BASIC ELECTROTECHNOLOGY.pdf

REED'S \SIC ELECTROTECHNOLOGY FOR ENGINEERS

BY

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E D M U N D G. R. KRAAL C E n g . . D F . H . (Hons.), M.I.E.E., M.1.Mar.E.

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Elecrrlcal Engmeering ond Radio Depvrtment S;:~eldcMarme and Technical College

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t'irst Edition - 1965 - 1973 (SI u n ~ t s ) Reprinted - I

j'rrt, Edition

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PREFACE F IRST E DITION This book is intended to cover the basic theoretical work in the syllabuses for Electrotechnology in Part B of the Department of Trade and Industry Examinations for Second and First Class Engineers and also Principles of Electricity and Electrical Engineering for Marine Engineer Cadets of the Alternative Training Scheme for Marine Engineers. It follows a similar pattern a s the previous yoiumes in this series which has already proved so successful, giving emphasis on first principles, referring to numerous illustrations. providing worked examples within the text, and supplying many problems for the student t o attempt on his own. The subject matter has been treated in the order and in the manner in which it would be taught at a college and the book is thuscomplementary to lecture notes taken at such a college. The typical examination questions at the end provide the student with the opportunity of'fin;illy testing himself thoroughly hcl'orc;~~lcrnplinp r h c c x ; ~ m ~ n ; ~ t 1:ullyworkctl-or11 ion. slcp-by-step solutions are given to cvcry p1.0blc111.t11115 \ ) C I I I ~I ~ I I . I I C I I I ; I I . I ~ , I I S C ful to the engineer at sea without a college tutor a t hand. The author wishes to acknowledge the assistance given by h ~ s College colleagues, M r . J . W . Powell for drawing the diagrams and Mr. T . E. Fox for assisting with the proof reading. Acknowledgement is also made to the Controller of Her Majesty's Stationery Office for permission to reproduce and use the specimen questions from "Examination of Engineers in the Mercantile Marine" as are made available by the Department of Trade and Industry. E. G. R. KRAAL

PREFACE This revision has been undertaken to meet the requirement of metrication and the up-dating of the Department of Trade and Industry examinations. In this connection, an additional chapter No. 15 has been devoted to the introduction of electronics and some extra material has been added to.preceding chapters. As for the first edition, the author has been assisted by his colleagues Messrs Powell and Fox. Their help is gratefully acknowledged. #-

E. G. R. KRAAL

CONTENTS ~I{APTF!W

I---THE

ELECTRIC

CIRC(JIT. EI~ECTRICAI

PAC;I.

'TERMS

Circuit conditions, Ohm's I ~ I w . Scsic:, and parallcl circui~s.Kirchhull's laws. Internal resistance of supply source. Electromotive force and terminal p.d. or voltage. The series-parallel circuit. Ammeters and voltmeters. Range extension of ammeters and voltmeters CHAPTER

1-1

7

2-THEELECTRIC CIRCUIT (CONTINUED). ELECTRICAL UNITS

The SI system. Mechanical units of force, work and energy, power. Electrical units of current, quantity, voltage and resistance. Examples relating mechanical and elecirical eneigy. Efficiency. Grouping of e l l s .. .. CHAPTER

18-36

3- CONDUCTORS AND INSULATORS

Resistance of a conductor,-variation with dimensions and material. Variation of conductor resistance with temperature. Temperature coefficient of resistance. Resistance of an insulator. --variation with dimensions and material. Variation 01' insulation resistance with temperature. Resistance of a semi-conductor,-variation with temperature. Heat and electrical energy. Relations between mechanical and heat energy. Relation between .. electrical and heat energy . .

C H A PT E R

37-57 LELECTROCHEMISTRY

Electrolysis. Electrolytic cells. Voltammeters (water,and copper). Quantitative laws of electrolysis ( ~ a r a d a ~ ' ~ ) . The electro-chemical equivalent, chemical equivalent, valency and atomic weight. Back e.m.f. of electrolysis. Resistance of electrolytes. Power expended during electrolysis. Primary and secondary cells. The simple voltaic cell,--cell e.m.f. Electromotive series. Polarisation. The primary cell,-

CHAPTER

L C o n t inued Daniell (energy and e.m.f.). The Leclanche cell (wet and dry types). The secondary cell,--capacity and efficiency. Charging procedure . . ..

CHAPTER

%MAGNETISM.

PAGE

58-86

ELECTROMAGNETISM

Natural and artificial magnets. The magnetic field.-flux and flux-density . Molecular theory of magnetism Electromagnetism. Fields due to long, straight, current-carrying conductor, loop and solen~id,-introduction of itn iron core. Force on a currentcarrylrtg conductor in a magnetic field, units of ampere, flux-density and flux. The magnetic circuit, magnetising force or magnetic field strength. Magnetising force of a current-carrying conductor. Permeability of free space

(Po>. . . CHAPTER

..

..

..

..

87-110

~ E L E C T R O M A G N E ~INDUCTION C

Flux-linkages. Faraday's and Lenz's laws of electromagnetic induction. Static induction,+.m.f. of self and mutual induction. Dynamic induction, -magnitude of e.m.f. The weber. Direction of induced e.m.f.-Fleming's right-hand rule. The simple magneto-dynamo. The simple d.c. generator, commutation, and practical requirements,-windings. A.C. and d.c. theory,-introduction . . . . 111-137 CHAPTER

CHAPTER

7- BASICA.C. THEORY

The a.c. waveform. Representati'on o f sinusoidal alternating quantities,trigonometrical and phasor represent:ttion. Addition ilnd subtraction of i l l ternating quantities,-graphical and m:~thcm;~tic;~l mcthotls R o o t mc;ln square and average values. 1;orlil factor. Peak factor . . .. . . 138--158 8-THE D.C. GENERATOR D.C. machine construction,-field system and armature. D.C. armature winding :\rr:lngcmcnts. The d.c. gcnerator,-4.rn.f. equation, no-load

CHAPTER

('lIAPT1:R

characteristics. Associated rnuynctic PAGE circuit effects. Generator characteristics. Types of d.c, generator,-perIllilYldIlt mayscl ul~dscpurutcly-cxcitcd types. The shunt-connected generator, -theory of sclf-excitation, 7'he magnetisation curve or O.C.C. and crhical resistance. Load characteristic. The series-connected generator, self-excitation and load characteristic. The compound-connected generator.Types of connection. Load characteristic . . 158-1 9 1 9-THEA.C. CIRCUIT (CONTINUED) Impedance, inductance, inductive reactance. Circuits with pure resistance, pure inductance and resistance and inductance in series,-power factor,-true and apparent power. C a p ~ i t a n c e , capacitive reactance. Circuits with pure capacitance, and resistance and capacitance in series. The series circuit,inductive impedances in series and inductive and capacitive impedances in series. The general series circuit,resonance .. .. .. . . 192-222 10 TFIE D.C. MOTOR

I>ircctinn of f;>rc.c, -l'lclil~rig's Icft-

hand rule. Magnitude of force. Back e.m.f. of a motor. Voltage, current and speed equations. Specd controlling factors. Types of d.c. motor,shunt, series and compound. The power and torque equations. Torque controlling factors. Motor characteristics. The shunt motor,--electrical characteristics (speed and torque), mechanical characteristic. The series motor, - electrical characteristics (speed and torque), mechanical characteristic. The compound motor,-xlectrical characteristics (speed and torque), mechanical characteristics. Cumulative and differential connection of fields,--strength of shunt and series fields. Motor starters. Speed control,-field and voltage control . . 2 2 L 2 4 6

CHAPTER

I

I I A I - I I I(

1 ~ A . C . C ' I K C ~ I T S ( C O N T I N L E D ) A N I ~ S Y S T E M S PAGE Power in the a.c. circuit. Act~veand reactive components. The parallel circuit. Inductive impedances in parallel. Inductive and capacitive impedances in parallel. Parallel resonance. Powerfactor improvement, advantages of p.f. improvement. k W , k V A and kVAr. Power-factor improvement ( kV A method). Polyphase working,-threephase systems. Star or Y connectiqn, -use of the neutral. Balanced and unbalanced loads. Delta or A (mesh) connection. Three-phase power Threephase k V A , k W and k VAr . . . . 247-275 I2

I,I-I;(,I ROMAC~NI~I.ISM (CONTINLJI~)

Permeability of free space 01,). Magnetising force due to a long, straight, current-carrying conductor, inside a solenoid and inside a toroid. Ferromagnetism. Relative permeability 01,). The B-H or magnetisation curve. Absolute permeability 01). Relilctance (S). The composite magnetic circuit,series and parallel arrangement. Magnetic fringing and leakage. Iron iosses, -the hysteresis loop. hysteresis and eddy-current losses. Pull of an electremagnet .. .. .. . . 276- 300 CHAPTER

1 3-THE ELECTRON THEORY, BASIC ELECTRONICS AND ELECTROSTATICS

Constitution of matter. The structure of the atom. Current flow as electron movement, ionisation. Electric field. The electroscope. Potential difference. Electrostatic charging,-induction. Distribution 01' charge. Electrostatic fields o f force. Electrostatic flux. ElecI ri(, lv)lrt\!i:~I 'l'lir (,:I p ; ~ ( . i t o r ( ' ; I ~ ; I ( . I tor jystcn1s.- \crle\ ; ~ n dpar:~llcl corlncction. c;\p:~citor currcnr Encrg! stored in an electr~cfield or dielectric Relative and absolute permittivity (s, and s). Permittivity of free space (e,,). c caslrr. lo LISC tI1c ~ I I I ~ I I I C I C w I . i [ l l ;I . V / I I I I I I in O I . C ~ C I .lo I I I C ; I S I I I . C the circuit current. A shunt 1s a spec~allyconstructed resistor ol' low ohmic value and, in order to make an ammeter capable of measuring a current greater than that which can be passed through it. a parallel arrangement of thc ammeter and the shunt is used. The ammeter is designed to carry a definite but small rr:~ctionof tiic main current : ~ n dthe rest of the current is made lo by-pass the ammeter through thc shunt, which is accurately

made a n d set to :I dcfinitc l-cs~sl;lncc\ ; I I I I ( . I t 15 C : I I I I ~ ~ : I I C \~4 I7 1 1 1 the ammeter instrument and r~iust;rl~iaghbc used w ~ t hi t Tlit' calibrated leads between Instrument ;rnd .;ll~rntI'orrn p;irt of tlic ;Irrangcmenl ; ~ n dnluhl not hc cut 01.~ I . I ~ > S ~ I I I .1'01.I ~ C13)~ j71cces 0 1 ordinary copper wlre. T h e diagram (Fig I I ) shows tile normal Lrrrangernent o f instrument and \bun! and tlie e.\ample hlious the form of' cnlcul:itlon necessary I t will tx 5ct.n that tlic calculation follow^ the piittern set I'or parallel rcsl.~ng ; I current 01' 7 5 t l ( ( 1 ;I l ~ ~ i I ~> dI ~ ,~ C ; I > L I I C [LI ~ I 1~ IOS SL' . I [ I I I C > \ + I I L ' ~ \ board. At the load. the voltage recorded is 10SV ;lnd ~ , h c n

the Ioxi is switched off' the volt;~gerises 10 I I O V . 1:11id~ l i c internal 'resistance of the generator, the resistance of thk supply cables and estimate the fault current if a 'shortcircuit' of negligible resistance occurred at the load terminals. 9.

The ammeter on a swltchboaid. scaled 0-300A is accidentally damaged. The associated shunt is marked 300A. 150mV A small ammeter, scaled 0-1A with a resistance of 0.12R, is available, and the possibility of using this is considered. Find if such an arrangement is possible, and if so, how it could be achieved using surplus resistors which are also available. t

10.

Five resistors AB. BC. C D , D E and EA are connected to form a closed ring ABCDEA. A supply of 90V is connected across A D , A being positive. The following is known about the resistors: AB is IOR, BC is of unknown value R , ohms, CD is of unknown value R , ohms, DE is 6R and EA is 9 0 . A high resistance voltmeter (taking negligible current) when connected across RE reads 34V with H positive and when c.orl~icc.~cd :IL.I.O.;~ ( ' 1 : rc;~tI\OV w ~ l l i I ! I I O ~ I I I L . 1~, 1 1 1 ~ 1 1 1 1 ~ ' valucs 01' K , 411d K,, thc current in branch ABCD and the

main supply current.

CHAPTER 2

THE ELECTRIC CIRCUIT (CONTINUED): ELECTRICAL UNITS All engineering studies stress theneed for units and an introduction to some of these will have been made when the subjects of mechanics and heat were being covered. Units allow measurements to be taken and calculations to be made. They are essential to the derivation of formulae from the basics of theory and enable presentations of related principles to be evolved. Thus for electrical engineering, even at the Chapter 1 stage, the ampere, volt and ohm were considered, and although these units have yet to be defined, their importance in rel~tionto the basic electric circuit will have been appreciated. The student will also recognise these units as being amongst those in common every-day usage. If, however, the impression has been given that the study of electrotechnology will involve the knowledge of an entire range of new units, then it is stressed that this is not the case. The whole modern concept of engineering technology is based on the universal adoption of SI units and, since some of these have been encountered in earlier work, it will not be long before, in this study of electrical engineering, common ground is being covered and the relevant relationships with associated units, already treated from the mechanical engineering aspect, are being stressed. Before proceeding with any further study of units of the S1 system, it w o u l d a e useful to introduce a historical note and consider the situation in engineering as it has developed. Towards the end of the last century two systems of units began to be employed in engineering; the British or foot-poundsecond (fps) system and a metric or centirnetre-gramme-second (cgs) system. The British or Imperial system had no merits since all units of the same kind, such as those of length, area, volume ther. indeed-there were also ric and horsepower which were ntly defined. The metric system t to industry and commerce but tages and it was adopted prior ical circles. In 1873 the British nt of Science selected the centiunits of !ength and mass for

h;~sc-unit of rime and the tio option of the second. for this purpose, gave the c e n t i m e t r e - g r a n i ~ ~ ~ c - s e c(cgb) o ~ ~ dsystrln. The metric system, in the cgs form, was adopted to a Iiirpt' cxtcnt li)s electrical cngiuccsing in thc curly dirys 01' dcvclopment. The system had the advantage that all the same kind of quantities are multiples of ten and it was also international. The sizes of the absolute unit of the centimetre and the gramme were found to give rise to difficulties for the desired electrical units which became either too large or too small for practical working. The use of these absolute units for essential formulae in engineering work also proved difficult and thus more workable or practical units had to be devised. Such practical units were to include the volt, ampere and ohm. In about 1900 practical measurement in metric units began to be based on the metre. kilogramme and the second and the aforementioned electrical practical units.These constituted the unrationalised MKS system. The next developmeqt came from a fact, which was repeatedly pointed out over the first half of the present century, that a system of units could be devised to make the practical units of the volt, ampere and ohm the absolute units of such a system. If, in addition, suitable adjustments of certain constants encountered in electromagnetism and electrostatics are accepted, then a more workable system of units would result. This system was known as the rationalised MKS system and its adoption W A S ~.ccommcndcdby tllc Internation:~l L i l ~ ~ t r ~ t ~ Ci)111cl~~~i~i~ mission of 1950. The change to the MKS caused some little inconvenience to the older electrical engineer and necessitated the revision of many of the better known works of reference and text-books. The student was required to appreciate however. that the new system did not upszt the course of learning in any way and that, if anything, the 11nits introduced made matters easier and formulae more manalr,eable. Prior to 1970, conditions d;3 exist when both the older Imperial and the newer MKS sydems of units were in use. The latest extension of.metric units .nto all branches of commerce and industry has enabled enginiering to evolve the SI system, the units of which are used throc ghout this book. Thus from the electrical viewpoint, it can be faid that the SI system is the rationalised M K S system with :nits in all the other fields of measurement being fully metrica..cd. T H E SI S 'STEM All measurement consists in cc mparison with some standard or unit. The three fundamental un ts are those of length, of mas.-

and of time. In the SI system the metre is taken as the fundamental unit of length, the kilogramme as the unit of mass and the second as the unit of time. From the fundamental units, can be built u p the derived units, which can be further classified as mechanical or electrical units. Thus Force I S a derived mechanical unit involving a fundamental unit and a derived unit, ie mass and acceleration. For the SI system, a unit of force, called the Nekton, has been introduced. Velocity is similarly a derived unit involving distance a n d - time. So also is acceleration a derived unit, involving velocity and time. Both velocity and acceleration are mechanical units. The ampere is really a derived unit involving force and length but as stated previously it is used as a fundamental electrical unit. Other electrical units are the volt i ~ n dt h o ohm which ore nlso dcrivcd units, but thc Joule and Wutt, although used principally in the past in connection with electrical engineering, are really derived mechanical units and will be defined under this heading. Once the units are recognised and understood, the reader is advised to discontinue their classification as mechanical o r electrical units and to accept them as general engineering units. This applies particularly to the units of work and power. Both the mechanical and electrical engineering fields are concerned with common practical appliances or associated problems and a ready use of the appropriate units, with a correct appraisal of the magnitudes of the quantities involved, is essential.to the modern engineer.

MEC HANIC AL* U N IT S The fundamental units require little definition since they are accepted standards. Thus the metre is the absolute standard, taken as the distance between two marks on a certain metal bar. Similarly the kilogramme is the mass of an accepted standard lump of'metal. The time unit is the second, which is defined as I of a mean solar day. 86 400 Most of the principal SI derived units will already have been introduced to the student but a revision is made here to allow an extension into the fieid of' electrical engineering. UNIT O F FORCE T H E N EWTON .

This is theforce required to accelerate a mass of' one kilogramme at the rate of one metre/second 2 . It has been fbund that the force of gravity acting on a mass of I kg is 9.81 ncwtons ; ~ n d ,since thc force on ;I body due to tlic ciirth's

THE ELECTRIC CIRCUIT

FLFCTKICAI, ONIT?

-- -- - --

- - - --- .

-

--

21

-

attraction is termed weight, i t follows that the weight of 1 k ~ l o gramme is 9.81 newtons. The symbol for force is F but any value in newtons can be represented by the letter N after the numerical value. Thus: lkg = 9.81N. UNIT OF WORK AND ENERGY T H E J O ULE . This can be defined

as the irork done or energ1 stored when a force of one newton acts through a d~stanceof one metre in the direction of the force. The symbol for work or energy is W but any value In joules can be represented by the latter J after the numerical value. From the definition, it follows that a force of F newtons. acting through a distance of s metres, does F x s newton metres of work o r Fs joules. Thus: W (joules) = F (newtons) x s (metres) *

UNIT OF POWER THE WA TT . Power

is the rate at which work is done or energj is converted and the unit is the watt. A watt is the power resulting, when a joule of energy is expended in a second. The symbol for power o r rate of doing work is P but any value in watts can be represented by the letter W after the numerical value. The definition can be more generally written as

P Other tbrms are

(wntts) =

I + ' (Joulc5)

-

r (seconds)

U' = Pr or

CI '

1 = -

P

The jaule and watt are the units orignally used in electrical engineering and they will be encountered constantly in electrical problems. Example 10 is set out here to serve as an introduction to electrolmechanical relationships. Example 10. A pump is required to lift 1200 litres of water through 10 metres in 6 minutes. Calculate the work done in joules and the power rating of the pump. Assume 1 litre of' water to have a mass of' 1 kilogramme. Work done = force of gravity x distance lifted = (1200 x 9.81) : . 10 = 12 x 9.81 x l C 3 newton metres = 1 17.72 x 10.' = 11 7 720N1n or 1 17 7103 1 7 720 - 1 I 772 - 327W Power = work done - -time 6x60 . 36 Power ratlng of' pump would t- 327 wntts

Note. In the above problem no account has been taken of machine efficiency. This will be introduced in due course, but for the example, the practical rating figure of the electric motor, assuming this to be the means of driving the pump, would be much larger. ELECTRICAL UNITS The same fundamental units are used as for the mechanical . primary units namely: the metre, kilogramme and s e ~ o n d The derived unit is the ampere, which has been adopted as the basic electrical unit of current and ;IS a I'ourth fundamental unit. Before considering the definition for the ampere, it is necessary at this stage, to describe two associated effects, which would be observed when a current flowed in a circuit. (1) If the resistance of the circu~twas concentrated in a short length of conductor, then a temperature rise of the wire in this region would t x noted, showing a conversion of energy into heat. (2) If the circuit was supplied through two wires laid together, then especially if the current is large and the wires flexible, a mechanical effect would be noted. When the current is switched on, the wires would be observed to move and this electromagnetic effict, as it is called, has been used to define the ampere for the SI system. The factors governing the magnitude and direction of the force on the wires will be described in the chapter on Electromagnetism. UNIT OF CURRENT THE AMPERE. This

is that current which, when maintained in each of two infinitely long, straight, parallel conductors situated in a vacuum and separated by a distance of one metre between centres, produces on each conductor a force of 2 x lo-' newton per metre length of conductor. As stated in Chapter 1, the symbol for current is I and any value in amperes is represented by the letter A after the nwnerical value. The reader is reminded that practical circuit currents may rilngc from thousands of amperes to minute values of' milli-amperes and attention is drawn to the Table of Prefixes of' Magnitudes as given at the front of this book. Full consideration must be given to the correct use of the abbreviation which follows the numerical value. When a current flows for n given tir~le,a quantity of electricity is said to be conveyed round the circuit. The quantity which passes ciin bc sliown to be related to thc work done in the circuit,

THE ELECTRIC CIRCUIT : ELECTRICAL UNITS - -

- --

--

-

23

but before this relationship is considered further, it is necessary to dcfinc qu:~ntityofclcctricity in terms of current : ~ n dtimr UNIT 0 1 . QUANTIT)' T HE COULOMB. The

usual unit-sometimes called the ampere second. For practical purposes a larger unit, for everyday electrical engineering is used. This is the Ampere hour as used in connection with the capacity of batteries and for accumulator charging. The symbol for quantity of electricity is Q and any value in coulombs can be represented by the letter C after the numerical value. Any value in ampere hours is represented by the letters A h. after the numerical value. Since the quantity ~f electricity which is conveyed round a circuit would vary with the strength of the flow of electricity and with time, a simple definition for the coulomb can be deduced thus: , A coulomb is the quantity of electricity cdnveyed by a steady current of one ampere flowing for a time of one second. Thus Q (coulombs) = I (amperes) x 1 (seconds) or Q (ampere hours) = I (amperes) x t (hours) From the above, the following can be deduced: 1 ampere hour = 1 ampere x 1 hour = 1 ampere x 3600 seconds = 3600 ampere seconds = 3600 coulombs. Thus 1A h = 3600C. Ex:~mplc I I . C'onsidct. I~?c:~mplr 5, w l ~ c ~:I ch:~ttcr.vof' c 1 n . f 42V and ~nternalresistance 752 is used to supply a clrcult of' three resistors 2, 4 and 8 0 in series. If the current is switched on for 30 minutes, find the quantity of electricity which would have been conveyed. Total resistance of circuit = 7 + 2 + 4 + 8 = 2152 42 Circuit Current = -- = 2A ;1 Quantity of Electricity = current x time in seconds = 2 x 30 2 60 = 3600C or Quantity of Electricity = current x time in hours - 2 x + = 1Ah. The passage of an electric current results in energy being expended. This energy may appear as the work done by the rotation of an electric motor, as the action of heating up a furnace element o r as the agency responsible for the electrolytic dissociation of a salt solution. The relation between conveying a quantity of electricity round a circuit by an applied voltage and the resulting work done can be used to derive the units of voltage

and resistance in terms of the coulomb and the joule which have already been defined. UNIT OF VOLTAGE THE VOLT. This is

the unit of electromotive force and potential difference and can be defined as the potential difference required between two points in a circuit, if one joule of work is to be done when passing one coulomb of electricity between the points. As stated in Chapter 1 , the symbol for voltage or e.m.f. is V and any value in volts is represented by the letter V after the numerical value. In accord with the remarks made concerning the representation of current, the reader's attention is drawn to the Table of Prefixes of Magnitudes, and to the correct use of the Abbreviations. From tho dofinition set out tibovc it is stated that the work done by part of an electrical circuit = quantity of electricity conveyed x voltage applied across that part of the circuit. Thus : fl Goules) = Q (coulombs) x V (volts) or W (joules7 = Z (amperes) x t (seconds) x V (volts). Other forms are W =

- $

v2t

or W = It(IR) = 1 2 ~ t Example 12. Consider Example 1 1 . A battery of e.m.f. 42V and internal resistance 7 0 is used to supply a circuit of three resistors, 2, 4 and 8 0 in series. If the current is switched on for 30 minutes, find the energy converted (as heat) by each resistor and inside the battery &self. Circuit current was found to be 2 amperes Using form W = 12Rt then energy converted in 2 ohm resistor = 22 x 2 x ' 30 x 60 = 14400 joules 4 ohm resistor = 22 x 4 x 30 x 60 = 28 800joules 8 ohm resistor = 2* x 8 x 30 x 60 = 57 600joules 7 ohm battery = 2' x 7 x 30 x 60 = 50400 joules Total energy converted by the circuit = 14 400 + 78 800 + 57 600 + 50 400 = 151 200 joules ( ' l / t ~ X The . total energy converted by the entire circuit may be Ii)utltl 1'1 orll Cj' =- I.'// joulcs = 42 x 2 x 30 x 60 = 151 200 joules. The definitions of' Power and Energy have already been con\~dered,but i t would be as well to summarise the points of' Importance, n:rnlely t h a t power is the rate at which work is done

1,'ro111 c i ~ * ( I i ~ ~ - t iqct o r ~ 0111 ~ : I \ N T \ C 11' - 1'11 : I I I ~ I 1 1 l o l l o \ ~ ~ ! I \I . I I P = VI or P (watts) = V (volts) x I (amperes).

The iibove is a most importi~nt reli~tinnship.I t c i ~ n;11so k expressed in the following f'orms: P

vZ

12R or P = R The attention of the reader is drawn to the following which must also be known. W Since P = - or W = Pt itfollows that t joules = watts x seconds and that a joule is one watt second. Now a joule is a small unit of energy and for practical purposes a much larger electrical unit of energy is used. This is the kilort,irtr holrr, abbreviated to k W h and is also know^ as the commerci:il unit of electricity or more commonly as 'a unit'. Since one kilowatt hour = one kilowatt x one hour = lO!N watts x 3600 seconds So one kilowatt hour = 3 600 000 joules. Example 13. A 220V electric fire is rated at 2kW. Find the current taken when the fire is switched on and also how much it would cost to use the fire for 5 hours with electricity being ~ unit. charged at 0 . 6 per =

X

I OOU

Current taken = L 7 = Y4YA 220 Electricity used = 2 x 5 = 10 kW h = 10 units Cost = 10 x 0.6 = 6p. UNIT OF RESISTANCE THE O H M . This was

defined in Chapter 1 as the unit of' sesislance and in terms of the volt and ampere thus:-a resistor has a value of one ohm resistance, if one ampere passes through i t when a potential difference of one volt is applied across ~ t ends s Now that the relations between the ampere, volt, joule and watt have been defined. i t is possible to give a l'ust1ie1-definition t'or the ohm which is associated with power or energy dissipation Thus the ohm can be defined as:--that resistance which when one ampere passes thrclugh it produces power at the rate. of one watt. Alternatively, the ohm is that resistance in which a current of one ampere flowing 'or one second generates a joule of energy. For a resistor the ensrgy produced by current flow appears a heat. and the followin: is of' importance. Since P = F I and I/ = I R then P = ( 1 R ) I or P = I'R as , leveloped earlier.

Power dissipated in a resistor is thus proportional to the current squared. If the current was doubled by raising the voltage, the power dissipation would be four times as large. The temperature would rise in proportion and assuming the resistor was capable of carryng its normal current only and had very little capacity for working at a higher temperature, then a 'burnout' would occur. The same limitations apply to cables, electrical machines and switchgear. Electrical equipment is assigned a rating which, on full load, enables it to operate with a safe temperature rise. An increase of the normal rated current. brought about by overloading or by an overvoltage, results in a temperature rise proportional to the new current squared. The total tcmpcruturc will rise very rapidly as the overcurrent occurs and if this is maintained then damage will result. Damage to electrical insulating materials can occur because of sustained overloads and overheating should be regarded as the main cause of failures of electrical machines. Example 14. A hot-plate of a ship's electric galley is fitted with a control marked Low, Mediurn and High. The heating element consists of two equal sections, lvhich are connected in parallel for High and in series for Low. Only one section is used for Medium. If the plate when set at high is rated at 2kW on 220V, find the wattage rating when the control is set at Low and at Medium. Current taken at High. Two sections in parallel

Current taken by 1 section

= 4.545A 220 220 = 48.41i2 Resistance of 1 section = 4.545 Resistance of 2 sections in series = 96.82i2 Current taken at Low. Two sections in series =

Power dissipated = 220 x 2.27 = 499-4W or wattage rating = 500W approximately. Current taken at Medium. One section only across 220V or I = 4.545A W i ~ t t i ~ grating c I S onc h ~ ~ lthat l ' of the High setting = IkW or wattage rating = 220 x 4.545 = 1OOO Watts = 1kW.

T H E E L E C TRI C C IR C U IT : ELECTRICAL UNITS

27

EXAMPLES RELATING MECHANICAL A N D EI,ECTRICAI> ENERGY The best understanding of the various units as discussed earlier, is achieved by considering examples where mechanical work is performed by electrical means o r vice versa. The relations between the units will need to be introduced a t all times, but it is necessary before proceeding to such examples, to stress that no m a c h n e is perfect and that its overall performance is measured by its efficiency. EFFICIENCY

The symbol usually used is q-the Greek letter eta. In all apparatus and machines, losses of energy occur due to bearing and brush friction, air turbulence, unwanted electrical currents. escape of heat, etc. These losses result in the output of such apparatus o r machines, when measured as work, being always less than the input when this is measured in th'e same work units. The ratio of the output to the input is termed the efficiency. Thus :

-

output input - losses or = input input output 0r output + losses Efliciency is usually expressed as a percentage. Example 15. A dicscl cnglnc I ~ ; I S ;I mc;~surctl~ntlicatcdpowcr 01' 7.5kW and a mechanical cllicicncy of' ti5 per cent. I t drlvcs a generator which supplies a lamp load at 110V. How many 60W lamps can be supplied, if the efficiency of the generator is measured to be 88 per cent? Find the total load current. The output of the engine = input x eficiency Efficiency

=

3 3

At the coupling betwees engine snd generator, i t can be assumed that there is no loss of energy, so the power input to the generator must be the powei output of'the engine and it followx that: Output of engine = i n p ~ , tto generator = 6.375kW Thus generator outpu! = input x efficiency

=

5610W

Number of lamp.; = 5610 - = 93-5 say 93 60

28

REED'S BASIC ELECTROTECHNOLOG'I'

Load current

93 x 60 - 558 110 - 11 = 50.73A say 51A

=

or alternatively ; 5610 Load current = - = 51A. 110 Example 16. A pump is required to lift 12 tonnes of' water through lorn in 2 minutes. Calculate the power required to drive the pump, the current taken if driven by a 220V motor and the cost of pumping at I p per unit. Assume the efficiency of the pump to be 60 per cent and the efficiency of the driving motor is 85 per cent. Work to be done = Force opposing gravity x distance lifted Note 1 tonne = 103kg Thus work to bc done = (12 x lo3 x 9.81) x lonewton metres = 117.72 x 10 4 Nm Also 1 177 200Nm = 1177.2kJ This is the oiltput of the pump. The input would be greater, ie 100 1 177.2 x - = 1962kJ 60 Since the pumping is to be accomplished in 2 minutes or 120 seconds, the power input during thls time = 1 962 000 120 = 16 350W Thus power required to drive the pump is 16.35kW The output power rating of the motor must be 16.35kW and 100 the input power would'be 16.35 x ---- = 19.24kW 85 19 240 - 962 Current taken by motor = ---- - -= 87.45A 220 11 2 19.24 = 0.641kw h Energy used = 19.24 x - = 60 30 Cost = 0.641 x 1 = 0 . 6 4 ~ . Example 17. The electric motor used to drive a ship's winch has an efficiency of 86 per cent. The winch can lift a mass of'0.5 tonnec through ;I distance of 22m in 22 seconds. The winding gear ol' ~ l winch ~ e has an efliciency o!' 60 per cent. Calculate the power rating of the motor and also the current taken from the 220V ship's mains. Work done by the winch = 500 x 9.81 x 22 newton metres = 107 910Nm or 107 910J 'This 1s tllc output 01' t11c winch or the output of the winding gcnr. T l i c input to the winding gear woi~ldbe .L

The input to the winding gear would also be the motor output = 179 8505

Since the lifting is done In 2 2 seconds, the motor would give out power during this time. - 179 850 = 8 1 7 5 ~ 22 For a motor output of 8175W. the input power would be

Thus power rating ot motol

Input or motor current

=

I \q

51 1\U

eo6 43 ? I A 220 =

Example 18. .4 storage batter! 15 providedJ'or emergency ubtr aboard a ship. The batter!, I S ;irr-,ingt.cIto suppi) certain essent~,ii services during the period of lime taken to start-up the 'jtand-b!' generator. The principal load to be supplied by the battery is the 'emergency' motor for an electric-I~ydr:~ulic steering gear. Thi, motor is rated a t 220V. 15kLV. and has an etlic~encyof' 88 per cent. The battery I S to be ol' a capacit) jufticient to oper:tte ti115 motor and an additional lighting load of'twenty 60W lamps f o r ;I period of 30 mlnutes. Fst~rnatethe size of' the battery and ;11\o l t h ~llscll~llgc ctlsrcl~~, Outpllr of' lnotol = 15hM' 100 187.5 = 17.04SkM' Input to motor. .= 15 r YK 11 Input current to motor = -17- 045 = 852 - - 3 = 7 7 5tl 130 11 L ~ g h t ~ nload g = 60 * 20 = 11WW - 60 = j 4 j A Lighting current = ''0° ----220 11 Total current = '7 5 + 5 45 = 82 '75A or D~scli,~rge curlent = 83 95.A S ~ z eot b,~trerj= 113 95 x = 41 (17514 h G R O U P I N G O F CELLS Ohm's law stales that the current in n circuit can be ~ncreased by raising the potential difference a ~ p l i e d3c1.c)~~' tlie C ~ I - C ' U oI ~r by decreasing the circuit resistance. I!' the supply source is a generator. the applied p.d. can be vari :d by controlling tiie e.m.f. being generated in the machine. but i.' a battery is the source ot' energ! then tiie applied voltage cann 11 be varied eabilk. Since a

battery consists of a group of cells and since the e.m.f. of any cell is fixed, being decided by its chemical composition, then a larger e.m.f. or a greater current can only be obtained by appropriate arrangement of the cells. The cells can be connected in series, parallel or series-parallel arrangements. SERIES CONNECTION. For this arrangement the -ve terminal of a cell is connected to the +ve terminal of the adjacent cell as shown in the diagram (Fig 13a). The arrangement is more simply depicted by Fig 13b. .4 battery of 3 cells in series is shown.

-44-F Fig 13(b)

Fig 13(a) From Kirchhoff s voltage law, the e.m.f.-of the source is equal to the sum of the e.m.f.s taken round the circuit and thus for a battery of n e l l s in series, the e.m.f. = e.m.f. of 1 cell x n. Also since this is a series circuit, then the current in any 1 cell is the circuit current. The internal resistances of the cells are also in series and should be treated in accordance with the deductions already made for the resistance of a series circuit. These points are illustrated by the example. Example 19. A battery consists of 4 cells in series, each of e.m.f. 1.5V and internal resistance 0.6R. Find the current flowing, if bVoLTs

2.4

onns

JPltlrt---t.

2 OHMS

0.6 OHM

Fig 14

THE ELECTRIC CIKC U I S .-

31

FLLC'TRICAL UNITS .-

-- - - -

-

-- -

thc br~ttcryis conncctcci to two rcsislors o f 2C;1 :~ncl0 , O C l wh~cll are connected in series. The arrangement is shown by the diagram (Fig 14). Total e.m.f. = 4 x 1.5 = 6V Total battery internal resistance = 4 x 0.6 = 2,452 Total circuit resistance = 2.4 + 2 + 0.6 = 5R 6 = 1.2.A So circuit current = 5 Other values of' interest would be Battery terminal voltage = 1.2 x 2.6 = 3.12V o r Battery ,, ,, = 6 - (1.2 x 2.4) = 6 - 2.88 = 3.12V Voltage drop across each resistance = 1.2 x 2 = 2.4V and 1.2 x 0.6 = 0.72V Current in 1 cell = circuit current = 1.24. PARAL LEL CONNECTION. For this arrangement, the t v e terminals of' all the cells are connected together as are all the - ve terminals. The arrangement is as shown in the diagram (Fig 15).

Fig 15

-

-

,.---'

From Kirchhoff's current law, the i-3tal current 1s the sum of the currents in each branch. Thus tti\: total current from the battery is equal to the sum of the curr.:nts available from each cell. For correct working, the e.m.f. o each cell should be the same. So also should the internal resist.ince although this is not essential. If n cells are in parallel, the toial current is n times that given by one cell, but the battery e.m.1'. is that of any one cell. This latter point can be reasoned from the fact that li' the r v e terminal of A is 3V abo1.e its -- ve ter-mi 131and t h c + \ e ter-rninal

of B is 2V above its -ve terminal, then the +ve connection between A and B is 2V above the -ve connection. If this is carried on for cell C and any further number of cells then it is seen that the whole +ve connection is 2V above the -ve connection, ie the battery voltage is 2V. The battery internal resistance is obtained from the parallel1 resistance formula, ie it is - th of a cell resistance. The battery n resistance once determined, is added to the external resistance to give the total circuit resistance as in the following example. Example 20. A battery consist of 4 cells in parallel, each of e.m.f. 1 -5Vand internal resistance 0.6R. Find the current flowing if connected to a resistance of 2,@2. The arrangement is shown in the diagram (Fig 16). 1.5 VOLTS I

OlOHH

L

1

Fig 16

,-. E.m.f, of'battery = e.m.f. of 1 cell = 1.5V t 0.6 = 0.1 5 0 b ~battery = Internal r e s i ~ t of 4 Total resisf'ahce of circuit = 2.6 + 0.15 = 2.750 1:. 57 Current = = 0,545A. Other informatien would be 2.75 ,-,Terminal voltage = 0.545 x 2.6 = 1.418V ' ~ ~ i , ,, = 1.5 - (0.545 x 0.15) = 1.5 - 0.082 _ _ i;,, '.i = 1.418V 0.545 C u ~ . ~ . c n01't 1 cell = --= 0,136A 6

I

I -1

I -1

-I -I --

t

f -1 -1

1

I.

4

S E K IE S - P A R A LLEL CONNECTION . TO build up this arrangement a number o f cells which are connected in series are then conncctcd In parallel w ~ t ha similar number of cells in series. The arrange~ l l r ' t l f 1s . I \ ~110'1\\'11 ill tllc' di:~gram(Fig 17) :lnd is used to providc 1>01li I I ~ C I ~ C ; I ~\.011:1gc C(~ ; ~ n dc t ~ r r c n1.01. ~ :I circ\~it.Tlic cells in

7'tIIi IiLEC'I'HIC' (.'IKC'IJI'I' : til l:(."I KI( ' A L C ' W I I S -.. ---. -

-

~~

--

.;cric\ provide the incre;i\c.cl c.1111' supl'ly file c.\[I.,Ic.urrcnt

33

-~

:1ri(1

thr p;ir.:~llclh : ~ n h \o f ( T I ~ \

Fig 17 The procedure for solving problems fhllows the reasoning already covered for the series and paralIe1 arrangements. but should be taken step by step. Example 21. Ten cells each of internal resistance 3Q and e.m.f. 2V are connected in two banks of 5 cells per bank. They are then connected to an external load resistance of 2851. Find the load current and the p.d. across the battery terminals. The arrangement is shown in the diagram (Fig 18a). TWO

BANKS

5 CELLS 1 VOLTS 3 OHMS PtR CELL

-

-L

L

Fig I 8 ( a )

E . m . f . of a bank = 5 x 2 = 10V = battery e.m.f Resistance of 1 bank = 5 x 3 = 1552 15 Resistance of battery = = 7,551

-,

Total c l r c u ~ tresistance = 7 5 + 20 = 27.551 10 Clrcult o r load current = --- = 0,364A 27 5 P.d. o r termlnal voltage = 0,364 x 20 = 7.3YV 0.364 Current per cell = current of 1 bank = ---- = 0.182A 2 Example 21 (continued). I f the battery 1s rearranged with 5 banks of'? cells in each. find the new current and voltage. The arrangement is shown in the diagram (Fig 18b). E . m . f .of a bank = 2 x 3 = 4V = battery e.m.f. Internal resistance o f a 1 bank = 3 x 3 = 6 R

3 CfLLS 3

20

OHMS

OHMS

PLO CfU

Fig 18(b) 6 Internal resistance of battery = -5 = 1.m Total circuit resistance = 20 + 1.2= 2 1 . m 4 Circuit or load current = -= 0.188A 21.2 Terminal voltrtgc 20 x 0.188 = 3.77V or 4 - 1.2 x 0.188 = 4 - 0.23 = 3.77V 0.188 = 0.0376A. Currentlcell = current of 1 bank = ----

-

5

THE ELECTRIC CIRCUIT. ELECTRICAL UNITS --

-

35

-

C'lIAPTER 2

PRACTICE EXAMPLES 1.

An electric hoist is required to lif't a load of'2 tonnes to a height of 30m. The cage has a mass of 0.25 tonnes and the lifting operation is timed to be completed in 1; minutes. If the 220V motor is metered to take a current of 50A, find the efficiency of the installation.

2.

Thirty cells each having an e.m.f. of 2.2V and an internal resistance of 0.3R are so, connected to give a supply e . n ~ . fol. 22V. If the arrangement is then connected to three 20V, 10W lamps in parallel, calculate (a) the terminal voltage of the battery, (b) the current taken by each lamp, (c) the power wasted in each cell. C

3.

A pump delivers 12 700 litres of water per hour into a boiler working at 15 bars. The pump whlch is 82 per cent efficient is driven by a 220V motor, having an efficiency of 89 per cent. Calculate the current taken by the motor. Assume 1 litre of water to have a mass of I kg and 1 bar = 1O5Nlm2.

4.

A resistor o f ' 5 f I I \ ~ ~ 1 1 1 r 1 r c 1l0r ( :I I h : ~ l t c r yIII:I(I(.1 1 1 7 o f ' f o ~ . . slrn~larccllx In seriex. IXacll cell has a n c.111.f'.ol' ? . ? V irncl llic

current which flows is 1.4A. If the cells were connected In parallel, find the current which would flow through the 5R resistor. 5.

A five-tonne cargo winch 1s required to lift a load of 5 tonnes at 36,5m/rnin. Calculate thc. power rating of the 220V driving motor if the efficiency o f t le winch gearing is 75 per cent and that of the motor can be taken as 85 per cent. Calculate also the current take1 from the ship's 220V mains.

6.

A 220V diesel-driven generator 1s required to supply the following on full load. (a) Light] ~g load comprising one hundred lOOW and two hundred 6 ) W lamps. (b) A heating load of 25kW. (c) Miscellaneous sn all loads taking a current of 30A. Calculate the required po ver output of the diesel e n g n e when the generator is supp ying all the loads at the same time. Assume a generator elf ciency of 85 per cent.

~ ~

7.

A battery is made up from three similar correctly connected dry cells in series. The open-circuit e.m.f. is measured to be 4.3V. When the battery is connected to an unknown resistor the current i s metered. to be 0.4A and the battery terminal voltage as 4.23V. If one of the cells of the battery is reversed and the circuit made up as before, estimate the new current value.

8.

A 150W, lOOV lamp is to be connected in series with a 40W, llOV lamp across a 230V supply. The lamps are required to operate at their rated power values. Determine the values of suitable resistors to be used wit11 the lamps and make a sketch showing h o b they would be connected.

9.

A resistor of 0.5250 is connected to the terminals of a battery consisting of 4 cells, each of e.m.f. 1.46V joined in parallel. The circuit current is found to be 0.8A. Find the internal resistance of each cell.

10.

Twelve cells, each of e.m.f. I.5V and internal resistance 0.225R, are arranged four in series per row or bank, u.itli three banks in parallel. The battery so formed is connected to a load consisting of a series-parallel resistor arrangerncnt. made up of a 2R resistor connected in parallel with a 712 resistor, these in turn being connected in series with a 2 . 2 ' 1 resistor. Find the battery terminal voltage, the power ratings of the resistors and the energy converted into he~itin the complete circuit if the arrangement is switched o n for 1 hour.

(

I I A I r I I I< 1

CONDUCTORS A N D INSI'I :ITORS The reasons as to why certain materials arc good conductors of' electricity whlle others are not. will be considered in det~iil later when the electron theory is studied; here i t can be stated that a substance which freely allows the passage ot' electricity is classed as a conductor. Examples :ire metals, carbon and certain liquids4hiefly solution of salts, acids or alkalis. An insulator can be defined here as a substance which will not allow the free passage of electricity. Examples are rubber, porcelain, slate, mica, some organic materials and certain liquids-notably oils. RESISTANCE O F A CONDUCTOR \.ARIATION O F C O N D U C T O R R E S I S T A N C E LVITH * D I ~ I ~ N S I O S SA N L ) MATERIAL The resistance or 'ohmic‘ value of n conductor. \uch a i a c o ~ l

of'wire, can be altered in different ways. Thus if'co~lsof'different lengths of' the same wire. lr jame material and same crosssectional area. are measured for resistrince. their ohmic ~ a l u e s would be fbund to varq in d ~ i e c tproportion to their lengths. Again if coils of wire of the same materia! and length, but o f tlifycrcnt cro\\-ccc~ior; ;Ire r n c ; ~ \ ~ ~ r ~hcir. r d . rc\isl,lnc.c L ; I ~ I I C \ would be I'ound to vary In inverse propcrrtlnn to tlie areas of' the wires with which they are wound. A similar series of comparative measurements witti coils 01' wire of the same length and cross-sectional area but different material, would show that the resistance value varied with the conductor material. The elementary tests described above, indicate that the resistance of a conductor or resistor can be altered by varying its dimensions or the nature of material used, and the relation of' these factors to the actual conductor resistance will now be examined in detail. l(a) DIME N S I O N S . Resistance of' a conductor is proportion,~lto its length or, more simply and using an example, i t can be said that the conductor resistance of a lOOm length of cable will be double that of a 50m length of the same cable. This can be readily shown as follows: Let Rq ohms = the resistance of a 50m length. Then two 5Om sections in series would have a resistance of R ohms Hence R = RA RA = 2RA

+

But the length has been doubled So 2 x Length = 2 x Resistance of original length. Summarising : Resistance is proportional to Length or R cc I. l(b) DIMENSIONS. Resistance of a conductor is inversely proportional to its area or, more simply and using an example, it can be said that the conductor resistance of a lmm 2 cross-sectional area cable will be twice that of the same length of cable, of the same conductor material but of 2mm2 o r twice the cross-sectional area. Thls can also be shown thus: I.ct the resistance of the l m m 2 cable be RA and suppose an identical cable to be connected in parallel with it. The resistance of the combination would be R ohms. R 0,. j( = A l lcllcc I - I i I R

KA

RA

5

2

Thus the resistance of the combination is the original cable resistance, but the area of the combination is twice that of the original cable or 2 x Area = x Resistance of original length. If the reasoning was repeated for four lrnrn 2 cables in parallel, the area would be four times the original and the new resi'stance would be R =

h. The resistance of a 4 4

x

1.0 o r 4mm2 cable

would be 4of the original cable resistance and the area would be 4 times that of the original cable. From the above, it follows that doubling the area halves the resistance of' a mnductor of' the same length and material and quadrupling the area reduces the resistance to a quarter of the original value. Thus: 1 Resistance is inversely proportional to Area or R a - . A

2. M A TE RI A L . The resistance of a conductor depends upon the material from which it is made. Thus the resistance of' a length of iron wire is approximately 7 times greater than the resistancc of :\ piccc of copper wire of exactly similar dimensions, ir same length and cross-sectional area. If therefore resistance v:~ric.;with tlic n;lturc of thc m:ltcrinl, it is necessary to dcfinc this property in such a way as to allow a comparison ol'resistance value for accepted standard dimensions of the conductor material. The term resistivitv or specific resistance (symbol p the Greek letter rha), is now introduced. This was originally expressed in ohms or niicrohms per cm 3 . The modem tendency 1s to slxcil'y i t ;IS---ohm-metreor--microhm-millimetre. ' l ' l ~ cI < c h ~ \ t ~ v 01. ~ l ySl>ccilic, IcL ~ I L ~ L ~ I I ~ :IINI I I ~ ~ c1 0I ; I ~ I I ; I Cpicc,cs ~ 01' ~ ; I ~ , C but if 3 similarly treated glass rod was buspendcd b a thread and brought near the original charged glass rod. a repulsion effect would be noted. Similarly an ebonite rod rubbed with f'ur. would be found to be charged and if'brought near the suspended charged glass rod, attraction would be noted. Summarising we can say that the electric charges acquired can be of two types, termed positive ( + v e ) and negative ( - v e ) charges and that like charges repel whereas unlike charges attract. The allocation of the designation +ve charge to the glass rod and - ve charge to the ebonite is purely arbitrary, but the general theory is that all uncharged bodies consist of +ve and - ve charges which neutralise each other. I!' these charges art. separated by some applied effort then their presence becomes detectable, and if they are caused to move from one body to another or through a circuit then t h e ~ rmovement is explained b\ the passage of' current. Returning to the subject of' this chapter. we see that for an electrolyte the molecules split up into charged ions which are extremely mobile. If two plates, termed the c.lec,trodrs, are

I

60

RFED'S BASIC ELECTROTECHNOLOGY

immersed in the electrolyte and a current is passed by means of them through the solution, then a potential difference exists across the electrolyte between the electrodes due to the applied potential. The +ve ions migrate, under the influence of the electric field of the potential difference, to the cathode, namely the electrode by which the current leaves the electrolyte. Such ions are called cations. The -ve ions, called anions, migrate to the anode, namely the electrode by means of which the current enters the solution. Metal or hydrogen ions always carry +ve charges and travel with the current to appear at the cathode, whcrcc~snon-mctnllic ions travel in the opposite direction to the current and may appear at the anode or may engage in secondary reactions, some of which will be described shortly. ELECTROLYTIC CELLS The whole arrangement consisting of electrodes and electrolyte as described above, is frequently called an electrolytic cell to distinguish it from a voltaic cell which will be described later in this chapter. Electrolysis does not occur with solids o r gases and is only possible for certain liquids Some, like oils, are nonconductors, whereas others, like mercury, conduct without decomposition. The remaining liquids are electrolytes, which can therefore be defined as liquids which decompose when current is passed through them. The electrolytic cell can be constructed to enable experiments and measurements to be made with great accuracy. In this form it is frequently referred to as a Voltalneter (Sulphuric acid solution) The diagram (Fig 24) shows the construction of the apparatus which is made of glass, with platinum electrode plates placed at A and C. The lead-in wires, passed through rubber corks, are not exposed to the solution to prevent corrosion, The voltameter is filled with acidulated water and the platinum electrodes are connected to a battery. Current passes from the anode to the cathode and bubbles of gas are given off which rise into the graduated tubes If care had been taken before passing current, to fill both tubes with the acidulated water by opening the taps and then closing them after all air had been expelled, then certain deductions can be made from the experiment. After a period of time, the gas collected would be found to be Hydrogen at the cathode and Oxvpcn : ~ tthe anode. The ratio of the volumes of H to 0 would be 2 : l and the amount 01' gas collected would be pro-

THE W A TER V O L T A MET E R

ELECTROCHEMISTRY

61

Fig 24 portional to the strength of the current and the time for which it flowed or more generally to the quantity of electricity passed. In the acidulated water there are considered to be sulphunc acid molecules which divide into three ions, two of hydrogen carrying ve charges H + , H + and one with - ve charge represented by SO, - . Note the total + ve hydrogen charge equals the - ve sulphate or sulphion charge, but the ions migrate under the influence of the electric field. Thus the H + ions give up their charges at the cathode and are liberated as hydrogen gas. The sulphions proceed to the anode, but as they cannot exist in a free state they combine with two hydrogen H + H + ions or atoms from the water thus liberating oxygen as in the equation. Thus: 2S04 + 2 H 2 0 = 2 H , S 0 4 + 0,. The oxygen rises from the anode and collects in the tube above it. The H , S 0 4 goes into solution and thus the electrolyte is decomposed in that the water appears to be used up, but the acid content remains the same and the solution gets stronger; that is, its specific gravity rises.

+

-

(Copper sulphate solution) The diagram (Fig 25) shows the usual arrangement. A copper sulphate solution, made from crystals dissolved in pure water, is contained in a glass or glazed earthenware tank. The electrodes are made from pure copper sheet. The CuSO, molecule is considered to split into two ions, C u + and SO4--. When a p.d. is applied to the electrodes and current is passed, the copper ions migrate to the copper cathode to combine with it and give u p

THE C O P PER VO LT AMET ER

+

'

Fig 25 their charges. The sulphions give up their charges at the anode and combine with the copper from this electrode to reform copper sulphate. Thus copper appears to be taken from one electrode and deposited on the other. The chemical equations for the electrodes are: (1) Cathode. CuSO, = Cu + SO,. (2) Anode. Cu + SO, = CuSO,. During electrolysis a certain amount of gassing may be noted at the plates. This would be due to decomposition of water in the solution as described for the water voltameter. Furthermore some complex action may occur in the electrolyte due to sulphions combining with hydrogen in the water to form H2S04. Oxygen from the water is then released to combine with anode copper to give copper oxide. This oxide will then dissolve in the H2S04 to give CuSO,. Irrespective of the action the final result is a simple'one, in that the loss in,mass of the anode equals the gain in mass of the cathode. Various forms of voltameter can be constructed to allow research into electrolysis. Thus a silver voltameter may be used consisting of silver (Ag) plates and a silver nitrate (AgNO,) solution. The cxumples described could be connected in series and the same quantity of electricity passed through all voltameters. If the electrodes were washed and carefully weighed bel'ore electrolysis LInd then washed and weighed again at'tcr electrolysis, certain conclusions would be reached which were first enunciated by Faraday in 1834 by his laws-of electrolysis. Q U A N I,ITATIVE L AW S O F EL E C TR O LYS IS (Farada y's Laws) 1 , The mHss of an element liberated from or deposited on an clcctrodc is proportional to the quantity of electricity which has pil sbccf.

ELECTROCHEMISTRY

63

2. The masses of elements liberated from or deposited on tn clcctrodcs by a given quantity o f electricjty are Atomic* thcir Chcmicul Equivalent it1 ---

-- ~ u b n c y

Consider the first law. It is found by experiment that the mass of any material deposited or liberated always depends,on the quantity of electricity which has passed. Thus in K Q coulombs or rn cc It. This proportion can be modified to: m = :It where z is a constant depending on the substance deposited. I is termed the el~ctrocheiniculequivulenr of the element. (E.C.E.) The mass in grammes or kllogrammes liberated by one coulomb of electricity is called the E.C.E. of asubstance. Thus, 10 amperes flowing through a copper voltameter for 1000 seconds would result in 10 000 coulombs having passed and 3.3g - -of ELECTROCHEMICAL EQUIVALENT

3.3

copper would be deposited. Thus the E.C.E. of cTpper = -10 000 = 0.00033g/C. Similarly that for h y dro g en would be 0.000 010 4, for oxygen 0.000 082 9 and for silver 0.001 118glC. In line with metrication and the use of SI units, it is more appropriate to think in terms of the kilogramme and the E.C.E. can be defined as the mass (in kilogrammes) 01' a substatlcc liberated by the passage of one coulomb. Thus the E.C.E. of copper would be 330 x 10-9kg/C. Since the milligrr~mmeis also a n accepted SI unit, the E.C.E. can be given as mg/C. Thus for copper it would be 0.33mgIC. The first law of electrolysis leads to a method of stating the unit of current, which was considered accurate enough to allow an original definition for the International Ampere. This then, was defined as, that unvarying current which deposits silver at the rate of 1 1 18 x ~ O - ~ kper g coulomb when passed through a solution of silver nitrate in water. The formula already deduced above allows the solution of problems associated with electrolysis and practical electroplating. The unit in which the E.C.E. is given should be noted. Example 32. Find the time taken to deposit 11.4g of copper when a current of 12A is passed through the copper sulphate solution contained in a copper voltameter. The E.C.E. of copper can be taken as 330 x kilogrammes/coulomb.

64

REED'S BASIC ELECTROTECHNOLOGY

m Since m = zit then t = -

Iz

= 2.88 x lo3 o r t = 2880 seconds = 48 minutes.

-

EQUIVALENT, VALENCY, ATOMIC WEIGHT The second law of electrolysis can also be deduced by examinIng the results of tests made with a number of different voltameters in series, having been subjected to the passage of the same quantity of electricity. The results of the experiment would show that the mass of' the substances deposited o r liberated at the electrodes would be proportional to the chemical equivalent of tlic S L I ~ ~ I ; I I I C C I\ t. w ~ t l l d;IISO hc concluded that if thc i~tomic weight of any substance is known, its E.C.E. can be found provided the valency is known and the E.C.E. of hydrogen is assumed. Thus if the chemical equivalents of hydrogen, oxygen, copper and silver were 1, 8, 3 1.8 and 107 respectively, the masses of H, 0, Cu a n d A g liberated by the same quantity of electricity would be in the same proportion and therefore the E.C.E. of a substance is the E.C.E. of hydrogen multiplied by the chemical equivalent of the substance. Thus taking the E.C.E. of hydrogen as 0.0104mg/C, that of silver would be 0.0104 x 107 = 1.1 18 mg/C. To conclude our deductions from the second law, it would be well to define the following terms. VALENCY. This can be described as the combining ratio of'a substance. As an example that of oxygen is 2, whereas that of hydrogen is 1, so water is represented by the symbol H,O. The valency of an elemental substance can also be defined as the number of atoms of' hydrogcn with which one atom of' the element can combine. Thus the valency of a sulphate is 2 since, for example, in sulphuric acid H,S04, 2 atoms of hydrogen are required to combine with the sulphate. CHEMICAL

ATOMIC' wuc;t{.r. Atomb arc extremely small and determination of their absolute masses present considerable difficulties. The Inass of' a n atorn of' hydrogcn is bclicvcd to bc 1.67 x 10- 14g and i t is still therefore customary, even in SI units, to refer to the relative weights of the atoms of various substances in terms of the atom of hydrogen. Thus the term atomic weight is still used and is the weight of an atom of the substance in relation to the rnnss 01' a n atom of hydrogen. Thus the value'for oxygen is 16, re i t has 16 times the weight of an atom of hydrogen.

ELEC TRO C HEMIS T RY

65

This is the ratio of atomic weight - to Atomic Wcighr. is thus valency or Chemical Equivalent = Valency weight of a substdnce which would combine wit11 one part by weight of hydrogen or eight parts by weight of oxygen. Thus the C.E. or combining weight of hydrogen is 1, that of oxygen is 8, silver 31.8 and so on. as would be fourid in chemical tables. From the second law we have Mass of Material X liberated Mass of Hydrogen liberated Chemical Equivalent of Material X Chemical Equivalent of Hydrogen 0,- % = C ' E of also from the expression rn = zlr InH C.E. of H zx-It - C.E. of X or zX = zH x C.E. of X z, It C.E. of H C.E. ofcH But the chemical equivalent of hydrogen = 1 so z,. - = z,- - (Chemical equivalent of substance X) o r zx = ZH x Atomic weight of substance X Valency of substance X .

CHEMICAL E Q U I VA L E N T .

Example 33. How many amperes would deposi' 2g of copper in 15 minutes, if the current is kept constant. Give11the E.C.E. of hydrogen as 0.0104mg/C, the Atomic Weight of Cu as 63.56 and rhc Vi~lcncv01'

Then z,,

< ' i:IS ~ 2.

=

63.57 zH x ---= 0.0104 x 3 1.8

2

= 0.33mg/C

2

rn

whence I = - = 21

0.33 x

x

15x60

Example 34. A voltameter consists of a solution of iinc sulphate and electrodes of' zinc and carbon If current is passed in at the carbon electrode, zinc is found to be deposited on the zinc electrode and oxygen is given 05at the carbon plate. If a current 3.5 amperes is passed for 1 hour, find the mass of zinc deposited and oxygen liberated from the solution. The E.C.E. of zinc can be taken as 338 x 10-gkg/C, the Atomic Weight as 65.38 and the Valency as 2. Take the Atomic Weight of oxygen as 16 and the Valency as 2.

From relationships already deduced: rnzN = zz, It or m,, = 338 x x 3.4 x 3600 = 338 x l o w 9 x 126 x lo2 or Zinc deposited = 425.88 x lo-' = 4.26 x 1 0 - ~ k go r 4.26g

-)

At wt of Zn and zo = z, Valency of Zn At Lt of 0 / A t 'wt of Zn -0 = zZN Valency o f 1 Valency of'Zn ol'Zn = At wr ol' 0 x Vatency -...-Valency 01'0 At wt of'. Z;

Also zzN = z , 50

Thus z , =

'ZN

A t w t o l ' o = 338 ~t wt of ~n

10-9

(Valency At ofO0) Wt

16 65.38

338 x 4.086 or 2, = 82.5 x 10-9kg/C o r 82.5 x 10-6g/C x 3.5 x 3600 and ,no = 82.5 x = 8.25 x x 126 x 10' = 8.25 x 1.26 x l o - ' = 1.05 grammes. -

BACK E.M.F. OF ELECTROLYSIS

The circuit laws enunciated in Chapter 1 govern the conditions fhr the majority of practical circuits. The loads of' such circuits are mainly resistive such as, the coils of' resistance wire in appliances like electric heaters and filament lamps. Such loads, termed pussive loads, are recognised by the fact that they conform to Ohm's Law. For other types of loads such as the electric motor and accumulator or storage battery when being charged, Ohm's law is not directly applicable and they represent uctive loads, of' which the electrolytic cell is also an example. The difference between pure ohmic resistance and that offered to the pass;igc ol'currcnt by a n clcctrolytic cell will now be considered. The diagram (Fig 26) shows a simple circuit for which the sourcc 01' supply i h ;I hattcry m ; d c u p of' thrce similar voltaic cells in series. Assume that the current through an electrolytic cell made up as shown, is adjusted and maintained at 3 amperes by the variable resistor provided for this purpose. IS the circuit is set and the supply potential is then reduced by removing one of' the voltaic cslls, i t mny well bc iissumed that, as the e.m.f. has hccn rcduccd to of' the original v;\lue, the current wil,l fall to 3 x 3 = 2 ;lmpc.I.cs. 111 acrual I'ucr the ncw current strength will

:

ELE C TR O C HE MI S T RY

67

Fig 26

be well below this value, and if the experiment is repeated by removing another voltaic cell so as to make the supply e.m.f. of the original, then the final current value will be well below that of 1 ampere, the value expected by an application of Ohm's law. The experiment shows that an extra current controlling factor is present in a circuit involving an electrolytic cell and the results can be explained by considering that a back e.mlf. is produced by the cell, so that the following equation represents the conditions :

4

v = 5' , +

I/se~.\~ed \VIIICII ~ O I I ~ I I I I I Lto' S gener;~te;III c.m.f. and mi~intainthc current. Tllc cIiemic:~laction ol'a voltaic cell results in a definite c.1~1.l'. wliic11 is the result of the action of both electrodes o r plates w ~ t l~~ l clcctr~oly~c. ~ c 11' lie voltl~lc~cr wits connected on ope11 circuit between the zinc and the solution, the former would be found to be negative to the latter by some 0.63 volts. When connected between the electrolyte and the copper plate, the voltmeter would record 0.47 volts, with copper positive to the sulphuric acid. With the instrument connected across both plates a reading of 1 . I V would result as could be expected. The production of a cell e.m.f. is explained by the electrolytic theory ;~ll-c:td! discussed under electrolysis. I t has been said earlier that when ;In electrolyte is made up, tlie molecules split into ions which arc clcctrici~llycharged and very mobile. An electrolyte is thus a n ionised solution and when a metal is immersed into it, some of'the metal appears to enter into solution, in that there is ;In irnmcdiate merging 01' surf'acc ions oftlie metal with the ions of' the electrolyte. Thus the barrier between the metal and electrolyte is not the surface of tlie former but along some layer of' electrical potential equilibrium, which has caused ion interchange to cease. The action on immersing iarious metals into an electrolyte differs for the different metals. Thus for zinc in dilute sulphuric acid there is a greater tendency for +ve ions to pass to the solution than ['or f v e ions of'the electrolyte to pass to the ~ i n cThc zinc plntc thus becomes deficient in +ve ions and becomes negative to the solution by a voltage of 0.63V. F o r copper in sulphuric acid, a different action takes place. This metal becomes positive to the solution because there is a greater tendency for the ve charged hydrogen ions of the electrolyte to move to the copper than for the +ions of the metal plate to pass to the solution and the copper rises to a potential of 0.46 \~oils\\,it11 respcct to thc clcctrolyte. ~l'lieinterchange of ions as described, results in a potential I)c~tigsc!L I bclwcen ~ tlic clcctrodcs and solution which gradually

+

opposes the interchange irntil this fini~llyce;lsch ;rnd cqirilihriurn is established. Thus the arrangenlent ot'zinc-copper clcctroclcs in sulphuric acid as described, results in ;I potentiiil ciiffcrencc of I. I V between the electrodes. Continuing our investigation of this type of'cell on open circu~t,we see that the initial Zn ions migrating to the solution combine with the sulphions to form zinc-sulphate, libera-ting + ve hydrogen ions which move to and accumulate on the copper plate. Thus the chemical action is explained by the formula Zn .+ H,SO, = ZnSO, + 2H. Thc potentials build up within the cell, quickly bring the ion thjgrations to an end and thus chemical action ceases. If now the open-circuit condition is changed to that of a closedcircuit, by joining the copper electrode to the zinc through an external circuit, the chemical action is immediately noted to recommence. Current is seen to flow and the formation of zinc-sulphate continues with the liberation of hydrogen at theccopper plate. The action of the cell on closed circuit conforms to the following reasoning. The open-circuit e.m.f. of l . l V , which is the result of the initial ion migration, can now cause a current, the magnitude of which is determined by the circuit resistance. A flow of current, as will be seen in Chapter 13, means a movement of negative charges has resulted; passing from the zinc cathode to the copper anode. The - ve charges move round the external circuit from the zinc cathode to neutralise the +ve iinode ch;irgcs, thus making way Ihr f'uriher migrcitions ol' t- vc hydrogen ion charges in the cell. The initial cell action, as described for the open-circuit condition, can now continue and the cell functions by maintained chemical action; provided polarisation is avoided. Polarisation has already been mentioned, and will be described again for the simple cell. We can now conclude our study of cell action by saying that all metal electrodes produce an e.m.f. as a result of the ion interchange action with the electrolyte, and that they can be arranged in a table, in order of the value of their e.m.fs.

I

ELE~TKOMOTIVESERIES . If any two elements shown in the table are used for a cell, the element lowest in the series is the +ve terminal, when considered w ~ t hrespect to the external circuit. The list comprises the more usual elements which are mainly metals, but hydrogen and carbon are found to behave like metals and are included. T o illustrate the use of the table, the ordinary dry cell as used for ----a hand-tor& -- can be considered. This cell uses carbon for the +ve -electrode- and -- zinc for the -ve electrode. w-

74

REED ' S BASIC ELECTROTECHNOLOGY

Materials well spaced apart in the table are usually used for practical cells. For the torch battery the e.m.f, is about 1.5 volts per cell. Aluminium Zinc Iron Nickel Lead Tin Hydrogen

Hydrogen Copper Carbon Mercury Silver Platinum Gold

POLARISATION. When the simple cell supplies current, polarisation occurs as described earlier. The circuit current gradually falls, even though the c h e m i c d action of the cell appears to proceed. Close examination will reveal that as the hydrogen bubbles make their way to the copper plate, not all are liberated here and rise to the surface. Some bubbles stick to the plate and this tendency increases until the whole plate is covered with bubbles to result in the cell becoming ineffective as a source of e.m.f. The layer of gas surrounding the +ve plate causes a polarising effect because (1) gas has a high resistance, so that any area of the plate covered with bubbles is almost insulated and cannot allow the passage of current. Thus the internal resistance of the cell rises as the gas layer increases and the circuit current mlls as a direct result (2) As hydrogen covers the copper plate, it begins to make its presence felt in that it effectively replaces the +ve copper electrode by a hydrogen electrode and thus reduces the e.m.f. of the cell: It will be seen from the table of the electromotive series that the spacing between zinc and hydrogen is smaller if compared with that for zinc and copper. The cell e.m.f. is thus much reduced giving the final result as described. Oncc thc cilusc ol'pol;~risationbccamc known i t Wiis appnrcnt that, in order to make the simple cell an effective source of' elccrricr~lcncrgy, ii mclliod of' preventing t l ~ ecollection of the hydrogen bubbles was necessary. The simplest forms of'cicpolrrrisrrs which have been developed, operate chemically, in that they combine with the liberated hydrogen to convert i t into water; thus preventing the gas from rcuching the + ve electrode and blitnketinp it. The methods by which this i s nccnmplished will he seen when the examples of prit,ttrry cells ;ire stucliccl.

ELECTROCHEMISTRY

75

I

THE PRIMARY CELL Utidct t l i ~ sIicnding nrc cur~stdcrcd ~)r.uct~cul cclls, wl~iclrutr

i

I I I

I

I

I

II I

sultable for providing a constant e.m.f. when operating under everyday conditions. They are however, cells which obtain their electrical energy from chem~calenergy, the active ater rial being used up in the.process. They differ from secondary cells in that the latter utilise materials which are not consumed when the cells provide electrical energy. The secondary cell can be electrically 'charged' so that its electrodes are chemically converted into materials which enable the cell to provide an active e.m.f. for supplying electrical energy. In this condition the cell discharges and the electrode materials again change chemically, reverting back to those of the uncharged state. The whole cycle of charge and discharge can then be repeated. Primary cells suffer from the two main disadvantages of the simple cell, (1) polarisation and (2) local action.,: Polarisation is overcome by the use of a su~tablechemical depolariser which is therefore an essential component of cell construction. Local action is minimised by using pure metal, such as zinc free from impurities like iron and lead. In its basic form the primary cell is a wet cell, which is not used to any extent nowadays. In the dry form the ~eclanchecell is the most common and particular attention should be paid to its construction and action. THE D A N IELL CELL. This cell is now not used, not even in the

laboratory. The only reason for describing it, is to illustrate the action of a depolariser, which is here a solution of copper sulphate (CuSO,). The cell consists o f . a porous (unglazed) earthenware pot which is placed in a copper vessel and contains the -ve electrode which is a zinc rod. The +ve electrode is the copper container, and the electrolyte is dilute sulphuric acid. The diagram (Fig 29) shows a cross-section of the arrangement and the action is as follows. The porous pot keeps the H2S04 separated from the CuSO, but allows the passage of ions from one liquid to the other. As for the simple cell, the zinc and sulphuric acid react to form zinc sulphate and hydrogen when the external circuit is made. The hydrogen ions make their way through the porous pot and enter the copper sulphate where they displace the copper ions to combine with the sulphions to form sulphuric acid. The copper ions arc displaced towards the +ve o r copper electrode and are deposikd on the copper vessel. There is thus no blanketing of the anodc with hydrogen and the cell gives a steady current without polari sing. The e.m.f. is about

POROUS

.

-.

.-. ...

-

----- -- -- - .

COPPf R CONTAINER COPPER SULPHATL

. .

-.

.-

-

.. .

:1

.. -..

~-~

U

--

L

P ~ ACID

U

R

K

Fig 29 1.1 volts, and the example set out below shows how the energy available and e.m.f. of a cell can be deduced. Of interest also are the chemical formulae which explain the action in the cell. Action at negative plate Zn + H,SO, = ZnSO, + H, Action at positive plate H, + CuSO, = H,SO, Cu

+

ENERGY AN D E . M . F . O F A D A N I E LL CELL. Any chemical action results in an e.m.f. For example that resulting in the formation of ZnSO, (zinc sulphate) by dissolving zinc in sulphuric acid gives rise to an e.m.f.which is exactly the same in value as the back e.m.f. produced when the compound is electrolysed. The same is true for any other metal such as copper dissolved in sulphuric acid. Now when I coulomb of electricity passes through the electrolyte 0400 338 gramme of zinc is dissolved or deposited. Also whcn 1 grarnme of Zn is dissolvcd in sulphuric acid 6846 joules 01' heat are liberated. This inl'ormation can bc obtiiincd from the appropriate chemical tables. Similarly 1 coulomb passing through Lllc clcclrolyle dcpositb or dissolves 0400 329 5 gramme of copper, and when 1 gramme of Cu is dissolved in sulplluric i~cid3700 j o u l ~of~ hci~t; ~ r cliheri~lcd. Thus the energy released during the passage ol' I coulomb which dissolves the zinc in a Daniell cell = 0.000 338 x 6846 = 2.32 joules. Similarly the energy utilised during the passage of I wulornb which deposits the copper in a Daniell cell = 0,000 329 5 x 3700 = 1.12 Joules. Therct'ore the energy avniliiblc 1.01. driving 1 coulomb through rtlc cell 2.32 - 1.22 :~1 . I ioulcs. 1

*fhe cflectivc ccll voltage will thercl'orc be I . I volts In order to C X I ) C I I ~1 ~' 1 , ~ ( l k l l ~1 h' 0 ~1 1 1 ~ ' l > ~ l h h ~ 01' l ~ C 1 ~ ~ ) l l ~ O111 l lCl~ ~C L~' I IL'II)'. I

Thus the efl'ective e.m.f. of a Daniel1 cell is 1 . 1 volts. T ~ I ILECLANCHI:. : CI:LL (Wet type). The diagram (Fig 30) shows a cell used for supplying small amounts of electricity in remote locations where cells of the dry type have no particular advantage. Up to quite recently such cells were used for railway signalling in places where no electricity mains were available.

The +ve electrode consists of a carbon rod which is placed i n a porous pot and surrounded by small pieces of carbon mixed with powdered manganese dioxide (MnO,) which serves as the depolariser. The pot is sealed with a layer of bitumen compound o r pitch. The negative electrode is a zinc rod and the electrolyte is a solution of ammonium chloride (NH,CI). The action of the cell may be summarised as follows. Zinc ions migrate into the electrolyte and when the external circu~tis made, the further chemical action results In zinc being dissolved by the electrolyte and zinc-chloride (ZnCl,), ammonia (NH,) and hydrogen are produced. It is the conversion of zinc into zinc chloride which provides the energy of the cell. The chloride and ammonia dissolve in the water of the electrolyte and the hydrogen as ions migrates through the porous pot and reaches the carbon. The hydrogen ions, while passing their charges to the carbon electrode, combine with the manganese dioxide, taking from i t some of its oxygen to form water. The manganese

dioxide is'thus reduced to a simpler oxide (Mn,O,), frequently known as sesquioxide of manganese. The solid depolariser is comparatively slow and polarisation takes place if the cell, is used continuously. When the circuit current is switched off, the depolarising action continues and the cell can be used again after a little while. It is best suited for intermittent duty such as bellringing. The e.m.f. is 1.55V, and the following chemical formula defines the chemical action. Action at negative plate : Z, + 2NH,Cl = ZnC1, 2NH, H, Action at positive plate: H, + 2Mn0, = Mn,O, H,O.

+

+

+

n i l : L.I:(.I.ANCIIII CEI.L. (Dry type). One form of construction is illustrated by the diagram (Fig 31) which is a cross-sectional view of a typical practical cell. BRASS CAP

/

prcw

Fig 31

The depol:\~.iscrof mongancsc dioxide is mixed with powdered carbon and packed round a central carbon rod. This whole ;~ssemblyis then placed in n linen bag which serves as the porous pot of the wet cell. The negative electrode is a piessed zinc cannister which contains the linen bag assembly and the electrolyte, which is made up as a paste of ammonium chloride (salammoniac), zinc chloride, flour and plaster of paris. One method of closing the cannister is by sealing it with pitch. Because the cell uses the same materials as the wet type its action is identical and the chemical formulae as already given also apply.

The form of cell as described, is in most general service: but otllcr li)r~nsl ~ i ~ vbrcrl c dcvclopccl I'or incorpuruling inlo lllz lityes type of battery, as is used for portable wireless sets. The reader should complement the information given here by referring to books specialising in the practical treatment of' electrical equipment.

THE SECONDARY CELL ( o r Accumulator) Because of the importance of this cell as a means of storing electricity ( i t is sometimes called a storage cell), the reader is again advised to consult a book giving more details of modern constructional methods, applications and maintenance requirements. The diagram (Fig 32) shows only the basic construction n out only the elementary principles. The and the d e ~ c n " ~ t i osets modem accumulator uses 'pasted' plates to allow the maximum use of the available material and the process involved in 'forming' the cell is somewhat involved for setting out in a book of basic theory. It is hoped however, that the information given below will provide sufficient knowledge to enable the action of . accumulator-to..be understood. Q e- -nickel-iron, -.... the lead-acid nickel-cadmium or alkaline battery also funcGons_.on s h i l a r p r i n c i p l ~ ~ a l t h o u gthe h plate materials and electrolyte 'differ. ThZ-Se of cell is also important and should be thoroughly investigated. ,-

1. CHASCE

2, DISCHAhCE

=

ot-

A

CHARGING SUPPLY

I-

!

-- -- -- Fig 32 The simple accumulator consists of' two lead (Pb) plates Immersed in dilute sulphuric a d d , the whole assembly being contained in a glass or moulded ebonite container. The cell has

to be worked into a suitable condition before it can be used for storing electricity and the process is carried out by alternatively 'cliar_eing' and then 'discharging' the cell. If a d.c. supply is connected to the plates as in the diagram (Fig 32, switch posltion 1 ) and the cell is subjected to electrolysis by passing current through it, oxygen and hydrogen gas is given off at the electrodes. As for the water voltameter, the first stage of the reaction would be decomposition of the acid (H2S04). A molecule of acid dissociates to produce hydrogen ions and sulphions (SO,). The sulphions move to the +ve plate, reacting with the water to li)rin sulphuric c~cid: ~ n doxygen Tllc Ik~ttcrntti\cks (lie -t- ve plate only to form lead-peroxide (PbO,), which causes the original Ic;id clcctrqdc to :Issume :r dark brown colour. The hydrogen 3s io115 I S discharged at the cathode and liberated in the gaseous state. The first chemical action is thus at the +ve plate only but if the supply is switched off (switch-intermediate position), the cell is now found to have the properties of a voltaic cell and will provide an electromotive force. If the cell is next short-circuited (switch-position 2), it will behave !ike a primary cell, passing a current for a short time during which period it discharges. The solution is electrolysed in the reversed direction and the original negative plate now acts as the anhde with its lead ions reacting with the sulphions of the electrolyte to form lead sulphate. Thus at - ve plate Pb SO, = PbSO, (lead sulphate). The hydrogen ions from the electrolysis during discharge, move to the original +ve plate, now the cathode. ~ h c h y d r o g e nnow reduces the lead peroxide to lead oxide which in turn reacts with the acid to form lead sulphate. Hz + Thus at the +ve plate the chemical action is PbO, H 2 S 0 , = PbSO, 2 H 2 0 . Both plates are converted into lead sulphate and assume a whitish colour. If the charging cycle is repeated (switch-position I ) , the direction of' current flow in the electrolyte is again reversed and tht. PhSO, on the - t v e plate becomes lead peroxide (PbO,). 'I'his i s a coti~plcxresult of' the electrolysis of thc acid. Sulpliions move to the +ve plate, react with the water to form H 2 S 0 4 and o a y g c ~'l'llc ~ . I;~ttc~. ; ~ t t ; ~ cthc k s +vc plate to lbrln I'bO, and morc sulphur~c acid. The chemical action at the f v e plate is SO, H 2 0 = HzS04 0 and 0 + H 2 0 PbSO, = I'bO, -t H,SO,. At the negative platc, lead is produced by the hydrogen ions liberated by the acid decomposition, moving to this clrctrorlr :lnd reducing tllc Ic:d sulphntc to spongy lead. Tlic t.licrnic.:~l:~ctinn; I I thc vc pl:~tci s : I'hSO, + H 2 =. H Z S 0 4 + Ph.

+

+

+

+

+

+

81

EL EC TRO C HEMIS TRY

After a number of cycles of charging and discharging the platcs bccomc porous und the ci~p:\cityo f the cell is increased. When a cell is fully charged, chemical conversions are colnpleted and hydrogen is freely given off, resulting in 'gassing'the accepted term indicating a full charge. The lead-acid accumulator in its practical form is provided with 'pasted' plates Here the active material is app!ied to plates in the form of a paste, the back-bone of the plate being a leadantimony grid. One 'forming' charge converts the paste into lead peroxide on the +ve plate and spongy lead on the -ve plate. Irrespective of the method of production the charge and discharge action can be summarised by the following chemical equation. Discharge ---------+ t Charge + ve Pole - ve Pole + ve Pole - ve Pole Lead Sulphuric Lead Lead Water Lead Peroxide Acid Sulphate * Sulphate PbO, + 2 ~ ~ 3 0 , + Pb = PbSO, 2H20 PbSO, It will be seen that during discharge water is formed, thus diluting and reducing the specific gravity of the electrolyte. During charge, acid is formed and the tests to check a fully charged cell include : (1) S.G. of cell charged (1.20 to 1.27); discharged (1.17 to I

+

+

1.18).

(2) ~ o l t i & eon open-circuit, cl?urged 252V txr cell or higher. (3) 'Gassing' on charge. (4) Positive Plate-rich dark brown colour. Negative Plateslate grey. CAPA C ITY OF A CELL. This is the ampere hour figure it can yield on a single discharge, until the e.m.f. falls to about 1.8V per cell (for the lead-acid cell). Generally the capacity is based on a 10 hour rate of discharge, since it decreases as the rate of discharge increases. Research has shown that the performance of a cell can also be improved by working the +ve plate at a higher current density than the negative. This is achieved by keeping the plates of equal area for convenience, and providing an extra -ve plate, ie always making the outside plites negative.

be ex ~ressedin terms of ( I ) the Ampere hour input and output, (2) the Watt hour input and o u t ~ u tThus: . Ampc re hours of discharge Ampere hour Efficiency = Am >erehours of charge

EFFICIENCY OF A CELL. T ~ Imay S

.

The ampere hour efficiency neglects the varying voltages during charge and discharge. Since this is important, we thus can have an energy efficiency compared to a quantity efficiency, and Watt hours of discharge Watt hour Efficiency = Watt hours of charge Average Discharge Volts x Amperes x Average Charge Volts x Amperes x Hours

ours

Example 36. A battery is charged with a constant current of 16 amperes for I I hours after which time it is considered to be fully charged, its voltage per, cell being recorded as 2.2V. Find ~ t amperc s hour efficiency if it is (1) discharged at a rate of 16 amperes for I0 hours, and (2) 28 amperes for 4 hours. In either C:I';C d~sclit~ rgc w:is d i s c o l l l i ~ l ~ r cwllcn d thc vol tagc per cell fell to 1.8V.

(1) Ampere hour Input = 16 x 1 1 = 176 Ampere hour output = 160 160 .: Efficiency = - = 0.91 o r 91 per cent. 176 (2) Ampere hour input = 16 x 11 = 176 Ampere hour output = 28 x 4 = 112 :. Efficiency = 112 -= 0.63 o r 63 per cent. 176 Example 37. A 12V accumulator is charged by means of a constant current of 16A passed for 11 hours. The p.d. during charging varies as shown. The battery is then discharged a t a constant current-of 16A for 10 hours, the p.d. again varying as shown. Calculate the Watt hour Efficiency of the battery. Start

Reading No 1 2 3

4 5 6 7 H

9 10 II

Time

Charge 10.8V 1I .ov 1 l.5V 1 1.8V

12-ov 12.2v 12.4V 12.6V 12.8V 13.0V 13.1V

Discharge 12.6V 12.4V 12.2v 12.0v 1 1.8V 1 1.6V 1 1.4v 1 1.2v 11.ov 10.9V 10.8V

83

ELECTROCHEMISTRY

Efficiency =

1162

l 6 lo = 12.20 x 16 x 11 1342

_

0.867 or 86.7%.

CHARGING PROCEDURE. British practice uses the 'constant current' method. This is also American practice but on the Continent the 'constant voltage' method is favoured. For this latter method the charging supply voltage is kept constant and is substantially higher than the battery e.m.f. for the discharged condition. The charging current is high initially but falls as the back e.m.f. of the battery rises. Thls method gives a lower charging time than the 'constant current' method, but due to the violent chemical action and heat generated in the battery there is danger of 'buckling' the plates, unless the battery is specially constructed. For the 'constant current' method arrangements must be provided for increasing the 'voltage applied to the battery as charging proceeds and the back e.m.f. rises. If a generator is used and Z is the charging current, R , the internal resistance of the battery and E, the battery e.m.f. at start of charge, then the applied voltage must be V = E b+ IR,. . . start of charge (1). If E,, is the battery e.m.f. at endof charge, the applied voltage would have to be V, = Eb,+ IR,. . . end of charge (2). Thus subtracting (1) from (2) vuriittion of voltage would bc V , - V = Ebl- Ebor thc applied voltage must be increased by an amount equal to the rise of the battery e.m.f. If a constant supply voltage is used for charging, then a variable resistor is required to obtain the neqssary current control, and its value will be reduced as charging proceeds. Let V be the supply voltage, Z the charging current, Ri the internal resistance of the battery, E, the battery e.m.f. at start of charge and R the control resistor. Then: V = E b + ZR,+ ZR . . . start of charge (1) If Eb,is the battery e.m.f, at end of charge and R , the new value of the control resistor. Then : V = Ebl ZRi+- JR, . . . end of charge (2) Subtracting (f) from (2) 0 = E,, - Eb + IR, - IR or (E,,- E,) = I(R - R,).Thus the control resistance must be reduced from R to R, as the battery voltage rises from Eb to Ebl. Example 38. A 24V enlergency battery IS to be charged from the 110V ship's mains v hen the e.m.f. per cell has fallen to a C

+

minimum value of 1.8V. The battery consists of 12 cells in series, has a capacity of lOOA h at a 1Qh rate and the internal resistance is 0.03R/celI. If charging continues until the voltage/cell rises to 2,2V, find the value of the variable resistor needed to control the charging. The charging current can be assumed to be equal to the maximum allowable discharge current. Discharge current =

10

= 10 amperes = charging current

At start of charge, battery voltage = 12 x 1.8 = 21.6V Battery internal resistance = 12 x 0.03 = 0.36Q Then 110 = 21.6

+ (10 x 0.36) -t (10 x

R)

At end of charge, battery voltage = 12 x 2.2 = 26.4V Then 110 = 26.4 + (10 x 0.36) + (10 x R,) or R1,= 110- 30 - 80 = 8 ohms. 10 10 Thus the variable resistor should have a value of 8.48 ohms and be capable of being reduced to 8 ohms. In practice a unit of 9 ohms would be used which would be reduced by adjusting the sliding contact un ti1 the charging ammeter recorded the correct current. Further adjustments would be made periodically as charging proceeds. It is important to note that besides the ohmic value of the resistor, the wattage rating must be specified. For the unit in the example, a rating of 1 2 =~ lo2 x 9 = 900 watts is required. The control resistor must be capable of dissipating up to this power as heat during the charging, although this waste of power will decrease slightly as charging proceeds. For example at the end of the charge the power wasted in the resistor would be lo2 x 8 = 800 watts. The most important point to stress is the correct connecting up of thc butrcry for clltrrging, io vc tcrminal of battery to vc of mains; -ve terminal of battery to -ve of mains. It is .;urpristng how tnc~ny1i111cslllis olclllcntctry rcquirctnenl is overlooked through carelessness. For incorrect cgnnection, no control of the current would be possible with the equipment provided and damage of the ammeter, control resistor, o r battery could result.

+

+

ELECTROCHEMISTRY

85

CHAPTER 4 PRACTICE EXAMPLES

!

1.

An accumulator is charged at the rate of 6 amperes for 18 hours and then discharged at the rate of 3.5 amperes for 28 hours. Find the ampere hour efficiency.

2.

The mass of the cathode of a copper voltameter before deposit was 14*52g,and after a steady current was passed through the circuit for 50rnin, its mass was 19.34g. The reading of the ammeter was 5.1A. Find the error of the ammeter, taking the E.C.E. of copper as 330 x 10-?kg/C.

3.

, A 90V d.c. generator is used to charge a battery of 40 cells in series, each cell having an average e.m.f. of 1.9V and an internal resistance of 0.0025R. If the fotal resistance o f the connecting leads is l R , calculate the value of the charging current.

4.

Nicke1.S~to be deposited on the curved surface of a shaft lOOmm in diameter and of length 150rnm. The thickness of deposit is to be 0.5mm. If the process takes 8h, calculate the current that must flow. The E.C.E. of nickel is 302 x 10-9kg/C. The density of nickel is 8600kg/m3.

5.

A nickel-alkaline battery is discharged at a constant current of 6A for 12h at an average terminal voltage of 1.2V. A charging current of 4A for 22h, at an average terminal voltage of 1.5V is required to re-charge the battery completely. Calculate the ampere hour and watt hour efficiences.

6.

A battery of 80 lead-acid cells in series is to be charged at a constant rate of 5A from a 230V, d.c. supply. If the voltage per cell varies from 1.8 to 2.4V during the charge, calculate the maximum and minimum values of the required control resistor. If the ampere hour capacity of the cells is 60, state the probable charging time required, assuming that the cells were in a completely discharged condition at the commencement of the charge.

7.

A metal plate measlrsring:5@nrn ,by 150rnm is to be copperplated in 30min. Calculate the current required to deposit a

thickness of 0.05mm on each side (ignore the edges). The E.C.E. of copper is 330 x 10-gkg/C and its density is 8800 kg/m3.

8.

A battery of 40 cells in series delivers a constant discharging current of 4A for 40h, the average p.d. per cell being 1.93V during the process. The battery is then completely recharged by a current of 8A flowing for 24h, the average p.d, per cell being 2.2V. Calculate the ampere hour and the watt, hour efficiencies for the battery.

9

Thirty lead-acid accumulators are to be charged at a constant current of 10A,.from a 200V d.c. supply, the e.m.f. per cell at the beginning and end of charge being 1.85Vand 2.2V respectively. Calculate the values of the necessary external resistor required at the beginning and end of charge, assuming the resistance of the leads, connections, etc to be 1R and that thz internal resistance is 0.01R per cell.

10.

When a current of 3.5A was passed through a solution of copper sulphate, 4.2g of copper were deposited. If the E.C.E. of copper is 330 x 1 0 - ~ k g / Cand the chemical equivalent of copper is 31-8, find the time for which the current was passed through the solution and also the mass of hydrogen liberated.

CHAPTER 5

MAGNETISM ELECTROMAGNETISM NATURAL MAGNETS

From very early times it was known to ancient civilisations, such as those of the Greeks and Chinese, that pieces of certain types of iron ore have magnetic properties. Pieces of the ore were known, not only to be capable of attracting and repelling other such pieces but could also pass on this property of magnetism. One further known fundamental property of a piece of the ore, called Magnetite or Lodestone, was that if it is freely suspended, as shown in the diagram (Fig 33), then it would come to rest in an approximate geographic North-South direction. The end pointing north is called a north-seeking or simply a North Pole, whilst the other end is a South Pole. The piece of ore constitutes a natural magnet and if brought into contact with a quantity of iron filings, these would be found to adhere mainly to its ends or poles.

Fig 33 Further simple investigations made with pieces of the magnetic ore would show that, if two such magnets are each suspended as described above and their polarities are determined and marked, then when the N pole of one suspended magnet is brought near the N pole of the other suspended magnet, repulsion of the poles will result. Two S poles brought near each other would behave

11

I

in a similar manner whereas, a S pole brought near the N pole of the other magnet will produce an attractive effect. Thus every magnet is seen to have two poles of unlike polarity and that like poles repel whereas unlike poles attract.

!

ARTIFICIAL MAGNETS

!

A piece of iron can be converted into a magnet and made to exhibit properties similar to that of the iron ore described above. Such a piece of iron is an artificial magnet and is said to be rnagnetised. A simple method of magnetising a specimen is by stroking it in one direction from end to end with one pole of an existing magnet, but the most effective method is by electromagnetism, which will be considered later in the chapter. Ccrltlin mi~tcrii~ls such us copper, aluminium, Icad, brass, wood, glass, rubber, etc cannot be magnetised. Thus all known materials can be classified under the heading of magnetic o r nonmagnetic substances. Some metals such as nickel, cobalt and magnesium exhibit very slight magnetic properties, but it is of' interest to record that, when alloyed with iron very strong magnetic properties result. An artificial magnet is usually made in bar o r horse-shoe form. When tested, the tips are found to constitute poles of opposite pojarity andjif suspended, a bar magnet will lie on an approxima,re N-S line. The magnetic compass makes use ofthis principle and ,consists of a short highly magnetised bar magnet which is pivoted at its centre. A card, calibrated in degrees and/or geogrdphic points, is mounted below and is used with the magnet to obtain a 'bearing'. It is necessary to mention here that the N-S direction as indicated by such a compass is not exactly geographic N and S. The angle between the lines of magnetic and geographic N- S, is called the 'variation' and varies for different parts of the world. If' the magnetic compass is being used, due allowance must be made for the variation, bef'ore a map can be truly oriented and used correctly. Ikl;rrr. o~*r.ccli~lg will\ I'UI.IIICI' s[udy 01' magnetism, i t would be as well to explain why a compass needle lies in the N-S dircction. Thc c;lrth itscll' hcl~;~vcs ;IS tl~oi~gli i l conl;lins ;I rnilgncl having 11sS pole in the region ol'the geographic north and its N pole near the geographic south. A compass needle placed on the earth's surface will lie so that its N pole will be attracted to the magnetic south (geographic north) pole of' the earth and its S pole will be attracted lo the magnetic north (geographic south). I:u~.ll~cs ~ ~ i e ~ i t01'i oI I~I C l ~ ; ~ l . t l l ' 111;1gnclisrn ~v will bc m;~dclater o n 111 t l i i x c~Ii;~ptcr

,

I I

89

M A G N E T I S M , ELEC T RO MAG N ETIS M

Summarising the facts deduced so far about natural or artiticial magnets, we know that every lnugncl has two poles 01' unlike polarity and that like poles repel whereas unlike poles attract. THE MAGNETIC FIELD

This is the space around a magnet where its magnetic effects can be felt. If a bar magnet is covered by a sheet of paper and iron filings are sprinkled on the paper, then on tapping the latter, the filings would be seen to align themselves as shown in the diagram (Fig 34). The filings would form a pattern which, if examined closely, would show that lines could be traced from the N pole of the magnet to the S pole through the space outside and from the S to N poles inside the mamet

L

-2 PATHS Of LlNtS Of FLUX 7RACkD OUT 81 IRON F I L I N G S

Fig 34 The field can also be plotted by using a small compass needle as shown in the diagram (Fig 35).

Fig 3 i

Field plotting with the aid of a compass needle is undertaken as follows. Place the magnet on a sheet of' paper and draw its outline. Set the compass needle against the N pole of the magnet and, with a pencil, mark a dot at the point in line with and adjacent to the N pole of the compass needle. Move the compass until the S pole of the needle is coincident with the original dot. Mark the new point in line with and adjacent to the N pole of' the needle. Repeat this procedure until the S pole of the magnet is reached. Join the dots together to give a 'line of force' or a 'line of J r t x ' which can be described as the line which, when drawn tllrougli any point in a magnetic field, shows the direction of' the ~ilagneticforce at that point. Using the compass needle the field can be mapped out for a considerable distance around :I IllilgllCt i1tlC1 t l ~ cI ' O I I U W I deductions II~ can bc made. 1 . Lines of' flux never cross. 2. The lines are always continuous. If various magnetic field arrangements are plotted as shown in the diagram (Fig 36) then other conclusions can be deduced. 3. Lines of flux are like stretched elastic threads and tend to shorten themselves. This explains the attractive effect between two unlike magnetic poles, which if fiee to d o so will move into contact, thereby reducing the length of the lines of flux.

MAGNETISM, ELECTKOMAGNETISM

91

4. Lines of' flux which are parallel ; ~ n din the same directton\ repel each other. 'This deduction is clearly seen tor the condition where two magnets are brought together, with like poles adjacent to each other. There would be a force of repulsion between the magnets and if the field is plotted between two like poles a neutral point would be found where the effects of the two repulsion forces balance each other and the total effect is as shown by the absence of control on a compass needle placed at this neutral point. The strength of the magnetic field around a magnet will vary from point to point,. but before this can' be measured and methods devised for making such measurements, a system of magnetic units and terms must be introduced. Faraday conceived the idea of the line of flux, as already introduced, and further suggested the use of these lines to depict the strength of the magnetic field. If a unit area at right angles to the lines of flux is considered then further definitions and terms can be introduced. A number of lines of flux collectively are said to constitute the magnetic Flux (symbol @-Greek letter phi) which is passing through the area being considered. Another unit of importance is Flux Density-and the value, at any point, is obtained from the expression: Flux Flux Density = -Area The diagram (Fig 37) illustrates the SI unit of flux or the Weber. Thus if' 50 lines of flux are shown passing through the p

area of 1 square metre, then for the plane being considered, the magnetic flux is 50 Webers. The symbol for flux density is B and the unit is the Tesla. Thus for any point P in the plane being considered, the flux density is 50 teslas. Note. The tesla is a new name introduced for the SI system. The original unit was the weber per square metre ie Wb/m 2 . We now have Flux = Flux Density x Area or @ (Webers) = B (teslas) x A. (square metres). The above relationship will be used continually when the study of electromagnetism and magnetic circuits is made and should be considered a basic and important formula. It is well to stress here the obvious, namely that lines of flux d o not exist but the properties of magnets and magnetic fields can best be assessed by assuming their existence and their having definite physical properties. It should also be remembered that the field of a magnet exists in all directions and is not confined to one plane. A fuller understanding of magnetic theory is rewarding to the engineer, but there is little space here for a more complete treatment of the subject. The basics however are summarjsed a s follows. MOLECULAR THEORY OF MAGNETISM

A molecule is defined as the smallest particle of a substance that can exist separately and in any magnetic material every molecule is thought to be a complete magnet. In a piece of un~llagnet~sed pagnetic material the molecules are considered to arrange themselves in closed magnetic chains o r circuits as shown in the diagram (Fig 38). Under this assumption i t js considered that each mo.lecular magnet is neutralised by adjacent

molecul;ir m:ignets so that no magnetism is apparent in the material. The process of magnetising the material is considered to be ;~cli~cvcd by simply arranging the molecular magnets so that their axes point in the direction of' the magnetising f'orcc. 1 ' 1 1 ~~ ~ ~ ~ o ~ ~I \l\ ' ~o ~l ' ~~ l ~li silYsc~ IsI ~~ I ~N ~ Iby~ ( t~l ~C e~lY)llo\+~i~~g know11 ohscrv;~lion\ ( I ) Tllcre i.; :I limit to the mount 01' magnetism Illat C ; I I I I>c I I I I ~ ~ I i tI1 . any ~ ~ ~orlc s;~mplcol' matcl.ial. l'I11s I S

MAGNETISM, ELECTROMAGNETISM

93

explained by the supposition that, once all the molecules had been 'lined up', no amount of extru mt~ynctising force ctln increase the strength of the magnet. (2) When a magnet is broken, the ends of' the moleculur mugnets are exposed und the broken pieces are found to be magnets themselves. (3) If heated to about 100°C and allowed to cool a magnet is weakened. If the.magnet is heated to red heat, the magnetic properties are lost altogether. Similarly if a magnetic material like hard steel, is cooled in a strong magnetic field then it will set as $ permanent magnet. It is considered that during heating, energy is transferred to the magnet which causes oscillations of the molecular magnets which tend to break the 'lining up' and results in these magnets taking up random directions. Similarly for the cooling process, as energy is passed from the hot material, the oscillations decrease in magnitude and violence and the molecular magnets are allowed to settle in the direction of the magnetising field. A more modern.theory of magnetism is b&ed on the electron theory and the conception of the atom. A chapter considering the electron theory in detail is introduced at a later stage but here it can be stated that an electron, the smallest known 've charge, when rotating in an elliptical path, constitutes a circular current which sets up a magnetic force along the axis of gyration. 'In a molecule the magnetic effects of the electrons of the atoms may neutralise each other giving little resultant effect. Again a spinning electron also sets u p a magnetic field along its axis of' spin. If the fields due to the etyects of spin balance out, due to electrons spillning in opposite directions, then the material is non-magnetic. A magnetic material is the result of' the fields not balancing out, but to explain the overall apparent effect, i t is thought that rather than single atoms or molecules being concerned, it -is a group of molecules which act together. Such a group is called a 'domain' and is considered to function like the more elementary molecular magnet already described. ELECTROMAGNETISM Earlier theory has referred to an association between magnetism and electricity and this was more specifically mentioned in Chapter 2 when the electrical units were defined. The discovery of a relation between an electric current and magnetism was made in 1820 by the scientist Oersted, when he accidentally noted that a wire arranged above and parallel to a compass needle, caused deflection of the latter when a current was passed through the wire. Reversal of the current caused a reversal of the deflection. Further experiments on the shape, direction and strength of the

MAGNETISM, ELECTROMAGNETISM

95

FIELD

Fig 40 mum on the circumftrence. Outside the wire the flux density varies inversely as the distance from it. The diagrams (Figs 40a and 40b). make use of: the conventional method of' indicating current direction. Consider an arrow ie current entering tlic surfiice of' the paper and receding I'rom tlic vicwcr, thcn tllc I'c;~tllct~ctl crltl w o t ~ l ~hc l 4ccn '1'I11\ would be shown with a cross. Similarly current flow low;il.db thc viewer would be shown with the tip ot' the iirrow it. a point 01. dot. The relation between the direction of' the lines of' f u x a n d the current is best summarised by Maxwell's 'Right-Hand Screw' Rule. This depicts the association that, if current flows in the direction in which a right-handed screw moves forward when turned clockwise, then the resulting field yill be in the direction of' turning the screw. I f the current is reversed, the screw should be unscrewed and the field would be reversed, o r would be in the direction of'turning the screw ie anti-clockwise.

(2)

FIELD D U E TO A C U R R E N T - C A R R ) ' I N G C O N D U C T O R BENT T O FORM A SINGLE L O O P

The diagrams (Figs 41a and 31b) show the loop,, tl3e current and the lines of flux which tend to encircle the conductor a s deduced from condition, ( I ) above. The resulting field can be plotted by locating the loop in a sheet of' cardboard as shown. The result can be considered as the field taken thrbugli the section XY of the loop and the similnl-ity wit11 the field of a s1iol.t

94

REED'S BASIC ELECTROTECHNOLOGY

magnetic fields associated with current-carrying conductors arranged in the form of loops and solenoids were the subject of much work by famous scientists such as Faraday, Maxwell and Gilbert. The lesults of the discoveries made then led to the deduction.of certain fundamental relationships which are now part of accepted basic theory. The shape of the magnetic fields due to simple arrangements of current-carrying conductors will now be considered.

( 1 ) FIELD DUE TO LONG STRAIGHT CURRENT-CARRYING CONDUCTOR The field associated with such a conductor may be determined by the use of' iron filings o r a compass needle as was described earlier in the section on magnetism-Figs 34 and 35. Assuming tliat thc currcnt is kept constant during such a test, a field consisting of concentric lines of' flux would be confirmed. The diagram (Fig 39) shows a vertical wire passing through a sheet of cardboard. The directions of' the current and lines should be particularly noted since this is fundamental knowledge.

A

CURRENT

Fig 39 Fk~rtlicrtests would s h o w that if' thc current is reversed, thc liclcl wuuld s c v c i x and I ( ~ l r chtrc~lg[ll01' ~ h cliolcl was ~ucasu~.ed

b) an appropriate sensitive instrument, then consideration of' the results would give a graph as illustrated by the diagram (Fig 40a), which shows 1-lux Density ( B ) plottcd to n base ofdistance ( A ) I'rnm thc centre 01' thc conductor. I t will be seen that inside the circular conductor, the strength 01' licld 01. Ilus dcnsi[> vuric.; I'~.omzc1.o at thc centre to ;I maxi-

, bar magnet will'be recognised. Thus the loop can be considered to set up a magnetic polarity which can be determined from first principles.

AXIS

U U t

4 f

*

CURRENT AWAY FROM OBSERVER

(b) Fig 41 .L

(3) FIELD

DUE TO A CURRENT-CARRYING CONDUCTOR WOUND AS A

SOLENOID

The next logical step in electromagnetic field investigations is !'or a coil of wire, which is basically a collection of' several loops. A solenoid is a form of a multi-turn coil where the axial length is much greater than its diameter. The turns of wire can be wound in an open spiral o r placed close together so that they totlcll, pl.oviJcJ i n s i ~ I ; ~ ~wire c d is used. The insu1;ition most commonly used is either one of the modern synthetic enamels o r ;I f i h r n ~ ~ tn:ltcri:ll \ s\~clias cotton o r silk in the f'orm o f thrc;ld. tape or braid. 'fhe turns ol'a solenoid may be arranged in several layers provided the current travels through the turns in the same direction. When the field is investigated by plotting with a compass, i t is found to be as shown in the diagram (Fig 42). I r will be seen th;~t; \ I 1 thc turns tend to produce ;\ mngnctic field i n tlie same direction. so that this can be deduced by considering tllc liclcl 01' ;I .sir~glclu1.11or lool>. TI1c t 1 l r . n ~ unitc lo send ;I

MAGNETISM, ELECTROMAGNETISM

97

straight field up the centre which comes out at the ends, opens and spreads out to return to the othcr cnd, giving the surnc distribution of lines of flux as would be obtained from a bar magnet.

J

L_

Y

-3

U

9 =,l,,oh~A'? -

t

attributed to the solenoid when carrying current. The polarity can be detemined by finding the direction of the lines for any one turn by applying the RightHand Screw Rule but additional aids are useful, the easiest of which being the Right-Hand Rule. This is explained as follows, and is shown in the diagram (Fig 43). Place the right hand on the coil with the fingers pointing in the direction in which current flows. Then the thumb will point in the direction of the N pole.

N

RIGHT HAND

Fig 43 INTRODUCTION OF AN IKON COKE

By iron is meant, at this stage, a magnetic substance. The iron, if made the core of a solenoid, strengthens the field by concentrating the flux and more clearly defining the poles. A magnetic core appears to allow the passage of flux more readily than does air. The reason for this will be introduced in later studies, but experiment shows that the most perfect type of flux path is where the whole of the magnetic circuit is formed from magnetic material. Where this is not practicable the air gaps o r

air paths are kept as short as possible and good examples are found in the electromagnetic paths for the flux in the electric bell and the electric motor or generator, as illustrated by the diagram (Figs 44a and 44b).

.Electromagnets are preferred to permanent magnets in industry f'or two main reasons. (1) They can be made more powerf'ul $anp_e_rmanent magnet% by providing the desired m m s i n g force, ir solenoid coils with sufficient turns and energis~ngcurrent. (2) The m x n e t i s m can be~controlled,ie i t C1n77c s\\;itil~cdon and off' or varied gradually by controlling the current. Altllough mucll more will be said about the material u w d for the core of' electromagnets. general practice can be ~ ~ ~ l l l l l l ll\L!~l .ll

~ l l l l ,\ ~

Permanent magnets are made of llurd s t e e l because tI11s nCl'C ~ ( > l l l d 2 X i0-' newtons. T h ~ is s known I'rom the definition of'the ampere. Also ( b ) the force on a metre length of this conductor is given by R I I newtons. o r is B (teslas) x l (ampere) x l (metre) = B (newtons). INIS IS equating ckpl.cssions ( ; I ) i ~ n d(h) f'or the t'orce on the condi~ctorwe sec t1ie:t the value o f ' B fbr the condition considered woulcl be 2 x I0 ' I I C W I O I I S i111d

r.

-

107

MAGNETISM, ELECTROMAGNETISM

Flux density at point P-

-B 4

MagnetisingforceatpointP

H

-.

~ -X I O - '

1 /2n= 4x x 10-'

SI units

of free space-symbol I(, Ampere-turnsirnetre for air = Flux Density 4x X lo-' Example 41. It is required to produce a flux of 0.018Wbacross an airgap 2.54mm long and having an effective area of 24 x 10square metres. Find the ampere-turns required. Area of Gap = 24 x

square metres -I a = 0-75T Required flux density B = 0.018 24 x 24,

The length of the air gap = 2.54mm = 0.254 x 1 0 - ~ m So total ampere-turns needed = 59.7 x lo4 x 0.254 x = 59.7 x 0.254 x lo2 = 15.15 x lo2 = 1515At. Example 42. A wooden ring having a mean diameter of 200 mm nnd u cross-scctionul urea o i 400mmJ is wound uniformly with a coil of 300 turns. If the current passed through the coil is 5A calculate the value of flux produced in the coil. The m.m.f. of the coil = 5 x 300 = 1500At The mean circumference = ~c x 200 = 628mm = 0-628m 1500 - 15 x lo2 The magnetising force H = Atlm = -0.628 0.628 The flux density B = p o H = 4 x x x lo-' x 2380 = 1.256 x 2.38 x l o p 3 = 0.003T Total flux @ = BA = 1.2 x 1 0 - W b = 0.003 x 400 x = 1.2pWb. Example 43. The magnet system of a moving-coil instrument provides a flux density in the air gap of 0.25T. The moving coil, of 120 turns, is carried on a former of (active side) length 25mm and width 18mm (between air-gap centres). If the coil carries a current of 2mA, calculate the turning moment on it.

F = BIl newtons = 0.25 x 2 x x 120 x 2 x 25 x = 0.25 x 2 x 2 x x 120 x 25 x = 30.0 x = 3 x 10-3N Torque = F x radius of' coil =3 10-3 9 10-3 = 27 x l o v 6 newton metres = 27pN m. The subject matter covered in this chapter has dealt with sufficient basic theory, terms and relationships to allow the general study of Electrotechnology to proceed. Regarding the magnetic circuit, u p to riow, only nir o r non-magnetic material has been considered for the medium of the magnetic field i ~ s s o c i i ~ l ~witti c l nn C ~ ~ C I ~ O I I I The ~ I ~ nccd ~ C ~ .for introducing il magnetic material into thc circuit to providc a well defined flux path, will require extensions of' our investigations into the magnetic circuit, which can be made more conveniently at a later stage. Chapter 12 will be found to cover the more chmposite magnetic circuit and the further aspects of ferromagnetism.

M A G N E T I S M . E L E C T R O M A G N E TIS M

109

PRACTICE EXAMPLES 1.

A conductor carrying a current of 100A is situated in and lying a t right angles to a magnetic field having a flux density of 0.25T. Calculate the force in newtonslmetre length exerted on the conductor.

2.

A coil of 250 turns is wound uniformly over a wooden ring of mean circumference 500mm and uniform crosssectional area of 400mm2. If the current passed through the coil is 4A find (a) the magnetising force (b) the total flux.

3.

A current of 1A is passed through a solenoid coil, wound with 3200 turns of wire. If the dimensions o f t h e air core are length 800mm, diameter 20mm, find the value of the flux produced inside the coil.

4.

Two long parallel busbars, each carry 2000A and are spaced 0.8m apart between centres. Calculate the force1 metre acting on the conductors.

5

A moving-coil pcrmi~ncnl rn;ignct inslrl~rncnt I l ; ~ s ;I resistance 01' IOR and the flux density in [lie gap is O ~ I ' I ' .' l ' h c coil has 100 turns of wire, is of mean width 30mm and the axial length of'the magnetic field is 25mm. Il'a p.d. of5OmV is required for full-scale deflection, calculate the controlling torque exerted by the spring.

6.

An air gap of length 3mm is cut in the iron magnetic circuit of a measuring device. If a flux of 0.05Wb is required In the air gap, which has an area of 650mm 2, find the ampere-turns required for the air gap to produce the necessary flux.

7.

A straight horizontal wire carries a steady current of 150A and is situated in a uniform magnetic field of 0.6T acting vertically downwards. Determine the magnitude of' the force per metre length and the direction in which it acts.

8.

An armature conductor has an effective length of400mm and carries a current of 25A. Assuming that the average flux

density in the air-gap under the poles is 0.5T, calculate the force in newtons exerted on the conductor.

9.

10.

In an electric motor the armature has 800 conductors each carrying a current of 8A. The average flux density of the magnetic field is 0.6T. The armature core has an effective length of 250mm and all conductors may be taken as lying on an effective diameter of 200mm. Determine the torque and mechanical power developed when the armature is rccolving nt 1 0 0 revlrn~n. Two long straight parallel busbars have their centres 25mm apart. IS each cnrrics a current of 250A. calculate the mutual fbrce/metre run.

I

1 I

CHAf''1'EK 6

ELECTROMAGNETIC INDUCTION

I

I

I I

1

,

j

U p to now, the only method considered' for producing an electromotive force has been that due to chemical action. In the progression of electrical knowledge, history shows that this also was the pattern followed and that electrochemistry was the first branch of the science which was developed and was to play its fhll part in subsequent electrical investigations of' the early nineteenth century. At that time electricity was a subject of interest to the scientist only and had not as yet been accepted as a medium which could be put to use for engineering processes. The -chemical cells as were then known, could not produce sufficient energy for practical purposes, nor kad any electromagnetic devices been evolved which could be put to engineering applications. As was mentioned in Chapter 5, it was only after the relation between current and magnetism was discove'red that attention was turned to various associated findings. These inevitably disclosed the related phenomena of electromagnetic induction and led to the development of machines which were subsequently to interest the engineer as means of producing either electrical o r mechanical energy. The initial electrotnagnctic ' i t r c l r ~ c - ~ i o t rinvcstiyntions i ~ r c attributed to Michael Faraday, who in 1821 showed that when the magnetic flux linked or associated with ;In electricnl circuit is changing, an e.m.f. is induced in the circuit. This e.m.f. is shown to last only whilst the change of flux is taking place and the faster the change the greater the e.m.f. The flux linked with a circuit, which invariably consists of a coil of insulated wire, may be changed in different yays. Thus: (a) A magnet could be moved in the vicinity of a coil of wire. This principle is used for the Alternator-a particular for12 of electrical generator. (b) A coil of wire could be moved in the vicinity of a magnet. This principle is used for the d.c. Dynamo o r Generatorthe more accepted modem term. (c) The flux could be changed by varying the cuirent in the energising coil of wire. The ampere-turns are thus varied and the flux produced varies accordingly. This principle is made use of in the operation o f t h e Transformer o r the Spark-coil of a petrol-engine ignition system.

For the above three ways of e.m.f. generation, it will be seen that cases (a) and' (b) involve relative physical movement between the magnet and the coil. Case (c) however, involves no such movement and the generated e.m.f. is achieved in a stationary coil with which the associated flux only is changing. Thus there are two distinct forms of e.m.f, generation o r inductionto use the alternative term. These forms are referred to under two basic headings: ( 1 ) Dynamic Induction, (2) Static Induction. Before these two methods are considered in detail it would be well to give some attention to >he meaning of an expression wllich will bc uscd frequently, namely Iflux-IitlX-trgc.~'. Earlier stcdies on magnetism -showed that the ficld of' ;I magnet can be represented by lines of' flux emanating f'rom the poles. The strength of the flux can be represented by the number of lines and is measured in webeis, while the flux density is measured in teslas. The flux lines make complete loops and the associated conductor or coil of wire in-which the e.m.f. is induced can also be considered to consist of' a number of turns. Since the rfumber of lines of' flux associated with the turns are ref'erred to as flux-linkages, then a magnet with poles of flux strength 3.4j1Wb associated or linking with a coil of 500 turns is said to result in a condition of 3.4 x x 500 = 04)017 weber-turns. The diagram (Fig 49) shows the basic idea of' the explanation. F L U X - LI N K A G E S .

CURRENT

F'AKADA) 'S L A b ' O F ELECTROMAGNETIC INDUCTION

This summarises the known relationship deduced for the generation of e.m.f. by electromagnetic induction and can be stated as: the magnitude of the e.m.f. produced, whenever there is; ; I cli;~ngcof' flus linkctl with a circuit, is proportional to the I.:klc ol' cllar~gc01' I l u x ~ - l ~ t ~ h a g c ~ .

r

EL E C T R O M A G N ETI C I N D U C T I O N

113

LENZ'S L AW

This identifies a phenomenon always noted for the e.m.f, produced by induction. The law can be stated as:-the direction of' the current due to the induced e.m.f. will always set up an effect tending to oppose the change which is causing it. This can be further explained in terms of the magnetic field which wo'uld be set up by the current caused by the induced e.m.f. Thus if the flux-linkages tend to increase, then the field produced by the induced current. resulting from the induced C.m.f., will tend to oppose this effect, ie it would tend to oppose the flux-linkage build up. Similarly, if the flux-linkages tended to be reduced, as when the current in a coil is switched off, then the induced e.m.f, will induce a .current which, if allowed to flow, would tend to keep up the,flux-linkages to their original value. It is stressed that the action of the induced current does not succeed in preventing the change, but would try to do so during * the period when the change is occurring. Faraday's law is capable of being expressed in mathematical form and thus formulae can be deduced for both static and dynamic electromagnetic induction. These will be considered under each appropriate heading. STATIC INDUCTION The generation of an e.m.f. by static induction is considered first, because it can be briefly dealt with here and then left to a more appropriate stage in our studies. It is of purticular jntcrcst in connection with ~nductanceand the theory of the a.c. circuit. The study of dynamic induction is however, of immediate importance, to allow electrical machines to be introduced as soon as possible and will be given most of the attention in this and subsequent chapters. b.M.F. DUE TO STATIC INDUCTION

Consider the diagram (Fig 50) which shows two coils A and B of insulated wire. Coil A can be connected to a battery through a switch, whereas B is wound over or placed adjacent to coil A and is connected to a sensitive centre-zero type voltmeter. This type of instrument is used because, as the pointer is positioned at the centre of the scale, a deflection to the left or right depending on the polarity of the supply can be registered. At the instant of switching on the current in coil A, the flux can be imagined to grow outwards and to cut the turns of coil B. The initial growing is shown by the dotted flux lines becoming fuller until the final.condition (full lines) is reached. The cutting

/-I - -

Fig 50 of coil B by the flux of A, results in an e.m.f. being induced, its magnitude and direction being governed by Faraday's and Lenz's laws. The flux-linkages, ie flux linking with the turns ( N , ) of coil B increase and if the linking flux grows to a value of @ webers from its original zero value, then the rate of change of flux-linkages will be the flux-linkages divided by the time ( t , ) taken for them to grow, ie the time taken for the current to reachrits final value I amperes in coil A. Thus if the resistance of coil A is RA ohms and V,, is the applied voltage to coil A then

I=

., Y

amperes. RA flux-linkages - NB@volts. and the e.m.f. induced in coil B = time tl For this equation NB = turns of coil B, @ is the flux in webers linking with it and t , is thc time taken for the energising current to reach its final value I. It could be assumed that value I is reached immediately the switch is closed;'because the flow of electricity is considered to be instantaneous, but here we have a condition wherc thc current takes an nppreciable time to reach its full value--due to the inductance of the arrangement. This action will hc considcrcd in detail later, but to revert to the lnit~uleflkcts being observed; it will be seen that when the switch for the primary coil A is closed, the voltmeter pointer gives a 'kick', say to the left, showing an e.m.f. to be induced in coil B--A

y.

the secondary circuit. The value of e.m.f. EB =

!+!&and the volt11

meter will show the polarity of coil B to be such that the current, which flowcd through the instrument, was in such a direction

ELECTROMAGNETIC INDUCTION

115

through the coil as to set up a secondary flux, which would be opposite to tlic originul flux cP und wotlld try to stop i t growins, A further point of importance to note would be that, although a kick of the voltmeter pointer would be seen, yet it would return to the zero position even though current in coil A was allowed to flow indefinitely. Thus an e.m.f. is induced only during the time when the flux-linkages were changing. Further experiments with coil B would show that if the number of turns of wire were doubled, then the induced e.m.f. would be twice as large, even though the flux 0 of coil A was the same. The flux-linkages have been increased and the induced e.m.f. rises in proportion. Consider next the instant of switching off the current in coil A. The voltmeter will again be seen to kick-to the right this time, showing an induced e.m.f. of reversed polarity. The flow of current in coil B is such as to try and maintain the flux to its where

original value @ and again EB =

t 2 is

the time taken

17

for 'switching off. It can be noted here that t , need not necessarily equal t,. If the switch is opened quickly, the current of A will be interrupted very quickly and EBcan be larger at switching off than at switching on; when the rate of growth of the flux is controlled by the inductance and resistance of the circuit. Up to now we have only considered the induced e.m.f. in coil B and this is said to be due to Mutuul Induction, ie the mutual action of cail A on 0. Wc now turn our rlttcntion to Sc~lflttrlriction, icv the conditions appertaining to coil A itself. At the instant of' switching on, the flux grows outwards and in so doing, cuts the turns of coil A-the primary circuit. An e.m.f. is thus induced N 0 Here N A is the turns ofcoil A, 0 is the linked given by EA = -A. t,

flux and 1 , the t&e taken for the current to reach its full value. As before, the direction of the Self-induced-e.m.f. EA will be such as to cause a current to flow in the opposirgdirection through the battery and will produce a flux which will try to oppose the build-up of flux @. We can now see the reason for the opposition to the growth of current in coil A at the instant of switching on and why the current I takes some little time to reach its full value. As before when the switch is opened, the flux collapses and in doing so, again cuts the turns of coil A, inducing a voltage of reversed polarity, which tries to keep the current flowing. Appreciable arcing will be observed at the switch contacts, but if the latter is operated quickly, the circuit will be interrupted quickly in spite of this and EA will be ineffective. It is stressed however, that this self-induced e.m.f. at 'switching off, can be

extremely large in some instances where a large number of turns of an energising winding are associated with a strong magnetic flux. An example would be the opening of the field circuit of a large alternator o r d.c. generator and special arrangements are necessary in order to limit the e.m.f, to a safe value and prevent 'break-down' of the insulation by this large induced voltage. Sufficient has now been said about static induction to allow consideration of simple practical problems. It will be apparent that the induced e.m.f, is proportional to the rate of change of flux-linkages and since this rate alters from instant to instant then, if instnntancous values arc not being considered, an average value must be taken and generally we can take the formula as set out to give the average e.m.f. and to serve all conditions. Thus: Induced e.m.f. E,,

=

("I

t

- @') volts,

where N is the number of turns of the coil with which the flux is linked, O, is the original value of flux in webers, @, is the final value of fhe flux in webers and t is the time in seconds during which the change is taking place. This general expression serves conditions of switching on and switching off and also intermediate conditions, when the flux is changed from one value t~ another,. Example 44. (Self-induction). A coil of 800 turns is wound o n a wooden former and a current of 5A is passed through it t o produce a magnetic flux of 200 micro-webers. Calculate the average value of e.m.f, induced in the coil when the current is (a) switched off in 0.08 seconds (b) reversed in 0.2 seconds. N@ - 800 x 200 x lob6 - 16 x lo4 x (a) E,, = t 0.08 8 x lo-' = 2 volts (b) E.,

=

("I

t

- *,)

here O, = @, but is in t h e reverse

direction. I'hus flux change is from @, to zero and then up again + a,) = 2@, webers. to 0,or a total change of (0, Thus E,, =

800[200 x l o L 6 - ( - 200 x 0.2

117

ELECTROMAGNETIC INDUCnON

Example 45. (Mutual-induction), I f the coil of the above example has a secondary coil of 2000 turnh wound onto it, find the e.m.f, induced in this second coil when the current of 5A is switched off in 0.08 seconds. ( I t can be assumed that all the flux of 200pWb created by the 5A current in the primary links with the secondary coil.) 2000(200 x 0.08 2 x 20 - -= 5 volts.

E." =

- 0)

=2

lo3 8 x

200

lo-2

8

5

Note how the e.m.f. of the secondary is - = 2.5 times the in2 duced e.m.f. value in the primary. It is in direct proportion to the 2000 = 2.5. This is the basic principle of the transturns ratio ---800 r former and ignition system spark-coil. It shows how a large voltage can be induced in a secondary coil by the flux associated with a low voltage primary coil. For a petrolengine ignition system, the e.m.f. in the secondary may be in the region of 8000 volts compared with the 12V applied to the primary. This is achieved by using the appropriate turns ratio for the primary and secondary coils, by providing an iron magnetic circuit to concentrate the flux fbr maximum linkage and by interrupting the primury circuit quickly by un cnginc-drjvcn cu~li-opcratcd switch. DYNAMIC INDUCTION As was mentioned earlier, this condition covers the cases where there is relative movement between a magnetic field and a conductor. Obviously this concerns either a stationary conductor and a moving field or a stationary field and a moving conductor. To avoid repetition of basic theory, the immediate explanations and diagrams will refer to a fixed field and moving conductor. The diagrams (Fig 51) show a field as produced by two permanent magnets and a conductor which is moved so as to cut the field, thus altering the flux-linkages. The reasoning can be seen as in interpretation of Faraday's and Lenz's laws and three cases are shown. For case (a) there is seen to be no change of flux-linkages, ie no cutting of the field. The conductor is merely moved at a velocity of v metreslsecond in the same direction of the lines of flux and no e.m.f. is recorded on the voltmeter. For case (b) the

Fig 51

conductor is moved at right angles to the field of flux-density B teslas and the voltmeter shows a constant deflection. The fluxlinkages can be copsidered to be changing since the flux lines arc cut or can be imagined to stretch and snap as the conductor passes through, to reform again behind the conductor. If' the c.onduclor i s movcd from left to right, a polarity i s notcd,

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119

!

ELECTROMAGNETIC INDUC~ON

i

which reverses if the conductor is moved from right to left. Alternatively, if the field is reversed so that the flux lines are considered to pass from a bottom N pole to a S pole a t the top of the diagram, and the conductor is moved from left to right, then a reversed polarity will again be indicated. The investigation will show further deductions. Thus: The magnitude of the induced e.m.f. varies with the speed of cutting the field or rate of change of flux-linkages. Hence E a V . Again, if the field being cut is varied by altering the density of the flux, then the e.m.f. will vary as B or E a B. Obviously also, the longer the conductor cutting a field, the greater will be the magnitude of the e.m.f. and E a I. Summarising these three conditions we see that E a Blv. Here 1 is the length of the conductor in metres. Case (c) of the diagram shows the conductor cutting the field at an angle 8. It is an intermediate condition between cases (a) and (b) and is best treated by resolving v iflo two component velocities at right angles to each other. Consider v cos 8 to be the component velocity in the direction of the flux lines, then v sin 8 will be the other component velocity a t right angles to the field. In accordance with the reasoning for cases (a) and (b) we see that velocity v cos 8 will be responsible for no induced e.m.f. whereas velocity v sin 0 will be responsible for such an e.m.f. and E K v sin 8. E w Rlv qin 0 will be A more general expression than those already deduced since i t will cover all conditions. For instance for the condition of case (a) 8 = 0" and since sin 0" = 0 :. Blv sin 0" = 0 or E = 0 as already stated Again for case (b). If 8 = 90" then sin 90" = 1 and Blv sin 90" = Blv giving E a Blv.

!

f

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E.M.F. DUE TO DYNAMIC INDUCTION

As explained above, the induced e.m.f. is found to be proportional to B, I, v and the sine of the angle made by the direction of' cutting, and the direction of the field. The actual magnitude of such an e.m.f. can however be deduced in more definite terms thus : Consider case (b) of Fig 51. In 1 second, the area cut by a conductor of length 1 metres and moving at a velocity of v metres/second is lv square metres. If the flux density in this area is B teslas, then the flux cut per second by the conductor = Blv webers. Using Faraday's law, we see that Blv can be used as a measure of the magnitude of the induced e.m.f. in volts or Induced e.m.f. E = Blv volts

If case (c) is considered the flux cut is proportional to the component of the velocity perpendicular to the field or induced e.m.f. E = Blv sin 8 volts. The above formula can also be deduced as follows; this approach may appeai- to the reader to be more satisfactory as a proof. The diagram (Fig 52) shows a conductor Q, carrying a current of I amperes in the direction shown. As before the flux density of the field is taken as B teslas and the length of the conductor as I metres. Using fundamentals already set out in the chapter on electromagnetism, it is known that a force is exerted on a current-carrying conductor in a magnetic field. Thus the conductor in the diagram experiences a force BII newtons urging it to the left. Accordingly a force of BIl newtons must be applied in the opposite direction to oppose movement of the conductor.

Fig 52 8 .

Consider the.conductor to move, from position Q to position P spaced x metres away. Then work done by the conductor in moving from Q to P = Force x distance = BIlx newton metres or joules. Let E volts be the e.m.f. induced in the conductor as a result of cutting the field, and t seconds the time taken to complete the operation.

Then Mechanical Power expended = 'I1' watts and if this t

appeared as electrical power, it would be El watts BIIx Blx or EI = and E = volts I t Obviously 5 = velocity of cutting v t or as before, E = Blv volts. From the above, the following deduction can also b e made.

.

ELeCIROXAGNETIC INDUCTION

12 1

Blx is (the fluxdensity x area) or the Bux Q, cut by the conductor in moving from position Q to P in time t seconds and since

0 E = -Blv then E = B x a r e a = t

t

t

Thus E (volts) = @(Webers) and we have an alternative t (seconds) formula for the e.m.f. generated in a conductor cutting a magnetic field. In this form, it is sunilar to that deduced for the NO statically induced e.m.f. namely E = -where N is the number t of turns of the coil, and O is the change of flux. flux-linkages or E = 0 since The formula can thus be E = t time the flux-linkages are numerically equal to @, there being only one conductor. ' cut and thus we have an alternative Important Note. - is t time way of stating Faraday's law, which can now be expressed as: The e.m.f. generated in a conductor is proportional to the rate of cutting lines of flux or is proportional to the flux cutlsecond.' The abo,ve form of Faraday's law is more applicable to dynamic induction and will be uscd scpeutcdly in connection with the Generator, Motor and Alternator. @

Example 46. A conductor is moved to cut a magnetic field at right angles. Find the e.m.f. induced in it, if the average density of the field is 0.45 teslas, the length of conductor is 80mm and the speed of cutting is 8.88 metresisecond. In the Formula E = Blv we have E = 0.45 x 0.08 x 8.88 = 0.32 volts. An alternate solution could be: Area swept by the conductor/second = 0.08 x 8-88 square metres. The flux in this area would be 0 = 0.45 x 0.08 x 8.88 webers and e.m.f. = flux cut per second 0.45 x 0.08 x 8.88 = 0.32 volts. or E = 1 Example 47. A four-pole generator has a flux of 12mWb/pole. Calculate the value of e.m.f. generated in one of the armature conductors, if the armature is driven at 900 rev/min. = 0.048Wb In 1 revolution a conductor cuts 4 x 12 x

1 Time of 1 revolution of the armature = -minutes

900

:. Rate of cutting flux =

Thus E =

0.048

--i- = l-5

Flux cut per revolution time taken to complete a revolution

0.048 x 15 = 0.72 volts/conductor.

Although this unit has been introduced, it has not a s yet been satisf'actorily defined, since the accepted .definition is based on the principle of electromagnetic induction. Thus we have : 'The weber is that magnetic flux which, when cut by a conductor in one second, generates in the conductor an e.m.f. of value equal to one volt.' Alternative ways of defining thc weber o r SI unit of flux are : 'An e.m.f. of one volt is generated when a conductor cuts flux a t the rate of one weberisecond', c+r'an e.rr,.f. of one volt is generated when the flux linked with cne turn changes at the rate of one weberlsecond.' THE WEBER .

DIRECTION OF INDUCED E.M.F.( H A N D RULES)

The direction of the induced e.m.f. can be deduced from first principles, using Lenz's law o r by the application of a rule which was first enuniinted by Professor T. A. Fleming and is now commonly known as F L E M IN G ' S R IG H T - H A N D RULE . It should be noted that there is also a Left-Hand Rule and to avoid later confusion in the students' mind, the point is made a t this stage and the following aid is suggested for memorising the appropriate rule to suit the circumstances. Thus the Generator is studied hqji~rethe Motor and for the average person, the use of the Right-hand I S prcf'crred hcfirc~ that o f t h e left. Therefore use the Ripht-11;lnd rule f'or the Generator and the Left-hand rule for tlic Motor. The (;enerator is n machine concerned with thc generated o r induced e.m.f. in armature conductors and thus for condittotis 01' elccl~.oclyli;~~llic i~l(lucliorl,wc ~ s ctllc t,ight-lli~ntl rule which can now be explained. R U L E (Fieming's). Consider a conductor in a magnetic field as shown in the diagram (Fig 53). Imagine the tnagnetic ficld to bc in tht: direction from Ief't to ~.it..Iil; ~ n dthc conductor to be moved at right Angles and upwards with a velocity of v rnetres/second. The e.m.f. across the ends ol'

R I G HT - H A N D

ELE C TR O M A G N ETI C I N D U C TI O N

123

Fig 53 the conductor is assumed as shown, ie the polarity is such that, if the ends of' the conductor are joined externally through an ammeter, current will flow as indicated. Its direction in the conductor is 5ccn :~ntlif' :tttcntion i s ylvcn to Fig 53b, it will be deduced that the field due to the conductor current is clockwise, to strengthen the field at the top and weaken the field at the bottom. Thus according to Lcnl's law, opposition is o f i r c d t o the motion of the conductor as one imagines the field lines concentrating or massing bef'ore the conductor, stretching and then snapping. A force of opposition to the direction of movement is apparent and the assumed polarity must be correct to confirm the action which takes place in practice. If a reversed polarity was assumed, the current would be in the opposite direction and field weakening would occur above the conductor and strengthening below. This would result in a driving force behind the conductor which would be a motoring rather than a generating condition. There is no opposition to moving the conductor and since such a condition is not possible, this alternative e.m.f. polarity assumption must be incorrect. Since the' original assumption actually is confirmed by practical conditions, the right hand can be drawn and used to find the direction of the induced current and therefore the induced polarity. This is shown in the diagram (Fig 54).

Fig 54 To use the rule, place the thumb, index finger and second finger of the right hand at right angles to each other. Point the index linger in the direction of' the flux lines and the thumb in the direction of moving the conductor. The current in the conductor, due to the induced e.m.f. would be in the direction indicated by the second finger. For the example being considered (Fig 53), current would be into the paper as deduced from first principles. T H E SIMPLE MAGNETO-DYNAMO Once the principles of electromagnetic induction were discovered, it soon became evident that the way was open to constructing a machine, in the true sense of the word, which could convert mechanical energy into electrical energy and thus generate electricity as a result of being driven by a prime-mover, such as a steam engine or water turbine. The idea of making insulated condu?tors move through a stationary magnetic field presented no difficulties for a small machine and so the basic construction of such a magneto-dynamo followed fundamental requirements. A typical machine is therefore, illustrated in the diagram (Fig 5 9 , and consists of permanent magnets to provide the field and a simple coil which is mounted on but insulated from a shaft which can be rotated. In order to allow contact to be made with the moving conductors, they are connected to slip-rings which are mountcd on but insulated from the shaft. Fixed 'brushes' in turn, contact the slip-rings to make sliding connections and allow an external circuit to be enexgised. It will be seen that the coil consists of two 'active' conductors which have been designated AB and CD. These are connected in series by the connection BC wbich, together with the front connections to the slip-rings, plays no part in the generation of e.m.f. but merely serves as a means of carqhng current to the external circuit. The load resistance of the external circuit has

Fig 55 been shown as concentrated in R and is connected to the terminals X, Y of the machine. Consider the operation of the machine as follows: As one conductor AB moves down through the field, the other CD moves up and the induced e.m.fs. will be such that A is +ve relative to B and C is + vc rclutivc to D. Thc induced currcnt, if allowed to flow, would be as shown by the arrows and, since it is from terminal Y to terminal X through the external circuit, Y is +ve with respect to X. It would be well for the student at this point to try the right-hand rule for himself and satisfy himself as to the polarity of the terminals for the half-revolution being considered. It should be noted that the right-hand rule as described, can be applied here to conductor AB, the condition being that AB is moving from the top vertical position round past the centre of the magnet pole and then onto the bottom vertical position. The position where it moves past the pole at right angles is of particular importance, being a condition of maximum e.m.f. After the coil has rotated a half revolution, conductor DC begins to move downwards and AB upwards. The polarity induced is now in reverse to that for the 1st half revolution, D being +ve relative to C and B is +ve relative to A. Terminal X is now the +ve terminal and Y is the negative. An alternating e.m.f. is generated, as shown in the diagram (Fig 56), which also

illustrates four positions of the coil viewed from the slip-ring end. For position 1, A and D are moving horizontally along the field and no e.m.f. is being generated. A sirnilar.condition exists for

Fig 56

'

position 3 , but for positions 2 and 4 maximum e.m.f. is being generated, since field-cutting at right-angles is taking place. For intermediate positions, the general condition of e.m.f. generation, ;IS rcprcscntcd by E a Blv sin 0, is followed, since ttlc conductors are cutting a uniform field at an angle but are moving a t a uniform velocity. Thus the e,m.f. generated a t a n y instant is not constant but varies and it is customary to use a small letter for what is termed the instantaneous value. The expression e = Blv sin 0 volts gives the magnitude of the voltage being generated, provided the correct units are used when substl.uting. If the voltage is plotted to a base of' revolutions, degrees or radians, a waveform such as:that illustrated will be obtained.

ELECTROMAGNETIC IN D U C T IO N

127

THE SIMPLE D . C . GFNRRATOR

The simple magneto-cly~rirrwn ~ u c l ~ ~u!,~ l dc\cr~hccl c, sbovc (Fig 55), or electrical generator, as it is now called-to use the modern term; is seen to have n uniform field arranged to be cut by conductors as shown. It will provide an e.m.f. whose magnitude varies sinusoidally, that is, the e.m.f. polarity and value follows a sine waveform. A sinusoidal waveform is desirable for a.c. working but for the d.c. generator, modifications are necessary to achieve a substantially constant unidirectional voltage magnitude and polarity. It is now apparent that a distinction is being made between the generation of direct current and alternating current and from here on the division between the two methods of generating, transmitting and utilising electrical energy will become marked. In this book, it is hoped that the study of both d.c. and a.c. theory will be followed simultaneously and the reader is asked to give the approprizte chapters equal attention. D.C. theory has up to now received more attention by the marine engineer, mainly because of the type of installation with which he is familiar. A.C. installations are however, for quite a considerable time now, being used to an increasing extent for ship work and are almost universal ashore. It is not proposed to enter, at this stage, into a discussion as to the relative advantages of alternating current over direct current or vice-versa; but it is stressed that the major portion of electrical theory is conccrned with u.c, circu~tsand milchines and that if Inter study difficulties are to be minimised, then full attention must be given to a.c. theory right from the beginning. The first of the modifications referred to above for the d.c. machine, involves the introduction of an iron or magnetic material into the armature or moving-coil part of the assembly. The coil made up from insulated wire or strip, is wound onto an iron armature which is carried on the shaft. The magnet system is also provided with iron pole-shoes or extensions as shown in the diagram (Fig 58). Since the length of the flux path through air is now reduced to two small air-gaps, the remainder being through the iron of the armature and field system; the flux density or B value in the air-gaps is increased and. the conductors will therefore cut a stronger field. Again as the air-gaps are now small and of constant width, the flux lines will cross them as shown and the field will be uniform over the pole-faces. The moving conductors thus pass from a small arc with substantially no magnetic flux into a large arc ~f constant flux density. The flux lines are seen to be radial in the gaps and are cut at right angles for most of the distance under the pole. The

Fig 57 e.m.f. waveform is now as shown in the diagram (Fig 57), ie it is proportional to the flux density through which it passes. COMM~JTATION.T O

obtain a constant unidirectional e.m.f, or to produce the true d.c. generator as is used in practice, the next step in modification is, to fit a form of automatic reversing switch.or

ELECT RO MAGNETIC IN D U C TI O N

129

oammutator which, even tl~ouyht l ~ dmovingcoil oontinwa to generate an alternating e.m.f., ensures that a unidirectional or 'rectified' e.m.f. appears at the terminals of the machine. The diagram (Fig 58) shows how a commutator is fitted. It consists of a metal slip-ring which is split into two parts, each insulated from the other and from the shaft. The ends of the coil are connected to each half or segment of the commutator. The stationary brushes are so adjusted that they bridge the gap in the slip-rings at the instant when the e.m.f. induced in the coil has zero value and is due to reverse. The diagram (Fig 59) shows the side view of the commutator and the reversing action of the switching arrangement can be more clearly seen. The diagrams can be considered t o b e complementary to those of Fig 56 although only conditions for positions 2 and 4 are shown. It is apparent that the obvious position for the brushes is on the 'magnetic geutral axis' a d that the brush Y will always be the +ve and brush X the -ve terminal. The new shape of the waveform is also shown.

Fig 59

A+ Position 2

7

1+ Position A

130

REED'S BASIC ELECTROTECHNOLOGY

For position 2 it will beTseen that - ve end D of conductor CD is connected to the +ve brush Y, whereas the +ve end A of AB is connected to the - ve brush X. For position 4 when the e.m.fs. have reversed in the conductors of the coil, end D which is now ve is connected to the -ve brush X and end A of AB is now - ve and connected to the +ve brush Y. It is well to remind the reader here, that the brush polarity is decided by the direction of current flow in the external circuit. Thus current flows from Y (the +ve brush) to X (the - ve brush) and then to A onto B, etc. The apparent anomaly of ve end A being connected to a - ve brush and so on is thus cxplaincd. Tlic resulting effect of thc commutating action is to produce a pulsating but unidirectional e.m.f, at the terminals of the generator.

+

+

PRACTICAL REQUIREMENTS.TO obtain a more uniform e.m.f., the two-part commutator and single coil can be repeated to give a n arrangement employing a greater number of segments and a larger number of coils. Each coil can consist of a number of turns to give a larger output voltage. The example shown in the diagram (Fig 60) is an armature

-

/-.

Fig 60 with two coils at right clngles. It follows thl~tfor this arrangcmcnt when coil A develops maximum e.m.f., coil B generates nq e.m.f. and whc11 tllc Irrmi\turc rot;~tcsIlirot~gli;I rlui~rtcro f ;I revolution, the conditions would be vice-versa. The accompanying diagram (Fig 61) shows the waveforms of the generated e.m.fs. The generator terminal voltage ncver falls to zero but it is obvious that two distinct disadvantages are still evident. Firstly, all the ~ x ~ ~ l d u c tarc o r s not used to mnximum ndvantage since only one coil ; ~ t;I time is being ernployed for supplying the external circuit. Secondly, but or prime importance is the new condition

131

ELECTROMAGNETIC INDUCTION

B

A

EM.WE 0 COILS AL 0

' \

6

A

A

r I \

I I

I

I I

I

f

I

:: n

2 ~i C

Fig 61 of commutation. Since the brushes must be placed in a position to contact the coil in which e.m.f. is being generated, it follows that if the generator is on load, ie supplying current, then at the instant when the connected segments leave the brush, since an e.m.f. still exists and current is flowing, arcing will take place at the brushes. If coil A is bcing considered : ~ n dFigs 60 and 61 :ire noted, it will be seen that at the instant when the gap between segments is being bridged by the brushes, coil A.is still cutting the field and coil B has only just entered the field. Thus coil A tends to be short-circuited by a coil in which the e.m.f. may not have risen to the required value and current will flow in the coil B. This current is diverted from the !oad current and also adversely affecting the commutation. If the number of coils is increased, the tendency would be to give a smoother output voltage but continued arcing at the brushes would persist. This arrangement Is obviously not satisfactory. In the early period of development of a satisfactory arrangement for the armature conductors many ideas were introduced. One such arrangement was incorporated in the Gramme-Ring Armature, which involved a special construction, in that the armature iron q c u i t was built up as a ring and the conductors were connected in series, with tappings being brought out to the commutator segments. The conductors were thus part of a continuous winding, but it will be noted that only the conductors on the outside of the ring are active, whereas that part of the

132

REED'SBASIC ELECIROTECHNOUXiY

winding which is on the inside and at the ends of the ring, cuts no flux, and is thus responsible for no e.m.f. This was a most uneconomical arrangement since it wasted conductor material.

Although the Gramme-Ring Armature is shown (Fig 62), it is stressed that this never was a commercial proposition. The only reason why it is being described here is that it shows the student how a continuous closed winding can be made to operate as a suitable source of generated e.m.f. Furthermore satisfactory commutation is possible and all 'active' conductors in the air-gaps arc used to advantage. It is not intended that the reader should give the arrangement undue attention nor should he consider the actual details worth remembering. All his attention should be given to the 'drum' type armature winding which is described next and dealt with more fully in books on the practical aspect of modem maohine construction and operation. The ring arrangement shows how a continuous winding is possibk, in which the induced e.m.f. automatically divides the armature into two parallel paths, the flow of current being as shown. The coils in which no e.m.f. is being generated are shortcircuited as they pass under the brushes and since there is no current, no sparking occurs and commutation is correct. All other active conductors are however used to advantage and are connected in series to supply the e.m.f. at the terminals. Even though the armature is rotating, the disposition of the active conductors in space can be considered to be stationary. As any one coil passes from one side of a brush to the other, it leaves the series circuit of one parallel branch of the armature winding to be replaced by a similar coil, which has just been cornmutated o r short-circuited by passing under a brush. Thus one coil enters into the series circuit of one of the parallel paths as one coil leaves and the total e.m.f. being generated is substantially constant.

E L E C TR O M A G N E T I C I N D U C T I O N

133

It should be noted that for a 2-pole machine the armature winding is i~utomiiticallydivided into two pnrnllcl poths. For a 4-pole machine the winding will be divided into four parallel paths and so on. The armature can now be looked upon as a battery made up from a number of similar cells. An armature with one hundred conductors, being wound for a 4-pole machine, will mean twenty-five conductors are in series for one parallel path and there are four such parallel paths. If 1V is generated in one conductor, the e.m.f. of 1 parallel path is 25V. This is also the e.m.f. of the machine, since as for a battery, the e.m.f. of the arrangement, is the e.m.f. of one parallel path. Again if the size of the conductor used for the winding is'suitable for carrying 10A, then the current carried bq'one parallel path is 10A and the total current that can be expected from the armature is 4 x 10 = 40A. THE DRUM W I N DI N G . The diagram (Fig 63) h o w s the basic

arrangement. This armatwg winding arrangement is accepted as the only modern method of connecting the active conductors together. The amount of 'copper' on the armature is used to maximum advantage since, except for the overhang at the back of a coil and the front connections to the commutator segments, the winding consists of lengths of copper conductors which are so placed to cut magnetic flux and thus generate e.m.f. The basic winding uses a number of coils in series between brushes, these coils being arranged at constant angles to each other. T h e resultant e.m.f, is thus more uniform and larger since many coils in series are employed. The connecting up of such coils presents difficulties; but any requirement can be realised by .proper choice of coil numbers, the span of a coil, number of parallel paths, etc. An example of a simple drum winding can now be considered.

Fig 63

i 1

I I

I i 1

A, B, C, D, etc are insulated conductors fitted into slots cut into the iron of the armature. There are also four commutator segments Nos 1 , 2 , 3 and 4. The conductors may be connected to each otlier and to the segments in a variety of ways and one possible arrangement is shown in the diagram. With rotation as shown, the e.m.f. in B, C and D are from front to back, while for F, G and H the e.m.f. are from back to front. The full lines show coqnection to segments and the dotted lines connections, which constitute the overhang of the coils, at the back. St:~rtinpat the -ve brush on segment 1 , current enters the armaturc from the external circuit and divides into two parallel paths. One path passes to A and flowing down this comes at the I l i ~ c l IO I,', IICIICC L I iICI.oSS ~ to scjilncnt 2 and Lhcn onto und down C to H, I'rom where it rises u p and goes to the brush on segment 3. This would be the +ve brush. The other current path is from segment 1 to conductors D, G, B, E and onto segment 3 or the +ve brush. There are thus two circuits in parallel and it will also be noticed that, as a brush passes from one segment to the next, one coil is short-circuited and the brush must be located so as to short the coil at the instant when its e.m.f. and resultant current is zero. Such an instant is shown in Fig 63b giving correct commutation conditions for the short-circuited coils between segments 1 and 2 and between segments 3 and 4. Example 48. A slow-speed d.c. generator has an armature of diameter 3.0m and active conductors of length 510mm. The average stseng4li of the field in the air-gap is 0.8T and the armature speed is 200 revimin. If the armature has 144 conductors arranged in 8 parallel paths, find the e.m.f. generated at the machine terminals.

Using formula E = Blv volts then R = 0.08 teslas. 1 = 510 x 10-3m and v is obtained t US :--

In I second the :ll-m;lhlrc rcvolvcs

2s or -10. t i ~ r e s .Also i n 60 3

1 revolution one conductor travels

xd = 3.14 x 3 = 9.42m

So in 1 second the conductor travels 9.42 x

10 = 3.14 x 3 =

10

31.4m

E . M . F . generated perconductor = 0.8 x 510 x 1 0 - 3 x 31.4 = 0.8 x 5 . 1 x 3.14 12.8 volts -2

Now

~ l i c r c;II.L.144

c o n t l ~ ~ c ( o111 r \ X 1~:11;1llcl I):IIII\ 144 = IS So the conductors in serles in each pnrnllel p;rtil = 13 'Thus the e.m.S. generated in 1 parallel path = 18 i( I3 8 = 230.4V -

A.C. A N D D.C. THEORY The reader will be required from now onwards to give equal attention to both the a.c. and d.c. theory appertaining to electrical technology. T o assist in this, the following chapters in Volume 6 will treat both aspects of theory alternatively, thereby ensuring correct progression along both channels of study. It should be noted that the amount of a.c. theory to be covered is considerable and will occupy the major part of Volume 7. Here in Volume 6, under the broad coverage of the d.c. theory, is included further basics on d.c. machines, the magnetic circuit, circuit conditions, methods of solution and thermionics, whereas the a.c. theory proceeds with basic fundamentals of circuit conditions and systems. Inductance and capacitance have also been introduced because of their importance to a.c. circuits and in connection with the capacitor, the treatment has been extended to include the electron theory, basic electronics and electrostatics. The following diagram will help to explain the course b a n g followed. C ' I 1Al''I'Iil~0 (Electromagnetic Induction)

CHAPTER 8 (The D.C. Generator)

CHAPTER 7 (Basic A.C. Theory) I

CHAPTER 10 (The D.C. Motor)

CHAPTER 9 (The A.C. Circuit)

CHAPTER 12 (Electromagnetism)

CHAPTER 1 1 (.4.C Circuits and Systems)

CHAPTER 14 (Miscellaneous Clrcult Cond1tlon5 dnd Method\ of Solut~on,Spec~,il ,4pplicCitlon\)

CHAPTER I? (The Electron The01 !, B'isic electronic^ ,rnd Electroctat~cs) CHAPTER 15 (Electron~c\)

CHAPTER 6 PRACTICE EXAMPLES 1.

Calculate the e.m.f. generated in the axles of a railway train when travelling at 100km/h. The axles are 1.4m in length and the component of the earth's magnetic field density is 40pT.

2.

Find the generated e.m.f./conductor of a 6-pole d.c. generator having a magnetic fluxlpole of 64mWb and a spccd of I000 rcvlmin. If there arc 4611 conductors, connected in six parallel circuits, calculate the total generated e.m.f. of the machine. Find also the total power developed by the armature when the current in each conductor is 50A.

3.

An iron-cored coil of 2000 turns produces a magnetic flux of 30mWb when a current of IOA is flowing from the d.c. supply. Find the average value of induced e.m.f. if the time of opening the supply switch iS 0.12 second. The residual flux of the iron is 2mWb.

4.

A one-turn armature coil has an axial length of 0.4m and a diameter of 0.2m. I t is rotated at a speed of 500 revlmin a field of uniform flux density of 1.2T. Calculate the magnitude of the e.m.f. induced in the coil.

5.

When driven at 1000 rev:min with a flux/pole of 20mWb, a d.c. generator has an e.m.f. of 200V. If the speed is increased to 1 100 rev/min and at the same time the flux/pole is reduced to I9rnWb/polc, what is thcn the induced e.m.f.?

0,

A coil 01' 2oU turns 1s rotated a t 1200 ~.sv/m;nbetween thc poles of an electromagnet. The flux density of the field is 0.02T i~ndthe xis of roti~tionis at right angles to the direction of the field. The eff'ective length of the coil is 0.3m and the mean width 0.2m. Assuming that the e.m,f. produced is sinusoidal, calculate (a) the maximum value of e.m.f. (b) the frequency.

7.

A coil of 19120 turns la wound on an iron core and wltn a ti1111 V ~ I I ~ 01'I CC L. I II . C I I ( [lowitip, 111 tl~cC I I . L . I ;II ~flttx I, (TS .lrnW\> i~ ~~rodiicrtl Wl1c.11rllr c ~ l ~ . c . uI.; ~ opcnctl. l lhc ll!~x I'ic!!.:

(TI

to ~ t sres~dual value of 1 5rnWb In 4 0 m s . Calculate the :lvcr:lgc V:IIIIC 01. t i i c I I H I I I < T ( I r 111 l'

8.

The armature ol' a four-pole generator rotates at 600 rev/ min. The area of each pole-face is 0.09m2and the flux density in the air-gap is 0.92T. Find the average e.m.f. induced in each conductor. If the armature winding is made up of 210 single-turn coils connected so as to provide four parallel paths between the brushes, find the generator terminal voltage.

9.

A solenoid 1.5m long is wound uniformly with 400 turns and a small 50 turns coil of lOmm diameter is plaaxj inside and at the centre of the solenoid. The axes of the solenoid and the coil are coincident. Calculate (a) the flux linked with the small coil when the solenoid carries a current of 6A and (b) the average e.m.f. induced in the sm51I coil when the current in the solenoid is reduced from 6A to zero in 50111s.

10.

Two coils A and B having 1000 and 500 turns respectively are magnetically coupled. When a current of 2A is flowing in coil A it produces a flux of 18mWb, of which 80 per cent is linked with coil B. If the current of 2A is reversed uniformly in O.ls, what will be the average e.m.f. in each coil?

CHAPTER 7

BASIC A.C. THEORY Introduction is made by quickly revising the relevant fundamentals of Chapter 6. The diagrams (Fig 64a and b) show an elementary form of a.c. generator in which a coil is rotated in a uniform magnetic field. The sides of the mil, ie the conductors, cut the magnetic flux and thus, an e.m.f. is induced which, from first principles is e = Blv volts. The letter e, for the value of induced e.m.f., has been introduced here because this value is not constunt but, as will be seen, vi~siesfrom instilnt to instent. Thus even though the w i l is rotated at a uniform velocity v, the rate of cutting i s not constant, but depends upon the angle at which the conductors cut flux. The velocity can be resolved into a cutting component ( v sin 8) and a noncutting component ( v cos 8). The cutting velocity component only is responsible for e.m.f. and can be used to give a general expression which gives the e.m.f. at any instant, as e = Blv sin 8 volts.

)8\j\ \

-. - - -(h) Fig 64

/ /

BASIC A.C: THEO RY

139

THE A.C. W A V E F O R M

In tlie expression t1 = Shl sin 0 , us for any alternator, 61. 1 and can be assumed to be const;lnt and made equal to K. The expression now becomes 61 = K sin U and a further value for K can be obtained if we cons~derthe instant when the coil sides are cutting the field at r ~ g h tangles. Velocity component v sin 0 generates a maximum e.m.f. which can be designated as Em. At this instance e = Em and we can write Em = K sin 8. But sin 8 = sin 90 = 1 . :. Em = K Substituting back in the expression, we have: e = Em sin 0 The above is an Important equation which shows that the generated e.m.f. varies sinusoidally. e is termed the instantaneous value and Em the maximum value. If attention is now turned to a waveforr~plotted to a time o r angle base, it will be remembered, from work dready done elsewhere on vectors, that a sine wave can be deduced from the vertical component of a rotating vector-for electrical work, such a rotating vector is called a phasor. If the length of the phasor is made to represent Em,then for any angle 8, the instantaneous value is the vertical projection and this also can be used as an ordinate for the waveform, when plotted to an angle or time base. The diagram (Fig 65) illustrates the procedure for deducing a waveform and the method is summariscd thus: I)r.;tw ; I c11.clcof' t a d 1 ~ 1m5 : ~ d cequ;tl

I O 1 1 1 In;tullrlillri ~

V;IIIIC oI'tlic

wave. Startlng from the h o r ~ ~ o n t amove l, the phasor through a known angle and project the vertical value onto an angle or time scale. Choose suitable scales so as not to distort the sinusoidal shape of the wave. The following remlnder shows the connection between the construction and the representation of a sinusoidally induced e.m.f. Since, from the triangle illustrated, e - = sin 8 so e = Em sln 0. Em

Fig 65

#

This expression is in the form deduced previously but can be further modified to suit the representation by a phasor. In accordance with accepted procedure, assume that the phasor Em rotates from the zero or horizontal position in an anticlockwise direction with an angular velocity of o radians/second (ois the Greek letter-small omega). Then 6 = wt where t is the time in seconds and the equation can now be written as e = Emsin at. The diagram (Fig 66) following below, shows some of the terms used in connection with a,c. theory. Periodic time = the time for 1 cycle. The frcqrienq- f of the wave = the number of complete cycles in the interval of' 1 second. In accordance with S1 recommendations, the name hertz (Hz) is now being adopted lor frequency Ineuuurctncnr. ,This rcpluccs tl~coldcr tcrm of cycles per second (CIS).Present marine practice uses either 50Hz or 60Hza.c. systems. The maximum value reached by the wave is also called its peak value, or its amplitude and, as mentioned earlier, the value at any instant is termed its instantaneous value and is denoted by a small letter such as e . In passing, we can observe that sinusoidal current conditions can also occur and that the expression for a current following a sine-wave law can similarly be written as i = I, sin cot.

LI CYC L E

---------C(

Fig 66 Following the introduction of the sine wave, as derived from a phasor, and the generation of a sinusoidal e.m.f. by a rotating coil, the treatment can be combined still further by the following deduction. The phasor is assumed to be rotating at a constant angular velocity of o radianslsecond and the waveform, if considered to have a frequency ol'j'hertz, will stretch out to cover in 1 second, an angle of 360fdegrees or 2nf radians. The phasor meanwhile will have passed through ro radians in 1 second and it l'ollows l l ~ i i t ( I J

C;III

1)c C~LI;IICLI to 2 ~ , / '01.

3 0 0 f ' . 1 . 1 ~c;~rIicr

14 1

B A S I C A . C . THEORY

dcclt~ccdcxprcs.;ions c:in r ~ o whc writtcn form, namely: ( > = E,,, sin 7nfr.

111

~ l i c i rmc;sr

\I\CTIII

The above is the first fundamental formula of a.c. theory and should be given full attention. It is important to note that if or 3.14 is substituted for rc, then the angle will be in radians and can be converted into degrees by multiplying by 57.3. The simpler method is to substitute 180- for rr, thus converting into degrees directly. Example 49. Find the instantaneous value of a 50Hz sinusoidai e.m.f. wave of maximum value IOOV, at an instant of time 0.003 seconds after the zero value. Substituting in e = Em sin 2rcfr we have e = 100 sin (2 x 180 x 50 x 0.003) o r e = 100 sin (18 x 3) = 100sin 54" = 190 x 0.809 and e = 80.9V. hnportant Note. A catch problem can occur when the instantaneous value is given and the time is required. Attention should be given to the following example, which illustrates the point being made. Example 50. Find the first time after zero, when the instantaneous value of a sinusoidal current wave is 6.8A. The maximum value is 12A and the frequency I S 50Hz. Find al.;o thc ,cconcl tirnc :il'lcr / c r o

Here i = I , sin 2x11 o r 6.8 = 12 sin (2 x 180 x 50 x r) 6.8 Thus- = sin (180 x 100 x r ) 12 and 0.566 = sin (18 x 103 x 1). Further :elution of this equation can only be made by reference to sine tables, from which an angle can be found whose sine equals 0.566. Thus let 0 = this angle, then sin 0 = 0.566. 0 is seen to be 34" 30' o r 34.5" *. s i n e = s i n ( l 8 x lo3 x ( ) o r 18 x lo3 x t = 34.5

34'5 = 1.9 x lo-' = 0.0019sor 1.9rns 18 x 10 The second time value required, is obtained by finding the time for a cycle and then subtracting the inverval, from a zero value, necessary for the instantaneous value to attain a height of 6.8A. Thus time for cycle = & seconds = 0.01s. So second time required = 0.01 0.0019 = 0.0081s or 8.1ms. and t =

4

-

REPRESENTATION O F SINUSOIDAL ALTERNATING QUANTITIES Earlicr it hiis been shown that an alternating voltage o r current can be represented by an expression such as e = Em sin 2xft o r i = I, sin 2xft and that this method of notation conveys all that is required to be known about the quantity, ie the fact that it follows a waveform whose amplitude, frequency and instantaneous value, at any particular time, can be found. This method of notation is called trigonometrical representation. TKIC;ONOMETRICAL REPRESENTATION .

This is useful for two quantitics which are alternating, but not necessarily in the same simultaneous manner. Thus an alternating voltage of 50Hz can C ~ I I I ~i111 C ill1~1.11ilting CuI.rcnt ill i\ circi~itwhich will altcmate a t 5 0 1 4 ~ .'l'he current need not however, be in phase with the 'voltage, which latter may reach its maximum value a little time before the current reaches its maximum value. The voltage is said to letrd the current or the current to lag the voltage. There is a plzusr diflrrence between the two quantities or between their waveforms and such a phase difference is shown by the inclusion of the phase angle (in radians). Thus if two current waveforms are represented by i, = I,, sin 2xft and i, = I,, sin (2xft + F), it means that the 3 180 second waveform leads the first by an angle of 2 radians o r 3 3

I,, sin (2rrfr - h) is 6 180 seen to lag the first or reference waveform by --- = 30". 6 The trigonometrical form of representation, being a mathematical expression, can be used for the usual trigonometrical o l > c . ~ ; ~ t ~\ oI I C~I ~i \ ;.I S multiplication.-division, expansion, etc and such ~ ~ p p l ~ c a t ~will o n sbe well illustrated in the course of a.c. tllcoc ;IS 11115IS ~ i c ~ l o ~ x d . = 60 . A third%aveform written as i3 =

This, commonly used for a.c. quantities bucll as current, volr;~gc,flux, ctc, Iins already hccn introduced in terms of vectors. In Volume I1 dealing with Mechanics a vector was introduced and defined. Since voltages or currents whose magnitudes and directions are known, they arc qu~~ntitics can be described by rotating bectors but since 'phase' is usually i~\\~ol\~txI, i~ is IIOW customary to represent these by phasors. T11\1\;;I volt;~gc~ I i ; ~ s c;ln o r be drawn to scale, its length reprcb c l l ~ i ~111c ~ gI I ~ ; I ~ , LI II ~~ 01'~C tlic voIt;~gc; I I I ~tl1c dircclion in wllich i t P I ~ A S O RREPRFSENTATION.

143

BASIC A . C . THEORY

acts can be shown by an arrow. This technique has already been I I I ~ I ~ O L I L I C ~~L;I I I . I I ~ I111 it115 C ' I I . I I ) ~ L; I I.I LI! , \vc L , , I I I I I O \ \ , OCL.L.C! tt) consider the accepted methods ol' phasor oper:\tlcln The rclation bctwccn and tllc gr;~pllicaldcduc~ionol', :I waveform from a phasor has already bccn covercd. S111cct'or I I I O S I common and practical a.c. work, waveforms and instantaneous values are of comparatively little im?ortance compared with magnitude and phase, the use of a phasor alone, as shown in t h e diagram (Fig 67), makes representation much simpler. Even further simplification and understanding can result I'rom the correct use of phasor diagrams. Such diagrams are used to illustrate a.c. circuit relationships and are particularly useful if more than one current and/or voltage 1s being considered at the same time. Note. For most practical work r.m.s. (root mean square) values are used in electrical engineering. The full mewing of the term will be dealt with later in this chapter, but it is mentioned here, because it is more convenient to make'a phasor equal to this value rather than the maximum value-since the representation for a maximum value holds good for an r.m.s. value. Thls modification will however only be introduced a t a later stage.

PHASOR

DIAGRAM

Fig 67 Phase difference can also be shown by phasors. Conslder two 50 hertz sinusoidal voltages represented by El, and E,,. The

phase angle 4 is known, the voltages being of the same frequency but out of phase. The voltages can be written as r , = El,, sln tot and e, = E,, sin (ot- 4 ) where the angle 4 can be assumed to be say 60" o r

X -

3

X

radians and e, = E,, sin (wt - - ) The wave-

3 '

forms can be drawn graphically as described earl~erand slnce the first leads the second then, if the instant when the first 14 going through its zero value is considered as the start of the angle or time scale, the first wave can be cons~deredas the

reference and the phasor diagram can be drawn as shown in the diagram (Fig 68). -*

- - -

Y

-1

. t*

--

I-A

8

Fig 68 11 will be YCCII t l l i ~ t1'01.llle phnuor diuyram, wc n~crelydepict the two phasors and their relation to each other. The first phasor has been taken as the reference and the second is seen to lag it by an angle 4. Direction of rotation is anticlockwise so E2, is behind E,, by the angle $. If an instant 8 degrees later in time has to be considered then the diagram can be drawn as shown (Fig 69), the horizontal being taken as the zero time or reference axis.

Fig 69 ADDITION AND SUBTRACTION O F ALTERNATING QUANTITIES When two or more sinusoidal voltages or currents act in a circuit the resultant can be obtained in either of the following ways ( 1 ) By Trigonometrical Methods (2) By Phasor Methods. (1) ~KIGONOMETKICALMETHODS. These methods require a good knowledge of trigonometrical identities and follow recognised procedures. Examples of their uses will occur 6n later studies. (2) PHASOR ME T H O D S . The resultant of two or more phasors may be obtained (a) Graphically or ( b ) Malhernalically. (:I) Thc Gtrrld~icrrlMrthod is performed by sctting 0111 thc

BASIC A . C . THEORY

145

pl1;isors to scale a t the given phase angles, completing the par;illelogram or polygon and ~ncasuring tlrc ~.esult:~~rt. 'l'lrc diagram (Fig 70) shows the method employed. Phasor addition is shown. T o subtract a phasor, reverse its direction and proceed as before, 1.1 the diagram, phasor addition is shown: ( I ) by completing the parallelogram to obtain the resultant of two phasors and then using this resultant with a third phasor to obtain the final resultant, (11) by completing the polygon as shown. Both methods are cumbersome and have the disadvantage that errors are cumulative.

PHASOR DIAGRAM

PHRSOR ADDITION

--I I PHASOR

ADDITION

Fig 70

If the resultant of two individual wz qeforms is required, then either of two procedures can be follov ed. The first procedure uses the known fact that the sum of any two sine waves of the same frequency 1; itself a sine wave. Thus any instantaneous value on the resulta , t wave is the sum of the individual instantaneous values taker from the other waves. Each waveform is drawn graphically in accordance with the method already outlined, care being ta.

Example 71. A coil of unknown inductance and resistance is connected in series with a 25R, non-inductive resistor across 250V, 50Hz malns. The p.d. across the resistor is found to be l5OV and across the coil 180V. Calculate the resistance and ~nductanceof the coil and also find its power factor. The circuit diagram and phasor diagram (Fig 123) are shown.

Fig 123 This example is important in that it involves basic fundamentals and yet has a simple solution. The phasor diagram is first explained with the various voltage drops considered in detail. V , is the voltage drop across resistor R = IR. V , is the voltage drop across the coil and is the resultant of two voltage drops, V , across the resistance of the coil = Ir and V, across the reactance of the coil = IX,. V , is in phase with current and V L is 90' ahead of the current. From the phasor diagram i t is apparent that V is the resultant of V , and V, and that the expression given for simple phasor summation can bc applicd .hcrc. .I~-. Thus V = \/I.', + 2VRV,cos4,, 01.250' 150' + 180' + 2 x 150 x 180 x cos gL :. 62 500 = 22 500 + 32 400 + 54 000 cos 4, or 54 000 cos b,. - = 62 500 - 54 900 7600 COS q L= ---- - - = 0,141 (lagging) 0'76 54 000 - 5.4

-

~

THE

A.C.

CIRCUIT ( CONTINUED )

The impcdilncc of thc coil

-

l80 0

;

219

?On

Resistance of coil = Z cos 4, = 30 x 0.141 = 4.23Q Reactance of coil = J302 - 4 ~ 2 = 3 ~29-7i2 29.7 Inductance of coil = = 0.0945H 2 x 3..14 x 50 Power factor of coil = cos 4, = 0.141 (lagging). Example 72. A moving-iron voltmeter with a resistance of 1732Q and an inductance of 0.625H registers 1 lOV with maximum deflection on a 50Hz, a.c. circuit. It is required to be placed in a 230V, 50Hz a.c, circuit in series with a non-inductive resistor. Find the value of R, the required resistor. Xofmeter = 2 x 3.14 x 50 x 0,625 = 314 x 0.625 = 196.250 Z of meter = J 1 7 3 2 ~ + 196.25' = 100J11.32' + 1.96' 110 Current for full-scale deflection = -= 0.063A 1 744 On 230V. New circuit impedance must be and 36502 = (1732 13 322 500 = (1732 ( 1 732

+ R)'

+ R)' + 196.252 + R)2 + 38 433.7

= 13 284 066

(1732 + R)2 = 13.28 x 106 or 1732 + R = 3.64 x lo3 = 3640 R = 3640 - 1732 = 1908R. Example 73. A coil of resistance 10R and inductance 0.1H is connected in series with a capacitor of capacitance 150pF, across a 200V, 50Hz supply. Calculate (a) the inductive reactance, (b) the capacitive reactance (c) the circuit impedance (d) the circuit current (e) the circuit power factor (f) the voltage drop across the coil (g) the voltage drog across the capacitor. (a) Inductive reactance = 2nfL = 2 x 3.14 x 50 x 0.1 = 3 1.4Q (b)

Capacitive reactance

- = - :

27tfC 2 x 3.14 x 50x 150 lo3 =21.2* 3.14 x 15 (c) Resultant reactance = 31.4 - 21.2 = 10.2R (inductive) = J102 + 1 0 . 2 ~ Impedance =

200 (d) Circuit current = -= 14A 14.28 lo - 0.7 (lagging)--since the circuit (e) Power factor = -14.28 reactance is net inductive (f)

Impedance ofcoil

= ,/lo2

+ 31.42 =

dl00

a14 x33R33 = 462V Voltage drop across coil

+ 986

=

=

=

(g) Voltage drop across capacitor = 14 x 21.2 = 296.8 = 297V. Note. The point already made in connection with resonance. Although resonance is not occurring here, the condition is working towards this and large voltages can be built up across coniponcnls. Thus thc fuct that thc voltt~gcsacross the coil and capacitor are larger than the supply voltage is in accordance with theory and all the values are in order.

I

CHAPTER 9

PRACTICE EXAMPLES 1.

A circuit has a resistance of 3R and an inductance of 0.01 H. The voltage across its ends is 60V and the frequency is 50Hz. Calculate (a) the impedance ( b ) the power factor (c) the power absorbed.

2.

A 100W lamp for a lOOV supply, is placed across a 220V supply. What value of resistance must be placed in series with it so that it will work under its proper conditions? If a coil is used instead of the resistor and if the resistance of the coil is small compared to its reactance, 'what is the inductance of the coil? The frequency is 5OHz.rWhat is the total power absorbed in each case?

3.

An inductive load takes a current of 15A from a 240V, 50Hz supply and the power absorbed is 2.5kW. Calculate (a) the power factor of the load (b) the resistance, reactance and impedance of the load. Draw a phasor diagram showing the voltage drops and the current components.

4.

Two inductive circuits A and B :\re conncctcd in series across 230V, 50Hz mains. The resistance values are A 1200 : B 100R. The inductance values are A 250mH; B 400mH. Calculate (a) the current (b) the phase difference between the supply voltage and current (c) the voltages across A and B (d) the phase difference between these voltages.

5.

Two coils are connected in series. When 2A d.c. is passed through the circuit, the voltage drop across the coils is 20V and 30V respectively. When passing 2A a.c. at 40Hz. the voltage drop across the coils is 140V and lOOV respectively. If the two coils in series are connected to a 230V, 50Hz supply, find the current flowing.

,6.'

' A simple transmission line has a resistance of 10 and A --'reactance at normal frequency of 2.5Q. It supplies a factory with 750kW, O$pf (lagging) at a voltage of 3.3kV. Determine the voltage at the generator and its power factor. Find also the output of the generator and draw the phasor diagram.

cr>TYT

-:--

A non-inductive resistor of 8 R is connected in series with an inductive load and the combination placed across a lOOV supply. A voltmeter (taking negligible current) is connected across the load and then across the resistor and indicates 48V and 64V respectively. Calculate (a) the power absorbed by the load (b) the powerabsorbed by the resistor (c) the total power taken from the supply (d) the power factors of the load and whole circuit.

'netted

A circuit, consisting of a resistor and a capacitor con-. in series across a 200V 4 0 H i supply, takes a current of 6.66A. When the frequency is increased to 5OHz and the \~olt;lpe maintained at 200V. the current becomes RA. ( ' i t l ~ ~ ~ 111r l i ~vitluc l ~ ol' r c ~ i s t a n c cand cupacirance and sketch a phasor diagram (not to scale) for either frequency.

9.

A coil, having an inductance of 0.5H and a resistance of 6012, is connected in series with a 10pF capacitor. The combination so formed is now connected across a sinusoidal supply and it is found that, at resonance, the p.d. across the capacitor is 100V. Calculate the current flowing in the circuit under this condition. Sketch the phasor diagram (not to scale).

10.

A certaln coll has a resistance of 400R and, when connected to a 60Hz supply, an impedance of 438R. If the coil is connected in series with a 40pF capacitor and a p.d. of 200V, 50Hz is applied to the circuit, find the current and the p.d. across the coil and the capacitor.

CHAPTER 10

THE D.C. MOTOR A d.c, machine will run ns i\ motor if its field i ~ n diirmuture nre connected to a suitable supply. The 'motoring' actlon is based on the fundamental law described in Chapter 5, which stated that a force is set u p on a conductor which lies in a magnetic field and carries current. The diagram (Fig 124) shows the basic arrangement for revision purposes.

F IS THE FORCE ON THC CONDUCTOPI TENDING TO TURN THL ARMATURE

-F DlPtCTlON OF FORCE FROM F I R I T PRlNClPLfI

Fig 124 DIRECTION 0 1 : 1:OUCE

The four small diagrams (Fig 125) show that, in order to reverse the direction of the force and thus the direction in which the armature will rotafe, it is necessary to reverse the current in the conductor with respect to the magnetic flux.

FOPCE O'WN (0

1

~ORCU EP ( CU R REN T R E V ER S ED )

(bl

FO RC E UP (FIELD R E V E R S ED ) (c)

F O R C E ' D O WN ( C U R R EN T L FI ELD R E V E R S ED )

(dl

Fig 125 The practical aspect of this rule should be remembered if a motor is found to.run in the incorrect direction when first connected up. Reversal of rotation can be obtained by interchanging the supply leads to the armature circuit. A hand rule

has been developed to help memorise motor action and is comparable with that enunciated in Chapter 6 for the generator. L EF T - H A N D P U L E (Fleming's). The diagram (Fig 126), shows the practical interpretation. The first and second fingers are made to represent the flux and current respectively, as for the right-hand rule. The direction of force o n the conductor will then be represented by the thumb. Note. As for the right-hand rule, the thumb, index finger and second finger must be placed at right angles to each other.

Fig 126 MAGNITUDE OF FORCE

From the first principles set out in Chapter 5, it was shown that the force acting on a conductor in a magnetic field, is proportional to the flux density, the current and the active length of the conductor ia the field. The law was summarised by the fundamental formula F = BIL newtons but, it is pointed out here that, the magnitude of the force also depends on the inclination of the conductor to the direction of the field. It is a maximum when they are at right angles. Example 74. Calculate the force in newtons, as established on ;I conductor, O.5m long, carrying a current of 500A in and at right angles to a magnetic field of uniform density 0.8T. S~ncr.I: = R I I . 'Then F = 0.8 x 500 x 0.5 = 8 x 5 x 5 = 200 newtons. I t should hc nc~rcd.tllat,i f thc conductor is sit~r;~tcd o n ;In a1'matul.c at a 1.irciiu5 I . metres, then the torque produced on the shaft can be expressed as F x r. newton metres. BACK E.M.F. OF A MOTOR

If the motor is i~llowedto ro~.;tedue to the torque produced hq' the armature conductors. then these same conductors will cut ;I mi~gnctic ficlti i ~ n dfrom Faracii~y'slaw i t is known tllat an e.m.f.

THE D.C. MOTOK

225

will be induced, the magnitude of-which will be given by the g c ~ ~ c l . c i lCo ~ S I, ~ I . C ~ ~ I iOl kI ~1 1,

111in

Ilcc~i tlcvclol~ctl 1 1 1 or;~tc~J in :I ~ ~ t l icl; ,I I I c ~ a '11101or starter' or lnorc simply a 'starter' and consists of a tapped resistor and a switching device W I I I L ~ I I c11~11ilc\ 1 1 1 ~ - IC.\I\I:II~L.L. to 1~ ~ > I . ; I C I I I ; I I I r~ c ~ l t ~ ;IIICI ~ ~ cf li ~ i i ~ l l y c u t out altogether. 'l'lle starter may also incorporate other special attachments which may be considered necessary for the safe operation of the motor. Thus it may include protective arrangements to safeguard the motor against the adverse effects of a rt~iuccdworkin3 i,oltagc or :In overcurrent. :lltliougli motor sti1rter.s will be studied later in more detail, i t I \ ' I I ) ~ Iop1.1;11~ l o I I I C I I I I ~ I 11c1.c I l I i i 1 1 llic Sor111of starter ncccssilry

24 1

THE D.C. MOTOR

for any particular machine is mainly decided by the duty for which the motor i s being used. Thus i t may be o f the manuallyoperated or automatie type. Ir trrlty be clealgt~otl far moroly starting and stopping the motor and this may require to be done only once a day. In contrast, the duty moy he such ns to require the motor to be started and stopped almost continually for long periods, as is necessary when working a winch or hoist. Such a starter is mure frequently referred to as a 'controller'. A further point of importance to note is that the starting resistor is not cut out in equal sections from the armature circuit, but that the resistance values of these sections follow a Geometrical Progression. The foregoing observations indicate that the starter is of sufficient importance to require detailed attention. It is an item of equipment which requires both careful and routine maintenance and a thorough knowledge cd its function is necessary both from a theoretical and practical point of view.*

SPEED CONTROL As for the starter, so for the full treatment of speed control, much additional study has yet to be made. It is proposed here. only to deal with the basic methods whereby the speed of a d.c. motor can be controlled and, in this connection, the reader is reminded of the basic deduction N a

4 or @

v

N a - (approx). @

.

Thus varying the voltage applied to the motor armature and keeping the flux constunt will vary the syccd i n direct proportion. This is termed 'Voltage Control'. Varying the flux of the machine and keeping voltage constant will vary the speed in inverse proportion and is termed 'Field Control'. F IE L D C O NTRO L . This is introduced first, since it is the most usual type of control. When a motor is loaded, its speed will vary with load. It may be desired to adjust the speed for any load condition ie keep it constant throughout the working range or to raise it above the normal running speed. Field control is used because its adaptation into the field circuit is easily achieved, control is smooth and effective and little energy is wasted as heat. It must be remembered that this type of control will give speed variation in an upward direction only. It is used for raising speed above normal and as flux is weakened, for the same driving torque, armature current will rise. Note. T a 01,. Thus the motor may be of larger dimensions, if speed variation is required and .interpoles (compoles) must be fitted to ensure good commutation throughout the working range.

VOLTAGE CONTROL. This is achieved in various ways for the different kinds of d.c. motor but the fundamental requirement is to reduce the voltage applied to the machine armature. Thus a large variable rheostat may be connected. in series with the armature o r the latter may be supplied from a variable voltage supply. The method is always used to lower speed and control is in a downward direction only. A wide range in the adjustment of motor speed can be obtained by combining field and voltage control and the methods of applying these are sufficiently important to require further detailed study. T o meet the requirements of the duty for which ~ h cnlotor is rcquircd, the starter and speed controller may be incorporated into one unit which, though simplifying the electrical circuit requirements, apprirs to complicate the theory 01' the control. Since the correct application and use of a motor is of prime importance to the practical engineer, it is hoped that the additional treatment, gven to the d.c. machine in the next book, will be regarded as a necessary continuation of theory and that too long a break is not introduced into the period of study before the necessary advancement is attempted. If such a break does occur in the student's studies of Electrotechnology then, he would be well advised to revise the work undertaken in this chapter..before proceeding to the work in Volume VII. Example 81. The armature of a motor has 660 conductors whose effective length is 410mm; of these, only 0.7 are sirnultaneously in the magnetic field. The flux density is 0.65T, the effective diameter of the armature is 300mm, and each conductor carries a curren't of 80A. If the armature speed is 800 revlmin calculate the output power developed. Force on one conductor is given by F = BZL newtons .: F = 0.65 x 80 x 410 x = 52.0 x 41 x = 2132 x lo-' = 21.32N Number of conductors in the field at any given instant = 0.7 x 660 :. Total force = 21.32 x 0.7 x 660 = 9.85kN

-

Tt~rqllc force

x

r ; ~ d i l or ~ s 7.

-

91150 x O!: newton metrcs 2

Thus T = 9850 x 0.15 = 1477.5Nm 2 x 3.14 x 800 x 9850 x 0.15 And power developed = 60 - 124kW Ex;implc 82. A shunt motor takes 180A. The supply voltage is 4OOV. the resistance of the shunt field is 200f2, and that of the

243

THE D.C. MOTOR

armature 0.02R. I f there i.; a voltage drop of 2V at the brushes. calculute ( 1 1 ) tile hrlck c , m , f ,o f llrr tncltor (h) thc outprlt powcr developed (c) the ellic~ency, neglecting all losses I'or which information is not given. Shunt-held current

=

4IK) -

-

=

212

200 Armature current = 180 - 2 = 178A Armature voltage drop = 178 x 0.02 = 3.56V (a) Back e.m.f. = 400 - 3.56. - 2 (voltage drop at brushes) = 400 - 5.56 = 39444V (b) Output power developed = 394.4

loo0

I

178 = 7 0 . 2 k ~

output - 394.4 x 178 - 70.2 Efficiency =- --4 x 18 input 400 x 180 17.55 or q = ---- = 0.975 or 97.5 per cent. 18 c Example 83. A four-pole d.c. motor with a lap winding is connected to 200V supply mains. The armature carries 600 conductors and has a resistance of 0.3Q. The resistance of the shuntfield circuit is 100R, the flux per pole is 0.02Wb. On no load, the armature current to 3A. If the normal full-load current in the armature is 50A, determine the drop in the speed of the motor from no load to full load. Neglect the effect of armature reaction. Back e.m.f. on no load Eb = 200 - IaoR,, Shunt-field current = = 2A I,,, = 3 - 2 = I A

?E 100

.'. Eb. = 200 - (1 x 0.3) = 200 - 0.3 = 199.7V No-load speed is given by No where: N P 600 x 0.02 x No 4 E~ = -Z@ - O O x - o r 199.7 = 60 A 60 4 Thus 199.7 = 0.2 x No Back e.m.f. E,, on full load is given by: E,, = 200 - I,, R, = 200 - (50 - 2) 0.3 Eb, = 200 - (48 x 0.3) = 200 - 14.4 or E,. = 185.6V Since Eb,= KO, N , and assuming a constant flux, then 0 ,= @, E = or A N, =N ! ! Ebo KG0 Ebo 998'5 185'6 - 5 x 185.6 Thus N , = 199.7 Full-load speed = 928 .revjrnin. "1

a

Example 84. Calculate the first resistance step of a starter for a 240V shunt motor having an armature resistance of OaSR, if the maximum current limit is 60A and the lower limit about 45A. Let R, = the total resistance of the series resistor put into the armature circuit. Then if 1,- is the armature current at start I,, = 60A 240 - 240 and also I,, = R, R, 0.5 R, -.s

+

+

As the motor starts and ;~cceleratesup to speed, the starter Ilandle is kept in position until the current falls to 45A. Thus the starting resistance is still in citcuit, but a back e.m.f. is building 10 ii !h;tl V U ~ yivcn U ~ by kqb, Here E,, = 240 - 45(3.5 + 0.5) = 240 - 45 x 4 = 240 - 180 = 60V At this stage the handle is moved and a seztion of the starting resistor is cut out. Let R, be the new value of the total starter resistance. The current rises to 60A but the back e.m.f. does not change until the motor speed changes. Thus at the instant of moving the handle 240 = Eb, I, (R, R1) o r 240 = 60 + 60 (0.5 + R , ) whence

+

+

or R, = 3 - 0.5 * = 2.5R. Thus the resistance removed during the first movement of the handle after switching on, is 3.5 - 2.5 = 1R The first resistance step is thus 1R.

THE D.C. MOTOR

245

CHAPTER 10

PRACTICE EXAMPLES 1.

A 1 10V series motor has a resistance of 0- 1 m . Determine its back e.m.F. when developing a shaft output of 7.5kW when the efficiency is 85 per cent.

L.

A 500V d.c. shunt motor has an input of 90kW when loaded. The armature and field resistances arc 0.ln and lOOS2 respectively. Calculate the value of the back e.m.f.

3.

A 460V, d.c. motor takes an armature current of 10A at no load. At full load the armature currmt is 300A. If the resistance of the armature is 0.025R, what is the value of the back e.m.f. at no load and full load.

4.

An armature w~ndingof a d.c. motor conslsts of 240 conductors arranged in four parallel paths on an armature whose effective length and diameter are 400mm and 300mm respectively. Assuming that the average flux density in the air gap is 1.2T and that thd input to the armature is &A, calculate (a) the force in newtons and the torque in newton metres developed by one conductor (b) the total torquc developed by the complete winding, assuming that all the conductors are effective (c) the power output of the armature in watts, if the speed is 800 revlmin.

5.

A maripe shunt motor is used for driving a3'freshwater' pump and is found to take an armature current of 25A at 220V, when running on full load. The speed is measured to be 725 revlmin and the armature resistance is 0.252. If the field strength is reduced by 10 per cent by means of the speed regulator and the torque remqins unchanged, determine the steady speed ultimately attained and the armature current.

6.

A shunt generator delivers 50kW at 250V and 400 rev/mm. The armature and field resistances are 0.02Q and 50R respectively. Calculate the speed of the machine when running as a shunt motor taking 50kW input at 250V. Allow 2V for brush-contact drop.

7.

A 105V,3kW d.c. shunt motor has a full-load efficiency of 82 per cent. The armature and field resistances are 0.25R and 90R respectively. The full-load speed of the motor is 1000 revlmln. Neglecting armature reaction and brush drop, calculate the speed at which the motor will run at no load if the line current at no load is 3.5A. Calculate the resistance to be added to the armature circuit, in order to reduce the speed to 800 revlmin, the torque remaining constant at fullload value.

8.

A shunt motor runs at 1000 ~ v / m i nwhen cold, taking 50A from a 230V supply. If the armature and field windings both increase in average temperature from 15°C to 60°C, as the motor warms up; determine the speed when the motor is warm, given that the armature resistance is 0 . 2 0 and the field resistance 20012 at 15°C and that the total current drawn from the supply remains constant. Neglect brush drop and armature reaction and assume the magnetic circuit to be unsaturated. (Resistivity temperature coefficient 0.40 per cent from and at 15°C.)

9.

A four-pole, shunt motor has a wave-wound'armature having 294 conductors. The flux per pole is 0.025Wb and the resistance of the armature is 0.35R. Calculate (a) the speed of the armature (b) the torque developed,'when the armature is. taking' a current of 200A from a 230V supply.

10.

A shunt motor runs at 600 revlmin from a 230V supply when taking a line current of %A. Its armature and field resistances are 0.4R and 104.5R respectively. Neglecting the effects of armature reaction and allowing a 2V brush drop, calculate (a) the no-load speed if the no-load line current is 5A. (b) the resistance to be placed in the armatulrecircuit in order to reduce the speed to 500 revlmin when taking a line current of 5OA. (c) The percentilge reduction ip the flux per pole in order that the speed may be 750 revlmin, when taking an ; % m a t u r ecurrent of 3OA with no added resistance in the arrnarure clrcuit.

4

CHAPTER 1 1

A.C. CIKCUL'I'S (contisued) ANL) SYSTEMS POWER IN THE A . C . ClRCUIT

From the various circuit conditions considered in Chapter 9, ie resistance in series with inductive reactance, resistance in

series with capacitive reactance and resistance in series with both inductive and capacitive reactance; it was seen that, the current flowing was sinusoidal and displaced from the applied sinusoidal voltage by an angle 4, termed the phase angle of the circuit. The general expressions were : For voltage v = Vmsin wt and for current i = I, sin (or - 4) a lagging phase angle being assumed for convenience. The instantaneous power p = vi = VmImsin o t sin(wt - 4) {cos $ - cos (*Of - 4) o r p = VmIm

2

= VIcos $ - VIcos ( 2 0 1 and average power P = VI cos 4 - 0

4)

1'

or P = VIcos 4 VI is frequently called the 'apparent power' of the circuit and P is referred to as the 'active power'. Then active power = apparent power x power factor. The reason for calling cos 4 the 'power factor' can now be readily seen. I t is the factor by which the apparent power must be multiplied to obtain the active value of power expended in u circuit. active power or cos 4 = P =I*R So power factor = apparent power VI I Z I = as deduced earlier. The following is also of interest. Since active power = apparent power x power factor then P (watts) = VI (volt amperes) x cos 4 Thus wattage is given by the volt amperes multiplied by the power factor. This can be expressed by W = V A cos 4 or k W = k VA cos 4. Note. The term kVA is an accepted method of.giving the rating of an a.c. generator, motor or transformer and it must be remembered that it does not indicate the power rating. More information is required before the latter can be deduced, and the power factor is usually specified at the same time. The volt

5

amperes o r V A of a circuit is a term in itself but more has yet to be said about its usage. It has been retained from the early days of electrical engineering, before standardising of terms and symbols was recognised as being beneficial and V A o r k V A , as a rating, is now used internationally for a x . circuits and machines. ACTIVE AND REACTIVE COMPONENTS

These terms are usually used in connection with current but under certain conditions can be applied 'to voltage and power. Consider the phasor diagram (Fig 136), for a simple a.c. circuit with current lagging the voltage by an angle 4. If tlie current I is considered to be split into its two quadrature components I, and I, as shown then I, = I cos 4 and I; = I sin 4 .

Fig 136 Since I c o s 4 is a current, in phase with the voltage V and we know VI cos 4 is the measure of the power expended in a circuit, then it appears that I cos 4 is the component of current which is responsible for pewer dissipation. Thus I c o s 4 is called the active power, wattful o r working component of current. Similarly I, = / sin 4, being always at right angles to voltage, is responsible for no power and is called the reactive, wattless o r idle component of current. The example further illustrates these terms. Example 85. A single-phase a.c. motor of 15kW and 90 per cent efficiency is run from a single-phase supply of 400V. Find the current taken from the mains, if the motor operates at 0.8 power factor (lagging). What is the value of the active current, the reactive current and the motor rating in volt amperes. Motor power output = 15k W = 15 x 1000 watts Motor power input = l5 low x 100 watts 90 I'llc \ult ;~tnpercrating

=

16'67 - 20.84kVA 0.8

Thc linc clirrent is ohtilined by dividing the volt-:~mpercv ; ~ l ~ ~ r 20.84 r 1OOO = 52.1A by the supply voltage. Thus I = 400Active component ol'cut.ret~lla= 52 1 x 0.8 = J1,7A, Reactive component of current = I, = 52.1 x 0.6 = 31.3A. Note. If 16 667 was divided by 400, then I, would have been obtained directly. 16 667 Thus 1, = -= 41.7A. I could then be, obtained by 400 4 1-7 - 52.1A and I. as before, by I sin 9 = 52.1 x 0.6. It is well 0.8 to point out the simple relation for sin 4 being 0.6 when cos 4 is 0.8. This is obviously referring to a right-angled triangle of sides 10, 8 and 6. Similarly for examples, cos 4 is frequently given as 0.707 or sometimes 0.7. This is referrin8 to a righ--angled isoscles triangle and sin 9 in this case is also 0.707 or 0.7 (approx). THE PARALLEL CIRCUIT

The parallel circuit is being treated under a separate heading, to remind the student that procedure is different to that for the series circuit. Nevertheless it will be seen that the method employed follows the familiar technique of phasor summation, ie that of resolving into the horizontal and vertical components or, to bc marc in line with the littest terms introduced in this chapter, into active and reactive components. The branches 01' the parallel circuit are made up of simple R, X, or .Y, values in series, and all work done in this connection will be in no ~ 3 4 ' altered. For a parallel circuit it is pointed out that the same voltage is applied to all branches and it is usual to work with V as the reference for the phasor diagram. The current condition is often written as I = 7, + f, + T3 etc. The dash above the I is to remind one that, this is a phasor summation and not an arithmetical one. Thus all correct operations for a phasor summation must be performed. INDUCTIVE IMPEDANCES IN PARALLEL

Assume two inductive impedances to be connected in parallel as shown in the diagram (Fig 137). Impedance Z1is made up of a resistance R, and inductive reactance X, whereas Z , is made up of resistance R, and inductive reactance X,. The phasor diagram and circuit relationships are also shown. Since V is common to both branches it is used as the reference phasor. The .problem is to find I where I = f1 + f2.

Fig 137

Here I, =

v and I, = v I

2 2

Resolving into active and reactive components, using arbitniry signs, we hnve I, = I, cos 4, + I2 cos and I, = - I , sin 4, - I , sin 4,. I t should be remembered that these phasors are vertically downwards. Then I =

d m 2and cos 4 = I;.Here cos 4 is the power

factor of the whole circuit. Example 86. In the circuit shown above, let R, = 3R and X, = 4while R, = 822 and X, = 642. If the applied voltage is 20V, find the total current supplied and the power factor of the complete circuit. Find also the total power expended. 2, = J i i 7 - T T 7 = 4 P T i 2 = @ i = 5 5 n 20 = 4A Then I, = 5 z2=4-= J m = m = l 0 ~ 20 and I, = - = 2A 10 3 4 cos 4, = - = 0.6 (lagging) sin 4, = - = 0.8 5 5 8 6 cos 4, = - = 0.8 (lagging) sin 4, = - = 0.6 10 10 Also I, = (4 x 0.6) + (2 x 0.8) = 2.4 + 1.6 = 4A I , = - (4 x 0.8) - (2 x 0.6) = - 3.2 - 1.2 = -4.4A Whence I

=

v-4T-+4T = fltT3

=

5 Y5A 4 C~rcuitpower factor cos 4 = --- = 0.67 (lagging) 5.95 Power expended = 20 x 5.95 x 0.67 = 80W The above can be checked thus: Power in b n n c h I = I I 2 R 1= 4, x 3 = 48W Power in bianch 2 = I Z 2 R 2= Z3 x 8 = 32W Total 8 0 ~ . =

A.C.CIRCUITS (continued) AND SYSTEMS

25 1

INDUCTIVE AND CAPAClTlVE IMPEDANCES IN PARALLEL

Thc prowdurc for solviny prablcma, rtmwcir~tcdwitti tliin lypc of circuit, follows that outlined above, except that due allowance is milde for the directions and signs when addin8 the rcrrctivc components. Thus in the diagram (Fig 138). impedance Z, is made up of resistance R, and capacitive reactance X , in series. The phasor for the reactive component of current is drawn vertically upwards and is allocated a +ve sign, whereas the reactive component of current for branch 1 is allocated a -ve sign. The total of the reactive components is thus a difference, as will be noted. The voltage is again used as the reference for the phasor diagram.

Fig 138 As before I, = I , cos 4 , + I , cos 4, and I, = - I , sin 4 , + I , sin 4,. I, will carry either a + ve or - ve sign, decided by the relative values of I, sin 4 , and I, sin 4,. Thus the resulting reactive component will act either upwards o r downwards and the resultant circuit current miiy kx lagging or leading .as shown ,by the example. As before I I = dm and cos 4 = . ; The qualifying term lagging o r leading is decided by the sign of I,. Example 87. A circuit consists of two branches in parallel. Branch A consists of a 20R resistor in series with a 0.07H inductor, while branch B consists of a 60pF capacitor in series with a 50R resistor. Calculate the mains current and the circuit power factor, if the voltage is 200V at 5OHz. Branch A. XA = 2 x f ~ =2 x 3.14 x 50 x 0.07 = 314 x 0.07 = 2252 RA = 20Q ... Z A = .JFT@ = = 29.7R 20 =200 - 6.74A and cos 4 'Thus IA= 29 7 A - ZA 29.7 = 0.674 (lagging) 22 sin 4, = ?A = - = 0.74 ZA 29.7

a

Branch B. X , =

cos

=

1

--

2xfC

1 o6 2 x 3.14 x 50 x 60

-

10 3 3.14 x 6

50 5 = -= 0.686 (leading) 2, 72.8

Then I, = (6.74 x-0.674)

+

(2.75 x 0.68) -t 1.885 = 6.43A I, = -(6.74 x 0.74)+(2.75 x 0.728) = 4.55

+

= - 5 2.005 = - 2.995A. Note. The mains current will lag, since the effect o f the inductive branch predominates. I= J I m 2 = .\/6.432 + 2.9952 = d 4 1 . 4 + 9 = = 7.1A

a

cos

4

I, =*6

= 0.902 (lagging). 1 7 . 1 The mains current is 7.1 amperes and the circuit operates at a lagging power factor of 0.9. =

P A R A LLEL RESONA~JCE

Before passing on to the more practical applications of parallel working, it would be well to point out that, a condition of resonance can occur for the parallel circuit. This condition is often termed 'current resonance' to distinguish it from 'voltage resonance' as dealt with for the series circuit. It will be seen, from the example set out above, that a condition can arise when I , sin 4, = I, sin 4, and as -these are the reactive components of cul.~.cntsin inductive and capacitive branches, then they will oppose each other tending to produce a total reactive component ol'zcto v;tluc Tlic rcnxtining itctivc colnponcnts will tot;~lto givc the line current, since I = +~-0= Ia and the. combined circuit will operate at unity power factor. The supply current will also be minimum especially if the resistance values in the two branches are small compared to the reactance values. This is illustrated by the pli:lsor diagram (Fig 139). from whicll it is .;ccn t I i : ~ t .sincc the power. filctors of both branches are low, the

A . C . C IR C U IT S

(continued) A N D SYSTEMS

----

253 --

phase angles 4 , and 4 , are large and I, cos 4, and IB cos & are small compared to the reactive components.

Fig 139

\

Large currents can flow in the chokecoil and capacitor branches, which are very much greater than the main supply current and these, are therefore, not supplied from the line. On examining the power waves for an inductor and capacitor, it w ~ l l be seen that they are directly opposite in phase, as are the current waves. It can be assumed that as the capacitor discharges, the power given out is absorbed by the choke in building up its field. When the field collapses, the power released charges the capacitor and there is a current due to oscillation of power between choke and capacitor. Apparatus using such a circuit is an oscillator and has many applications in radio and electrical filter circuits. If no supply is available the current is not maintained, due to energy loss in the circuit resistance which, however small. cannot be neglected. To maintain the oscillatory current, the resistance loss must be supplied at the correct frequency from the external supply source. POWER-FACTOR IMPROVEMENT The full meaning and advantages of this technique, which IS much used in practical electrical engineering work, is best illustrated by the use of an example, as now considered.

Example 88. (a) Two inductive coils of resistance values 5R and 8R and inductance values of 0.02H and 0.01H respectively are connected in parallel across a 240V, 50Hz supply. Find the coil currents, the circuit current and its power factor. The arrangement is shown by the diagram (Fig 140). A phasor diagram is also drawn.

F'ig 140 Branch A. X, = 2xfL = 2 x 3.14 x 50 x 0.02 = 3.14 x 2 = 6.280 Z, = = 1/25 39.4 = = 8.02R

Jm2 +

240 = 29.8A cos 1, = 8.02

-

4,

\644

5 80.2 0.622 (lagging)

= =

6.28 -= 0.78 sin 4, 8.02 Branch B. X, = $ that of b r a n c H , since L is halved = 3.14R -- z,*= = \/64 + 9.9 = \rnn3 = 8.60

,/8r+.nl

= 0.93 (lagging) 3.14 sin 4, = --- = 0.366 8.6 Then I, = (29.8 x 0.622) (27.9 x 0.93) = 18.6 + 26 44.64 1, (29.8 x 0.78) + (27.9 x 0.366) = 23.3 + 10.2

+

-

-

33.5.

It will be noted that the arbitrary - ve sign has not been used here since, both branches are inductive and there is no doubt as to, the resultant current being laggin Then I = w6-m' = IO&G'-ST = 10\/19.8 11.2 = 1OJ31 or I = 55.6A cos 6, = --44'6 - 0.801 (lagging).

+

55.6

A.C. CIRCUITS

(continued) AND SYSTEMS

255

Example (contd) (b) Find the effect on the main circuit cumnt and power factor, if a capacitor of 400pF,was connected across the supply in parallel with the coils. The phasor diagram (Fig 141), shows the new conditions.

Fig 141 1

Branch C . Reactance of capacitor X c = 2xfC 1o6 - lo' ohms Thus Xc = 2 x 3.14 x 50 x 400 - 3.14 x 4

capacitive and there being no resistance in branch C,-nly reactance, then cos & = 0 and sin $c = 1 Again I , acts at 90" to the voltage and is wholly reactive, there being no active component. Then I, as before = 446A and I, = - 23.3 - 10-2+ 30.25 or I, = -33.5 + 30.25 = -3-25A It will be seen that the arbitrary signs have been introduced here, because the reactive current of branch C acts in the opposite direction to that of branches A and B. The circuit current is now: I = d m 2= J44.6' 3 . 2 ~ ~ = 10d4.46' + 0,325' = 10d19.8 +0.106 = 10JI!DC% = 446A

+

44.6

and cos 4 = -= 1.0 ie unity. 44.6 From the above example it is seen that, by connecting a capacitor in parallel with the inductive loads, the total line

-

current is reduced from 55.6 to 44.6 amperes and the overall circuit power factor is improved from 0.8 (lagging) to unity. The resulting advantages of the arrangement are now considered in detail. ADVANTAGES OF P.F. IMPROVEMENT

For the majority of commercial loads, the current lags behind the voltage, due to the inductance of the apparatus or the operating characteristics of motors and control gear. Typical values of power factor are: System supplying lighting loads only: power factor (lagging) = 0.95. System supplying lighting and power loads: power factor (litgglng) = 0 7 5 to 0.85. System supplying power loads: power factor (lagging) = 0.5 to 0.7. The lower the p w e r factor, the greater the line c u r r e n l t x d be__fpr-a9 given load k W or output power rating and the attendant d & a d V a n t a g G E (a) the transmission losses in the supply cables or power lines are increased; these being given by PR, where R is the cable or line resistance. Thus for a given amount of power transmitted, the

----

1

current at 0.7 power factor is - x current at unity power 0.7 factor = 1.43 x current at unity power factor. Also the transmissiorrloss at 0.7 power factor is (1.43)2 x loss at unity power factor = 2 x loss at unity power factor. (b) Because of the larger currents resulting from a low power factor, there will be a greater voltage drop in thc supply lines resulting in a lower voltage a t the load. Conversely the size of the conductors must be increased to keep the voltage drop figure to an acceptable value. (c) Again because of the larger current resulting from a low power factor, the size of the current-carrying conductors in transformers, control-gear and alternators must be larger than need be. This means that the physical dimensions of the equipment must be larger and that advantage is not taken of good design. The equipment is.als.0 more costly. (d) The 'Regulation ,-a term used for the 'sittingdown' of the voltage of generating and transmitting plant, is adversely affected by a low power factor. The lower the power factor, the greater the internal voltage drop in this 1

equipment ie armature reaction and attendant effects are worscnctl. Electricity supply authorities encourage good power factor oprntion by offering ndvnntngcous tnrith, Power fi~ctorurn Bc improved by adding to existing loads, apparatus which will take sufficient leading current to minimise the lagging current of the load. Static capacitors are frequently used for this purpose. Example 89. A 40kW load, operating at 0.707power factor (lagging), is supplied from 500V. 50Hz mains. Calculate (a) the capacitor value required to raise the line power factor to unity (b) the capacitance required to raise the power factor to 0-95 (lagging). kW 40 (a) Load k V A = -cos d 0.707 46000 - 80 800 = 113.15A Load current = 0.707 x 500 - 0.707 = Active component of load current I , = I , cos 4, = 113.15 x 0.707 = 79.997A = 80A Reactive component of load current I , = I, sin 4, = 113.15 x 0.707 = 80A. To nullify this reactive current, a capacitor can be fitted to operate in parallel with the load. This capaoitor must pass a similar value of reactive current as shown by the phasor diagram (Fig 142a). Thus I, must be 80A. Reactance X, of capacitor must be = 6.25n 80

Fig 142

258

REED'S BASIC E L E C T R O T E C H N O ~ Y

Since X ,

(b)

=

1

t

, ; ,

For this part of the problem it will be seen that I, sin 4, is not to be cancelled completely since, the line phase angle is only to be reduced from 4, to 4, and line current to a ncw vnluc I,. This is illustriitcd by the' diagram (Fig 142b). Sincc thc power or active component remains the same, then for this condition V I , cos 4, = 40 000 as before. Also s i n k cos 4, = 0.95 (lagging) sin I , sin 4, = 84.2 x 0.312 = 26.1A.

4,

=

0.3 12 and

The line reactive current component has now to be reduced from 80A to 26.1A = 53.9A. This then must be the new value of I , or a capacitor must be used which takes a current of 53.9A. Thus X

- --500 - lo6 (C being in microfarads) - 53.9 - 2Tcfc - 53.9 x lo2 lo6 x 53.9 orC = 500 x 2 x 3.14 x 50 - 5 x 3.14 - 539 lo' = 343pF. 15.7 It will be noted that 343pF will bring the line power factor to 0.95 and that a further (510 - 363) = 167pF would be needed to bring the value to unity. Since the cost of a capacitor depends on its capacitance value and little advantage is gained by improving the power factor above 0.95, it is not always necessary to achieve unity power factor working. In this way some saving to thc consumer can be clt'ected. I t is important to note that, although power factor is improved, an increased power output is 1101 obti~ined fro111 tllc lontl. Students frcqucntly hnve the erroneous idea that, if for example, the power-factor working ol' a circuit supplying a 5kW motor is improved, then the motor will then give iln output greater than 5kW. This is not the case. All that is achieved is that, by connecting an additional item of . ; t ; ~ t ~ c;\pp;Ir'ntus ' cross the motor. the total line current is rcduccd, ic a conditiori tan be attained when the minimum \ul)l)lv c.t~r.r.crllrc~lt~irctl 101. ; I sl~ccificd powcr output is uscd.

A.C.CIRCUITS(continued)

AND SYSTEMS

259

This minimum supply current reduces all the disadvantages already enumerated hut the motor crlrrent itself remains unaltered.

kW, k V A and kVAr As mentioned earlier in the chapter, the above terlllinology is much used in practical electrical engineering and some revision is necessary, before the full treatment is considered. The diagram (Fig 143a and b) should be considered with the text. For a circuit, where the current and voltage are out of phase, the phasor diagram is as shown. Current I can be resolved into an in-phase o r active component I cos 4 and an out-of-phase o r reactive component I sin 4. I cos 4 is responsible for all the power dissipated by the circuit, since P = VI cos 4, and is also called the power o r wattful component, whilst I sin 4 is responsible for n o power, being at right angles to the voltage, and is called the C wattless o r reactive component.

From the expression P = V I cos 4 it is seen tha.t P'can be the 'active' component of VI (symbol S--see Note), and if the term volt amperes or kilovolt amperes is used for VI then the k V A (kilovolt amperes) can be regarded as being resolved into two components, one of which is the power component. The term W or k W (kilowatts) can be used to describe this component and the other component termed the 'volt amperes reactive' or 'reactive kilovolt amperes' can be designated by V A r or k V A r . If the current phasors of Fig 143a are multiplied by V , the new condition becomes more apparent and leads to a power diagram. The product VI (S) is shown as the volt amperes VI ( V A ) or -= k V A and is referred to as the 'apparent power'. 1000 Since VI cos 4 = P then V A cos 4 = W and k V A cos 4 = k W . k W is a measure of the 'active power', in line with, original definition for power factor. ie the ratio of active power to apparent power.

Thus: power factor or cos

kW 4=-

kVA s&ilarly VI sin 4 (Q-see Nore), or k V A sin 4 is the 'reactive power' o r volt amperes reactive designated by k V A r and from the power diagram (Fig 143b), we have: Apparent Pow'er = .\/True power2 + Reactive power2 Summarising k W = k V A cos 4. k VAr - = k V A sin 4 kVA = z / k W + k ~ ~ r ' kw k VAr cos 4 = - sin 6 = k VA k VA Note. The symbols S , P and Q are recommended as subptitutes for V I , VI cos 4 and VI sin 4 but it is probable that the units :' kilovolt amperes, kilow:ttts nnd kilovolt amperes reactive will continue to be used and shown on the phasor diagrams, since this is the older, though basically inconsistent, practice of the electrical power engineer. The appropriate alternative has been introduced and shown where it is considered to be appropriate in this chapter. For summarising we have: P = VI cos 4 Q = VI sin 4 and S = VI Q = S sin 4 Thus S = d m ' and P = S cos 4 P Q cos 4 = - and sin 4 = -. S S It must be r e m e h e r e d that the k V A values of various loads are not in phase and therefore cannot be added arithmetically. kW values are all active components, are in phase and can be added. k V A r values are reactive components, they can be inpbase or in anti-phase and can be added, provided due allowance is made for the sign. This is shown by the following examples. -'

!

!

-

I

I

I

\

I I

#'

Example 90. Two loads are connected in parallel. Load A 1s 800kVA at 0.6 (lagging). Load B is 700kVA at 0.8 power

I

A . C . C IR C U IT S

(continued)

A N D SYSTEMS

26 1

factor (lagging). Find the total k W, k V A and overall power factor of the joint loads. Scc t l ~ ctlic~yro~n (Fig 144), For load A. cos 4, = 0.6 sin 4, = 0.8 Active power, PA = VIAcos 4, 800 x 0.6 480kW Reactive power, QA = VIAsin 4, = 800 x 0.8 = 640kV Ar For load B. cos 4 , = 0.8 sin 4, = 0.6 Active power, PB'= VIBcos 4B= 700 x 0.8 = 560kW Reactive power, QB = VIBsin 4, = 700 x 0.6 = 420kV Ar Total active power, P = 480 + 560 = 1040kW Total reactive power, Q = 640 + 420 = 1060kVAr 7 .Total apparent powq, S = 1060 = lo3 J1@12 1.06~ = 1 0 3 m 5 = lo3 x 1.485 = 1485kVA * 1040 Overall power factor = -- = 0.7 (lagging). I t85 Problems involving a number of loads, may be best treated by setting out the power-diagram components in tabular fashion as shown. An arrow has been drawn in to illustrate the phasor direction and to remind the student as to which columns can be added arithmetically.

-

-

+

Example 91. A 220V, single-ghusc nltcrnator supplica thc following loads : (a) 20kW at unity power factor for lighting and heating. (b) A 75kW induction motor having an efficiency of 90.5 per cent operating at a power factor of 0.8 (lagging). (c) A synchronous motor taking 50kV A at a power factor of 0.5 (leading). Find the total k V A , current and the power factor of the combined load. Load (a) can be set into the columns directly as shown.

Load (h) Motor power output = rating as given = 75kW 75 = 82-9kW Motor input active power = 0.905 82.9 Apparent power = - = 103.6kVA 0.8 Load (c) can also be set into the columns directly.

262

R EE D

Lord

h IfA 1 h lJA cos (SIX (P)

I

1

1 0 3 . 6 ~ ~ 82.9 --

1

a I

b

4

-

I

I

B A S I C ELECTROTECHNOLOGY

k V ~ r o r 11

kWor-j

i

1

s

20

20-+

I

k V A sin

I

d

(Q,

cos

4

-+

-

I

127.9

+

1-

1

11

62.16 1 ,

1

43 3 -

-

sln

I$ 1

I

1

1

0

I

1 -4

0.8

0.6

0.5

0,866

I

25

/

I

o -

I I

--

1

J

- 18.86 i

Total apparent power (S) = 2/127.9'+

18.9' = 129kV A

Total current = 129

loo' = 588A 220 127.9 Resultant power factor = ---- = 0.99 (lagging). 129 Note reactive component of inductive load predominates, hence the resultant lagging power-factor condition.

(k V A method). Treatment of problems follows lines, similar to those set Out for the 'current method'. The diagram for the load condition is built up by splif?ing the original load k V A into its k W and k V A r components. Since the k W remains the same, then for a new power-factor condition for the supply, the final k V A r value is obtained by reducing the original k V A r by an amount equal to the XC'Ar of the apparatus being added. Such apparatus must use no power and the static capacitor is such an item of equipment. The added k V A r being leading, will reduce the lagging X 1'.4r of the supplv. I t should be noted that if a synchronous rnotc)r. 1s used to obtain a better overall power factor, then this also contributes output power which must be taken into account. . I ' I I I ~W ; I \ I I I U S ~ I ~ ; I I C ( I h y 1 1 1 ~prcvioi~sC X ; I I I I ~ ~ C . Example 92. A 400V, SOHz, 20kW, single-phase induction motor has a full-load efficiency of 91.15 per cent and operates at ;I powcl- factor of 0.87 (lagging). Find the k V A r value of the capacitor to be connected in parallel to improve the circuit pcnver klctor to 0.95 (I;~gpinp).Find also the capaciti~ncevalue 01' tllis c;~p;~citor. Thc di;~gram(Fig 145), illustrates the problem ;111cI ~ o ~ l l t l o l l .

P O W E R - F A C T O R IMPROVEMKNT

A.C. CIRCUITS (conrinltcd) AND SYSTEMS

2 63

Fig 145 Motor output = motor rating as given = 20kW 20 Motor active input power P , = ------ = 21.94kW 0.9115 21'94 ---- 25.22kVA 0.87 Also, since cos 4, = 0.87 then, from tables, sin 4, = 0.493 Thus Q, or S, sin 4, = 25.22 x 0.493 = 12.44kVAr Although the power factor of the circuit is to be improved to cos 42,the power of the circuit is not altered :. P, = P, or S , cos 4, = S , cos 4, whcnce S, cos ., S, = cos b, Motor apparent power S ,

I

I

, I

Again cos $, = 0.95 therefore, from tables, sin 4, = 0.3123 and Q, = 23.1 x 0.3123 = 7.21kVAr Required Q value = 12.44 - 7.21 = 5.23kVAr. This therefore must be the rating- of the capacitor. -5230 Capacitor current I, = -= 13.75A 400 Capacitor reactance

Ii

=

400 - 30.59i2 X, = -

or X,

13.75

=

30.59 =

lo6 2 x 3.14 x 50 x C

where C is the value in microfarads. lo4 lo3 - 104pF. Hence C = 30.59 x 3.14 - 961 -

POLYPHASE WORKING The student who intends to have a good practical knowledge of electrotechnology, must make himself thoroughly conversant with the terms, relationships and theory of polyphase working. The importance of the work now to be covered cannot be too strongly stressed. ~ x p e r i e n c ehas shown that most students consider this part of theory to be 'that little extra, which breaks the camel's back' and accordingly give it insufficient attention a t first. The result is that much hasty revision is necessary when the v;~riouso.c. m;ichines iire to be studied later. Detailed attention to fundamentals will bring long-term advantages and, although the next book will be devoted to more advanced a.c. [ c c l ~ ~ ~ o l o ~g lyi,csul~jcclIlliltIcr now to bc considcrcd m u s ~bc treated as bas~cand essential to such further studies. I t must be both understood and rnemorised. THREE-PHASE SYSTEMS

Universal practice has established 3-phase systems to be the most advantageous for polyphase working. A single-phase supply, as is usual for small installations, can always be obtained

A.C. C IR C U IT S .

(continued) -

AN D SYSTEMS - - -- -

265

-

from a 3-phase system and in this way the relative advantage of either system is available. The 2-phase system is only rarely used und its unimportc~nccdocs not wi~rritnlits study u l this stuyc. More than 3-phase arrangements, such as 6-phase, have relatively fewer and even more specinlised applications and here we confine our investigation to 3-phase working only. Consider a 2-pole magnet, as shown in the diagram (Fig 146). to be rotated inside an external stationary armature or stator. Three coils are shown equally displaced, with 'starts' and 'finishes' marked symmetrically to make .a regi1ar arrangement. Induced e.m.fs. will result in each coil, which are identical in magnitude but displaced in phase by 120 electrical degrees. A phase sequence R-Y-B is assumed ie the rotor turns so that the red-phase voltage reaches its maximum 120' before the yellowphase voltage reaches its maximum and the latter 120" before the blue-phase voltage, as shown by the waveform diagram. A sinusoidal distribution of rotor flux is assumed and that sinewave e.m.f.s are induced. The methods by which this is achieved will be considered under the detailed study of the alternator. The three separate coils can be used to supply three independent single-phase loads, but advantages are obtained by interconnecting the coils or phase windings and the two important methods are described as either the STAR or DELTA connection. STAR OF Y CONNECTION

The diagram (Fig 147). shows the arrangement and it will be seen that here, the three coils or phase windings are connected so that either all the starts or,finishes are joined together to form the star-point ie 'corresponding ends' are connected together. Similarly the supply lines are connected to the free ends, remote from the star-point.

Fig I47 The phasor diagram has been drawn in terms of voltage with the red-phase voltage (V,) used as the reference. The notation being used from now on should be observed. It will be seen that the small letter suffix denotes the phase value, while the capital

letter denotes the line value. The lines have been identified withthe colours of the phases to whose 'starts' they have been connected. The double suffix such as V , - , denotes the voltage between lines, the example being the Red to Yellow line voltage. Assume the condition when the red-phase voltage wave is positive and that the 'start' of the red-phase winding is +ve with respect to the 'finish' or neutral point. Current will flow through the lines and load as shown. For the example and condition being considered this is possible because, for the yellow phase at the same instant, its start will be - ve with respect to its finish, since the yellow phase wnveform is in its -ve half-cycle. Thus for the phasor diagram, the voltage between the red and yellow lincs is,obtained by the phasor difference of Vr and V , . Since a phasor din'erence is being considered, thc resultant I S obtained by reversing one phasor with respect to the other and completing the parallelogram. From the deduction set out below, it will be seen that the line voltage is d 3 times a phase voltage. This relation also holds for the other lines and the associated phases. A further point of importance for the star connection, is that the line current equals the phase current or

IL =

Ipp

Consider the phasor diagram. Let the line voltage V R - , = 2x x d3 But - = cos 30" :. x = --- Vr or 2x = 43 V , vr 2 Hence V R - , = .\/3Vr or the voltage between lines = .\/3 x a phase voltage. Thus V , = *d3 x V,,. For a star connection the following must therefore be remembered. Line voltage = .\/3 Phase voltage or V = V' 3 V,, = 1.732V,, Line current = Phase current or I = I,, It will be noted that the subscript L , as in VLand IL, is omitted when generalising. This is usual and both V and I can be assumed to be line values. Ag;r~nthe relations deduced have been derived for an alternator or source of supply but they also rcl;itc to a stiir-conncctcd load a s the cxnmplc shows. Example 93. Three 50R resistors are connected in star across 41 5V, 3-phase mains. Calculate the line and phase currents and the power taken from the supply. Since the load is balanced. the voltage across each resistance is tllc correct phase volt;ipc. 'Thus V,,

=

415 - 240V v'3

A.C. CIRCUITS

(continued) AND SYSTEMS

267

Phase current = line current or I,, = 1

?4!! 50

-

4,HA

Power dissipcited by one phusc of load =

Iph1 Rph

= 4.82 x 50 = 2.4 x 4.8 x 100 = 24 x 48 = 1152W or 1.152kW

and 3-phase power from the supply = 3 x 1.152 = 3.456kW = 3.5kW. USE OF THE NEUTRAL

One obvious use of the star-connection is for distribution, since two voltages are available to the consumer, one for lighting and the other for power. Either I-phase or Iphase loading is also possible and this is shown by the diagram (Fig 148). c

Ibuu~

Fig 148 BALANCED LOAD

A 3-phase load is said to be balanced when the currents in all three phases are equal and their phase angles are the same. If an instant in time is considered, as shown, on the diagram (Fig 149).

I

Fig 149

it will be seen that the sum of instantaneous values of the currents i, i, i, = 0. Since these currents meet at the load neutral point an'd the resultant flows tHrough the neutral line, then the neutral carries no current and need not be used for balanced loading,

+ +

UNBALANCED LOAD

A neutral must be used if the load phase currents are unequal

or if their phase angles are different. The neutral line will carry the unbalanced current ie the resultant of the three line currents. Since this ncutral current is a phasor sum, it can be obtained graphically o r mathematically, as shown by the example. Example 94. Thc loads of n 4:wirc. 3-phase system are: Red line to neutral current = 50A, power factor of 0.707 (lagging). Yellow line to neutral current = 40A, power factor = 0.866 (lagging) Blue line to neutral current = MA, power factor = 0.707 (leading) Determine the value of the current in the neutral wire. The solution is worked with reference to the diagram (Fig 150). I, = 50A lagging the voltage by 45" since cos 45 = 0.707 Iry = 40A lagging the voltage by 30" since cos 30 = 0.866 I, = 40A leading the voltage by 45" since cos 45 = 0.707 Resolving into horizontal and vertical components. I,, = (50 x cos 45) - (40 x cos 30) - (40 x cos 15) = (50 x 0.707) - (40 x 0.866) - (40 x 0.966) = 35.35 - 34.64 - 38.64 = -37.93A I, = - (50 x sin 45) - (40 x sin 30) + (40 x sin 15) = - (50 x 0.707) - (40 x 0.5) + (40 x 0.259) - - 35.35 - 20 + 10.36 = - 44.99

4

NOT TO SCALE

Fig I50

Current in the neutral is the resultant

cos 8 = - 37'93 = - 0.65'3 58.9 0 = 49.5". Note. The - ve sign gives the quadrant in which I, lies. This is shown on the diagram. A (MESH) CONNECTION The arrangement is shown in the diagram (Fig 151). For this connection, the three-phase windings are arranged to form a closed circuit by connecting 'uncorresponding ends' ie the start of one phase to the ,finish of another phase. Thus R start is connected to B finish, Y start to R finish, etc. T h e same reasoning as was introduced for Fig 148, is applied here, except that the voltages are considered. Thus if the diagram represented the three equal phase voltages, then it would be seen that, for any instant of time, the sum of the instantaneous values of voltages v, + v, + v, = 0. For the mesh or closed winding, since the sum of the instantaneous voltages is zero, no circulating current flows round the mesh. The lines are taken from the junction points and for this connection, i t is obvious that the voltage developed across a phase is the voltage provided for the conncctcd lines. Thus V , or V = V,, It will be noted that the lines have been identified with the colours of the phases to whose starts they have been connected. Assume the condition when the red-phase voltage is positive ie DELTA OR

I

I

I

I

1

Fig 151 that the start of the red-phase winding ; + ve with respect to the finish. Current will flow through fron R phase into R line as shown. At this same instant the voltagc in B phase is negative ir its finish is +ve with respect to its star . Thus it is also correctly connected for feeding current into the i line and a line current

is thus obtained by considering the phasor d~ferenceof two phase currents. The resultant line current is obtained by reversing a phase current (I,) and combining it with I , as shown in the diagram (Fig 151 ). As before x = I , cos 30' = I,, cos 30" or x = 1 . 3 Iph and as I, = 2 x = 2 t/31,h 2 Hence I, or I = t/ 31ph Thus for a delta connection, Line voltage = Phase voltage and Line current = 1 3 Phase current. This rci;~rionshipcan bc deduced for any line and the connected phases and will give the same result. As before V and I ;IIT wed for line v;rl\lcs and I,' and I,, for phase values. L

Example 95. Three 5 0 0 resistors are connected in delta across 41 5V, 3-phase lines. Calculate the line and phase currents and the power taken from the mains. Voltage across I-phase resistor = 415V 41 5 Current in one phase of load = - = 8.3A 50 Since the load is balanced, line current = 11 3 x 8.3 amperes = 1.732 x 8.3 = 14.78A Power in one phase of load = 8.32 x 50 = 3445W Power in three phases of load = 3 x 3,445 = 10.3kW. &

THREE- PHASE P O W E R

For a Star-connected load. V = t'3Vph and I = I,, The power expended in one phase = V p hI,, cos and the power expended in three phases = 3 v,, I,, cos 4 ('onvert~ngto line values. the above becomes:

v

4 3 V I cos 4

Three-phase power = 3 --- I cos \

3

or P = \ Dc>lt(l-conncctcdlo;~d1,' =. V,, and I = t'31,h 'l'he power expended in one phase = I.', I,, cos 4 and the power expended in three phases = 3 v,, I,, cos 4 Converting to line values the above becomes: I Thrcc-ptlasc 1x)wcr = 3 1,' - - cos 4 b 3 Fclr

;I

4

(continued) AND SYSTEMS

A.C. CIRCUITS

---

-- -

-

-

--

- -

-

2,

-

Thus the general expression holds, irrespective of the type of connection. nilmely for either ;I st;lr or delt;l connection ' I I I I C C . ~ ) ~ I ~ I ~ C~ ) t ) w o t i~ * ~ I V C I Ijy I \ -1 I''/ L ' L I ~(1 Example 96. A 75kW, 4MV, 3-phi\se, deltil-connected induction lilotor has u I'ull-load ellic!ency ol' 91 per cent : ~ n d operates at a power factor of 0.9 (lagging). Calculate the line and phase currents at full load. Output power = 75 x lo3 watts 75 103 100 wafts Input power = 91 also P = 4 3 V I c o s 4 So 4 3 x 400 x I x 0.9 = 75 x lo5 91 75 lo4 750 lo' amperes and I = 1 . 7 3 2 ~ 4 ~ 9 ~ 19 . 17 3 2 ~ 3 6 ~ 9 1 *

or I = 7500 lo' = 132.2A 5674 132.2 Motor phase current = ---- = 76.3A. v'3

k V A , k W AND k VAr Since power factor can be defined as the ratio of true power to apparent power, this can be applied to three-phase workin g . Thus : active power power factor = ---nppnrcnt p O W s r Again from earlier deductions, we see that irrespective of star or delta connection P = 4 3 V I cos d. Accordingly: THREE-PHASE

cos

4

P

=

4 3 VI It follows that for three-phase working, in order that the definition for power factor should apply, apparent power ( S ) = 4 3 VI Note the introduction of d3,--distinguishing this condition from single-phase working. P or k W and, in exactly the Again it is known that cos 4 = S kVA ' vr 3 same way, it follows that; Three-phase k VA = ~loo0 This final deduction is most important. Example 97. A 3-phase, 400V motor takes a current of 163A when the output is 9kW. Calculate (a) the k V A input, ( b ) rhc power factor, if the efficiency at this load is 89 per cent.

3 VI = 1 3 x 400 x 16.5 loo0 loo0 = 6.6 x 1,732 = 1 l.43kV A ( b ) Output power = 9kW 9.0 True active input power = -= 10.11kW 0.89 active power - 10.1 1 So power factor = apparent power 11.43 = 0.88 (lagging). Esnmple 97a. A 3-phase, star-connected alternator supplies a delta-connected induction motor at 600V. The current taken is 40A. Find ( a ) the phase voltage of the alternator (b) the current i n c i ~ c hp h i l ~ cof thc motor. Refer to the diagram (Fig 152). ( a ) k V A input

=

--

Fig 152

--

( a ) For a star connection V = V,, 600 .'. V,, = v = --= 346V. This is the alternator phase v3 \ 3 voltale I -(b) For a delta connection I,, = - 40 = 23.1A 4 3 v'3 This is the current in each motor phase. (c) I f the motor operates at a power factor of 0-8 (lagging) and an efficiency of 88 per cent, find the k V A rating of the t the motor. alternator and nower o u t ~ u of

Motor ilpparent input power

~ ~ V l' 3 loo0 - 1.732 x 600 x 40 1000 = 41.6kV A = ----

True active input power Output power

=

-

41.6 x 0.8 = 33.28kW 33.28 x 88 - = 2 9 , 2 1 k W IotK)

Example 98. A delta-connected load is shown by the diagram (Fig 153). 11'thc supply volti~peis 40OV, 50111. ~ i ~ l ~ tthe ~ lredi ~ t line current. Assume the currents as shown and maintain the correct phase sequence of R, Y and B. If the red-line current is assumed to feed current into the red phase of the load, the blue-phase current would be in the opposite sense or a phasor difference is involved. " b ~ b.

Fig 153

#YB

400 Here I, = - = 4A in phase with V,, 100 400 Also I, = - = 3.7A leading V,, by 90" 106 Reversing I, and using the modified Cosine Rule 21Jb CoS 30' = dl,*

+

+

~

CHAPTER 11 PRACTICE EXAMPLES 1.

A coil consumes 300W when connected into a d.c. circuit of 60V. It consumes 1200W when connected into an a.c. circuit of 130V. What is the reactance of the coil?

2.

A circuit consists of two branches A and. B in parallel. Branch A has n resistance of I 2 8 and a reactance of 3R, whilst the values of branch B are 8R and 20R respectively. The circuit is supplied at 100V. Calculate the current in each hrnnch ;tnd t l supply ~ current.

3.

An inductive circuit of resistance SOQ and inductance 0.02H is connected in parallel with a capacitor of value 25pF across a 200V, 50Hz supply. Find the total current taken from the supply and its phase angle.

4.

Two coils of resistances 8R and 10R and inductances 0.02H and O.05H respectively, are connected in parallel across IOOV, 50Hz mains. A capacitor of capacitance value 80pF in series with a resistor of 20f2 is then connected in parallel with the coils. Find the total current taken from the mains and its phase angle with respect to the applied voltage.

5.

A single-phase motor has an input of 50.6A at 240V, the power input being IOkW, and the output 9kW. Calculate the value of the apparent power, power factor and of the efficiency.

6.

A slngle-phase motor running from a 230V, 50Hz supply takes a current of 11.6A when giving an output of 1.5kW, the eflicicricy k i n g 80 per cent. Calculate the capacitance required to bring the power factor of the supply current to 0.95 (1;1gging) C;llctrlntc ;~lso thc X-l'Ar rating of tllc capacitor.

7.

The load taken from a slngle-phase supply eonsists of: (a) Filament lamp load of lOkW at unity power factor. ( b ) Motor Io;~dof' 80kV ,A a t 0.8 power factor (lagging). (c) hlotor load of 4OkV A at 0.7 power factor (leading).

a.

A . C . C IR C U IT S

(continued)

AND SYSTEMS

275

Calculate the total load taken from the supply in k W and V A a n d the power factor of the comh~nedload Find the '111111113' CUlI'CI11 I f 11\12 YU})I)!Y V O I L L ~ ~ 15 C 250V In

8. ,-,bf

Three equal impedances of 10SI, ench with u phusc i~nplc 30" (lagging), constitute a load on a 3-phase alternator, giving lOOV per phase. Find the current per line and the total power when connected as shown. (a) Alternator in star, load in star. (b) Alternator in star, load in delta. (c) Alternator in delta, load in delta. (d) Alternator in delta, load in star.

9.

A 500V,3-phase, star-connected alternator supplies a starconnected induction motor which develops 45kW. The efficiency of the motor is 88 per cent and the power factor is 0.9 (lagging). The efficiency of the alternator at this load is 80 per cent. Determine (a) the line current, (b) the power output of the alternator, (c) the output power of the primemover.

10.

' A 400V, 3-phase system takes 40A at a power factor of 0.8 (lagging). An over-excited synchronous motor is connected to raise the power factor of the combination to unity. If the mechanical output of the motor is 12kW and the efficiency is 91 per cent, find the k VA input to the motor and its power factor. Find also the total power taken from the supply mains.

CHAPTER 12

ELECTROMAGNETISM (continued) Earlier treatment of the electromagnetic circuit has introduced fundamental relationships between magnetomotive force-F, F l u x 4 and Reluctance-S. It will also be remembered that, the m.m.f. per metre length is called the magnetising force, magnetic field strength or intensity-H, whilst B (in teslas) is the flux density. F For revision, we write Q, = -, and F = H x I where I is the 3 0 1 lc~igtllof tlic magnetic circuit. Also B = - and S = - where p A DA is the absolute permeability and A is the area of the mcgnetic circuit PERMEABILITY OF FREE SP A CE ( p , ) . Earlier

work defined the term permeability as the ratio of flux density in a medium to the magnetising force producing it. Fur free space, a vacuum, air or any other non-magnetic material, a value for the permeability can be deduced as 4x x lo-' in the appr.opriate SI units or p, = 4x x microhenrys per metre (pH/m). Also since p, = B : H then H =

4x x 10-

, amperes o r ampere-turns

per metre.

Thus the amperesjmetre o r ampere-turnslmetre for air flux density permeability of free space' From the foregbing i t can also be deduced that BIH = a constant or B ac H. If therefore for air, values of B are plotted to a base of H , a graph, as-shown in the diagram (Fig 154), would be obtained. This would obviously be a straight line and if measurements of flux density B were made, at a point outside, but near to, a long straight current-carrying conductor, for various values of magnetising force H , by varying the current; i t being remembered that I

I

H = -- where r 1s the radius from the point to the centre of the 2xr c . o ~ l d ~ ~ i . l then o r . tllc s l r ; ~ ~ g l ~ ~ - 11/11 l i r l c rcli~lior~ship would bc

confirmed. I . MAGNET!SIKG FORCE DIJE TO A LONG, STRAIGHT CURRENT('AHHYING CONDUCTOR As mentioned above, the magnetising force outside, but

;~d~;iccnt to. ;I current-cilr~.yingconductor, is given by /-I =

I 2xr

-- .

r

Fig 154

I

This expression was deduced earlier (in Chapter 5) and should be revised. It should also be remembered that, H is the m.m.f.1 metre length. Magnetomotive force-F, is measured in ampereturns and thus the total m.m.f. for any magnetic circuit outside the conductor can be found from F = HI.

I

2. MAGNETISING FORCE INSIDE SOLENOID If a parallel field of flux lines is assumed inside a solenoid as illustrated by the diagram (Fig 155), then its length can be taken as I metres, the number ot'turns on the coil as N nnd tlic c u r r c ~ i l passed as I amperes. The lines of flux are known to fan out at the ends and for their return path they spread out into space. This external return path has negligibly small magnetic reluctance and the whole m.m.f. of the coil can be regarded as being utilised in setting up the field inside the solenoid. Thus the m.m.f. per unit length is, by definition, H-the magnetising force F IN Thus H = - = - ampere-turns per metre. I 1

I

I--AMPERES

I

1- METRE $ N TURNS

Fig 155

I

I I l I I

3. MAGNETISING FORCE INSIDE A TOROID The diagram (Fig 156) shows this simple electromagnetic arrangement. f t consists of a solenoid bent back upon itself so that the lines of flux are confined to the inside of the coil. We can now consider a non-magnetic ring to be wound uniformly with a coil of N turns, carrylng a current of I amperes. The mean circumference is 1 metres and since the flux is confined to the inside and the path is uniform, the magnetising force or m.m.f. per unit length is given by: IN H = - ampere-turns per metre.

I

Fig 156

Example 99. A wooden ring having a mean circumference of' 300mm and a uniform cross-sectional area of 400mm2, is wound uniformly overau with 300 turns of insulated wire. If the current is 3A, calculate (a) the magnetising force (b) the flux density inside the toroid and (c) the total flux produced. (a) The total m.m.f. produced = 3 x 300 = 900At The mean circumference is 300mm = 0.3 metres F 900 .: The magnetising force H = - = 1 0.3 = 3000Atlm (h)Thc flus density is given by B = p, H = 4 ~ lo-' 3000 12.56 x lo-' x 3 x 103 1.256 x 3 x 10-3 = 3.768 x = =

=

(c) The total flux produced

0.003 77T

= B x A = 0.003 77 x 400 x = 01. (D =

1.5 x 1.5uWb

webers

FERROM AGNETISM I t is known that, when iron is used as tlie core of an electrom;lpnet. thc ficld is intensified or that a very much greater flux results from tllc ~n;ignctising~ I I I ~ ~ ~ ~ C 01' -the I Ucncryisiny I . I I ~ coil. must bedue to the S' reluctance S and if the dimensions oP the core I and A are kept the same as for the air path, it follows that the permeability of the iron must be very much greater than that of air. Thus we can now make reference to the permeability of a magnetic material which is termed relative peJmeabiliry.

Since the only ckilnye in the reliition

d,

=

(p,) This is the ratio of the flux densit) which is produced in a magnetic material to tlie flux density which would be produced in air by the same magnetmotive force. For materials such as iron, nickel, cobalt, etc this value of pr can be very large, ranging from 1000 to 200d or even more for some special electrical steels. It can be quoted as the permeability figure for the material but is not constant and varies with the flux-density value at which the material is being worked. It is of interest to mention that materials such as bismuth have a relative permeability value of less than 1. RE L A T I V E PERM E A BILIT Y

THE B-H OR MAGNETISATION CURVE If a specimen of magnetic material is made up in the form of a ring and is wound with an energising coil, measurements of flux density for various values of magnctising rorcc can hc m:~tlchy winding on a secondary coil and using the principle of transformer action. I t is not proposed at this stage to discuss the manner in which the tests are made but it can be stated that this is an accepted industrial method for determining the magnetic properties of various materials. It has already been seen thht; if the flux density B is plotted against the magnetising force H for air, a straight line is obtained, but for magnetic materials, curves as shown in the diagram (Fig 157) will result. It will be noticed that, at first, the graphs.are approximately straight lines, showing B to be proportional to H. Then the curves begin to turn over forming a 'knee' and finally they become horizontal and exhibit'little increase in B for a large increase in H. In this state, the material is said to 'saturate.'an explanation for this has already been given when dealing with the molecular theory of magnetism. If the permeability (p,) is plotted to a base of B, curves as

-

MACNITISINC

FORC t . H (A S / m )

Fig 157 shown in the diagram (Fig 158) would result. The permeability curve has a peak corresponding to the point on the B-H curve where the tangent goes through the origin. Beyond this peak the permeability value drops off fairly rapidly. An examination of the B-H and p,-B curves shows how the properties o f various magnetic materials differ. The effect on machine design can also easily be seen,-lower working B values necessitate larger section and greater mass to obtain a required flux \,slue. The effect of high permeability materials is also

;lpparent and the shape of the B-H curve with the saturation cll'cc.1, ~ I I O W lllc ~ I I I I I ~ ~ ~ Ior~ I IIIIIC'IIIIIC ~ I ~ ficI(I ' I ~ % ~ C I I I R The magnetic properties are dependent upon the artu.CII COIIIpositiou ol' tllc subsrancc. T l l ~ s~n:~ng:~ncsc-steel i a p17i\cti~i\ll\ non-magnetic, but small quantities of carbon or silicon wlicn added to steel vary the shape of the B-H curve and sheets of commercial steel marketed under trade names, like Stalloy or Lohys, are available to suit different design requirements. ( p ) .Since permeability is the ratio of the flux density in a medium to the magnetising force producing it. and for air it is known that B = p a , it follows, for a material with a relative permeability of p,, that B = prp,H o r B = p H where p = prpo. Summarising, we can define absolute permeability as the ratio of flux density in a substance (in teslas) to the magnetising force (in ampere-turns per metre) which produces that flux density. AB S O L U T E PERME A BI L IT Y

I i

I I

I I I I 1

S). This term has been mentioned earlier. It has been likened to the resistance of an electrical circuit. Since flux is proportional to the m.m.f. and is restricted by the reluctance, further investigation will show that reluctance in turn is proportional to the length 1of the magnetic circuit and is inversely proportional to its area A . Furthermore it must be inversely proportional to the permeability. since the greater the permeability of the material the greater the flux and hence the sm;lllcr

R EL U C T A N C E (SYMBOL

the reluctance. We can therefore write, S =

I

and point out uA hcrc tllac ;~hsolutcpcrmci~hilityis hcillg uscri. i~ being I . C I I I ~ I I ~ bered that p = p,po. Calculations on magnetic circuits using magnetic materi:~ls are now possible, but it should be noted that, unlike electrical V circuit calculations which use I = -, it is not always necessary ---

R F to use the comparable relationship of @ = - The solution of S most problems associated with the magnetic circuit can be made without always determining the reluctance, and experience will show the best method of solution. The following examples indicate the alternative way of treating typical simple problems. Example 100. A solenoid is made up from a coil of 2000 turns. carries a current of 0.25A a n d is Im long. An iron rod of diameter 20mm. forms the core for the solenoid and is also 1.0m long. Calculate the total flux produced if the Iron has :I permeability of 1000. Note. Here relative permeability is implied. '

-

M . M . F . of coil is given by F = HI =

I

x

I = IN

xd2 3.14 x 400 x Area of iron = - = 4 4 = 3.14 x square metre 1 I Reluctance of iron = --- = --~ LA

Flux 0 = - = S

'0°

PrPJ

;'944 = 5 x 3.944 x l o - ' weber 10

ALTERNATIVE SOLUTION

,

M . M . F . of coil. F = HI so magnetising force H = F Also B =

j1H = ,LI,~,H = 1000 x 4 x rt x lo-' x 500 =4 x 5 lo-' x 10' = 20 x x = 0i628T Total @ = BA

0.628 x 3.14 x

webers 197.2pWb. Example 101. A cast-steel ring has a cross-section of 400mm2 and a mean diameter of 240mm. I t is wound with a coil having 200 turns. What current is required to produce a flux of400pWb, i f the permeability of the steel is 1000? : \ I . ~ : I of'sttcl = 400 :, IO\squarc lnctrc =

= 1.972 x

=

I 104 -ampere-turns per 104 4 x x metre I o4 x x x 240 x 10- 3 R1 hl.1- 01' rlrlg = 111 = - -4 x x

so H

=

--

4

-

B

- -

EL EC T RO MAG NET IS M

283

(continued)

600

If the coil turns are 200, then the current would be -= 3A. 200 ALTERNATIVE SOLUTlON

Iicluctancc of ring S = or S =

I

-- ~rjlLV4 -

n x 240 x lo3 x 4~ x lo-' x 400 x

6 4 x 10-~ = 1.5 x lo6 ampere-turnslweber Since 0 = 400pWb = 4 x weber then required m.m.f. = 5'0= 1.5 x lo6 x 4 x = 6 x lo2 = 600At -

600

Required current = - = 3A.

200

C

THE COMPOSITE MAGWETIC CIRCUIT THE SERIES ARRANGEMENT

Consider a magnetic circuit built up as shown in the diagram (Fig 159). It is obvious that the sections are in series and that the same flux passes through them.

Fig 159 Then total m.m.f. = m.m.f. across section 1 + m.m.f. across section 2. If the flux is Q, then @S = OS! -- OS, where S is reluctance of the composite circuit. Thus: S = S , + S,. Summarising: Total reluctance = the sum of the individual reluctances of the sections for a ser: : lrrangement.

THE PARALLEL ARRANGEMENT

Such a magnetic circuit is not frequently encountered but is considered here, being complementary to the series circuit. The arrangement is shown In the diagram (Fig 160).

Fig 160 If the different paths of the magnetic circuit are in parallel. then the necessary m.m.f. is that which will produce the required flux in each part of the circuit considered by itself. Let F = the rn.1n.f. required to produce fluxes @,, @, @, etc. F also produces total flux @. .-. f14 -So @ = Total Reluctance of circuit S and since @ = 0 ;+ cP2

.,@

=FA Sl

+ 0,-

+F,+F,

S2 S3 But F,, F,, F, are the m.rn.f. across the same points of the magnetic circuit and are equal to F. . .

I .

-

S

1s referred to as the perrneance of a magnetic circuit and the

above can bc s u ~ i l ~ n a ~ . ~bys esaying d tllal; the reluctance ol' a

divided magnetic circuit (sections in parallel), can be found by knowing that its permeance is equal to the sum of the permeances of the individual circuits. Example 102. A n iron ring has a mean diameter of 200mm , ~ n d;I C ' I ~ L ~ S ~ - S C C o~ I O' ~I O~ O l i l ~ ~ i.A11 ' . a i l . gap of0.4rnm is made by ; I r:~tli;~l .;;lu.-c.trl ;1cr-oss(lie r i n g . A s c ~ ~ m i n;I gpcnnc;~hilitvof 3000

(ronfinue~d)

E L E C TR O M A G N ET I S M

285

for the iron. find the current rewired to ~ r o d u c ea tlus of 250pWh, if tile cncrgising coil is wbund with'^^^) turns. (I( Z(K) n 10 ' ) ( 0 4 * I 0 ') Keluctance ol' Iron S = 3000 x 4 x n x 1 0 m fi

'

62.76 36 x rr .x = 555.2 x lo3 ampere-turns/wekr 0.4 104 Reluctance of air gap S , = 4 x x 1 0 - ' x 3 x 10-4

-

= 1061.5 x lo3 ampere-turnslweber S, + SA = lo3 (555.2 + 1061.5) lo3 (1616.7) ampere-furns/weber Total m.m.f. = @S = 2.5 k l o p 4 x 161.67 x lo4 = 2.5 x 161.67 = 404.15At 404.19 = 0.674A. Current = 600

Total Reluctance S

= =

ALTERNATIVE SOLUTION

2.5 x 3 x 10-~ = 0.833T Now H for air is given by: 0,833 - 0,833 x 10' 0.833 HA= -u, 4 lo-'' 4 x n

Since 0 = 250 x

weber

then B =

Length of air gap = 0.4 x l o e 3 metre Ampere-turns for air = 66 32 x 104 x 4 x - 265.28At H for iron is given by: 0.833 0.833 - - 66.32 x lo4 H I= py, 4 x x 1 0 - ~x 3 x loT a x lo3 -663'2 - - 221.066At/rn 3 Now length of iron path = (628 - 0.4) = 627.6 x metre .Ampere-turns for iron 5 221.066 x 0.6276 = 138.74At Total ampere-turns = 265.28 138.74 = 404.02 404.02 = 0.673A Current = ------

+

600

I I 1

For the three previous examples, alternative solutions have been given in which the reluctances, as such, for the various sections of the magnetic circuit being considered, have not been found. This alternative' method of solution is especially of value when the relevant B and H data for the magnetic material is given in tabular or graphical form. The relative permeability is not given as a specific value and would have to be found before the reluctance could be calculated. Obviously any solution along these lines would be tedious and the following example is recommended to the reader as a n instruction on how to solve the type of problem being discussed. Exan~plc103. An iron ring of square cross-section has an external diameter of 140mm, and an internal diameter of 100mm. A r;ldi;~ls;lw-c~~t through the cross-section of thc ring forms nn alr gap ol I lnm. 11' the ring is unil'ormly wound with 500 turns of wire, calculate the current required to produce a flux of 0.35mWb in the gap. Magnetic data of the material of the ring are given below and shown in the diagram (Fig 161). Take p, as 47t x 1 0 - ' p H / m .

GAP 0.351Wb

Fig 161 0.65 0.89 1.06 1.18 Flux density (T) Magnetising Force ( A P m ) 200 300 400 500 Solution uses the graph (Fig 162) obtained from the above d;1t;1 : \ r c . ~ i ol'iron a n d air gap = 20 x 20 x = 4 x IV4square metre 1.c11gtl1ol l1.011 - n >. ( I I I C ~ I I cl~;rr~~etcr) I - a l r gap = ( n x 120 x l K 3 )- ( 1 x 10-3)metre = (376.8 - 1.0) = 375.8 x metre 6 ,cngrli oi' ;ilr gap =- I x 10 ' metre

I

ELECTROMAGNETISM

287

(continued)

MACNETISINC FORCE

-

H (A Jrn)

Fig 162

From graph for the iron H = 290Atlm So for Iron, m.m.f. F , = 290 x 375.8

x

lW3

= 108.88At

For air, since H

B -

=,11"

-

and for air, m.m.f. F A =

0.875

4 x .n x 10-7

ampere-turns metre

0.875 x lo7 x 1 x 4 x x

Total m.m.f. F, + FA= 108.4 + 696.7 = 805.6At 805.6 Current is deduced from = -- 1.6112A 500 Energising current = 1.6A (approx.). 8s-

Tlie diagram (Fig 163) illustrates the manner in which flux is known to bridge an air gap, especially if the latter is comparatively large MAGN ETIC F RI N G I N G

Fig 163 I t will be seen t h ; t t the flux in air tends to occupy a larger area than that of the iron, and the flux density is thus reduced. An allowance can be made for this effect in problems when required,. but unless a directive to this effect is given, the area of the air gap can be taken as the area of the iron. M A G N E T I C L E A K A G E . For some magnetic circuits, due to the shape of the iron core and the positioning of the energising coil, a small amount of flux leakage occurs as shown by the diagram (Fig FLUX IN POLE

= VSIfVL

FLUX

X

If AKACf COfFFlC IENT

164). Somc lines o f flux arc not confined to the iron and coni1)lrtr ! I I ~ I I ' 1>itt115 [li~,ougiiair. I-'or. pr.;rctic:tl purposes, a l'i~ctor known as the In~X-rrgc.r,ocjSc,iettt may be given which, when used, increases the required working flux value by a n amount sufiicient to allow !.or' this leakage. Thus: the required total flus = the useful o r working flux x lea kagc coeff~cient. Tlic Ic;~h;~gc ~ o c l i i ~ i c nm;ly t hc hctwccn 1 . 1 ; ~ n d1.3.

Example 104. (a) A magnetic circuit has an iron path of length 500mm and ;In air g a p o f lcnpth 0 , 5 m m , the whole being 0l' llnlf'0l'Ill SqULIrt: Cross-SCCIIOII, 100()111111' i l l ill'Cil. CilI~llIillt:lilt: number of ampere-turns necessary to produce a total flux of Irl~Wbi n the : ~ i r piip. Igllorc frilrpinp, :l[rrf :ISSUIIIC :I lc:~L:~yc coefficient of 1.3. The B-H curve for the iron is given by the following table : H ( A t / m ) 100 200 300 400 600 800 1000 1200 B (T) 0.42 0.8 0.98 1.08 1.22 1.3 1.36 1.4 (b) A conductor is passed through the air gap at a speed of 1100mls. If the length of the conductor is greater than the length of the side of the gap, calculate the e.m.f. induced. The solution uses the graph, obtained from the above data, a s shown in the diagram (Fig 165).

Fig 165 Area of iron and air gap

= 1000

x

= 10- square metre

Length of iron = 500 x Flux density (B) for air

= =

I

0.5m =

I tesla.

B 1 For air, since N = - = p0 4 x 10-7 - - l o 7 ampere-turns,metre 4 x x lo' x 0.5 x lo-' and m.m.f., FA = 4 x x - - =1-o4 = lo' 3988At 8x11

2.512

From graph f o r the iron, H = 800At;m for a B value of 1.3T and m.m.f., F, = 800 x 0.5 = 400At 1 o t ; i I ;iliil~cre-turli~ required 398.8 + 400 = 798.8At ( b ) Induced e.m.f. given by E = Blv volts S ~ n c ethe area of the alr gap is that of a square, the side of the square is L. 1000 x = v'10 x = 3.162 K metre Thus E = I x 3,162 x x lo2 = 3.162 volts \'oic In the above B = 1 tesla. I = 3,162 x lo-' metre and v = 1OOmls.

-

I R O N LOSSES The efficiency of electrical machines and transformers is lo~vered by tlg losses which occur in them. Apart from the . C l c c ~ / ~ n t ~ Lasses icnl such as Friction and Windage and the Copper Lo.\.srs, due to the resistance of the conductors, an additional 1qs.s occurs when 3 magnetic material is taken through a cyclic iiiriation of magnetisation. This loss I S termed the Iron i o . s s :ind 1s itself made up of two component losses which are ( 1 ) the I!i~stc~rc~.si.r L o x and (2) the Eddy current Los.s. The cause of :hc\e two sources of power loss will next be considered. r t l l - H Y S T E RE S I S L OO P

II'the m;lpnetlslnp force applied to an iron ci~rnpleis increased exactly the sariic way ;I> when making the test for a B-H curve, and is then reduced to zero, ~t will be found that the new B-H curve, for decreasing L.;IIUCS 01' 11, will lie ~ I ~ O Vthe C original ascending curve, and that when H is zero B is {eft at some value. The effect of the descendIns LXII.\C be111g ; I ~ o \ . ctlie ascending one. is said to be due to ' I ~ \ . ~ I ~ I . C \ I \ ' . since tllc R v:rlucs lag behind those for tlie corr.c.\l~o~icl~ng 11 \ll.engtll when increasing. T h e word hysteresis 11.0111 L ~ ' I . O to S O I I ~ C1ii;1~11iiu1ii v;~lue,in

. -

ELE C T R O MA G N ETI S M (continued) .-

-

29 1

--

Fig 166 comes from the Greek meaning 'to lag'. The diagram (Fig 166) shows the effects being discussed. The value of B when H is zero is the 'remanence' and is a measure of the residual magnetism. In order to demagnetise the iron it is necessary to apply a negative magnetising force, known as the 'coercive force'. If now, H is increased in the negative direction to its previous maximum value, the curve will reach a value equal to the previous maximum B and if H is next gradually rccluccd to zero, rcvcrscd und incrcuscd to its arigini~l~nuxinlunl, a closed loop will be traced. This is a hysteresis loop and is a measure of part of the iron loss. T o take the iron through the various stages represented by the loop, an alternating magnetising force has to be applied. One method of achieving this is by connecting the energising coil to an a x . supply, when theiron will continue to go through the same series of changes or magnetic cycles. To confirm that energy is being expended, it will be found that the iron core will register a temperature rise. Although it is not proposed, at this stage, to prove ttie fact that the area of the loop is a measure of the powel loss due to hysteresis, the loop can be regarded as an indicator diagram. More advanced studies will show that the energq absorbed per cubic metre per cycle, due to hysteresis, is given in joules by the area of the loop, provided the scales used for the graph are in the appropriate SI units. During the development of the proof, it would be stated that the energy stored in the magnetic field is represented by the area OABCDO (Fig 166).

When the field collapses, energy is returned to the s u p ~ l ywhich is represented by the area DBCD. The area of the loop OABDO represents the energy lost as heat through hysteresis and is obviously the difference between the energy put into the magnetic circuit when setting up the field and that recovered when the field decays.

Fig 167

If the iron sample was non-magnetic, ie air, then the B-H curve would be a straight line, as is shown in the diagram (Fig 165), and the energy stored in the field when it is set up, is represented by the area of the triangle OBC. This energy is recovered wlierb, the field collapses. No energy is wasted when air is'qhe m e d ~ u mof a pulsing magnetic field. For alr. area of triangle OBC = -1 O C x CB 2

B

x

H

where B, is thc maximum Ilux density value, which has been attained for the H value which was impressed. I 1 Bm - -Bm But B = p,M. :. Area of triangle = - x Bm x -

2

Po

~

P

O

Sincc thc ;irc;i of thc tri;~nplcrcprcscnts thc cnergy stored in ; ~ r rper cubrc rnetlt ( i n joules), i t Ibllows that: B",~ F o r air. Energy stored per cubic metre = - joules 2

~

0

The types of hysteresis loop, a s obtained from various m;lgnctic m:rteri:~ls.can be grouped into one of three classes a s sIio1~11by the diagr;rni (Fig 168). I ("'1' I i \ I'or I I ; I I ( I 51ccl.Tlic I;irgc v:~luco f tllc cocrcivc h r c c

ELECTROMAGNETISM

(('Or?litlll('(/)

20 3

Fig 168

indicates that the material is suitable for permanent magnets. The area however is large, showing that hard steel is not suitable for rapid reversals of magnetism. Loop 2 rises sharply showing a high p and a good retentivity (large intercept on B axis). The loop is typical of cast-steel and wrought-iron, w b c h are suitable materials for cores of electromagnets and yokes of electrical machines. Loop 3 has a small arca and a high p . The material (mainly alloyed sheet-steels) is suitable for rapid reversals of ma g netism and is used for mnchine armatures, transformercores, etc 1. HYSTERESIS LOSS. Since this. is a function of loop area, the effect of varying B on the area requires consideration. When the value of H is increased eg doubled, the value of B is not doubled and consequently the ratio of the loop area is not quadrupled. 1 1 is li)t~ntllo incrci~sc: ~ b o t 3~ ,ll 1i111csi ~ t ~ il'wc tl nsstlrllc Arci~crl' Loop to be proportional to BmX then x lies between 1 and 2here B, is the maximum value to which the flux density has k e n taken. Steinmetz originally found x to be practically constant for all specimens of iron and steel and to equal 1.6 (approximately). This figure is called the Steinmetz Index and we can now write: power loss per cubic metre per cycle as PH cc Bm1''6 o r PH = constant x Bm"6 x f joules per second. Thus P H = K H x B,"~ x f watts per cubic metre, where KH 1s a hysterrsls coeficicnt, depending on the material being con'sidered, and f is the cyclic frequency of magnetic alternation. Summarising : Hysteresis loss P, = K,B,' ' 6 f watts per cubic metre-an empirical formula.

2. EDDY - C U RRE N T LOSS. When an armature rotates in a magnetic field, an e.m.f. is induced in the conductors. Since the conductors are let into slots, it is obvious that the armature teeth can also be

looked upon as conductors and that e.m.fs. will be induced across them. Moreover. since the electrical circuit is complete for these e.m.fs., currents will flow from one end of a tooth through the armature end-plate. along the spider or shaft and back to the other end of the tooth through the opposite armature end-plate. These currents. known as 'eddy currents.' produce a power losr, due to the resistance of the iron circuit, which is a PR This eddy-current loss is also dependent on a &n R other factors.-which are considered below. Every attempt 3' r ~ ~ ; ~ hy c l crn;iclline designers to keep such loss to a minimum. I llc prlncrpal methods by which this is achieved are (a) by laminating the iron circuit and insulating the laminations from each other by \,amish. cellulosr. or paper ( b ) b!, using iron with a high specific resistance and (c) by keeping the frzquency of the magnetic alternations or cycles to a minimum. Since a generated voltage is proportional to flux and speed. then E K Oh' or E K B,f where B, is the maximum. flux density and f is the frequency of alternation. E2 Again since power loss cr - we can write: R Power loss rc ~ , ' , f ~ o r PE = K ~ f Bwatts~ per~cubic metre, whcrc K,: is a n ( ~ l ( l . v - ~ . i i r (r.(o, (~~~f i c i e n which [ is dependent upon the type of material belng used. its thickness and other dimensions.

4 ??'

'

P U L L OF A N ELECTROMAGNET It has been seen that the energy stored in a magnetic field in BZ air is given by - joules per cubic metre, where B is in teslas 2/43

Consider two poles arranged as shown in the diagram (Fig 169),

i?r -X

Each has an area A square metres and let F be the force

ot'

r ~ l t r r ~ c t i o(in n ncwtonu) hctwc.~ri t l i r ~ l c ~ l r u

Let one pole be m o v ~ da s l n i ~ l l~ I S ~ ~ I I .\I C( CI ~ w ~ I . c . \ ) i l g i l l l l h t [ I I C force F. Then the work dorlc i s F\- nrw~on1nrlrt.s 01. io\~lrs. The volume of the magnetic field has been increased by ,4.v cubic metres and therefore the energy stored in the field is InB2 creased by -- x . 4 . ~joules. This is obviously equal to the worA 2 ~ 0 B 2 A.u done In separating the poles so that Fu = 2/10

or F = B 2A newtons. where A is in square metres and B in 2/10 teslas. Example 105. An electromagnet is wound with 500 turns. The air gap has a length of 2mm and a cross-sectional area of 1000mm2. Assuming the reluctance of the iron70 be negligible compared with that of the air gap. and neglecting magnetlc leakage and fringing, calculate .he magnetic pull when the current is 3 amperes. N.M.F. ofcoil = 500 x 3 = 1500At This m,m.f. 1s used to pass the flux through the air gap, since the reluctance of the iron is negligible. The magnetising force for the air, is glven by 'the ampere-turns per metre' o r Also the flux density in air is B where

= 0.942T

B2A Now pull F = --2p0

~ h u Fs =

-

-

0.942' x 1000 x 2 x 4 x x lo-'

newtons

z:f

-newtons

and the pull would be 353.3N. Example 106. A four-pole d.c. generator has a cast-steel yoke and poles and has a laminated steel armature. The dimensions of the component parts of the magnetic circuit are as follows:

Yoke. Total mean circumference 3 . 0 4 ~ 1 Pole Total mean length 0.24m Air gap. Total mean length 2mm Armature. Total mean path between poles 0.4m

'

CSA 0.04m2 CSA 0.065m2 CSA 0.065m2 CSA 0.025m2

Fig 170

The magnetisation curves are: H(At m ) 400 800 1200 1600 2000 2400 1.37 1.43 1.3 1.2 Cast Steel B(T) 0.45 1 1.63 Laminated S t e d B(T) 1 1.34 1.48' 1.55 1.6 Calculate the ampere-turns per pole. for a flux per pole of 0.08Wb in the air gap. The diagram (Fig 170) illustrates the problem and the appropriate magnetic characteristics are shown by the graphs of the diagram (Fig 171). A I K (;AI'

Length 2 x 1W3 metre Ar-e.10.065 = 6 .:

square metre

, ,.

Total ampere-turns o r M . M . F . for air gap IS given by: 1: = ..- 1 23-.- -- x 2 x 10 ampere-turns A 4 x 1 1 4 ~10.'

(Cast Steel) Length 24 x metre Area 0.065 = 6 . 5 x 1 0 r,ciu;lrc rnc!rr

P O LE

From graph, Up = 1370At/m or Total Fp for pole = 1370 x 24 x

=

330At

(Cast Steel) 3.04 Length -= 0.76 = 76 x metre (between poles) 4 or 0.38m magnetic length Area = 2 x 0.04 = 0.08 square metre. (Note the doubling of the area since full pole area has been taken for the flux) = 8 x 10-'m2

Y O KE

From graph H , = 800At rn Total F, for pokc 800 x 38 x

=

304At

(Laminations) 0.4 Length - = 20 x metre (magnetic length) 2 Area = 2 x 0.025 = 0.05 = 5 x lo-' sauare metres

A RM A T U RE

#

From graph H L = 2000At m Total FL for armature = 2000 x 20 x lo-' = 400At Total M.M.F. per pole = 1955 + 330 + 304 = 9989~r.

+ 400

CHAPTER 1 2

PRACTICE EXAMPLES 1.

A brass rod of cross-section 1000mm2 is formed into a closed ring of mean diameter 300mm. It is wound uniformly with a coil of 500 turns. If a magnetising current of 5A flows in the coil, calculate (a) the magnetising force (b) the flux density and (c) the total flux.

2

An electromagnetic contactor has a magnetic circult of length 250mm and a uniform cross-sectional area o f 4 h 2 Calculate the number of ampere-turns required to produce a flux of 500pWb, given that the relatlve permeability of the material under these conditions is 2500. ALSO po = 4x x lo-' henrylmetre.

3.

In a certain magnet~ccircuit having a length of l m and a uniform cross-section of 500mm2, a magnetising force of 500 ampere-turns produces a magnetic flux of 400pWb. Calculate (a) the relative permeability of the material and (b) the reluctance of the magnetic circuit. po = 4x x lo-' H/m.

4.

An iron ring having a mean circumference of 1.25m and a cross-section;tI are;i o f 1 500mm2, i5 wound with 400 turns of wire. An exciting currenr of' 2.5.4 produces a flux df' 0.75mWb in the iron ring. Calculate (a) the permeability (relative) of the iron (b) the reluct:tnce of the iron (c) the m.m.f. of the exciting winding.

5.

A U-shaped electromagnet has an armature separated from each pole by an air gap of 2mm. The cross-sectional area of both the electromagnet and the armature is 1200mm2 and the total length of the iron path is 0.6m. Determine the ampere-turns necessary to produce a total flux in each air gap of I . l3mWb neglecting magnetic leakage and fringing. The magnetisation curve for the iron is given by: B(T) 0.5 0.6 0.7 0.8 0.9 1.0 1.1 H(At/m) 520 585 660 740 820 910 1030

6.

A clrcular ring of iron of mean diameter 0.2m and crosssectional area 600mm 2 has a radial air gap of 2mm. It 1s magnetised by a coil having 500 turns of wire. Neglecting

magnetic leakage and fringing, estimate the flux density in the air gap. when a current of 3A flows through the coil. Use the magnetlc characteristics as given by the graph of Q5.

7.

A built-up magnetic circuit without an air gap, consists of two cores and two yokes. Each core is cylindrical, 5 h m diameter and 160mm long. Each yoke is of square crosssection 47 x 47mm and is 180mm long. The distance between the centres of the cores is 130mm. Calculate the ampere-turns necessary to obtain a flux density of 1.2T in tlic cores. Ncglcct magnctic lenkngc. The magnetic charactcristics of' tlic tnatcrial arc: 1.15 1.2 1.0 1.05 1 . 1 R(T) 0.9 310 380 470 f/(Ariln). 200 260 050

8.

A n iron rod 15mm diameter, is bent into a semi-circle of 50mm inside radius and is wound uniformly with 480 turns of wire so as to form a horse-shoe electromagnet. The poles are faced so as to make good magnetic contact with an iron armature 15 x 15mm cross-section and 130mm long. ( a ) Calculate the current required to produce a pull of 196.2N between the armature and pole faces. Neglect magnetic leakage. ( b ) If the armature is fixed so as to leave uniform air gaps O.5mm wide at each pole face, calculate the ampereturns necessary to obtain a flux density of 1 . 1 5T in the air gap. Neglect leakage and fringing. Use the magnetic characteristics as given by the graph of 47.

9.

Two coaxial magnetic poles each lOOmm in diameter are separa ed by an air gap of 2.5mm and the flux crossing the air gap is 0,004Wb. Neglecting fringing calculate (a) the energy In joules stored in the air gap (b) the pull in newtons between the poles.

10.

Calculate the ampere-turns per field coil required for the air yap. the armilturc tcctli and the pole, 01.3 d.c. machine working with a useful flux of O.O5Wb/pole, having given: Illl'cctivc ;Ircq o S ; ~ igr ; ~ pM OOOmmZ.Mean Icnpth'of air gap 5mm. Ell'ec*tive area 01' pole 40 O m r n ' . Mean length ol' pole 250mm. Effective area of teeth 25 000mm2. Mean lehgth of teeth 45mm. Magnetic leakage coefficient = 1.2. Magnetic characteristics 01' tllc materials i\rc: R(T) 1.3 1.4 1.5 1.6 1.8 2.0 / / ( A t ' I I I ) 1200 IS00 2000 3000 HSOt) 24 OW

THE ELECTRON THEORY, B A S I C ELEC'I'KON IC'S A N L) ELEC"I'I~OS'1'A'I'IC'S Electrical engineering is concerned with the passage of electricity round conducting circuits and with the resulting effects. The passage of electricity is referred to as t5e flow of current but this in turn, is due to the movement of chargcs round the circuits. Electricity is considered to be of two kinds, namely positive and negative and in its smallest quantity consists of +ve and - ve charges. CONSTITUTION OF MATTER Matter may be de'fined as anything that occu5es space. I t may be'in solid, liquid o r gaseous form but basically consists of molecules of the substance. A molecule is the smallest particle of a substance that can exist by itself. Thus molecules have the properties of the substance which they form but themselves ccnsist of groups of atoms. As an example, a molecule of water. written H,O, consists of 2 atoms of hydrogen and one of oxygen. The atom is defined as the smallest particle that can enter into chemical action, but is itself a complex structure consisting of charges of electricity. A substance that contains only atoms with tlie samc propcrtics is called iln clcmcnr, but one conluining atoms of different properties is called a compound. THE STRUCTURE OF THE ATOM

According to the electron theory, as propounded by eminent scientists like Rutherford and Bohr, each atom has a core or nltcleus surrounded by planetary c~lc~crrons. The nucleus conslsts of +vely charged particles or protons and uncharged particles called neutrons but the arrangement is such that the nucleus has a net +ve charge which is equal and opposite to that of the planetary electrons. These electrons are considered to have - ve charges and to revolve round the nucleus so constituting a miniature solar system. The nucleus thus represents the sun and the electrons represent the planets. Under normal conditions, when the atom is said to be stable or unexcited, the planetary electrons together neutralise the +ve charged particles of the nucleus, so the complete atom itself has no electrical charge. The diagram (Fig 172) gives an idea of the atomic structure as

it is assumed. The simplest atom is that of the element hydrogen, consisting of a nucleus with one proton (having a +ve charge), around which travels one electron in an orbit. The electron with its - ve charge neutralises that of the proton. In the diagrams, the electrons are denoted by circles, with their charges shown, and are considered to be moving on dotted orbits. The nucleus is shown with a full circle, has a net positive charge attributed to the protons contained therein and these are shown by + marked circles. The neutrons are shown by small circles with no charge sign. * - -

-.

ILtCTl.ON

a4

-

\ \

fltCTRON #

I

NfUIRON /

1

/

\ \

/ /

''

-4'-

\-or

,'

/

/',/

'.**'

\

-

HYDROCE N ATOM

* - * -=\ -

.wc

Hf LlUM

\

f'. -

_

#

.e'

OXlC f N

ATOM

ATOM

Fig 172

The next element considered is helium. This has 2 planetary electrons and the nucleus consists of 2 protons and 2 neutrons. The planetary electrons of most atoms are associated with the nucleus in a definite manner ie the electrons are in groups termed 'shells,' such that the planetary path of each shell is difierent. This% shown if an oxygen atom is considered. This has a nucleus of eight protons and eight neutrons. The planetary electrons are eight in two orbits or shells-six in the outer shell and two in the inner shell. For any one atom, the electrons in a shell can be less than, but never more than a definite number. Thus the first shell cannot have more than two electrons and the second shell more than eight. .',-* -9-- - - *\ ,B' / ,/ -. \

.

/ ; l a \ a\!\ '

,

\

/

,

/

/o-.*

\

1

\

/'

\

' 4 -

~\

\

_

' -

1

/

I

,o.-

-

\*%\'\

/

\

/ /6 /

I

\

'\

' \ \

\

\

I

\ \

*\ ..

\

\

1 t0 \ 1

(

/

-

,d

\

1I

'

,

'

\

- - - D'

/

-..&-'

I#

,' /

P , /

/'

I

ELECTRON THEORY, BASIC ELECTRONICS, ELECTKOSTATI(3

303

-

Diagram (Fig 173) represents the atom structure of tivo metals-lithium and sodium. In each case, and if other metals ;Ire considcrctl, i t will bc Fccn th:~ti111 I i i i v r cltic or- I W O c-lrc I r o r i ~ 111 IIIC

OLIICI'IIIL)SI

511~11, 1'1115 I c ~ I I u I ' c

15

COII~ILICICLI

it>

~ I I C IC,I~OII

for metals hrtv~nggood cnnducting properties. sinc-c ~hracOUICI.most electrons are loosely held by the electrostatic field and can be easily removed. All the examples with the diagrams have shown atoms in their neutral state; ie the total - ve charges on the planetary electrons is equal to and balanced by the total +ve charge on the protons, so that the complete atom has no charge. The neutral atom has no electrical characteristics-it neither attracts nor repels other atoms o r electrons. However, since the nlolecules of any substance are continually in motion, the atoms, which are part ol'tlic molecules, also move about inside the molecules and such movement can result in an upsetting of the atomic structure. IONS (Positive and Negative). An atom may lose or gain an electron as the result of a disturbing action. It then becomes electrically unbalanced, having acquired a charge and is called an ion. Thus an atom minus an electron, exhlbits a +ve charge and is a +ve ion. Similarly an atom w h c h gains an electron. exhibits a - ve charge and is a - ve ion. P

CURRENT FLOW AS ELECTRON MOVEMENT

Current flow, according to the electron theory, is due to the movement of electrons from one atom to the next, each electron c;~rryinpwith i t ;I - vc cll:~ryc.In s \ ~ l ~ s t ; 5~\ 1~c ~~l l: I~> c~si l c l : ~ l \ . which are classified as good conductors of' electricity, the outer orbital electrons can move comparatively freely between onc atom and its neighbours. Such electrons, called 'mobile electrons,' move in a random manner from atom to atom, so that, the transference of charge and therefore passage of electricity, in any one particular direction, does not occur and n o current is considered t o flaw. If an electrical force, in the form of a pressure or potential difference or electric field, as it is sometimes called, is applied across a good conductor then the mobile electrons will move under the influence of this force towards the higher potential or +ve terminal. The required electrical force can be produced by a battery or generator which can be regarded as a pump moving the electrons round the circuit. A stream or movement of electrons is said to constitute an electric current but. it is stressed here that, attention must be paid to the difference between the direct~onof conventional current flow and electron flow. Thus if a length of wire is connected to two terminals.

between which an electrical pressure or potential difference ( p . d . ) exists, then a current will flow from the +ve terminal through the wire to the - ve terminal. Electron flow will be, however, from the - ve terminal to the +ve terminal. This fundamental difference between conventional current and electron flow must always be remembered and is illustrated by the diagram (Fig 174a and b). It is also stressed here that, the electrical generator or battery which maintains a pressure or potential difference (p.d.) between the ends of a conductor, does not itself make electricity but merely causes a movement of the charges o r clcctrons wllicll arc ;~lrendyprcsent in the circuit. FREE ELECTRONS I N RANDOM M O T I O N

V CURR€NT

yo; -qi,+?+~.ls.r-b-++;tb, . 05 \ .

4

d ..

~

- = 105 - 33.7 ( R + 0.25) Also since Eb a @ and N, since flux is constant 800 then % = 800 o r E,, = 96.57 x Eb. lo00

Back e.m.f. (at reduced speed) = 77.26V

SOLUTIONS TO PRACTICE EXAMPLES

441

Note. As for No 6. Flux being proportional to field current, If is substituted for @. 8. Cold condition Iao = 50 - 11.15 = 48.85A E,, = 230 - (48.85 x 0.2) = 230 - 9.77 = 220.23V Hot condition Temp rise = 60 - 15 = 45°C

r

Similarly R,,= 200 = =

200 200

+ (200 x 45 x 0.004) + (90 x 0.4) = 200 + 36 = 236Q

/,I = 50 - 0.975 = 49,025A E,, = 230 - (49.025 x 0.236) = 230 - 1 1.57 = 2 18.43V Again since E, cc CD and N and since CD K If we can write E,,o = kZfoNO and Eb, = k l f ,N 1 i r N I = Eb, X Ifo XNo or!%=

SO

& & If,

No

and N ,

Ebo

=

Ebo X

218.43 x 1.15 x loo0 220.23 x 0.975

= 1 160 revlmin

9. (a) V = E,

Ifl

Ans.

+ ZaRa

o r E, = V - IaRa = 230 - (200 x 0.35) = 230 - (35 x 2)

and E, = 230

-

70

=

160V

Z@N P E, 60,4 Also since E, = -x - then N = - x 60 A Z P

(b) Again Torque is given by: P T = 0.159 x 2 0 1 , - newton metres A

230 10. I , = -- = 2.2A I, = 5 - 2.2 = 2.8A 104.5 Also since I f I = 2.2A then I,, = 50 - 2.2 = 47.8A Again Ebl = 230 - (47.8 x 0.4) - 2 = 230 - 19.12 - 2 = 208.88V And Ebo = 230 - (2.8 x 0.4) - 2 = 230 - 1.12 - 2 = 226.88V O

Eb2= k@l Nl ( a ) Since Eb,, k@O'VO Thcn Vd = Ebo N~ asrumlng constant flux. Eh .

(b) At 600 revimin E,, = 208.88V 24ssumrnga constant flux. then for 500 revimin E,, = 208.88 x 2.

.:

o r Eh, = 1044'4 -- - 174.07V 6 The voltage across the armature has to be reduced by 230 - 173.07 = 55.93V or since V = c, + I , ( N ,+ R ) + 2 then V - E, = I; ( R , + R ) + 2 o r 55.93 = I, R, + 1,R + 2 SO I,R = 55.93 - 2 - (47.8 x 0.4) = 53.93 - 19.12 = 34.81

SOLUTIONS TO PRACTICE EXAMPLES

443

(c) Under the new condition, @ is to be altered,-hence If E kcP N or E,, = 230 - (30 x 0.4) - 2 . . -!a = -2

Eb,

= 228 - 12 = 216V

Thus current-and therefore flux, is to be reduced to 1.82 = 0.827 = 82.7 per cent Ans. 2.2

CHAPTER 11 1 For the d.c. clrcuit, P = VI or 300 = 60 x I 300 so d.c. current = --- = 5A

60 L' 60 and d.c. resistance = - = - = 12R I 5 For the a.c. c~rcuit,P = 12R :. 1200 = I2 x 12 o r l2 = 100 and I = 10A S~ricccurrcnt taken hv the a.c. circuit F 10A

Thus X = f i 9 - 144 = f l 5R Reactance of coil = 5 ohms Ans. 2. Impedance of Branch A . ZA= Current IA

v

= -

= --

ZA

100

--

12.4

=

a= 12.4i2J F12K Z 9 =

8.08A. Also cos $, =

--

12.4

= 0.968 (lagging)

-

and sin .1

Impedance af Branch B. Z , =

d m i

-= ( X . 0 8

7N 1

=

+ 400

0.172)

X

@B

(8.08 x 0.242) - (4.64 x 0,928) - 1.96 - 4.3 = - 6.26A - .7 - - - -L/'9.53- + 6.26' = J91 39 = I IA Arls.

-

+

-

t4c11c.cI

d64

dlB

0,968) + (4.64 1 7 3 = 0 53A

React~\,ecomponent of current: ir = - I,, sin qb,, - 1" sin = =

3 12.4

= - = 0.242

= 21.652

m 4

. . \ c ~ i \ c cornponcnts o f current. 1 , -: 1, cclh d)A t 1,)

4,

SOLUTIONS

445

M PRACTICE EXAMPLES

3. Let the inductive circuit be circuit A, then: XA = 2xfL = 2 x 3.14 x 50 x 0.02 = 6.28R RA = 50n J~Q -t- '6.28' d250 i- 398.5 - 10JSS ZA = 10 x 5.05 = 50.5R

-

..

RA c~~+A= - --- =50 --

ZA

50.5

I0 - 0.99 (lagging) 10.1

Let the capacitive circuit be circuit B, then 127R r 200 I - - = 1.575 cos 4, = 0 sin 4, = 1 ' -127 Then I, = (3.96 x 0.99) + (1.575 x 0) = 3.92 + 0 = 3.92A I, = - ( 3 . 9 6 ~ 0 . 1 2 4 ) + ( 1 . 5 7 5 ~ 1 = ) -0.491+1.575 =

= 4.075A 3'92 = 0.962 (leading) Total current = 4.075A cos 4 = 4.075 Phase angle 4 = 15" 50' Ans.

4. Let the branches be A , B and C respectively. Then: X, = 2 n f L = 2 x 3.14 x 50 x 0.02 = 6.1HR Z A= Jg2 6.28' = 5- J =5 -

+

= 10.2R

and cos

4,

8 10.2

= - = 0.785 (lagging)

sin

4,

6-28 10.2

= - = 0.616

'10 15.7 cos 4, = = 0.537 (lagging) sln 4, = --- = 0.845 18.6 18.6

446

REED'S BASIC ELECTROTECHNOLOGY

Then Z , = J202 + 39.8' = J400 + 1584 = Ji-984 = 44.54R 20 39'8 = 0.894 cos ip, = ---- = 0.449 (leading) sin 4, = 44.54 44.54 Aclding the active and reactive current components. I, = I, cos ip, + I, cos ip, + I, cos 4, = (9.8 x 0.785) + (5.37 x 0.537) + (2.24 x 0.449) = 7.7 + 2.88 + 1.01 = 11.59A I , = -- I , sin $, - lBsin 4, + I , sin @ c = - (9.8 x 0,616)-(5.37 X 0.845)+(2.24 x 0.894) = - 6.04 - 4.54 + 2.04 = - 8.54A l'hrrl I = v/li2 + 1,' = dl ] ' . S O 2 + = d7747TT7293 = = 14.38A Ans. cos 4 =

--

14.38

0.805 (lagging)

4

=

36' (approx) Ans.

5. Apparent power = VI = 240 x 50.6 x l o F 3 kilovolt amperes = 12.144kV A Ans. true power 10 Power factor = 12.144 apparent power .. = 0.823 (lagging) Ans 9kW output (power) -- P(tic.~cnc.!= Input ( p o i i r ) lOkW or q = 90 per cent Ans .1

6. Output from motor = 1.5kW

Efficiency = 80 per cent

"0° Input to motor = - x 100 = 1875W 80 Also power input.to motor, V l cos q5 = 1875W : 1875 = 230 x 1 1.6 cos d Po\\.cr. colnpc7ncnt of input current. I cos rf, = 1 1.6 x 0.7 = X.12A I s m ip = 11.6 x 0,714 = 8.28A A t the neu power fi~ctor.the power component of current S I?.\ = I , c o \ (),. Also since I, cos 4, = I cos (p 8.12 I , x 0.05 8 . 12 a n d I , =- - - - = 8.55.A

..

React~ve ..

;=

.,

,

,

0.95

447

SOLUTIONS M PRACTICE EXAMPLES

Note. cos 4, = 0.95 sin 4; = 0.327 The reactive component of input current, at the new power factor = I, sin 4, = 8.55 x 0.327 = 2.8A So reduction of reactive current = 8.28 - 2.8 = 5.48A and capacitor current = 5.48A 230 1 O6 Capacitor reactance = ---- ohms = 5.48 3x1C

Rating of capacitor, = 230 x 5.48 x = 1.26kV Ar Ans. active power 10 -7. Load (a) Apparent power, S, = power fdctor 1

= lOkVA

C

Reactive power, Q, = S, x sin

4 = 10 x 0

= OkV Ar Load (b) Apparent power, Sb =: 80kVA at a power factor of 0.8 (lagging) Active power, Pb = 80 x 0.8 = 64kW Reactive power, Qb = 80 x 0.6 = 48kV Ar (kwng) Load (c) Apparent power, S, = 4OkVA at a power factor of 0.7 (leading) Active power, PC = 28kW ~ e a c t i v epower,-^, = 40 x 0.7143 = 28.57kV Ar Total power tnken from the supply, P = 10 + 64 + 28 = lO2kW Ans. Total reactive power, Q = 0 - 48 + 28.57 = - 19.g Total apparent power from supply, S = v'1022 IS

+

P

Power factor of combined load, - = S

Mains current

= 104kVA Ans. 102 - = 0.98 (lagging) 104 Ans.

"O0 - 416A Ans. 250 \ $(a) ha& voltage = I ~ O V1rnped;ince per phase of load = !OR =

e

100

a.

:. Load current p r phase = -- = 10A

*

10 Line current = ~ d a s current e = 10A Ans. Total power, P = 3 VI cos q5

1/

But V = d3V,, = 1.732 x 100 = 173.2Vand I = 10A v'3 x 173.2 x 10 x cos 30 :. P = loo0

or total power, P (b) L ~ n evoltage = t/3 x 100 Voltage per phase of load

=

2.598kW Ans.

3 x 100 volts x 100 ZPh 10 = t 3 x 10 amperes L ~ n current e = phase current = x \ 3 x 10 = t

v,-- v 3 Current per phase of load = -1 =

l ot,tl power, I' =

30A

v.3

An\ 1

1 ~ 0 11) 4

t 3 x z 3 x 100 x 30 x 0 8 6 6 1 000 = 3 x 3 x 0.866 kilowatts = 7 794kW Ans. -

(c)

Line voltage = 1OOV Voltage per phase of load = l00V Current ,,

.

- -=

10A 10 Line current = t / 3 I,,, = 1,732 x 10 = 17.32A Ans. Total power. P = \, 3 V I cos 4 3 100 \ 3 10 x 0.866 1000 = 3 x 0.866 kilowatts = 2.598kW Ans. ,

,,

4

(d)

Line Voltage = lOOV 100 Voltage per phase of load = t'3 100 - 10 10 - 7 3 Z'mPeres Current ,, .. ,. \,

3

Line current = Phase current = 10 1.732 Total power, P = v 3 VI cos 4 -

=

-=

0.866kW Ans,

10 ~ / 3

-

5.77A Ans.

SOLUTIONS TO PRACTICE EXAMPLES

449

9. Output from motor = 45kW = 45 OOOW Efficiency of motor = 88 per cent 100 Input to motor = 45 000 x 88

Since P =

(a)

3 VIcosc#J 51 140 Line current, I = t 3 x 500 x 0.9 \

or I =

51 = 65.6A Ans. 1.732 x 450 (b) Output from the alternator = input to motor = 51.14kW Ans. 51 140 x 100 watts (c) Input to alternator = 80. . 51 140 x 5 or output power of prime-mover = 4 ~1000 = 6 4 k ~Ans.

10. Input power to the system, P, = v/3 VI cos c#J % / 3 x 400 x 40 x 0.8 1000 = 22.2kW Power output from the motor at 91 per cent efficiency = 12kW :.Power input to the motor, P,,, = 12 x 100 13.18kW 91 ~ ' x3 400 x 40 Apparent power of system, S, = 1000 = 2 7 . 7 1A ~ Reactive power of system. Q, = S, sin 4 = 27.7 x 0.6 = 16.62kV Ar T o improve the power factor to unity, the reactive power of the motor must be equal to the reactive power of the system, so reactive power of motor, Q, = 16.62kV Ar. Apparent power to motor, S,,, = /-l = J13.18' + 1 6 . 6 2 ~ = d173.8 276.3 = 1= 21.21kVA Ans, 113.18 =Power factor of motor, cos &,, = s, 21.21 = 0.62 (leading) Ans. Total power taken from the mains = power supplied to the system power supplied to the motor = P, + P, = 22.2 13.18 = 3538kW Ans.

+

+

+

The phasor diagram shows the method of solution. Even though the problem is a 3-phase one, the diagram, as drawn, can be a ppl~ed.slnce balanced conditions can be assumed. L V A O f MOTOR

I /

I

TOTAL kW

Fig 25

CHAPI'EII 12

Total m.m.f., F = 5 x 500 = 25OOAt Mean circumference = xd = 3.14 x 300 x = 0.942m F = --2500 So magnetising force, H = 1 0.942 = 2654Atlm Ans.

1. (a)

(b) Since po =

B

-

H

then B = po H and B = 4 x x x x 2654 = 4 x 3.14 x 2.654 x = 12.56 x 2.654 x = 33.33 x tesla = 0.0033T = 3.3mT Ans.

(c) Total flux,

Q, = BA = 0.0033 x 1000 x = 3.3 x 10-6Wb o r =

weber 3.3pWb Ans

5 B = -Q,= 500 x 7 = - = 1.25T A 400x104 Also since B = p H = / L , 11, H then

2.

Length of iron = 250 x 10-3 metres X 1.25 So total m.m.f., F = 250 X x

o r F = .250

X

10-3

ampere-turns = 99.7At

Required ampere-turns = 99.7, say 100. Ans. 400 x 400 = 0,8T 500 x = %% total magnetomotive force Also, as H is given by length F 500 Then H = - = -- = 500Atlm i I Q,

3. (a) B = - then B = A

452

REED'S BASIC ELECTROTECHNOLOGY

Also. since B = p H = p, p , H then

or relative permeability = 1275 Ans.

(b) Reluctance

=

Length Arzi~ p,,/l, x ,+f

ampere-turnrlweber

/I Y

I

=

1.25MAiWb Ans

F = 400 x 2.5 4. (a) H = ampere-turns per metre = Thus H =

B = 0.5 x 1.25 pHorp = H loo0 p = , ~ ~ ~ p "

Again B Also

I 1.25 Ioo0 - 8OOA t/m 1.25

=

I = ( h ) Rcl\lct;lncc, .S -- -/,,,l

--

-

0 027

I --25x

I 500

-

0.625 loo0

- --

103 v

1 (I

(I

ampcrclurns/weker

453

SOLUTIONS TO PRACTICE EXAMPLES

square metres 11.3 = - teslas 12

5. Area of air gap = 1200 x 1.13 X .'. BA = 12 also since B

=I

then H for air =

p,H

B = p 12

11.3 x

4x x

lo-7

- 113 x

lo6

48x metre

The magnetomotive force for the air gaps is given by: 75 x lo4 x 2 x 2 x ampere-turns = 300 x 10 = 3000At Area of iron is the same as for the air gaps .: B value of the iron is the same Using the graph of Fig 26, we see that, for atflux density of 0,942T. the ampere-turns per metre length of the iron = 850.

I. 0

<

-I

w fi

0.5 0

0

XX,

Fig 26 Since length of iron path = 0.6m :. M.M.F. for iron = 0.6 x 850 = 6 x 85 = 510At Total magnetomotive force required = 3000 510 or 3510At Ans.

+

6. Circumference of flux path = rr x 0.2 = 0.628m = 0402m Length of air gap = 2 x . .. iron = 0,6261~1 This problem I S best solved by trial and error thus: Assume a flux density of 0.5T in the air gap and iron since these ;ire of the same cross-sectional area. Then using the graph of Fig 26: M . M . F . for iron = 520 x 0.626 = 326At

Total m.m.f. would be (326 + 795) = 1121At. Thus too low a flux density has been assumed. Again. assume ;I flux density 0.6T. then: M . M . F . for iron = 585 x 0.626 = 365.21At

M.M.F. for air

E6*4x

x

l 10-

3 or

55

of tl~.at required for 0.5T 6 = - x 795 = 1.2 x 795 = 954At 5 would be 365.2 + 954 = 1319At-still too =

Total n1.m.f low Assume 3 flu3 density of 0.7T. T h e n : M . M . F . for iron = 660 x 0.626 = 411.16At 7 M . M . F . for air = - x 795 = 1.4 x 795 = 1113At 5 1526At Total m.m.f. would be 413 + 1 1 13 Thus for an exciting ampere-turn value of 3 x 500 = 1500. the estimated flux density in the air gap would be a little less than 0.7T Ans.

-

7 . Since the B value in the cores is to be 1.2T. rlicn tlie A t l n l requlred u.ill be 650. This is seen from the graph of Fig 27. Tlic total ~ n a g n e t o ~ n o t i vI'orce e for thc cores will be 2 x 160 x l o - ) x 650 = 32 x 6.5 = 208At. In the \.ekes, the flux is the same a s t h a t for the cores a n d the flux density will therefore be different. as the areas are different.

Thus flux,

Q,

= 1.2 x

rr x 50' x

-

4

weber

SOLUTIONS TO PRACTICE EXAMPLES

455

Fig 27

From the graph, the At/m for a density of 1.066T is 330 Mean lcngth of flux path in yokes = ( 2 x 130) + (2 x 47) = 354mm = 0.354m Total m.m.f. for the yokes = 330 x 354 x = 3.3 x 35.4 = 116~8At Total m.m.f. for complete magnetic circuit = 208 + 116.8 = 324.8 say 325At Ans. 8. (a) Pull of magnet = 196'2N or 98,1 N per contact face

B ~A B*A Also Pull = newtons 2 ~ , 2 4~ x l o - ? B~A . ~ 98.1 , = 8~ 10-~ 98.1 x 8n x lo-' x 4 Whence B~ = x x 15 x 15 x

f i3

and B = \ m 6= 1.185T From the graph of Fig 27, for a B value of 1.185T, the H value = 560Atlm. n square Area of one contact face = - x 1 5 x~ 4 metre n Flux, @ = BA = 1.185 x - x 225 x lo-' weber 4 = 0.296 x 3.14 x 225 x lo-" = 0.296 x 3.14 x 2.25 x = 2.09 x 10-4Wb Since B for horse-shoe magnet :I I X51' tlic11,llro1111:ig 27, the i f vulue = 560 At/m Length of magnet path 115 = 180.5mm = 0.1805m. = rt x &

Note. Mean circumference of ring = 27t x (radius of ring) = 2x x (50 + radius of rod) = 2rt x (50 + 7.5) = rt x 2 x 57.5 = n x 1 1 5 millimetres Then m.m.f. for magnet = 560 x 0.1805 = 101.1At Flux - 2.09 x 2.09 B value for armature = -6= -Area 1j2 x 10 2.25 d F 0.9321 and H value = 215At/m (from Fig 27) Length of armature path = (115 15) = 130mm = 130 x metre So m.m.f. required = 215 x 130 x = 13 x 2.15 = 27.95At Total magnetomotive force required = 101.1 27.95 = 129 129At Cusrcnt = -- = 0.268A Ans. 480 In the air gap B = 1.15 teslas (h)

+

+

:. Flux, @ = 1.15 x

rt

-

4

2.03 x 2.03 x Flux density in core = n x-. 152 -4

x 1 5 x~ lo-.'

weber

=

= x

lo-"

457

SOLUTIONS TO PRACTICE EXAMPLES

Length of core path = x x 57.5 = 180.5mm = 0.1805m M.M.F. for core = 0.1805 x 470 = 84.8At 2.03 Flux density in armuture 152-4--j-6-6 - x lo-* a 2.03 -j:j5 3 :

= 0.905T

From curve H value = Length of armature path = M.M.F. for armature = Flux density in I air gap = 1.15 B ... H = - = 4~ lo-'

205At/m 130mm = 0.13m 0.13 x 205 = 26.65At I.15T, But B = p,H 1.15 lo7 ampere12.56 turnslmetre

M.M.F. for 2 air gaps - 2 x 0.5 x -

x 1.15 x lo7 12.56

r

9.125 x lo2 ampere-turns = 912.5At Total m.m.f. for circuit = 84.8 26.65 + 912.5 = 1023.95, say 1024At Ans. =

9, (a)

+

10-6

n1m2 Area of air gap = 4

- 314 -

Volume of air gap =

l o ' square metre

4 3.14 x 10''

4

2.5 x 10-3

= -3'14 x lob4 cubic metre

16

~ i u dinsity x in gap

=

T[ X

0.004 x 4 loo2 X

-

Energy stored in joules =

x B~

-x 2

~

lo' -

tesla = 0.508T lt

Volume

0

0.5082 3.14 10-4 X 2 4~ lo-' 16 - 0.508' x 10' - 0.258 x 103joules 8 x 16 128 258 -= 25 Ans. 128

-

458

REED 'S BASIC ELECTROTECHNOLOGY

( b ) Pull (newtons) =

=

B 2 -~ 0.508' ----- -

x 3.14 x X 10- 'X 4

2p,

2 x 477

806N

Ans.

Useful flux = 0.05Wb/pole 0.05 Flux density in air gap = 60000 x

10. .Air gap.

Also B = p,H B . . H value lor alr = - =

M.M.F. for air gap

= =

0.833

- 833 x 104 ---

-

4nx10-~ 471 66.2 x lo4 ampere-turnslmetre 66.2 x lo4 x 5 x 10-3

SOLUTIONS TO PRACTICE EXAMPLES

459

Total flux = 0.05 x 1.2 = 0.06Wb 0.06 Flux density in pole = - 6 teslas 40 000 x -4 o r B .= 1.5T From the magnetlc characteristic, plotted for Fig 28. a tlus density ( B ) of 1.5T gives an H value of 2'CKX)~tlm. Thus m.m.f. required for pole = 2000 x 250 x lo-' = 500At Pole.

Teeth. Total flux = 0.05Wb (same as the gap) 0.05 Flux density in teeth = 25 000 x = 2T From the characteristic, a flux density of 2T, gives an H value of 24 000At/m. Thus m.m.f. required for teeth = 24 000 x 45 x = 24 x 45 = 1080At Total field-coil m.m.f. = 3310 500 1080+ = 4890 say 4500At Ans.

+

+

CHAPTER 13 1. For a series combination, the equivalent capacitance is given 1 1 1 2+13 by C, where - = -- + --- = -- C 0.02 0.04 0.04 0.04 0.04 o r C = ---- = 0.0133pF Also Q = CV 3 Q = 0.0133 x x lo2 coulombs 1 3 3 x lo-" = 6 6 ~ 7 Vrind V , = -.-.-1.33 -.-. x l'I1c11 I = 0 4 2 ' ~lo-'; 0.04 x lij-" = 33.3v. 'I'he voltage drops are respectively 66.7V and 33.3V Ans.

',

2. The final two parallel 5 p F capacitors are equivalent to one unit of 10pF. The capacitance C of the branch, consisting of 20pF, 1OpF and 20pF in series, is given by: 1 + 2 + 1 = - o 4r C = 5 p F -- - + - + - = c - 2 0 10 20 20 ?O This series circuit is in parallel with a 5pF capacitor, making the equivalent capacitance = 10pF The final arrangement between A and B is now equivalent to a 20pF, 10pF and 20pF capacitor in series. The equivalent capacitance is g v e n by:

3. Since Q = CP'. :. quantity of electricity received initially is given by Q = 1000 x x 100 = lo5 x = 10- coulombs. Since the plates are separated by an insulated rod there is no loss o f charge a n d hence Q remains the same. Under the new condition since, as before, Q = CV then lo-' lo5 - 10' -- -1 0 0 = 333 ' I 1 r 1 I' - Q 3oux 10--~=300- 3 3 Thus the potential difference will have ~ncreasedby 333 3 100 = 233.3V Ans

c

,"

4. 1-he c;ipacltor \vould be made up from 10 plates In p;~r:~llcl, rnahlng one nsscmblq. interleaved with 9 plates in parallel li)r.~~llrlg tllc olllcl. plalc assembly: Tlicrc will be I S I I I ~ C ; Ihcp;i,--

46 1

SOLUTIONS TO PRACTICE EXAMPLES

ators or 18 electr~cfields and the total capacitance will be 18 times the capacitance between one pah. ,f 7lates. E,E, A Thus C of one pair of plates = EA - =1 I

= 88.5 x 7 x 2.58 x 10-l2 = 1.6 x farads or with eighteen units in parallel C = 18 x 1.6 x = 28.8 x farads = 28.8 x l o w 9 x lo6 microfarads = 0.0288pF Ans.

5. Since Q = CV, then Q = 3 x x 1 0 i 6 x 10 x lo3 = 3 x coulombs 3 x - 3 Q :. Flux density, D = A 10OOOx10-~-~ = 3 x l o v 4 coulomb per square metre Ans. electric flux density D -Also, permittivity, E = electric force d And electric force, d =. - = lo lo: = 10 x 106volts per 1 1 x 10metre 3 - ~1 0 - ~ Hence E = ~ I S O E = E , x E, = -C =

and

&

or

E, =

6.

E,

C

+ 3

30 = 3-39 8.85 EA

= - where

6

!= 3.5 x

and Hence

E = E,

x

x 1 0 4

1 0 x lo6 x 8 . 8 5 ~lo-''

- 3 x /OH 8 . 8 5 ~10'

Ans.

A = 6 x lo4 x = 6 x metres

square metres

square metres

E,

= 8 . 8 5 x lo-'' x 3 8.85 x 10-l2 x 3 x 6 x C= 3.5

Energy. W = ;Cv2 joules

=

1,264 x 16.2 x

= 20.477 x

joules = 2 0 . 4 8 ~ 5 Ans.

7. A 10-plate capacitor I S made up from two 5-plate assemblies intr.l.lr.;~\cdwith ei\ch other and separated by the dielectric. I I ~ r arc c tl1115 nine electric tields or the final capacitance is nine tlrnc4 t h ; ~ tof one plate arrangement. t lcrc

('

=

1,

/I

I

wliere A

=

I500 r 10 - " square metres

I

=

0.3 x

metres

F: = Eo X 6 ,

8.85 x

lo-'*

x 4 x 15 x 3 10-4 = 2.95 x 1 0 - l 2 x 60 = 2.95 x 0.6 x 1 0 - l o = 1.77 x lo-'' farads = 1.77 x microfarads Total capacitance = 9 x 1.77 x = 15.93 x = 0 ~ 0 0 1 6 g F Ans. (' =

4

8. Let

of' the series arrangement. 1 1 1 3 + 2 5 then- = - + - = - = - o r C = 12pF C 20 30 60 60 The clii~rgestored is given by Q = C V = 12 x x 600 = 72 x coulombs. 7 , 2 ~ 1 0 - ~ 3.6 f'.L). across 20pF capacitor A = --------20 x = = 360V Ans. I'.D. across 3OpF capacitor B = 600 - 360 = 240V Atis. If' P . D ;Icrc>ssF3 i.; 400V thcn P.D ;Icross p;~r;~llcl ;lrr;lnpcmcnt ivould Ilc 7 0 0 V . A l w tllc cquiv:~lentcapacitance must bc 60lrI: (double) since the voltage is half that across B. (' mlrst be 30jrF, bcinp in parallel wit11 A A i w the energy stored, W = -fCV2 = x 40 x x 20O2 = 4 x 2 x 10- joules 'Thus LC' = 0.85 Ans. (' = capacitance

F

4

'

SOLUTIONS TO PRACTICE EXAMPLES

463

9. Since Q = CV and Q = Ir then I1 = CV V o r I = C - where V = the voltage change I

50 amperes 1 x 10= 2 x lo-' x lo3 = 2A Ans.

ii. I = 40 x

x

iii. I = 40 x

x

O 1 x 10-

iv. I = 40 x

i

lM) amperes = 4A Ans. 1 x 'lo-

v. I = 40 x

x

50 amperes = 2A Anr. 1 x 10-

amperes = OA

Ans

,

The accompanying graph (Fig 29) shows the current and voltage conditions.

Fig 29

1 1 000 = 1 I 0000 10. Reactance of cable per kilometre = ----1 6 lo6 I 0 Also X, = ---- ohms 2nfC 2 x 3.14 x 50 x C

whence I I x lo3 =

lo' andC= lo microfarads 3.14 x C 1 1 x 3.14 or C = 0.289pF electric flux density Again since permittivity = electric force

Bur Q I)

= C'l' r

'I'

-.

/1

O

"Y

lo-'

4

II

lo'

v

coulomb per \qu.,rc metre

1 1 ~ 1 0 ~ 10 x 1 0 - ~ = I l x lo5 \,olts per metre The mean diameter of the insulation = 10 + 12 = 22mm The area of the dielectric = mean circumference x length = rrd x 1000 square metres = 3.14 x 22 x x lo3 = 3.14 x 22 = 69.1m2 Also, from the above, x l l x 10' coulomb per n = CV = 0.289 x ,.A 3.14 x 22 square metre - 3.179 x lo-' 69.1 D 0~289x11~10-~ Again E = - = 8 3.14 x 22 x 1 1 x 10' - 2.89 X 3.14 x 22 Also E = E , x E , dnd & =

.'

.

F '

- =

1

=- -

2.89 x

lo-'

3.14 x 22 x 8.85 x lo-"

I . Draw tile circuit diagram Let I 3 amperes = current in battery towards junction C ,, ,, C,D away from ,, ,, ,. I , ., ,, ,. CA ,. . 12 I\'otr. There I S no current between A D .: I , flows in A B and I , in D B . Applying Kirchhoff s laws l3 = I l + 1, Also point A is at the same potential as point D :. Voltage drop in A B = .oltagc drop in DB or 31, = 61, giving I2 = 21, Also voltage d r o p in CA = voltage drop in CD o r 81, = . y I , Substituting in the above 8 x 21, = .uI, o r s = 16.Q Ans. Also, if R = the resistance between points C and B with no current in Y

T h u s 1,

.

3,

3,

=

1

10 7.33

+

- -l o

8.33

= 1.2A

Ans.

2 . Let I , in current in battery X and I , = current in battery Y Applying K i r c l ~ h o f f slaws, we can build up the equation\ . . . (a) 8 = 1 . 5 / , + 6 ( 1 , + 1,) 7 . 5 1 , + Of2 . . . (b) and 4 = 31,. -t 6 ( 1 , + 1 2 ) = 61, + 91, Solving (:I) and (b) we have 24 = 22.51, + 181, 8 = 121, + 1812 16 Subtracting, 16 = 1 0 ~ 5 1o, r 1, = -= 1.524A 10.5 Also from ( b ) 91, = 4 - 61, = 4 - 6 x 1.524 = 4 - 9.144 = - 5.144

-

Thus the assumed direction for current 1, was wrong and battery Y is being charged. Current in battery X = 1 - 5 2 4 A (discharge) Ans. ,, ,, Y = 0 . 5 7 1 A (charge) Ans. Total current = 1,524 - 0,571 = 0 . 9 5 3 A in 6R resistor So terminal voltage = 6 x 0.953 = 5 72V Ans.

3. ( 1 ) By Maxwell's Method. Refer to diagram below (Fig 30).

Fig 30 Equatlng voltage drops to e.m.fs. then: 21, + 3(I, - 12)= 10 - 12 o r 51, - 312 = - 2 o r - 31, + 1 3 1 2 = 12 3(12 - I,) + 1012 = 12 giving 151, - 91, = - 6 and - 151, + 6512 = 60 54 Adding, 5612 = 54 or I2= - = 0.964A 56 So 51, = - 2 + 3 x 0.964 = - 2 + 2.892 = 0.892 0,892 o r I, = -= 0.1784A 5

Current I, in 10Q resistor = 0.964A Ans. .. I, in 10V battery = 0.1784A = 0.18A Ans. . I I in I , = 0,964 - 0,1784 = 0.7876A = 0.79A Ans.

(2) By Superposition of Current Method. See the attached diagrams (Figs 31a and 31 b). 2 OHMS

- - - ---

J OtcMI

I,

2 OHHI

i

I

I

I 0 OHMS

I, A

467

SOLUTIONS TO PR A C TI C E EXAMPLES

(a) Only IOV battery is effective. If R = resistance of parallel circuit 1 = --1 0 + 3 13 30 t h e n -1 = -1 + = - o r R = - = 2.3150 R 3 10 30 30 13 Total resistance = 2.3 15 + 2 = 4.31 5 0 So current I, = -- 2.325A 4.315 Voltage drop across R = 2.315 x 2.325 2.315 x 2,325 - 5.35 = Current I2= --- - 3 3 ,, 1 3 = - =5.35 0.535A 10 (b) Only 12V battery is effective. If R = resistance of parallel circuit Total resistance

=

3

+ 1.666 = 4.666R

l2 =-

2.575A 4 66 Voltage drop in parallel section = 2.575 x 1.666 = 4.29V Current I S = 4'29 - 0.429A 10

So current I4

4'29 - 2.145A Current I6 = 2 So current in 12V battery = I, - I, = 2.575 - 1.783 amperes = 0.79A Ans. ,, ,, IOV .. = I, - z6 = 2.325 - 2.145 amperes = 0.18A Ans. Current in 10R resistor = I, + I, = 0.535 + 0.429 amperes = 0.9644 Ans. 4. Let the currents be as shown in the diagram (Fig 32)

-

SOHMS

12 OHM$

1 1 0 OHMS

OHMS

VOLTS

Fie 72

WLTS

-

Taking .the mesh formed by the 4V battery, the 5R. IOQ and 12R resistors and the 6V battery. 4 - 6 = 51, + IOI, - 121, (a 1 o r - 7 = 51, + 10lz - 1113 Taliins the mesh formed bl, the 4V battery, the 5R reslstor n ~ i d15R resiqtor. 4 = 51, + I S ( / , - I*) . . . (b) o r 4 = 201, - 1512 T a k ~ n gthe mesh formed by the 6V battery, the 12R resistor and the 852 resistor. - 6 = - 1213 - 8(12 + 13) or 0 0 1 , t - 812 . . . (c) - 8 = 201, + 401, - 481, Multipl! ( a ) by 4. \ I I I I I I ,IC.I (11) 4 ?Or, - 1 512 glLlng - I2 = 551, - 4XI3 . . . (d) M u l t ~ p l >( c ) by 7.4. 14.4 = 4813 $ 19.21, - 12 = - 481, + 551, Adding ( d l ..- -- -2.4

=

Thus 1,

=

74.21, 3.4 - = 0.0324A 74.2

S u b \ t ~ t u t ~ for n s I, In ( c ) then h = 701, + 8 x 0 0324 o r 6 = 201, + 0 2592 6 - 0.2592 = glvlng 1, = 20 Current in 8 R resistor = I, + I, = 0.0324 + 0.287 = 0.3194A o r 0.32A Ans. .d

5. W ~ t hon11 the 4V battery eff'ective as hhown by the d ~ a g n i m ( F I 33) ~ S OHHI

10OHHS

w

=

T

A

I2 OHMS

7

8OHMS

VOLTS

I5 O H M S 'vvvvv

B

Let R be the resistance of the 12Q and 8 R resistors In ptirallel

S O L U ~ I O N STO PRACTICE EXAMPLES

469

Resistance of branch A = 10 + 4.8 = 14.8C2 Resistance of branches A and B in parallel = R .A.R 1 1 1 29.8 222 = 7.43R Then -= - + - = --- or RAB= RAo 14.8 15 222 29.8 Total resistance of circuit = 5 + 7.43 = 12,4352 4 Circuit current = --- = 0-322A 12.43 Voltage drop across A = 0.322 x 7.43 = 2.59V 2.59 Current in A = --- = 0.1615A 14.8 Voltage drop across 12 and 8 0 resistors = 0.161 5 x 4.8 = 0.775V 0.775 Current in 8C2 resistor = -= 0.097A

8

With only the 6V battery effective, as shown by the diagram (Fig 34). p

5 OHMS

A

10OHMS

12 OHMS

Fig 34

Let R be the resistance of the 5R and 1 5 0 resistors in ~arallel Resistance of branch A

=

10 + 3.75 = 33.750

Resistance of branches A and B in parallel = RAB li = 1 + -1 = 8 + 13.75 21.75 -Then --RAB 13.75 8 110 110 Total resistance of circuit = 12 f 5.06R = 17.06R 6 Circcit current = -= 0.352A 17.06 Voltage drop across branch A = 0.352 x 5.06 = 1.78V 1.78 Current in 8C2 resistor = - = 0.2225A 8

Currpnt due to 4V battery = 0.097A from junction of IOR and 12R resistors to batter) - ve tcrminal Current due to 6V battery = 0,2225A from junction of ]OR and 12R resistors to battery - ve terminal These currents can be added. thus: Current In 8R resistor = 0.3195A or 0.32A Ans. 6. Consider thc top junction or a p e s of the network (Fig 35).

rl

Fig 35

Let 1, amperes flow in the 2R reslstor towards this junction Let(, .. ,. resistor A away from this junction Let 1, ,, . ., 6R resistor in the direction left to right Then using Kirchhotl's laws, we can build up the following equations. ~~~~~~~~~~I I S I 21, t 41, - ()I2 = O (a) Top circuit 21, + 4(11 - 1,) = 12 . . (b) I l l l - l l l l l l 1 1 1 1 I I , ) ( I t I , ) - 4 = 0 . (c) C s ~ n g( a ) and ( b ) 21, + 41, - 612 = 0 6 - 41, = 12

47 1

SOLUTIONS TO PRACTICE EXAMPLES

Substituting in (a) 211 + 41, -'6(1.3311 - 2) = 0 21, + 41, - 7.981, + 12 = 0 41, - 5.9811 = - 12 Substituting in ( c ) 41, - 41, - 812 - 81, - 41, = 0 411 - 812 - 1613 = o 41, - 8(1.331, - 2) - 161, = 0 41, - 10.641, + 16 - 161, = O - 6.6411 - 161, = - 1 6 or 6.641, + 161, = 16

.

(d)

.

.

. . (e)

Multiplying (d) by 4 and solving with (d) and (e) 161, - 23,921, = - 48 161, + 6.641, = 16 Subtracting - 30,561, = - 64 or I, = 2.09A using (b) 611 - 12 = 41, or I, = 1.51, - 3 thus I, = (1.5 x 2.09) - 3 = 3.14 - 3 = 0.14.A Current is 0.14A (downwards) Ans, 7. XA = 27tfL = 2 x 3.14 x 50 x 0.1 = 31.4R Z A 2 = 30' + 31.42 = 900 + 986 = 1886 A'so '

~

1o6 - 1o3 - ---- ohm x~ 3.142 x 50 x 30 9.42

6

9.447

Joint G = 0.0159s Joint B = - 0.007 035 Y = d 0 . 0 1 5 9 ~+ 0.007 032 = I O - J~ 1 . 5 9 ~+ 0.703' = 1 0 - ~ J 2 . 5 2 5 + 0.50 = Jm = 10- x 1.74 siemens

'

z= lo' = 57.6Q 1.74

Joint impedance = 57.6R Ans.

The equivalent resistance = G x z2= 0.0159 x 57.6' = 52.8R

The equivalent reactance = B x z2= 0.007 03 x 57.62 = 33.18 x 0.703 = 23.320 230 Circuit current = - = 3.99A 57.6 Power = 3.992 x 52.8 = 844W Ans. 8. Branch 1 of section AB ,ind Z,' = 602 + 63 5 2 = 3600 + 4050 = 7650 Branch 2 of cection AB 7 x 314 x 50 x O 2 5 = 3 1 4 x 25 = 7 X 5 R \, 'ind L,' = 60' 78 S Z = 3600 + 6150 = 9750

+

Total (; = 0.01 4S 63.5 Also B, = - = 0,008 31s 7650 78 5 B2 = - -= - 0.008 06s. (Note the 9750

-

I-e sign)

Total B = 0.000 25s [ r B = 0 s + 0 02s2 = lo-' x 1.4 = 0.014s

and

1. = 10-2J].42 1

Z =

= 71.5R 8.014 OR Equivalent R = 71.5R Equivalent X Total R = 100 + 71.5 -. 171.50 X = 94.2 + 0 = 9 4 . 2 0 . , Note ,'A = 2 x 3.14 x 50 x 0.3 = 94.20 .. Z = fi71.5' + 94.2' = 1 0j 1 7 . 1s2 + 9.322 = 10~'?94.!2 + 88.74 = 10\/?83-06 = 10 x 19.7 = 197R 8

z

-

,

,

Impedance of scctlon BC' = v/ 10O2 + 94.1' i H 840 137.4 or ZBc = 1.17.4C)and \olt:~gcd r o p = 2.54 = 348.99V 7'hu\ 3 4 0 V will I>c tlic vo1t;lgc ;Icro\s \cctiorl I%('

-

SOLUTIONS TO PRACTICE EXAMPLES

a n d ZA2=

+

j2

9.422 = 9

+ 88.74

473

= 97.74

+

a n d ZB2= 10O2 7.8' = 10 OOO + 60 = 10 060 Branch C X = X, - X C 1o6 = (2X 3.14X 5OX 0.02) 2X 3.14X 50X 3 0 Zc2 = 72 + 4.322 = 49 + 18.66 = 67.66 3 -9.42 = -0.0964s Then GA - ---- = 0,0307s B A = --- 97.74 97.74

Total G = 0.1436s Y = =

~ lo-"

Total B = - 0.031 8 s o - ' J 10-'J2,062 ~ o + 0,101 T = l o - ' X 11.471 = 0.1471s

z=--1

- 6.798R 0.147 1 T h u s circuit impedance Z = 6.798R Ans. E q u i v a l e n t resistance R = G x Z 2 = 0.1436 x 6.79g2 = 0.1436 x 46.44 = 6.5R An5 Equiv;~lcntrc;~ct;incc(intluctive, sincc total R i s -- v t ) X 2 ' = 0.0318 X 6.798' = X, = = 0.03 18 x 46.44 = 1.48R Ans.

%, = R, = 2 0 0 X A = OR X = 2 x 3.14 x 50 x 0.05 o r XB = 15.70R Z R- 2 = 52 15.7' = 25 + 246.49 = 271.49 1o6 Circuit C X = 2 X 3.14 X 50 X 50

10. C i r c u ~ A t Circuit B

Then

+

GA = -R~ 7 -

Z,

-,

20i = - = 0.05s 20 20

Thus total G = 0.0684s

~ =

474

R E E D ' S BASIC ELECTROTECHNOLOGY

0 20 15.7 = - 0'0578s ---271.5 1 -= 0,0157S 63.65

BA = 7 XA - - T =OS

Similarly

z,

X7 B BB = = ZB B

-

Xc ZC2

-

-

Total B = - 0.0421s Hence Y = 10-2J6,842 + 4 . 2 1 = ~ 1 0 - ~ 2 / ~ m 2 x 8.05 = 0.0805s - 10-~\/64_51 = I . . I'} IOO h 0.0805 :H,OSA A I ~ s . -

(-.irctilt p o w c kictor is plvcn by tllinu

)

;lnd is lagging. since 6 is '

o r riel ~ n d u c l ~ v c :

: cos

4

=

0.0684 = 0.85 (lagging) A n s 0.0805

CHAPTER 15 1 . Anode current (amperes) =

power d i s y y t e d (watts) --. anode to cathode voltage (voli,)

Thus anode current = 2.1mA Ans. 2. The d.c. resistance of the valve, for the condition, is given - anode to cathode voltage anode current 58 = 9.666 x lo3 ohms thus d.c. resistance, R = or9.67kQ Power dissipated is given by: P = V,I, Thus P = 58 x 6 x l o p 3 = 348 x

or P = I , ~ R watts or 348mW Ans. o r P = (6 x 10-3)2 x 9.666 x lo3 = 36 x x 9.66 = 347.76 x watts o r 348mW Ans.

3. Resistance, r, =

1 conductance

or r, = 1 =

1.64 x

-

- 8.13 -

x lo3 1.64-

4.957 x lo3 ohms = 4.96kQ

Ans

- change in anode voltage

4. A.C. resistance =

change in anode current 129-75 - 54 x 103 o r ra = (22 - 12)io-%10 = 5.4 x lo3 ohms o r 5.4kQ Ans. D.C. resistance, R (for 75V condition) 75 = 6.25 x l a 3 ohms 12 x 10= 6.25kQ Ans. D.C. resistance, R (for ,129V condition)

-

129 = 5.864 x 1030hms 22 x 10= 5.86kR Ans.

5 . T h e c h ~ ~ r a c t e r ~ sat pi cp e a r s t o be straight line between t h e 10.5 a n d 4.2rnA values a n d , o v e r this region, t h e 2.c. reclstance \vould he 95 - 15 rS i ,' ra = (51, (10.5-4.2)lO-"

- .l o 6.3

lo'

=

1 5 9 L R An,.

Fig 36 6 Tlic c l i ; ~ r ; ~ c t e r ~ 1s \ t ~plotted c a s s h o w n a n d the load line 1s d r a \ \ n In t I ~ u \ . I A h \ r ~ ~ nI:,c = O T h e n voltage o n a n o d e ~ v o u l dbe 6 o V t o ! ? I \ C poll11 ,\ II

. \ \ \ L I I ~ ~il:' I~

\.II\c'

resl\l;lnce 10 be o t ' r n l n ~ m u balue. ~ ~ i Thc

I \ l l l ~ l \ o l ~ t < ~ l l l, 1c1 1~ k I l 111% g l \ c \ tI1c lo:l(l I l l l C 1'01., I l o a d - r e \ ~ \ t a n c e v:~lueof' 30OR. Tlic p o ~ n to l ' ~ n t c r s e c l i o nwith t l ~ ec l i a r a c t e r ~ s t ~Ic\ I' a n d rhc standing current 15 IOOmA. A n s . T h e power dissipated in t h e load resistor = (100 x 1 0 - 3 ) 2 x 300 wiitts = I0 r 300 = 3W Anh.

l J 0 l l ~ lI{

'

477

SOLUTIONS TO PRACTICE EXAMPLES

- - ...~.~ . ~

-

.. .

-

.

.

-. . -.-. -

- -

- -.

.

.

200

IS0

,-.5

S

-z

$100 3 V 4

cl z

0

4

50

0

A

1 0

K,

X)

ANODt V O L l A C t

43 VO(VOLTS)

Y,

-

Fig 37 Alternatively (from the load); for polnt P, V , = 30V . . Voltage dropped across load resistor = V - Va = 60 - 30 = 30V Power dissipated = voltage across resistor x anode current = 30 x 100 x = 3W Ans. 7 . Tablc of results.

Sincc I, = Va3i2 thcn: Anode voltage ( V ) 0 1 2 3 4 5 Anode current ( m A ) 0 1 ,. ,, (mA) 0 1 2.83 5 . 2 8 1 1.2 Load voltage (V) 0 1 ' 2.83 5.2 8 11.2 Supply voltaze (V) 0 2 4.83 8.2 12 16.3 Specimen calculation, for the 2V anode voltage condition : - 1 8 = 2.831nA 1, = V a 3 , ' 2 , So I 33 2 - (23 ) 1 ' 2 - The supply voltage for this condition = load voltage drop + valve voltage d r o p = /,R + 5 = (2.83 x x lo3) + 2 o r V = 2.83 + 2 = 4.83 volts. The deduced valve static and dynarn~ccharacteristics :\re plotted,, as shown below. with I, to a lia and V base. Fro111 the latter, the value of anode current, for an applicd voltage o f 8V, 1s seen to be 5.1mA. Note that this result could have been obtained by drawing

i I

1

-

--

-

--

VOL1 A C k

Fig 38 a load line on the static characteristic for a resistor of I kR and a supply voltage of 8V. This is shown dotted and the current is seen to be 5.1mA. For this problem, since the dynamic characteristic has been deduced the load-line method is unnecessary.

8. The problem is solved by plotting the valve characteristic and drawing the load line for the resistive circuit. Solution will be assisted, if the circuit is drawn out. Consider the valve on open-circuit ie non-conducting. Then the current through the resistive circuit would be ,111~1 [tic

\oI[,~gedrop across the anodc rcslstor woulci hc 4 10 lo3 = 40v Thc p.d. between anode and cathode 31' the calve, l o r zero I, would be 240 - 40 = 200V This gives a kalue for point A on the load line. Next consider the valve 'short-circuiting' i c ~witli no p . d . across it. The full 240V would be applied across the anode ',

resistor and the current would be --i40 10 x 1 F = 24mA This is

479

SOLUTIONS TO PRACTICE EXAMPLES

for a zero anode-voltage condition and gives point B on the load line. Join AB and the intersection point shows a value of 60V applied to the valve and an anode current of some 17mA Ans. A s a check, we have: For 60V across the valve, ii current

*

= I.2rnA through the 50k22 shunt circuit. With 50 x 10 180V would be dropped 60V across the valve, 240 - 60 180 across the anode resistor ie a current of 10 lo-" =8mA :. Valve current = 18 - 1.2 = 16.8mA. a s indicated by the graph. Ans.

of

-

t

2 6 0

J t-

IY 3

40

a V

B

:2 0

0

z 4

100 ANODE

-

200 VOLTAGE

250

v~(v)

Fig 39 9. P = V I , where V is the forward voltage of the diode for ;I current of 5A. :. P = 0.7 x 5 = 3.5W k n s . The rate of heat dissipation per 'C rise is 0.1 J, s o r 0.1W Thus the rate of heat dissipation for a 70°C rise is 0.1 x 70 = 7W If the voltage drop is constant a t 0.7V then the diode

7 = 10A Ans. current is given by I, = 0.7 10. Since a number of examples involving the use of a load line have now been illustrated. this problem should present no difficulties. T h u s :

( a ) F o r a n o anode-current condition, 1, = 0 amperes and V , = 1 volt. Thus P o ~ n A t is obtained. F o r a zero diode-reblstance condition. a p p l ~ e dvoltage I, = load resistance 1 = O.0lA o r lOmA SOI. IPa = 0 ~ 0 1 t h 100 (acrohs dlode Point R is then o b t a ~ n e d Tile rcsulr~nglo,td liric Intersects w ~ t htile c u r \ c s to g l \ e points X a n d Y with tile r e q ~ ~ i r eanswers d At 25'C the I o ; ~ dcurrent is 3.4mA Ar 00' C' rllc Ir):~dcut.t.cnr is 4. 1 t11A The current \ariation would be 0.7rnA Ans. 0:

1, =

(1)) I1.C'. rc.si\tancc I'or point X

=

=

-D.C. resistance So: point Y = = =

I

Fig 40

0.66- - - - -

3.4 x 10-3 0,194 x lo3 !94R Ans. 0.59 4.1 0.144 x lo3 1440 Ans.

SELECTION O F TYPICAL. EXAM INATION

QUES'TIONS SECOND CLASS 1.

A thin rectangular plate 350mm by 250mm 1s totally covered on both sides with nickel 0.12mm thick.in 8.25h. The current required would, if supplied to a voltameter. cause 0.0805kg of silver to be deposited in lh. If the E.C.E. of nickel is 304 x 10-9kg/C, calculate the density of nickel in kg/m 3 , if the E.C.E. of silver is taken as 1 1 18 x 1 0 - ~ k g / ~ . C

2.

A 1 IOV d.c. lighting system comprises six 150W and forty 60W lamps. Calculate the inductance of a choke-coil of negligible resistance which, when placed in series with this system, would enable it to be operated on 230V, 50Hz mains.

3.

The resistances of the armature, field coils and starter of a 220V shunt motor are 0.2, 165 and 9.8Q respectively, the field being connected across the first stud of the starter and an armature terminal. Calculate (a) the field current a t the instant of starting; (b) the field current when running, and (c) the total current taken by the motor at the instant of starting, considering the armature as stationary.

4.

If the instantaneous value of a current is represented by i = 70.7 sin 5201, calculate the current's (a) maximum value,

(b) r.m.s. value, ( c ) frequency, (d) instantaneous value 0.001 5s after passing through zero. 5.

A 220V furnace uses a current of 4.5A to melt 5.5kg of lead in 12 min. If the initial and melting temperatures of the lead are 16°C and 327°C respectively, its specific heat is 0-1278kJ/kg°C, and its latent heat is 22.72kJ/kg, calculate the furnace efficiency.

6.

The maxiniurn value of a sinusoidal current wave is 170A. Find g r a ~ h i q l l y the instantaneous current value after 0-OOls, 0.003s, 0.006s, and 0.008s after zero and increasing positively. Take the frequency as 50Hz.

7.

The open-circuit voltage of a cell, as measured by a voltmeter of 10052 resistance. was 1.5V. and the p.d. when supplying current to a 10R resistance W L I S I.25V. me:isured by the same voltmeter. Detennine the e.m.f. and internal resistance of the cell.

8

An nlternat~ngcurrent series circuit consists of a coil '4 that has an inductance of 0.3H and negligible resistance and ,I resistor B of 100R. Tile supply voltage is 200V with a frequency of 50Hz. Determine ( a ) the impedance of the circuit. ( b ) the current flowing, ( c ) tlic power factor.

0

Wli;~ric ~nr:lnthy rllc lrnii 'h;~chc . n ~ . f .; I' S ;~pplicdto ;In electric motor? A 40kW, 220V shunt motor has a full-loiid efficiency of 90 per cent, an armature resistance of 0.075R and a shunt-field resistance of 5 5 0 . When 'at starting', the starter handle is moved onto the first stud, it is desired to limit the current through the armature to 1.5 times the value which i t has when the motor is on full load. What must be the total value of the starting resistance? If, on overload, the speed falls to 90 per cent of its normal full-load value. what would be the armature current? Neglect the effect of armature reaction.

10

Determine the instantaneous value of a sinusoidal e.m.f. of frequendy 50Hz and lOOV maximum value, 0401 seconds after it has passed through its zero value. Determine also the time that elapses before the voltage reaches 50 per cent of its maximum value.

11.

State Lenz's law. An iron ring is wound with a coil of 84 turns and carries 0.015rnWb of residual magnetism. When the coil is excited. the magnetic flila increases to 0.3mWt-1in 0.12s. C';~IcuI~te tlie ; I V C ~ ~ Iviilue ~ C ot'tlie e.111.l'.wliicli will bc self-induced in the exciting coil wliile the fluu is incrcasin~. and state the direction in wliicli ~t u~llact rcliitivc to tlic supply voltage, giving reasons.

12.

Find the impedance and power fiictor of an 21.c. clrcuit consisting of 1\vn picccs of :lpp:rratus in series. Piece A Iias a resist;ince of 2R i ~ n dinductive reactance of 14R. a n d piece R II;IS ;I rcsi5t;incc 01' 100 and ;I c ; ~ p ; ~ c i t ireactance \~c of 6i2.

SELECTION OF TYPICAL EXAMINATION QUESTIONS

483

13.

Define the temperature coefficient of regstance of a conductor. Name a conductor which has a negative temperature coefficient. When first switched on, the field winding of a 200V shunt motor takes 2A. After running for two hours the field current is observed to have decreased to 1.7A. tllc supply voltage having remained constant and the setting of' the shunt regulator not having been altered. It' the ambient air temperature is 15"C, calculate the average temperature rise within the windings. Temperature coefficient of resistance of copper is 0.004 28 at 0°C.

14.

A 440V single-phase motor is rated at 7.5kW and operates at a power factor of 0.8 (lagging) with an efficiency of 88 per cent. Find the current taken from the supply.

15.

Explain how the value of a current can 'be measured by the deposition of a metal from a copper sulphate solution. Find the quantity of electricity (in coulombs) which will deposit O.Olkg of copper from a solution of copper sulphate. What current would be required if this process took one hour? What mass of silver would be deposited from a silver nitrate solution if the same current flowed for the same length of time? E.C.E. of copper = 330 x 10-9kg/C. E.C.E. of silver = 1118 x 1OV9kg/C.

16.

A coil consumes 300W when the voltage is 60V d.c. On a n i1.c. circuit the consumption is 1200W whcn Ihc volt;~gc1 5 130V. What is the reactance of the coil?

17.

A motor has four poles, its armature is 0.36m in diameter and has 720 conductors whose effective lengths are 0.3m. The flux density of the field under the poles is 0.7T. Each conductor carries 30A. If the armature is turning at 680 rev/ min, find the torque in newton metres and the power developed if only two-thirds of the conductors are effective.

18.

Define the average value and r.m.s. value of an alternating quantity. Calculate the average r.m.s. value for the stepped half wave given. Time (ms) 0-10, 10120, 20-30, 30-40, 40-50, 50-60, 60-70 Steady current (A) 2 4 6 8 6 4 2

d4

REED'S BASIC ELECTKOTECHNOLOGY

19.

T w o 200V lamps are connected in series across a 400V supply. O n e lamp is 75W. the other is 40W. Wliat resistance would have to be connected in the circuit so that each Iainp gives its correct i l l u m i n ; ~ t ~ o nW? h a ~would be the power loss in the iesistance? Wliat is tlie extra cost per week. if electricity cost 0 3 p per un:t?

20.

A series circuit consists of a capacitor of 5 0 p F a n d a coil of inductance I.5H a n d resistance 3000. Find the total impedance when u o r k i n g on a 50Hz supply. Find whether tllc cu~.rcntIc:~d\01. I : I ~ tllc \oltagc.

21.

A OV battery is connected to polnts Anc 1$1,1cIgcA l i ~ ' l > Side\ , A l i , LIC' ;11ii1 A I ) 01' t l bridge ~ Iii~~~c re\l\ti~nce value5 oI' I ,511. 212 a n d IR respcctivcly. If tllc gi~lvanometer is connected across BD a n d shows no deflection, find tlie \,slue of resistance of side DC.

22.

Explain the term 'power factor'. A n alternator supplies 560kW a t a power factor o f 0 . 7 (lagging). W h a t extra power would be available if the power factor is increased to 0.8 (lagging) for the same kV A output'?

33

1 lie ;~rln:itur-ewinding o l ' a 5 1 - p o l e . Ii~p-woundgcncr;r[or 1s made u p from wlre 250m long and 7n1m2 cross-section:~i ;ire;\. It' tllc >lxxlfic resistance ol' cnppcr is 1.7 x 1 0 % ~ m , find the rexlstancc of' tile armature.

24.

IOOV a.c. I S applied to a circult of 3R resistance a n d 4R rc:Ictiince. F ~ n d( a ) the current in the circuit, ( b ) the a c t i x e.m.f. (resistive voltage d r o p ) of the circuit, (c) the e.m.f. of self-inductance (re:tctive voltage d r o p ) .

25.

A p c n c r : ~ t o ~is. o v e r - c o i n l ~ o i ~ ~ i d;111d e d designed to 4~11~1~1y tlic cll.cuir 1oi1J ; ~ t770C'. 11' tllc 111;1clliiletcrlnin;il ~ o l t ; ~ gI \ c 270 2 V u llcn tllc l o ; ~ di~~irrcrit1 5 1 OA. wI1;1t will be the r ~ ~ a c l t r rtcririrnal ~c voltage w11c1it l ~ cIoi~dI \ WOA'!

36

A coll ol' 115R ~ m p e d a n c eIi;is :I resistance of 1 0 0 0 wllcn conncctcd ,\cross :I SOH/ hupply. Find its inductance. II' tlic impcdiince fall5 to 1 3 0 . 6 0 when tlle I'requcncy is varied, finl{ tllc IlCLk I'l-ci~ucllc),L ,1Iuc.

SELECTION O F TYPICAL EXAMINATION QUESTIONS --- ---- - --

--

485

27.

4.5 litres of fresh water at 17 C is heated to boiling point in 15 min. If the h a t e r is 80 per cent ellicient and the supply voltage is 220V, find tlie current taken I'rom the mains and the resistance of the heater. Take the density of water a s I kgllitre and thc specific Ileal c a p ~ ~ c i t;IS y 4.2kJ:kg ( '

28.

An alternating voltage of r.m.5. value IOOV is applied to :I circuit with negligible resistance and an inductive reactance of 25R. Determine the r.m.s. value of the current flowing. Show graphically the var~ationof current and voltage durlng one cycle of applied voltage. What is the value of the current when the voltage is at its maximum value?

29.

A generator has eight brush-arms, eirch with six brushes. 30mm long and cnch with :I bearing surb~ceof 301nm b 20mm. The current density is 0 . 0 5 4 ~ : m min~ tlie brushes Find the sectional area of tlie cables, if the leads to the switchboard are each 9.2m and the current density must not exceed 1.0A/mm2. Find the power lost in the brushes and cables. The resistivity for carbon and copper is 2550 x and 1.7 x 10-sQm respectively.

30.

A coil is connected in series with a capacitor of 60pF across a 200V, 50Hz supply. The current is 3A and the power absorbed is 144W. Calculate ( a ) the p.d. across the c;~p;lcitor,( b ) thc rcsistilncc and induct;lnce of the coil. ( c ) the power I'uclor 01' l l i u coil, and ( d ) l l ~ epower I'i~clor01 111c whole circuit.

31.

An iron conductor and an aluminium conductor are connected in parallel to a supply. The iron conductor is 10 per cent longer than, and half the diameter of, the aluminium conductor. Given that the ratio of the resistivities of iron to aluminium is 40 to 13, find the ratio of the currents in the two conductors.

32.

Two currents I, = 14.14A and I, = 8.5A with a phi~se difference of 30G,are fed into a common conductor. Find the resultant current and the heating effect in joules when it passes through a resistor of value 4S1 for a period of 2 minutes.

48 6

REED'S BASIC ELECTROTECHNOLOGY

33

A six-pole d.c. generator has 498 conductors. the e.m.f per conductor being 1 3 V and tlie current in each IOOA. Find the e.ni.l'. and current ciutput of tlie annature. if it is ( a ) lap wound. ( b ) \+:a\,e\\,ound.

34

A c~rcuittakes a current of 10A from 220V. 50Hz malns at a power f'actor of 0,866 (lagging). Find the value of the current when tlic ~.oltageis ( a ) passing through its maximum ~ , ; ~ l and u e ( b ) 0.005 seconds later.

3

Tlie filament of an electric lamp is 600mm long and 0.04 in cii;1liic1cr.Tlic resistivity of the material when cold is 65 x 1 0 - ~ R mIf. tlie resistance of the wire when hot is 5 tirncs ~ l i crccij~;~ricc when i t is cold, wlli~t would he tllc work~ngcurrent taken by tlie lamp when placed in a circuit having a supply voltage of 1 IOV?

111111

36.

An inductance coil has a resistance of 19.50 and when connected to a 220V, 50Hz supply, the current passing 1s 10A. Find the inductance of the coil.

37.

State F a r a d a ) ' ~ and Lenz's Laws of Electromagnetic Induction. A 4-pole. 250V motor has its armature removed in order to test the continuity of the field windings which are connected in series and consist of 2000 turns each. What is tlie average e.m.f. induced when the current is switched off. if tlie flux .falls from 0.026Wb to 0.001 Wb in 0.2s?

38.

A choke when connected across 206V a.c. mains, passed a current of 10A and dissipates 500W. If it is connected in series with apparatus having an impedance of 5 0 and a capac~tivereactance of 4R, find the impedance and power f'rtctor of the complete c~rcuit.

39

;\ p~ecco I ' c o 1 ~ ~ wire ~ e r I S belit 1 0 t'os~ii;I circle and ;~notlicr. plece ol' the same wire is placed across to Sorm a diameter. ; ; I 1 I lic 111r1ct ori is hclng clcctric;~llyconncclcd. I T t hc rcslbL ; I I I C C 01 t l ~ ch [ r i ~ ~ gw l ~~ts cI \ 211. find [!it LO{;II cusre1it 110~411ig when a p.d. of 220V is applied across the Junctions.

40.

Tlie average value of a sinusoidal waveform is 125A irnd the frequency is 60Hz. Calculate tlie first time from zeri, when the instantaneous value of the current is 95A. Find ~ I I ~ C tilt ) r.111.s.\ , i l I t ~ t 01' t l i t C L I I . ~ C I I ~ ,

SELECTION OF TYPICAL EXAMINATION QUESTIONS

487

41.

An 15kW motor of efficiency 90 per cent ic supplied ar 240V by a 2-wire system. The supply cables are 500m long and are of diameter 5mm. Find the current taken by the motor, the voltage at the supply point and the effic~encyof the distribution system. Take the specific resisrancc 01' copper as 1.7 x 10-8Rm.

42.

If the impedance of a circuit is 20R, the resistance I S 16R and the inductance 0.047 75H, find the frequency of the supply.

43.

Differentiate clearly between the kilowatt and the kilowatt-hour. A heater with an efficiency of 85 per cent develops lOMJ in 30 minutes at 200V. Find the energy consumption in kilowatt hours and the current taken. Find also the length of wire in the element if its resistamce is 0.26 ohm per metre.

44.

An electric heater of resistance 6 . 3 2 is connected in series with a choke of inductance value 0.1H. If the mains Srequency is 50Hz, find the vol?age to be applied to the arrangement in order to maintain' 110V across the heater. I f the frequency was increased by 5 per cent, keeping the applied voltage constant, find the voltage across the heater.

45.

An ammeter is tested with a silver voltameter. The welght of the cathode before deposit was 04238kg and after a steady current was passed for half' an hour, the we~ghrW;I\ 0.039 25kg. The reading on the ammeter was 7.3A. Find the error of the ammeter, taking the E.C.E. of silver as 1 1 18 x 10-~kg/C.

46.

A resistor of ohmic value 3R is connected in series with a coil of inductance 0.1H and resistance 1R. If lOOV at a frequency of 50Hz is applied to the circuit, find the current flowing.

47.

A 500V, d.c. shunt motor has a full-load armature current of 20A. Three per cent of the input power is dissipated a s heat in the armature. What would be the current on starting if.5OOV is applied across the armature. Find also the value of starting resistance required to limit the starting current to twice the full-load current.

-

48.

Find the total efyective reactance of a 50Hz circuit made u p from a coil of inductance 100mH, in series with a capacitor of 20jtF. I f the coil has a resistance of 10R, find the impedance of the circuit.

49

The nrmature resistance of a 200V. shunt motor is 0.4Q a n d the no-load annature current is 2A. When fully loaded and taking an armature current of 50A. the speed is 1200 revlrnin. Find tlie no-load speed and state the assumption rn:~dcin tlie caIcu1;1t1(7n. *

50.

A c o ~ lti~hes5A wllen connected across a 220V. 59Hz .;t~pplv.The power tahen is 420W. Calculate ( a ) the p o ~ v c ~ . I:tctol.. ( 1 ) ) tile I I I ~ ~ > C ( I~Y ~ ~ I I\ [I~~ I IC ) CC Crci1cti1ncc ,. :111d1nc1~1cLance of' thc ccil.

SOLUTIONS TO TYPICAL SECOND CLASS EXAM INATION QUESTIONS 1 . When used ,in the voltameter i m = z l t

Wllcn used Ibr pli~ting: Area of coating = 350 x 250 x 2 = 175 000mm 2 Volume ,, = 175 x lo3 x 12 x = 2 1 000mm3 Also mass of nickel deposited = 304 x x 20 x 8.25 x 3600 = 0.304 x 16.5 x 36 x kilogrammes 0.304 x 16.5 x 36% So density = 21 000 IO-~ - 304 x 16.5 x 36 21 = 8600kgjm3 Ans ( 6 x 150) + (40 x 60) = 900 + 2400 3300W System current = --3300 = 30A 110 230 I m p e d a ~ c erequired on 230V a.c. = - = 7.660

2. Totalwattage

= =

30

Keslstance ol'lamps =

l lo

---

30

= 3,660

Reactance of choke coil = J7q66' - 3.662 7= J 5 8 . 6 8 - 13.4 = d 4 5 . 2 8 6.73 - 6.73 x lo-' 3.14 2 x 3.14 x 50 = 2.14 x = 0.0214H Ans.

Inductance =

220 1.35A Ans 165 (b) When running, the starter resistance is inserted into the field circuit by vi,rtue of the position of the contact ami. Field-circuit resistance = 165 + 9.8 = 174.80 220 Field current when running = -= 1.26A Ans. 174.8

3. ( a ) Field current at instant of starting =

--- =

( c ) Armature-circuit tesistance, at instant of starting = 0.2 + 9.8 = 10R 220 Annature current = - = 22A 10

..

Total current taken by motor = 22

+ 1.33 = 23.33A

Ans.

4. ( a ) Maximum value of current = 70.7A Ans ( b ) The current is sinusoidal :. r.m.s. value = 0.707 x maximum value = 0,707 x 70.7 = 49.94A Ans. (c) i = 70.7 sin (520 x 0.0015) is in the form i = I,,, sin ot wllcrc 1 , ) = 520 r;~di;ins/sccond.Also (11 cquals 2rtf o r 2n/' = 520 520 - 520 = 82.8Hz Ans. Thusf = -2 x rr 2 x 1.14

( d ) Again i = 70.7 sin 2xft. and if degrees are used for the angle, then i = 70.7 sin (2 x 180 x 82.8 x 0.001 5) = 70.7 sin (3.6 x 8.28 x 1.5) = 70.7 sin 44.65 = 70.7 x 0.703 = 49.65A Ans.

220 12 k .~ l o w a t thour 4'5 x loo0 60 = 0.198kW h = 0.198 x 3600 kilojoules = 712.8kJ Energy received (by lead) = (5.5 x 31 1 x 0.1278) + (5.5 x 22.72) kilojoules = 218 7 + 125 = 343.7kJ Energq put out by furnace into lead = 343 7kJ 343.7 Etticiency = = 0.482 = 48.2 per cent Ans 7 12.8

5. Energy input to furnace = d

6

Plot

;I

sinc u.;I\.e on gr;tpli paper with

;I

masimum v ; ~ l r ~ e

I>;I\L* 01' 0 01 5 . ,A I'I~CLILICI~L~!01. 5 0 1 I / give\ 1 1 1 ~ time of a hall' wave a s 0.01seconds. The wave can be plotted from a phasor of length equal to

01. l70A

, I I I ~;I

170A o r by usc of tables to obtain ordinates. Thus'for 30 the insti~ntanenusi.alue o r ordinate would be 170 x sin 30 = I 70 x 0.5 = 85A. SO for 45 i = I 70 x n.707 = 120.2A 1 0 0 ' i 147,09A. F o r 00 i T 170A. C I C .

SOLUTIONS TO TYPICAL EXAMINATION QUESTIONS

49 1

Answers from the deduced waveform: When time t = 0.001s. i = 57.5A. When time t = 0.003s, i = 132A: When time t = 0.006s, I = 162A. When time r = 0 . 0 0 8 ~ ~ i = 102A.

7. Let E = the e.m.f, of the cell and R , = the internal rcslstance. With the voltmeter across the cell terminals only, 1.5 Current taken by voltmeter = -. = 0.01 5A 100 . . . (a) Then E = 1.5 + (0.015) x Ri) With the voltmeter and resistor across the cell terminals, 1.25 Current taken by resistor = - = 0.125A 10 1.25 Current taken by voltmeter = -= 09125A 100 Current supplied by cell = 0.1 25 + 0.0125 = 0.1 375A . . . (b) Thus E = 1.25 + 0.1375Ri Solving (a) and (b) then E = 1.25 + 0.1 375Ri and E = 1.5 + 0.015Ri Subtracting 0 = - 0.25 + 0.,1225Ri 0.25 o r R i = -= 2.04R Ans. 0.1225 and E = 1.5 + (0.015 x 2.04) = 1.5 + 0.031 Thus cell e.m.f. = 1,531V Ans. 8. R A = OR

XA

2xfL. ~2 x 3.14 x 50 x 0.3 = 31.4 X 3 = 94.2R

R, = 1000 Z = d 1 0 0 2 + 94.2' =. 1 0 . J M 2 = IOdlOO + 88-74 = l o d m 4 = 10 x 13.74 = 137.4R Ans. 200 Current I = -= 1.45A Ans. 137.4 R 100 Power factor, cos 4 = - = - = 0.73 (lagging) Ans Z 137.4 9. Output = 40 x 1000 watts 100 400 x lo3 watts Input = 40 x lo3 x - = 90 9 Input current =

44 444 - 2222.2 - -= 202.02A 220 11

492

R EE D ' S B A S I C E L E C T R O TE C H N O L OG Y

220 - = 4A 55 Armature current = 202.02 - 4 = 198.2A Armature starting current = 198.2 x 1.5 = 297.3A 220 Resistance of armature circuit = ---- = 0.74R 297.3 Resistance to be added = 0.74 - 0,075 = 0,665Q Ans. On normal load E, = 220 - (198.2 x 0.075) volts = 220 - 14.87 = 205.13V On 90"" speed E,, = 0.9 x 205.13 = 184.62V /\s~ii;~rurc \olr;~gedrop = 220 - 184.62 = 35.3XV 35.38 Annaturecurrent = ----- = 471.7A Ans

Shunt-field current =

0.075

10 \ u h \ t ~ t u t ~ ning r = I', sin 2nlr I = 100 x s i n ( 2 x 180 r 50 x 0 0 0 1 ) = I00 s ~ n (18 x 1 ) = 100 x sln 18 = 100 x 0 3 0 9 = 30 9V Ans. If = SOV, then 50 = 100 sin ( 2 x 180 x 50 x r ) 50 = sin (2 x 180 x 50 x r ) and --100 4 = sin 0, whence 0 = 30' or 1 3 = _ x 10-2 Thus 30 = 18 x 10'1 or r = 18x10' 6 = 0.166 x 10- seconds Tlic i.cquired tinie = 166nis A n s

--

84 -

x

--

lo-' 12

(2 85)

=

7

2 85

I()

-2

The induced voltage will oppose the appl~cdsilpply voltage. ~ I I L I S ~.cducirig[tic I . ; I [ C 01' C L I I . I . C I I ( g1.0w111 12. Total resistance of circuit = 2 + 10 = 12f2 Total reactance .. ,, = 14 - 6 = 8R (inductive) From impedance relationsliip: -7 = \ 122 t X 2 = "'144 + 6 4 = b/20t( = 14.4R I'hc c ~ ~ . c impedance u~t is 14.40 A n s

--

~

SOl,llTIONS TO TYPICAI. EXAMINATION Qt1I:Sl'IONS - -----

493

~

T h e power factor = 0.832 (lagging) Ans

13. Carbon has a negative temperature cocllicient. 200 Resistance of field (cold) = R , = - 3.- = IOOR

-

'00 Res~stanceof' field (hot) = H , = - = 117.64R 1.7 Then 100 = R, ( l + 0.004 28 x 15) and 117.64 = Ro ( 1 + 0.004 28 x 7J o r I ,1764 x 1.0642 = 1 + 0.004 28 T and 1,2519 - 1 = 0.004 28T Temperature rise of winding = 58.85 - I5 C = 43.85'C Ans.

14. M o t o r output = 7.5 x 1000 = 7500W 100 M o t o r input = 7500 x --- = 3750 x 25 = 8523W 88 II P 8523 : . / = -----amperes But V I cos @ = P 440 x 0.8 V cos q5 8523 or I = - = 24.21 A 352 kilogranirnes o f copper are deposited b!, I 15. 330 x coulomb 0.0 1 - -l o x 103 :. 0.01 kg of copper will require 330 x 0.33 = 30.303 x 103 = 30 303C Ans. If time was 1 hour, since Q = It Q - 30 303 - 303.03 ... I = - - ----- - --t 3600 36 o r current required = 8.418A Ans. Mass of silver would be in proportion to the E.C.E 0.01 x l l I8 x 0.01 x 1 1 1.8 So mass = 330 x = -- 33 = 0.01 x 3.387 = 0,033 87kg o r 33.87g Ans.

V2 V2 60 x 60 R P 300 When on a.c. P = 1200W also P = 1*R

16. When on d.c. P = - :. R = - = ------ = 1 2 0

V 130 The impedance Z of the circuit = - = - = 13R I I0 So reactance X

v'132 - 12' = d m 4 4 = f l 5 5R T h i ~ sreactance of coil = 5R Ans. =

=

17. Force on 1 conductor is pven by F = Bll newtons or F = 0.7 x 30 x 0.3 = 6.3N

2 No of conductors in the field at any instant = - x 720

3

= 480

Total force = 480 x 6.3 = 3024N Torque = force x radius = 3024 x 0.18 newton metres = 544.32 So torque exerted = 544Nm Ans. 2x)YT 2 x 3.14 x 680 x 544 Power developed = --- 60 60 = 3.14 x 68 x 181.33 watts = 38.7kW Ans. d

18.

I f plotted, t h ~ swavefonn w ~ l lbe found to be made up of seven rectangular blocks. the mid-ordinates of which are 2. 4. 6. etc. as given

= 4.57A

Ans.

SOLUTIONS TO TYPICAL EXAMINATTON QUESTIONS

495

75 19. 75W lamp. I = - = 0.375A 200 40 40W lamp. I = - = 0.2A 200 With lamps in serics 40W lamp will only p;~ss0.2A .: (0,375 - 0.2) amperes must be piissed through il sliunt resistor connected across the 40W lamp. This resistor is i i l > r ) to be suitable for 200V, and its resistance value must - -200 = 1143R Ans. 0,175 Power loss in this resistor = 200 x 0.175 = 35W Ans 35 - 4 9 x 1 2 Units used per week = 7 x 24 x ---- lo00 100 Cost = 588

0'5p = 2.941, Ans, 100 *

20. Here L = 1.5H so X, = 2 ~ f L= 2 x 3.14 x 50 x' 1.5 = 471R

X, - X, .'. 471

407.3R (inductive) = J30o2 + 407.3~ Impedance o f circuit, Z = = 1 0 0 J 3 ~ + 4.0732 say 100/3~ + 42 = 500R The circuit impedance is 500 ohms . Ans. Curre,nt will lag the voltage because the circuit is inductive. Ans. Total reactance X

21.

=

-

63.7

=

With no current in BD, points B and D are a t the same potential, and since A is common, then: P.D. across AB = p.d. across A D . Similarly since C is common, then: P.D. across BC = p.d. across DC. Let I, = current in tog branch and I, = current In bottom branch Then I, x 1.C =: I, x 1 and I, x 2 = I , x Rx 2 x Il Written as a rat o Rx x IZ1x1, 1.5~1, 2 4 ' Thus Rx = - = - = 1.33R Ans. 1.5 3

49 6

RE ED ' S BASIC ELECTROTECHNOLOGY

or hilo\oll iilnperes ( S ) = Thus S

14

kilowatts (P) power kictor (cos 4)

560 0.7

.= ---

=

800kV A

This would bc the rehistancc of'~ l l cleng~h01' wirc. H u t ~ 1 1 t h ;I 1'1p-bound generator. u.1tl1six parallel patha in thc ;irm;i-

0"07 - 0.101 16R ture. -Phuh re\ict;~nceof 1 parallel path = ---6 H u t there are 6 p;~thbIn parallel so the equivalent resistance 0.101 16 - 10.116 Ic-2 is one-sixth of thc above = ----------

6

6

= 1.686 x

= 0.01686R

Ans.

( a ) Current I

1' 100 =- = = 20A I

Ans.

Z 5 ( b ) R e i i s t ~ ~i.o!t:~gedrop c I,', = I R = 20 x 3 = @JV Ans. ( c ) liei~ctivevolt:cge drop b', = I X = 20 x 4 = 80V 4 n s .

25

M'llcn t l ~ cloact current is 10A. t l ~ ctc~lt:rgc d r o p in tllc ( . ; ~ t > I v \ I>c.~ir,c,c*n . ni;~chineterminals and load = 220.2 - 210 = 0.2v

By propnrtlon the voltirge drop for 600 ampere\ 600 = 0 2 x - - = 0.2 x 60 = 12V 10

SOLUTIONS TO TYPICAL EXAMINATION QUESTIONS 497 -- - -- - - - -.- . - - -- ---. --

26. Reactance X, at 50Hz

Also X, = 2rcfL .: L

~t

So

= =

= 1

K

0 0 d 1 = loom6

75 XL - -2 x 3.14 x 50 100 x 3.14 -0.75 = 0.24H Ans. 3.14 =

new frequency X, =

new frequency 50 o r new frequency

d-2 100 J

;/m~@

= 100 J1.206~ - I = 100JFEC-1 = l o = 100 x 0.674 = 67.4Cl 67.4 --

o m 4

75 50. 4 67.4 x - = 33.7 x 75 3 = 11.233 x 4 = 44.93 hertz The new frequency value would be 45Hz I n s . =

27. Mass of water = volume x density = 4.5 x 1 = 4.5kg Heat received by water = 4.5 x 4.2 (100 - 17) = 19.9 x 83 kiloioules = 1569kJ 100 Electrical energy supplied to heater = 1569 x 80 - l 5 690 - 1961kJ

8 1961 Power rating of heater = - -time 15 x 60 196 1 - -= 2.18kW 9W2180 - I09 - 9.gA Current taken from mains = -- - 220 11 Mains current = IOA (:ipprox) Ans. Resistance of heater = ,-220 = 22.252 Ans. 9.9 energy _-

28. Here R = O R and Z

=

X

=

25l2

V 100 :. r.m.s. value of current I = = --- = 4A

Z 25 The graphical solution consists of a sinusoidal voltage wave with a sinusoidal current wave lagging it by 90G,since the circuit is wholly inductive. Thus when voltage is maximum current is zero. When voltage has fallen to zero, the currer-t has risen to its maximum value and as voltage rises to its negative maximum the current falls to zero. When V is a maximum, current value is zero Ans.

~

~

29.

e 1 brush = 30 x 20 square millimetres Bearing s ~ r f a c of = 600mm2 With a current density of 0 . 0 5 4 ~ / m mthe ~ , current carried by one brush = 600 x 0.054 = 6 x 5.4 = 32.4A With 8 brush arms, there are 4 positive and 4 negative brush arms. Also since there are 6 brushes per arm, the number of brushes in parallel carryng current into o r out of the annature = 4 x 6 = 24 brushes, o r total current carried by brushes = 32.4 x 24 = 777.6A. Current density in the cable is limited to 1.6A:mm2 777.6 = 486mm2 Anh. . . ('able nrc to carry 777.6A = 1.6

3.4 x 9.2 x - 31.28 x 486 486 = 0.643 x ohms Voltage drop in cable = 777.6 x 0.643 x 10- = 0.499 = 0.5V Power loss in cable = 777.6 x 0.5 = 388.8W Ans. P 1 where' 1 is the Resistance of the brushes is given by R =-

A

length of a +ve plus a - ve brush and ,4 is the area of half the total number of brushes. 2550 2 x 30 x - 25.5 x 60 x W 3 Thus R 4 108 600 x 1 0 - 5 24 600 x 24 - 255 x = 0.106 x ohm 24 Power loss In the brushes is given by 1 2 R = 777.6' x 0.106 x = 7.776 x 7.776 x 1.06 watts Thus power loss in brushes = 64.1 W Ans.

v 200 If Z is the circuit impedance then Z = - = - = 66.6R I 3 Now only resistance is responsible for power dissipation. 1 44 = 16R then P = I' R or 144 = 3 ' ~ .Thus R =

SOLUTlONS TO TYPICAL EXAMINATION QUESTIONS

499

For this circuit X is made up of inductive reactance X, and capacitive reactance X , which nullify each other. The resultant reactance X = d m 2= d66.6* - 1 62 = 44436 - 256 = m 0 = 647R But X, alone is 53R :. X, = 64.7 + 53 = 117.7R Thus ( a ) p.d. across capacitor = 53 x 3 = 159V Ans. (b) Resistance of coil = R = 16R Ans. 1 17.7 - 1.177 Inductance of coil = % = 2nf 1 0 0 ~ 3 . 1 4 3.14 = 0.38H Ans. (c) Impedance of coil z,-=J 162T1 17.72 = J14109 = 119R Ans. R -16 Power factor of coil = cos $, = 119 3 = 0.1346 (lagging) Ans. (d) Power faotor of circuit = cos

4

R

=-

z

-16 = 0.245 (lagging) 66.6 Ans. Note. Both power factors are lagging, the circuit being net inductive.

31.

The equations for t h e iron and aluminium conductors can P ili

be written as R i= - and R, Ai

and di=

@a

.: A , =

A

-2since

4

pa'a =-

Aa

area

K

diameter 2

40 x 1.1 x I, x A, x 4 13 x I, x A, Ra 160 x 1.1 176 - 13.54 13 13 1 Thus, since the resistance ratio of the iron to the aluminium wires are 13.54 to 1, and as the wires are in parallel. the currents in the wires are in the ratio Iron :Aluminium = 1 : 13.54. Ans.

R i= So-

+ (2 x I1x I2 x cos 30) 32. Resultant current I = JII2 + =J14.14~+8.5'+(2~ 14.14~8.5~0.866) = d20O 72.25 208.17 = J48042 = 21.92A Resultant current = 21.92A Ans. Power dissipated = 1 2 R = 21.92' x 4 watts = I2Rr joules Energy at heat = 2 1 , 9 2 2 x 4 x 2 x 60 = 480.42 x 480 = 230 6025 Ans.

+

+

total conductors - 498 = 83 -parallel paths 6 1I.1n.l'.ol' I parallel p;~thr- c.rn.1'. ol'machinc = 1.5 x 83 = 124.5V Ans. Current per parallel path = current in 1 conductor = 160.4 Current of 6 paths in parallel '= 6 x 100 = 600A Ans. ( a ) WaveJwound. A = 2 498 = 249 total conductors - Conductors in series = ---parallel paths 2 E.m.f. of' 1 parallel path = e.m.f. of machine = 1.5 x 249 = 373.5V Ans Current per parallel path = current in 1 conductor = IOOA C u r r e n ~o f 2 paths in parallel = 2 x 100 = 200A Ans. Conductors in series =

34.

The power factor of this circuit is 0.866 (lagging) o r cos 4 4 = 30G,where 4 is the angle of lag between the voltage and the current-the latter lagging the former. The values of 10A and 220V a s given, can be assumed to be r.m.s. valucs. So the maximum value of current is given by I = 0,707 I,,, (here I,,, is the maximum value). Sine-wave working is ;~ssumcd = Q.866 (lagging) and

Also V

=

0.707V , (maximum value). Sine-wave working

15

assumed.

Thc voltage and current can be written as L' = L', sin cot and i = I, sln ((!)I - rb) 4 15 In radians

1 50 (a) When the voltage is at a maximum, the time is for; cycle 1 or t = ------ - 0.005 seconds 4 x 50 Current at this instant is given by substituting in i ,= 14.14 sin (2x50 x 0.005 - 4) or i = 14.14 sin (2 x 180 x 50 x 0.005 - 30). x and 4 in degrees = 14.14 sin (90 - 30) = 14.14 sin 60" = 14.14 x 0.866 = 12.25A Ans. (b) At an instant 0.005s later t would be 0-01s :, i = 14.14 sin (2 x 180 x 50 x 0.01 - 30) or i = 14.14sin (180 - 30) = 14.14sin 150" i = 14.14 sin 30" = 14.14 x = 7.07A Ans.

At a frequency of 50Hz, time for 1 cycle =

- seconds

35. Resistance of filament (cold) 65 x lo-'' x 600 x 7 x 4 - 273mohms 22 x 16 x lo-'' 88 273 x 5 x 10' ohms :. Resistance when hot = 88 110x88 . 968 Working current = 273 x 50 273 x 5 x 10' --- - 0.0709 = 0.071A Ans. 273 36. From the information given, the impedance Z of the . v 220 circuit is = - or Z = - = 22R I 10 The resistance R is 19.5n. Therefore the reactance X is obtained from X = d m 2= J22' - 19.5' = J484 - 380.25 = J 1 i 5 = lO.lsR 10.15 - 0.1015 Also X = 2xfL :. L = 2 x 3.14 x 50 - 314 = 0.032H Ans. 37. From Faraday's law. E*" = N ( @ I - @,) or R =

a

t

-2000 x 4(2.6 x lo-' - 0.1 x 0.2 - 80 x (2.6 - 0.1) - 80 x 2.5 = ,, 0.2 0.2 Induced e.m.f. = lOOOV or 1 kV Ans.

25 volts

500 Also since P = 1 2 R . then R of coil = --r = 5R 10 Thus reactance X of coil = V'Z* - , R ~ = Jw-Ti o r X = \;424.36 - 25 = d m 6 = 19.98R Z ofadditional apparatus = 5R X , ,, = 4R (capacitive) .: R ,, = = = 3n Total resistance of circuit = 5 3 = 8R Tor;ll reactance ,, = 19.98 - 4 = 15.98R Note. The inductive and capacitive reactances have been 9

Jm2

+

st1 ht r a c ~ c d . *I otal impedance ofcircuit

=

J82

= J64 =

Since cos Thus power factor inductive Ans 39.

=

+ 1 5.5)g2 + 255.36 = 17.880 Ans.

= 0.44 (lagging) 4=17.88

0.44 (lagging), slnce circuit

15

net

Since the circuit is built up from wire of the same material and area, then the resistance of various parts of the . sectional . clrcult are proportional to length. The resistance of the diameter = 2 ohms :. The resistance of the circumference = x d = 2x ohms. i'he resistance of circumference = x ohms The circuit is made up of a diameter and two circumferences in parallel. :. if R is the circuit resistance

With 220V applied across R , the current would be

SOLUTIONS TO TYPICAL EXAMINATION QUESTIONS

503

40. For a sine wave: Average value = 0.6365 x maximum value av value - 125 :. Maximum value of current I,,, = -0.6365 0.6365 or I, = 196.4A Also as current is sinusoidnl, then i = I,,, sin 2 x f t or 95 = 196.4 sin (2 x 180 x 60 x t ) 95 and sin (18 x 12 x lo2 x t ) =- = 0.483 196.4 Now from sine tables, the sine of 28.9" is 0.483 Thus t = 0.134 x lo2 = 0.001 34s. First time from zero = 0.001 34s or 1.34rn.s A n s . R.M.S. value of current = Q.707 x 196.4 = 138.86A Ans. r

41. Output of motor = 15kW 15 . Input to motor =- kilowatts = 16 666W 0.9 16 666 input current o r current in cables = 240 = 6 9 4 A Ans pl Resistance of cable is given by A

Voltage drop in cable = 69.44 x 0.866 = 60.14V Input voltage at supply cables = 240 + 60.14 = 300.14V Ans. Power output from cables Efficiency of distribution = Power input to cables 240 x 69.44 - 240 300.14 x 6944 - 3oO14 = 0.799 = 79.9 per cent Ans 42., Here Z = 20Q and R

=

16Q

:. X = J2iGF-S' =6-4J =

m =1 2 n

Also X = 2nfL = 2 x 3.14 x f x 0.047 75

= 12

- 40Hz Ans. 29.987 The frequency of the supply is 40Hz Ans -

43 Output of heater = l 0 M J 'ilso IkLV h = 3600 x 1000 = 36 x lo5 loule5 10 x lo6 100 Nou c.ncrg\ o u t p u ~o f Ilc,rtcr = -= 17s 36 x lo5 = 36 = 2 78kW 11 O L I I ~ L I I - 7, 711 1-rlcryy 1111)111 =: 3 27h W 11 / \ I ] \ ellic~ency 0 85 energy - -Power lnput = ---27 = 6 54kW time 05 I C2 Also. since P = 1 2 R = I x R x I = I/ x - = -R R '-

L '2 R =P - 40000--- - 6 1 2 Q TIius R = 2002 6340 6540 resl51,lnce - 6 12 Length c ~ elenlent l , ~ tO 26R m = -- -- ohms Der metre 0.26 = 23.54; Ans. P 6540 Current taken = - = -- = 32.7A An\. v 200 4

44. If 110V is maintained across the heater. the current would be

In~ped~ince Z of complete clrcult

=

&?-+ 3142

+

= J42 25 985 96 = dl028 Z I =

10fi0-

3 207 = 32 07Q Appl~ed\ o l t L y efor 16 92A = 16 92 x 32 07 - 542 6V An5 If frequency rljes 5 per cent. reactance rlses 5 per cent N e u ~ c ~ ~ c t . ~ r l c7 cl 4 r 1 0 5 1 2 07R = 10 x

+ 32.97'

= J42.35 1 0 \ M 3 10 x 3.36 = 33.6R 542.6 Circuit current = ----- 16.15A 33.6 Voltage across heater = 16.15 x 6.5 = 104.97 = 105V (i~pprox) Ans.

New impedance = J6.52 = =

J1129j

+ 1087

=

45. Mass deposited = 39.25 x 10- - 23.8 x 10 = 15.45 x kilogramrnes True current is obtained from m = zlr = 7.68A Error of ammeter

=

7.68 - 7.3 = 0.38A (low) Ans.

o r as a percentage error

0.38 ---- = 0.049 7.68 = 4.9 per cent low

C

=

Ans.

46. Total resistance of circuit = 3 + 1 = 4 0 Reactance of circuit is given by X = 2njL or X = 2 x 3.14 x 50 x 0.1 = 31.40 = d l 6 -+ 985.96 Circuit impedance Z = = d m 6 = 1 0 m 2 = 31.650 C~lrrcntflowing = --100 = 3.16A Ans. 3 1.65 47. Neglecting the field current, since i t will bc sm;~ll Input power = V I = 500 x 20 = 1OOOOW 3 3 per cent of the input power = - x 10 000 = 300W 100 This is dissipated as heat in the armature, ir it is a copper o r r2R loss. :. I ~ R ,= 300 o r 2 0 2 ~ = , 300 300 3 and R, = - = - = 0.75a 4 400 The starting current with only armature resistance to limit 500 the armature current = IaS= --- = 666.66A Ans. 0.75 Twice full load current = 20 x 2 = 40A

:. Total resistance required in the armature circuit to limit

starting current to 40A =

500 = 12.50 40

----

S ~ n c cfi, = 0.75R Series resistance \vc>uld be 12.5 - 0.75 = 1 1 , 7 5 0 Ans 48 Induct~vereactance Y, = 2xJL = 2 x 3 14 x 50 x 100 x or Y, = 3 14 x lo4 x = 31 4 0 lo6 1O6 C , ~ p a c l t ~ reactance ve X - -'=2xfC 2 x 3 1 4 ~ 5 0 ~ 2 0

Total e f e c t ~ c ere'lctance X = 159.23 - 31 4 or \ 127 X3R ( c . ~ ~ l , ~ c ~ t (111, ~\c)

\/'I 0-+- 16 340

~/16440 = 1 0 2 d 1 . W= lo2 x 1,283 = 128.30 Ans. =

=

49. On no-load: Ebo = I . - I,R, = 200 - (2 x 0.4) = 200 - 0.8 = 199.2V On full-load: E,, = 200 - (50 x 0.4) = 200 - 20 = 18OV As this is a shunt motor, constant field current and therefore the same constant flux can be assumed for the no-load ltnd full-load conditions. Since Ebo = k@,Y0 and E,, = k @ , N , and 0,= 0,

' 9 9 2- = ---3984 - 1328 revimin T h ~ sgives N o = 1200 x 1 80 3 Ans

AlsoPowerP

=

1 2 R o rR

=

P J2

=

450

450 =

2T

Hence I