to produce a resultant field which acts on the rotor. The rotor turns because the magnetic field rotates!

the AC field excitation makes this type of motor somewhat unattractive! The amount by which the rotor lags the main field is dependent on the load. If the load is increased too much, the angle between the rotor and the field will increase to a value which causes the linkage of flux to break. At this point the rotor speed will rapidly decrease and the motor will either burn out due to excessive current or the circuit protection will operate in order to prevent damage to the motor.Key point
The synchronous motor is so-called because its rotor is synchronized with the rotating field set up by the stator. Its construction is essentially the same as that of a simple AC generator (alternator).

5.19.3 Synchronous motors We have already shown how a rotating magnetic field is produced when a three-phase AC is applied to the field coils of a stator arrangement. If the rotor winding is energized with DC, it will act like a bar magnet and it will rotate in sympathy with the rotating field. The speed of rotation of the magnetic field depends on the frequency of the three-phase AC supply and, provided that the supply frequency remains constant, the rotor will turn at a constant speed. Furthermore, the speed of rotation will remain constant regardless of the load applied. For many applications this is a desirable characteristic however one of the disadvantages of a synchronous motor is that it cannot be started from a standstill by simply applying three-phase AC to the stator. The reason for this is that, the instant AC is applied to the stator, a high-speed rotating field appears. This rotating field moves past the rotor poles so quickly that the rotor does not have a chance to get started. Instead, it is repelled first in one direction and then in the other. Another way of putting this is simply that a synchronous motor (in its pure form) has no starting torque. Instead, it is usually started with the help of a smallinduction motor (or with windings equivalent to this incorporated in the synchronous motor). When the rotor has been brought near to synchronous speed by the starting device, the rotor is energized by connecting it to a DC voltage source. The rotor then falls into step with the rotating field. The requirement to have an external DC voltage source as well as

Key point
Synchronous motors are not self-starting and must be brought up to near synchronous speed before they can continue rotating by themselves. In effect, the rotor becomes “frozen” by virtue of its inability to respond to the changing field!

5.19.4 Three-phase induction motors The induction motor derives its name from the fact that AC currents are induced in the rotor circuit by the rotating magnetic field in the stator. The stator construction of the induction motor and of the synchronous motor are almost identical, but their rotors are completely different. The induction motor rotor is a laminated cylinder with slots in its surface. The windings in these slots are one of two types. The most common uses so-called squirrel cage construction (see Figure 5.202) which is made up of heavy copper bars connected together at either end by a metal ring made of copper or brass. No insulation is required between the core and the bars because of the very low voltages generated in the rotor bars. The air gap between the rotor and stator is kept very small so as to obtain maximum fieldstrength. The other type of winding contains coils placed in the rotor slots. The rotor is then called


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Figure 5.202 Squirrel cage rotor construction. Figure 5.204 Force on the rotor of an induction motor.

Figure 5.203 Typical stator construction.

a wound rotor. Just as the rotor usually has more than one conductor, the stator usually has more than one pair of poles per coil, as shown in Figure 5.203. Key point
The induction motor is the most commonly used AC motor because of its simplicity, its robust construction and its relatively low cost. These advantages arise from the fact that the rotor of an induction motor is a self-contained component that is not actually electrically connected to an external source of voltage.

From Lenz’s law we know that an induced current oppose the changing field which induces it. In the case of an induction motor, the changing field is the rotating stator field and so the force exerted on the rotor (caused by the interaction between the rotor and the stator fields) attempts to cancel out the continuous motion of the stator field. Hence the rotor will move in the same direction as the stator field and will attempt to align with it. In practice, it gets as close to the moving stator field but never quite aligns perfectly with it! Key point
The induction motor has the same stator as the synchronous motor. The rotor is different in that it does not require anexternal source of power. Current is induced in the rotor by the action of the rotating field cutting through the rotor conductors. This rotor current generates a magnetic field which interacts with the stator field, resulting in a torque being exerted on the rotor and causing it to rotate.

5.19.5 Slip, torque and speed Regardless of whether a squirrel cage or wound rotor is used, the basic principle of operation of an induction motor is the same. The rotating magnetic field generated in the stator induces an e.m.f. in the rotor. The current in the rotor circuit caused by this induced e.m.f. sets up a magnetic field. The two fields interact, and cause the rotor to turn. Figure 5.204 shows how the rotor moves in the same direction as the rotating magnetic flux generated by the stator. We have already said that the rotor of an induction motor is unable to turn in sympathy with the rotating field and, in practice, a small difference always exists. In fact, if the speeds were exactly the same, no relative motion would exist between the two, and so no e.m.f. would be induced in the rotor or this reason the rotor operates at a lower speed than that of the rotating magnetic field. This phenomenon is known as slip and it becomes more significant as the


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The percentage slip is given by: synchronous speed − rotor speed × 100% Percentage slip = synchronous speed = AB − BC × 100% AB

Figure 5.205 Relationshipbetween torque and slip.

rotor develops increased torque, as shown in Figure 5.205. From Figure 5.205, for a torque of A the rotor speed will be represented by the distance AC whilst the slip will be represented by distance AD. Now: AD = AB − AC = CB For values of torque within the working range of the motor (i.e. over the linear range of the graph shown in Figure 5.205), the slip is directly proportional to the torque and the per-unit slip is given by: slip AD Per-unit slip = = synchronous speed AB Now since AD = AB − BC, slip = synchronous speed − rotor speed thus: synchronous speed − rotor speed Per-unit slip = synchronous speed = AB − BC AB

The actual value of slip tends to vary from about 6% for a small motor to around 2% for a large machine. Hence, for most purposes the induction motor can be considered to provide a constant speed (determined by the frequency of the current applied to its stator) however one of its principal disadvantages is the fact that it is not easy to vary the speed of such a motor! Note that, in general, it is not easy to control the speed of an AC motor unless a variable frequency AC supply is available. The speed of a motor with a wound rotor can be controlled by varying the current induced in the rotor but such an arrangement is not very practical as some means of making contact with the rotor windings is required. For this reason, DC motors are usually preferred in applications where the speedmust be varied. However, where it is essential to be able to adjust the speed of an AC motor, the motor is invariably powered by an inverter. This consists of an electronic switching unit which produces a high-current three-phase pulse-width modulated (PWM) output voltage from a DC supply, as shown in Figure 5.206. Key point
The rotor of an induction motor rotates at less than synchronous speed, in order that the rotating field can cut through the rotor conductors and induce a current flow in them. This percentage difference between the synchronous speed and the rotor speed is known as slip. Slip varies very little with normal load changes, and the induction motor is therefore considered to be a constant-speed motor.

Example 5.104 An induction motor has a synchronous speed of 3600 rpm and its actual speed of rotation is measured as 3450 rpm Determine (a) the perunit slip and (b) the percentage slip.


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Figure 5.206 Using an inverter to produce a variable output speed from an AC induction motor.

Solution (a) The per-unit slip is found from: Per-unit slip = 3600 − 3450 3600 150 = = 0.042 3600

Now: sN = N − Nr from which: Nr = N − sN = N(1 − s) and: Nr = N(1 − s) = f (1 − s) p

(b) The percentage slip is given by: Percentage slip = 3600 − 3450 × 100% 3600 150 = × 100% = 4.2% 3600

where Nr is the speed of the rotor (in revolutions per second), f is the frequency of the applied AC (in Hz)and s is the per-unit slip. Example 5.105 An induction motor has four poles and is operated from a 400 Hz AC supply. If the motor operates with a slip of 2.5% determine the speed of the output rotor. Solution Now: Nr = f 400 (1 − s) = (1 − 0.025) p 2

Inside an induction motor, the speed of the rotating flux, N, is given by the relationship: N= f p

where N is the speed of the flux (in rev/s), f is the frequency of the applied AC (in Hz) and p is the number of pole pairs. Now the per-unit slip, s, is given by: s= N − Nr AB − BC = AB N

= 200 × 0.975 = 195 Thus the rotor has a speed of 195 revolutions per second (or 11,700 rpm).

where N is the speed of the flux (in revolutions per second) and Nr is the rotor speed.


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Example 5.106 An induction motor has four poles and is operated from a 60 Hz AC supply. If the rotor speed is 1700 rpm determine the percentage slip. Solution Now: Nr = from which: s=1− =1− Nr p =1− f 1700 ×2 60 60 f (1 − s) p

When the rotor is stationary, the expanding and collapsing stator field induces currents in the rotor which generate a rotor field. The opposition of these fields exerts a force on the rotor, which tries to turn it 180a—Š from its position. However, this force is exerted through the center of the rotor and the rotor will not turn unless a force is applied in order to assist it. Hence some means of starting is required for all single-phase inductionmotors. Key point
Induction motors are available that are designed for three-, two- and single-phase operation. The three-phase stator is exactly the same as the three-phase stator of the synchronous motor. The two-phase stator generates a rotating field by having two windings positioned at right angles to each other. If the voltages applied to the two windings are 90a—Š out of phase, a rotating field will be generated.

56.7 = 1 − 0.944 = 0.056 60 Expressed as a percentage, i.e. 5.6% 5.19.6 Single- and two-phase induction motors In the case of a two-phase induction motor, two windings are placed at right angles to each other. By exciting these windings with current which is 90a—Š out of phase, a rotating magnetic field can be created. A single-phase induction motor, on the other hand, has only one phase. This type of motor is extensively used in applications which require small low-output motors. The advantage gained by using single-phase motors is that in small sizes they are less expensive to manufacture than other types. Also they eliminate the need for a three-phase supply. Single-phase motors are used in communication equipment, fans, portable power tools, etc. Since the field due to the single-phase AC voltage applied to the stator winding is pulsating, single-phase AC induction motors develop a pulsating torque. They are therefore less efficient than three- or two-phase motors, in which the torque is more uniform. Single-phaseinduction motors have only one stator winding. This winding generates a field which can be said to alternate along the axis of the single winding, rather than to rotate. Series motors, on the other hand, resemble DC machines in that they have commutators and brushes.

Key point
A synchronous motor uses a single- or threephase stator to generate a rotating magnetic field, and an electromagnetic rotor that is supplied with DC. The rotor acts like a magnet and is attracted by the rotating stator field. This attraction will exert a torque on the rotor and cause it to rotate with the field.

Key point
A single-phase induction motor has only one stator winding; therefore the magnetic field generated does not rotate. A single-phase induction motor with only one winding cannot start rotating by itself. Once the rotor is started rotating, however, it will continue to rotate and come up to speed. A field is set up in the rotating rotor that is 90a—Š out of phase with the stator field. These two fields together produce a rotating field that keeps the rotor in motion.

5.19.7 Capacitor starting In an induction motor designed for capacitor starting, the stator consists of the main winding together with a starting winding which is connected in parallel with the main winding and spaced at right angles to it. A phase difference between the current in the two windings


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Figure 5.207 Capacitor starting arrangement.

isobtained by connecting a capacitor in series with the auxiliary winding. A switch is included solely for the purposes of applying current to the auxiliary winding in order to start the rotor (see Figure 5.207). On starting, the switch is closed, placing the capacitor in series with the auxiliary winding. The capacitor is of such a value that the auxiliary winding is effectively a resistive–capacitive circuit in which the current leads the line voltage by approximately 45a—Š . The main winding has enough inductance to cause the current to lag the line voltage by approximately 45a—Š . The two field currents are therefore approximately 90a—Š out of phase. Consequently the fields generated are also at an angle of 90a—Š . The result is a revolving field that is sufficient to start the rotor turning. After a brief period (when the motor is running at a speed which is close to its normal speed) the switch opens and breaks the current flowing in the auxiliary winding. At this point, the motor runs as an ordinary single-phase induction motor. However, since the two-phase induction motor is more efficient than a singlephase motor, it can be desirable to maintain the current in the auxiliary winding so that motor runs as a two-phase induction motor. In some types of motor a more complicated arrangement is used with more than one capacitor switched into the auxiliary circuit. For example, a large value of capacitor could be used in order to ensure sufficienttorque for starting a heavy load and then, once the motor has reached its operating speed, the capacitor value can be reduced in order to reduce the current in the auxiliary winding. A motor that employs such an arrangement, where two different capacitors

are used (one for starting and one for running) is often referred to as capacitor-start, capacitorrun induction motor. Finally, note that, since phase shift can also be produced by an inductor, it is possible to use an inductor instead of a capacitor. Capacitors tend to be less expensive and more compact than comparable inductors and therefore are more frequently used. Since the current and voltage in an inductor are also 90a—Š out of phase, inductor starting is also possible. Once again, a starting winding is added to the stator. If this starting winding is placed in series with an inductor across the same supply as the running winding, the current in the starting winding will be out of phase with the current in the running winding. A rotating magnetic field will therefore be generated, and the rotor will rotate. Key point
In order to make a single-phase motor selfstarting, a starting winding is added to the stator. If this starting winding is placed in series with a capacitor across the same supply as the running winding, the current in the starting winding will be out of phase with the current in the running winding. A rotating magnetic field will therefore be generated, and therotor will rotate. Once the rotor comes up to speed, the current in the auxiliary winding can be switched-out, and the motor will continue running as a single-phase motor.

5.19.8 Shaded pole motors A different method of starting a single-phase induction motor is based on a shaded-pole. In this type of motor, a moving magnetic field is produced by constructing the stator in a particular way. The motor has projecting pole pieces just like DC machines; and part of the pole surface is surrounded by a copper strap or shading coil. As the magnetic field in the core builds, the field flows effortlessly through the unshaded segment. This field is coupled into the shading coil which effectively constitutes a short-circuited loop. A large current momentarily flows in this loop and an opposing field is generated as a consequence. The result is simply that the unshaded segment initially experiences a larger


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Figure 5.208 Action of a shaded pole.

magnetic field than does the shaded segment. At some time later, the fields in the two segments become equal. Later still, as the magnetic field in the unshaded segment declines, the field in the shaded segment strengthens. This is illustrated in Figure 5.208. Key point
In the shaded pole induction motor, a section of each pole face in the stator is shorted out by a metal strap. This has the effect of moving the magnetic field back and forth across the pole face.The moving magnetic field has the same effect as a rotating field, and the motor is self-starting when switched on.

9. Describe a typical capacitor starting arrangement for use with a single-phase induction motor. 10. With the aid of a diagram, explain the action of a shaded pole motor.

5.20 Multiple choice questions The example questions set out below follow the sections of Module 3 in the JAR 66 syllabus. Note that the following questions have been separated by level, where appropriate. Several of the sections (e.g. DC circuits, resistance, power, capacitance, magnetism, inductance, etc.) are not required for category A certifying mechanics. Please remember that ALL these questions must be attempted without the use of a calculator and that the pass mark for all JAR 66 multiple-choice examinations is 75%! Electron theory 1. Within the nucleus of the atom, protons are: [A, B1, B2] (a) positively charged (b) negatively charged (c) neutral 2. A positive ion is an atom that has: [A, B1, B2] (a) gained an electron (b) lost an electron (c) an equal number of protons and electrons 3. Within an atom, electrons can be found: [A, B1, B2] (a) along with neutrons as part of the nucleus

Test your understanding 5.19
1. Explain the difference between synchronous AC motors and induction motors. 2. Explain the main disadvantage of the synchronous motor. 3. Sketch the construction of a squirrel cage induction motor. 4. Explain why theinduction motor is the most commonly used form of AC motor. 5. An induction motor has a synchronous speed of 7200 rpm and its actual speed of rotation is measured as 7000 rpm Determine (a) the per-unit slip and (b) the percentage slip. 6. An induction motor has four poles and is operated from a 400 Hz AC supply. If the motor operates with a slip of 1.8%, determine the speed of the output rotor. 7. An induction motor has four poles and is operated from a 60 Hz AC supply. If the rotor speed is 1675 rpm, determine the percentage slip. 8. Explain why a single-phase induction motor requires a means of starting.


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(b) surrounded by protons in the center of the nucleus (c) orbiting the nucleus in a series of shells 4. A material in which there are no free charge carrier is known as: [A, B1, B2] (a) a conductor (b) an insulator (c) a semiconductor 5. The charge carriers in a metal consist of: [A, B1, B2] (a) free electrons (b) free atoms (c) free neutrons Static electricity and conduction 6. Two charged particles are separated by a distance, d. If this distance is doubled (without affecting the charge present) the force between the particles will: [B1, B2] (a) increase (b) decrease (c) remain the same 7. A beam of electrons moves between two parallel plates, P and Q, as shown in Figure 5.209. Plate P has a positive charge whilst plate Q has a negative charge. Which one of the three paths will the electron beamfollow? [B1, B2] (a) A (b) B (c) C

9. Two isolated charges have dissimilar polarities. The force between them will be: [B1, B2] (a) a force of attraction (b) a force of repulsion (c) zero 10. Which one of the following gives the symbol and abbreviated units for electric charge? [A, B1, B2] (a) Symbol, Q; unit, C (b) Symbol, C; unit, F (c) Symbol, C; unit, V Electrical terminology 11. Which one of the following gives the symbol and abbreviated units for resistance? [A, B1, B2] (a) Symbol, R; unit, (b) Symbol, V; unit, V (c) Symbol, R; unit, A 12. Current can be defined as the rate of flow of: [A, B1, B2] (a) charge (b) resistance (c) voltage 13. A current of 3 A flows for a period of 2 min. The amount of charge transferred will be: [B1, B2] (a) 6 C (b) 40 C (c) 360 C 14. The volt can be defined as: (a) a joule per coulomb (b) a watt per coulomb (c) an ohm per watt [B1, B2]

Figure 5.209

8. The force between two charged particles is proportional to the: [B1, B2] (a) product of their individual charges (b) sum of their individual charges (c) difference between the individual charges

15. Conventional current flow is: [A, B1, B2] (a) always from negative to positive (b) in the same direction as electron movement (c) in the opposite direction to electron movement 16. Conductance is the inverse of: [A, B1, B2] (a) charge (b) current (c) resistance


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Generation of electricity 17. A photocellproduces electricity from: [A, B1, B2] (a) heat (b) light (c) chemical action 18. A secondary cell produces electricity from: [A, B1, B2] (a) heat (b) light (c) chemical action 19. A thermocouple produces electricity from: [A, B1, B2] (a) heat (b) light (c) chemical action 20. Which one of the following devices uses magnetism and motion to produce electricity? [A, B1, B2] (a) a transformer (b) an inductor (c) a generator 21. A small bar magnet is moved at right angles to a length of copper wire. The e.m.f. produced at the ends of the wire will depend on the: [B1, B2] (a) diameter of the copper wire and the strength of the magnet (b) speed at which the magnet is moved and the strength of the magnet (c) resistance of the copper wire and the speed at which the magnet is moved DC sources of electricity 22. The e.m.f. produced by a fresh zinc–carbon battery is approximately: [A, B1, B2] (a) 1.2 V (b) 1.5 V (c) 2 V 23. The electrolyte of a fully charged lead– acid battery will have a relative density of approximately: [A, B1, B2] (a) 0.95 (b) 1.15 (c) 1.26

24. The terminal voltage of a cell falls slightly when it is connected to a load. This is because the cell: [B1, B2] (a) has some internal resistance (b) generates less current when connected to the load (c) produces more power without the load connected 25. The electrolyte of a conventional lead–acid cell is: [A, B1, B2] (a) water (b) dilute hydrochloric acid (c) dilute sulphuricacid 26. The anode of a conventional dry (Leclanché) cell is made from: [A, B1, B2] (a) carbon (b) copper (c) zinc 27. A junction between two dissimilar metals that produces a small voltage when a temperature difference exists between it and a reference junction is known as a: [A, B1, B2] (a) diode (b) thermistor (c) thermocouple 28. A photocell consists of: [A, B1, B2] (a) two interacting layers of a semiconductor material (b) two electrodes separated by an electrolyte (c) a junction of two dissimilar metals 29. The materials used in a typical thermocouple are: [A, B1, B2] (a) silicon and selenium (b) silicon and germanium (c) iron and constantan DC circuits 30. The relationship between voltage, V, current, I, and resistance, R, for a resistor is: [B1, B2] (a) V = IR (b) V = R I (c) V = IR2 31. A potential difference of 7.5 V appears across a 15 resistor. Which one of the


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following gives the current flowing: [B1, B2] (a) 0.25 A (b) 0.5 A (c) 2 A 32. A DC supply has an internal resistance of 1 and an open-circuit output voltage of 24 V. What will the output voltage be when the supply is connected to a 5 load? [B1, B2] (a) 19 V (b) 20 V (c) 24 V 33. Three 9 V batteries are connected in series. If the series combination delivers 150 mA to a load, which one of the following gives the resistance of the load? [B1, B2] (a) 60 (b) 180 (c) 600 34. The unknown current shown in Figure 5.210 will be: [B1, B2](a) 1 A flowing towards the junction (b) 1 A flowing away from the junction (c) 4 A flowing towards the junction

(a) 3.75 V (b) 1.9 V (c) 4.7 V 36. Which one of the following gives the current flowing in the 60 resistor as shown in Figure 5.212? [B1, B2] (a) 0.33 A (b) 0.66 A (c) 1 A

Figure 5.212

Resistance and resistors 37. A 20 m length of cable has a resistance of 0.02 . If a 100 m length of the same cable carries a current of 5 A flowing in it, what voltage will be dropped across its ends? [B1, B2] (a) 0.02 V (b) 0.1 V (c) 0.5 V 38. The resistance of a wire conductor of constant cross section: [B1, B2] (a) decreases as the length of the wire increases (b) increases as the length of the wire increases (c) is independent of the length of the wire 39. Three 15 resistors are connected in parallel. Which one of the following gives the effective resistance of the parallel combination? [B1, B2] (a) 5 (b) 15 (c) 45 40. Three 15 resistors are connected in series. Which one of the following gives the effective resistance of the series combination? [B1, B2]

Figure 5.210

35. Which one of the following gives the output voltage produced by the circuit shown in Figure 5.211? [B1, B2]

Figure 5.211


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Aircraft engineering principles

(a) 5 (b) 15 (c) 45 41. Which one of the following gives the effective resistance of the circuit shown Figure 5.213? [B1, B2] (a) 5 (b) 6 (c) 26

(a) 2.24 A (b) 5 A (c) 10 A 47. An aircraftcabin has 110 passenger reading lamps each rated at 10 W, 28 V. What is the maximum load current imposed by these lamps? [B1, B2] (a) 25.5 A (b) 39.3 A (c) 308 A 48. An aircraft fuel heater consists of two parallel-connected heating elements each rated at 28 V, 10 A. What total power is supplied to the fuel heating system? [B1, B2] (a) 140 W (b) 280 W (c) 560 W 49. An aircraft battery is being charged from a bench DC supply that has an output of 28 V. If the charging current is 10 A, what energy is supplied to the battery if it is charged for 4 h? [B1, B2] (a) 67 kJ (b) 252 kJ (c) 4.032 MJ 50. A portable power tool operates from a 7 V rechargeable battery. If the battery is charged for 10 h at 100 mA, what energy is supplied to it? [B1, B2] (a) 25.2 kJ (b) 252 kJ (c) 420 kJ Capacitance and capacitors 51. The high-voltage connection on a power supply is fitted with a rubber cap. The reason for this is to: [B1, B2] (a) provide insulation (b) concentrate the charge (c) increase the current rating 52. Which one of the following gives the symbol and abbreviated units for capacitance? [B1, B2] (a) Symbol, C; unit, C (b) Symbol, C; unit, F (c) Symbol, Q; unit, C

Figure 5.213

42. A 10 wirewound resistor is made from 0.2 m of wire. A second wirewound resistor is made from 0.5 m of the same wire. The second resistor will have a resistance of: [B1, B2] (a) 4 (b) 15 (c) 25 Power 43. The relationship between power, P, current, [B1, B2] I,and resistance, R, is: (a) P = I × R (b) P = R I (c) P = I 2 × R 44. A DC generator produces an output of 28 V at 20 A. The power supplied by the generator will be: [B1, B2] (a) 14 W (b) 560 W (c) 1.4 kW 45. A cabin reading lamp consumes 10 W from a 24 V DC supply. The current supplied will be: [B1, B2] (a) 0.42 A (b) 0.65 A (c) 2.4 A 46. A generator delivers 250 W of power to a 50 load. The current flowing in the load will be: [B1, B2]


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53. A capacitor is required to store a charge of 32 µC when a voltage of 4 V is applied to it. The value of the capacitor should be: [B1, B2] (a) 0.125 µF (b) 0.25 µF (c) 8 µF 54. An air-spaced capacitor has two plates separated by a distance, d. If the distance is doubled (without affecting the area of the plates) the capacitance will be: [B1, B2] (a) doubled (b) halved (c) remain the same 55. A variable air-spaced capacitor consists of two sets of plates that can be moved. When the plates are fully meshed, the: [B1, B2] (a) capacitance will be maximum and the working voltage will be reduced (b) capacitance will be maximum and the working voltage will be unchanged (c) capacitance will be minimum and the working voltage will be increased 56. A 20 µF capacitor is charged to a voltage of 50 V. The charge present will be: [B1, B2] (a) 0.5 µC (b) 2.5 µF (c) 1 mC 57. A power supply filter uses five parallelconnected 2200 µF capacitors each rated at 50 V. What singlecapacitor could be used to replace them? [B1, B2] (a) 11,000 µF at 10 V (b) 440 µF at 50 V (c) 11,000 µF at 250 V 58. A high-voltage power supply uses four identical series-connected capacitors. If 1 kV appears across the series arrangement and the total capacitance required is 100 µF, which one of the following gives a suitable rating for each individual capacitor? [B1, B2] (a) 100 µF at 250 V (b) 25 µF at 1 kV (c) 400 µF at 250 V

59. Which one of the following materials is suitable for use as a capacitor dielectric? [B1, B2] (a) aluminium foil (b) polyester film (c) carbon granules 60. The relationship between capacitance, C, charge, Q, and potential difference, V, for a capacitor is: [B1, B2] (a) Q = CV C (b) Q = V (c) Q = CV 2 61. The material that appears between the plates of a capacitor is known as the: [B1, B2] (a) anode (b) cathode (c) dielectric

Magnetism 62. Permanent magnets should be stored using [B1, B2] (a) anti-static bags (b) insulating material such as polystyrene (c) soft iron keepers 63. Lines of magnetic flux: [B1, B2] (a) originate at the south pole and end at the north pole (b) originate at the north pole and end at the south pole (c) start and finish at the same pole, either south or north 64. The magnetomotive force produced by a solenoid is given by: [B1, B2] (a) the length of the coil divided by its crosssectional area (b) number of turns on the coil divided by its cross-sectional area (c) the number ofturns on the coil multiplied by the current flowing in it 65. An air-cored solenoid with a fixed current flowing through it is fitted with a ferrite core. The effect of the core will be to: [B1, B2] (a) increase the flux density produced by the solenoid


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(b) decrease the flux density produced by the solenoid (c) leave the flux density produced by the solenoid unchanged 66. The permeability of a magnetic material is given by the ratio of: [B1, B2] (a) magnetic flux to cross-sectional area (b) magnetic field intensity to magnetomotive force (c) magnetic flux density to magnetic field intensity 67. The relationship between permeability, µ, magnetic flux density, B, and magnetizing force, H, is: [B1, B2] (a) µ = B × H B (b) µ = H (c) µ = H B 68. The relationship between absolute permeability, µ, relative permeability, µr , and the permeability of free-space, µ0 , is given by: [B1, B2] (a) µ = µ0 × µr (b) µ = µ0 µr
µ (c) µ = µ r
0

Inductance and inductors 71. Which one of the following gives the symbol and abbreviated units for inductance? [B1, B2] (a) Symbol, I; unit, L (b) Symbol, L; unit, H (c) Symbol, H; unit, L 72. Which one of the following materials is suitable for use as the coil winding of an [B1, B2] inductor? (a) brass (b) copper (c) steel 73. Which one of the following materials is suitable for use as the laminated core of an inductor? [B1, B2] (a) brass (b) copper (c) steel 74. Lenz’s lawstates that: [B1, B2] (a) the reluctance of a magnetic circuit is zero (b) an induced e.m.f. will always oppose the motion that created it (c) the force on a current-carrying conductor is proportional to the current flowing 75. The inductance of a coil is directly proportional to the: [B1, B2] (a) current flowing in the coil (b) square of the number of turns (c) mean length of the magnetic path 76. The inductance of a coil can be increased by using: [B1, B2] (a) a low number of turns (b) a high permeability core (c) wire having a low resistance DC motor and generator theory 77. The commutator in a DC generator is used to: [B1, B2] (a) provide a means of connecting an external field current supply (b) periodically reverse the connections to the rotating coil winding

69. The relative permeability of steel is in the range: [B1, B2] (a) 1 to 10 (b) 10 to 100 (c) 100 to 1000 70. The feature marked X on the B–H curve shown in Figure 5.214 is: [B1, B2] (a) saturation (b) reluctance (c) hysteresis

Figure 5.214


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(c) disconnect the coil winding when the induced current reaches a maximum value 78. The core of a DC motor/generator is laminated in order to: [B1, B2] (a) reduce the overall weight of the machine (b) reduce eddy currents induced in the core (c) increase the speed at which the machine rotates 79. The brushes fitted to a DC motor/generator should have: [B1, B2] (a) low coefficient of friction andlow contact resistance (b) high coefficient of friction and low contact resistance (c) low coefficient of friction and high contact resistance 80. A feature of carbon brushes used in DC motors and generators is that they are: [B1, B2] (a) self-lubricating (b) self-annealing (c) self-healing 81. Self-excited generators derive their field current from: [B1, B2] (a) the current produced by the armature (b) a separate field current supply (c) an external power source 82. In a series-wound generator: [B1, B2] (a) none of the armature current flows through the field (b) some of the armature current flows through the field (c) all of the armature current flows through the field 83. In a shunt-wound generator: [B1, B2] (a) none of the armature current flows through the field (b) some of the armature current flows through the field (c) all of the armature current flows through the field 84. A compound-wound generator has: [B1, B2]

(a) only a series field winding (b) only a shunt field winding (c) both a series and a shunt field winding

AC theory 85. Figure 5.215 shows an AC waveform. The waveform is a: [A, B1, B2] (a) square wave (b) sine wave (c) triangle wave

Figure 5.215

86. Figure 5.216 shows an AC waveform. The periodic time of the waveform is: [B1, B2] (a) 1 ms (b) 2 ms (c) 4 ms

Figure 5.216

87. Figure 5.217 shows an AC waveform. The amplitude of the waveform is: [B1, B2] (a) 5 V (b) 10 V (c) 20 V


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(a) 60a—Š (b) 90a—Š (c) 120a—Š 92. An aircraft supply has an r.m.s value of 115 V. Which one of the following gives the approximate peak value of the supply voltage? [B1, B2] (a) 67.5 V (b) 115 V (c) 163 V
Figure 5.217

88. Figure 5.218 shows two AC waveforms. The phase angle between these waveforms is: [B1, B2] (a) 45a—Š (b) 90a—Š (c) 180a—Š

93. The peak value of current supplied to an aircraft TRU is 28 A. Which one of the following gives the approximate value of r.m.s. current supplied? [B1, B2] (a) 10 A (b) 14 A (c) 20 A

Resistive, capacitive and inductive circuits 94. A circuit consisting of a pure capacitance is connected across an AC supply. Which one of the following gives the phase relationship between the voltage and current in this circuit? [B1, B2] (a) The voltage leads the current by 90a—Š (b) The current leads the voltage by 90a—Š (c) The current leads the voltage by 180a—Š 95. An inductor has an inductive reactance of 50 and a resistance of 50 . Which one of the following gives the phase relationship between the voltage and current in this circuit? [B1, B2] (a) The current leads the voltage by 45a—Š (b) The voltage leads the current by 45a—Š (c) The voltage leads the current by 90a—Š 96. A capacitor having negligible resistance is connected across a 115 V AC supply. If the current flowing in the capacitor is 0.5 A, which one of the following gives its reactance? [B1, B2] (a) 0 (b) 50 (c) 230

Figure 5.218

89. An ACwaveform has a frequency of 400 Hz. Which one of the following gives its period? [B1, B2] (a) 2.5 ms (b) 25 ms (c) 400 ms 90. An AC waveform has a period of 4 ms. Which one of the following gives its [B1, B2] frequency? (a) 25 Hz (b) 250 Hz (c) 4 kHz 91. Which one of the following gives the angle between the successive phases of a three[A, B1, B2] phase supply?


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97. A pure capacitor having a reactance of 100 is connected across a 200 V AC supply. Which one of the following gives the power dissipated in the capacitor? [B1, B2] (a) 0 W (b) 50 W (c) 400 W 98. The power factor in an AC circuit is defined as the: [B1, B2] (a) ratio of true power to apparent power (b) ratio of apparent power to true power (c) ratio of reactive power to true power 99. The power factor in an AC circuit is the same as the: [B1, B2] (a) sine of the phase angle (b) cosine of the phase angle (c) tangent of the phase angle 100. An AC circuit consists of a capacitor having a reactance of 40 connected in series with a resistance of 30 . Which one of the following gives the impedance of this circuit? [B1, B2] (a) 10 (b) 50 (c) 70 101. An AC circuit consists of a pure inductor connected in parallel with a pure capacitor. At the resonant frequency, the: [B1, B2] (a) impedance of the circuit will be zero (b) impedance of the circuit will be infinite (c) impedance of the circuit will be the same as at all other frequencies

linkage willoccur between the coils when the relative angle between them is: [B1, B2] (a) 0a—Š (b) 45a—Š (c) 90a—Š 104. The primary and secondary voltage and current for an aircraft transformer is given in the table below: [B1, B2] Primary Voltage (V) Current (A) 110 2 Secondary 50 4

Which one of the following gives the approximate efficiency of the transformer? [B1, B2] (a) 63% (b) 85% (c) 91% 105. The “copper loss” in a transformer is a result of: [B1, B2] (a) the I 2 R power loss in the transformer windings (b) the power required to magnetize the core of the transformer (c) eddy currents flowing in the magnetic core of the transformer Filters 106. The frequency response shown in Figure 5.219 represents the output of a: [B1, B2] (a) low-pass filter (b) high-pass filter (c) band-pass filter

Transformers 102. A transformer has 2400 primary turns and 600 secondary turns. If the primary is supplied from a 220 V AC supply, which one of the following gives the resulting secondary voltage: [B1, B2] (a) 55 V (b) 110 V (c) 880 V 103. Two inductive coils are placed in close proximity to one another. Minimum flux

Figure 5.219


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107. The frequency response shown in Figure 5.220 represents the output of a: [B1, B2] (a) low-pass filter (b) high-pass filter (c) band-pass filter

(a) connecting an external circuit to a rotating armature winding (b) supporting a rotating armature without the need for bearings (c)periodically reversing the current produced by an armature winding 112. Decreasing the field current in a generator will: [B1, B2] (a) decrease the output voltage (b) increase the output voltage (c) increase the output frequency 113. A single-phase AC generator has 12 poles and it runs at 600 rpm. Which one of the following gives the output frequency of the generator? [B1, B2] (a) 50 Hz (b) 60 Hz (c) 120 Hz 114. In a star-connected three-phase system, the line voltage is found to be 200 V. Which one of the following gives the approximate value of phase voltage? [B1, B2] (a) 67 V (b) 115 V (c) 346 V 115. In a delta-connected three-phase system, the phase current is found to be 2 A. Which one of the following gives the approximate value of line current? [B1, B2] (a) 1.2 A (b) 3.5 A (c) 6 A 116. In a balanced star-connected three-phase system the line current is 2 A and the line voltage is 110 V. If the power factor is 0.75 which one of the following gives the total power in the load? [B1, B2] (a) 165 W (b) 286 W (c) 660 W AC motors 117. The rotor of an AC induction motor consists of a: [B1, B2]

Figure 5.220

108. Signals at 10 kHz and 400 Hz are present in a cable. The 10 kHz signal can be removed by means of an appropriately designed: [B1, B2] (a) low-pass filter (b) high-pass filter (c) band-pass filter 109. Signals at 118, 125 and 132 MHz are present in the feeder to an antenna. The signals at 118 and 132 MHz can be reduced by means of:[B1, B2] (a) low-pass filter (b) high-pass filter (c) band-pass filter 110. The circuit shown in Figure 5.221 is a: (a) low-pass filter (b) high-pass filter (c) band-pass filter

Figure 5.221

AC generators 111. The slip rings in an AC generator provide a means of: [B1, B2]


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(a) laminated iron core inside a “squirrel cage” made from copper or aluminium (b) series of coil windings on a laminated iron core with connections via slip rings (c) single copper loop which rotates inside the field created by a permanent magnet 118. The slip speed of an AC induction motor is the difference between the: [B1, B2] (a) synchronous speed and the rotor speed (b) frequency of the supply and the rotor speed (c) maximum speed and the minimum speed 119. When compared with three-phase induction motors, single-phase induction motors: [B1, B2]

(a) are not inherently “self starting” (b) have more complicated stator windings (c) are significantly more efficient 120. The use of laminations in the construction of an electrical machine is instrumental in reducing the: [B1, B2] (a) losses (b) output (c) weight 121. A three-phase induction motor has three pairs of poles and is operated from a 60 Hz supply. Which one of the following gives the motor’s synchronous speed? [B1, B2] (a) 1200 rpm (b) 1800 rpm (c) 3600 rpm


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Chapter

Electronic fundamentals

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6.1 Introduction If you havepreviously studied Chapter 5, you will already be aware of just how important electricity is in the context of a modern aircraft. However, whereas Chapter 5 introduced you to the fundamentals of electrical power generation, distribution and utilization, this section concentrates on developing an understanding of the electronic devices and circuits that are found in a wide variety of aircraft systems. Such devices include diodes, transistors and integrated circuits, and the systems that are used to include control instrumentation, radio and navigation aids. We will begin this section by introducing you to some important concepts starting with an introduction to electronic systems and circuit diagrams. It is particularly important that you get to grips with these concepts if you are studying electronics for the first time! for them to operate correctly; there is a requirement for them to have their own supply and bias voltages. We will explain how this works later in this section when we introduce transistors and integrated circuits but, for the moment, it is important to understand that most electronic circuits may often appear to be somewhat more complex than they are, simply because there is a need to supply the semiconductor devices with the voltages and currents that they need in order to operate correctly. In order to keep things simple, we often use block schematic diagrams rather than full circuit diagrams in order to helpexplain the operation of electronic systems. Each block usually represents a large number of electronic components and instead of showing all the electrical connections we simply show a limited number of them, sufficient to indicate the flow of signals and power between blocks. As an example, the block schematic diagram of a power supply is shown in Figure 6.1. Note that the input is taken from a 400 Hz 115 V alternating current (AC) supply, stepped down to 28 V AC, then rectified (i.e. converted to direct current (DC)) and finally regulated to provide a constant output voltage of 28 V DC. Key point
Electronics is based on the application of semiconductor devices (such as diodes, transistors and integrated circuits) along with components, such as resistors, capacitors, inductors and transformers, that we met earlier in Chapter 5.

6.1.1 Electronic circuit and systems Electronic circuits, such as amplifiers, oscillators and power supplies, are made from arrangements of the basic electronic components (such as the resistors, capacitors, inductors and transformers that we met in Chapter 5) along with the semiconductors and integrated circuits that we shall meet for the first time in this section. Semiconductors are essential for the operation of the circuits in which they are used, however,

Figure 6.1 A block schematic diagram of a power supply.


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Figure 6.2 A selection of symbols used inelectronic circuit schematics.


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6.1.2 Reading and understanding circuit diagrams Before you can make sense of some of the semiconductor devices and circuits that you will meet later in this section it is important to be able to read and understand a simple electronic circuit diagram. Circuit diagrams use standard conventions and symbols to represent the components and wiring used in an electronic circuit. Visually, they bear very little relationship to the physical layout of a circuit but, instead, they provide us with a “theoretical” view of the circuit. It is important that you become familiar with reading and understanding circuit diagrams right from the start. So, a selection of some of the most commonly used symbols is shown in Figure 6.2. It is important to note that there are a few (thankfully quite small) differences between the symbols used in American and European diagrams. As a general rule, the input should be shown on the left of the diagram and the output on the right. The supply (usually the most positive voltage) is normally shown at the top of the diagram and the common, 0 V, or ground connection is normally shown at the bottom. This rule is not always obeyed, particularly for complex diagrams where many signals and supply voltages may be present. Note also that, in order to simplify a circuit diagram (and avoid having too many lines connected to the same point), multiple connections tocommon, 0 V, or ground may be shown using the appropriate symbol (see the negative connection to C1 in Figure 6.3). The same applies to supply connections that may be repeated (appropriately labelled) at various points in the diagram. Three different types of switch are shown in Figure 6.2: single-pole single-throw (SPST), single-pole double-throw (SPDT) and doublepole single-throw (DPST). The SPST switch acts as a single-circuit on/off switch whilst the DPST provides the same on/off function but makes and breaks two circuits simultaneously. The SPDT switch is sometimes referred to as a changeover switch because it allows the selection of one circuit or another. Multi-pole switches are also available. These provide switching between

many different circuits. For example, one-pole six-way (1P 6W) switch allows you to select six different circuits. Example 6.1 The circuit of a simple intercom amplifier is shown in Figure 6.3.
12V R3 180 T1 C1 470u R1 8.2k TR1 2N3053 LS1

Input

C2 10u

R2 2k2

R4 220 0V

Figure 6.3 Intercom amplifier – see Example 6.1.

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

What is the value of C1 ? What is the value of R1 ? Which component has a value of 220 ? Which component is connected directly to the positive supply? (e) Which component is connected to the circuit via T1 ? (f) Where is coaxial cable used in this circuit?

Solution (a) (b) (c) (d) (e) 470 µF 8.2 k R4 R3 (the top end of R3 is marked “+12 V”) LS1 (theloudspeaker is connected via a stepdown transformer, T1 ) (f) To screen the input signal (between the “live” input terminal and the negative connection on C2 ).


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Key point
Circuit diagrams use standard conventions and symbols to represent the components and wiring used in an electronic circuit. Circuit diagrams provide a “theoretical” view of a circuit, which is often different from the physical layout of the circuit to which they refer.

a resistor whilst Figure 6.4(b) shows a similar graph plotted for a non-linear device such as a semiconductor. Since the ratio of I to V is the reciprocal of resistance, R, we can make the following inferences: 1. At all points in Figure 6.4(a) the ratio of I to V is the same showing that the resistance, R, of the device remains constant. This is exactly how we would want a resistor to perform. 2. In Figure 6.4(b) the ratio of I to V is different at different points on the graph; thus, the resistance, R, of the device does not remain constant but changes as the applied voltage and current changes. This is an important point since most semiconductor devices have distinctly non-linear characteristics!

6.1.3 Characteristic graphs The characteristics of semiconductor devices are often described in terms of the relationship between the voltage, V, applied to them and the current, I, flowing in them. With a device such as a diode (which has two terminals)this is relatively straightforward. However, with a three-terminal device (such as a transistor), a family of characteristics may be required to fully describe the behaviour of the device. This point will become a little clearer when we meet the transistor later in this section but, for the moment, it is worth considering what information can be gleaned from a simple current/voltage characteristic. Figure 6.4(a) shows the graph of current plotted against voltage for a linear device such as

Key point
Characteristic graphs are used to describe the behaviour of semiconductor devices. These graphs show corresponding values of current and voltage and they are used to predict the performance of a particular device when used in a circuit.

Figure 6.4 I/V characteristics for (a) linear and (b) non-linear device.


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Example 6.2 The I/V characteristic for a non-linear electronic device is shown in Figure 6.5. Determine the resistance of the device when the applied voltage is:

Test your understanding 6.1
1. Identify components shown in Figure 6.6(a)–(o).

Figure 6.6 See Question 1.
2. Sketch the circuit schematic symbol for: (a) a PNP transistor (b) a variable capacitor (c) a chassis connection (d) a quartz crystal 3. Explain, with the aid of a sketch, the operation of each of the following switches: (a) SPST (b) SPDT (c) DPDT Questions 4–8 refer to the motor driver circuit shown in Figure 6.7.Figure 6.5 I/V characteristic – see Example 6.2.

TR1 TIP141 28V R1 1.2k D1 15V 0V C1 100u D2 Red R2 1k M1

(a) 0.43 V (b) 0.65 V Solution (a) At 0.43 V the corresponding values of I is 2.5 mA and the resistance, R, of the device will be given by: R= V 0.43 = = 172 I 2.5

Figure 6.7
4. What type of device is: (a) D 1 , (b) D 2 and (c) TR 1 ? 5. Which components have a connection to the 0 V rail? 6. Which two components are connected in parallel? 7. Which two components are connected in series? 8. Redraw the circuit with the following modifications: (a) TR 1 is to be replaced by a conventional NPN transistor,

(b) At 0.65 V the corresponding values of I is 7.4 mA and the resistance, R, of the device will be given by: R= V 0.65 = = 88 I 7.4


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(b) an SPST switch is to be placed in series with R 1 , (c) the value of C 1 is to be increased to 220 µF, (d) the light emitting diode (LED) indicator and series resistor are to be removed and replaced by a single fixed capacitor of 470 nF. 9. Corresponding readings of current, I, and voltage, V, for a semiconductor device are given in the table below: V (V) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 I (mA) 0 0.2 0.5 1.5 3.0 5.0 8.5 13.0 20.0 Plot the I/V characteristic for the device. 10. Determine the resistance of the device in Question 9, when the applied voltage is: (a) 0.35 V (b) 0.75 V

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diode, varactor diode, varistor, rectifier diodes, Zenerdiode. Knowledge level key
A – B1 – B2 2

6.2 Semiconductors 6.2.1 Diodes Syllabus Diode symbols; Diode characteristics and properties; Diodes in series and parallel; Main characteristics and use of silicon (Si) controlled rectifiers (SCRs), LED, photo-conductive diode, varistor, rectifier diodes; Functional testing of diodes. Knowledge level key
A – B1 2 B2 2

Syllabus Materials, electron configuration, electrical properties; P- and N-type materials: effects of impurities on conduction, majority and minority carriers; P–N junction in a semiconductor, development of a potential across a P–N junction in unbiased, forward- and reverse-biased conditions; Diode parameters: peak inverse voltage (PIV), maximum forward current, temperature, frequency, leakage current, power dissipation; Operation and function of diodes in the following circuits: clippers, clampers, full- and half-wave rectifiers, bridge rectifiers, voltage doublers and triplers; Detailed operation and characteristics of the following devices: SCR, LED, Schottky diode, photo-conductive

Semiconductor materials This section introduces devices that are made from materials that are neither conductors nor insulators. These semiconductor materials form the basis of many important electronic components, such as diodes, SCRs, triacs, transistors and integrated circuits. We shall start with a brief introduction to the principles of semiconductors and then go on to examine thecharacteristics of each of the most common types that you are likely to meet. You should recall that an atom contains both negative charge carriers (electrons) and positive charge carriers (protons). Electrons each carry a single unit of negative electric charge while protons each exhibit a single unit of positive charge. Since atoms normally contain an equal number of electrons and protons, the net charge present will be zero. For example, if an atom has 11 electrons, it will also contain 11 protons. The end result is that the negative charge of the electrons will be exactly balanced by the positive charge of the protons. Electrons are in constant motion as they orbit around the nucleus of the atom. Electron orbits are organized into shells. The maximum number of electrons present in the first shell is two, in the second shell eight and in the third, fourth and fifth shells it is 18, 32 and 50, respectively. In electronics only the electron shell furthermost from the nucleus of an atom is important. It is important to note that the movement of electrons between atoms involves only those present in the outer valence shell (Figure 6.8). If the valence shell contains the maximum number of electrons possible the electrons are rigidly bonded together and the material has the properties of an insulator. If, however, the valence shell does not have its full complement of electrons, the electrons can be easily detached from their orbital bonds,and the


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Figure 6.8 Electrons orbiting a nucleus.

Figure 6.10 Effect of introducing a trivalent impurity.

Figure 6.9 Effect of introducing a pentavalent impurity.

material has the properties associated with an electrical conductor. In its pure state, Si is an insulator because the covalent bonding rigidly holds all of the electrons leaving no free (easily loosened) electrons to conduct current. If, however, an atom of a different element (i.e. an impurity) is introduced that has five electrons in its valence shell, a surplus electron will be present (see Figure 6.9).

These free electrons become available for use as charge carriers and they can be made to move through the lattice by applying an external potential difference to the material. Similarly, if the impurity element introduced into the pure Si lattice has three electrons in its valence shell, the absence of the fourth electron needed for proper covalent bonding will produce a number of spaces into which electrons can fit (see Figure 6.10). These spaces are referred to as holes. Once again, current will flow when an external potential difference is applied to the material. Regardless of whether the impurity element produces surplus electrons or holes, the material will no longer behave as an insulator, neither will it have the properties that we normally associate with a metallic conductor. Instead, we call the material asemiconductor – the term simply indicates that the substance is no longer a good insulator or a good conductor but is somewhere in between! Examples of semiconductors include germanium (Ge) and silicon (Si). The process of introducing an atom of another (impurity) element into the lattice of an otherwise pure material is called doping. When the pure material is doped with an impurity with five electrons in its valence shell (i.e. a


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Aircraft engineering principles Table 6.1 Leading digit – number of P–N junctions 1 Diode 2 Transistor 3 SCR or dual gate MOSFET 4 Optocoupler Letter – origin N North American JEDEC-coded device Serial number – the serial number does not generally have any particular significance Suffix – some transistors have an additional suffix that denotes the gain group for the device (where no suffix appears the gain group is either inapplicable or the group is undefined for the device in question) A Low gain B Medium gain C High gain

pentavalent impurity) it will become an N-type (i.e. negative type) material. If, however, the pure material is doped with an impurity having three electrons in its valence shell (i.e. a trivalent impurity) it will become P-type material (i.e. positive type). N-type semiconductor material contains an excess of negative charge carriers and P-type material contains an excess of positive charge carriers. Key point
Circuit diagrams use standard conventions and symbols to represent thecomponents and wiring used in an electronic circuit. Circuit diagrams provide a “theoretical” view of a circuit which is often different from the physical layout of the circuit to which they refer.

Semiconductor classification Semiconductor devices are classified using a unique part numbering system. Several schemes are in use including the American Joint Engineering Device Engineering Council (JEDEC) system, the European Pro-Electron system and the Japanese Industrial Standard (JIS) system (which is Japanese based). In addition, some manufacturers have adopted their own coding schemes. The JEDEC system of semiconductor classification is based on the following coding format: leading digit, letter, serial number, suffix (optional) The leading digit designates the number of P–N junctions used in the device. Hence, a device code starting with 1 relates to a single P–N junction (i.e. a diode) whilst a device code starting with 2 indicates a device which has two P–N junctions (usually a transistor) (Table 6.1). The letter is always N (signifying a JEDEC device) and the remaining digits are the serial number of the device. In addition, a suffix may be used in order to indicate the gain group. The European Pro-Electron system for classifying semiconductors involves the following coding format (Table 6.2). first letter, second letter, third letter (optional), serial number, suffix (optional)

Table 6.2 First letter – semiconductor material A Ge BSi C Gallium arsenide, etc. D Photodiodes, etc. Second letter – application A Diode, low power or signal B Diode, variable capacitance C Transistor, audio frequency (AF) low power D Transistor, AF power E Diode, tunnel F Transistor, high frequency, low power P Photodiode Q LED S Switching device T Controlled rectifier X Varactor diode V Power rectifier Z Zener diode Third letter – if present this indicates that the device is intended for industrial or professional rather than commercial applications Serial number – the serial number does not generally have any particular significance Suffix – some transistors have an additional suffix that denotes the gain group for the device (where no suffix appears the gain group is either inapplicable or the group is undefined for the device in question) A Low gain B Medium gain C High gain


Electronic fundamentals Table 6.3 Leading digit – number of P–N junctions 1 Diode 2 Transistor 3 SCR or dual gate MOSFET 4 Optocoupler First and second letters – application SA PNP high-frequency transistor SB PNP AF transistor SC NPN high frequency SD NPN AF transistor SE Diode SF SCR SJ P-channel field effect transistor (FET)/MOSFET SK N-channel FET/MOSFET SM Triac SQ LED SR Rectifier SS Signal diode ST Diode SV Varactor SZ Zener diode Serial number – the serial number does not generally have any particular significance Suffix – some devices have a suffix that denotes approval of the device for use by certainorganizations

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Figure 6.11 A P–N junction diode.

(c) MOSFET (JEDEC-coded) (d) Ge low-power signal diode (Pro-Electron coded) (e) Transistor (JEDEC-coded) (f) PNP high-frequency transistor (JIS-coded). The P–N junction diode When a junction is formed between N- and P-type semiconductor materials, the resulting device is called a diode. This component offers an extremely low resistance to current flow in one direction and an extremely high resistance to current flow in the other. This characteristic allows diodes to be used in applications that require a circuit to behave differently according to the direction of current flowing in it. An ideal diode would pass an infinite current in one direction and no current at all in the other direction (Figure 6.11). Connections are made to each side of the diode. The connection to the P-type material is referred to as the anode while that to the N-type material is called the cathode. With no externally applied potential, electrons from the N-type material will cross into the P-type region and fill some of the vacant holes. This action will result in the production of a region on either sides of the junction in which there are no free charge carriers. This zone is known as the depletion region. If a positive voltage is applied to the anode (see Figure 6.12), the free positive charge carriers in the P-type material will be repelled and they will move away from the positive potential

The JISis based on the following coding format (Table 6.3). leading digit, first letter, second letter, serial number, suffix (optional) The JIS coding system is similar to the JEDEC system. Example 6.3 Classify the following semiconductor devices: (a) (b) (c) (d) (e) (f) 1N4001 BFY51 3N201 AA119 2N3055 2SA1077

Solution (a) Diode (JEDEC-coded) (b) Si high-frequency low-power transistor (Pro-Electron coded)


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Figure 6.12 A forward-biased P–N junction diode.

Figure 6.13 A reverse-biased P–N junction diode.

towards the junction. Likewise, the negative potential applied to the cathode will cause the free negative charge carriers in the N-type material to move away from the negative potential towards the junction. When the positive and negative charge carriers arrive at the junction, they will attract one another and combine (recall from Chapter 5 that unlike charges attract). As each negative and positive charge carriers combine at the junction, new negative and positive charge carriers will be introduced to the semiconductor material from the voltage source. As these new charge carriers enter the semiconductor material, they will move towards the junction and combine. Thus, current flow is established and it will continue for as long as the voltage is applied. In this forward-biased condition, the diode freely passes current. If a negative voltage is applied to the anode (see Figure 6.13),the free positive charge carriers in the P-type material will be attracted and they will move away from the junction. Likewise, the positive potential applied to the cathode will cause the free negative charge carriers in the N-type material to move away from the junction. The combined effect is that the depletion region becomes wider. In this reverse-biased condition, the diode passes a negligible amount of current.

Key point
In the freely conducting forward-biased state, the diode acts rather like a closed switch. In the reverse-biased state, the diode acts like an open switch.

Diode characteristics Typical I/V characteristics for Ge and Si diodes are shown in Figure 6.14. It should be noted from these characteristics that the approximate forward conduction voltage for a Ge diode is 0.2 V whilst that for a Si diode is 0.6 V. This threshold voltage must be high enough to completely overcome the potential associated with the depletion region and force charge carriers to move across the junction. Key point
The forward voltage for a Ge diode is approximately 0.2 V whilst that for a Si diode is approximately 0.6 V.

Example 6.4 The characteristic of a diode is shown in Figure 6.15. Determine: (a) the current flowing in the diode when a forward voltage of 0.4 V is applied;


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Figure 6.14 Typical I/V characteristics for Ge and Si diodes.

(b) the voltage dropped across the diode when a forwardcurrent of 9 mA is flowing in it; (c) the resistance of the diode when the forward voltage is 0.6 V; (d) whether the diode is a Ge or Si type. Solution (a) When V = 0.4 V, I = 1.9 mA. (b) When I = 9 mA, V = 0.67 V. (c) From the graph, when V = 0.6 V, I = 6 mA. Now V 0.6 R= = = 0.1 × 103 = 100 I 6 × 10−3 (d) The onset of conduction occurs at approximately 0.2 V. This suggests that the diode is a Ge type. Maximum ratings It is worth noting that diodes are limited by the amount of forward current and reverse voltage

Figure 6.15 See Example 6.4.


462 Table 6.4 Device code 1N4148 1N914 AA113 OA47 OA91 1N4001 1N5404 BY127 Material Maximum reverse voltage 100 V 100 V 60 V 25 V 115 V 50 V 400 V 1250 V Maximum forward current 75 mA 75 mA 10 mA 110 mA 50 mA 1A 3A 1A Maximum reverse current 25 nA 25 nA 200 µA 100 µA 275 µA 10 µA 10 µA 10 µA

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Application

Si Si Ge Ge Ge Si Si Si

General purpose General purpose Radio frequency (RF) detector Signal detector General purpose Low voltage rectifier High voltage rectifier High voltage rectifier

they can withstand. This limit is based on the physical size and construction of the diode. In the case of a reverse-biased diode, the P-type material is negatively biased relative to the N-type material. In this case, the negative potential to the P-type material attracts the positive carriers, drawing them away from the junction. This leaves the area depleted;virtually no charge carriers exist and therefore current flow is inhibited. The reverse bias potential may be increased to the breakdown voltage for which the diode is rated. As in the case of the maximum forward current rating, the reverse voltage is specified by the manufacturer. Typical values of maximum reverse voltage or PIV range from 50 to 500 V. The reverse breakdown voltage is usually very much higher than the forward threshold voltage. A typical general-purpose diode may be specified as having a forward threshold voltage of 0.6 V and a reverse breakdown voltage of 200 V. If the latter is exceeded, the diode may suffer irreversible damage. Diode types and applications Diodes are often divided into signal or rectifier types according to their principal field of application. Signal diodes require consistent forward characteristics with low forward voltage drop. Rectifier diodes need to be able to cope with high values of reverse voltage and large values of forward current; consistency of characteristics is of secondary importance in such applications. Table 6.4 summarizes the characteristics of some common semiconductor diodes.

Figure 6.16 Various diodes (including signal diodes, rectifiers, Zener diodes, LEDs and SCRs).

Diodes are also available as connected in a bridge configuration for use as a rectifier in an AC power supply. Figure 6.16 shows a selection of various diode types (including those that we will meet later in thissection) whilst Figure 6.17 shows the symbols used to represent them in circuit schematics. Zener diodes Zener diodes are heavily doped Si diodes that, unlike normal diodes, exhibit an abrupt reverse breakdown at relatively low voltages (typically 2 VDD 3 2 V 100,000) but is liable to considerable variation from one device to another. Open-loop voltage gain may thus be thought of as the “internal” voltage gain of the device: AVOL = VOUT VIN

Closed-loop voltage gain The closed-loop voltage gain of an operational amplifier is defined as the ratio of output voltage to input voltage measured with a small proportion of the output fed back to the input (i.e. with feedback applied). The effect of providing negative feedback is to reduce the loop voltage gain to a value that is both predictable and manageable. Practical closed-loop voltage gains range from 1 to several thousand but note that high values of voltage gain may make unacceptable restrictions on bandwidth, seen later. Closed-loop voltage gain is the ratio of output voltage to input voltage when negative feedback is applied, hence: AVCL = VOUT VIN

where AVCL is the closed-loop voltage gain, VOUT and VIN are the output and input voltages, respectively, under closed-loop conditions. The closed-loop voltage gain is normally very much less than the open-loop voltage gain. Example 6.18 An operational amplifier operating with negative feedback produces an output voltage of 2 V whensupplied with an input of 400 µV. Determine the value of closed-loop voltage gain. Solution Now: AVCL = thus: AVCL = 2 2 × 106 = 5000 = 400 × 10−6 400 VOUT VIN

where AVOL is the open-loop voltage gain, VOUT and VIN are the output and input voltages, respectively, under open-loop conditions. In linear voltage amplifying applications, a large amount of negative feedback will normally be applied and the open-loop voltage gain can be thought of as the internal voltage gain provided by the device. The open-loop voltage gain is often expressed in decibels (dB) rather than as a ratio. In this case: VOUT AVOL = 20 log10 VIN Most operational amplifiers have open-loop voltage gains of 90 dB, or more.

Input resistance The input resistance of an operational amplifier is defined as the ratio of input voltage to input current expressed in ohms. It is often expedient to assume that the input of an operational


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amplifier is purely resistive though this is not the case at high frequencies where shunt capacitive reactance may become significant. The input resistance of operational amplifiers is very much dependent on the semiconductor technology employed. In practice, values range from about 2 M for common bipolar types to over 1012 for FET and CMOS devices. Input resistance is the ratio of input voltage to input current: RIN = VIN IIN

in ohms. Typical values of output resistance range from