THE MAGNETIC CIRCUIT
WORKS BY THE SAME AUTHOR
Published by McGRAW=HILL BOOK COMPANY
The Electric Circuit
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(New Edition in preparation)
The Magnetic Circuit
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Published by JOHN WILEY & SONS
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Engineering Applications of Higher Mathe
matics
Part I. MACHINE DESIGN. Small 8vo,
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THE
MAGNETIC CIRCUIT
BY
V. KARAPETOFF
M
MCGRAW-HILL BOOK COMPAiSTY
239 WEST 39TH STEEET, NEW YORK
6 BOUVERIE STREET, LONDON, E.G.
1911
TK /.T3
Engineering
Library
Copyriglit, 1911
BY
MCGRAW-HILL BOOK COMPANY
PREFACE
THIS book, together with the companion book entitled " The
Electric Circuit," is intended to give, a student in electrical
engineering the theoretical elements necessary for the correct
understanding of the performance of dynamo-electric machinery,
transformers, transmission lines, etc. The book also contains
the essential numerical relations used in the predetermination of
the performance and in the design of electrical machinery and
apparatus. The whole treatment is based upon a very few funda-
mental facts and assumptions. The student must be taught to
treat every electric machine as a particular combination of electric
and magnetic circuits, and to base its performance upon the
fundamental electromagnetic relations rather than upon a sepa-
rate " theory " established for each kind of machinery, as is some-
times done.
The book is not intended for a beginner, but for a student
who has had an elementary descriptive course in electrical engi-
neering and some simple laboratory experiments. The treat-
ment is somewhat different from that given in most other books
dealing with magnetic phenomena. It is based directly upon
the circuital relation, or interlinkage, between an electric current
and the magnetic flux produced by it. This relation, and
the law of induced electromotive force, are taken to be the
fundamental phenomena of electro-magnetism. No use what-
ever is made of the usual artificial concepts of unit pole, magnetic
charge, magnetic shell, etc. These concepts of mathematical
physics, together with the law of inverse squares, embody the
theory of action at a distance, and are both superfluous and
misleading from the modern point of view of .a continuous action
in the medium itself.
The ampere-ohm system of units is used throughout, in
accordance with Professor Giorgi's ideas, as is explained in the
254299
vi PREFACE
appendices. Those familiar with Oliver Heaviside's writings
will notice his influence upon the author, in particular with
regard to a uniform and rational nomenclature. The author trusts
that his colleagues will judge his treatment and nomenclature
upon their own merits, and not condemn them simply because
they are different from the customary treatment.
In the first four chapters the student is introduced into the
fundamental electromagnetic relations, and is made familiar with
them by means of numerous illustrations taken from engineering
practice. Chapters V to IX treat of the flux and magneto-
motive force relations in electrical machinery, first at no load, and
then under load when there is an armature reaction. The remain-
ing four chapters are devoted to the phenomena of stored magnetic
energy, namely inductance and tractive effort. The subject is
treated entirely from the point of view of an electrical engineer,
and the important relations and methods are illustrated by
practical numerical problems, of which there are several hundred
in the text. All matter of purely historical or theoretical interest
has been left out, as well .as special topics which are of interest
to a professional designer only. An ambitious student will find
a more exhaustive treatment in the numerous references given in
the text.
Many thanks are due to the author's friend and colleague,
Mr. John F. H. Douglas, instructor in electrical engineering in
Sibley College, who read the manuscript and the proofs, checked
the answers to the problems, and made many excellent sugges-
tions for the text. Most of the sketches are original, and are the
work of Mr. John T. Williams of the Department of Machine
Design of Sibley College, to whom I am greatly indebted.
CORNELL UNIVERSITY, ITHACA, N. Y.,
September, 1911.
CONTENTS
PAGE
PREFACE v
SUGGESTIONS TO TEACHERS xi
CHAPTER I. THE FUNDAMENTAL RELATION BETWEEN FLUX AND MAG-
NETOMOTIVE FORCE 1
A simple magnetic circuit. Magnetomotive force. Magnetic flux.
The reluctance of a magnetic path. The permeance of a magnetic
path. Reluctivity and permeability. Magnetic intensity. Flux
density. Reluctances and permeances in series and in parallel.
CHAPTER II. THE MAGNETIC CIRCUIT WITH IRON 20
The difference between iron and non-magnetic materials. Mag-
netization curves. Permeability and saturation. Problems involv-
ing the use of magnetization curves.
CHAPTER III. HYSTERESIS AND EDDY CURRENTS IN IRON 32
The hysteresis loop. An explanation of saturation and hysteresis
in iron. The loss of energy per cycle of magnetization. Eddy cur-
rents in iron. The significance of iron loss in electrical machinery.
The total core loss. Practical data on hysteresis loss. Eddy cur-
rent loss in iron. The separation of hysteresis from eddy currents.
CHAPTER IV. INDUCED E.M.F. IN ELECTRICAL MACHINERY 55
Methods of inducing e.m.f. The formulae for induced e.m.f.
The induced e.m.f. in a transformer. The induced e.m.f. in an
alternator and in an induction motor. The breadth factor. The
slot factor ks. The winding-pitch factor kw. Non-sinusoidal vol-
tages. The induced e.m.f. in a direct-current machine. The ratio
of A.C. to D.C. voltage in a rotary converter.
CHAPTER V. EXCITING AMPERE-TURNS IN ELECTRICAL MACHINERY. ... 80
The exciting current in a transformer. The exciting current in a
transformer with a saturated core. The types, of magnetic circuit
occurring in revolving machinery. The air-gap ampere-turns. The
method of equivalent permeances for the calculation of air-gap
ampere-turns.
vii
viii CONTENTS
PAGE
CHAPTER VI. EXCITING AMPERE-TURNS IN ELECTRICAL MACHINERY.
(Continued) 100
The ampere-turns required for saturated teeth. The ampere-
turns for the armature core and for the field frame. Magnetic
leakage between field poles. The permeance and reluctance of irreg-
ular paths. The law of flux refraction.
CHAPTER VII. MAGNETOMOTIVE FORCE OF DISTRIBUTED WINDINGS . . . 121
The m.m.f. of a direct-current or single-phase distributed wind-
ing. The m.m.f. of polyphase windings. The m.m.fs. in a loaded
induction machine. The higher harmonics of the m.m.fs.
CHAPTER VIII. ARMATURE REACTION IN SYNCHRONOUS MACHINES. . . . 139
Armature reaction and armature reactance in a synchronous
machine. The performance diagram of a synchronous machine
with non-salient poles. The direct and transverse armature reac-
tion in a synchronous machine with salient poles. The Blondel
performance diagram of a synchronous machine with salient poles.
The calculation of the value of the coefficient of direct reaction. The
calculation of the value of the coefficient of transverse reaction.
CHAPTER IX. ARMATURE REACTION IN DIRECT-CURRENT MACHINES .... 163
The direct and transverse armature reactions. The calculation
of the field ampere-turns in a direct-current machine under load.
Commutating poles and compensating windings. Armature reac-
tion in a rotary converter.
CHAPTER X. ELECTROMAGNETIC ENERGY AND INDUCTANCE 177
The energy stored in an electromagnetic field. Electromagnetic
energy expressed through the linkages of current and flux. Induc-
tance as the coefficient of stored energy, or the electrical inertia of a
circuit.
CHAPTER XI. THE INDUCTANCE OF CABLES AND OP TRANSMISSION
LINES 189
The inductance of a single-phase concentric cable. The mag-
netic field created by a loop of two parallel wires. The inductance
of a single-phase line. The inductance of a three-phase line with
symmetrical and semi-symmetrical spacing. The equivalent reac-
tance and resistance of a three-phase line with an unequal spacing of
the wires.
CHAPTER XII. THE INDUCTANCE OF THE WINDINGS OF ELECTRICAL
MACHINERY 208
The inductance of transformer windings. The equivalent leak-
age permeance of armature windings. The leakage reactance in
induction machines. The leakage reactance in synchronous machines.
The reactance voltage of coils undergoing commutation
CONTENTS ix
PAOE
CHAPTER XIII. THE MECHANICAL FORCE AND TORQUE DUE TO ELEC-
TROMAGNETIC ENERGY 240
The density of energy in a magnetic field. The longitudinal
tension and the lateral compression in a magnetic field. The deter-
mination of the mechanical forces by means of the principle of virtual
displacements. The torque in generators and motors.
APPENDIX 1 262
APPENDIX II.. . . . 266
SUGGESTIONS TO TEACHERS
(1) THIS book is intended to be used as a text in a course
which comprises lectures, recitations, computing periods, and
home work. Purely descriptive matter has been omitted or only
suggested, in order to allow the teacher more freedom in his
lectures and to permit him to establish his own point of view.
Some parts of the book are more suitable for recitations, others
as reference in the computing room, others, again, as a basis for
discussion in lectures or for brief theses.
(2) Different parts of the book are made as much as possible
independent of one another, so that the teacher can schedule
them as it suits him best. Moreover, most of the chapters are
written according to the concentric method, so that it is not
necessary to finish one chapter before starting on the next. One
can thus cover the subject in an abridged manner, omitting the
last parts of the chapters.
(3) The problems given at the end of nearly every article are
an integral part of the book, and should, under no circumstances,
be omitted. There is no royal way of obtaining a clear under-
standing of the underlying physical principles, and of acquiring
an assurance in their practical application, except by the solution
of numerical examples. It is convenient to assign each student
the complete specifications of a machine of each kind, and ask
him to solve the various problems in the text in application to
these machines, in proportion as the book is covered. Numer-
ous specifications and drawings of electrical machines will be
found in the standard works of E. Arnold, H. M. Hobart,
Pichelmayer and others, mentioned in the footnotes in the
text. A first-hand acquaintance with these classical works on
the part of the student is very desirable, however superficial this
acquaintance may be.
xi
Xif SUGGESTIONS TO TEACHERS
(4) The book contains comparatively few sketches; this gives
the student an opportunity to illustrate the important relations
by sketches of his own. Making sketches and drawings of electric
machines to scale, with their mechanical features, should be one
of the important features of an advanced course, even though it
may not be popular with some analytically-inclined students.
Mechanical drawing develops precision of judgment, and gives
the student a knowledge of machinery and apparatus that is
tangible and concrete.
(5) The author has avoided giving definite numerical data,
coefficients and standards, except in problems, where they are
indispensable and where no general significance is ascribed to
such data. His reasons are: (a) Numerical coefficients obscure
the general exposition. (6) Sufficient numerical coefficients and
design data will be found in good electrical hand-books and pocket-
books, one of which ought to be used in conjunction with this
text, (c) The student is liable to ascribe too much authority to
a numerical value given in a text-book, while in reality many
coefficients vary within wide limits, according to the conditions
of a practical problem and with the progress of the art. (d)
Most numerical coefficients are obtained in practice by assuming
that the phenomenon in question occurs according to a definite law,
and by substituting the available experimental data into the corre-
sponding formula. This point of view is emphasized throughout
the book, and gives the student the comforting feeling that he
will be able to obtain the necessary numerical constants when
confronted by a definite practical situation.
(6) The treatment of the magnetic circuit is made as much
as possible analogous to that of the electrodyamic and electro-
static circuits treated in the companion book. The teacher will
find it advisable to make his students perfectly fluent in the use
of Ohm's law for ordinary electric circuits before starting on the
magnetic circuit. The student should solve several numerical
examples involving voltages and voltage gradients, currents and
current densities, resistances, resistivities, conductances, and
conductivities. He will then find very little difficulty in master-
ing the electrostatic circuit, and with these two the transition
to the magnetic circuit is very simple indeed. The following
table shows the analogous quantities in the three kinds of cir-
cuits.
SUGGESTIONS TO TEACHERS
Xiii
Electrodynamic,
r Voltage or e.m.f.
| Voltage gradient (or
I electric intensity)
Electric current
Current density
r Resistor
j Resistance
<• Resistivity
r Conductor
•j Conductance
I Conductivity
Electrostatic,
Voltage or e.m.f.
Voltage gradient (or
electric intensity)
Dielectric flux
Dielectric flux density
Elastor
Elastance
Elastivity
Permittor (condenser)
Permittance (capacity)
Permittivity (dielec-
tric constant)
Magnetic.
Magnetomotive force
M.m.f. gradient (or
magnetic intensity)
Magnetic flux
Magnetic flux density
Reluctor
Reluctance
Reluctivity
Permeator
Permeance
Permeability
LIST OF PRINCIPAL SYMBOLS
The following list comprises most of the symbols used in the text. Those
not occurring here are explained where they appear. When, also, a symbol
has a use different from that stated below, the correct meaning is given
where the symbol occurs.
Symbol. Meaning. .
a Air-gap ................................................... 90
a Width of commutator segment .............................. 237
A Area .................................................. . . 11
Aa Area of flux per tooth pitch in the air ........................ 101
Ai Area of flux per tooth pitch in the iron ...................... 101
(AC) Number of ampere-conductors per centimeter ................ 165
b Thickness of transformer coil .............................. 211
b Width of brush ........................................... 236
b' Thickness of mica ____ , .................................... 236
B Flux density .............................................. 14
Bm Maximum value of the flux density .......................... 81
C2 Number of secondary conductors ........................... 133
Cpp Conductors per pole per phase .............................. 219
d Duct width .............................................. 94
e, E Electromotive force ...... ................................ 39, 65
/ Frequency ............................................... 48
F Mechanical force .......................................... 274
Ft Tension per square centimeter .............................. 243
FC Compression per square centimeter .......................... 244
H Magnetic intensity .................... .................... 13
Hm Maximum value of the magnetic intensity .................... 81
i, I Electric current ....................................... 39, 205
z'o Magnetizing current ............................ ........... 81
/i Current per armature branch ............................... 233
k, k' Transformer constant ................................. 215, 221
ka Air-gap factor .......................... ........ .......... 90
kb Breadth factor ............................................ 65
ks Slot factor .............................. .' ................ 68
kw Winding-pitch factor ...................................... 68
I Length ................................................... 11
xv
xvi LIST OF PRINCIPAL SYMBOLS
Symbol. Meaning. I*. wb«dg,«»l
la, lg Gross armature length 90, 102
li Semi-net armature length 220
ln Net armature length 102
L Inductance 184
m Number of phases 69
M Magnetomotive force 7
Ma M.m.f. of armature 144
Ma M.m.f. of direct reaction 153
Mf M.m.f. of field 144
Mn Net m.m.f 144
Mt M.m.f. of transverse reaction 153
MI Demagnetizing m.m.f 165
M 2 Distorting m.m.f 166
n Number of turns 127, 211
TOI, NI Number of turns in primary of a transformer 62, 81
N2 Number of turns in secondary of a transformer 62
N Total turns in series 65
Om Mean length of turn 211
p Number of poles 133
P Power 48
(P Permeance 9
(Pa Permeance of air-gap 89
(Pa Equivalent permeance around conductors in the ducts per cm. . 220
(Pc Permeance of the path of the complete linkages 182
(Pe Equivalent permeance around the end-connections per cm 220
(Peg Equivalent permeance 184
(Pi' Equivalent permeance around embedded conductors per cm. . . 220
(Pp Permeance of the path of the partial linkages 182
(Ps Permeance of simplified air-gap . . 90
(Pz Zig-zag permeance 223
q Number of sections in a transformer 213
q Number of turns per commutator segment 235
r, R Resistance 51, 177
(R.P.M.) Speed in revolutions per minute 260
(R Reluctance 7
s Distance 244
s Number of coils short-circuited by a brush 236
s Slot width 93
S Number of slots per pole per phase 68
S Surface area 244
t Thickness of laminations 51
t Time : 9
t Tooth width 93
t' Tooth width corrected 93
T Time of one cycle 64
T Torque 253
LIST OF PRINCIPAL SYMBOLS xvii
Symbol. Meaning. Pa^^«edjfined
v Velocity 59
v Volts per ampere turn 155
V Volume 39
w, wp Pole width 90, 166
W Energy 39
W Density of energy 240
Wm Mechanical work done 250
W8 Energy stored in the magnetic field 250
x Reactance 144
a Angle 69
a Coefficient 220
P Phase angle 150
T Angle 75
d Brush shift in cm 163
£ Eddy-current constant '. 51
£ Winding pitch 71
Ty Hysteresis coefficient 48
0 Angle 253
A Tooth pitch 93
/* Permeability 11
v Reluctivity 11
T Pole pitch 64
<f) External phase angle 144
f/V Internal phase angle 144
0 Flux 7
$m Maximum value of the flux 81
@p Flux in the path of the partial linkages 182
$1 Primary leakage flux 86
$3 Secondary leakage flux 86
X Form factor 65
Amplitude factor 84
Angle between current and no-load e.m.f 144
Angle 197
THE MAGNETIC CIRCUIT
CHAPTER I
THE FUNDAMENTAL RELATION BETWEEN FLUX
AND MAGNETOMOTIVE FORCE
1. A Simple Magnetic Circuit. The only known cause of
magnetic phenomena is an electric current, or, more generally,
electricity in motion.1 The fundamental relation between an
electric current and magnetism can be best studied with the simple
arrangement shown in Fig. 1. A coil, CC, of very thin wire is uni-
formly wound in one layer on a spool made in the shape of a circu-
lar ring (toroid) . The tubular space inside of the ring is filled with
some " non-magnetic " material, so called; for instance, air, wood,
etc. When a direct current is sent through the coil, the space
inside the coil is found to be in a peculiar state, called the magnetic
state. This magnetic state can be experimentally proved by
various means, such as a compass needle, iron filings, etc. A
region in which a magnetic state is manifested is called a magnetic
field. Thus, in Fig. 1, the tubular space inside the coil is the mag-
netic field excited by the current in the coil CC.
No magnetic field is found in the space outside the coil upon
exploring it with a magnetic needle or with iron filings. For
reasons of symmetry, the field inside the ring is the same at all
the cross-sections. Thus, a uniformly wound ring constitutes
1 Werner v. Siemens, Wiedemann's Annalen, Vol. 24. (1885), p. 94; Lar-
mor, Ether and Matter (1904), p. 108. The magnetism of a permanent magnet
is probably due to molecular currents produced by some orbital motion of
electrons within the atoms of iron. The older concepts of magnetic charges
and free poles are summarily dismissed in this book as inadequate and
artificial.
THE MAGNETIC CIRCUIT
[ART. 1
the simplest magnetic circuit, because the field is uniform and is
entirely confined within the winding.
Iron filings orient themselves within the coil in directions indi-
cated in Fig. I by the concentric lines with arrow-heads. These
lines show that the medium is " magnetized " along circles con-
centric with the ring. Lines which show the directions in which a
medium is magnetized are generally called magnetic lines of force.1
They are analogous to the lines of electrostatic displacement,
though their directions and physical nature are entirely different ;
SECTION A-A C
FIG. 1. — A simple magnetic circuit.
see the chapter on the electrostatic circuit, in the author's
Electric Circuit.
The positive direction of the lines of force is purely conven-
tional, and is defined as that in which the north-seeking end of a
compass moves. Its relation to the current is, by experiment, that
given by the right-hand screw rule. Namely, if the direction of
the flow of a current is that of the rotation of a right-hand screw,
the lines of force point in the direction of the progressive move-
ment of the screw. Reversing the current reverses the direction
1 For actual photographs, showing iron filings which map out the magnetic
field inside of coils of various shapes, see Dr. Benischke, Die Wissenschaft-
lichen Grundlagen der Elektrotechnik (1907), p. 126; also his Transformatoren
(1909), pp. 4, 6, and 57.
CHAP. I] FLUX AND MAGNETOMOTIVE FORCE 3
of the field; this fact can be demonstrated by a small compass
needle. The positive direction of the lines of force is indicated
in Fig. 1 by arrow heads. The direction of the current is shown
in the conventional way by dots and crosses; namely, a dot
indicates that the current is approaching the observer, while a
cross indicates that the current is receding.
The magnetic state within the coil can also be explored by a
small test-coil inserted into the field and connected to a galvanom-
eter. When this coil is properly placed with respect to the field
and then turned about its axis by some angle,, the galvanometer
shows a deflection, because a current is induced in the coil by the
magnetic field. There are also other means for detecting a mag-
netic field, for which the reader is referred to books on physics.
The total magnetic field produced by a current is called a
magnetic circuit, by analogy with the electric and the electrostatic
circuits. Experiment shows that the magnetic lines of force are
always closed curves like the stream lines of an electric current, or
like the lines of electrostatic displacement (when these are com-
pleted through the conductors).
Fig. 1 exemplifies a fundamental law of electromagnetism ;
namely, an electric current creates a magnetic field in such direc-
tions that the lines of force are linked with the lines of flow of the
current, in the same manner that the consecutive links of a chain
are linked together. This law admits of no theoretical proof, and
must be accepted as a fundamental experimental fact. Wherever
there is an electric circuit there is also a magnetic field linking
with it. The two are inseparable, and increase and decrease
together. Each form of an electric circuit with a certain strength
of current in it corresponds to a definite form of magnetic field.
It is possible that the electric current and the magnetic field
are but two different ways of looking upon one and the same
phenomenon.
The linkages of magnetic lines with a current are seen more
clearly in Fig. 11, which shows the magnetic field produced by a
loop of wire, aa. It will be seen that the arrangement in Fig. 1
is more suitable for an elementary study, because the field is much
more uniform, especially if the radial thickness of the ring is
small as compared to its mean diameter, so that all the lines of
force are of practically the same length.
The same right-hand screw rule applies in the case of Fig.- 11
4 THE MAGNETIC CIRCUIT [ART. 2
as in Fig. 1. When the current in the loop of wire circulates in
the direction of rotation of a right-hand screw (toward the reader
on the left), the lines of force within the loop point in the direc-
tion of the progressive movement of the screw (upward) . The
rule can be reversed by saying that when the direction of the
lines of force around a wire is that of the rotation of a right-hand
screw, the current in the wire flows in the direction of the pro-
gressive movement of the screw. The first statement is con-
venient in the case of a ring winding, the second in the case of a
long straight conductor. Both rules can be combined into one
by considering the exciting electric circuit and the resulting mag-
netic circuit as two consecutive links of a chain. When the arrow-
head in one of the links (no matter which) points in the direction
of rotation of a right-hand screw, the arrow-head in the other link,
as it passes through the first, must point in the direction of the
progressive movement of the screw.
2. Magnetomotive Force. Experiment shows that the mag-
netic field within the ring (Fig. 1) does not change if the current
and the number of turns of the " exciting " winding vary so that
their product remains the same. That is to say, 500 turns of
wire with a current of 2 amperes flowing through each will pro-
duce the same field as 1000 turns with 1 ampere, or 200 turns with
5 amperes, because the product is equal to 1000 ampere-turns in
all cases. Even one turn with 1000 amperes flowing through it
will produce the same effect, provided that the turn is made of
a wide sheet of metal spread over the whole surface of the ring,
so as to make its action uniform throughout.
The reason for the above can be seen by considering 1000
separate turns with a current of 1 ampere flowing through each
turn, and each turn supplied with current from an independent
electrical source, say a dry cell. Connecting all the cells and all
the turns in series gives 1000 turns with one ampere flowing
through each. Connecting the cells and the turns in parallel
results in one wide turn with 1000 amperes of current in it.
Such changes in the electrical connections cannot affect the action
of each current outside the wire, because the value of the current
and the position of the turn is the same in both cases. Hence,
the magnetic action depends only upon the number of turns each
carrying 1 ampere, in other words, it depends upon the num-
ber of ampere-turns.
CHAP. I] FLUX AND MAGNETOMOTIVE FORCE 5
The number of ampere-turns of the exciting winding is called
the magnetomotive force of the magnetic circuit, because these
ampere-turns are the cause of the magnetic field. One ampere-
turn is the logical unit of magnetomotive force. In the example
above, the magnetomotive force is equal to 1000 ampere-turns.
In electric machines the field excitation often reaches several
thousand ampere-turns, and the magnetomotive force is for con-
venience sometimes measured in kiloampere-turns, one kilo-
ampere-turn being equal to 1000 ampere-turns.
3. Magnetic Flux. The magnetic disturbance at each point
within the ring has not only a direction, but also a magnitude. The
disturbance is said to be in the form of a flux, for the following
reason : One may think of the magnetic state as being due to the
actual displacement of some hypothetical incompressible sub-
stance along the lines of force; in this case the flux represents the
amount of this substance displaced through each cross-section of
the ring, and is analogous to total electrostatic displacement. Or,
as some modern writers think, there is an actual flow of an incom-
pressible ether along the lines of force. In that case the flux may
be thought of as the rate of flow of the ether through a cross-sec-
tion. The viewpoint common to these two explanations gave
rise to the name flux which means flow.
Some physicists consider the magnetic circuit as consisting
of infinitely subdivided (though closed) whirls or vortices in the
ether, the rotation being in planes perpendicular to the lines of
force. Each line of force is considered, then, as the geometric
axis of an infinitely thin fiber or tube of force, and the ether within
each tube in a state of transverse vortex motion. The line of
force represents the direction of the axis of rotation, and the flux
may be thought of as the momentum of the rotating substance
per unit length of the tubes of force. According to any of these
three views, the energy of a current is actually contained in the
magnetic circuit linked with the current.
Whichever view is adopted, the magnetic flux can be defined as
the sum total of magnetic disturbance through a cross-section per-
pendicular to the lines of force. Experiment shows that the total
flux is the same through all complete cross-sections of a magnetic
circuit. This could have been expected from the point of view
of a displacement or flow along the lines of force; each tube of
force being like a channel within which the displacement or the
6 THE MAGNETIC CIRCUIT [ART. 3
flow of an incompressible substance takes place. For this reason
the magnetic flux is said to be solenoidal (i.e., channel-shaped).
The familiar law of electromagnetic induction discovered by
Faraday is used for the definition of the unit of flux. Namely, when
the total magnetic disturbance or flux within a turn of wire
changes, an electromotive force is induced in the turn. By experi-
ments in a uniform field, the fact is established that the value of
the induced electromotive force is exactly proportional to the rate
of change of the flux linking with the test loop. This fact is
used in the definition of the unit of flux.
With the volt and the second as the units of e.m.f. and of
time respectively, the corresponding unit of flux is called the
weber } and is defined as follows : A flux through a turn of wire
changes at a uniform rate of one weber per second when the e.m.f.
induced in the turn remains constant and equal to one volt. Such
a unit flux can be also properly called the volt-second, though as
yet neither name has been recognized by the International Electro-
technical Commission. The weber or the volt-second is too large
a unit for most practical purposes. Therefore a much smaller
unit, called the maxwell,1 is used, which is equal to one one-hun-
dred-millionth part of the weber, or
one maxwell = one weber XlO~8.
The lines of force in Figs. 1 and 11 can be made to represent
not only the direction of the field, but its magnitude as well, if they
be drawn at suitable distances from each other. That is, such
that the total number of lines passing through any part of a
cross-section of the ring is equal numerically to the number of
maxwells in the flux through the same part. With this conven-
tion, each line stands symbolically for one maxwell; some engi-
neers and physicists speak of the number of lines of force in a flux
when they mean maxwells.
While the weber is too large a unit, the maxwell is too small for
many practical purposes. Therefore two other intermediate units
1 The origin of the maxwell becomes clear when one remembers that the
volt was originally established as 108 electromagnetic C.G.S. unit of electro-
motive force. The maxwell is related to the C.G.S. unit of e.m.f. or the so-
called abvolt in the same way in which the weber is related to the ordinary
volt. In other words, when the flux within a coil varies at the rate of cne
maxwell per second, one abvolt is induced in each turn of the winding.
CHAP. I] FLUX AND MAGNETOMOTIVE FORCE 7
are used, namely the kilo-maxwell, equal to one thousand max-
wells, and the mega-maxwell, equal to one million maxwells.
These two units are sometimes called the kilo-line and the mega-
line, the word " line " being used for the word maxwell, as
explained above.1
Prob. 1. The flux within the coil (Fig. 1) is equal to 63 kilo-maxwells.
A test coil of five turns is wound on the exciting coil so as to be linked
with the total flux. What voltage is induced in this test coil when the
current in the main (exciting) coil is reduced to zero at a uniform rate
in seven seconds? Ans. 0.45 millivolt.
Prob. 2. At what rate must the flux be varied in the preceding
problem in order to induce one volt in the test coil?
Ans. ' 0.2 weber (20 megalines) per second.
Prob. 3. When the flux varies at a non-uniform rate show that the
voltage induced in the test coil at any instant is equal to (d<P /dt) X 10-,
where t is time in seconds, and 0 is the flux in maxwells. Show that
the exponent of 10 must be —2 instead of —8 if the flux is expressed in
megalines.
4. The Reluctance of a Magnetic Path. Experiment shows that
the total flux within the coil (Fig. 1) is proportional to the applied
magnetomotive force, when the space inside is filled with air.
Therefore, a relation similar to Ohm's law holds, namely,
M = (R0, ........ (1)
where M is the magnetomotive force in ampere-turns, $ is the flux
in maxwells, and (ft is the coefficient of proportionality between
the two, called the reluctance of the magnetic circuit. Script (ft is
used to distinguish reluctance from electric resistance. The mag-
netomotive force M is the cause of the flux; or, with . reference to
an electric circuit, M is analogous to the applied electromotive
force. 0 is analogous to the resulting current, and the reluctance
(ft takes place of the electric resistance. Therefore, eq. (1) is
known as Ohm's law for the magnetic circuit. Of course, the
1 This possibility of creating new units of convenient size is a great
advantage of the metric or decimal system of units. New units are gener-
ally understood, by the use of Latin and Greek prefixes, signifying their
numerical relation to the fundamental unit. For instance, it is perfectly
legitimate to use such units as deci-ampere and hecto-volt, in spite of the
fact that they are not in general use. Anyone familiar with the agreed
prefixes will know that the units spoken of are equal to one-tenth of one
ampere, and to one hundred volts. See Appendix I on the Ampere-Ohm
System.
8 THE MAGNETIC CIRCUIT [ART. 4
analogy is purely formal, the two sets of phenomena being entirely
different. An equation similar to (1) can be written for the flow
of heat, of water, etc. It merely expresses the experimental fact
that, for a certain class of phenomena, the effect is proportional
to the cause.
If the space within the coil be filled with practically any known
substance, solid, liquid, or gaseous, the reluctance (R remains
within less than ± 1 per cent of the value which obtains with air.
The notable exceptions are iron, cobalt, nickel, manganese, chro-
mium, and some of their oxides and alloys.1 When the circuit
includes one of these so-called " ferro-magnetic " substances, a
much larger flux is produced with the same m.m.f., that is, the
reluctance of the circuit is apparently reduced to a considerable
extent. Moreover, this reluctance is no longer constant, but
depends upon the value of the flux. The behavior of iron and
steel in a magnetic circuit is of great practical importance, and is
treated in detail in Chapters II and III.
The definition of the unit of reluctance follows directly from
eq. (1). A magnetic circuit has a unit reluctance when a magneto-
motive force of one ampere-turn produces in it a flux of one
maxwell.2 No name has been given to this unit so far. The author
ventures to suggest the name rel, and he uses it provisionally in this
book. Granting that reluctance is a useful quantity in magnetic
calculations, one must admit that it should be measured in some
units of its own ; unless one chooses to use the cumbersome nota-
tion " ampere-turns per maxwell." The name rel is simply the
beginning of the word reluctance. Thus, a magnetic circuit has
a reluctance of one rel when one ampere-turn produces one
maxwell of flux in it. Tha unit rel is analogous to the ohm in the
electric circuit, and to the-ffar^|m the electrostatic circuit.
Prob. 4. What is the reluctance of the magnetic circuit in Fig. 1
if 47,600 ampere-turns produce a flux of 2.3 kilo-maxwells?
Ans. 47,600/2300 = 20.7 rels.
Prob. 5. How many ampere-turns are required to establish a flux
of 1.7 megalines through a reluctance of 0.0054 rel? Ans. 9180.
Prob. 6. A wooden ring is temporarily wound with 330 turns of
wire; when a current of 25 amperes is flowing through the winding the
^ee Dr. C.P. Steinmetz, Magnetic Properties of Material Electrical
World, Vol. 55 (1910), p. 1209.
2 See Appendix I at the end of the book.
CHAP. I] FLUX AND MAGNETOMOTIVE FORCE 9
flux is found to be equal to 21 kilo-maxwells. The permanent winding
on the same ring must produce a flux of 65.1 kilolines at a current of
9.3 amperes. How many turns will be required? Ans. 2750.
5. The Permeance of a Magnetic Path. In calculations per-
taining to the electric circuit it is convenient to deal with the recip-
rocals of resistances when conductors are connected in parallel.
The reciprocal of a resistance is called a conductance and is meas-
ured in mhos if resistance is measured in ohms. Similarly, a die-
lectric is characterized sometimes by its elastance, at other times
by the reciprocal of its elastance, which is called permittance.
When permittance is measured in farads, elastance is measured in
darafs (see the chapter on the Electrostatic Circuit in the author's
Electric Circuit) .
Analogously, when two or more magnetic paths are in parallel
it is convenient to use the reciprocals of the reluctances. The
reciprocal of the reluctance of a magnetic path is called its per-
meance; eq. (1) becomes then
$ = (PM, . . / (2)
where
(P=1/(R (3)
A script (P is used for permeance in order to avoid confusing it
with power. For the unit of permeance corresponding to the
rel, the author proposes the name perm. A magnetic path has a
permeance of one perm when one maxwell of flux is produced for
each ampere-turn of magnetomotive force applied along the path.
The unit "-perm " has been in use among electrical designers
for some time, although no name has been given to it. Notably
Mr. H. M. Hobart has used it extensively in his writings,
in the calculation of the inductance of windings. He speaks of
" magnetic lines per ampere-turn per unit length " (of the
embedded part of a coil) . This is equivalent to perms per unit
length.
In the ampere-ohm system the internationally accepted unit
of permeance is the henry.1 Therefore, if in eq. (2) M is measured
in ampere-turns and $ in webers, (P is in henrys, and no new unit
for permeance is necessary. In this case the reluctance (R in eqs.
1 Although the henry is defined as the unit of inductance, it is shown in
Art. 58 below that permeance and inductance are physically of the same
dimensions and hence measureable in the same units.
10 THE MAGNETIC CIRCUIT [ABT. 6
(1) and (3) is in henrys"1; or spelling the word henry backwards,
as in the case of mho and daraf, the natural unit of reluctance in
the ampere-ohm system gets the euphonious name of yrneh (to be
pronounced earney).
Since, however, the maxwell is used almost exclusively as the
unit of flux, it seems advisable to introduce the rel and the perm
as units directly related to it. Should engineers gradually feel
inclined to use the weber and its submultiples as the units of flux,
then the henry, the yrneh, and their multiples and submultiples
would naturally be used as the corresponding units of permeance
and reluctance.
We have, therefore, the two following systems of units for
reluctance and permeance, according to whether the maxwell or
the weber is used for the unit of flux (one ampere-turn being the
unit of m.m.f. in both cases) :
Unit of flux
Unit of permeance
Unit of reluctance
Maxwell
Weber
Perm
Henry
Rel
Yrneh
L/V/A _E— t-V/J-JLJ- V JLAAAV^i.
One perm = 10~ 8 henry ; one rel = 10 8 yrnehs.
Prob. 7. What is the permeance of the magnetic circuit in prob. 4?
Ans. 0.0483 perm. =4.83 X 10-10 henry.
Prob. 8. What is the permeance of the ring in prob. 6?
Ans. 2.545 perm. = 0.02545 microhenry.
Prob. 9. How many ampere-turns are required to maintain a flux
of 2.7 megalines through a permeance of 750 perms? Ans. 3600.
6. Reluctivity and Permeability. The reluctance of a magnetic
path varies with the dimensions of the path according to the same
law as the resistance of an electric conductor or the elastance of a
dielectric. That is to say, the reluctance is directly proportional
to the average length of the lines of force and is inversely propor-
tional to the cross-section of the path. This relationship can be
verified by measurements on rings of different dimensions (Fig. 1) .
When the diameter of the ring is increased twice, keeping the same
cross-section, the length of the path of the flux is also increased
twice. Experiment shows that the new ring requires twice as
many ampere-turns as the first one for the same flux ; or, only one-
half of the flux is produced with the same number of ampere-
CHAP. 1] FLUX AND MAGNETOMOTIVE FORCE 11
turns. If the diameter of the ring is kept the same but the cross-
section of the path is increased twice, the flux is doubled with the
same magnetomotive force. These and similar experiments show
that the reluctance and the permeance of a uniform magnetic path
obey the same law as the resistance and the conductance of a con-
ductor, or the elastance and the permittance of a prismatic slab
of a dielectric.
We can therefore put
(ft = vl/A, . ... . . . . . (4)
where I is the mean length of the path, A is its cross-section, and
v is a physical constant. By analogy with resistivity and elastiv-
ity, v is called the reluctivity of a magnetic medium. If (R is in rels,
and the dimensions of the circuit are in centimeters, v is in rels per
centimeter cube. In other words, the reluctivity of a magnetic
medium is the reluctance of a unit cube of this medium when the
lines of force are parallel to one of the edges. For air and all other
non-magnetic substances the experimental value of v is 0.8 rel per
centimeter cube,1 or 0.313 rel per inch cube.
The expression for permeance corresponding to eq. (4) is
(5)
where the coefficient /* is called the permeability of the magnetic
medium. It corresponds to the electric conductivity ? and the
dielectric permittivity K. Since the permeance of a path is the
reciprocal of its reluctance, the permeability of a medium is the
reciprocal of its reluctivity, or
(6)
When the perm and the centimeter are used for the units of per-
meance and length, permeability is expressed in perms per centi-
meter cube. For all non-magnetic materials //=1.25 perms per
centimeter cube (more accurately 1.257). With the henry and the
centimeter as units /*= 1.257 X10~8 henries per centimeter cube.
In the English system ^=3.19 perms per inch cube for non-
lMore accurately 0.796 rel per centimeter cube. As a rule, magnetic
calculations are much less accurate than electrical calculations, because there
is no "magnetic insulator" known, so that there is always some magnetic
leakage present, which is difficult to take into consideration. For this reason
the value 0.8 is sufficiently accurate for most practical purposes.
12 THE MAGNETIC CIRCUIT [Aux. 7
magnetic materials. For magnetic materials JJL is considerably
larger than for non-magnetic, and varies with the field strength.
In calculations either reluctivity or permeability is used, according
to the conditions of the case and the preference of the engineer.
The student has probably heard before that the permeability of
air is assumed equal to unity. The discrepancy between this com-
monly accepted value and the value 1.25 given above, is due to a
different unit of magnetomotive force, called the gilbert, which is
sometimes employed. The author considers the gilbert to be
of doubtful utility, for reasons stated in Appendix II ; hence no
use is made of it in this book.
Prob. 10. Assuming the value of /* = 1.257 to be given, check the
value of // = 3.19 in the English system, and also the values of y in the
metric and the English systems, as given above.
Prob. 11. In prob. 4 the reluctance of a ring was 20.7 rels. If the
cross-section of the ring is 120 sq. mm., what is the average diameter of
the ring? Ans. 9.9 cm.
Prob. 12. How many ampere-turns are required to establish a flux
of 47 kilolines in a ring of rectangular cross-section, made of non-magnetic
material; the radial thickness of the ring is 8 cm., the axial width 11 cm.
and the average radius 16 cm? Ans. About 43 kiloamp ere- turns.
Prob. 13. How many ampere-turns would be required in the preced-
ing problem for the same flux if the ring were made of iron, the relative
permeability of which (with respect to air) is 500?
Ans. 86 ampere-turns.
7. Magnetic Intensity. In order that the student may better
appreciate the significance of the concept of magnetic intensity,
it is advisable to refresh in his mind the corresponding quantity
used in the electric circuit, viz., the electric intensity. Namely, in
problems on the electric and the electrostatic circuit it is some-
times desirable to consider not only the total voltage, but also
the voltage used up or balanced per unit length of the path along
which the electricity flows or is displaced. This quantity, the
rate of change of voltage along the circuit, is known as the electric
intensity, or the voltage gradient. It is denoted by F (see the
Electric Circuit), and is measured in volts per linear centimeter.
When the voltage drop is uniform along a conductor or a dielectric,
F=E/l, where E is the voltage between the ends of the part of the
circuit under consideration, and I is the corresponding length.
When the voltage drop is not uniform, F is different for different
points along the path, and for each point F=dE/dl.
CHAP. I] FLUX AND MAGNETOMOTIVE FORCE 13
In a similar way, the magnetomotive force of a magnetic circuit
is used up bit by bit in the consecutive parts of the circuit. One
can speak not only of the total magnetomotive force of a closed
circuit, but also of the magnetomotive force acting upon a certain
part of the circuit, and of magnetomotive force per unit length of
the lines of force. Thus, for instance, if 1000 ampere-turns is con-
sumed in a uniform magnetic circuit 4 cm. long, the magnetomo-
tive force per unit length of path is 250 ampere-turns.
The magnetomotive force per unit length of path is called the mag-
netic intensity at a point, or the m.m.f. gradient, and is denoted by
H. Thus, if the circuit is uniform, the magnetic intensity at any
point is
H-M/l, . . . .... (7)
where M is the magnetomotive force acting upon the length I of
the circuit. If the magnetic circuit is non-uniform, for instance, if
the cross-section of the ring is different at different places, or if the
permeability is different at some parts of the circuit due to the
presence of iron, the m.m.f. gradient is different at different points,
and at each point it is expressed by the equation
H=dM/dl, . .... ....... (8)
where dM is the m.m.f. necessary for establishing the flux in the
length dl of the circuit. If M is in ampere-turns, and I is in centi-
meters, H is in ampere-turns per centimeter.
Eqs. (7) and (8) can be also written in the form
M=m, ......... (9)
and
- C2Hdl.
Jh
(10)
These formulae, expressed in words, simply mean that the magneto-
motive force acting upon a certain part of a magnetic circuit is the
line integral of the magnetic intensity along the path, or the sum of
the m.m.fs. used up in the elementary parts of the path. The rela-
tion between M and H will become clearer to the student in the
various applications that follow.
Prob. 14. What is the magnetic intensity in prob. 12?
Ans. About 425 ampere-turns per cm. of path.
14 THE MAGNETIC CIRCUIT [ART. 8
8. Flux Density. It is often of importance to consider the flux
density, or the value of a flux per unit of cross-section perpendicular
to the direction of the lines of force. Flux density is usually
denoted by B, and is measured in maxwells (or its multiples) per
square centimeter.1 When the flux is distributed uniformly over
the cross-section of a path, the flux density
....... (11)
where A is the area of the cross-section of the path. If the flux is
distributed non-uniformly, an infinitesimal flux d(D passing through
a cross-section dA must be considered. In the limit, the flux den-
sity at a point corresponding to dA is
(12)
The areas A and dA are understood to be at all points normal
to the direction of the field. Solving these two equations for the
flux we find
•-B-A, (is)
or
,oAB-dA, . . . . . . (14)
the integration being extended over the whole cross-section of the
path. Expressed in words, these last two formulae mean that
the total flux passing through a surface is equal to the sum of the
fluxes passing through the different parts of that surface.
Magnetic flux density is analogous to current density C7, and
to dielectric flux density D treated in the Electric Circuit. The
student will find no difficulty in interpreting eqs. (11) to (14)
from the point of view of the electric and electrostatic circuits.
The relation between B and H is obtained from eq. (1) in which
the value of (R is obtained from eq. (4). Namely, we have
or
1 Some writers express flux density in gausses, one gauss being equal to
one maxwell per square centimeter. The unit kilogauss, equal to one kilo-
maxwell per square centimeter, is also used. While the terms gauss and
kilogauss are convenient abbreviations, no use is made of them in this book
in order to keep the relation between a flux and the cross-section of its path
explicitly before the student.
CHAP. I] FLUX AND MAGNETOMOTIVE FOR E 15
The last expression, according to eqs. (7) and (11), can be written
simply as
H=Bv, (15)
or, since v=l/p,
B = tiH :-.'.. ,( . (16)
Eqs. (15) and (16) state Ohm's law for a unit magnetic path, for
instance, a path one centimeter long and one square centimeter in
cross-section. H is the magnetomotive force between the oppo-
site faces of the cube, fi is the permeance of the cube, and B is the
flux passing through it. The reader will remember similar equa-
tions U=fF and D = KF for the unit electrical conductor and the
unit prism of a dielectric respectively.
Instead of beginning the theory of the magnetic circuit with
eq. (1) and developing it into eq. (16), it is possible to begin it with
eq. (16). Namely, the known magnetic phenomena show that
at each point in the medium there is a magnetic intensity H which
is the cause of the magnetic state, and that the effect is measured
by the flux density 5; /* is the physical constant which shows the
proportionality between H and B. The magnetic circuit is then
assumed to be built up of infinitesimal tubes of flux in series and
in parallel, and finally eq. (1) is obtained.
Prob. 15. What is the flux density in prob. 12?
Ans. 534 maxwells per square centimeter (534 gausses).
Prob. 16. How many ampere-turns per pole are required to establish
a flux density of 7 kilolines per square centimeter in the air-gap of a
machine, the clearance being 3 mm.? Solution: According to eq. (15)
H = 7000/ 1 .25 = 5600 ampere-turns per centimeter of length. Hence the
required m.m.f. is 5600X0.3 = 1680 ampere-turns.
9. Reluctances and Permeances in Series and in Parallel. In
practice, one has to deal mostly with magnetic circuits of irregular
form, for instance, those of electric machines (Fig. 24) in which
the flux is established partly in air and partly in iron, each of vary-
ing cross-section. The circuit consists in this case of several reluc-
tances in series. One may say, for instance, that the total mag-
netomotive force required in this machine, per magnetic circuit,
is 8000 ampere-turns, of which 6000 are used in the air-gap, 1500
in the field frame, and 500 in the armature. This is analogous to
distinguishing between the total e.m.f . of an electric circuit, and the
voltage drop in the various parts of the circuit.
16 THE MAGNETIC CIRCUIT [ART. 9
In some cases two or more magnetic paths are in parallel, for
instance, when there is magnetic leakage (see below). In most
cases the engineer has to consider complicated magnetic circuits
which consist partly of paths in series, partly of paths in parallel.
Thus, in the same machine, the m.m.f . or the difference of mag-
netic potential between the pole-tips is 6500 ampere-turns. This
m.m.f. maintains a useful flux of say 2.5 megalines through the
armature, and say 0.5 megaline of leakage or stray flux between
the pole-tips. Thus the total flux in the field frame is 3 mega-
lines.
The fundamental law of the magnetic circuit, as expressed by
eq. (1), is analogous to Ohm's law for the simple electric circuit.
Therefore magnetic paths in series and in parallel are combined
according to the same rule that ele'ctrical conductors are combined
in series and in parallel. Namely, when two or more magnetic
paths are in series, their reluctances are added ; when two or more
magnetic paths are in parallel their permeances are added. Or,
for a series combination,
........ (17)
and for a parallel Combination
(18)
It will be remembered that similar relations hold also for impe-
dances and admittances in the alternating current circuit, and
for elastances and permittances in the electrostatic circuit.
The proof of formulae (17) and (18). is similar to that usually
given for the combination of electric resistances in series and in
parallel. Namely, when reluctances are in series the total mag-
netomotive force is equal to the sum of component m.m.f .s., or
. . . •.-"-. . . (17a)
Dividing both sides of this equation by the common flux 0 eq.
(17) is obtained. When permeances are in parallel, the total flux
is the sum of the component fluxes, or
eq
(18a)
Dividing both sides of this equation by the common M, eq. (18) is
obtained.
CHAP. I] FLUX AND MAGNETOMOTIVE FORCE 17
One of the reasons for which calculations are as a rule more
involved and less accurate in the magnetic than in the electric cir-
cuit is that there is no magnetic insulation known, and therefore the
paths of the flux in a great majority of cases cannot be shaped and
confined at will. The student will appreciate, therefore, the reason
for selecting a toroidal ring as the simplest magnetic circuit. If
the winding is distributed uniformly there is no tendency for mag-
netic leakage, except for a very small amount in and around each
wire. With almost any other arrangement of a magnetic circuit
there is a difference of magnetic potential, or an m.m.f. between
various parts of the circuit, and part of the flux passes directly
through the path of the least resistance, in parallel with the useful
path. A familiar example of this is the magnetic leakage between
the adjacent pole-tips oian electrical machine (Fig.29), or between
the coils of a transformer (Fig. 50) .
The conditions in a magnetic circuit are similar to those
in an imperfectly insulated electric circuit, when it, together with
its sources of e.m.f., is immersed in a conducting liquid. Part of
the current finds its path through the liquid instead of through
the conductors; the current is different in different parts of the
circuit, and the calculations are much more involved and less
accurate, because the paths of the current in an unlimited medium
can be estimated only approximately.
In order to prevent or to minimize leakage the exciting ampere-
turns should be distributed over the whole magnetic circuit, to
each part in proportion to its reluctance. Then the m.m.f. is con-
sumed where it is applied, and no free m.m.f. is left for leakage.
Unfortunately, such an arrangement is impracticable in most
cases, though it ought to be approached as nearly as possible (see
Prob. 17 below).
If there were a magnetic insulator, that is, a substance or a
combination the permeability of which was many times lower than
that of the air, it would be a great boon to the electrical industry.
It would then be possible to avoid magnetic leakage by insula-
ting magnetic circuits as perfectly as electric circuits are insulated.
The absence of leakage would allow a reduction in the size of the
field frames and exciting coils of direct- and alternating-current
machines. It would also permit us to improve the voltage regula-
tion of generators and transformers, to raise the power factor of
induction motors, and to increase considerably their overload
18 THE MAGNETIC CIRCUIT [ART. 9
capacity; it would also largely eliminate sparking in commutating
machines.
Prob. 17. A long iron rod, having a cross-section of 9.3 sq.cm., is
bent into a circular ring so that the ends almost touch each other. The
ring is wound with 500 turns of wire, the winding being concentrated
around the gap to minimize the leakage. When a current of 2.5 amperes
is sent through the winding a flux of 74.9 kilo-maxwells is established
in the circuit. Assuming the reluctance of the iron to be negligible,
calculate the clearance between the ends of the rod.
Ans. Between 1.9 and 2.0 mm.
Prob. 18. What is the length of the air-gap in the preceding problem
if the estimated reluctance of the iron part of the circuit is 2 milli-rels?
Ans. 1.7 mm.
Prob. 19. A magnetic circuit consists of three parts, the reluctances
of which are (R^ 0.004 rel, (R2 = 0.005 rel, and (R3 = 0.013 rel. The
paths (R2 and (R3 are in parallel with each other and are in series with
(Ha. What is the total permeance of the circuit? Ans. 131.4 perms.
Prob. 20. In the preceding problem let (Hi be the reluctance of the
steel frame of an electric machine, (R2 be that of two air-gaps, and the
armature, and (R3 the leakage reluctance between two poles. The ratio
of the total flux in the frame to the useful flux through the armature
is called the leakage factor of the machine. What is its value in this
case? Ans. 1.38.
Prob. 21. Referring to the two preceding problems let the air-gap be
reduced so as to reduce the leakage factor to 1.2. How many ampere-
turns will be required to produce a useful flux of 2.1 megalines in the
magnetic circuit under consideration? Ans. 12,950.
Prob. 22. An iron ring having a cross-section" of 4 by 5 cms. is placed
inside of a hollow ring. This ring has a mean diameter of 32 cm., an
axial width of 11 cm., and a radical thickness of 8 cm. How many
ampere-turns are required to produce a total flux of 47 kilolines (count-
ing that in the air as well as that in the iron), if the estimated relative
permeability of the iron is 1400? Hint: Let the average flux density in
the air be Ba, and that in the iron be B{. We have two simultaneous
equations: 20£; + (88-20)#o=47, and £;/£a = 1400. Ans. 134.
Prob. 23. What per cent of the total flux in the preceding problem
is in the air? Ans. 0.24 per cent.
Prob. 24. Show that in a ring, such as is shown in Fig. 1, the flux
density, strictly speaking, is not uniform, but varies inversely as the
distance from the center. Solution: Take an elementary tube of flux
of a radius x. The magnetic intensity at any point within the tube is
H=M/2nx, and the flux density, according to eq. (16), B = /i.M/2nx.
Prob. 25. What is the true permeance of a circular ring of rectangular
cross-section, the outside diameter of which is Dlt the inside diameter D2,
and the axial width ht Solution: The permeance of an infinitesimal
tube of radius x is cKP =/j.hdx/2xx. The permeances of all the tubes
CHAP. I] FLUX AND MAGNETOMOTIVE FORCE 19
are in parallel and should be added; hence, integrating the foregoing
expression between the limits %Dl and %D2 we get : (P = (/j.h/2rc)'Ln(Dl/D2) .
Prob. 26. Show that, when the radial thickness b of a ring is small as
compared to its mean diameter D, the exact expression for permeance,
obtained in the preceding problem, differs but little from the approxi-
mate value, fthb/xD, used before. Solution: Using the expansion,
.. and putting
we get ($> = (iJLhb/7:D)[l+%(b/Dy+%(b/D)4 + . . . ]. When the ratio of
b to D is small, all the terms within the brackets except the first one,
can be neglected.
Prob. 27. Show that the answer to prob. 11 is 2.1 per cent high on
account of the density being assumed there as uniform throughout the
cross-section of the ring.
CHAPTER II
THE MAGNETIC CIRCUIT WITH IRON
10. The Difference between Iron and Non-Magnetic Materials.
Steel and iron differ in their magnetic properties from most other
known materials in the following respects :
(1) The permeability of steel and iron is several hundred and
even thousand times greater than that of non-magnetic materials.
(2) The permeability of steel and iron is not constant, but
decreases as the flux density increases.
(3) Changes in the magnetization of steel and iron are
accompanied by some sort of molecular friction (hysteresis) with
the result that the same magnetomotive force produces a different
flux when the exciting current is increasing than when it is de-
creasing (Fig. 7).
Besides iron, the four adjacent elements in the periodic system,
viz., cobalt, nickel, manganese, and chromium, are slightly mag-
netic. Some alloys and oxides of these metals show considerable
magnetic properties. Heusler succeeded in producing alloys of
manganese, aluminum, and copper which are strongly magnetic.
These alloys have not been used in practice so far.1
11. Magnetization Curves. The magnetic properties of the steel
and iron used in the construction of electrical machinery are shown
in Figs. 2 and 3. These curves are called magnetization curves, or
B — H curves] sometimes also the saturation curves of iron.
The flux density, in kilolines per square centimeter of cross-sec-
tion, is plotted, in these curves, against the ampere-turns per
centimeter length of the magnetic circuit as abscissae.
The student may conveniently think of these curves as represent-
1 For the preparation and properties of Heusler's alloys see Guthe and
Austin, Bulletins of Bureau of Standards, Vol. 2 (1906), No. 2, p. 297; Dr. C.
P. Steinmetz, Electrical World, Vol. 55 (1910), p. 1209; Knowlton, Physical
Review, Vol. 32 (1911), p. 54.
20
CHAP. II]
MAGNETIC CIRCUIT WITH IRON
21
ing the results of tests on samples of iron in ring form, as in Fig. I.1
The current in the exciting coil is adjusted to a certain value, and
the corresponding value of the flux in the iron ring is determined
by any of the known means, for instance, by a discharge through
50 100 150 Scale "B" 200
10 20 30 Scale "A" 40
H = AMPERE-TURNS PER CENTIMETER
Fig. 2 — Magnetization in steel and iron — castings and forgings.
1 For an experimental study of the magnetic circuit with iron and for
practical testing of the magnetic properties of steel and iron see Vol. 1,
Chapters 6 and 7, of the author's Experimental Electrical Engineering.
22 THE MAGNETIC CIRCUIT [ART. 11
a secondary coil connected to a calibrated ballistic galvanometer.
The exciting ampere-turns divided by the average length of the
path give the magnetic intensity H. The total flux divided by the
cross-section of the iron path gives the value of the flux density B,
which is plotted as an ordinate against H for an absissa. Similar
tests are made for other values of H and B\ the results give the
magnetization curve of the material. In other words, a magneti-
zation curve gives the relation between the magnetomotive force
and the flux for a unit cube of the material. By combining unit
cubes in series and in parallel a relationship is established
between flux and ampere-turns for a circuit of any dimensions,
made of the same material.
The curves shown in Fig. 2 refer to the following materials: (a)
Cast iron, which is used as the magnetic material in the stationary
field frames of direct-current machines, and in the revolving-field
spiders of low speed alternators. It is evident from the curves
that cast iron is magnetically much inferior to steel ; but it is used
on account of its lower cost and ease of machining. (6) Cast steel,
which is used for pole pieces, plungers of electromagnets, etc.
It is used also for the field frames of such machines in which
economy of weight or space is desired, for instance, in railway
and crane motors, and in machines built for export, (c) Forged
steel, which is used for the revolving fields of turbo-alternators,
on account of the considerable mechanical stresses developed in
such high speed machines by the centrifugal force.
The curves in Fig. 3 refer to carbon-steel laminations and to
silicon-steel laminations. The former is used in the armatures of
direct and alternating-current machines, the latter mainly in trans-
formers. There is not much difference between the two kinds with
regards to their B — H curves, but silicon steel shows a much lower
loss of energy due to hysteresis and eddy currents (see Art. 20
below). A material of much higher permeability is used for
armature cores, when it is desired to use very high flux densities
in the teeth. A magnetization curve for such steel laminations is
shown in Fig. 28.
For convenience and accuracy the lower part of each curve in
Fig. 2 is plotted separately to a larger scale, " A," while the upper
parts are plotted to a smaller scale, " B." Thus, Fig. 2 contains
only three complete magnetization curves. The curve for silicon-
steel laminations in Fig. 3 is also plotted to two different scales,
CHAP. II] MAGNETIC CIRCUIT WITH IRON 23
1000 2000 3000 Scale"C" 4000
CQ
1000 2000 3000 Scale "C" 4000
100 200 300 Scale "B" 400
10 20 30 Scale "A" 40
H = AMPERE-TURNS PER CENTIMETER
FIG. 3. — MagnetizaticJb. in steel laminations.
24 THE MAGNETIC CIRCUIT [ART. 12
while three different scales are used for the carbon-steel curve.
The values of H at very low flux densities are unreliable because
in reality each curve has a point of inflexion near the origin, not
shown in Figs. 2 and 3 (see Fig. 7).
The curves given in Figs. 2 and 3 represent the averages of many
curves obtained from various sources. The iron used in an indi-
vidual case may differ considerably in its magnetic quality from the
average curve. The value of B obtainable with a given H depends
to a large degree upon the chemical constitution of the specimen,
impurities, heat treatment, etc. As a rule, the soft and pure grades
of steel are magnetically better, that is to say, they give a higher
flux density for the same magnetizing force, or, what is the same,
they possess a higher permeability. Annealing improves the
magnetic quality of iron, while punching, hammering, etc., lowers
it. Therefore, the laminations used in the construction of elec-
trical machinery are usually annealed after being punched into
their final shape. This annealing also reduces hysteresis loss.
12. Permeability and Saturation. Permeability is defined in
Chapter I as the permeance of a unit cube, or, according to eq.
(16), as the ratio of B to H. The two definitions are, of course,
identical. Therefore, the values of permeability for various values
of B are easily obtained from the magnetization curves. For
instance, for cast steel, at B= 15 kilolines per sq. cm. the magnetic
intensity H is 26 ampere-turns per cm., so that JJL= 15000/26 = 577
perms per cm. cube. This is the value of the absolute permeabil-
ity in the ampere-ohm-maxwell system. In most books the relative
permeability of iron is employed, referring to that of the air as
unity. Since in the above-mentioned system //=1.25 for air,
the relative permeability of cast steel at the selected flux density
is 577/1.25 = 461.
In practice, the calculations of magnetic circuits with iron are
arranged so as to avoid the use of permeability /* altogether, using
the B — H curves directly. In some special investigations, how-
ever, it is convenient to use the values of permeability, and also
an empirical equation between PL and B. For instance, see the
Standard Handbook for Electrical Engineers ; the topic is indexed
" permeability — curves/' and " permeability — equation/' These
fji—B curves show that there must be a point of inflection in
the B—H curves at low densities, because the values of /* reach
their maximum at a certain definite density instead of being con-
CHAP. II] MAGNETIC CIRCUIT WITH IRON 25
slant for the lower part of the curves. Such would be the case if
the lower parts of the B—H curves were straight lines, as shown
in Figs. 2 and 3, because then the ratio of B to H would be con-
stant. However, in ordinary engineering work the lower parts of
magnetization curves are usually assumed to be straight lines, and
the permeability constant.
Three parts can be distinguished in a B — H or magnetization
curve: the lower straight part, the middle part called the knee of
the curve, and the upper part, which is nearly a straight line. As
the magnetic intensity H increases, the corresponding flux density
B increases more and more slowly, and the iron is said to approach
saturation. With very high values of the magnetic intensity H,
say several thousand ampere-turns per centimeter, the iron is com-
pletely saturated and the rate of increase of flux density with H is
the same as in air or in any other non-magnetic material. That is
to say, the flux density B increases at a rate of 1.257 kilolines for
each kilo-ampere-turn increase in H. Such is the slope of the upper
curve in Fig. 3.
In view of this phenomenon of saturation the total flux density
in iron can be considered as consisting of two parts, one due to the
presence of iron, the other independent of it, as if the paths of the
lines of force were in air. These two parts are shown separately
in Fig. 4. The part OA, due to the iron, approaches a limiting
value B8) where the iron is saturated. The part OC, not due to the
iron, increases indefinitely in accordance with the straight line
law, B = fj.H, where /*= 1.257. The curve OD of total flux density
resembles in shape that of OA, but approaches asymptotically
a straight line KL parallel to OC.
While it is customary to speak of the saturation in iron as being
low, high, or medium, the author is not aware of any generally
recognized method of expressing the degree of saturation numeri-
cally. It seems reasonable to define per cent saturation in iron with
respect to the flux density B8} so that, for instance, the per cent
saturation at the point N is equal to the ratio of PN' to Bs. This
method of defining saturation, while correct theoretically, pre-
supposes that the ordinate B8 is known, which is not always the
case.
The percentage saturation of a machine is defined in Art. 58 of
the Standardization Rules of the American Institute of Electrical
Engineers (edition of 1910) as the percentage ratio of OQ to PN,
26
THE MAGNETIC CIRCUIT
[AKT. 13
QN being a tangent to the saturation curve at the point N under
consideration. An objection to this definition is that according to it
the per cent saturation does not approach 100 as N increases
indefinitely; on the contrary, the per cent saturation gradually
decreases to zero beyond a certain value of N. This is, of course,
absurd. Moreover, the foregoing definition of the Institute refers
explicitly to the " percentage of saturation of a machine," and it
is not clear whether magnetization curves of the separate materials
are included in it or not. The practical advantage of this definition
as compared to that given above is that it is not necessary to
know the value of B8.
FIG. 4. — A magnetization curve analyzed.
13. Problems Involving the Use of Magnetization Curves.
The following problems have been devised to give the reader a clear
understanding of the meaning of magnetization curves, and to
develop fluency in their use. These problems lead up to the
magnetic circuit of electric machines treated in Chapters V and
VI. With almost any arrangement of a magnetic circuit there
is some leakage or spreading of the lines of force, which is difficult
to take into account theoretically. This leakage is neglected in
most of the problems that follow, so that the results are only
approximately correct. Leakage is considered more in detail in
Art. 40 below, though practical designers are usually satisfied with
CHAP. II] MAGNETIC CIRCUIT WITH IRON 27
estimating it from the results of previous tests, rather than to
calculate it theoretically.
Prob. 1. Samples of cast steel are to be tested for their magnetic
quality up to a density of 19 kilolines per square centimeter. They are
to be in the form of rings, 20 cm. average diameter, and 0.75 sq.cm.
cross-section. For how many ampere-turns should the exciting winding
be designed, and what is the lowest permeance of the circuit, if some
specimens are expected to have a permeability 10 per cent lower than
that according to the curve in Fig. 2 ?
Ans. 10.4 kiloampere-turns; 1.37 perm.
Prob. 2. Explain the reason for which it is not necessary to know the
cross-section of the specimens in order to calculate the necessary ampere-
turns in the preceding problem.
Prob. 3. Some silicon steel laminations are to be tested in the form of
a rectangular bunch 20 by 2 by 1 cm., in an apparatus called a permeam-
eter. The net cross-section of the iron is 90 per cent of that of the
packet. It is found for a sample that 336 ampere-turns are required to
produce a flux of 25.2 kilo-maxwells, the ampere-turns for the air-gaps and
for the connecting yoke of the apparatus being eliminated. How does
the quality of the specimen compare with the curve in Fig. 3?
Ans. The permeability of the sample at B = 14 is about 5 per cent
lower than that according to the curve.
Prob. 4. What are the values of the absolute and the relative per-
meability and reluctivity of the sample in the preceding problem?
Ans. n v
relative 663 (numeric) 0.00151 (numeric)
absolute 833 perms per cm. cube 0.00120 rels per cm. cube.
Prob. 5. What is the maximum permeability of cast iron according to
the curve in Fig. 2? Ans. About 600 perms per cm. cube.
Prob. 6. Mark in Figs. 2 and 3 vertical scales of absolute and relative
permeability, so that values of permeability could be read off directly by
laying a straight edge between the origin and the desired point of the
magnetization curve.
Prob. 7. What is the percentage of saturation in carbon steel lamina-
tions at a flux density of 20 kilo-maxwells per square centimeter,
according to both definitions given in Art. 12? Ans. 92.5; 88.5.
Prob. 8. An electromagnet has the dimensions (in cm.) shown in
Fig. 5; the core is made of carbon steel laminations 4 mm. thick, the
lower yoke is of cast iron. The length of each air-gap is 2 mm.; each
exciting coil has 450 turns. What is the exciting current for a useful
flux of 2.2 megalines in the lower yoke? Neglect the magnetic leakage
between the limbs of the electromagnet (this leakage is taken into consid-
eration in the next problem). £eKMe^:(With laminations 4 mm. thick
the space occupied by insulation between stampings is altogether negligi- i/
ble ; therefore the flux density in the steel is the same as in the air-gap,)
and is equal to 17.2 kl/sq. cm.; in the cast iron the flux density is 11.5
'
28
THE MAGNETIC CIRCUIT
[ART. 13
kl/sq. cm. One-half of the average length of the path in the steel is
37.3 cm., and in the cast iron 20.5 cm. Hence, with reference to the
magnetization curves, we find for one-half of the magnetic circuit (the
other half being identical, it is sufficient to calculate for one-half) :
amp.-turns for steel core 65 X 37.3 = 2425
amp .-turns for one air-gap 0.2 X 0.8 X 17200 = 2752
amp.-turns for the cast-iron yoke 180 X 20.5 = 3690
Total
8867.
Ans. The exciting current is 8867/450 = 19.7 amperes.
r Prob. 9. In the solution of the preceding problem the effect of leakage
is disregarded. It is found by experiments on similar electromagnets
FIG. 5. — An electromagnet (dimensions in centimeters).
that the leakage factor is equal to about 1.2, that is to say, the flux in the
upper yoke is 20 per cent higher than that in the lower one. This means
that out of every 1200 fines of force in the upper yoke 1000 pass through
the lower yoke as a part of the useful flux, and 200 find their path as a
leakage through the air between the limbs, as shown by the dotted lines.
Calculate the exciting current required in the preceding problem, assuming
(a) that the total leakage flux is concentrated between the two air-gaps
along the line aa; (6) that it is concentrated along the line bb, at one-
third of the distance from the bottom of the exciting coil, that is 6.33
cm. from the air-gaps.
Ans. (a) 44.2 amperes; (6) 40 amperes.
Prob. 10. Show that it is more correct in the preceding problem to
assume the leakage flux concentrated at one-third of the distance from
the bottom of the exciting coils, than at the center of the coils.
CHAP. II] MAGNETIC CIRCUIT WITH IRON 29
Prob. 11. A ring of forged steel has such dimensions that the average
length of the lines of force is 70 cm. The ring has an air-gap of 1.5 mm., and
is provided with an exciting winding concentrated near the air-gap so as
to minimize the leakage. What is the flux density at an m.m.f. of 4000
ampere- turns? First Solution : Assume various values of B, calculate the
corresponding values of the ampere-turns, until the value of B is found, for
which the required excitation is 4000 ampere-turns (solution by trials).
Second solution: Let the unknown density be B and the corresponding
magnetic intensity in the steel be H. The required excitation for the steel
is then 7QH, and for the air-gap 0.15X0.8X10005 = 1205 ampere-turns.
Therefore,
70# + 1205 =4000.
The values of B and H must satisfy this equation of a straight line,and
besides they must be related to each other by the magnetization curve
for steel forgings (Fig. 2). Hence, B and H are determined by the intersec-
tion of the straight line and the curve. The straight line is determined by
two of its points ; for instance, when H = 40, B = 10 ; when H = 24, B = 19.3.
Drawing this line in Fig. 2 we find that the point of intersection corre-
sponds to B = 16.3. : Ans. 16.3 kilolines per sq. cm.
Prob. 12. Solve the preceding problem, assuming the ring to be made
of silicon steel laminations: 10 per cent of the space is taken by the
insulation between the laminations.
Ans. Flux density in the laminations is 15.2 kl/sq. cm.
Prob. 13. In a complex magnetic circuit, an air-gap 3 mm. long and
26 sq. cm. in cross-section is shunted by a cast-iron rod 14 cm. long and
10 sq. cm. in cross-section. WTiat is the number of ampere-turns neces-
sary for producing a total flux of 215 kilolines through the two paths in
parallel, and what is the reluctance of the rod per centimeter of its length
under these conditions? Ans. 1160 ampere-turns; 0.933 milli-rel.
Prob. 14. The magnetic flux in a closed iron core must increase and
decrease according to a straight-line law with the time, then reverse and
increase and decrease according to the same law in the opposite direction.
Show the general shape of the curve of the exciting current, neglecting
the effect of hyteresis.
Prob. 15. Show that if in the preceding problem the flux varies accord-
ing to the sine law the curve of the exciting current is a peaked wave.
Show how to determine the shape of this curve from a given magnetiza-
tion curve of the material. This problem has an application in the calcu-
lation of the exciting current in a transformer.
Prob. 16. In the magnetic circuit shown in Fig. 6 the useful flux
passes through the air-gap between the two steel poles; a part of the flux
1 The student will see from the solution of this problem that in the case
of a series magnetic circuit it is much easier to find the m.m.f. required for
a given flux than vice versa. On the other hand, in the case of two mag-
netic paths in parallel (such as in prob. 13), it is easier to find the flux for a
given m.m.f.
30
THE MAGNETIC CIRCUIT
[ART. 13
is shunted through the cast-iron part of the circuit. At low saturations a
considerable part of the total flux is shunted through the cast-iron part,
but as the flux density increases the cast iron becomes saturated, and a
larger and larger portion of the flux is deflected into the air-gap. What
percentages of the total flux in the yoke are shunted through the cast iron
when the flux density in the air-gap is 1 kl/sq. cm. and 7 kl/sq. cm.
respectively? Solution : When the flux density in the air-gap is 1 kilo-
line per sq. cm. the m.m.f. across the gap is 1000X0.8X0.5 = 400 ampere-
turns. The flux density in the steel poles is 2 kl/sq. cm., and the required
m.m.f. in them is about 16 ampere-turns. Therefore, the total m.m.f.
across AC and consequently across the cast-iron part is 416 ampere-turns,
Sq. Cm.
35 Sq. Cm.
Cas.t Sleel
FIG. 6. — A complex magnetic circuit.
or H = 24.5 ampere-turns per centimeter of length of the path in the cast
iron. Thisvalue of //corresponds on the magnetization curve toB = 6kl/sq.
cm. ; hence, the total n\ux in the cast iron is 72 kl. The flux in the yoke is
60 + 72 = 132 kl., and the percentage in the cast-iron shunt is 72/132 or
about 55 per cent. Similarly, it is found that, when the flux density in the
air-gap is 7/kl sq. cm., about 25 per cent of the flux is shunted through
the cast-iron part. The foregoing arrangement illustrates the principle
used in some practical cases, when it is desired to modify the relation
between the flux and the magnetomotive force, by providing a highly
saturated magnetic path in parallel with a feebly saturated one.
Ans. 55 per cent and 25 per cent approximately.
Prob. 17. Indicate how the preceding problem can be solved if the
cast-iron part were provided with a small clearance of say 1 mm. Hint:
See the second solution to problem 1 1 .
CHAP. II] MAGNETIC CIRCUIT WITH IRON 31
Prob. 18. What is the length of the yoke in Fig. 6 if the exciting cur-
rent increases 12 times when the flux density in the air-gap increases from
1 to 7 kl/sq. cm.? Hint: If Hx and H7 are the known magnetic intensi-
ties in the yoke, corresponding to the two given densities, and x is the
unknown length of the yoke, we have, using the values obtained in the
solution of problem 15: (H1x+416)12 = H7x+3090.
Ans. About 1.2 m.
CHAPTER III
HYSTERESIS AND EDDY CURRENTS IN IRON
14. The Hysteresis Loop. Steel and iron possess a property
of retaining part of their magnetism after the external magnetomo-
tive force which magnetized them has been removed. Therefore,
the magnetization or the B-H curve of a sample depends some-
what upon the magnetic state of the specimen before the test. This
property of iron is called hysteresis. The curves shown in Figs. 2
and 3 refer to the so-called virgin state of the materials, which state
is obtained by thoroughly demagnetizing the sample before the
test. A piece of iron can be reduced to the virgin state by placing
it within a coil through which an alternating current is sent, and
gradually reducing the current to zero. Instead of changing the
current, the sample can be removed from the coil.
Let a sample of steel or iron to be tested be made into a ring
and provided with an exciting winding, as in Fig. 1. Let it be
thoroughly demagnetized; in other words, let its residual mag-
netism be removed ; then let the ring be magnetized gradually or in
steps to a certain value of the flux density. Let OA in Fig. 7 rep-
resent the virgin magnetization curve, that is to say the relation
between the calculated values of B and H from this test, and let
PA be the highest flux density obtained. If now the magnetizing
current be gradually reduced, the relation between B and H is no
more represented by the curve OA, but by another curve, such as
AC; this is because of the above-mentioned property of iron to
retain part of its magnetism. When the current is reduced to
zero, the specimen still possesses a residual flux density OC. Let
the current now be reversed and increased in the opposite direc-
tion, until H reaches the negative value OF, at which no magnetic
flux is left in the sample. The value of H=OF is called the coer-
cive force. When the magnetic intensity reaches the negative value
of OP' = OP, experiment shows that the magnetic density P'A' in
the sample is equal and opposite to PA.
32
CHAP. Ill]
HYSTERESIS AND EDDY CURRENTS
33
Let now the exciting current be again decreased, reversed and
increased to its former maximum value corresponding to H=OP.
It will be found that the relation between B and H follows a differ-
ent though symmetrical curve, A'C'F'A, which connects with the
upper curve at the point A. The complete closed curve is called
the hysteresis loop-, a sample of iron which has been subjected to a
varying magnetomotive force as described before, is said to have
undergone a complete cycle of magnetization. If the same cycle
FIG. 7. — A hysteresis loop.
be repeated any number of times, the curve between B and H
remains the same, as long as the physical properties of the sample
remain unchanged.
The lower half of the hysteresis loop is identical with the
inverted upper half, so that the residual flux density OC' = OC,
and the coercive force OF' = OF. The shape of the loop for a
given sample is completely determined by the maximum ordinate
AP, or the maximum excitation OP. If the excitation be carried
further, for instance, to the point D on the" virgin curve, the
hysteresis loop would be larger, beginning at the point D, and
would be similar in its general shape to the loop shown in Fig. 7.
34 THE MAGNETIC CIRCUIT [ART. 15
A piece of iron can also be carried through a hysteresis cycle
mechanically. Thus, instead of changing the excitation, the
sample may be moved to a weak field, reversed, and returned to
its original location. The relation between B and H, however,
will be the same in either case.
An important feature of the hysteresis cycle is that it requires
a certain amount of energy to be supplied by the magnetizing
current, or by the mechanism which reverses the iron with
respect to the field. It is proved in Art. 16 below that this energy
per cubic unit of iron is proportional to the area of the hysteresis
loop. This energy is converted into heat in the iron, and therefore
from the point of view of the electromagnetic circuit represents
a pure loss. If the cycles of magnetization are performed in
sufficiently rapid succession, for instance by using alternating
current in the exciting winding, the temperature of the iron rises
appreciably.
The phenomenon of hysteresis is irreversible; that is to say,
it is impossible to make a piece of iron to undergo a cycle of mag-
netization in the direction opposite to that indicated by arrow-
heads, in Fig. 7. If it were reversible the loss of energy occasioned
by performing the cycle in one direction could be regained by
performing it in the opposite direction. In this respect the
hysteresis cycle differs materially from the theoretical reversible
cycles studied in thermodynamics, and reminds one of an irre-
versible thermodynamic cycle, in which friction or sudden expan-
sion is present.
15. An Explanation of Saturation and Hysteresis in Iron.
While the physical nature of magnetism is at present unknown,
there is sufficient evidence that the magnetization of iron is
accompanied by some kind of molecular change. Let us assume,
in accordance with the modern electronic theory, that there is an
electric current circulating within each molecule of iron, due to the
orbital motion of one or more electrons within the molecule. Each
molecule represents, therefore, a minute electromagnet acted upon
by other molecular electromagnets. In the neutral state of a piece
of iron, the grouping of the molecules is such that the currents
are distributed in all possible planes, and the external magnetic
action is zero. Under the influence of an external magnetomotive
force the molecules are oriented in the same way that small mag-
netic needles are deflected by an external magnetic field. With
CHAP. Ill] HYSTERESIS AND EDDY CURRENTS 35
small intensities of the external field, the molecules of iron return
into their original stable positions as soon as the external m.m.f.
is removed; when, however, the external magnetic intensity
becomes considerable some of the molecules turn violently and
assume new groupings of stable equilibrium. Therefore, when
the external m.m.f. is removed, there is some intrinsic magneti-
zation left, and we have the phenomenon of residual mag-
netism.
With an ever-increasing external m.m.f., more and more of the
molecules are oriented so that their m.m.fs. are in the same direc-
tion as the external field, the iron then approaching saturation.
Any further increase in the flux density is then mainly due to the
flux between the molecules, the same as in any non-magnetic
medium.
According to the foregoing theory, an external m.m.f. turns
the internal m.m.fs. into more or less the same direction; these
m.m.fs. then help to establish the flux in the intermolecular spaces
which are much greater than the molecules themselves. There-
fore, the higher flux density in iron is not due to a greater permea-
bility of the iron itself, but to an increased m.m.f. It is never-
theless permissible, for practical purposes, to speak of a higher
permeability of the iron, disregarding the internal m.m.fs., and
considering the permeability, according to eq. (16), as the ratio
of the flux density to the externally applied magnetic intensity.
The foregoing theory explains . also the general character of
the permeability curve of iron. With very small values of H
the molecules of a piece of iron are oriented but very little, but
are rapidly oriented more and more as H is increased. There-
fore, for small values of H, /j. must be expected to increase with H.
On the other hand, when the saturation is very high, an increase
in H changes B but little, because practically all of the available
internal m.m.fs. have been utilized. Therefore, for large values
of H, jj. decreases with increasing H. Consequently, there is a
value of H for which /* is a. maximum. This is the actual shape of
permeability curves (see for instance the reference to the Standard
Handbook given in Art. 12 above).
The phenomenon of magnetization is irreversible because the
changes from one stable grouping of molecules to the next are
sudden. Each molecule, in changing to a new grouping, acquires
kinetic energy, and oscillates about its new position of equilib-
36
THE MAGNETIC CIRCUIT
[ART. 15
rium until the energy is dissipated by being converted into heat.
This heat represents the loss of energy due to hysteresis.
This theory of saturation and hysteresis is due originally
to Weber, and has been improved by Ewing, who has shown
experimentally the possibility of various stable groupings of
a large number of small magnets in a magnetic field. By varying
the applied m.m.f. he obtained a curve similar to the hysteresis
loop of a sample of iron. For further details of this theory see
Ewing, Magnetic Induction in Iron and other Metals (1892),
Chapter XI.
The following analogy is also useful, Let a body Q (Fig. 8),
rest on a support and be held in its central position by two springs
FIG. 8. — A mechanical analogue to hysteresis.
S, S, which can work both under tension and under compression.
Let this body be made to move periodically to the right and to
the left of its central position, under the influence of an alterna-
ting external force H. Call B the deflections of the body from
its middle position. The relation between B and H is then similar
to the hysteresis loop in Fig. 7, provided that there is some
friction between the body Q and its support, and provided that
the springs offer in proportion more resistance when distorted
greatly than when distorted slightly.
Starting with the neutral position of the body let a gradually
increasing force H be applied which moves the body to the right.
This corresponds to the virgin curve in Fig. 7, except that this
simple analogy does not account for the inflection in the virgin
curve near the origin. Let then the force H be gradually reduced,
allowing the springs to bring Q nearer the center. When the
CHAP. Ill] HYSTERESIS AND EDDY CURRENTS 37
external force is entirely removed, the body is still somewhat to
the right of its central position, because the friction balances part
of the tension of the springs. Here we have something analogous
to residual magnetism and to the part AC of the hysteresis loop.
A finite force H is required in the negative direction to bring Q
to the center. This force corresponds to the coercive force of a
piece of iron.
By following this analogy through the complete cycle one
can show that a loop is obtained similar to a hysteresis loop.
Also, it can be shown that the phenomenon is irreversible, and
that total work done by the force H is equal to the work of friction.
Moreover there is a periodic interchange of energy between the
springs and the source of the force H} and the net loss of energy
is represented by the area of the loop corresponding to Fig. 7.
Prob. 1. An iron ring is thoroughly demagnetized, and then the cur-
rent in the exciting winding is varied in the following manner: It is
increased gradually from zero to 1 ampere and is then reduced to zero.
After this, the current is increased to 2 amperes in the same direction, and
again reduced to zero. Then the current is increased to 3 amperes again in
the same direction, and reduced to zero, etc. Draw roughly the general
character of the B-H curve, taking the hysteresis into consideration.
Hint : First study a similar process on the mechanical analogy shown in
Fig. 8.1
Prob. 2. A piece of iron is made to undergo a magnetization process
from the point A (Fig. 7) to a point between F and Af such that, when
subsequently the exciting circuit is opened, the ascending branch of the
hysteresis curve comes to the origin. Show that such a process does not
bring the iron into the neutral virgin state, in spite of the fact that 5 = 0
for H = Q. Hint : Consider the further behavior of the iron for positive
and negative values of H .
Prob. 3. A millivoltmeter is connected to the high-tension terminals
of a transformer, and the current in the low-tension winding is varied in
such a way as to keep the voltage constant : Show that the curve of the
current plotted against time is proportional to the hysteresis loop of the
core. Hint: Since d@/dt is constant, 0 is proportional to the time.
Prob. 4. The magnetic flux density in an iron core is to vary with the
time according to the sine law. Plot to time as abscissae the instantane-
ous values of the exciting ampere-turns per centimeter length of the core
from an available hysteresis loop, and show that the wave of the exciting
current is not a sine wave and is unsymmetrical. Note: This problem
has an application in the calculation of the exciting current of a trans-
former; see Art. 33 below.
1 A solution of this and of the next problem will be found in Chapter V
of Ewing's Magnetic Induction in Iron and other Metals, 1892.
38 THE MAGNETIC CIRCUIT [ART. 16
16. The Loss of Energy per Cycle of Magnetization. When a
magnetic flux is maintained constant the only energy supplied
from the source of electric power is that converted into the i2r
heat in the exciting winding; no energy is necessary to maintain
the magnetic flux. This is an experimental fact, fundamental in
the theory of magnetic phenomena. When, however, the flux
is made to vary, by varying the exciting ampere-turns or the
reluctance of the magnetic circuit, electromotive forces are
induced in the magnetizing, winding by the changing flux.
A transfer of energy results between the electric and the magnetic
circuits.
Beginning, for instance, at the point A of the cycle (Fig. 7),
and going toward C, the flux is forced to decrease. According
to Faraday's law, the e.m.f . induced by this flux in the magnetiz-
ing winding is such as to resist the change, i.e., it tends to main-
tain the current. Therefore, during the part AC of the cycle
energy is supplied from the magnetic to the electric circuit.
This shows that energy is stored in a magnetic field. During the
part CFA' of the hysteresis loop energy is supplied from the
electric to the magnetic circuit, because at the point C, the
current is reversed and becomes opposed to the e.m.f. The
other half of the cycle being symmetrical, with the flux and the
current reversed, energy is returned to the electric circuit during
the part A'C' of the cycle, and is again accumulated in the
magnetic circuit during the part C'F'A.
If the part AC of the cycle were identical with C'F'A, and the
part A'C' were identical with CFA', the amounts of energy trans-
ferred both ways would be the same, and there would be no net
loss of energy at the end of the cycle. In reality the two parts
are different; the amounts of energy returned from the magnetic
circuit to the electric circuit in the parts AC and A'C' are smaller
than the amounts supplied by the electric circuit in the parts
CFA' and C'F'A. This is because the last two parts of the curve
are more steep than the first two, and consequently the induced
e.m.fs. are larger for the same values of the current. The net
result is therefore an input of energy from the electric into the
magnetic circuit, this energy being converted into heat in the iron.
No such effect is observed with non-magnetic materials, because
the two branches of a complete B-H cycle coincide with a straight
line passing through the origin.
CHAP. Ill] HYSTERESIS AND EDDY CURRENTS 39
To prove that the energy lost per cubic unit of iron per cycle
of magnetization is represented by the area of the hysteresis loop,
we first write down the expression for the energy returned to the
electric circuit during an infinitesimal change of flux in the
part AC of the cycle. Let the flux in the ring at the instant under
consideration be 0 webers, and the magnetomotive force ni amp-
ere-turns, where i is the instantaneous value of the current, and n
is the total number of turns on the exciting winding. The instan-
taneous induced e.m.f ., due to a decrease of the flux by d<P during
an infinitesimal element of time dt seconds, is e= —ndtf>/dtvolt.
The sign minus is necessary because e is positive (in the direction
of the current) when dd> is negative, that is to say, when the flux
decreases. The electric energy corresponding to this voltage is
dW = eidt= —nid$ watt-seconds (joules).
Hence, the total energy returned to the electric circuit during the
part AC of the cycle is-
W= - C'nidQ,
•/ A
or, interchanging the limits of integration,
W= C nid$.
^c
Since all the parts of the ring undergo the same process, and
the curve in Fig. 7 is plotted for a .unit cube of the material, it is
of interest to find the loss of energy per cubic centimeter of mate-
rial. If & is the cross-section and I the mean length of the lines
of force in the iron, we have that the volume
V = Sl cubic centimeters.
Dividing the expression for the energy by this equation, we find
that the energy hi watt-seconds per cubic centimeter of iron is
W/V-.f
(19)
where H is in ampere-turns per centimeter, and B is in webers
per square centimeter.
But HdB is the area of an infinitesimal strip, such as is shown
by hatching in Fig. 7. Consequently, the right-hand side of eq.
40 THE MAGNETIC CIRCUIT [ART. 17
(19) represents the area of the figure ACQ, which is therefore a
measure for the energy transferred to the electric circuit, per cubic
centimeter. In exactly the same way it can be shown that the
energy supplied to the magnetic circuit during the part C'A of the
cycle is represented by the area AC'Q. Hence the net energy
loss for the part of the cycle to the right of the axis of ordinates
is represented by the area ACC'A. Repeating the same reasoning
for the left-hand side of the loop it will be seen that the total
energy loss per cycle of magnetization per cubic centimeter of
material is represented by the area AC A' C'A of the hysteresis
loop. For a given material, this area, and consequently the loss,
is a function of the maximum flux density PA, and increases with
it according to a rather complicated law. Two empirical formulae
for the loss of energy as a function of the density are given in Art.
20 below.
In the problems that follow the weight of one cubic decimeter
of solid carbon steel is taken to be 7.8 kg., and that of the alloyed
or silicon steel 7.5 kg. The weight of one cubic decimeter of
assembled carbon steel laminations is taken as 0.9X7.8 = 7 kg.,
and that of silicon steel laminations as 0.9X7.5 = about 6.8 kg.
Prob. 5. A hysteresis loop is plotted to the following scales : abscissae
1 cm. = 10 amp .-turns /cm. ; ordinates, 1 cm. = l kilo-maxwell/sq. cm. ; the
area of the loop is found by a planimeter to be 72 sq. cm. What is the
loss per cycle per cubic decimeter of iron?
Ans. 7. 2 watt-seconds (joules).
Prob. 6. The hysteresis loop mentioned in the preceding problem was
obtained from an oscillographic record at a frequency of 60 cy., with a sam-
ple of iron which weighed 9.2 kg. What was the power lost in hysteresis
in the whole ring? Ans. 510 watts.
Prob. 7. The stationary coil of a ballistic electro-dynamometer is con-
nected in series with the exciting electric circuit (Fig. 1) ; the moving coil
is connected through a high resistance to a secondary winding placed on
the ring. The exciting current is brought to a certain value, and then the
current is reversed twice in rapid succession, in order that the iron may
undergo a complete magnetization cycle. Show that the deflection of the
electro-dynamometer is a measure for the area of the hysteresis loop.
Hint: HdB=H(dB/dt) ctt=Const. X iedt.1
17. Eddy Currents in Iron. Iron is an electrical conductor;
therefore when a magnetic flux varies in it, electric currents are
1 Searle, "The Ballistic Measurement of Hysteresis," Electrician, Vol.
49, 1902, p. 100.
CHAP. Ill] HYSTERESIS AND EDDY CURRENTS 41
induced along closed paths of least resistance linked with the flux.
These currents permeate the whole bulk of the iron and are called
eddy or Foucault currents. Eddy currents cause a loss of energy
which must be supplied either electrically or mechanically from
an outside source. Therefore, the iron cores used for variable fluxes
are usually built of laminations, so as to limit the eddy currents
to a small amount by interposing in their paths the insulation
between the laminations. Japan, varnish, tissue paper, etc., are
used for this purpose. In many cases the layer of oxide formed
on laminations during the process of annealing is considered to be
a sufficient insulation against eddy currents.
The usual thickness of lamination varies from 0.7 to 0.3 mm.,
according to the frequency for which an apparatus is designed,
the flux density to be used, the provision for cooling, etc. The
more a core is subdivided the lower is the loss due to eddy currents,
but the more expensive is the core on account of the higher cost
of rolling sheets, and of punching and assembling the laminations.
Besides, more space is taken by insulation with thinner stampings,
so that the per cent net cross-section of iron is reduced. The
net cross-section of laminations is usually from 95 to 85 per cent
of the gross cross-section, depending upon the thickness of the
laminations, the kind of insulation used, and the care and pres-
sure used in assembling the core. For preliminary calculations
about ten per cent of the gross cross-section is assumed to be
lost in insulation.
Fig. 9 shows two iron cores in cross-section, one core solid,
the other subdivided into three laminations by planes parallel
to the direction of the lines of force. The lines of force are shown
by dots, and the paths of the eddy currents by continuous lines.
Eddy currents are linked with the lines of force, the same as the
current in the exciting winding. In fact, eddy currents are similar
to the secondary currents in a transformer, inasmuch as they
tend to reduce the flux created by the primary current. The
core must be laminated in planes perpendicular to the lines of
flow of the eddy currents, so as to break up their paths and at the
same time not to interpose air-gaps in the paths of the lines of
force.
An iron core can be further subdivided by using thin iron
wires in place of laminations. Such cores were used in early
machines and transformers, but were abandoned on account of
42
THE MAGNETIC CIRCUIT
[ART. 18
expense and poor space factor. Iron-wire cores are used at present
in only high-frequency apparatus, in which eddy currents must
be carefully guarded against; for instance in the induction coils
(transformers) employed in telephone circuits.
It will be seen by an inspection of Fig. 9 that eddy currents
are much smaller in the laminated core because the resistance of
each lamination is increased while the flux per lamination and con-
sequently the induced e.m.f . is considerably reduced. It is proved
in Art. 21 below that the power lost in eddy currents per kilogram
of laminations is proportional to the square of the thickness of
FIG. 9. — Eddy currents in a solid and in a laminated core.
the laminations, the square of the frequency, and the square of
the flux density.
Prob. 8. Show that the armature cores of revolving machinery must
be laminated in planes perpendicular to the axis of rotation.
Prob. 9. Show that assuming the temperature-resistance coefficient of
iron laminations to be 0.0046 per degree Centigrade the eddy current loss of
a core at 70° C. is only about 82.5 per cent of that at 20° C.
Prob. 10. Explain the reason for which the hysteresis loss in a given
core and at a given frequency depends only on the amplitude of the excit-
ing current, while the eddy-current loss depends also upon the wave-form
of the current.
18. The Significance of Iron Loss in Electrical Machinery.
The power lost in an iron core on account of hysteresis and eddy
currents, taken together, is called iron loss or core loss. It is of
importance to understand the effect of this loss in the iron cores
CHAP. Ill] HYSTERESIS AND EDDY CURRENTS 43
of electrical machinery and apparatus: First, because they bring
about a loss of power and hence lower the efficiency of a
machine; Secondly, because they heat up the iron and thus
limit the permissible flux density, or make extra provisions for
ventilation and cooling necessary; Thirdly, because they affect
the indications of measuring instruments. The effects of hystere-
sis and eddy currents in the principal types of electrical machinery
are as follows:
(a) In a transformer an alternating magnetization of the iron
causes a core loss in it. The power thus lost must be supplied from
the generating station in the form of an additional energy com-
ponent of the primary current. The core is heated by hysteresis
and by eddy currents, and the heat must be dissipated by the
oil in which the transformer is immersed, or by an air blast.
(b) In a direct-current machine the revolving armature is
subjected to a magnetization first in one direction and then in
the other; the heating effect due to the hysteresis and eddy
currents is particularly noticeable in the armature teeth in which
the flux density is usually quite high. The core loss, being sup-
plied mechanically, causes an additional resisting torque between
the armature and the field. In a generator this torque is sup-
plied by the prime mover; in a motor this torque reduces the
available torque on the shaft.
(c) The effect of hysteresis and of eddy currents in the armature
of an alternator or of a synchronous motor is similar to that in a
direct-current machine.
(d) In an induction motor the core loss takes place chiefly
in the stator iron and teeth, where the frequency of the magnetic
cycles is equal to that of the power supply; the frequency in the
rotor corresponds to the per cent slip, so that even with very
high flux densities in the rotor teeth the core loss in the rotor is
comparatively small. At speeds below synchronism the necessary
power for supplying the iron loss is furnished electrically as part
of the input into the stator. At speeds above synchronism this
power is supplied through the rotor from the prime mover.
(e) In a direct-current ammeter, if it has a piece of iron as
its moving element, residual magnetism in this iron causes inac-
curacies in its indications. With the same current the indication
of the instrument is smaller when the current is increasing than
when it is decreasing; this can be understood with reference to
44 THE MAGNETIC CIRCUIT [ART. 19
the hysteresis loop. With alternating current the effect of hystere-
sis is automatically eliminated by the reversals of the current
which passes through the instrument.
From these examples the reader can judge as to the effect of
hysteresis in other types of electrical apparatus not considered
above.
Prob. 11. Show that in an 8-pole direct-current motor running at a
speed of 525 r.p.m. the armature core and teeth undergo 35 complete
hysteresis cycles per second.
Prob. 12. Show that for two points in an armature stamping, taken
on the same radius, one in a tooth, the other near the inner periphery of
the armature, the hysteresis loops are displaced in time by one-quarter
of a cycle.
19. The Total Core Loss. In practical calculations on electrical
machinery the total core loss is of interest, rather than the hystere-
sis and the eddy current losses separately. For such computations
empirical curves are used, obtained from tests on steel of the same
quality and thickness. The curves of total core loss given in
Fig. 10 have been compiled from various sources, and give a
fak idea of the order of magnitude of core loss in various grades
of commercial steel laminations. The specimens were tested in
the Epstein apparatus, which is a miniature transformer (see
the author's Experimental Electrical Engineering, Vol. 1, p. 197),
and the values given can be used for estimating the core loss in
transformers and in other stationary apparatus with a simple
magnetic circuit.
In using the curves one should note that the ordinates are watts
per cubic decimeter of laminations, hence the gross volume and
not the volume of the iron itself is represented. On the other hand,
the abscissae are the true flux densities in the iron. In choosing a
material the following points are worthy of note : (1) Silicon steel
is now used for 60-cycle transformers, almost to the exclusion of
any other, on account of its lower core loss ; it is sometimes used
for 25-cycle transformers also. (2) The material called " Good
carbon steel " is that which is used for induction motor stators,
and in general for the armatures of alternating and direct-
current machinery ; also, sometimes for the cores oHow frequency
transformers. (3) The material called " Ordinary carbon steel "
should be used only in those cases for which the core loss is of
small importance.
CHAP. Ill]
HYSTERESIS AND EDDY CURRENTS
45
46 THE MAGNETIC CIRCUIT [ART. 19
The thickness of lamination to be used in each particular
case is a matter of judgment based on previous experience, and
no general rule can be laid down, except what is said in Art. 17
above, in regard to the factors upon which the eddy -current loss
depends. The gauges 26 to 29 are representative of the usual
practice. If it should be necessary to estimate the core loss for a
different thickness and at another frequency than those given in
Fig. 10, the method explained in Art. 22 below may be used.
The core loss in the armatures and teeth of revolving machinery
is found from tests to be considerably above that calculated from
the curves of loss on the same material when tested in stationary
strips. This is probably due in part to the fact that the conditions
of magnetization are different in the two cases. In the one case
the cycles of magnetization are due to a pulsating m.m.f., which
simply changes its magnitude; in the other case to a gliding m.m.f.,
with which the magnetic intensity at a point changes its direction
as well. Besides, the distribution of flux densities in teeth and in
armature cores is very far from being uniform. Therefore,
when using the curves given in Fig. 10, for the calculation of iron
loss in generators and motors, it is necessary to multiply the results
by certain empirical coefficients obtained from the results of tests
made on similar machines. Mr. I. E. Hanssen recommends add-
ing 30, 35, and 40 per cent to the loss calculated from the curves
obtained on stationary samples when estimating the iron loss in
an armature back of its teeth, at 25, 40, and 60 cycles respectively.
For teeth he recommends adding 30, 60, and 80 per cent, at the
same frequencies.1 These values are quoted here merely to give
a general idea of the magnitude of the excess of core loss in revolv-
ing machinery; a responsible designer should compile the values
of such coefficients from actual tests made on the particular class
of machines which he is designing.
Some engineers do not use for revolving machinery values of
core loss obtained on stationary samples, but plot the curves of core
loss obtained directly from tests on machines of a particular kind,
for various frequencies and flux densities. This is a reliable and
convenient method provided that sufficient data are available to
separate the core loss in the teeth from that in the core itself. Mr.
H. M. Hobart advocates this method, and curves of core loss
1 Hanssen, " Calculation of Iron Losses in Dynamo-electric Machinery,"
Trans. Amer. Inst. Elec. Eng., Vol. 28 (1909), Part II, p. 993.
CHAP. Ill] HYSTERESIS AND EDDY CURRENTS 47
obtained directly from actual machines will be found in his several
books on electric machine design.
It is customary now to characterize a lot of steel laminations
with respect to its core loss by the co-called figure of loss (Verlust-
ziffer) , which is the total core loss in watts per unit of weight, at a
standard frequency and flux density. In Europe the figure of loss
is understood to be the watts loss per kilogram of laminations, at
50 cycles and at a flux density of 10 kilolines per square centimeter;
the test to be performed in an Epstein apparatus under definitely
prescribed conditions.1 Sometimes a second figure of loss is
required, referring to a density of 15 kilolines per square centi-
meter, when the laminations are to be used at high flux densities.
In this country a figure of loss is sometimes used which gives the
watts loss per pound of material at 60 cycles and at a flux density
of 60 kilolines per square inch (or else at 10 kilolines per square
centimeter; see the paper mentioned in problem 20 below).
In some cases it is required to estimate the hysteresis and the
eddy current losses separately ; also it is sometimes necessary to
separate the two losses knowing a curve of the total loss. These
calculations are explained in the articles that follow.
Prob. 13. The core of a 60-cycle transformer weighs 89 kg.; the
gross cross-section of the core is 8 by 10 cm., of which 10 per cent is taken
by the insulation between the laminations. The total flux alternates
between the values of ±0.49 megaline. If the core is made of gauge 26
good carbon steel, what is the total core loss according to the curves in
Fig. 10? Solution: The flux density is 490/(8X 10X0.9) =6.8 kl/sq. cm.
The core loss per cubic decimeter at this density and at 60 cycles is,
according to the curve, equal 11.5 watt. The volume of the laminations,
including the insulation, is 89/7 = 12.7 cu. dm. The total loss is 11.5X
12.7 = 146 watts. Ans. 146 watts.
Prob. 14. What flux density could be used in the preceding problem
if the core were made of silicon-steel laminations, gauge 29, provided that
the total core loss be kept the same in both cases?
Ans. About 9 kl/sq. cm.
Prob. 15. Calculate the core loss in the stationary armature of a 60-
cycle 450-r.p.m. alternator of the following dimensions: bore 180 cm.;
gross axial length 24 cm.; two air-ducts 0.8 cm. each; radial width of
stampings back of the teeth, 15 cm.; the machine has 144 slots, 2 cm.
wide and 4.5 cm. deep. The core is made of 26-gauge good carbon steel;
the useful flux per pole is 4.65 megalines, and two-thirds of the total num-
ber of teeth carry the flux simultaneously. Use Mr. Hanssen's coefficients.
1 See Ekktrotechnische Zeitschrift, Vol. 24 (1903), p. 684.
48 THE MAGNETIC CIRCUIT [AKT. 20
Note: All parts of the core and all the teeth are subjected to complete
cycles of magnetization in succession ; therefore, in calculating the core
loss the total volume of the core and of the teeth must be multiplied by
the loss per cubic decimeter, corresponding to the maximum magnetic
density in each part. "The density in a tooth varies along its length,
being a maximum at the tip. The average density may be assumed to be
equal to that at the middle of the teeth. Ans. About 9 kw.
•20. Practical Data on Hysteresis Loss. The energy lost in
hysteresis per cycle per kilogram of a given material depends only
upon the maximum values of B and H} and does not depend upon
the manner in which the magnetizing current is varied with the time
between its positive and negative maxima. It is only at very high
frequencies, such as are used in wireless telegraphy, that the par-
ticles of iron do not seem to be able to follow in their grouping the
corresponding changes in the exciting current. With such high
frequencies iron cores are not only useless, but positively harmful.
However, at ordinary commercial frequencies the loss of power
Ph due to hysteresis is proportional to the number of cycles per
second and can be expressed as
Ph=f'V-F(B) watt.,
where / is the number of magnetic cycles per second, V is the vol-
ume of the iron, and F(B) is a function of the maximum flux den-
sity B. F(B) represents the loss per cycle per cubic unit of
material, and is therefore equal to the area of the hysteresis loop
in Fig. 7.
One can assume empirically that the unit loss per cycle, F(B),
increases as a certain power n of B} this power to be determined
from tests. The preceding formula becomes then
(20)
where ^ is an empirical coefficient which depends upon the quality
of the iron and upon the units used. Dr. Steinmetz found from
numerous experiments that the exponent n varies between 1.5 and
1.7, and proposed for practical use the formula
ph= 77/7^1 -e x 10~7 watt, .... (21)
where the factor 10~7 is introduced in order to obtain convenient
values for y when B is in maxwells per square centimeter, and V
is in cubic centimeters. It is more convenient for practical calcula-
CHAP. Ill] HYSTERESIS AND EDDY CURRENTS 49
tions to use B in kilolines per square centimeter, and V in cubic
decimeters. In this case the constant 10~7 is not necessary (see
Prob. 17 below) ; but the student must now remember to multiply
by 6.31 the values of T? found in the various pocketbooks.
Hysteresis loss cannot be represented always with sufficient
accuracy by formula (21) or (20) over a wide range of values of B,
because the exponent n itself seems to increase with B. Where
greater accuracy is required at medium and high flux densities the
following formula may be used:
) ...... (21a)
In this formula the term with B2 automatically becomes of more
and more importance as B increases. By selecting proper values
for r/ and if' a given experimental curve of loss can be approxi-
mated more closely than by means of formula (21). On the other
hand, formula (21) is more convenient for comparison and analysis.
Curves of hysteresis loss and values of the constant rj will be
found in various handbooks and pocketbooks. It is hardly worth
while giving them here, because hysteresis loss varies greatly with
the quality of iron and with the treatment it is given before use.
Moreover, the quality of the iron used in electrical machinery is
being improved all the time, so that a value of r) given now may
be too large a few years from now.
Considerable effort is being constantly made to improve the
quality of the iron used in electrical machinery so as to reduce its
hysteresis loss. The latest achievement in this respect is the pro-
duction of the so-called silicon steel, also called alloyed steel, which
contains from 2.5 to 4 per cent of silicon. This steel shows a much
lower hysteresis loss than ordinary carbon steel. Incidentally,
the electric resistivity of silicon steel is about three times higher
than that of ordinary steel, so that the eddy -current loss is reduced
about three times. The advantage that silicon steel has over car-
bon steel is clearly seen in Fig. 10. Silicon steel is largely used
for transformer cores because it permits the use of higher flux
densities, and therefore the reduction of the weight and cost of a
transformer, in spite of the fact that silicon steel itself costs more
per kilogram than carbon steel.
Another great advantage of silicon steel is that it is practically
non-aging-, this means that the hysteresis loss does not increase
with time. An increase in the hysteresis loss of a transformer
50 THE MAGNETIC CIRCUIT [ART. 20
during the first few years of its operation used to be a serious
matter in the design and operation of transformers, because of the
subsequent overheating of the core and of the coils. Silicon steel
shows practically no increase in its hysteresis loss after several
years of operation. Moderate heating, which considerably in-
creases the hysteresis loss in ordinary steel, has no effect on silicon
steel.
Impurities which are of such a nature as to produce a softer
iron or steel and a material of higher permeability, are as a rule
favorable to the reduction of the hysteresis loss, and vice versa.
Mechanical treatment and heating are also very important in their
effects on hysteresis loss. In particular, punching and hammering
increases hysteresis loss, while annealing reduces it. Therefore
laminations are always annealed carefully after being punched
into their final shape.
The requirements for the steel used in permanent magnets are
entirely different from those for the cores of electrical machinery.
In permanent magnets a large and wide hysteresis loop is desired,
because it means a high percentage of residual magnetism (ratio
of CO to AP, Fig. 7) and a large coercive force, OF. Both are
favorable for obtaining strong permanent magnets of lasting
strength. Combined carbon is particularly important for obtain-
ing these qualities, as is also the proper heat treatment after mag-
netization.
Prob. 16. In the 60-cycle transformer given in prob. 13, the core
weighs 89 kg. and is made of 26 gauge good carbon steel. The maxi-
mum flux density is 6.8 kl./sq. cm. What is the hysteresis loss assuming
T? to be equal to 0.0012? Ans. About 124 watt.
Prob. 17. What is the constant in formula (21) in place of 10~7, if,
with the same >?, the density B is in kilo-maxwells per sq. cm., and the
volume is in cubic decimeters? Ans. 6.31.
Prob. 18. Show how to determine the values of y and nin eq. (20),
knowing the values W i and W2 of the energy lost per cycle at two given
values of maximum flux density, J5t and B2.
Ans. n = (log TF2-log WJ/(log B,— log BJ.
Prob. 19. The following values of hysteresis loss per cu. decimeter
have been determined from a test at 25 cycles (after eliminating the eddy
current loss) :
Flux density in kl/sq.cm., B = 5. 0 I 6 . 5
Hysteresis loss in watts, Ph = 1 . 30 I 2 . 00
8.0
2.8*
10.0
4.11
What are the values of TJ and n in formula (20)? Suggestion;
Use logarithmic paper to determine the most probable value of n, by
CHAP. Ill] HYSTERESIS AND EDDY CURRENTS 51
drawing the straight line log P/i=n log B+ log Const. See the author's
Experimental Electrical Engineering, Vol., 1, p. 202.
Ans. Ph = 0.00368 fVB1'™.
21. Eddy Current Loss in Iron. With the thin laminations
used in the cores of electrical machinery the eddy -current loss in
watts can be represented by the formula
Pe=£V(tfB)2, . . /..'. . . (22)
where e is a constant which depends upon the electrical resistivity
of the iron, its temperature, the distribution of the flux, the wave
form of the exciting current, and the units used. V is the volume
or the weight of the core for which the loss is to be computed ; t is
the thickness of laminations, / the frequency of the supply, and B
the maximum flux density during a cycle. If B is different at
different places in the same core, the average of these should
be taken, (B is the time maximum but the space average).
Sometimes formula (22) contains also 10 to some negative power
in order to obtain a convenient value of e.
Formula (22) can be proved as follows : The loss of power in a
lamination can be represented as a sum of the i2r losses for the
small filaments of eddy current in it. But i2r=e2/r-} it can be
shown that the expression in parentheses in formula (22) is pro-
portional to the sum of e2/r per unit volume. When the frequency
/ increases say n times, the rate of change of the flux, dQ/dt, and
consequently the e.m.fs. induced in the iron are also increased n
times. Therefore, the loss which is proportional to e2 increases n2
times. In other words, the loss is proportional to the square of
the frequency. Similarly, the induced voltage is proportional to
the flux density J5; and consequently, the loss is proportional to B2.
To prove that the loss is proportional to the square of the
thickness of laminations one must remember that increasing the
thickness n times increases the flux and the induced e.m.f . within
any filament of eddy current also n times. But the resistance of
each path is reduced n times (neglecting the short sides of the rect-
angle). Consequently, the expression e2/r is increased n3 times.
However, inasmuch as the volume of the lamination is also
creased n times, the loss per unit volume is only n2 times larger. In
other words, the loss per unit volume increases as t2. A more
rigid proof of this proposition is given in problem 21 below.
52 THE MAGNETIC CIRCUIT [ABT. 22
For values of e the reader is referred to pocketbooks; the
numerical values given there must, however, be used cautiously,
because the eddy -current loss depends on some factors such as the
care exercised in assembling, and the actual distribution of the
flux, which factors can hardly be taken into account in a formula.
As a matter of fact, formula (22) is used now less and less in prac-
tical calculations, the engineer relying more upon experimental
curves of total core loss (Fig. 10).
Prob. 20. According to the experiments of Lloyd and Fischer [Trans-
Amer. Inst. Eke. Engs., Vol. 28 (1909), p. 465] the eddy-current loss in
silicon-steel laminations of gauge 29 (0.357 mm. thick) is from 0.12 to 0.18
watts per pound at 60 cycles and at B = 10,000 maxwells per sq. cm. What
is the value of the coefficient £ in formula (22) if P is in microwatts, V is
the weight of the core in kg. (not the volume, as before) ; if also t is in mm.,
and B is in kilolines per sq. cm.?
Ans. From 5.78 to 8.67 ; 7.2 is a good practical average.
Prob. 21. Prove that the loss of power caused by eddy currents, per
unit volume of thin laminations, is proportional to the square of the thick-
ness of the laminations. Solution : The thickness t of the sheet (Fig. 9)
being by assumption very small as compared with its width a, the paths
of the eddy current may be considered to be rectangles of the length a and
of different widths, ranging from t to zero. Consider one of the tubes of
flow of current, of a width 2x, thickness dx, and length I in the direction
of the lines of magnetic force. Let the flux density vary with the time
between the limits ±B. Then the maximum flux linking with the tube
of current under consideration is approximately equal to 2axB ; therefore,
the effective value of the voltage induced in the tube can be written in the
form e = CaxBf, where C is a constant, the value of which we are not con-
cerned with here. The resistance of the tube is p(2a +4x)/(ldx), or very
nearly 2ap/(ldx). Thus we have that the i2r loss, or the value of e*/r
for the tube under consideration, is dPe = C2ax2B2f2ldx/2p. Integrat-
ing this expression between the limits 0 and t/2 we get Pe= C2at3B'*/2l/ 48p.
But the volume of the lamination is V=atl. Dividing P by V we find
that the loss per unit volume is proportional to (t/B}2.1
Prob. 22. Prove that the loss of power by eddy currents per unit
volume in round iron wires is proportional to the square of the diameter of
the wire. The flux is supposed to pulsate in the direction of the axes of
the wires, and the lines of flow of the eddy currents are concentric circles
Hint : Use the method employed in the preceding problem.
22. The Separation of Hysteresis from Eddy Currents. It is
sometimes required to estimate the total core loss for a thickness
^or a complete solution of this and the following problem, including
the numerical values of C, see Steinmetz, Alternating Current Phenomena
(1908), Chap. XIV.
CHAP. IIIl HYSTERESIS AND EDDY CURRENTS 53
of steel laminations other than those given in Fig. 10, or at a differ-
ent frequency. For this purpose, it is necessary to separate the
loss due to hysteresis from that due to eddy currents, because the
two losses follow different laws, expressed by eqs. (20) and (22)
respectively.
In order to separate these losses at a certain flux density it is
necessary to know the value of the total core loss at this density,
and at two different frequencies. For a given sample of lamina-
tions, the total core loss P at a constant flux density and at a
variable frequency /, can be represented in the form
P=Hf+Ff2, ...... (23a)
where Hf represents the hysteresis loss, and Ff2 the eddy or Fou-
cault current loss. H is the hysteresis loss per cycle, and F is the
eddy -current loss when /is equal to one cycle per second. Writing
this equation for two known frequencies, two simultaneous equa-
tions are obtained for H and F, from which H and F can be deter-
mined.
In practice the preceding equation is usually divided by /, be-
cause in the form
P/f=H + Ff, ....... (236)
it represents the equation of a straight line between P/f and /.
This form is particularly convenient when the values of P are
known for more than two frequencies. In this case the values of
P/f are plotted against / as abscissae, and the most probable
straight line is drawn through the points thus obtained. The
intersection of this straight line with the axis of ordinates gives
directly the value of H. After this, F is found from eq. (236).
Knowing H and F at a certain flux density, the separate losses
Hf and Ff2 can be calculated for any desired frequency. For the
same material, but of a different thickness, the hysteresis loss per
kilogram weight is the same, while the eddy -current constant F
varies as the square of the thickness, according to eq. (22). Thus,
knowing the eddy loss at one thickness it can be estimated for any
other thickness.
It is sometimes required to estimate the iron loss at a flux den-
sity higher than the range of the available curves ; in other words,
the problem is sometimes put to extrapolate a curve like one of
those in Fig. 10. There are two cases to be considered.
54 THE MAGNETIC CIRCUIT [ART. 22
(A) If two or more curves for the same material are available,
taken at different frequencies, the hysteresis is first separated from
the eddy -current loss as is explained before, for several flux densi-
ties within the range of the curves. Then the exponent, according
to which the hysteresis loss varies with the flux density is found,
by plotting the hysteresis loss to a logarithmic scale (see problems
18 and 19 above). Finally the two losses are extrapolated. In
extrapolating, the hysteresis loss is assumed to vary according to
the same law, and the eddy current loss is assumed to vary as the
square of the flux density; see eq. (22).
(B) Should only one curve of the total loss be available for
extrapolation, this curve may be assumed to be a parabola of the
form P=aB + bB2. Dividing the equation throughout by B we
get
(24)
This is the equation of a straight line between P/B and B. Plot-
ting P/B against the values of B as abscissae, a straight line is
obtained which can be easily extrapolated. In some cases the
values of P/B thus plotted give a line with a perceptible curva-
ture. Nevertheless, the curvature is much smaller than that of
the original P curve, so that the P/B curve can be extrapolated
with more certainty, especially if the lower points be disregarded.1
Prob. 23. From the curves in Fig. 10 calculate the core loss per cubic
decimeter of 29-gauge silicon-steel laminations, at a flux density of 10
kl/sq.cm. and at 40 cycles. Ans. About 10 watts.
Prob. 24. Using the data obtained in the solution of the preceding
problem calculate the figure of loss of 26-gauge laminations at 60 cycles.
Ans. 2.7 watt /kg.
Prob. 25. Check the curve of total core loss for the ordinary carbon
steel at 40 cycles with the curves for 25 and 60 cycles.
Prob. 26. Extrapolate the curve of core loss for the silicon steel at 25
cycles up to the density of 20 kl/sq.cm. Which method is the more
preferable? Ans. 22 watts per cu.dm. at B = 20.
Prob. 27. Show that the core loss curve for ordinary carbon steel, at
60 cycles, follows closely eq. (24).
1 If the P/B curve should prove to be a straight line, then it is probable
that the hysteresis loss follows eq. 2 la more nearly than eq. 20. In this case,
even if we had data for two frequencies, method (B) would be both more
accurate, and more simple than method (A).
CHAPTER IV
INDUCED E.M.F. IN ELECTRICAL MACHINERY
23. Methods of Inducing E.M.F. The following are the prin-
cipal cases of induced e.m.f. in electrical machinery and apparatus:
(a) In a transformer, an alternating magnetizing current in the
primary winding produces an alternating flux which links with
both windings and induces in them alternating e.m.fs. A similar
case is that of a variable current in a transmission line which
induces a voltage in a telephone line which runs parallel to it.
(6) In a direct-current machine, in a rotary converter, and in
a homopolar machine electromotive forces are induced in the
armature conductors by moving them across a stationary magnetic
field.
(c) In an alternator and in a synchronous motor, with a sta-
tionary armature and a revolving field, electromotive forces are
induced by making the magnetic flux travel past the armature
conductors.
(d) In a polyphase induction motor the currents in the stator
and in the rotor produce together a resultant magnetomotive force
which moves along the air-gap and excites a gliding (revolving)
flux. This flux induces voltages in both the primary and the sec-
ondary windings.
(e) In a single-phase motor, with or without a commutator, the
e.m.fs. induced in the armature are partly due to the " transformer
action," as under (a), and partly to the " generator action," as
under (6).
(/) In an inductor-type alternator both the exciting and the
armature windings are stationary; the pole pieces alone revolve.
The flux linked with the armature coils varies periodically, due to
the varying reluctance of the magnetic circuit, because of the
motion of the pole pieces. This varying flux induces an alternating
e.m.f. in the armature winding. Or else, one may say that the
55
56 THE MAGNETIC CIRCUIT [ART. 23
flux travels along the air-gap with the projecting poles, and cuts
the armature conductors.
(g)' Whenever the current varies in a conductor, e.m.fs. are
induced not only in surrounding conductors but also in the con-
ductor itself. This e.m.f. is called the e.m.f. of self-induction.
Such e.m.fs. are present in alternating-current transmission lines,
in the armature windings of alternating-current machinery, etc.
While the e.m.f. of self-induction does not differ fundamentally
from the transformer action mentioned above, its practical aspect
is such as to make a somewhat different treatment desirable.
Inductance and its effects are therefore considered separately in
chapters X to XII below.
All of the foregoing cases can be reduced to the following two
fundamental modes of action of a magnetic flux upon an electrical
conductor:
(1) The exciting magnetomotive force and the winding in which
an e.m.f. is to be induced are both stationary, relatively to one
another; in this case the voltage is induced by a varying magnetic
flux. Changes in the flux are produced by varying either the
magnitude of the m.m.f ., or the reluctance of the magnetic circuit.
This method of inducing an e.m.f. is usually called the transformer
action.
(2) The exciting magnetomotive force and the winding in which
the e.m.f. is to be induced are made to move relatively to each
other, so that the armature conductors cut across the lines of the
flux. This method of inducing an e.m.f. is conventionally referred
to as the generator action.
By analyzing the transformer action more closely it can be
reduced to the generator action, that is to say to the " cutting " of
the secondary conductor by lines of magnetic flux or force. This is
so, because in reality the magnetic disturbance spreads out in all
directions from the exciting winding, and when the current in the
exciting winding varies the magnetic disturbance travels to or
from the winding in directions perpendicular to the lines of force
(Fig. 11). This traveling flux cuts the secondary conductor and
induces in it an e.m.f. However, the question as to whether an
e.m.f. is induced by a change in the total flux within a loop, or by
the cutting of a conductor by a magnetic flux is still in a somewhat
controversial state;1 although Bering's experiment is a strong
1 Carl Hering, " An Imperfection in the Usual Statement of the Funda-
CHAP. IV]
INDUCED E.M.F.
57
argument in favor of the theory of " cutting " of lines of force.
He showed that no e.m.f. is induced in an electric circuit when a
flux is brought in or out of it without actually cutting any of the
conductors of the electric circuit. For practical purposes it is
convenient to distinguish the transformer action from the genera-
tor action, so that the matter of unifying the statements (1) and
(2) into one more general law is of no immediate importance.
24. The Formulae for Induced E.M.F. In accordance with the
definition of the weber given in Art. 3; we have
e=-d$/dt, (25)
where e is the instantaneous e.m.f. in volts, induced by the trans-
FIG. 11. — E.M.F. induced by transformer action.
former action in a turn of wire which at the time t is linked with a
flux of fl> webers. The value of e is determined not by the value of
$ but by the rate at which 0 varies with the time. In the case of the
generator action d® in formula (25) represents the flux which the
conductor under consideration cuts during the interval of time dt.
It can be shown that the two interpretations of dd> lead to the
same result. Namely, in the case of the transformer action (Fig. 11),
the new flux, d@, is brought within the secondary turn by cutting
through the conductor of this turn. Therefore, in the case of the
mental Law of Electromagnetic Induction, Trans. Amer. Inst. Elec. Engs.,
Vol. 27 (1908), Part. 2, p. .1341. Fritz Emde, Dag Induktionsgesetz, Elek-
trotechnik und Maschineribau, Vol. 26 (1908); Zum Induktionsgesetz, ibid.,
Vol. 27 (1909) ; De Baillehache, Sur la Loi de 1'Induction, Bull. Societe Inter-
nationale des Ekctriciens, Vol. 10 (1910), pp. 89 and 288.
58 THE MAGNETIC CIRCUIT [ART. 24
transformer action d@ can also be considered as the flux which cuts
the loop during the time dt, the same as in the generator action.
On the other hand, the moving conductor in a generator is a part
of a turn of wire, and any flux which it cuts either increases or
decreases the total flux linking with the loop. Consequently, in the
case of the generator action d® can be interpreted as the change of
flux within the loop, the same as in the transformer action. Thus,
the mathematical expression for the induced e.m.f. is the same
in both cases, provided that the proper interpretation is given to
the value of d@.
The sign minus in formula (25) is understood with reference to
the right-hand screw rule (Art. 1), i.e., with reference to the direc-
tion of the current which would flow as a result of the induced
electromotive force. Namely, the law of the conservation of
energy requires that this induced current must oppose any change
in the flux linking with the secondary circuit. If this were other-
wise, a slight increase in the flux would result in a further indefinite
increase in the flux, and any slight motion of a conductor across a
magnetic field would help further motion.
The positive direction of the induced e.m.f. is understood to be
that of the primary current which excites the flux at the moment
under consideration. If the flux linked with the secondary circuit
increases, d@/dt in formula (25) is positive, but the secondary
current must be opposite to the primary in order to oppose the
increase. Thus, the secondary current is negative, and by assump-
tion the induced e.m.f. e is also negative. Therefore, the sign
minus is necessary in the formula. When the flux decreases,
d$/dt is negative, but the secondary current is positive, because it
must oppose the reduction in flux. Hence, in order to make e a
positive quantity, the sign minus is again necessary.
The following two special cases of formula (25) are convenient
in applications. Formula (25) gives the instantaneous value of
the induced e.m.f.; it is once and a while required to know the
average e.m.f. induced during a finite change of the flux from fl>i
to tf>2- By definition, the average e.m.f. is
*
edt,
where h is the initial moment and t2 the final moment of the
interval of time during which the change in the flux takes place.
CHAP. IV] INDUCED E.M.F. 59
Substituting in this equation the value of e from (25) , and integ-
rating, we get
-*i) (26)
This shows that the average value of an induced e.m.f. does
not depend upon the law according to which the flux changes with
the time, and is simply proportional to the average rate of change
of the flux.
As another special form of eq. (25) consider a straight con-
ductor of a length I centimeters moving at a velocity of v centi-
meters per second across a uniform magnetic field of a density of B
webers per sq. cm. Let B, I, and v be in three mutually perpendic-
ular directions. The flux dtf> cut by the conductor during an infini-
tesimal element of time dt is equal to Blvdt. Substituting this
value into eq. (25) we get, apart from the sign minus,
e=Blv (27)
Should the three directions, B, I, and v, be not perpendicular to
each other, I in eq. (27) is understood to mean the projection of
the actual length of the conductor, perpendicular to the field, and
v is the component of the velocity normal to B and 1. Both B and
v may vary with the position of the conductor, in which case eq.
(27) gives the value of the instantaneous voltage. If, at a cer-
tain moment, the various parts of the conductor cut across a field
of different density, eq. (27) must be written for an infinitesimal
length of the conductor, thus: de=Bv-dl, and integrated over
the whole length of the conductor.
Besides the rule given above, the direction of the e.m.f. induced
by the generator action can also be determined by the familiar
three-finger rule, due to Fleming, and given in handbooks and ele-
mentary books on electricity. This rule is useful beause it empha-
sizes the three mutually perpendicular directions, those of the
flux, the conductor, and the relative motion. In applying this
rule to a machine with a stationary armature one must remember
that the direction of the motion in Fleming's rule is that of the
conductor, and therefore is opposite to the direction of the actual
motion of the magnetic field.
Problem 1. A secondary winding is placed on the ring (Fig. 1) and is
connected to a ballistic galvanometer. Let the number of turns in the
secondary winding be n, the flux linking with each turn be 0 webers, and
60 THE MAGNETIC CIRCUIT [ART. 25
the total resistance of the secondary circuit be r ohms. Show that when
the current in the primary winding is reversed, the discharge through
the galvanometer is equal to 2nti> coulombs.
Prob. 2. A telephone line runs parallel to a direct-current trolley
feeder for 20 kilometers. When a current of 100 amperes flows through
the feeder a flux of 2 kilo-maxwells threads through the telephone loop,
per meter of its length. What is the average voltage induced in the tele-
phone line when the current in the trolley feeder drops from 600 to 50 amp.
within 0.1 sec.? Ans. 22 volts.
Prob. 3. Determine the number of armature conductors in series in a
550 volt homopolar generator of the axial type, running at a peripheral
speed of about 100 meters per sec., when the length of the armature iron
is 50 centimeters, and the flux density in the air-gap is between 18 and 19
kilolines per sq. cm. Note: For the construction of the machine see the
Standard Handbook, index, under " Generators, homopolar."
Ans. 6.
Prob. 4. Draw schematically the armature and the field windings of a
shunt-wound direct-current generator, select a direction of .rotation, and
show how to connect the field leads to the brushes so that the machine
will excite itself in the proper direction.
Prob. 5. From a given drawing of a direct-current motor predict its
direction of rotation.
Prob. 6. In an interpole machine the average reactance voltage per
commutator segment during the reversal of the current is calculated to be
equal to 34 volts. What is the required net axial length of the commu-
tating pole to compensate for this voltage if the peripheral speed of the
machine is 65 meters per second, and the flux density under the pole is
6 kl. sq.cm.? The armature winding has two turns per commutator seg-
ment. Ans. 22 cm.
25. The Induced E.M.F. in a Transformer. The three types
of transformers used in practice are shown in Figs. 12, 13, and 14.
Considering the iron core as a magnetic link, and a set of primary
and secondary coils as an electric link, one may say that the core-
type transformer has one magnetic link and two electric links;
the shell-type has one electric link and two magnetic links; the
combination or cruciform type has one electric and four magnetic
links. Still another type, not used in practice, can be obtained
from the core-type by adding two or more electric links to the same
magnetic link. Each electric link is understood to consist of two
windings: the primary and the secondary.
When the primary winding is connected to a source of constant-
potential alternating voltage and the secondary winding is con-
nected to a load, alternating currents flow in both windings and an
alternating magnetic flux is established in thef iron core. If the
CHAP. IV]
INDUCED E.M.F.
61
Core ,f
primary electric circuit, that is, the one connected to the source of
power, were perfect, that is, if it possessed no resistance and no
reactance, the alternating magnetic flux in the core would be the
same at all loads. It would have such a magnitude that at an}*-
instant the counter-e.m.f. induced by it in the primary wind-
ing would be practically equal and opposite to the impressed
voltage. In reality the resistance and the leakage reactance of
ordinary commercial transformers are so low that for the purposes
of calculating the magnetic circuit the primary impedance drop may
be disregarded, and the mag-
netic flux considered constant
and independent of the load.
If the primary applied volt-
age varies according to the sine
law, which condition is nearly
fulfilled in ordinary cases, the
counter-e.m.f., which is practi-
cally equal and opposite to it,
also follows the same law.
Hence, according to eq. (25),
the magnetic flux must vary
according to the cosine law,
because . the derivative of the
cosine is minus the sine. In
other words, both the flux and
the induced e.m.f. vary accord- FIG. 12.— A core-type transformer.
ing to the sine law, but the two
sine waves are in time quadrature with each other. When the
flux reaches its maximum its rate of change is zero, and therefore
the counter-e.m.f. is zero. When the flux passes through zero its
rate of change with the timers a maximum, and therefore the
induced voltage at this instant is a maximum.
Let 0m be the maximum value of the flux in the core, in webers,
and let / be the frequency of the supply in cycles per second.
Then the flux at any instant t is $ = $mcos 2nft, and the e.m.f.
induced at this moment, per turn of the primary or secondary
winding is
e= -
sn
Thus, the maximum value of the induced voltage per turn is
62
THE MAGNETIC CIRCUIT
[ART. 25
27z/$m; hence, the effective value is 2;r/0TO/v2 = 4.44/0m. Let
there be NI primary turns in series; the total primary voltage is
then equal to NI times the preceding value. Expressing the flux
in megalines we therefore obtain the following practical formula
for the induced voltage in a transformer:
10-2 (28)
Coils
Coils
FIG. 13.— A shell-type
transformer.
Mica
Shields
FIG. 14. — A cruciform-type
transformer.
In practice, E\ is assumed to be equal and opposite to the applied
voltage (for calculating the flux only, but not for determining the
voltage regulation of the transformer). Formula (28) holds also
for the secondary induced voltage E% if the number of secondary
turns in series N2 be substituted for NI. The voltage per turn is
the same in the primary and in the secondary winding; therefore,
the ratio of the induced voltages is equal to that of the number
CHAP. IV] INDUCED E.M.F. 63
of turns in the primary and secondary windings: that is, we
have El:E2 = N1:N2.
Prob. 7. A 60-cycle transformer is to be designed so as to have a flux
density in the core of about 9 kl./sq.cm.; the difference of potential
between consecutive turns must not exceed 5 volts. What is the required
cross-section of the iron? Ans. 210 sq.cm.
Prob. 8. The transformer in the preceding problem is to be wound for
6600 v. primary, and 440 v. secondary. What are the required numbers
of turns? Ans. 1320 and 88.
Prob. 9. Referring to the transformer in the preceding problem, what
are the required numbers of turns if three such transformers are to be used
Y-connected on a three-phase system, for which the line voltages are 6600
and 440 respectively? Ans. 765 and 51.
Prob. 10. In a 110-kilovolt, 25-cycle transformer for Y-connection the
net cross-section of the iron is about 820 sq.cm. and the permissible maxi-
mum flux density is about 10.7 kl/sq.cm. What is the number of turns
in the high-tension winding? Ans. 6500.
Prob. 11. The secondary of the transformer in the preceding problem
is to be wound for 6600 v., delta connection, with taps for varying the
secondary voltage within i5 per cent. Specify the winding.
Ans. 709 turns; taps taken after the 34th and 68th turn.
Prob. 12. Explain the reason for which a 60-cycle transformer usually
runs hot even at no load, when connected to a 25-cycle circuit of the same
voltage. Show from the core-loss curves that the voltage must be
reduced to from 75 to 85 per cent of its rated value in order to have the
normal temperature rise in the transformer, at the rated current.
Prob. 13. Show graphically that the wave of the flux, within a trans-
former, becomes more and more peaked when the wave of the applied e.m.f.
becomes more and more flat, and vice versa. Hint : The instantaneous
values of e.m.f. are proportional to the values of the slope of the curve of
flux.
Prob. 14. The wave of the voltage impressed upon a transformer has
a 15 per cent third harmonic which flattens the wave symmetrically. Show
analytically that the corresponding flux wave has a 5 per cent third har-
monic in such a phase position as to make the flux wave peaked.
26. The Induced E.M.F. in an Alternator and in an Induction
Motor. Part of a revolving field alternator is shown in Fig. 15.
The armature core is stationary and has a winding placed in slots,
which may be either open or half closed. The pole pieces are
mounted on a spider and are provided with an exciting winding.
When the spider is driven by a pime-mover the magnetic flux
sweeps past the armature conductors and induces alternating
voltages in them.1 In order to obtain an e.m.f. approaching a sine
1 For details concerning the different types of armature windings see the
author's Experimental Electrical Engineering, Vol. 2, Chap. 30.
64
THE MAGNETIC CIRCUIT
[ART. 26
wave as nearly as possible the pole shoes are shaped as shown
in the sketch, that is to say, so as to make a variable air-gap and
thus grade the flux density from the center of the pole to the edges.
In high-speed turbo-alternators the field structure often has a
smooth surface, without projecting poles (Fig. 33), in order to
reduce the noise and the windage loss. Such a structure is also
stronger mechanically than one with projecting poles. The grad-
ing of the flux is secured by distributing the field winding in slots,
so that the whole m.m.f . acts on only part of the pole pitch.
Consider a conductor at a during the interval of time during
which the flux moves by one pole pitch r. The average e.m.f.
FIG. 15. — The cross-section of a synchronous machine.
induced in the conductor is, according to eq. (26), equal to
where 4> is the total flux per pole in webers, and T is the time of
one complete cycle, corresponding to 2r the space of two pole
pitches. But T=l/f, so that the average voltage induced in a
conductor is
,= 2/0.
(29)
The value of eave thus does not depend upon the distribution of the
flux 0 in the air-gap.
If the pole-pieces are shaped so as to give an approximately
sinusoidal distribution of flux in the air-gap, the induced e.m.f. is
also approximately a sine wave, and the ratio between the effect-
CHAP. IV] INDUCED E.M.F. 65
ive and the average values of the voltages is equal to
or 1.1 1.1 If the shape of the induced e.m.f. departs widely from the
sine wave the actual curve must be plotted and its form factor
determined by one of the known methods (see the Electric Cir-
cuit). Let the form factor in general be 7 and let the machine
have N armature turns in series per phase, or what is the same,
2N conductors in series. The total induced e.m.f. in effective
volts is then
(30)
This formula presupposes that there is but one slot per pole per
phase, so that the e.m.fs. induced in the separate conductors are all
in phase with each other, and that their values are simply added
together. In reality, there is usually more than one slot per pole
per phase, for practical reasons discussed in the next article. It
will be seen from the figure that the e.m.fs. induced in adjacent
slots are somewhat out of phase with each other, because the crest
of the flux reaches different slots at different times. Therefore,
the resultant voltage of the machine is somewhat smaller than that
according to the preceding formula. The influence of the dis-
tribution of the winding in the slots is taken into account by mul-
tiplying the value of E in the preceding formula by a coefficient kb,
which is smaller than unity and which is called the breadth factor.
Introducing this factor, and assuming 7 = 1.11, which is accurate
enough for good commercial alternators, we obtain
(31)
where $ is now in megalines. Values of kb are given in the articles
that follow.
Formula (31) applies equally well to the polyphase induction
motor or generator. There we also have a uniformly revolving
flux in the air-gap, the flux density being distributed in space,
according to the sine law. This gliding flux induces e.m.fs. in the
stator and rotor windings. The only difference between the two
kinds of machines is that in the synchronous alternator the field is
made to revolve by mechanical means, while in an induction
machine the field is excited by the polyphase currents flowing in
1 For the proportions of a pole-shoe which very nearly give a sine wave
see Arnold, Wechselstromtechnik, Vol. 3, p. 247.
66 THE MAGNETIC CIRCUIT [ART. 27
the stator and rotor windings. The formulae of this chapter also
apply without change to the synchronous motor, because the con-
struction and the operation of the latter are identical with those of
an alternator; the only difference being that an alternator trans-
forms mechanical energy into electrical energy, while a synchro-
nous motor transforms energy in the reverse direction. In all
cases the induced voltage is understood and not the line voltage.
The latter may differ considerably from the former, due to the
impedance drop in the stator winding.
Prob. 15. A delta-connected, 2300-v., 60-cycle, 128.5-r.p.m. alternator
is estimated to have a useful flux of about 3.9 megalines per pole. If the
machine has one slot per pole per phase how many conductors per slot are
needed? Ans. 8.
Prob. 16. A 100,000-cycle alternator for wireless work has one conduc-
tor per pole and 600 poles. The rated voltage at no load is 110 v. What
is the flux per pole and the speed of the machine?
Ans. 82.5 maxwells; 20,000 r.p.m.
Prob. 17. It is desired to design a line of induction motors for a per-
ipheral speed of 50 met. per sec., the maximum density in the air-gap to be
about 6 kilolines per sq.cm. What will be the maximum voltage induced
per meter of active length of the stator conductors? Hint: Use formula
(27). Ans. 30 volt.
Prob. 18. Formula (31) is deduced under the assumption that each
armature conductor is subjected to the " cutting " action of the whole
flux. In reality, almost the whole flux passes through the teeth between
the conductors, so that it may seem upon a superficial inspection that
little voltage could be induced in the conductors which are embedded in
slots. Show that such is not the case, and that the same average voltage
is induced in the conductors placed in completely closed slots, as in the
conductors placed on the surface of a smooth-body armature. Hint:
When the flux moves, the same amount of magnetic disturbance must
pass in the tangential direction through the slots as through the teeth.
Prob. 19. Deduce eq. (31) directly from eq. (27). Can eq (31) be
derived under the case of the transformer action?
27. The Breadth Factor. Armature conductors are usually
placed in more than one slot per pole per phase, for the following
reasons :
(a) The distribution of the magnetic field is more uniform,
there being less bunching of the flux under the teeth ;
(b) The induced e.m.f. has a better wave form;
(c) The leakage reactance of the winding is reduced;
(d) The same armature punching can be used for machines with
different numbers of poles and phases;
CHAP. IV] INDUCED E.M.F. 67
(e) The mechanical arrangement and cooling of the coils is
somewhat simplified.
The disadvantage of a large number of slots is that more space
is taken up by insulation, and the machine becomes more expen-
sive, especially if it is wound for a high voltage. The electromo-
tive force is also somewhat reduced because the voltages induced
in different slots are somewhat out of phase with one another.
The advantages of a distributed winding generally outweigh its dis-
advantages, and such windings are used almost entirely. Thus,
it is of importance to know how to calculate the value of the
breadth factor kb for a given winding.
| Phase 1
I » 2
D » 3
FIG. 16. — A fractional-pitch winding.
In the winding shown in Fig. 15 each conductor is connected
with another conductor situated at a distance exactly equal to the
pole pitch. It is possible, however, to connect one armature con-
ductor to another at a distance somewhat smaller than the pole
pitch (Fig. 16). Such a winding is called & fractional-pitch wind-
ing, in distinction to the winding shown in Fig. 15; the latter
winding is called a full-pitch or hundred-per cent pitch winding.
It will be seen from Fig. 16 that, with a two-layer fractional-pitch
winding, some slots are occupied by coils belonging to two different
phases. The advantages of the fractional-pitch winding are :
(a) The end-connections of the winding are shortened, so that
there is some saving in armature copper.
(6) The end-connections occupy less space in the axial direc-
tion of the machine, so that the whole machine is shorter.
68 THE MAGNETIC CIRCUIT [ART. 28
(c) In a two-pole or four-pole machine it is necessary to use a
fractional-pitch winding in order to be able to place machine-
wound coils into the slots.
A disadvantage of the fractional-pitch winding is that the
e.m.fs. induced on both sides of the same coils are not exactly in
phase with each other, so that for a given voltage a larger number
of turns or a larger flux is required than with a full-pitch winding.
Fractional-pitch windings are used to a considerable extent both
in direct- and in alternating-current machinery.
Thus, the induced e.m.f . in an alternator or an induction motor
is reduced by the distribution of the winding in more than one
slot, and also by the use of a fractional-winding pitch. It is
therefore convenient to consider the breadth factor A;& as being
equal to the product of two factors, one taking into account the
number of slots, and the other the influence of the winding pitch.
We thus put
kb=kskw, (32)
where ks is called the slot factor and kw the winding-pitch factor.
For a full-pitch winding kw=l, and kb=k8') for a fractional-
pitch unislot winding ks = 1, and Afc = kw. The factors ks and kw
are independent of one another, and their values are calculated
in the next two articles.
28. The Slot Factor ks. Let the stator of an alternator (or
induction motor) have two slots per pole per phase, and let the
centers of the adjacent slots be displaced by an angle a, in electri-
cal degrees, the pole pitch, T, corresponding to 180 electrical
degrees. If E (Fig. 17) is the vector of the effective voltage
induced in the conductors in one slot, the voltage Er due to the
conductors in both slots is represented graphically as the geometric
sum of two vectors E relatively displaced by the angle a. We see
from the figure that %E' = Ecos %a, or E' = 2E cos i«. If both sets
of conductors were bunched in the same slot we would then have
E'= 2E. Hence, in this case the coefficient of reduction in
voltage, or the slot factor, fcs=cosja.
Let now the armature stamping have S slots per pole per phase,
the angle between adjacent slots being again equal to a electrical
degrees. Let the vectors marked E in Fig. 18 be the voltages
induced in each slot; the resultant voltage E' is found as the geo-
CHAP. IV]
INDUCED E.M.F.
69
metric sum of the E's. The radius of the circle r = %E/sm%a, and
\E' = r sm^Sa . Therefore,
E'/SE=(sm
sn
(33)
When S=2, the formula (33) becomes identical with the expres-
sion given before.
FIG. 17. — A diagram illustrating the slot FIG. 18. — A, diagram illustrating
factor with two slots. the slot factor with several slots.
The angle a depends upon the number of slots and the number
of phases. Let there be ra phases; then aSm= 180 degrees, and
. . .' '. . . . (34)
The values of ks in the table below are calculated by using the for-
mulae (33) and (34).
VALUES OF THE SLOT FACTOR ks
Slots per Phase
per Pole.
Single-phase
Winding.
Two-phase
Winding.
Three-phase
Winding.
1
1.000
1.000
1.000
2
0.707
0.924
0.966
3
0.667
0.911
0.960
4
0.653
0.907-
0.958
5
0.647
0.904
0.957
6
0.643
0.903
0.956
Infinity
0.637
0.900
0.955
In single-phase alternators part of the slots are often left empty
so as to reduce the breadth of the winding and therefore increase
the value of ks. For instance, if a punching is used with six slots
per pole, perhaps only three or four adjacent slots are occupied.
In this case, it would be wrong to take the values of ks from the
first column of the table. If, for instance, three slots out of six
70 THE MAGNETIC CIRCUIT [ABT. 29
are occupied, the value of ks is the same as for a two-phase wind-
ing with three slots per pole per phase.
Prob. 20. Check some of the values of ks given in the table above.
Prob. 21. The armature core of a single-phase alternator is built of
stampings having three slots per pole; two slots per pole are utilized.
What is the value of kst Ans. 0.866.
Prob. 22. A single-phase machine has S uniformly distributed slots per
pole, of which only S' are used for the winding. What is the value of /bs?
Ans. Use S' in eq. (33) instead of S', preserve S in eq. (34).
Prob. 23. A six-pole, 6600 v., Y-connected, 50-cycle turbo-alternator is
to be built, using an armature with 90 slots. The estimated flux per pole
is about 6 megalines. How many conductors are required per slot?
Ans. 20.
Prob. 24. What is the value of ks when the winding is distributed uni-
formly on the surface of a smooth-body armature, each phase covering /?
electrical degrees? Solution : Referring to Fig. 18, ks is in this case equal
to the ratio of the chord E' to the arc of the circle which it subtends. The
central angle is /?, and' we have ks = (sin^/?) / (%px/ 180°) . In a three-phase
machine /? = 60 degrees, and therefore ks = 0.955. This is the value given
in the last column of the table above.
Prob. 25. Deduce the expression for ks given in the preceding problem
directly from formula (33). Solution: Substituting $« = /?; S = 00 and
a =0, an indeterminate expression, O.oo, is obtained. But when the
angle a approaches zero its sine is nearly equal to the arc, so that the
denominator of the right-hand side of eq. (33) approaches the value
iS.ia = i/?, where /? is in radians. Changing /? to degrees, the required
formula is obtained.
29. The Winding-pitch Factor kw. Let the distance between
the two opposite sides of a coil (Fig. 16) be 180 —7- degrees, where
Y is the angle by which tfye winding-pitch is shortened. The volt-
ages induced in the two sides of the coil are out of phase with
each other by the angle 7-, so that if the voltage induced in each
side is e, the total voltage is equal to 2e Cosjf (Fig. 17). Fig. 17.
will apply to this case if we read f for the angle a. Hence, we
have that
kw= cos \Y (35)
In practice, the winding-pitch is measured in per cent, or as a frac-
tion of the pole pitch r. For instance, if there are nine slots per
pole and the coil lies in slots 1 and 8, the winding-pitch is 7/9, or
77.8 per cent. If the coil were placed in slots 1 and 10 we would
have a full-pitch, or a 100 per cent pitch winding. Let in gen-
CHAP. IV]
INDUCED E.M.F.
71
eral the winding-pitch be £, expressed as a fraction. Then
Y= (1 — Q180°. Substituting this value of f into formula (35) we
obtain
/cu,=cos[900(l-Q]
(36)
The values of kw given in Fig. 19 have been calculated according
to this formula.
0.90
0.80
0.70
50
70
90
100
Per Cent Winding Pitch
FIG. 19. — Values of the winding-pitch factor.
In applications, one takes the value of k8 from the table, assum-
ing the winding pitch to be one hundred per cent, and multiplies it
by the value of kw taken from the curve (Fig. 19). With frac-
tional-pitch two-layer windings the value of ks corresponds to the
number of slots per layer per pole per phase, and not to the total
number of slots per pole per phase. This is clear from the explana-
tion given in the preceding paragraph. Thus/for instance, in Fig.
16, k8 must be taken for three slots and not for five slots. If one
has to calculate the values of kb often, it is advisable to plot a set
72 THE MAGNETIC CIRCUIT [ART. 30
of curves, like the one in Fig. 19, each curve giving the values of kb
for a certain number of slots per pole per phase, against per cent
winding pitch as abscissae.
Prob. 26. In a 4-pole, 72-slot, turbo-alternator the coils lie in slots 1
and 13. What is the per cent winding-pitch and by what percentage is
the e.m.f. reduced by making the pitch short instead of 100%?
Ans. 66.7 per cent; l—kw = 13.4 per cent.
Prob. 27. What is the flux per pole at no load in a 6600-volt, 25-cycle,
500-r.p.m., Y-connected induction motor which has 90 slots, 36 conduc-
tors per slot, and a winding-pitch of about 73 per cent?
Ans. 7.26 megalines.
Prob. 28. Show that for a chain winding kw is always equal to unity,
in spite of the fact that some of the coils are narrower than the pole pitch.
Prob. 29. Draw a sketch of a single-layer, fractional-pitch winding,
using alternate slots for the overlapping phases. Show what values of ka
and kw should be used for such a winding.
30. Non-sinusoidal Voltages. In the foregoing calculations
the supposition is made that the flux density in the air-gap is dis-
tributed according to the sine law so that sinusoidal voltages are
induced in each conductor. Under these circumstances the
resultant voltage also follows the sine law, no matter what the
winding-pitch and the number of slots are. The flux is practically
sinusoidal in induction motors because the higher harmonics of
the flux are wiped out by the secondary currents induced in the
low-resistance rotor. But in synchronous alternators and motors
with projecting poles the distribution of the flux in the air-gap is
usually different from a pure sine wave. For instance, when the
pole shoe is shaped by a cylindrical surface concentric with that of
the armature, the air-gap length and consequently the flux density
are constant over the larger portion of the pole; therefore, the
curve of the field distribution is a flat one. This shape is improved
to some extent by chamfering the pole-tips or by shaping the pole
shoes to a circle of a smaller radius, so that the length of the air-gap
increases gradually toward the pole-tips.
When a machine revolves at a uniform speed, the e.m.f. induced
in a single armature conductor has exactly the shape of the field-
distribution curve, because in this case the rate of cutting the flux
is proportional to the flux density (see eq. 27 above). There-
fore, when a machine has but one slot per pole per phase (which
condition is undesirable, but unavoidable in low-speed alternators,
or in those designed for extremely high frequencies), the shape of
CHAP. IV] INDUCED E.M.F. 73
the pole-pieces must be worked out very carefully in order to have
an e.m.f. approaching the true sine wave. With a larger number
of slots this is not so necessary because the. em.fs. induced in differ-
ent slots are added out of phase with each other, and the undesir-
able higher harmonics partly cancel each other. The voltage
wave is further improved by a judicious use of a fractional-pitch
winding. These facts are made clearer in the solution of the prob-
lems that follow.1
Prob. 30. The flux density in the air-gap under the poles of an alterna-
tor is constant for 50 per cent of the pole pitch, and then it drops to zero,
according to the straight-line law, on each side in a space of 15 per cent
of the pole pitch. Draw to scale the curves of induced e.m.f. for the fol-
lowing windings : (a) Single-phase, one slot per pole ; (b) Single-phase,
nine slots per pole, five slots being occupied by a one-hundred per cent
pitch winding ; (c) The same as in (b) only the winding-pitch is equal to
7/9; (d) Three-phase, Y-connected full-pitch winding, two slots per pole
per phase; in the latter case give curves of both the phase voltage and the
line voltage. On all the curves indicate roughly the equivalent sine wave,
in order to see the influence of the number of slots and of the fractional
pitch in improving the wave form.
Prob. 31. A three-phase, Y-connected alternator has three slots per
pole per phase, and a full-pitch winding. The field curve has an 8 per
cent fifth harmonic, that is to say, the amplitude of the fifth harmonic is
0.08 of that of the fundamental sine wave. What is the magnitude of the
fifth harmonic in the phase voltage and in the line voltage. Solution : In
formula (33) the angle a between the adjacent slots is 20 electrical degrees
for the fundamental wave. For the fifth harmonic the same distance
between the slots corresponds to 100 electrical degrees. Hence, for the
fundamental wave
fcs = sin 30°/(3 sin 10°) =0.96;
while for the fifth harmonic
ks6 = sm 150°/(3 sin 50°) =0.217.
This means that, due to the distribution in three slots, the fundamental
wave of the voltage is reduced to 0.96 of its value in a unislot machine,
while the fifth harmonic is reduced to only 0.217 of its corresponding value.
Therefore, the relative magnitude of the fifth harmonic in the phase
voltage is 8 X 21. 7/96 = 1.8 per cent, which means that the fifth harmonic
is reduced to less than one-fourth of its value in the field curve. In
calculating the line voltage the vectors of the fundamental waves in a
three-phase machine are combined at an angle of 120 degrees. Conse-
1 For further details see Professor C. A. Adams' paper on " Electromotive
Force Wave-shape in Alternators," Trans. Amer. Inst. Elec. Engs., Vol.
28 (1909), Part II, p. 1053.
74 THE MAGNETIC CIRCUIT [ART. 31
quently, the vectors of the fifth harmonic are combined at an angle
of 120X5=600 degrees, or what is the same, — 120 degrees. Therefore
the proportion of the fifth harmonic in the line voltage is the same as
that in the phase voltage.
Prob. 32. Solve the foregoing problem when the winding pitch is 7/9.
Ans. 0.33 per cent. This shows that by properly selecting the
winding pitch an objectionable higher harmonic can be
reduced to a negligible amount.
Prob. 33. Show that the line voltage of a Y-connected machine can
have no 3d, 9th, 15th, etc. harmonics, that is to say, harmonics the num-
bers of which are multiples of 3, no matter to what extent such harmonics
are present in the induced e.m.fs. in each phase.
Prob. 34. Prove that in order to have even harmonics in the induced
e.m.f . of an alternator two conditions are necessary : (a) the flux distribu-
tion under the alternate poles must be different; (b) the distribution of
the armature conductors under the alternate poles must also be different
from one another. Indicate pole shapes and an arrangement of the arma-
ture winding particularly favorable for the production of the second har-
monic. Note: In spite of a different distribution of flux densities the
total flux is the same under all the poles. Therefore, the average voltages
for both half cycles are equal (see Art. 24), though the shape of the two
halves of the curve may be different, due to the presence of even har-
monics. This shows that there is no " continuous-voltage component "
in the wave, or rather that the voltage is in no sense unidirectional, and
that a direct-current machine cannot be built with alternate poles without
the use of some kind of a commutating device.
31. The Induced E.M.F. in a Direct-current Machine. The
e.m.f. induced in the armature coils of a direct-current machine
(Fig. 20) is alternating, but due to the commutator, the voltage
between the brushes of opposite polarity remains constant.
This voltage is equal at any instant to the sum of the instantaneous
e.m.fs. induced in the coils which are connected in series between
the brushes. When a coil is transferred from one circuit to
another, a new coil in the same electromagnetic position is intro-
duced into the first circuit, and in this wise the voltage between
the brushes is maintained practically constant, except for the small
variations which occur while the armature is coming back to a
symmetrical position. These variations are due to the coils short-
circuited by the brushes and to the fact that the number of
commutator segments is finite.
Thus, to obtain the value of the voltage between the brushes,
it is necessary to find the sum of the e.m.fs. induced at some
instant in the individual armature coils which are connected in
CHAP. IV]
INDUCED E.M.F.
75
series between the brushes. Each e.m.f. represents an instanta-
neous value of an alternating e.m.f.; the e.m.fs. induced in different
coils differing in phase from one another, because they occupy
different positions with respect to the poles. The voltages induced
in the extreme coils of an armature circuit differ from one another
by one-half of a cycle.
Instead of adding the actual instantaneous voltages, it is suffi-
cient to calculate the average voltage per coil, and to multiply it
by the number of coils in series, because the wave form of the
e.m.fs. induced in all the coils is the same, and their phase differ-
FIG. 20. — The cross-section of a direct-current machine.
ence is distributed uniformly over one-half of a cycle. According
to eq. (26) the average voltage per turn per half a cycle is 24>/%T,
where %T is the time during which the coil moves by one pole
pitch, and $ is the flux per pole, in webers. Substituting I// for
T, the average voltage per turn is equal to 4/0. Let there be N
turns in series between the brushes of opposite polarity; then the
induced voltage of the machine is
1-2
(37)
where 0 is now in megalines. Thus, in a direct-current machine
the induced voltage between the brushes depends only upon the
total useful flux per pole, and not upon its distribution in the air-
gap.
76 THE MAGNETIC CIRCUIT [ART. 31
The relation between the number of turns in series and the total
number of turns on the armature depends upon the kind of the
armature winding.1 If the armature has a multiple winding N is
equal to the total number of turns on the armature divided by the
number of poles. For a two-circuit winding the number of turns
in series is equal to one-half of the total number of turns. The num-
ber of poles and the speed of the machine do not enter explicitly
into formula (37), but are contained in the value of /.
Prob. 35. A 90-slot armature is to be used for a 6-pole, 580-r.p.m., 250-
v., direct-current machine with a multiple winding. How many conduc-
tors per slot are necessary if the permissible flux per pole is about 3 mega-
lines? Ans. 10.
Prob. 36. A 550-v., 4-pole railway motor has a two-circuit armature
winding which consists of 59 coils, 8 turns per coil. The total resistance
of the motor is 0.235 ohm. When the motor runs at 675 r.p.m. it takes in
81 amp. What is the flux per pole at this load? Aris. 2.5 ml.
Prob. 37. Show that in a direct-current machine the use of a fractional-
pitch winding has no effect whatever upon the value of the induced e.m.f .,
as long as the winding-pitch somewhat exceeds the width of the pole shoe.
Prob. 38. Prove that formula (37) is identical with the expression
(38)
where C is the total number of armature conductors, p is the number of
poles, and p' is the number of circuits in parallel.
Prob. 39. Show that the induced e.m.f. is the same when the armature
conductors are placed in open or in closed slots as when they are on the
surface of a smooth-body armature. See Prob. 18, Art. 26.
Prob. 40. Considerable effort has been made to produce a direct-cur-
rent generator with alternate poles, and without any commutator. One of
the proposals which is sometimes urged by a beginner is to use an ordinary
alternator, and to supply the exciting winding with an alternating current
of the synchronous frequency. The apparent reasoning is that the field
being reversed at the completion of one alternation the next half wave of
the induced e.m.f. must be in the same direction as the preceding one, thus
giving a unidirectional voltage. Show that such a machine in reality
would give an ordinary alternating voltage of double the frequency.
Hint : Make use of the fact that an alternating field can be replaced by two
constant fields revolving in opposite directions. Or else give a rigid
mathematical proof by considering the actual rate at which the armature
conductors are cut by the field, which field is at the same time pulsating
and revolving.
1 For details concerning the direct-current armature windings see the
author's Experimental Electrical Engineering, Vol. 2, Chapter 30.
CHAP. IV.] INDUCED E.M.F. 77
Prob. 41. Prove that if in the preceding problem the frequency of the
rotation of the poles is flt and the frequency of the alternating current in
the exciting winding is/2, that the voltage induced in the armature is a com-
bination of two waves having frequencies of fi +/2 and fi — /2 respectively.
32. The Ratio of A.C. to B.C. Voltage in a Rotary Converter.
A rotary converter resembles in its general construction a direct-
current machine, except that the armature winding is connected
not only to the commutator, but also to two or more slip rings.1
When such a machine is driven mechanically it can supply a direct
current through its commutator, and at the same time an alter-
nating current through its slip rings. It is then called a double-
current generator. But if the same machine is connected to a
source of alternating voltage and brought up to synchronous
speed it runs as a synchronous motor and can supply direct current
through its commutator. It is then called a rotary converter. It
is also sometimes used for converting direct current into alter-
nating current, and is then called an inverted rotary.
Both the direct and the alternating voltages are induced in the
armature of a rotary converter by the same field, and our problem
is to find the ratio between the two voltages for a given arrange-
ment of the slip rings. Consider first the simplest case of a single-
phase converter with two collector rings connected to the arma-
ture winding, at some two points 180 electrical degrees' apart. If
the armature has a multiple winding each slip ring is connected
to the armature in as many places as there are pairs of poles. In
the case of a two-circuit winding each collector ring is connected
to the armature in one place only.
If the machine has p poles then p times during each revolution
the direct-current brushes make a connection with the same arma-
ture conductors to which the slip rings are connected. At these
moments the alternating voltage is a maximum, because the direct-
current brushes are placed in the position where the induced volt-
age in the armature is a maximum. Thus, with two slip rings,
connected 180 electrical degrees apart, the maximum value of the
alternating voltage is equal to the voltage on the direct-current
side. If the pole shoes are shaped so that the alternating voltage
is approximately sinusoidal, the effective value of the voltage
between the slip rings is I/ /v/2 = 70.7 per cent of that between the
1 See the author's Experimental Electrical Engineering, Vol. 2, Chapter 28.
78 THE MAGNETIC CIRCUIT [ART. 32
direct-current brushes. The same ratio holds true for a two-
phase rota,ry, for the voltages induced in each phase.
Let now two slip rings be connected at two points of the arma-
ture winding, a electrical degrees apart. In order to obtain the
value of the alternating
voltage the vectors of the
voltages induced in the
individual coils must be
added geometrically, as in
Fig. 18. With a large num-
ber of coils the chords can be
FiG.21.-Relationbetweenthealternating ^placed by the arc, and in
voltages in a rotary converter. tms Wa7 FlS- 21 1S obtained.
The diameter MN = ei of the
semicircle represents the vector of the alternating voltage when
the points of connection to the slip rings are displaced by 180
electrical degrees, while the chord MP=ea gives the voltage
between two slip rings when the taps are distant by a electrical
degrees. It will thus be seen that
(39)
But we have seen before that ei = Q.7Q7E, where E is the voltage
on the direct-current side of the machine. Hence, for sinusoidal
voltages,
ea= 0.707# sin Ja ....... (40)
The following table has been calculated, using this formula.
Number of slip rings ....................... 2 3 4 5 6
Angle between the adjacent taps in electrical
degrees ................................ 180 120 90 72 60
Ratio of alternating to continuous voltage, in
percent ............................... 70.7 61.2 50 41.5 35.3
The foregoing theory shows that the ratio of the continuous to
the alternating voltage is fixed in a given converter, and in order
to raise the value of the direct voltage it is necessary to raise the
applied alternating voltage. This is done in practice either by
means of various voltage regulators separate from the converter,
or by means of a booster built as a part of the converter. Another
method of varying the voltage is by using the so-called split-pole
converter. In this machine the distribution of the flux density in
CHAP. IV] INDUCED E.M.F. 79
the air-gap can be varied within wide limits, and consequently
that component of the field which is sinusoidal can be varied. The
result is that the ratio of the direct to the alternating voltage is
also variable. Namely, we have seen before that the value of the
continuous voltage does not depend upon the distribution of the
flux, but only upon its total value, while the effective value of the
alternating voltage depends upon the sine wave or fundamental
component of the flux distribution.1
Prob. 42. Check some of the values given in the table above.
Prob. 43. A three-phase rotary converter must deliver direct current
at 550 v. What is the voltage on the alternating-current side?
Ans. 337 v.
Prob. 44. The same rotary is to be tapped in three additional places so
as to get two-phase current also. How many different voltages are on
the alternating-current side and what are they?
Ans. 389, 337, 275, 195, 100.
Prob. 45. The table given above holds true only when the flux density
is distributed approximately according to the sine law. Show how to
determine the ratio of alternating to continuous voltage in the case of two
collector rings connected to taps 180 electrical degrees apart, when the
curve of field distribution is given graphically. Solution : Divide the pole
pitch into a sufficient number of equal parts and mark them on a strip of
paper. Place the strip along the axis of abscissae. The sum of the ordi-
nates of the flux-density curve, corresponding to the points of division, at
a certain position of the strip, gives the instantaneous value of the alter-
nating voltage. Having performed the summation for a sufficient number
of positions of the strip, the wave of the induced e.m.f. is plotted. The
scale of the curve is determined by the condition that the maximum ordi-
nate is equal to the value of the continuous voltage. The effective value is
found in the well-known way, either in rectangular or in polar coordinates
(see the Electric Circuit} .
Prob. 46. Apply the solution of the preceding problem to the field dis-
tribution specified in Prob. 30, Art. 30. Ans. 81.5 per cent.
Prob. 47. Extend the method described in Prob. 45 to the case when
the distance between the taps connected to the slip rings is less than
180 degrees. Show how to find the scale of voltage.
Prob. 48. How does a fractional pitch affect the values given in the
table above, and the solution outlined in Prob. 45?
Prob. 49. Show how to solve problems 45 to 48 when the field curve
is given analytically, as B = F(a), for instance in the form of a Fourier
series. Hint: See C. A. Adams, " Voltage Ratio in Synchronous Conver-
ters with Special Reference to the Split-pole Converter," Trans. Amer.
Inst. Elec. Engs., Vol. 27 (1908), part II, p. 959.
1 See C. W. Stone, " Some Developments in Synchronous Converters,"
Trans. Amer. Inst. Elec. Engs., Vol. 27 (1908), p. 181.
CHAPTER V.
THE EXCITING AMPERE-TURNS IN ELECTRICAL
MACHINERY
33. The Exciting Current in a Transformer. The magnetic
flux in the core of a constant-potential transformer is determined
essentially by the primary applied voltage, and is practically
independent of the load (see Art. 25). When the terminal
voltage is given, the flux becomes definite as well. The ampere-
turns necessary for producing the flux are called the mag-
netizing or the exciting ampere-turns. When the secondary cir-
cuit is open the only current which flows through the primary
winding is that necessary for producing the flux. This current is
called the no-load, exciting, or magnetizing current of the trans-
former. When the transformer is loaded, the vector difference
between the primary and the secondary ampere-turns is practi-
cally equal to the exciting ampere-turns at no-load.
The exciting current is partly reactive, being due to the
periodic transfer of energy between the el'ectric and the magnetic
circuits (see Art. 16 above), partly it represents a loss of energy
due to hysteresis and eddy currents in the core. Some writers call
the reactive component of the no-load current the magnetizing
current, and the total no-load current the exciting current.
Generally, however, the words magnetizing and exciting are used
interchangeably to denote the total no-load current. The
components of the current in phase and in quadrature with the
induced voltage are called the energy and the reactive components
respectively.
The no-load or exciting current in a transformer must usually
not exceed a specified percentage of the rated full-load current ;
it is therefore of importance to know how to calculate the
exciting current from the given dimensions of a transformer.
Knowing the applied voltage and the number of turns, the maxi-
mum value of the flux is calculated from eq. (28). We shall
80
CHAP. V] EXCITING AMPERE-TURNS 81
assume first that the maximum flux density in the core is so
low, that it lies practically on the straight part of the mag-
netization curve of the material (Fig. 3). The case of high flux
densities is considered in the next article.
Since by assumption the instantaneous magnetomotive forces
are proportional to the corresponding flux densities, the magnetiz-
ing current must vary according to the sine law. It is sufficient,
therefore, to calculate the maximum value of the magnetomotive
force, corresponding to the maximum flux. Knowing the ampli-
tude ®m of the flux and the net cross-section of the core, A, the
flux density Bm becomes known ; from the magnetization curve of
the material (Fig. 3) the corresponding value of Hm, or the ampere-
turns per unit length of path, is found. The mean length I of the
lines of force is determined from the drawing of the core, so that
the total magnetizing ampere-turns Mm = Hml can be calculated.
The mean magnetic path around the corners is somewhat shorter
than the mean geometric path.
Let HI be the number of turns in the primary winding, and i0
the effective value of the reactive component of the exciting cur-
rent. We have then
2 = Mm ........ (41)
From this equation the quantity which is unknown can be calcu-
lated.
It is presupposed in the above deduction that the joints
between the laminations offer no reluctance. In reality,
the contact reluctance is appreciable; its value depends upon
the character of the joints, and the care exercised in the
assembling of the core. This reluctance of the joints can be
expressed by the length of an equivalent air-gap having the same
reluctance. Thus, experiments show that each overlapping joint
is equivalent to an air-gap 0.04 mm. long. A butt joint, with
very careful workmanship, is equivalent to an air-gap of about
0.05 mm.; in practice, a butt joint may offer a reluctance of from
50 to 100 per cent higher than the foregoing value.1 Knowing the
1 H. Bohle, "Magnetic Reluctance of Joints in Transforming Iron," Journal
(British) Inst. Electr. Engs,, Vol. 41, 1908, p. 527. It is convenient to
estimate the influence of the joints in ampere-turns at a standard flux
density. For each lap joint 32 ampere-turns must be added at a density
of 10 kilolines per square centimeter, while a butt joint requires at the same
82 THE MAGNETIC CIRCUIT [ART. 33
length a of the equivalent air-gap, the number of additional ampere-
turns is calculated according to the formula aBm/ /*, and then is
multiplied by the number of joints in series (usually four). This
number of ampere-turns must be added to Mm calculated above.
The energy component i\ of the exciting current is determined
from the power lost in hysteresis and eddy currents in the core.
Having calculated this power P as is explained in Article 19, we
find ii = P/Ei, where E\ is the primary applied voltage. Knowing
i0 and ii, the total no-load current is found as their geometric sum,
The watts expended in core loss depend only upon the volume
of the iron, the frequency, and the flux density used. It can be
also shown that the reactive volt-amperes required for the excita-
tion of the magnetic circuit of a transformer depend only upon the
volume of the iron, the frequency, and the flux density. Namely,
neglecting the influence of the joints, eq. (41) can be written in the
form
Eq. (28) in Art. 25 can be written as
where A is the cross-section of the iron, and Bm is the maximum
flux density, in kilolines per square centimeter. Multiplying these
two equations together, term by term, and cancelling HI we get,
after reduction,
Eiio/V-nfBnHnXlQ'*, .... . (42)
where V=Al is the volume of the iron, in cubic centimeters.
The left-hand side of eq. (42) represents the reactive magnetizing
volt-amperes per unit volume of iron; the right-hand side is a
function of / and Bm only, because Hm can be expressed through
Bm from the magnetization curve of the material.
Formula (42) can be plotted as a set of curves, one for each
commercial frequency. These curves are quite convenient in the
design of transformers, because they enable one to estimate
directly either the permissible volume of iron, or the permissible
flux density, when the reactive component of the exciting cur-
density from 60 to 80 ampere-turns. At other flux densities the increase
is proportional.
CHAP. V] EXCITING AMPERE-TURNS 83
rent is limited to a certain percentage of the full -load current. In
practice, such curves are sometimes plotted directly from the
results of tests on previously built transformers. These experi-
mental curves are the most secure guide for predicting the exciting
current in transformers; formula (42) shows their rational basis.
Prob. 1. Prove that if there were no core loss the exciting current
would be purely reactive, that is to say, in a lagging phase quadrature
with the induced voltage.
Prob. 2. The core of a 22-kv. 25-cycle transformer, like the one
shown in Fig. 12, has a gross cross-section of 4500 sq.cm.; the mean
path of the lines of force is 420 cm.; the material is silicon steel; the
maximum flux is 36 megalines. The expected reluctance of each of the
four butt joints is estimated to be equivalent to an 0.08 mm. air-gap.
What are the two components of the exciting current, and what is the
total no-load current? Ans. 1.8; 0.4; 1.85.
Prob. 3. In what respects does the calculation of the magnetizing
current in a shell-type or cruciform-type transformer differ from that
in a core-type transformer?
Prob. 4. Show that for flux densities up to 10 kl. /sq.cm. the mag-
netizing volt-amperes per kilogram of carbon steel at 60 cycles are
approximately equal to 7.3 (5m/ 10) 2.
Prob. 5. Show that the influence of the joints can be taken into
account in formula (42) by adding to the actual volume of the iron
the volume of the air-gaps multiplied by the relative permeability of
the iron.
Prob. 6. A shell-type 1000-kva., 60-cycle transformer is to have a
core made of silicon-steel punchings of a width w = l7 cm. (Fig. 13);
the average length of the magnetic path in iron is 180 cm. ; the reactive
component of the no-load current must not exceed 2 per cent of the
full-load current. Draw curves of the required height of the core per
link, and of the total core loss in per cent of the rated kva., for flux
densities up to 10 kl. /sq.cm.
Ans. £2/i=5200; at 5=9, P=0.51 per cent.
34. The Exciting Current in a Transformer with a Saturated
Core. In the preceding article the flux density in the core is
supposed to be within the range of the straight part of the satura-
tion curves (Fig. 3), so that, when the flux varies according- to the
sine law, the magnetizing current also follows, a sine wave. We
shall now Consider the case when the flux density rises to a value
on or beyond the knee of the magnetization curve. Such high flux
densities are used with silicon steel cores, especially at low frequen-
cies. In this case the magnetizing current does not vary according
to the sine law, but is a peaked wave, because at the moments when
84
THE MAGNETIC CIRCUIT
[ART. 34
the flux is approaching its maximum, the current is increasing
faster than the flux, on account of saturation. The amplitude
factor of the current wave, or the ratio of the amplitude to the
effective value is no more equal to \/2, but is larger. Let this
ratio be denotedby /a. Then eq. (41) becomes
(43)
where to is as before the effective value of the reactive component
of the exciting current. The value of %a is obtained by actually
3.0
2.5
.2.0
il.5
1.0
0.5
0.0
10
15
Flux density in Kl. per Sq. Cm.
FIG. 22. — Ratio of the amplitude to the effective value of the magnetizing
current.
plotting the curve of the magnetizing current from point to point
and calculating its effective value. Since the procedure is rather
long, it is convenient to calculate the values of #a once for all for
the working range of values Bm. This has been done for the mate-
rials represented in Fig. 3, and the results are plotted in Fig. 22.
Strictly speaking, the exciting current is unsymmetrical, due
to the effect of hysteresis, and the values of ya ought to be cal-
culated, using the hysteresis loops of the steel. However, it is very
nearly correct to calculate %a from the magnetization curve, and to
CHAP. V] EXCITING AMPERE-TURNS 85
calculate the energy component of the exciting current separately,
from the core loss curves (Fig. 10). The magnetizing current
required for the joints is calculated separately, using ya= \/2,
according to eq. (41). The total effective magnetizing current is
found by adding together the values of IQ for the iron and for the
joints. The loss component, i\, is added to this value in quadra-
ture, to get the total no-load current. As is mentioned above, it
is preferred in practice to estimate the total exciting current of new
transformers from the curves of no-load volt-amperes per kilogram
of iron, the values being obtained from tests on similar trans-
formers.
Prob. 7. The core of a 25-cycle cruciform type transformer (Fig. 14)
weighs 265 kg.; the mean length of the magnetic path is 170 cm.; the
material is silicon steel. The 4400-v. winding of the transformer has
1100 turns in series. What is the reactive component of the no-load
current? Ans. 8.4 amperes.
Prob. 8. Check a few points on the curves in Fig. 22.
Prob. 9. Show that in formula (42) the coefficient n is a special
case of the more general factor 4.44/£a, when the magnetizing current
does not follow the sine wave.
Prob. 10. What are the reactive volt-amperes per kilogram of carbon
steel at 40 cycles and at a flux density of 16 kl./sq.cm.? Ans. 56.4.
Prob. 11. Show how to calculate the exciting ampere-turns required
for a given flux in a thick and short core in which the flux density is
different along different paths.
35. The Types of Magnetic Circuit Occurring in Revolving
Machinery. The remainder of this chapter and the next chapter
have for their object the calculation of the exciting ampere-turns
necessary for producing a certain useful flux in the principal types
of electric generators and motors. In direct-current machines,
in alternators, and in rotary converters it is necessary to know the
exciting or field ampere-turns in order to plot the no-load satura-
tion curve, to predict the performance of the machine under vari-
ous loads, and to design the field coils. In an induction motor one
wants to know the required excitation in order to determine the
no-load current, or to calculate the number of turns in the stator
winding, when the limiting value of the no-load current is pre-
scribed. The general procedure in determining the required
number of ampere-turns for a given flux is in many respects the
same in all the types of electrical machinery, so that it is possible
to outline the general method before going into details.
86
THE MAGNETIC CIRCUIT
.[ART. 35
In direct-current machines and in synchronous generators,
motors, and rotary converters, the magnetic flux (Figs. 15 and 20)
from a field pole passes into the air-gap and the armature teeth.
In the armature core the flux is divided into two halves, each half
going to one of the adjacent poles. The magnetic paths are com-
pleted through the field frame. Part of the flux passes directly
from one pole to the two adjacent poles through the air, without
going through the armature. This part of the flux is known as
the leakage flux. The closed magnetic paths and the field coils of a
machine may be thought of as the consecutive links of a closed
chain. While in a transformer the chain is open, in generators
FIG. 23. — The paths of the main flux and of the leakage fluxes in an
induction motor (or generator).
and motors the chain must be closed on account of the continuous
rotation.
In induction machines, both generators and motors (Fig. 23),
the flux at no load is produced by the currents in the stator wind-
ings only. When the machine is loaded, the flux is produced by
the combined action of the stator and rotor currents, the rotor cur-
rents opposing those in the stator, the same as in a transformer.
Therefore, the flux in the loaded machine may be regarded as the
resultant of the following three component fluxes: The main or
useful flux, $, which links with both the primary and the secondary
windings; the primary leakage flux, $1; which links with the stator
winding only ; and the secondary leakage flux, $2 which is linked
with the rotor winding alone. The leakage fluxes not only do not
CHAP. V] EXCITING AMPERE-TURNS 87
contribute to the useful torque of the machine, but actually reduce
it. In reality, there is of course but one flux, the resultant of the
three, but for the purposes of theory and computations the three
component fluxes can be considered as if they had a real separate
existence. In this and in the following chapter the main flux
only will be discussed for this type of machinery. Considera-
tion of the leakage flux will be reserved to Art. 66.
The total magnetomotive force per magnetic circuit is equal
to the sum of the m.m.fs. necessary for establishing the required
flux in the separate parts of the circuit which are in series, viz., the
pole-pieces, the air-gap, the teeth, and the armature core. All the
necessary elements for the solution of this problem have been dis-
cussed in the first two chapters. It remains here to establish some
semi-empirical " short-cut " rules and formulae for the irregular
parts of the circuit, for which, although close approximations can
be made, the exact solution is either impossible or too complicated
for the purposes of this text. The following topics are considered
more in detail in the subsequent articles of this and of the follow-
ing chapter.
(a) The ampere-turns necessary for the air-gap when it is
limited on one side or on both sides by teeth, so that the flux den-
sity in the air-gap is not uniform.
(6) The ampere-turns necessary for the armature teeth when
they are so highly saturated that an appreciable part of the flux
passes through the slots between the teeth.
(c) The ampere-turns necessary for the highly saturated cores
in which the lengths of the individual paths differ considerably
from one another, with a consequent lack of uniformity in the flux
density.
(d) The leakage coefficient and the value of the leakage flux
which passes directly from pole to pole. This leakage flux
increases the flux density in the poles and in the field frame of the
machine, and consequently increases the required number of
ampere-turns.
All of the m.m.f. calculations that follow are per pole of the
machine, or what is the same, for one-half of a complete magnetic
circuit (cdfg in Figs. 15, 20, and 23), the two halves being identical.
This fact must be borne in mind when comparing the formulae with
those given in other books, in which the required ampere-turns are
sometimes calculated for a complete magnetic circuit.
88
THE MAGNETIC CIRCUIT
[ART. 36
Prob. 12. Inspect working drawings of electrical machines found
in various books and magazine articles; indicate the paths of the main
and of the leakage fluxes; and make clear to yourself the reasons for the
use of different kinds of steel and iron in the frame, the core, the pole-
pieces, and the pole shoes.
Prob. 13. Make sketches of the magnetic circuit of a turbo-alternator
with a distributed field winding, of a homopolar machine, of an inductor-
type alternator, and of a single-phase commutator motor. Indicate the
paths of the useful and of the leakage fluxes.
36. The Air-gap Ampere Turns. The general character of the
distribution of the magnetic flux in the air-gap of a synchronous
and of a direct-current machine is shown in Fig. 24, the curvature
of the armature being disregarded. The principal features of this
flux distribution are as follows :
Armature
4-X-4
^Air-duct
mature
FIG. 24. — The cross-section of a direct-current or synchronous machine,
showing the flux in the air-gap.
(a) The flux per tooth pitch A is practically the same under all
the teeth in the middle part of the pole, where the air-gap has a
constant length, and is smaller for the teeth near the pole-tips
where the air-gap is larger.
(b) On the armature surface the flux is concentrated mainly
at the tooth-tips; very few lines of force enter the armature
through the sides and the bottom of the slots.
(c) There is a considerable spreading, or fringing, of the lines
of force at the pole-tips.
(d) In the planes passing through the axis of the shaft of the
machine there is also some spreading or fringing of the lines of
force at the flank surfaces of the armature and the pole, and in
the ventilating ducts.
CHAP. V] EXCITING AMPERE-TURNS 89
This picture of the flux distribution follows directly from the
fundamental law of the magnetic circuit, the flux density being
higher at the places where the permeance of the path is higher.
The actual flux distribution is such that the total permeance of all
the paths is a maximum, as compared to any other possible distri-
bution. In other other words, the flux distributes itself in such a
way, that with a given m.m.f . the total flux is a maximum, or with
a given flux the required m.m.f. is a minimum. This is confirmed
by the beautiful experiments of Professor Hele-Shaw and his col-
laborators,1 who have obtained photographs of the stream lines of a
fluid flowing through an arrangement which imitated the shape
and the relative permeances of the air-gap and of the teeth in an
electric machine.
Let (Pa be the total permeance in perms of the air-gap between
the surface of the pole shoe and the teeth, and let 0 be the useful
flux per pole, in maxwells, which is supposed to be given. Then,
according to eq. (2), Art. 5, the number of ampere-turns required
for the air-gap is
(44)
The problem is to calculate the permeance of the gap from the
drawing of the machine.
One of the usual practical methods is to calculate (P under cer-
tain simplifying assumptions and then multiply the result by an
empirical coefficient determined from tests on similar machines.
The simplest assumptions are (Fig. 25) : (a) that the armature has
a smooth surface, the slots being filled with iron of the same per-
meability as that of the teeth; (6) that the external surface of the
pole shoes is concentric with that of the armature; (c) that the
equivalent air-gap aeq is equal to two-thirds of the minimum air-
gap plus one-third of the maximum air-gap of the actual machine ;
(d) that the ventilating ducts are filled with iron ; (e) that the paths
of the fringing flux at the edges of the pole shoe are straight lines,
and extend longitudinally to the edge of the armature surface and
laterally for a distance equal to the equivalent air-gap on each
side.
1 For a detailed account of the experimental and theoretical investigations
on this subject, with numerous references, see Hawkins and Wallis, The
Dynamo (1909), Vol. 1, Chapter XV.
90
THE MAGNETIC CIRCUIT
[ART. 36
With these assumptions, the permeance of the " simplified "
air-gap is
...... (45)
where w8 is the average width of the flux, and ls is its average
axial length. Or
ws = i (wa + Wp) = wp + a' ;
The lateral spread a! of the lines of force at each pole-tip is
taken to be approximately equal to a^.
f
1/t
«>
1
i 1
I
II
I
!
Mimm
j/ a*<l
t
IMMMMMIM
— lp !
FIG. 25. — Magnetic flux in the simplified air-gap.
The permeance of the actual air-gap is smaller than that of the
simplified gap, so that we have
....... (46)
where ka is a coefficient larger than unity, called the air-gap factor.
Substituting the value of (P from eq. (46) into (44) gives
(47)
so that ka is the factor by which the ampere-turns for the simplified
air-gap must be multiplied in order to obtain the ampere-turns
required for the actual air-gap. The value of ka usually varies
between 1.1 and 1.3, depending on the relative proportions of the
teeth, the slots, and the air-gap, and on the shape of the poles.
The numerical values of fcaare calculated from the results of tests
on machines of proportions similar to that being computed. Let
the no-load saturation curve of a machine be available from test ;
this is a curve which gives the relation between the induced voltage
CHAP. V] EXCITING AMPERE-TURNS 91
and the field current of the machine. From the known specifica-
tions of the machine this curve can be easily converted into one
which gives the useful flux per pole against the ampere-turns per
pole as abscissae. The lower part of such a curve is always a
straight line, there being then practically no saturation in the iron.
On this part of the curve, practically the whole m.m.f . is consumed
in the air-gap, so that the actual permeance of the air-gap is found
by dividing one of the ordinates by the corresponding abscissa.
The permeance of the simplified air-gap is calculated from eq. (45),
and the ratio of the two gives the value of the coefficient ka. This
value is then used in the design and calculation of the performance
of new machines with similar proportions. Engineering judg-
ment and practical experience are factors of considerable impor-
tance in estimating the values of ka for new machines.
The same method of calculating the air-gap ampere-turns is
applicable to induction machines (Fig. 23). The ampere-turns
are computed, assuming both the rotor and the stator to have
smooth iron surfaces, without slots; the result is then multiplied
by a factor ka larger than unity, determined from tests upon
machines of similar proportions.
A more accurate, though more elaborate, method for calcula-
ting the air-gap ampere-turns is explained in the next article.
Prob. 14. Calculate the air-gap ampere-turns per pole for a 6600 v.,
25-cycle, 375-r.p.m. alternator to be built according to the following
specifications: The bore 2.4 m.; the gross axial length of the armature
core 55 cm.; seven air ducts 9 mm. each; the minimum air-gap is 15
mm.; the maximum air-gap is 30 mm. The poles cover 66 per cent of
the periphery; the axial length of the pole shoes is 53 cm. The useful
flux per pole at no-load and at the rated voltage is 19.1 megalines. The
air-gap factor is estimated to be about 1.15. Ans. 10,600.
Prob. 15. The no-load characteristic obtained from the test upon
the machine specified in the preceding problem has a straight part such
that at a field current of 52 amp. the line voltage is 4000 v. Each field
coil has 120 turns. What is the true value of the air-gap factor?
Ans. 1.12.
Prob. 16. A pole shoe is so shaped that the minimum air-gap is a0
and the maximum air-gap is a1=a0+Ja, the increase in the length being
proportional to the square of the distance from the center of the pole.
What is the length aeq of the equivalent uniform. air-gap such that its
total permeance is the same as that of the given air-gap? Assume a
smooth-body armature, and neglect the fringing at the pole-tips. Solu-
tion: Let the peripheral width of the pole be 2w; then the length of
92 THE MAGNETIC CIRCUIT [ART. 37
the air-gap at a distance x from the center is a = a0 +da(x/w}2. The per-
meance of an infinitesimal path of the width dx is proportional to dx/ax.
Hence we have the relation
x.w
I dx/[a0 +Aa(x/w) 2]= w/aeq,
JQ
from which
aeq = Va^a/tan"1 ( \/Ja/a0) .
Prob. 17. What is the length of the equivalent air-gap in the preced-
ing problem if the clearance at the pole-tips is twice the clearance at
the center of the pole? Ans. 1.273a0.
Prob. 18. Show that, when the air-gap is non-uniform, the length
of the equivalent uniform gap can be determined approximately, accord-
ing to Simpson's Rule, from the equation
where a0, at, and am are the lengths of the gap at the center, at the tip of
the pole, and midway respectively. If the air-gap is uniform under the
major portion of the pole, but the pole shoe is chamfered, more terms
must be taken in Simpson's formula in order to obtain aeq with a
sufficient accuracy.
Prob. 19. What is the length of the air-gap required in problems
16 and 17, according to the formula given in problem 18?
Ans. 1.276a0.
37. The Method of Equivalent Permeances for the Calculation
of Air-gap Ampere-turns. An inspection of Fig. 24 will show that
the total permeance of the air-gap is made up of a number of per-
meances in parallel. It is equal therefore to the sum of these
permeances. For the purpose of calculation two kinds of per-
meances are considered separately: those from the teeth to
the pole surface proper, and those from the teeth to the pole-tips.
The former can be calculated quite accurately, the latter are to
some extent estimated.
The permeance per tooth pitch in the part of the air-gap near
the center of the pole can be divided into two parts, that under the
tooth-tip, and the fringe from the sides of the slots and in the ven-
tilating ducts. The permeance of the paths which proceed from
the tooth-tip constitutes the larger portion and is made up of
nearly parallel lines; this permeance is therefore easily computed.
The values of the permeance of the fringe from the flank of the
tooth to the perpendicular surface of the pole have been deter-
CHAP. V] EXCITING AMPERE-TURNS 93
mined theoretically by Mr. F. W. Garter.1 Only the numerical
results are given here, in a somewhat simplified practical form;
the solution itself presupposing a knowledge of the properties of
conjugate functions.2
Consider the permeance of two tooth fringes, such as opqr and
o'p'q'r' (Fig. 24), perpendicular to the plane of the paper. This
permeance depends only upon the ratio of the slot width s to the
length a of the air-gap, for let both s and a be increased say twice :
The length and the cross-section of each elementary tube of force
is also increased twice, hence its permeance remains the same.
The permeance of each fringe can be replaced by the permeance
of an equivalent rectangular path of the length a and of a width
%At (Fig. 26). This is the same as increasing the width of the
tooth by the amount At and assuming all the lines of force to be
parallel to each other in the air-gap. The permeance of the path
which replaces the two fringes is equal to pAt/a. From what has
been said above follows that the ratio At /a depends only upon the
ratio of s/a; the relationship between the two ratios is plotted in
Fig. 26, from Carter's calculations. For the sake of convenience
and accuracy, the curve is drawn to two different scales, one for
large the other for small values of s/a.
The curve in Fig. 26 may be interpreted in two ways: It may
be said to represent the "geometric permeance'' of the fringe (for
/*=!); or else it may be said to give the correspondings sets of
values of s and At, measured in the lengths of the air-gap as the
unit. With a given a, At increases with s, because the maximum
width of the actual fringe is Js. With a given s the width At
increases toward the pole-tip (if the air-gap is variable), because
with a longer air-gap the fringing lines of flux fill a larger part of
the air-gap under the slot.
The corrected width of the tooth is t' = t+ At and the permeance
of the air-gap, in perms per tooth pitch, is
<Pat=135(M/ax+t/ax)lrt, .... (48)
^ote on Air-gap Induction, Journ. Inst. Electr. Eng. (British), Vol. 29,
(1899-1900), p. 929; Air-gap Induction, Electrical World, Vol. 38, (1901)
p. 884; See also Hawkins and Wallis, The Dynamo (1909), Vol. 1, p. 446;
E. Arnold, Die Gleichstrommaschine (1906), Vol. 1, p.^266.
2 J. C. Maxwell, Electricity and Magnetism, Vol. 1, p. 284; J. J. Thomson,
Recent Researches in Electricity and Magnetism, Chapter III; Horace Lamb,
Hydrodynamics (1895), Chapter IV.
94 THE MAGNETIC CIRCUIT [ART. 37
where ax is the length of the air-gap at the center of the tooth, and
leff is the effective axial length of the machine (see below) . The
value of At/ax must be taken from Fig. 26 for the corresponding
ratio s/ax.
The permeance of the pole fringe, hmn in Fig. 24, cannot be cal-
culated by the foregoing method, because this permeance depends
upon the irregular shape of the pole-tip. The pole-fringe permeance
is usually estimated graphically by drawing lines of force, taking
Fig. 24 as a guide1 ; the permeance of each tube of flux between the
pole and the armature is. fiA/l, where A is the mean cross-section,
and I is the mean length of the tube. The fringe permeance is of
the order of magnitude of 10 per cent of the total permeance of
the air-gap, so that some error in its estimation does not seriously
affect the total required ampere-turns. Careful designers some-
times calculate the air-gap permeance for two positions of the
pole, differing from each other by one-half of the tooth pitch, and
take the average of the two results.
Carter's curve could be used directly for calculating the pole-
fringe permeance, if the pole waist were of the same width as
the pole shoe (line mm' in Fig. 24), and if the armature had no
slots. In this case the space between the adjacent poles could be
considered as a big slot, and the curve in Fig. 26 could be directly
applied to it. On account of a smaller width of the pole core and
because of the armature slots the mean length of the lines of force
in the fringe is increased, so that the actual permeance of the pole-
fringe is somewhat smaller than that according to Carter's curve.
By practice and experience one can acquire a judgment as to what
fraction of Carter's permeance to take in a given case.
The length le^ is a sum of the parts such as l\, 1'2, etc. (Fig. 24),
on which the lines of force are parallel, and of small additional
lengths which take account of the fringing in the air-ducts and at
the pole flanks. These additional lengths are again estimated
from Carter's curve (Fig. 26). The fringe Ad in an air-duct of
the width d is practically the same as that in a slot of the width
s = d. The additional length Af for the pole flanks is found by con-
sidering the two fringes as due to a slot of the width /. When the
stationary and the revolving parts are of the same axial length so
that/=0, there still remains some fringe permeance between the
1 See Art. 41 below in regard to the drawing of the lines of force by the
judgment of the eye.
CHAP. V]
EXCITING AMPERE-TURNS
95
96 THE MAGNETIC CIRCUIT [ART. 37
flank surfaces of the two iron structures. This permeance is, how-
ever, very small, and has to be estimated empirically, if at all.
Strictly speaking, leff is different for each tooth, if the air-gap
is variable, because the amount of fringing in the air-ducts and at
the flanks is different. However, it is hardly worth the effort
in ordinary cases to calculate leff for each tooth. It is sufficient to
take an average leff for some intermediate value of the air-gap.
In some high-speed alternators, and usually in induction
motors, air-ducts are provided in both the stationary and the
revolving parts, in the same planes. The flux fringe in an air-duct
is then of such a shape that the lines of force are parallel to one
another in the middle of the air-gap, between the stator and the
rotor. Therefore, when using the curve in Fig. 26 for such a case,
the cylindrical surface midway between the stator and the rotor
must be taken to correspond to that of the solid iron surface
assumed in the deduction of the curve. Hence, \ax must be used
instead of ax in determining Ai.
Having calculated the permeances of the several paths per pole
pitch the total permeance of the air-gap is found as their sum, or
(49)
Then, the required number of ampere-turns is determined from
eq. (44). The method gives quite correct results, especially with
some experience in estimating the permeances of irregular paths.
Each designer usually modifies slightly the empirical factors which
are indispensable in this method, and devises short cuts good for
the particular kind of machine in which he is interested.
Instead of calculating the permeance of each tooth separately,
some engineers replace the actual variable air-gap by an equiva-
lent constant air-gap aeg, either by the judgment of the eye, or as
in prob. 18 above. The actual peripheral length of the pole arc is
increased by from one to one and one-half aeq on each side to take
into account the fringing at the pole-tips. This gives the number
of teeth under the pole. The permeance of each tooth is calcu-
lated from eq. (48) for ax = aeq, and is then multiplied by the num-
ber of teeth. With some practice, one can obtain in this manner
quite accurate results at a considerable saving in time.
The method outlined above is not directly applicable to induc-
tion machines which have slotted cores on both sides of the air-
CHAP. V] EXCITING AMPERE-TURNS 97
gap (Fig. 23) : At each instant some stator teeth are opposite rotor
teeth, others bridge over some rotor slots, and vice versa. The
amount of overlap varies from instant to instant, causing periodic
fluctuations in the air-gap reluctance.
Assume first that both the stator and the rotor have smooth
surfaces facing the air-gap. Let the permeance of such a machine
be (P8. If now the armature be slotted, the cross-section of
the paths in the air-gap (neglecting the fringe) is reduced in the
ratio ti/Ai where ti and AX are the stator tooth width and tooth
pitch respectively. The permeance (Ps is also reduced in the same
ratio. Let the rotor be also provided with slots; the average
cross-section of the path is thereby further reduced in the ratio
(£2/^2), where t% and \2, are the tooth width and the tooth pitch on
the surface of the rotor. Thus, disregarding the spread of the
flux, the average air-gap permeance of an induction motor is
the symbol (Pa being put in parentheses to indicate that a further
correction for the tooth fringe is necessary.1
In order to take the fringe into consideration, an empirical cor-
rection is made in this formula. Namely, it is assumed that the
actual permeance of the fringes of the stator teeth is the same as
if the rotor had a smooth core, and vice versa. Accordingly, in the
preceding formula, the values ti and t% of the tooth widths are
corrected for the fringe, using Carter's curve (Fig. 26). The
formula becomes then
<Pa=«l'A)(«2'/A2)tf>,, ..... (50)
where £/ and tj are the corrected widths of the stator and rotor
teeth respectively. This formula has been found to be in a satis-
factory agreement with experimental results.2
1 For a more rigorous proof of this formula see C. A. Adams, "A Study
in the Design of Induction Motors," Trans. Amer. Inst. Electr. Engs., Vol. 24
(1905), p. 335.
2 T. -F. Wall, The Reluctance of the Air-gap in Dynamo-machines, Journ.
Inst. Electr. Engrs. (British), Vol. 40 (1907-8), p. 568. E. Arnold in his
Wechselstromtechnik, Vol. 5, Part 1, pp. 42, 43, calculates the value of ka
for an induction machine in a somewhat different way. With open slots
in the stator, Arnold's method gives lower values of ka than they are in reality.
See Hoock and Hellmund, Beitrag zur Berechnung des Magnetizierungs-
98 THE MAGNETIC CIRCUIT [ART. 37
Referring to Art. 36, eq. (50) may be interpreted as follows:
Let ka' be the air-gap factor for the slotted stator and a smooth-
body rotor; let ka" be the same factor for the slotted rotor and a
smooth-body stator. Then the air-gap factor of the actual
machine
ka=ka'Xka". . .' ..... (51)
In an induction motor the magnetic flux is distributed in the
air-gap approximately according to the sine law, due to the dis-
tributed polyphase windings. Therefore the value of M deter-
mined from eq. (44) gives only the average value of the m.m.f.
required for the air-gap. With a sine-wave distribution of the
flux the maximum m.m.f. is x/2 times larger than the average
value.
Prob. 20. What is the permeance of the air-gap of a 16-pole direct-
current machine, the armature of which has a diameter of 250 cm. and
is provided with 324 slots, 1 2 by 1 5 mm . ? The gross length of the armature
is 23 cm., and it is provided with three ventilating ducts, 10 mm. wide
each. The axial length of the poles is 21.5 cm. The pole shoes cover
05 per cent of the periphery, and are not chamfered. The length of the
air-gap is 10 mm. Ans. About 900 perm.
Prob. 21. The machine mentioned in problem 14 has 120 slots, 3 by
6.5 cm. The pole-shoes are shaped according to the arc of a circle of a
radius equal to 90 cm. and subtending 36 degrees; the pole-tips are formed
by quadrants of a radius equal to 2.5 cm. Check the value of the field
current (52 amp.) given- in problem 15, by the method of equivalent
permeances.
Prob. 22. What is the maximum m.m.f. across the air-gap of an
induction motor, if the gross average flux density in the air-gap (total
flux. divided by the gross area of the air-gap, not including the vents)
is 3 kl./sq.cm., and the clearance is 1.2 mm.? The bore is 64 cm.; the
stator is provided with 48 open slots, 22 by 43 mm. The rotor has 91
half-closed slots, the slot opening being 3 mm. The machine has a vent
7 mm. wide for every 9 cm. of the laminations.
Ans. 820 amp. turns.
Prob. 23. Show that At/a = 1.2 +2.93 log (s/2a), if the fringing
lines of force are assumed to be concentric quadrants (Fig. 27, to the left)
with the points c as the center; the average length of path in the part
6cc' is estimated to be equal to 1.2a, and the average width 0.72a. Hint:
The permeance of an infinitesimal tube of force of a radius x and of a
width dx is iL-dx/(%nx). Integrate this expression between the limits
stromes in Induktionsmotoren, Elektrotechnik und Maschinenbau, Vol. 28
(1910), p. 743.
CHAP. V]
EXCITING AMPERE-TURNS
99
of a and }s. See Adams, loc. cit., p. 332. The formula can be used only
when s is larger than 2a.
Prob. 24. Show that J*/a = 2.93 log (1 +\*s/a), if the fringing lines
of force are assumed to consist of concentric quadrants (Fig. 27, to
c' ]f
a
FIG. 27.— Two simplified paths for the fringing flux.
the right), with the point cr as a center, continued as straight lines.
See Arnold, Die Gleichstrommachine, Vol. 1, p. 269.
Prob. 25. Show that formula (50) applies to synchronous and direct-
current machines with salient poles as well, if tf is the width of the
pole shoe, corrected for the fringe, and A2 is the pole pitch.
CHAPTER VI
EXCITING AMPERE-TURNS IN ELECTRICAL
MACHINERY— (Continued)
38. The Ampere-turns Required for Saturated Teeth. The teeth
and the slots of an armature, under the poles, are magnetically in
parallel (Fig. 24) ; hence, part of the flux passes from the pole into
the armature core through the slots between the teeth. But, with
a moderate saturation in the teeth, say below 18 kilolines per
square centimeter, the amount of the flux which passes through the
slots is altogether negligible. If the taper of the teeth is slight,
the required ampere-turns are found for the average flux density
in the tooth, taking the value of H from the curves in Fig. 3.
Should the taper of the teeth be considerable, as is the case in
revolving armatures of small diameter, the flux density should be
determined in say three places along the tooth, viz., at the root,
in the middle part, and at the crown. Let the corresponding
values of magnetic intensity from the magnetization curve of the
material be H0, Hm, and HI . Assuming H to vary along the tooth
according to a parabolic law, we have, according to Simpson's
rule in the first approximation, that the average intensity over the
tooth is
(52)
If a greater accuracy is desired, the values of H can be determined
for more than three cross-sections of the tooth and Simpson's
rule applied.1 For instance, let the length be divided into n equal
1 A designer who has to calculate ampere-turns for teeth frequently
will save time by plotting curves for the average H against the flux density
BQ at the root of the teeth. Each curve would be for one taper, and these
curves would cover the usual range of taper in the teeth. See A. Miller Gray
"Magnetomotive Force in Non-uniform Magnetic Paths," Electrical World,
Vol. 57 (1911), p. 111.
100
CHAP. VI] EXCITING AMPERE-TURNS 101
parts, where n is an even number. Then, we have that
n-2)]. • (53)
When the flux density in the teeth is considerable, say between
18 and 24 kilomaxwells per square centimeter, an appreciable part
of the total flux passes through the slots between the teeth, also
through the air-ducts, and in the insulation between the lamina-
tions. Dividing, therefore, the flux per tooth pitch by the net
cross-section of the tooth, one gets only the so-called apparent flux
density in the tooth, which density is higher than the true density.
With highly saturated teeth, a small difference in the estimated
flux density makes an appreciable difference in the required number
of ampere-turns; it is therefore of importance to know how to
determine the true density in a tooth, knowing the apparent
density.
Consider first the case of a machine with a large diameter, in
which the taper of the teeth can be neglected. Assume the con-
centric cylindrical surfaces at the tips and at the roots of the teeth
to be equipotential surfaces, and the lines of force to be all parallel
to each other, in the slots as well as in the iron. In reality, some
lines of force enter the teeth on the sides of the slots (Fig. 24), so
that the foregoing assumptions are not quite correct ; but they are
the simplest ones that can be made. Any other assumptions
would lead to calculations too complicated for practical use.
Let Breal be the true flux density in the iron of the tooth, and
let Bapp be the apparent flux density in the tooth under the assump-
tion that no flux passes through the slots, air-ducts, or insulation
between the laminations. Then, denoting the actual flux density
in the air by Ba, we have the following expression for the total
flux per tooth pitch:
== A ^n
reai
where Ai and Aa are the cross-sections in square centimeters of the
paths per tooth pitch, in the iron and air respectively. Since the
iron and the air paths are of equal length, and are in parallel, the
m.m.f. gradient is the same in both. Let H be this gradient, in
102 THE MAGNETIC CIRCUIT . [ART. 38
kiloampere-turns per centimeter. Then, if all the flux densities
are in kilomaxwells per square centimeter,.
Substituting this value of Ba into the preceding equation we
obtain, after division by At-,
BapP = Breal-^\^(Aa/A^H ..... (54)
The ratio Aa/Ai can be expressed through the dimensions of
the machine as follows: Ai^tln where t is the width of the tooth,
and ln is the net axial length of the laminations, without the air-
ducts and insulation. Aa=XLg—tln, where X is the tooth pitch
(Fig. 20), and lg is the gross length of the armature core. Hence,
(55)
In eq. (54) the flux density Bapp and the ratio AJAi are known
in any particular case, and the problem is to find Breal and H:
The other equation which connects Breal and H is the magnetiza-
tion curve of the material, and the problem can be solved in a simi-
lar manner to problem 1 1 in chapter II (see also problem 4 below) .
Professional designers use curves like those shown in Fig. 28,
which give directly the relation between Bapp and Breai within the
range of values of Aa/Ai which occur in practice. The curves are
plotted point by point by assuming certain values of Breal and cal-
culating the corresponding Bapp from eq. (54). For instance, for
jBreaj=24, the saturation curve shown in Fig. 28 gives H= 1.33, so
that for Aa/Ai=2, we have: £opp = 24 + 1.25X2X1.33 = 27.33.
This determines one point on the curve marked " Ratio of air to
iron = 2." In using these curves one begins with the known value
of Bapp on the lower axis of abscissae, and follows the ordinate to
the intersection with the curve for the desired ratio Aa/ ' AI\ this
gives the value of Breal. By following the horizontal line from the
point so located to the intersection with the B-H curve, the corre-
sponding value of H is read off on the upper axis of abscissae.
The curves in Fig. 28 are completely determined by the shape
of the B-H curve, so that, if the material to be used for the arma-
ture core differs considerably from that assumed in Fig. 28, new
curves of Breal versus Bapp ought to be plotted, or else the method
CHAP. VI]
EXCITING AMPERE-TURNS
103
jL
\T\
\
\
\
\
\\
S
3
%
g
I
S
^r
^H
104 THE MAGNETIC CIRCUIT [ART. 38
may be used which is suggested in problem 4 below. A comparison
of the B-H curve with those in Fig. 3 shows that a much better
quality of steel is presupposed in Fig. 28. Such is usually the case
when it is desired to employ highly saturated teeth, for otherwise
it might be practically impossible to get the required flux. The
curves in Fig. 3 refer to an average quality of electrical steel.
Formula (54) and the curves in Fig. 28 presuppose that the
teeth have no taper, or that the taper is negligible. If the taper of
the teeth is quite considerable the tooth and the slot are divided
by equipotential cylindrical surfaces into two or more parts, and
H is determined separately for each part. Then the effective value
of H is calculated according to Simpson's rule, using either formula
(52) or (53).
Prob. 1. A four-pole direct-current armature has the following
dimensions : diameter 45 cm. ; gross length of core 20 cm. ; two air-ducts
7 mm. each; 67 open slots 1 by 3 cm. The poles are of such a shape
that the flux per pole is carried uniformly by 11.5 teeth. How many
ampere-turns per pole are required for the teeth when the flux per pole
is 3 megalines? Use the saturation curve for carbon-steel laminations
in Fig. 3. Ans. 148 amp .-turns.
Prob. 2. How many ampere-turns are required in the preceding
problem when the flux per pole is 4.4 megalines?
Ans. Between 2400 and 2500.
Prob. 3. The machine, in problem 22 of the preceding chapter, had
a gross average flux density in the air-gap of 3 kl./sq. cm. The bore
was 64 cm. The stator was provided with 48 slots 22 by 43 mm. The
machine has a vent 7 mm. wide for every 9 cm. of the laminations.
What is the maximum m.m.f . required for the stator and rotor teeth, if
the size of each of the 91 rotor slots is 14 by 30 mm. below the overhang?
Ans. between 2400 and 2500 amp. -turns.
Prob. 4. Instead of drawing the curves shown in Fig. 28, the relation
between Breaz and Bapp can be found by the following construction:
Disregard the lower scale marked " Apparent Flux Density "; extend
the left-hand scale to the division 34 and mark the scale " Real and Appar-
ent flux density." Cut out a strip of paper, and copy the left-hand scale
on the left-hand edge of the strip. On the right-hand edge of the strip
mark the scale for Aa/Ai as follows: division 26 of the flux density to
correspond with zero, division 27 with 0.4, division 28 with 0.8, etc.
Apply the left-hand edge of the strip to division 2.0 on the upper horizon-
tal scale, and to division 26 on the lower horizontal scale. Move the
strip up and down until the upper horizontal scale coincides with the
desired value of Aa/At marked on the strip. Lay a straightedge on the
divisions of the two vertical scales corresponding to the given apparent
flux density. The intersection of the straightedge with the B-H curve
CHAP. VI] EXCITING AMPERE-TURNS 105
will give the required values of Breai and H. Check this construction
for a few points with the values obtained from the curves, and give a
general proof. Hint: This construction amounts to considering the
B-H curve and Eq. (54) as two simultaneous equations with two unknown
quantities Breai and H. See problem 11 in chapter II, Art. 13.
39. The Ampere-turns for the Armature Core and for the
Field Frame. In many machines the m.m.f. required for the air-
gap and the teeth are large as compared to those required for the
armature core and the field frame; in such cases the latter are
either altogether neglected, or are estimated roughly, by increas-
ing the ampere-turns calculated for the rest of the magnetic cir-
cuit by say five or ten per cent. Where this is not permissible, the
usual procedure is to estimate the maximum flux density in the
core or frame under consideration and to measure from the drawing
of the machine the length of the average path of the lines of force
in it. The assumption is made that the same flux density is main-
tained on the whole length of the path, and the required ampere-
turns are calculated from the magnetization curve of the material
(Figs. 2 and 3). While the ampere-turns determined in this way
are usually larger than those actually required, the method is per-
missible if the total amount of the m.m.f. for the parts under con-
sideration is small as compared to the total m.m.f. of the magnetic
circuit. If a greater accuracy is desired, the path is subdivided
into two or more parts in series, and the average density deter-
mined for each part; and then the ampere-turns required for each
part are added.
The tendency now is to increase the flux density in the arma-
ture cores of alternators and induction motors so as to reduce the
size of the machine. This is made possible through a better quality
of laminations, which show a smaller core loss, and also through the
use of a more intensive ventilation. With these high densities
and with the comparative large values of the pole pitch necessary in
high-speed machinery, the ampere-turns for the core constitute an
appreciable amount of the total m.m.f. of the machine, and it is
therefore desirable to calculate them more accurately.
The flux density in the core is a minimum opposite the center
of a pole, and is a maximum in the radial plane midway between
two poles (Fig. 15) . At each point the flux density has a tangen-
tial and a radial component. The latter is comparatively small
and can be neglected ; the tangential component can be assumed
106 THE MAGNETIC CIRCUIT [ART. 39
to vary according to the sine law, being zero opposite the center of
the pole and reaching its maximum between the poles. With
these assumptions, knowing the maximum flux density in the core,
the flux density at all other points is calculated, and the corre-
sponding values of H are determined from the B-H curve of the
material. The average value of H for one-half pole pitch is then
found by Simpson's rule, eqs. (52) and (53). With the sine-wave
assumption, the average H depends only upon the maximum
flux density, so that for a given material a curve can be compiled
from the B-H curve, giving directly Have for different values of
/? i
•L-'max'
Should a still greater accuracy be required, the following
method can be used: Draw the assumed or the calculated curve
of the distribution of flux density in the air-gap, and indicate to
your best judgment the tubes of force in the armature core, say
for each tooth pitch. The flux in the radial plane midway between
the two poles can be assumed to be distributed uniformly over the
cross-section, and this fact facilitates greatly the determination of
the shape of the tubes of flux. The m.m.f . required for each tube
is calculated by dividing it into smaller tubes in series and in paral-
lel; thus, either the average m.m.f. for the whole flux can be found,
or the maximum m.m.f. for one particular tube.2
The frame to which the poles are fastened in direct-current and
in synchronous machines is usually made of cast iron; in some
cases the frame is made of cast steel; in high-speed synchronous
machines the revolving field is made of forged steel. The magneto-
motive force required for such a frame is found in the usual way
from the magnetization curve of the material, knowing the area
and the average length of the path between two poles; the length
is estimated from the drawing of the machine. In figuring out the
flux density in a field frame one must not forget that (1) only one-
half of the flux per pole passes through a given cross-section of the
frame (Fig. 20) ; (2) the total flux in the frame and in the poles is
larger than that in the armature by the amount of the leakage flux
between the poles. This leakage is usually estimated in per cent
1 This method is due to E. Arnold. See his Wechselstromtechnik, Vol. 5,
(1909), part 1, p. 48.
2 For details of this method see Hoock and Hellmund, Beitrag zur Berech-
nung des Magnetiziemngsstromes in Induktionsmotoren, Elektrotechnik und
Maschinenbau, Vol. 28 (1910), p. 743.
CHAP. VI] EXCITING AMPERE-TURNS 107
of the useful flux, from one's experience with previously built
machines, or it can be calculated by the methods explained in the
next article. Thus, a leakage factor of 1.20 means that the flux in
the field poles is 20 per cent higher than that in the armature, the
leakage flux constituting 20 per cent of the useful flux. The usual
values of the leakage factor vary between 1.10 and 1.25, depending
upon the proximity of the adjacent poles, the degree of saturation
of the circuit, and the proportions of the machine.
The ampere-turns required for the pole-pieces are calculated in
a similar way, assuming the whole leakage to take place between
the pole-tips, so that the flux density in the pole-waist corresponds
to the total flux; including the leakage flux. In exceptional cases
of highly saturated pole-cores this method may be inadmissible, on
account of too large a margin which it would give as compared to
the ampere-turns actually required. In such cases part of the
leakage may be assumed to be concentrated between some two
corresponding points on the waists of two adjacent poles, or it may
be assumed to be actually distributed between the two pole-waists.
See probs. 9 and 10 in chapter II.
In some machines the joint between the pole and the frame
offers a perceptible reluctance, like the joints in the transformer
cores discussed in Art. 33. Some designers allow a certain
fraction of a millimeter of air-gap to account for this reluctance,
and add the number of ampere-turns required to maintain the
flux in this air-gap to those for the pole-piece. The length of
this equivalent air-gap is found by checking back no-load
saturation curves obtained from experiment. As a usual rule,
it is advisable to increase the total calculated ampere-turns of the
magnetic circuit by about 5 to 10 per cent. This increase covers
such minor points as the reluctance of the joints, omitted .in
calculations, as well as certain inaccurate assumptions; it also
covers a possible discrepancy between the assumed and the
actual permeability of the iron. With a liberally proportioned
field winding and a proper regulating rheostat a designer can
rest assured that the required voltage will be obtained, though
possibly at a somewhat different value of the field current than
the estimated one.
Prob. 5. The stator core of a six-pole induction motor has the
following dimensions: bore 112 cm.; outside diameter 145 cm.; gross
length 55 cm. ; the slots are 2 cm. X4.5 cm. ; the machine is provided with
108 THE MAGNETIC CIRCUIT [ART. 40
8 ventilating ducts 9 mm. wide each. What is the maximum m.m.f.
required for the stator core per pole if the flux per pole is 0.15 weber?
Ans. 190 using Arnold's method.
Prob. 6. Draw a curve between the average H and maximum B
in the core, assuming a sinusoidal distribution of the flux density in the
tangential direction, for the carbon steel laminations in Fig. 3.
Ans. Have = 26.5 for Bmax = 18.
Prob. 7. The cross-section of the cast-iron field yoke of a direct-
current machine is 370 sq.cm.; the mean length of path in it between
two consecutive poles is 85 cm. The length of the lines of force in each
pole-waist is 21 cm.; its cross-section 420 sq.cm. The poles are made
of steel laminations 4 mm. thick, so that the space lost between the
laminations is negligible. The reluctance of the joint between a bolted
pole and the yoke is estimated to be equivalent to 0.1 mm. of air. What
is the required number of ampere-turns for the pole-piece and the yoke,
per pole, when the useful flux of the machine is 5 megalines per pole?
The leakage factor is estimated to be equal to 1.20.
Ans. About 930.
40. Magnetic Leakage between Field Poles. It is of impor-
tance in modern highly saturated machines to know accurately the
leakage flux between the poles, in order to estimate correctly the
ampere-turns required for the field poles and the frame of the
machine. Moreover, the design of the poles can be improved,
knowing exactly where the principal leakage occurs and how it
depends upon the proportions of the machine. The value of the
leakage factor also affects the voltage regulation of the machine,
because at full load the m.m.f. between the pole-tips has to be
larger than at no-load, on account of the armature reaction.
For new machines of usual proportions the value of the leakage
factor can be estimated from tests made upon similar machines.
But in new machines of unusual proportions the designer has to
rely upon his judgment, assisted if necessary by crude compara-
tive computations of the permeance between adjacent poles. In
this and in the next article some examples of such computations
are given, not so much in order to give a definite method to be fol-
lowed in all cases, as to show the student a possible procedure and
to train his judgment in estimating the permeance of an irregular
path.
Four principal paths of leakage can be distinguished between
two adjacent poles (Fig. 29) : (a) between the sides of the pole
shoes which face each other; (6) between the sides of the pole
cores (waists) parallel to the shaft of the machine; (c) between the
CHAP. VI]
EXCITING AMPERE-TURNS
109
flanks or sides of the pole shoes perpendicular to the shaft; (d)
between the flanks of the pole cores. In the calculations which
follow, the permeances are computed between a pole and the planes
of symmetry, MN, between the two poles, the permeance of the
other half of each path being the same. All these leakage paths
are in parallel with respect to the pole, so that the total leakage
(c)
FIG. 29.— The leakage flux between field poles.
permeance is equal to their sum. Knowing this total permeance
and the m.m.f. between the pole and the plane MN the leakage
flux is found, and knowing this flux and the useful flux per pole
the leakage factor is easily calculated.
We shall now estimate the permeances of each of the four above
mentioned paths of leakage.
(a) Between the adjacent pole-tips. Estimate the average
cross-section A of the path, in square centimeters, and the aver-
110 THE MAGNETIC CIRCUIT [ART. 40
age length I between the pole-tip and the plane MN, being guided
by Fig. 29. (Do not encroach upon the fringe to the armature.)
Then the permeance of the path is, in perms,
. . .... . (56)
If a greater accuracy is desired, subdivide the total path into
smaller paths in series and in parallel, and calculate the permeance
or the reluctance of each separately. Then the total permeance is
found according to the well-known law of combination of reluc-
tances and permeances in series and in parallel (Art. 9). When
mapping out the lines of force in the air, begin them nearly at right
angles to the surface of the pole (see Art. 41 a below) and draw
them so as to make the total permeance of the path a maximum,
that is, reducing as far as possible the length and increasing the
cross-section of each elementary tube of flux. The medium may
be said to be in a state of tension along the lines of force, and of
compression at right angles to their direction, by virtue of the
energy stored in the field. Hence, there is a tendency for the tubes
of force to contract along their length and expand across their
width.
(b) Between the opposite pole-cores. In this part of the leakage
field each elementary concentric path is subjected to a different
m.m.f., that between the roots of the poles being practically zero,
while the m.m.f. between the points p and p' is equal to that
between the pole-tips. In most cases it is permissible to consider
the whole leakage flux as if passing through the whole length of the
pole core, and then crossing to the adjacent poles at the pole-tips.
Therefore, it is convenient to add the permeance between the pole
cores to that between the pole-tips. But the average m.m.f.
between the waists is only about one-half of that between the tips,
so that the equivalent permeance between the pole cores, reduced
to the total m.m.f., is equal to one-half of the actual permeance.
If the actual permeance, calculated according to formula (56) is (P,
the effective permeance is \(P. The average length and cross-
section of the path are easily estimated from the drawing of the
machine.
(c) Between the flanks of the pole shoes. The path extends indefi-
nitely outside the machine, and the lines of force are twisted
curves, so that it is difficult to estimate the permeance graphically.
As a rough estimate, this permeance can be reduced to that of the
CHAP. VI]
EXCITING AMPERE-TURNS
111
path in the air between two rectangular poles of an electromagnet
(Fig. 30). Assume the paths of the flux to consist of concentric
quadrants with the centers at c and c', joined by parallel straight
lines, and let the width of the poles in the direction perpendicular
to the plane of the paper be h. Then the permeance of an infin-
itesimal layer of thickness dx, between one of the poles and the
plane MN of symmetry, is
d(P
Integrating this expression between the limits o and b we find
(P= IMh log (1.57&/Z + 1) perms . . . . (57)
(compare with prob. 24 in Chapter V, Art. 37).
In applying this formula and Fig. 30 to the case of the flank
leakage between the pole shoes, h is the average radial height of
the pole shoe, 6 is equal to one-
half the width of the pole shoe,
and 21 is the distance between
the two opposing pole-tips.
While the method evidently
gives only a crude approxi-
mation to the actual perme-
ance, formula (57) at least
fixes a lower limit to the per-
meance in question.
(d) Between the flanks of
the pole cores. The conditions
are similar to those under (c),
so that the permeance is esti-
mated again on the basis of
formula (57) . The sides of the
two rectangles in Fig. 29 are
not parallel to each other as in
Fig. 30, but this difference is
taken into account by mentally
turning^ them into a parallel
position, and estimating the equivalent distance 21 between the
edges of the opposing poles. The dimension h is in this case the
radial height of the pole-waist, and b is one-half of the width of
the pole-waist. The flank leakage is smaller than that between
FIG. 30.— The magnetic path between
the poles of an electromagnet.
112 THE MAGNETIC CIRCUIT [ART. 40
the opposite side, so that one may be satisfied with a lesser degree
of accuracy. The equivalent permeance, reduced to that between
the pole-tips, is again equal to one-half the actual permeance, for
the same reason as under (6) above.
The total leakage permeance between a pole and the two planes
of symmetry is equal to the sum of the permeances calculated as
above. In summing them up it will be seen from Fig. 29 that the
permeances (a) and (b) must be taken twice, and also that (c) and
(d) must be taken four times. The leakage flux per pole is
obtained by multiplying the total leakage permeance by the m.m.f .
between the pole-tip and the plane of symmetry. This m.m.f. is
equal to that required to establish the useful flux, along the path
qrs, through the air-gap and the armature of the machine, and
consequently it is known before the pole-piece and the field wind-
ing are computed in detail. Knowing the leakage flux and the
useful flux, the leakage factor is figured out according to the defi-
nition given above.
When calculating permeances as indicated above, one is advised
to make liberal estimates of the same, for two reasons : In the first
place, the true permeance of a path is always the largest possible, so
that, whatever assumptions one makes, the calculated permeance
comes out smaller than the actual. In the second place, in design-
ing a new machine it is better to be on the safe side and rather
underestimate than overestimate the excellence of the perform-
ance. Some writers give more elaborate rules and formulae for the
calculation of the leakage permeance which are useful in the
design of machines of special importance.1
The leakage factor remains practically constant as long as the
flux density in the armature core and teeth is moderate, so that
the reluctance of the useful path qrs is nearly constant. This is
because the reluctance of the leakage paths is constant, and, if the
reluctance of the useful path is also constant, the useful flux and
the leakage flux increase in the same proportion when the m.m.f.
between the pole-tips is increased. When the armature iron is
approaching saturation, the leakage factor increases with the field
1 For a more detailed treatment of the leakage between poles see the
following works: E. Arnold, Die Gleichstrommaschine, Vol. 1 (1906), pp.
284-294; Hawkins and Wallis, The Dynamo, Vol. 1 (1909), pp. 469-484;
Pichelmayer, Dynamobau (19Q8), pp. 127-131; Cramp, Continuous-Current
Machine Design (1910), pp. 42-47 and 226-230.
CHAP. VI] EXCITING AMPERE-TURNS 113
current, because the leakage flux increases the more rapidly than
the useful flux. This increase is partly offset by the fact that the
pole-tips also become gradually saturated by the leakage flux, so
that the leakage factor does not increase as rapidly as it would
otherwise. The practical point to be observed is, that for the
higher flux densities, if accuracy is desired, the leakage should
be estimated separately for a few points on the no-load saturation
curve.
For a given terminal voltage, the leakage factor of a machine is
somewhat higher at full-load than at no-load, because the required
m.m.f . between the pole-faces is higher, due to the armature reac-
tion and to the voltage drop in the armature. In comparatively
rare cases, when the armature reaction assists the field m.m.f., for
instance, in the case of an alternator supplying a leading current,
the leakage factor decreases with the increasing load. The fol-
lowing example illustrates the influence of the load upon the value
of the leakage factor.
Let the useful flux per pole in an alternator, at the rated
voltage and at no-load, be 5 megalines, and let 6000 amp. -turns
per pole be required for the air-gap and the armature core. Let
the permeance of the leakage paths between a pole and the neutral
planes be 120 perms, so that the leakage flux is 0.72 megaline, and
the leakage factor is (5.00 +0.72)/5.00= 1.14. Let a useful flux
of 5.5 megalines be required at the same voltage and at full load,
an increase of 10 per cent being necessary to compensate for the
internal drop of voltage due to the armature impedance. If the
teeth and the armature core were not saturated at all, an m.m.f.
of 6600 amp.-turns would be required. In reality, the m.m.f. is
higher, say 7500 amp.-turns. Let the armature reaction be equal
to 1500 demagnetizing ampere-turns per pole. To compensate
for its action, 1500 additional ampere-turns are required on each
field coil. Thus, the difference of magnetic potential between
a pole-tip and the adjacent plane of symmetry MN (Fig. 29)
is now 9000 amp.-turns, and the leakage flux is increased to
1.08 megalines. Therefore, the leakage factor at full load is
(5.50 + 1. 08) /5.50= 1.20. Similar relations hold for the direct-
current- machines.
In calculating the performance of a synchronous or a direct-
current machine one has to use the relation between the field cur-
rent and the voltage induced in the armature. Ordinarily, the
114 THE MAGNETIC CIRCUIT [ART. 40
no-load saturation curve is used for this purpose, assuming that
the leakage factor is the same at full load as at no load. However
careful designers sometimes plot a separate curve, using a higher
leakage factor, for use at full load.
Prob. 8. Assume in the illustrative example given in the text the
armature current to be leading, so that the voltage drop in the armature
is negative and the armature reaction strengthens the field. Show that
with the same value of the armature current the leakage factor is about
1.09.
Prob. 9. Draw rough sketches of the magnetic circuits of two
machines, one possessing such proportions, number and shape of poles
as to give a particularly low leakage factor, the other markedly deficient
in this respect.
Prob. 10. Calculate the leakage factor and the leakage permeance
per pole of a six-pole turbo-alternator of the following dimensions; the
bore is 1.2 m.; the axial length of the poles 0.6 m.; minimum air-gap
1 cm.; maximum air-gap 2 cm.; total height of the pole 23 cm.; the
height of the pole-waist 18 cm.; the breadth across the pole-waist 25 cm.;
that across the pole-tips 36 cm. The reluctance of the useful path in
the air-gap and in the armature is estimated to be about 0.57 millirel
per pole. Ans. 1.115; about 200 perms.
Prob. 11. The leakage factor of the machine specified in the pre-
ceding problem was found from an experiment to be 1.13, at no-load,
when the total flux per pole was 20.35 megalines. What is the true
leakage permeance if 20 kiloampere-turns were required at that flux for
the air-gaps and the armature, per pair of poles ? Ans. 234 perms.
Prob. 12. The machine specified in the two foregoing problems
requires at full load 20 per cent more ampere-turns for the air-gap and
armature, on account of the induced voltage being 12 per cent higher
than at no-load. The armature reaction amounts to 4000 demagnetizing
ampere-turns per pole. What is the leakage factor at full load, according
to the calculated leakage permeance and according to that obtained
from the test? Ans. 1.16; 1.19.
Prob. 13. A closed electric circuit consisting of a battery and of
a bare conductor is immersed in a slightly conducting liquid, so that part
of the current flows through the liquid. Indicate the common points
and the difference between this arrangement and a magnetic circuit
with leakage. Using the electrical analogy, show that armature reaction
increases the leakage factor; also explain the fact that, in order to com-
pensate for the action of M demagnetizing ampere-turns on the armature,
more than M additional ampere-turns are required on the pole-pieces.
Prob. 14. In some books the permeance between two pole-faces
(Fig. 30) is calculated by assuming the lines of force to be concentric
semicircles as shown by the dotted lines. Show that such a permeance
is smaller than that according to formula (57) and therefore should not
be used. Hint: Compare the lengths of two corresponding lines of
force.
CHAP. 'VI]
EXCITING AMPERE-TURNS
115
Prob. 15. Let AB and CD (Fig. 31) represent the cross-sections
of two opposite pole-faces of an electromagnet, inclined at an angle 26
to one another. Show that of the three assumptions with regard to the
shape of the lines of force in the air between the poles (a) is more correct
than (6) and (6) is more correct than (c) ; in other words, the assumption
(a) gives a higher permeance than (6) or (c). Hint: tan 0 > 0 > sin 0.
Prob. 16. Show that the permeance according to Fig. 31a, between
one of the faces and the plane MN of symmetry, is equal to
(ph/fyLn(9w/l+l),
where h is the width of the pole-faces perpendicular to the plane of the
paper, and that formula (56) and (57) are special cases of it.
Prob. 17. The formula given in the preceding problem is deduced
under the assumption that the same m.m.f. is acting on all the lines of
FIG. 31. — The magnetic paths between the poles of an electromagnet (three
assumptions).
force. Let now Fig. 31a represent a cross-section of two opposing pole
cores in an electric machine, the m.m.f. between A and C being zero,
and uniformly increasing to a value M between the points B and D.
Show that the equivalent permeance of the path, referred to the m.m.f.
M is equal to (tdi/d}[l-(l/6w)Ln(dw/l + l}].
NOTE. If it is desired to use regularly the foregoing formula in esti-
mating the leakage factor, the values of the expression in the brackets
[ ] can be plotted as a curve for the values of (l/Ow) as abscissae. Similar
formulae can be deduced and curves plotted for the permeance of the
flank leakage between adjacent poles. The paths of the lines of force
over the poles can be assumed to be concentric quadrants and between
the poles to have a shape similar to that indicated in Fig. 3 la.
41. The Permeance and Reluctance of Irregular Paths. In
using the methods described above for the calculation of the.
ampere-turns for the air-gap, the teeth, and the cores, and in esti-
116 THE MAGNETIC CIRCUIT [ART. 41
mating the leakage factor, the reader has seen the difficulties
involved in the computation of the permeance of an irregular path.
In the parts of a magnetic field not occupied by the exciting
windings, the general principle applies that the lines of force and
the equipotential surfaces assume such shapes and directions that
the total permeance' becomes a maximum, or the reluctance a
minimum. When this condition is fulfilled, the energy of the
magnetic field becomes a maximum, as is explained in Art. 57.
When the field needs to be considered in two dimensions only,
that is, in the case where we have long cylindrical surfaces the
properties of conjugate functions can be used for determining the
equations of the lines of force and of the equipotential surfaces;
see the references in Art. 37 above. However, the purely mathe-
matical difficulties of the method are such as to make the analytical
calculation of permeances feasible in the simplest cases only.
In most practical cases, especially in three-dimensional prob-
lems, recourse must be had to the graphical method of trial and
approximation, in order to obtain the maximum permeance.
The field is mapped out into small cells by means of lines of force
and equipotential surfaces, drawing them to the best of one's
judgment; the total permeance is calculated by properly com-
bining the permeances of the cells in series and in parallel. Then
the assumed directions are somewhat modified, and the permeance
is calculated again, etc., until by successive trials the positions of
the lines of force are found with which the permeance becomes a
maximum.
The work of trials is made more systematic by following a pro-
cedure suggested by Lord Rayleigh. Imagine infinitely thin sheets
of a material of infinite permeability to be interposed at intervals
into the field under consideration, in positions approximately
coinciding with the equipotential surfaces. If these sheets exactly
coincided with some actual equipotential surfaces, the total
permeance of the paths would not be changed, there being no
tendency for the flux to pass along the equipotential surfaces. In
any other position of the infinitely conducting sheets, the total
permeance of the field is increased, because through these sheets
the flux densities become more uniformly distributed. Moreover,
these sheets become new equipotential surfaces of the system,
because no m.m.f . is required to establish a flux along a path of
infinite permeance. Thus, by drawing in the given field a system
CHAP. VI] EXCITING AMPERE-TURNS 117
of surfaces approximately in the directions of the true equipo-
tential surfaces, and assuming these arbitrary surfaces to be the
true equipotential surfaces, the true reluctance of the path is
reduced. In other words, by calculating the reluctances of the
laminae between the " incorrect " equipotential surfaces and adding
these reluctances in series, one obtains a reluctance which is lower
than the true reluctance of the path. This gives a lower limit for
the required reluctance (or an upper limit for the permeance) of
the path.
Imagine now the various tubes of force of the original field
wrapped up in infinitely thin sheets of a material of zero permeabil-
ity. This does not change the reluctance of the paths, because
there are no paths between the tubes. But if these wrappings are
not exactly in the direction of the lines of force, the reluctance of
the field is increased, because the densities become less uniform,
the non-permeable wrappings forcing the lines of force from their
natural positions. Thus, by drawing in a given field a system of
surfaces approximately in the directions of the lines of force, cal-
culating the reluctances of the individual tubes, and adding them
in parallel, a reluctance is obtained which is higher than the true
reluctance of the path. This gives an upper limit for the reluc-
tance (or a lower limit for the permeance) of the path under
consideration.
Therefore, the practical procedure is as follows: Divide the
field to the best of your judgment into cells, by equipotential
surfaces and by tubes of force, and calculate the reluctance of the
field in two ways: first, by adding the cells in parallel and the
resultant laminae in series; secondly, by adding the cells in series
and the resultant tubes in parallel. The first result is lower than
the second. Readjust the position of the lines of force and of the
equipotential surfaces until the two results are sufficiently close to
one another; an average of the two last results gives the true
reluctance of the field.
One difficulty in actually following out the foregoing method
is that the changes in the assumed directions of the field that will
give the best result are not always obvious. Dr. Th. Lehmann has
introduced an improvement which greatly facilitates the laying out
of a field.1 We shall explain this method in application to a two-
1 " Graphische Methode zur Bestiminung des Kraftlinienverlaufes in der
Luft"; Ekktrotechnische Zeitschrift, Vol. 30 (1909), p. 995.
118 THE MAGNETIC CIRCUIT [ART. 41
dimensional field, though theoretically it is applicable to three-
dimensional problems also. According to Lehmann, lines of force
and level surfaces are drawn at such distances that they enclose
cells of equal reluctance. Consider a slice, or a cell, in a two-
dimensional field, v centimeters thick in the third dimension
(where v =!//*), and of such a form that the average length Z of
the cell in the direction of the lines of force is equal to its average
width w in the perpendicular direction. The reluctance of such a
cell is always equal to one rel, no matter whether the cell itself is
large or small. This follows from the fundamental formula for
the reluctance, which in this case becomes (R = vl/(vXw) = 1.
The judgment of the eye helps to arrange cells of a width equal
to the length, in the proper position with respect to each other and
to the adjoining iron; the next approximation is apparent from
the diagram, by inspecting the lack of equality in the average
width and length of the cells. Lord Rayleigh's condition is
secured by this means, since the combination of cells of equal
reluctance leads to but one result, whether they are combined
first in parallel or first in series. After a few trials the space is
properly ruled, and it simply remains to count the number of cells
in series and in parallel. Dr. Lehmann shows a few applications
of his method to practical cases of electrical machinery, and the
reader is referred to the original article for further details.
The foregoing methods apply only to the regions outside the
exciting current, because only in such parts of the field the maxi-
mum permeance corresponds to the maximum stored electromag-
netic energy. Within the space occupied by the exciting windings
the condition for the maximum of energy is different (see Art. 57),
and is of a form which hardly permits of the convenient application
of a graphical method. However, in most practical cases the
directions of the lines of force within the exciting windings are
approximately known a priori: or else, the windings themselves
can be assumed, for the purposes of computation, to be concentrated
within a very small space. For instance, the field winding can be
assumed to consist of an infinitely thin layer close to the pole-waist.
Then the condition that the permeance is a maximum is fulfilled in
practically the whole field, and the field is mapped out on this basis.
Prob. 18. Sketch the field between the armature and a pole-piece
or some proportion of tooth, slot, and air-gap and determine the lower
and upper limits of the reluctance by Lord Rayleigh's method.
CHAP. VI]
EXCITING AMPERE-TURNS
119
Prob. 19. For some ratio of slot width to air-gap draw the tooth-
fringe field to the perpendicular surface of the pole, adjust the number
and spacing of the lines of force by Dr. Lehmann's method, and see how
closely you can check the corresponding point on Carter's curve (Fig. 26).
Prob. 20. From the given drawing of a machine, determine the
permeance of the fringe from the pole-tip to the armature by Lehmann's
method; consult, if necessary, Dr. Lehmann's original article.
Prob. 21. Map out the leakage field between the opposing pole-tips
and cores of a given machine, and determine its equivalent permeance by
Lehmann's method, assuming the field coils to be thin and close to the core.
41a. The Law of Flux Refraction. When mapping out a field
in air, the lines of force must be drawn so as to enter the adjoining
iron almost normally to its surface, even if they are continued in
the iron almost parallel to its surface. The lines of force change
their direction at the dividing surface suddenly (Fig. 32), and in
so doing they obey the so-called laio of flux refraction; namely,
tan di
(58)
Equipotential \
surfaces *\
Iron
Air
Since /za is many times smaller than /*;, the angle 6a is usually very
small, unless Oi is very nearly 90 degrees. It may be said in gen-
eral that the lower the permeability
of a medium the nearer the lines
of force are to the normal at its
limiting surfaces. In .this way, the
path between two given points is
shortened in the medium of lower
permeability and is lengthened in
the medium of higher permeability.
Thus, the total permeance of the
circuit is made a maximum.
To deduce' the above-stated law
of refraction, consider a tube of
flux between the equipotential sur-
faces ab and cd, the width of the
path in the direction perpendicular
to the plane of the paper being one
centimeter. Let Bi and Ba be the
flux densities, and Hi and Ha the
corresponding magnetic intensities in the two media. Two
conditions must be satisfied, namely, first, the drop of m.m.f.
FIG. 32.— The refraction of
a flux.
120 THE MAGNETIC CIRCUIT [ART. 41
along ac is the same as that along bd, and secondly, the total
flux through cd is equal to that through ab. Or
and
Dividing one equation by the other, and rearranging the terms,
eq. (58) is obtained.
Prob. 22. Show that the total refraction which is in some cases
experienced by rays of light is impossible in the case of magnetic lines
of force.
Prob. 23. Part of a flux emerges from the flank of a tooth into the
slot at an angle of 1° to the normal. What is the angle which the lines
of force make with the side of the slot in the iron, assuming the relative
permeability of the iron to be 1000? Ans. 90° - 0; = 3° IT.
CHAPTER VII
THE MAGNETOMOTIVE FORCE OF DISTRIBUTED
WINDINGS
42. The M.M.F. of a Direct-current or Single-phase Dis-
tributed Winding. In the two preceding chapters it is shown how
to calculate the ampere-turns required for a given flux in an elec-
tric machine. When the exciting winding is concentrated, that is,
when all the turns per pole embrace the whole flux, the number of
ampere-turns is equal to the product of the actual amperes flowing
through the winding times the number of turns. Such is the case
hi a transformer, in a direct-current machine, and in a synchronous
machine with salient poles. In some cases, however, the exciting
windings are distributed along the air-gap, so that only a part of the
flux is linked with all the turns, and the actual ampere-turns have
to be multiplied by a factor in order to obtain the effective m.m.f .
Such is the case in an induction motor, and in an alternator with
non-salient poles. Moreover, one has to consider the m.m.f. of
distributed armature windings when calculating the performance
of a machine under load, because the armature currents modify the
no-load flux. In this chapter the m.m.fs. of distributed windings
are treated mainly in application to the performance of the induc-
tion motor; in particular, to the calculation of the no-load current
and the reaction of the secondary currents. The armature reaction
in synchronous and hi direct-current machines is analyzed in the
next two chapters.
Distributed Winding for Alternator Field. A cross-section of a
four-pole field structure with non-salient poles for a turbo-alterna-
tor is shown in Fig. 33a. The flux is graded (Fig. 336) in spite of
a constant air-gap, because the total ampere-turns act only upon
the part a of a pole ; two-thirds of the ampere-turns act upon the
parts 6, b and one-third upon the parts c, c. The m.m.f. and the
flux in the parts d, d are equal to zero. Thus, theoretically, the
flux density in the air-gap should vary according to a " stepped "
121
122
THE MAGNETIC CIRCUIT
[ART. 42
FIG. 33a. A four-pole revolving fieldstructure with non-salient poles.
curve (Fig. 336) ; in reality, the corners are smoothed out by the
fringes. The total number of ampere-turns per pole must be such
as to create the assumed
maximum flux density
Bmax in the air-gap
under the middle part
of the pole. The slots
are placed with due re-
gard to the mechanical
strength of the struc-
-Pole-pitch-
ture, and so as to get a
flux-density distribu-
tion approaching a sine
wave. The middle part of the curve is left flat, because very little
FIG. 336. The flux-density distribution for
the field shown in Fig. 33a.
CHAP. VII] M.M.F. OF DISTRIBUTED WINDINGS
123
flux would be gained by placing a narrow coil near the center of
the pole, at a considerable expense in copper, and in power loss for
excitation. The total flux, which is proportional to the area of the
curve in Fig. 336, must be of the magnitude required by eq. (31)
for the induced e.m.f . If greater accuracy is desired, the curve in
Fig. 336 is resolved into its fundamental sine wave and higher
harmonics; the area of the fundamental curve must then give
the flux 0 which enters into eq. (31).
Single-phase Distributed Winding. Let us consider now the
stator winding of an induction motor, and in particular the m.m.f .
created by the current in one phase. We begin with the simplest
case of a winding placed in one slot per pole per phase (Fig. 34).
The reluctances of the
stator core and of the
rotor core are small as
compared with that of
the air-gap and the
teeth, and are taken
into account by in-
creasing the reluctance
of the active layer of
the machine (air-gap
and teeth). If P and
Q are the centers of
the slots in which the
opposite sides of a coil
are placed, the m.m.f.
distribution along the air-gap is that shown by the broken line
RPP'Q'QS. In other words, the m.m.f. across the active layer,
at any instant, is constant over a pole pitch, and is alternately
positive and negative under consecutive poles.
Let n be the number of turns per pole, and i the instanta-
neous current ; then the height PP' of the rectangle is equal to ni.
It is understood of course that such an m.m.f. acting alone does
not produce the sinusoidal distribution of the flux density assumed
in the previous chapters: In a single-phase motor the sinusoidal
distribution is due to the simultaneous action of the stator and
rotor currents, and also to the fact that the windings are distrib-
uted in several slots per pole. In a polyphase machine the simul-
taneous action of the two or three phases also helps to secure a
FIG. 34. — The m.m.f. of a single-phase unislot
winding resolved into its harmonics.
124 THE MAGNETIC CIRCUIT [ART. 42
sinusoidal distribution. As long as the coil PQ acts alone, the
m.m.f. has a " rectangular " distribution in space, and, if the cur-
rent in the coil varies with the time according to the sine law, the
height of the rectangle, or the m.m.f. across the active layer, also
varies according to the sine law. In what follows it is important
to distinguish between variations of the m.m.fs. in space, i.e., along
the air-gap, and those occurring in time, as the current in a winding
varies.
For the purposes of analysis the rectangular distribution of the
m.m.f. can be replaced by an infinite number of sinusoidal distribu-
tions (Fig. 34), according to Fourier's series.1 The advantages of
such a development over the orginal rectangle PP'Q'Q are as fol-
lows:
(a) The sine wave is a familiar standard by which all other
shapes of periodic curves are judged.
(6) When adding the m.m.fs. due to the coils in different slots,
or belonging to different phases, it is much more convenient to add
sine waves than to add rectangles displaced in space and varying
with the time.
(c) In the actual operation of an induction motor or generator
the higher harmonics in the m.m.f. wave are to a considerable
extent wiped out by the corresponding currents in the rotor, so that
the rectangular distribution is actually changed to a nearly sinu-
soidal one (see Art. 45 below).
Let h be the height of the rectangle; we assume that for all the
points along the air-gap the sum of the ordinates of all the sine
waves is equal to h; or
Ti = Ai sin x+A3 sin 8^ + ^.5 sin 5z+etc. . . (59)
Here x is the angle in electrical degrees, counted along the air-gap,
and AI, A 3, A 5, . . . are the amplitudes of the waves, to be deter-
mined as functions of h. No cosine harmonies enter into this for-
mula, because the m.m.f. distribution is symmetrical with respect
to the center line 00' of the exciting coil. To determine the ampli-
tude of the nth harmonic An, multiply both sides of eq. (59) by
sin nx d(nx), and integrate both sides between the limits 2=0 and
1 For the general method of expanding a periodic function into a series of
sines and cosines, see the author's Experimental Electrical Engineering, Vol. 2,
pp. 222 to 227.
CHAP. VII] M.M.F. OF DISTRIBUTED WINDINGS 125
x = x. All the terms on the right-hand side vanish, except the one
containing sin2 nx, and we have
from which
An=4h/(ra) ....... . (60)
Thus, the required series is
h = 4/0r (sin x + J sin 3x + £ sin 5z + etc.) . . (61)
This means that the amplitude of the fundamental wave is 4/7:
times larger than the height h of the original rectangle; the ampli-
tude of the third harmonic is equal to one-third of that of the fun-
damental wave ; the amplitude of the fifth harmonic is one-fifth of
that of the fundamental wave, etc. In practical applications the
fundamental wave is usually all we desire to follow, but in some
special cases a few of the harmonics are important.1
Let now the winding of a phase be distributed in S slots per
pole (Figs. 15 and 16), the distance between the adjacent slots
being a electrical degrees. The conductors in every pair of slots
distant by a pole pitch produce a rectangular distribution, of the
m.m.f . like the one shown in Fig. 34, or, what is the same, an equiv-
alent series of sine-wave distributions. The m.m.fs. produced
by the different coils are superimposed, and, since a sum of sine
waves having equal bases is also a sine wave, the resultant m.m.f.
also consists of a fundamental sine wave and of higher harmonics.
The fundamental waves of the m.m.fs. of the several coils are dis-
placed by an angle of a electrical degrees with respect to one
another, so that the amplitude of the resultant wave is not quite
S times larger than that of each component wave. The reduction
coefficient, or the slot factor, k8, is the same as that for the induced
e.m.f. (Art. 28), because in both cases we have an addition of sine
waves displaced by a electrical degrees, (see also prob. 20 in Art.
1 This method of treating the m.m.fs. of distributed windings by resolv-
ing the rectangular curve into its higher harmonics is due to A. Blond el.
See his article entitled " Quelques proprietes gene"rales des champs magne"-
tiques tournants," L'Eclairage Electrique, Vol. 4 (1895), p. 248. Some authors
consider the actual " stepped " curves of the m.m.f. or flux distribution, a
procedure rather cumbersome, and in the end less accurate, in view of the
fact that the higher harmonics are to a considerable extent wiped out by the
currents in the rotor.
126 THE MAGNETIC CIRCUIT [ART. 42
28.) For the same reason, the value of the winding-pitch factor,
kw, deduced in Art. 29, holds for the m.m.fs. as well as for the
induced e.m.fs.
When adding the waves of the higher harmonics due to several
coils, one must remember that an angle of a electrical degrees for
the fundamental wave is equivalent to 3a electrical degrees for the
third harmonic, 5c* for the fifth harmonic, etc. Therefore, when
using the formula (29) and Fig. 19, different values of a and of per
cent pitch must be used for each harmonic, and in this connection
the reader is advised to review Art. 30. In the practical problems
given below the higher harmonics of the armature m.m.f. are dis-
regarded altogether. The results so obtained are in a sufficient
agreement with the results of experiments to warrant the great
simplification so achieved. For the completeness of the treatment,
and as an application of the general method, an analysis of the
effect of the higher harmonics of an m.m.f. is given in Art. 45
below. However, this article may be omitted, if desired, without
impairing the continuity of the treatment in the rest of the book.
Resolution of a Pulsating m.m.f. into Two Gliding m.m.fs. The
reader is aware from elementary study that the pulsating m.m.fs.
produced by two or three phases combine into one gliding (revolv-
ing) m.m.f. in the air-gap. It is therefore convenient to consider
even a single-phase pulsating m.m.f. as a combination of m.m.fs.
gliding along the air-gap in opposite directions. In this wise, the
m.m.fs. due to different phases are later combined in a simple
manner. This method of treatment is similar to that used in
mechanics, when an oscillatory motion is resolved into two rotary
motions in opposite directions. Also in the analysis of polarized
light a similar method of treatment is used.
Take the first harmonic of the m.m.f. (Fig. 34) and assume the
current in the exciting coil to vary with the time according to the
sine law; then the amplitude of the m.m.f. wave also varies with
the time according to the sine law. Imagine two m.m.f. waves, of
half the maximum amplitude of the pulsating wave, gliding uni-
formly along the air-gap in opposite directions; the superposition
of these waves gives the original pulsating wave. One can see
this by drawing such waves on two pieces of transparent paper and
placing them in various positions over a sketch showing the pul-
sating wave. It will be found that the sum of the corresponding
ordinates of the revolving waves gives the ordinate of the pulsating
CHAP. VII] M.M.F. OF DISTRIBUTED WINDINGS 127
wave at the same point. Or else, represent the two gliding waves
by two vectors of equal magnitude M, revolving in opposite direc-
tions. The resultant vector is a pulsating one in a constant direc-
tion, and varies harmonically between the values ±2M.
The analytical proof is as follows: Let the exciting current
reach its maximum at the moment t= 0. Then, if the amplitude of
the m.m.f. wave at this instant is equal to A, the amplitude at any
other instant t is equal to A cos 2nft. Therefore, the m.m.f. cor-
responding to a point distant x from P and at a time t is equal to
A cos 2xft sin x. By a familiar trigonometrical transformation
we have
A sin z cos 2rft=%A sin (x+2xft) +%A sin (x -2nft). (62)
The right-hand side of this equation represents two sine waves, of
the amplitude %A, gliding synchronously along the air-gap, that is,
covering one pole pitch during each alternation of the current.
The wave %A sin (x+2nft) glides to the left, because, with increas-
ing t, the value of x must be reduced in order to get the same phase
of the m.m.f. wave, that is, to keep the value of (x+2nft) constant.
The other wave glides to the right, because, with increasing t, the
value of x must be increased in order to obtain any constant value
of (x—lnft). A similar resolution into two gliding waves can be
made for each higher harmonic of the pulsating m.m.f. wave; the
higher the order of a harmonic the lower the linear speed of its two
gliding wave components.
In practice it is usually 'required to know the relationship
between the effective value i of the magnetizing current, the num-
ber of turns n per pole per phase, and the crest value of one of the
gliding m.m.f. waves. From the preceding explanation this rela-
tionship for the fundamental wave is
0.9A;fem, .... (63)
where M is the amplitude of each of the two gliding m.m.fs., niV2
represents the maximum height h of the original rectangle, and the
factor J is introduced because the amplitude of each gliding wave
is one-half of that of the corresponding pulsatingjwave. The breadth
factor kb is the same as that used for the induced e.m.fs. (Arts. 27 to
29) . Similar expressions can be written for each higher harmonic,
remembering that their amplitudes decrease according to eq. (61),
128 THE MAGNETIC CIRCUIT [ART. 43
and that a different value of fa must be used for each harmonic.
The value of M is calculated so as to produce the required revolv-
ing flux, as is explained in Chapters IV, V, and VI. From eq. (63)
either n or i, or their product can be determined.
Prob. 1. A single-phase four-pole induction motor has 24 stator slots,
two-thirds of which are occupied by the winding; there are 18 con-
ductors per slot. The average reluctance of the active layer is 0.09 rel.
per square centimeter. What current is necessary to produce a pulsating
flux of such a value that the maximum flux density due to the first
harmonic is 5 kl./sq.cm., when the secondary circuit is open?
Ans. 8.3 amp.
Prob. 2. Show that in the preceding problem the difference between
the actual flux per pole and its fundamental is less than 2 per cent.
Prob. 3. Show that, if in Fig. 34 the angle x is counted from the
crest of the first harmonic, the expansion into the Fourier series is similar
to eq. (61), except that cosines take place of the sines, and the terms
are alternately positive and negative.
43. The M.M.F. of Polyphase Windings. Consider a two-
phase winding of the stator of an induction motor (Fig. 35a) ; let
f 2 Armature i~
(66| _ ftfj [66] JJ61
p->- °2^ slots ^Ol "*-; ; — •
r\ i—
FIG. 35a. — A two-phase winding.
the current in phase 1 lead that in phase 2 by \T, or by 90 electrical
degrees. A little reflection will show that the resultant m.m.f. of
the two phases glides from right to left : Let the current in phase 1
reach its maximum at the instant £=0; at this instant the current
in the coil 2 is zero, and the m.m.f. wave is distributed uniformly
under the coil 1 ; at the instant t= \T the current in phase 1 is zero,
and the m.m.f. is distributed under the coil 2. At intermediate
instants both coils contribute to the resultant m.m.f., so that its
maximum occupies a position intermediate between the centers
Oi and 02 of the coils 1 and 2.
The actual rectangular distribution of the m.m.f. due to each
phase can be replaced by a fundamental sinusoidal one and its
higher harmonics, as in Fig. 34. The pulsating fundamental m.m.f.
of each phase can be replaced by two waves of half the ampli-
tude, gliding synchronously in opposite directions. Let the wave
CHAP. VII] M.M.F. OF DISTRIBUTED WINDINGS 129
due to phase 1, and gliding to the left, be denoted by LI, and that
due to phase 2 by L2. Let the corresponding waves gliding to the
right be denoted by RI and R2. Disregarding the higher har-
monics, the resultant m.m.f. is due to the combined action of the
four gliding waves L1? L2, RI and R2. At the instant t = Q the
crest of the wave LI is at the point 0\; at the instant t=\T the
crest of the wave L2 is at the center 02 of the coil 2. Consequently,
at the instant t=0 the crest of the wave L2 is 90 electrical degrees
to the right of 02, or it is at 0\. Thus, the waves LI and L2 actu-
ally coincide in space, and form one wave of double the amplitude.
The crest of the wave RI is at the point Oi when t=0; the crest
of R2 is at the point 02 when t=%T. Therefore, at t=0 the crest
of R2 is 90 electrical degrees to the left of the point 02, and the
waves RI and R2 travel at a distance of 180 electrical degrees from
each other. But twro such waves cancel each other at all points
and at all moments, so that there is no resultant R wave. Thus
the resultant fundamental wave of m.m.f. in a two-phase machine
is gliding. Its amplitude is twice as large as that of either of the
component gliding m.m.fs. of the two phases, which components
_ ,3 _ Armature 2 .\
|5S] ifrjjj J5S1 |S3| |£fi~ J66] [M] [HT~
R
FIG. 356. — A three-phase winding.
are expressed by eq. (63). If the current in phase 2 were leading
with respect to that in phase 1, the L fluxes would cancel each
other and the resultant flux would travel from left to right.
Consider now a three-phase winding (Fig. 356) and call the
m.m.fs. which glide to the left, and which are due to the separate
phases, by LI, L2, and L3 respectively. Let the waves which
travel to the right be denoted by RI, R2, and R3. Assume the cur-
rent in phase 2 to be lagging by 120 electrical degrees, or by J77,
with respect to that in phase 1, and the current in phase 3 to be
lagging by JT with respect to that in phase 2. By a reasoning
similar to that given for the two-phase winding above it can be
shown that the three L waves coincide in their position in space,
and give one gliding wave of three times the amplitude of each
wave. The three- R waves are relatively displaced by 240 elec-
130 THE MAGNETIC CIRCUIT [ABT. 43
trical degrees, or, what is the same, by 120 electrical degrees;
hence, their m.m.fs. mutually cancel at each point along the
air-gap. This can be proved by .drawing three sine waves dis-
placed by 120 degrees and adding their ordinates point by point;
or else one can replace each wave by a vector, and show that the
sum of the three vectors is zero because they form an equilateral
triangle.
The reasoning given for the two- and three-phase windings can
be extended to any number of symmetrical phases, say m, pro-
vided that the windings are displaced in space by 360/m electrical
degrees, and also provided that the currents in these windings are
displaced in time by 1/mth of a cycle. The gliding fundamental
waves due to each phase which go in one direction are in phase
with each other, and, when added, give a wave m times larger
than that expressed by eq. (63) ; while the fundamental waves
going in the opposite direction are displaced in space by 720/m
electrical degrees, and their combined m.m.f. is zero. The direc-
tion in which the resultant m.m.f. travels is from the leading to
the lagging phases of the winding. Thus, for any symmetrical
m-phase winding
....... (64)
w'here M denotes the amplitude of the fundamental sine wave of
the resultant gliding m.m.f., n is the number of turns per pole per
phase, and i is the effective value of the current in each phase.
Prob. 4. It is desired to build a 60 horse-power, 550-volt, 4-pole,
Y-connected induction motor, using a stator punching with 4 slots per
pole per phase, and a winding pitch of one hundred per cent. The required
maximum m.m.f. per pole is estimated at 1550 ampere-turns. What is
the total required number of stator turns (for all the phases) if the mag-
netizing current must not exceed 25 per cent of the full-load current?
The estimated full-load efficiency is 92 per cent, the power factor at
full load is about 90 per cent? Ans. Not less than 504.
Prob. 5. What is the required number of turns in the preceding
problem, if the stator winding is to be delta-connected and to have a
winding pitch of about 75 per cent? Ans. Not less than 936.
Prob. 6. What is the amplitude of the first harmonic of the total
armaturere action in a 100-kva, 440- volt, 6-pole, two-phase alternator
with non-salient poles? The stator has 72 slots; the coils lie in slots 1
and 9 ; the number of conductors per slot is Cs> In practice, the arma-
ture reaction must not exceed a certain limit, and this helps to determine
the permissible value of Cs Ans. 4800(7S amp. -turns.
CHAP. VII] M.M.F. OF DISTRIBUTED WINDINGS 131
Prob. 7. Plot the actual " stepped " curves of m.m.f. distribution
for a two-phase winding with three slots per pole per phase, for the fol-
lowing instants: 2 = 0, t=^$T, and t = %T. Compare the maximum and
the average m.m.f. of the actual distribution with those of the first
harmonic.1
Prob. 8. Solve the preceding problem for a three-phase winding
with 2 slots per pole per phase, and a winding pitch of £. Take two
instants, t=0 and t=^T, and show that for the instants t=^sTf
etc., the m.m.f. distribution is the same as for t=Q, while for
etc., the m.m.f. distribution is the same as for t = ^T.
Prob. 9. Prove directly that two equal pulsating sine waves of m.m.f.
or flux, displaced by 90 electrical degrees in space and in time relatively
to each other, give a gliding sine wave, the amplitude of which is equal
to that of each pulsating wave. Solution: The left-hand side of eq.
(62) gives the value of the m.m.f. at a point x and at an instant t, due
to phase 1 ; the m.m.f. produced at the same point and at the same instant
by the phase 2 is A sin (x + %K) cos (2xft— %K). Adding the two expressions
gives A sin (x+2xft), which is a left-going wave of amplitude A.
Prob. 10. Prove, as in the preceding problem, that the three pulsating,
sine waves of m.m.f. produced by a three-phase winding, give together
a gliding m.m.f., the amplitude of which is 50 per cent larger than that
of each pulsating wave.
Prob. 11. Prove by the method given in problem 9 above that m
pulsating m.m.f. waves displaced in space and in tune by an electrical
angle In/m produce a gliding m.m.f. the amplitude of which is %m times
larger than that of each pulsating wave. See Arnold, Wechselstrom-
technik, Vol. 3 (1908) p. 302.
44. The M.M.Fs. in a Loaded Induction Machine.2 Eq. (64)
gives the magnetizing current i of an induction motor at no-load,
i.e., when the rotor is running at practically synchronous speed, so
that the secondary currents are negligible. When the motor is
loaded, the useful flux which crosses the air-gap is due to the com-
bined action of the primary and the secondary currents. In com-
mercial motors the flux at full load is but a few per cent below that
at no load, the difference being due to the impedance drop in the
1 Problems 7 and 8 are intended to acquaint the student with the usual
method of calculation of the m.m.fs. of distributed windings and to show the
advantage of Blondel's method used in the text. For numerous stepped
curves and calculations, see Boy de la Tour, The Induction Motor, Chapter IV.
2 The treatment in this article presupposes a general knowledge of the
equivalent performance diagram of induction machines; the purpose of
the article being to deduce the exact numerical relations. This article and
the one following can be omitted without impairing the continuity of treat-
ment in the rest of the text.
132 THE MAGNETIC CIRCUIT [ART. 44
primary winding, the same as in a transformer. Therefore, the
net number of exciting ampere-turns, M, is approximately the
same as at no load. This means that the geometric sum of
the m.m.fs. produced by the primary and the secondary currents at
any load is nearly equal to the m.m.f . due to the primary winding
alone at no load. In this respect the induction motor is similar to
a transformer.
(a) Calculation of the Secondary Current. Knowing the- primary
full-load current, the secondary full-load current can be calcu-
lated from the required counter-m.m.f . ; the procedure can be best
illustrated by an example. In the motor given in prob. 4 above,
the full-load current is estimated at 57 amp. ; taking the direction
of the vector of the applied voltage as the axis of reference, the
full-load current can be represented as 51.3 — /24.8 amp. The
magnetizing current, 0.25X57= 14.25, is practically in quadrature
with the applied voltage, because it is hi quadrature with the
induced counter e.m.f ., the same as in a -transformer. The full-
load current of 57 amp. contains a component which supplies the
iron loss in the stator; we estimate it to be equal to about 1.1 a'mp.
(2 per cent of the input). Thus, the component of the primary
current, the action of which must be compensated by the second-
ary currents, is '(51.3 -/24.8) - (1.1 -/14) = 50.2 -j'10.8 amp., or its
absolute value is 5.14 amp. This is called the current transmitted
into the secondary, or the secondary current reduced to the primary
circuit. This current produces a maximum m.m.f. of 0.9X3 XO. 958
X42X51.4 = 5580 amp.-turns.
Let the rotor be provided with a three-phase winding, with 5
slots per pole per phase, and let the winding pitch be 13/15. The
number of slots is selected so as to be different from that in the sta-
tor, in order to insure a more uniform torque, and to reduce the
fluctuations in the reluctance of the active layer. We have,
according to eq. (64), that 5580 = 0.9X3X0.935 X (ni), from which
m = 2210 amp.-turns. Certain practical considerations, for
instance, the value of the induced secondary voltage, usually
limit the choice of one of these factors; then the other factor
also becomes definite. If, for instance, the rotor is to have 10
conductors per slot, the secondary current will be about 89 amp.
The secondary i2r loss is determined by the desired per cent slip;
knowing the secondary current and the number of turns, the
necessary size of the conductor can easily be calculated.
CHAP. VII] M.M.F. OF DISTRIBUTED WINDINGS 133
Sometimes the secondary winding consists of coils individually
short circuited; this is an intermediate type of winding between
an ordinary squirrel-cage winding and a three-phase winding
such as is used with slip-rings. Let the foregoing motor be
provided with such a winding, of the two-layer type, and let the
rotor have 71 slots, 6 conductors per slot, the coils being placed
in slots 1 and 14. In formula (64) m stands for the number of
symmetrically distributed phases, the current in each phase being
displaced in time by 2x/m with respect to that in the next phase.
In the winding under consideration, each coil represents a phase,
and one has to go over a pair of poles until one finds the next coil
with the current in the same phase. Thus, in this case, the num-
ber of secondary phases is equal to the number of slots per pair of
poles, or m=35.5. Each coil has 3 turns, but there is only one coil
per pair of poles, so that n=1.5. Substituting these values into
eq. (64), and also M = 5580, kb= 0.912, we find i= 128 amp. As a
matter of fact, in this case it is not necessary to decide what the
values of n and m are, because eq. (64) contains only the product
mn, which is the total number of turns per pole. Thus, in our case
mn=(7lX3)/4.
Formula (64) holds also for a squirrel-cage winding, the number
of secondary phases being equal to the number of bars per pair of
poles. Since there is but one bar per phase, each bar can be con-
sidered as one-half of a turn, and in formula (64) n=0.5 and fc&= 1,
so that it becomes
(64o)
where €2 is the total number of rotor bars, and p is the number of
poles. Or else, one may say that the total number of turns per
pole is equal to one-half the number of bars per pole, so that
mn = %C2/p. This again gives eq. (64a). For a direct proof of
formula (64a) see problem 15 below. Applying this formula to
the same rotor with 71 slots we find that the current per bar is
700 amp.
(b) The Equivalent Secondary Winding Reduced to the Primary
Circuit. When investigating the general theory of the induction
motor or calculating the characteristics of a given motor, it is con-
venient to replace the actual rotor winding by an equivalent wind-
ing identical with the primary winding of the motor. In this case
the primary current transmitted into the secondary is equal to the
134 THE MAGNETIC CIRCUIT [ART. 44
actual secondary current (one to one ratio of transformation), and
the primary and the secondary voltages induced by the useful flux
are also equal. Each electric circuit of the stator then can be
combined with the corresponding rotor circuit. In this manner
the so-called " equivalent diagram " of the induction motor is
obtained,1 a way of representation which greatly simplifies the
theory of the machine.
Let 12 be the secondary current in the coils or bars of the actual
rotor, and ir2 that in the equivalent rotor. The counter-m.m.f.
of both rotors must be the same, this being the condition of their
equivalence, so that
from which
) ..... (65)
This is the ratio of current transformation in an induction motor.
The ratio of transformation of the voltages is different, namely,
(66)
In an ordinary transformer e2'/e2 = 12/^2 ==ni/n2, because there
kbi = kb2=l, and m2 = mi = l. For this reason, the induction
motor is sometimes regarded as a generalized transformer.
Taking the product mie for the actual and the equivalent rotor
it will be found that the total electric power input is the same in
both, provided that the same phase displacement is preserved in
the equivalent rotor as in the original one. The latter condition is
essential in order that the operating characteristics of the two
machines be the same. This means (a) that the total i2r loss of
the equivalent rotor must be equal to that of the original rotor, in
order to preserve the same slip, and (b) that the leakage react-
ances of the two rotors must affect the power factor of the
primary current in the same way.
Let r2 and r2 be the resistances of the actual and of the equiva-
lent rotor, per pole per phase. We have the condition that
(67)
'Chas. P. Steinmetz, Alternating Current Phenomena (1908), p. 249;
Elements of Electrical Engineering (1905), p. 263.
CHAP. VII] M.M.F. OF DISTRIBUTED WINDINGS 135
Substituting the ratio of i2/i2 from eq. (65) we find
r2/r2=(mi/m2)(kbinl/kb2n2).2 .... (68)
For a transformer this equation reduces to the familiar expression
The ratio of the inductances is the same as that of the resist-
ances ; this can be proved as follows : In order that the equivalent
winding may have the same effect on the power factor of the motor
as the actual winding, the equivalent winding must draw from the
line an equal amount of reactive volt-amperes, due to its leakage
inductance. The magnetic energies stored in the two rotor wind-
ings must therefore be equal, and we have, according to eq. (104)
in Art. 58,
ml.tf2'2L'2, ...... (69)
where L2 and Z/2 are the leakage inductances of the real and the
equivalent rotor windings, per pole per phase. The form of this
equation is the same as that of eq. (67) ; therefore, substituting
again the ratio of i'2/i2 from eq. (65) an expression is obtained for
the ratio of L'2/L2 identical with that given by eq. (68), namely
L27L2=(m1/m2)(/c61n1//c62n2)2 ..... (70)
This result could also be foreseen from the fact that the reactances
and the resistances enter symmetrically in the equivalent diagram,
and relation (68) holds therefore for the reactances x2 and x2.
But in the equivalent diagram the secondary and the primary fre-
quency is the same, so that the ratio of the inductances is equal to
that of the reactances; this gives eq. (70).
It must be clearly understood that the expressions (68) and (70)
refer to the resistances and inductances per pole per phase. When
the windings of a phase are all in series, both in the stator and in the
rotor, the same ratio holds of course for the resistances per phase ;
otherwise the actual connections must be taken into consideration,
keeping in mind that the total i2r loss must be the same in the
equivalent winding as in the actual one. Having obtained the re-
sistance of the equivalent winding per pole, the turns are connected
in the same way as the stator turns. This fact must be remembered
in particular when dealing with individually short-circuited coils
1 See the author's Experimental Electrical Engineering, Vol. 2, p. 77.
136 THE MAGNETIC CIRCUIT [ART. 45
in the rotor, or with a squirrel-cage winding. In these two cases
the individual coils or bars in the rotor are all in parallel, while the
stator coils of a phase are usually all in series, or in two parallel
groups. In the case of a squirrel-cage winding the resistance r2
includes that of a bar, of two contacts with the end-rings, and of
the equivalent resistance of a section of the two end-rings.1
Prob. 12. In a 300 horse-power, Y-connected, 14-pole induction
motor the full-load current is estimated to be 310 amp. The primary
winding consists of 336 turns placed in 168 slots; the winding-pitch is
0.75. What is the minimum number of bars in the squirrel-cage second-
ary winding, if the current per bar must not exceed 800 amp.? The
secondary counter-m.m.f. is equal to about 90 per cent of the primary
m.m.f. Ans. 208.
Prob. 13. What must be the resistance of each secondary bar in the
preceding problem (including the equivalent resistance of the adjoining
segments of the end-rings and also of the contacts) if the slip at full load
is to be about 4 per cent.? Hint: The per cent slip is equal to the i2r
loss in the rotor, expressed in per cent of the power input into the second-
ary. If x is the i2r loss in the rotor, expressed in horse-power, we have
that x = 0.04 (300 + x) . Ans. 70 microhms.
Prob. 14. The motor with the individually short circuited second-
ary coils, that is used as an illustration in the text above, is to be
investigated with respect to its performance. By what factor must the
actual resistance and inductance of each secondary coil be multiplied
in order to obtain the equivalent resistance and inductance per primary
phase? Also by what factor must the equivalent current be multiplied
in order to obtain the actual current in each secondary coil?
Ans. 297; 0.402.
Prob. 15. Prove formula (64o) directly, by considering the m.m.f s.
of the individual bars. Solution : At any instant the currents in the bars
under a pole are distributed in space according to the sine law, because
the gliding flux which jnduces these currents is sinusoidal. The average
current per bar is *A/2X(2/*) =0.9i. The number of turns per pole is
C2/2p, and all these turns are active at the crest of the m.m.f. wave.
Therefore, Jlf = 0.9i(C2/2p).
45. The Higher Harmonics of the M.M.FS. In the preceding
study, the effect of the higher harmonics in the m.m.f. wave was dis-
regarded. In fact, these harmonics usually exert a negligible
influence upon the operation of a good polyphase induction motor,
under normal conditions. These m.m.f. harmonics move at lower
speeds than the fundamental field ; therefore, the fluxes which they
1 See the author's Electric Circuit; also E. Arnold, Die Wechselstromtechnik,
Vol. 5, Part I (1909), p. 57.
CHAP. VII] M.M.F. OF DISTRIBUTED WINDINGS 137
produce cut the secondary conductors at comparatively high rela-
tive speeds ; thus, secondary currents are induced which wipe out
these harmonics to a considerable degree. There are practical cases,
however, in which some one particular harmonic becomes of some
importance, and affects the operation of the machine, particularly
at starting. For this reason the following general outline of the
properties of the higher harmonics in the m.m.f . is given.1
In a single-phase machine (Fig. 34) all the higher harmonics of
the m.m.f. are pulsating at the same frequency as the fundamental
wave, but the width of the nth harmonic is only I/nth of that of
the fundamental wave. Each pulsating harmonic can be replaced
by two gliding harmonics of half the amplitude, one left-going, the
other right-going. The linear velocity of these gliding m.m.fs. is
only I/nth of that of the fundamental gliding waves, because they ~
cover in the time %T a distance equal only to their own base, PQ/n \
(180 electrical degrees) . With one slot per pole, the amplitudes of
the higher harmonics decrease according to eq. (61), but with more
than one slot, or with a fractional-pitch winding they decrease
more rapidly, because different values of kb must be taken for each
harmonic (see Art. 30 above).
In a two-phase machine, consider (Fig. 35a) the gliding waves
Ln and Rn, of the nth harmonic. For this harmonic, the distance
between 0\ and 0% is equal to \nn electrical degrees. At the
instant t—0 the crest of the wave Lnl is at the point 0\\ at the
instant t=\T the crest of the wave Ln2 is at the point 0^. There-
fore, the two waves travel at a relative distance of \n(n — 1) elec-
trical degrees, considering the base of the nth harmonic as equal
to its own 180 electrical degrees. In a similar manner, the distance
between the crests of the two right-going waves is found to be
equal to %x(n + 1) electrical degrees. We thus obtain the following
table of the angular distances between the waves due to the two
phases :
Order of the harmonic 1 3 5 7 9 11 13
Distance between the two Ln waves 0 it In %K IK 5;r GTT
Distance between the two Rn waves K 2^ STT 4rc 5?r 6n IK
The waves which travel at a distance 0, 2n, 4;r, etc., are simply
added together, while those at a distance TT, 3nf 5n, etc., cancel each
1 For a more detailed treatment see Arnold, Wechselstromtechnik, Vol.
3 (1904), Chapter 13, and Vol. 5, part I (1909), Chapter 9.
138 THE MAGNETIC CIRCUIT [ART. 45
other. Thus, in a two-phase machine, the 3d, 7th, llth, etc.,
harmonics travel against the direction of the main m.m.f., while the
5th, 9th, 13th, etc., harmonics travel in the same direction as the
fundamental m.m.f., though at lower peripheral speeds.
Applying a similar reasoning to a three-phase winding (Fig.
356) we find that the three Ln waves travel at a relative distance of
|7r(n — 1), while the relative distance between the three Rn waves is
f x(n+ 1) electrical degrees. We thus obtain the following table of
the angular distances between the waves due to the three phases :
Order of the harmonic 1 3 5 7 9 11 13 15
Distance between the three Ln waves. .. 0 IT: |TT 0 tyc *px 0 ^-n
Distance between the three Rn waves. . . |TT f n 0 V^ *£* 0 ^-K ^TT
The component waves, of any harmonic, which travel at a distance
zero from each other, are simply added together, and give a resul-
tant wave of three times the amplitude of the component. The
three waves which travel at an angular distance of f n or one of
its multiples from each other give a sum equal to zero. Thus, in a
three-phase machine, the 1st, 7th, 13th, etc., harmonics travel in
one direction, while the 5th, llth, 17th, etc., harmonics travel
against the direction of the fundamental m.m.f. The higher the
, order of a harmonic the lower its peripheral speed. The harmonics
of the order 3, 9, 15, etc., are entirely absent.
Prob. 16. What are the amplitudes of the fifth and the seventh
harmonics, in percentage of that of the fundamental wave, for a three-
phase winding placed in 2 slots per pole per phase, when the winding-
pitch is 5/6? Ans. 1.4 and 1.0 per cent respectively.
Prob. 17. Show that, in order to eliminate the nth harmonic in
the m.m.f. wave, the winding-pitch must satisfy this condition ; namely,
f/i: = (2q + 1) /n, where f is defined in Fig. 16, and q is equal to either 0,
1, 2, 3, etc. Hint: Cos \fn must be = 0.
Prob. 18. Investigate the direction of motion of the various har-
monics of the m.m.f. in a symmetrical w-phase system.
Prob. 19. Show that only the nth harmonic in the m.m.f. wave,
due to the nth harmonic in the exciting current, moves synchronously
with the fundamental gliding m.m.f., and therefore distorts it perma-
nently.
Prob. 20. A poorly designed 2-phase, 60-cycle induction motor has
4 poles, 1 slot per phase per pole, and a winding pitch of 100 per cent.
At what sub-synchronous speed is it most likely to stick? Hint: The
torque due to any harmonic reverses as the motor passes through the
corresponding sub-synchronous speed. Anp. 360 r.p.m.
CHAPTER VIII
ARMATURE REACTION IN SYNCHRONOUS
MACHINES
46. Armature Reaction and Armature Reactance in a Syn-
chronous Machine. When a synchronous machine carries a load,
either as a generator or as a motor, the armature currents, being
sources of m.m.f., modify the flux created by the field coils, and
thus influence the performance of the machine. Fig. 36 shows an
•<* Direction of Rotation <m&
FIG. 36 — The flux distribution in a single-phase synchronous machine
under load.
instantaneous flux distribution in the simplest case of a single-
phase alternator, with one slot per pole; the armature conductors
are marked a and b. With the directions of the armature and field
currents indicated in the sketch, the flux is crowded toward the
right-hand tips of the poles. In order to show this, imagine two
fictitious conductors of and V with currents equal and opposite
139
140 THE MAGNETIC CIRCUIT IART. 46
to those in the actual conductors a and b respectively. The addi-
tion of these fictitious conductors does not modify the armature
m.m.f. because they neutralize each other. The conductor a'
may be considered as forming a turn with a, while bf forms a turn
with b. It will be seen that the m.m.f. of the coil aaf assists that
of the field coil A, while the m.m.f. of the coil W is opposite to that
of the field coil B.
The armature current in the coil ab not only distorts the no-
load field, but also reduces the total flux per pole. This may be
seen by considering the flux in the four parts of the air-gap, marked
x, y, x', and y', where x = x' and y = y'- The sum of the fluxes in
the portions y and yr is the same as without the armature current,
because the flux density in the part y is increased by the same
amount by which it is reduced in the part y' (neglecting satura-
tion) . But in the parts x and xf the flux is reduced by the arma-
ture m.m.f., so that the total result over the pole-pitch is. a reduc-
tion in the value of the no-load flux. The position of the armature
conductors and the direction of the armature currents have been
selected arbitrarily. They can be chosen so that the flux will be
crowded toward the left-hand tips of the poles, or so that the total
flux will be increased by the armature m.m.f., instead of being
reduced. The influence of the armature currents, in modifying the
value of the field flux and distorting it, is called the armature reac-
tion. The armature reaction is measured in ampere-turns, since
it is a magnetomotive force.
In addition to the general distortion of the field by the arma-
ture currents, there is a local distortion around each armature
conductor. This distortion does not extend into the pole shoes,
but is limited to the slots and the air-gap; it is indicated in Fig. 36
by ripples in the flux around a and b. These ripples may be
regarded as a result of the superposition upon the main flux of the
local fluxes 4>a and <P& excited by the armature currents. While
these local fluxes, shown by the dotted lines, have no real existence,
except around the end connections of the armature conductors,
it is convenient to consider them separately. They are purely
alternating fluxes, in phase with the currents with which they are
linked, so that they induce in the armature windings alternating
e.m.fs. in a lagging phase quadrature with the currents.
The effect of these local fluxes upon the voltage of the machine
is represented by a certain armature reactance, because the effect is
CHAP. VIII] REACTION IN SYNCHRONOUS MACHINES 141
the same as if the armature winding created no leakage fluxes
around it, but a separate reactance coil were connected in series
with each armature lead. The calculation of the armature react-
ance, or of the local fluxes, is treated in Art. 67, the subject of this
chapter being armature reaction only, that is, the effect of the load
upon the main magnetic circuit. In the numerical problems of this
chapter, for the solution of which it is necessary to know the value
of the armature reactance, this value is given. It is not quite
correct, strictly speaking, to separate the local distortion of the
main flux as a phenomenon by itself; moreover, the separation is
somewhat indefinite and arbitrary. However, the flux so separated
is comparatively small, and the treatment of the armature reaction
proper is thereby greatly simplified.
The distribution shown in Fig. 36 varies from instant to instant
because the relative position of the armature changes with refer-
ence to the poles, as well as the value of the armature current.
Besides, there are usually two or three armature phases, and sev-
eral slots per pole per phase. It would be out of the question to
calculate the actual fluxes for each instant and to take into account
their true influence upon the e.m.f. induced in the armature. In
practice, certain approximate average values of armature reaction
and of armature reactance are employed, which permit one to
predict the actual performance of a machine with a sufficient
accuracy.
In the case of a synchronous generator (alternator) the problem
usually presents itself in the following form : It is required to pre-
determine the field ampere-turns necessary for a prescribed ter-
minal voltage at a given load. Knowing the resistance and the
leakage reactance of the armature, the voltage drop in the arma-
ture is added geometrically to the terminal voltage; this gives the
induced voltage in the machine. Knowing from the no-load satura-
tion curve the required net excitation at this voltage, and correct-
ing it for the effect of the armature reaction, the necessary field
ampere-turns are obtained. The results of such calculations for
different values of the armature current and for various power
factors, plotted as curves, are called the load characteristics of the
alternator.
In the case of a synchronous motor the terminal voltage is usu-
ally given, and it is required to determine the field excitation such
that, at a given mechanical output, the input to the armature be at
142 THE MAGNETIC CIRCUIT [ART. 46
a given power-factor; a leading power-factor is usually prescribed,
in order to raise the lagging power-factor of the whole plant. The
problem is solved in like manner to that of the generator, by
taking into account the proper signs when calculating the reactance
drop and the armature reaction. The results, plotted in the form
of curves, are called the phase characteristics, or V-curves of a
synchronous motor.1
It will be seen from Fig. 36 that the crowding of the flux to one
pole-tip, by the armature currents, is primarily due to the fact that
the pgles shown there are projecting or salient, so that the reluc-
tance along the air-gap is variable. With non-salient poles the
flux is simply shifted sidewise without being distorted. Therefore,
before going into the details of the calculation of armature reaction
in machines with salient poles we shall first consider (in the next
article) the case of a machine with non-salient poles.
^r
Prob. 1. Draw the distribution of the flux, similar to that shown
in Fig. 36, when the armature conductors are opposite the centers of the
poles, and when they are somewhere between the adjacent pole-tips.
Prob. 2. Explain the details of the flux distribution in Fig. 36, by
means of a hydraulic analogy, assuming A and B to represent two main
centrifugal pumps, and a and b to be two smaller pumps placed in the
stream.
Prob. 3. Let each field coil in Fig. 36 have N turns, and let the
exciting current be /; let the number of conductors at a be Cs, and the
instantaneous value of the armature current i. What is the total flux
per pole, if the average permeance of the machine per pole is (P perms
per electrical radian, and the angles 6 and x are in electrical radians?
Ans. (2NI6-Csix)(P. in maxwells.
Prob. 4. Let a synchronous machine be loaded in such a way that
the armature current reaches its maximum when the conductors a and b
(Fig. 36) are opposite the centers of the poles, in other words, the current
is in phase with the e.m.f. which would be induced at no load. Prove
that (neglecting saturation) the average flux per pole during a complete
cycle is the same as without the armature reaction, but is crowded to
the leading tip of the pole, i.e., in the direction of rotation in the case of
a motor, and to the trailing tip, or against the direction of rotation when
the machine is working as a generator. Hint: The flux is weakened as
much in the position x of the conductors as it is strengthened in the
symmetrical position x'', the distortion is in the same direction in both
positions.
!See the author's " Experimental Electrical Engineering," Vol. 2, p. 121;
also his " Essays on Synchronous Machinery," General Electric Review,
1911, p. 214.
CHAP. VIII] REACTION IN SYNCHRONOUS MACHINES 143
Prob. 6. Let a synchronous machine be loaded in such a way that
the armature current reaches its maximum when the conductors a and 6
(Fig. 36) are midway between the poles, in other words, when the current
is displaced by 90 electrical degrees with respect to the e.m.f. induced
at no load. Prove that the average distortion during a complete cycle
is zero, but that the flux is weakened if the armature current lags behind
the induced e.m.f., and is strengthened by a leading current. Hint:
The flux is weakened in both of the symmetrical positions, x and xr, of
the conductors, but the distortion is in opposite directions.
Prob. 6. In a single-phase synchronous machine the armature
current reaches its maximum when the armature conductors are dis-
placed by an angle ^ with respect to the centers of the poles;1 prove that
the field is distorted by the component i cos </> of the current and is
weakened or strengthened by* the component i sin ^.
47. The Performance Diagram of a Synchronous Machine with
Non-Salient Poles. Let, in a machine with non-salient poles, the
field winding be placed in several slots per pole, so that the field
m.m.f . in the active layer of the machine is approximately distrib-
uted according to the sine law. Consider the machine to be a
polyphase generator supplying a partly inductive load. The ampli-
tude of the first harmonic of the armature reaction has the value
given by eq. (64) in Art. 43, and revolves synchronously with the
field m.m.f., as is explained there. Since the sum of two sine waves
is also a sine wave, the resultant m.m.f. is also distributed in the
active layer of the machine according to the sine law.
To deduce the phase displacement, in space, between the two
sine waves, consider the coil a b (Fig. 36) to be one of the phases of
the polyphase armature winding. For reasons of symmetry, the
maximum m.m.f. produced by a polyphase winding is at the center
of the coil in which at that particular moment the current is at a
maximum. Assume first that the current in the phase a b reaches its
maximum when the conductors a and b are opposite the centers of
the poles. The maximum armature m.m.f. at that instant is dis-
placed by 90 electrical degrees with respect to the center lines
of the poles. The direction of the armature current is determined
by the well-known rule, and it is found to be such that the arma-
ture m.m.f. lags behind that of the pole, considering the direction
of rotation of the poles as positive. Since both m.m.fs. revolve
synchronously, this angle between the two m.m.f. crests is pre-
1 The angle $ is different from the external phase-angle 0 between the
current and the terminal voltage; see Fig. 37.
144 THE MAGNETIC CIRCUIT [ART. 47
served all the time. Thus, in a polyphase generator, the armature
m.m.f. lags behind the field m.m.f. by 90 electrical degrees in
space, when the currents are in phase with the voltages induced at
no-load. This statement is in accord with that in problem 4 in the
preceding article, because, if each phase shifts the flux against the
direction of rotation, all the phases together simply increase the
result.
Let now the currents in the armature windings be lagging
90 electrical degrees behind the corresponding e.m.fs. induced
at no-load. This simply means that the armature m.m.f. is shifted
further back by 90 degrees as compared to the case considered
before; therefore, the angle between 'the field m.m.f. and the
armature m.m.f. is 180 electrical degrees, and the two m.m.fs. are
simply in phase opposition. This is in accord with the statement
in prob. 5.
From the two preceding cases it follows that, when in a syn-
chronous machine with non-salient poles the currents lag by an
angle </> electrical degrees (Figs. 37 and 38) with respect to the
induced voltage at no-load, the armature m.m.f. wave lags by an
angle of 90 + ^ electrical degrees behind the field m.m.f. wave. In
the case of a generator with leading currents the angle (p is negative ;
in a synchronous motor <p is larger than 90 degrees.
Let, in Fig. 37, i be the vector of the current in one of the phases,
and let e be the corresponding terminal voltage, the phase angle
between the two being <£. Adding to e in the usual way the ohmic
drop ir in the armature, in phase with i, and the reactive drop ix
in leading quadrature with i, the induced voltage E in the same
phase is obtained.1 The resultant useful flux, $, which induces
this e.m.f. leads E by 90 degrees in time; 0 is in phase with the
net or resultant m.m.f. Mn which produces it. The m.m.f. Mn is a
sum of the field m.m.f. Mf and of the armature reaction Ma
1 On account of skin effect and eddy currents in the armature conductors,
the effective resistance r to alternating currents is considerably higher than that
calculated or measured with direct current. The actual amount of increase
depends upon the character of the winding, the size of the conductors, the
shape of the slots, the frequency, etc., so that no definite rule can be given.
Fortunately, the ohmic drop constitutes but a small percentage of the voltage
of a machine, so that a considerable error committed in estimating the value
of the ir drop affects the voltage relations but very little. See A. B. Field,
"Eddy Currents in Large Slot-wound Conductors," Trans. Amer. Inst.
Elect. Engrs., Vol. 24 (1905), p. 761.
CHAP. VIII] REACTION IN SYNCHRONOUS MACHINES
145
expressed by eq. (64) . The triangle OFG represents the relations in
space, while the figure OABD is a time diagram. Therefore, the
two figures are independent of one another; but it is convenient
to combine them into one, by using the common vectors 0 and i.
FIG. 37. — The performance diagram of a synchronous generator, with
non-salient poles.
With respect to the triangle OFG, the vector i represents the posi-
tion of the crest of the armature m.m.f . relatively to the crest OG
of the field m.m.f., the angle between the two being 90 + </>, as is
explained above. Thus, the vector Ma is in phase with i.
146 THE MAGNETIC CIRCUIT [ART. 47
When i and e are given, the vector E is easily found if the
resistance and the reactance of the armature winding are known.
The required net excitation, Mn, is then taken from the no-load
saturation curve of the machine, and Ma is figured out from eq.
(64). Then the required field ampere-turns, Af/, are found from
the diagram, either graphically or analytically.
The diagram shown in Fig. 37 is known as the Potier diagram.
Strictly speaking, it is correct only for machines with non-salient
poles, but as an approximate semi-empirical method it is some-
times used for machines with projecting poles, in place of the more
correct diagram shown in Fig. 40. Fig. 37 represents the condi-
tions in the case of a generator with lagging currents. When the
current is leading the vector i is drawn to the left of the vector e,
with the corresponding changes in the other vectors.
A similar diagram for a synchronous motor which draws a
leading current from the line is shown in Fig. 38. The vector e'
represents the line voltage, and e is the equal and opposite voltage
which is the terminal voltage of the machine considered as a gen-
erator. The rest of the diagram is the same as in Fig. 37. A lead-
ing current with respect to the line voltage e' is a lagging current
with respect to the generator terminal voltage e, so that the field
is weakened by the armature reaction in both cases (Mn < Mf in
both figures). The energy component i\ of the current is reversed
in the motor, therefore the field is shifted in the opposite direc-
tion; Mn leads Mf in the motor diagram and lags behind it in the
generator diagram. The case of a synchronous motor with a
lagging current can be easily analyzed by analogy with the above-
described cases.
In practice, it is usually preferred to represent the relations
shown in Figs. 37 and 38 analytically, rather than to actually con-
struct a diagram. The following relations hold for both the gen-
erator and the motor. Projecting all the sides of the polygon
OABD on the direction e and on the direction perpendicular to
and leading e by 90 degrees, we have
E cos(f)g=e+ir cos<f)+ix sin <£, . . . (71)
E sin <j)K=ix cos<£— ir sin <£, .... (72)
where cf>e is the angle between the vectors e and E, counted positive
when E leads e, as in Fig. 37. The subscript z suggests that the
CHAP. VIII] REACTION IN SYNCHRONOUS MACHINES
147
angle </>2 is due to the impedance of the armature. The expressions
i cos $ and i sin 0 represent the energy component and the react-
ive component of the current respectively; they are designated in
FIG. 38. — The performance diagram of a synchronous motor, with
non-salient poles.
Figs. 37 and 38 by ii and i2. Denoting the right-hand sides of the
eqs. (71) and (72) by ei and e2 for the sake of brevity, we have:
62 = i\x —i2r.
(73)
(74)
148 THE MAGNETIC CIRCUIT [ART. 47
Squaring eqs. (71) and (72) and adding them together gives
'~ (75)
Dividing eq. (72) by (71) results in
tan</>2=e2Ai ....... (76)
Consequently, the angle between E and i becomes known; namely,
....... (76a)
where </>' is called the internal phase angle. Knowing E, the cor-
responding excitation Mn is taken from the no-load saturation
curve of the machine ; from the triangle OFG we have then :
asm<f>', . . . (77)
where </>' is known from eq. (76a) . In numerical applications it is
convenient to express all the M's in kiloampere-turns.
The diagram shown in Fig. 38 and the equations developed
above can be used for determining not only the phase characteris-
tics of a synchronous motor, but its overload capacity at a given
field current as well. This latter problem is of extreme importance
in the design of synchronous motors. The input into the machine,
per phase, is — ei cos <£> ; the part ir of the line voltage is lost in the
armature, the part ix corresponds to the magnetic energy which is
periodically stored in the machine and returned to the line, without
performing any work. The remainder, E, corresponds to the use-
ful work done by the machine, plus the iron loss and friction. If
the armature possessed no resistance and no leakage reactance the
terminal voltage would be equal to E in magnitude and in phase
position. Thus, the expression — Ei cos <f>', corrected for the core
loss in the armature iron, represents the input into the revolving
structure, per phase. The overload capacity of the machine is
determined by the possible maximum of this expression.
The problem is complicated by the fact that the relation
between E and Mn is expressed by the no-load saturation curve,
which is difficult to represent by an equation. The problem is
1 In numerical applications it is more convenient to use the approximate
formula
l .......... (75a)
obtained by the binomial expansion of expression (75) ; since all other terms
can be neglected when e2 is small as compared to et.
CHAP. VIII] REACTION IN SYNCHRONOUS MACHINES 149
therefore solved by trials, assuming a certain reasonable value of
E, and calculating the expression — Ei cos (/>', until a value of E
is found, for which this expression, corrected for the core loss, fric-
tion, and windage, becomes a maximum. The problem of finding
i and (f>f for an assumed E is a definite one, because the four equa-
tions (71), (72), (76a) and (77) contain only four unknown quanti-
ties, i, <f>, <j>zj and </>'. Instead of solving the problem by trials, an
analytical relation can be assumed between E and Mn, on the use-
ful part of the no-load saturation curve, for instance a straight line
(not passing through the origin), a parabola, etc. The problem is
then solved by equating the first derivative of the product —
Ei cos ft to zero, having previously expressed E, i, and cos <j>'
through some one independent variable. Both methods have
been worked out for a synchronous motor with salient poles.1 The
relations are simplified for a machine with non-salient poles.
The foregoing theory of the armature reaction does not apply
directly to single-phase machines. The pulsating armature reac-
tion in such a machine can be resolved into two revolving reactions,
as in Art. 42. The reaction which revolves in the same direction
with the main field is taken into account as in a polyphase machine.
The inverse reaction is partly wiped out by the eddy currents pro-
duced in the metal parts of the revolving structure ; it is therefore
difficult to express the effect of this reaction theoretically. The
treatment in this book is limited to polyphase machines, which are
used in practice almost exclusively.2
Prob. 7. In the 100 kva., 440-volt, 6-pole, two-phase alternator,
given in Problem 6, Art. 43, the amplitude of the first harmonic of the
armature reaction was 4800(7S ampere-turns. What is the per cent
voltage regulation of the machine at a power-factor of 80 per cent
lagging, if C« = l, that is if the armature has one conductor per slot?
The armature reactance is 0.038 ohm, and the armature resistance is
0.008 ohm, both per phase. The no-load saturation curve of the
machine is as follows :
e = 400 440 490 525 550 volts.
Mn=6.7 8.0 10.0 12.0 14.0 kiloamp. -turns.
Ans. 22 per cent.
1 See the author's "Essays on Synchronous Machinery," General Electric
Review, 1911, July and September.
2 In regard to the armature reaction in single-phase machines, see E.
Arnold, Die Wechselstromtechnik, Vol. 4 (1904), pp.~ 32-39; Pichelmayer,
Dynamobau (1908), pp. 251-259; Max Wengner, Theoretische und Expert-
mentelle Untersuchungen an der Synchronen Einphasen-Maschine (Oldenbourg,
1911.)
150 THE MAGNETIC CIRCUIT [ART. 48
Prob. 8. The machine specified above is to be used as a synchronous
motor. Determine graphically the required field excitation when the
useful output on the shaft is to be 700 kw., and in addition the machine
must draw from the line 600 leading reactive kva. The efficiency of the
machine at the above-mentioned load is estimated to be about 91 per
cent. Ans. 12.5 kiloampere-turns.
Prob. 9. Draw to the same scale as the diagram shown in Fig. 37,
another similar diagram, for the same value of the current and of the
phase angle 0, except that the current is to be leading. Assume a
reasonable shape of the saturation curve in determining the new value of
Mn. Show that a much smaller exciting current is required with the
same kva. output, than in the case of a lagging current.
Prob. 10. Solve problem 9 for the motor diagram shown in Fig. 38,
assuming the current to be lagging with respect to the line voltage.
Prob. 11. For a given alternator, show how to determine the voltage
e (Fig. 37), analytically or graphically, when Mf, i, and ^ are given;
explain when such a case arises in practice.
Prob. 12. For a given synchronous motor, show how to determine
the reactive component i2 of the current (Fig. 38), analytically or graphic-
ally, when Mf, e and it are given ; explain when such a case arises in
practice.
Prob. 13. Work out the details of the above-mentioned method
for the determination of the overload capacity of a synchronous- motor
by trials. Hint: Introduce the components of fc and i, in phase and in
quadrature with E; rewrite eqs. (71) and (72) by projecting the figure
OABD on the direction of E and on that perpendicular to E. Use no
angles in the formulae, and neglect the small terms containing r, where
they lead to complicated equations of higher degrees.
48. The Direct and Transverse Armature Reaction in a Synchro-
nous Machine with Salient Poles. In a machine with non-salient
poles the armature reaction shifts the field flux but hardly distorts
its shape. In a machine with projecting poles the flux, generally
speaking, is both altered in value and crowded toward one pole-
tip (Fig. 36). It is convenient, therefore, to resolve the traveling
wave of the armature m.m.f . into two waves, one whose crests coin-
cide with the center lines of the poles, the other displaced by 90
electrical degrees with respect to it. The first component of the
armature m.m.f. produces only a " direct " effect upon the field
flux, that is, it either strengthens or weakens the flux, without dis-
torting it. The second component produces a " transverse "
action only, viz., it shifts the flux toward one or the other pole-tip,
without altering its value (that is, neglecting the saturation) .
We have seen before that an armature current, which reaches
its maximum when the conductor is opposite the center of the
CHAP. VIII] REACTION IN SYNCHRONOUS MACHINES 151
pole, distorts the flux; while a current in quadrature with the
former exerts a direct reaction only. It is natural, therefore, to
resolve the actual current in each phase into two components, in
time quadrature with each other, and in such a way that each
component reaches its maximum in one of the above-mentioned
principal positions of the conductor with respect to the field-poles.
Let the current in each phase be i, and let it reach its maximum at
an angle </> after the induced no-load voltage is a maximum (Fig.
40) . Then, the two components of the current are
id=i sin (p
and
it — i cos <[>.
The component id produces a direct armature reaction only,
and the component it a transverse reaction only.1
For practical calculations, and in order to get a concrete picture
of the armature reaction, it is convenient to represent the armature
reaction as shown in Fig. 39. Namely, the direct reaction, due to
the components id of the armature currents, is replaced by an equiv-
alent number of concentrated ampere-turns Md on the pole. The
value of Md is selected so that its action in reducing or strength-
ening the flux is equal to the true action of the armature currents.
The transverse reaction, due to the component it of the armature
currents, is replaced by a certain number of ampere-turns, Mt, on
the fictitious poles, (£), (N), shown by dotted lines between the real
poles. For simplicity, and for other reasons given in Art. 51, the
fictitious poles are assumed to be of a shape identical with that of
the real poles. The number of exciting ampere-turns Mt is so
chosen, that the effect of the fictitious poles is approximately the
same as that of the distorting ampere-turns on the armature.
The flux of the fictitious poles strengthens the flux of the real
poles on one side and weakens it by the same amount on the other
side, so that the fictitious poles actually distort the main flux
without altering its value. Strictly speaking, the complete action
of the distorting ampere-turns on the armature cannot be imitated
1 The resolution of the armature reaction in a synchronous machine
into a direct and a transverse reaction was first done by A. Blondel. See
V Industrie Electrique, 1899, p. 481 ; also his book Moteurs Synchrones (1900),
and two papers of his in the Trans. Intern, Electr. Congress, St. Louis, 1904,
Vol. 1, pp. 620 and 635.
152
THE MAGNETIC CIRCUIT
[ART. 48
by fictitious poles of the same shape as the main poles, because
harmonics of appreciable magnitude are thereby neglected. How-
ever, actual experience shows that the performance of a machine,
calculated in this way, can be made to check very well with the
observed performance, by properly selecting the coefficients of the
direct and the transverse reaction. In a generator, the flux is
crowded against the direction of rotation of the poles (Fig. 36) ;
consequently, the fictitious poles lag behind the real poles, as
shown in Fig. 39. In a synchronous motor they lead the real
poles by 90 electrical degrees.
If the ratio of the pole arc to pole-pitch were equal to unity,
as with non-salient poles, the whole wave of the demagnetizing
Actual distribution
of transverse flux
Flux distribution
due to fictitious pole
FIG. 39. — The direct and transverse armature reactions in a synchronous
machine, represented by fictitious poles and field windings.
m.m.f. of the armature would be acting upon the pole, and the
equivalent concentrated m.m.f. Md on the pole would have to be
equal to the average value of the actual distributed armature
m.m.f. We would have then
(78)
where the maximum armature m.m.f. is determined by eq. (64),
Art. 43, and 2/7r=0.637 is the ratio of the average to the maximum
ordinate of a sine wave. In reality, only a part of the armature
m.m.f., the one near its amplitude, acts upon the poles, the action
of lower parts of the wave being practically zero because of the
gaps between the poles. Therefore, the ratio between the maxi-
mum m.m.f. M sin <p and the average equivalent m.m.f. M^ is
CHAP. VIII] REACTION IN SYNCHRONOUS MACHINES 153
larger than 0.637. For the ordinary shapes of projecting poles,
experiment and calculation (see Art. 50 below) show that this
ratio varies between 0.81 and 0.85. Using an average of these
limits instead of 2/7r in eq. (78) and substituting for M its expres-
sion from eq. (64) we obtain the following practical formula for
estimating the armature demagnetizing ampere-turns per pole in a
synchronous machine with projecting poles:
(79)
In this formula i sin (j> is the component id of the armature current,
per phase. In actual machines the numerical coefficient in this
formula varies between 0.73 and 0.77, depending on the shape of
the poles and the ratio of pole-arc to pole-pitch.
By a similar reasoning, if the ratio of pole-arc to pole-pitch were
equal to unity, the equivalent number of exciting ampere-turns on
the fictitious poles would be
3f«=(2/7r)Mcos# ...... (80)
Since the ratio of pole-arc to pole-pitch on the fictitious poles
is less than unity, the numerical coefficient should be larger than
2/7T. But, on the other hand, the permeance of the air-gap under
the fictitious poles is much higher than the actual permeance of
the machine in the gaps between the poles, so that a much smaller
number of ampere-turns Mt is sufficient to produce the same
distorting flux. The combined effect of these two factors is to
reduce the coefficient in formula (80) to a value considerably
below 2/7T. For the usual shapes of projecting poles, experiment
and calculation (See Art. 51 below) show that this ratio varies
between 0.30 and 0.36. Using an average of these limits instead
of 2/7T hi eq. (80), and substituting for M its expression from eq.
(64), we obtain the following practical formula for estimating
the distorting ampere-turns per pole, in a synchronous machine
with projecting poles:
(81)
In this formula i cos ^ is the component it of the armature cur-
rent, per phase. In some actually built machines the coefficient
in this formula comes out lower than 0.30, but in preliminary cal-
154
THE MAGNETIC CIRCUIT
[ART. 49
ta?
B
eolations it is advisable to use at least 0.30. When a synchronous
motor is working near the limit of its overload capacity, the influ-
ence of the distorting ampere-turns is particularly important, and
in estimating the overload capacity of a synchronous motor it is
better to be on the safe side and to take the value of the numerical
coefficient in eq. (81) somewhat higher than 0.30. The value of
this coefficient varies within wider limits than that of the corre-
sponding coefficient
in formula (79) ;
but, fortunately, it
affects the perform-
ance to a lesser de-
gree (see Art. 51).
49. The Blondel
Performance Dia-
gram of a Syn-
chronous Machine
with Salient Poles.
Having replaced the
actual armature
reaction by two
m.m.fs. Md and Mt
(Fig. 39) the elec-
tromagnetic rela-
tions in the machine
become those indi-
cated in Figs. 40 and
41. Fig. 40 refers
to a generator and
is analogous to Fig.
37; Fig. 41 refers
to a motor and is
analogous to Fig.
38. The polygon OABD, which represents the relation between
the terminal and the induced voltages, is the same as before, but
the induced voltage E is now considered as a resultant of the
voltages En and Et induced by the real and the fictitious poles
respectively.1 In the generator the fictitious poles lag behind
1 The subscript n stands for net, to agree with the m.m.f. Mn used later
on; the subscript t stands for transverse.
FIG. 40. — The performance diagram of a synchro-
nous generator, with salient poles.
CHAP. VIII] REACTION IN SYNCHRONOUS MACHINES
155
the real ones, in a motor they lead the real poles. Hence, in the
generator diagram, Et lags 90 degrees behind En, while in the
motor diagram it leads En by 90 degrees.
In the case of a generator the problem usually is to find the
field excitation Mf neces-
sary for maintaining a
required terminal voltage
e, with a given current i
and at a given power-
factor cos <£. First, the
figure OABD is con-
structed, or else the
values of E and <f>' are
determined from eqs.
(75), (76), and (76o).
In order to find the
ampere-turns required on
the main poles it is neces-
sary to determine the
voltage En induced by
them. For this purpose
the angle /? must first be
known, for
En = E cos /?. . (82)
As an intermediate step,
it is necessary to express
Et through the ampere-
turns Mt, which are the
cause of Et. The m.m.f .
Mt is small as compared
to the total number of
ampere-turns on the real
poles; hence, the lower
straight part of the no-
load saturation curve of
. . FIG. 41. — The performance diagram of a syn-
the machine can be used chronous ^^ with salient poles> '
to express the relation
between Mt and Et. Let v be the voltage corresponding to
one ampere-turn on the lower part of the no-load saturation
156 THE MAGNETIC CIRCUIT [ART. 49
curve; then Et =Mtv. Substituting the value of Mt from eq. (81),
we have
#,=#/ cos (<£'+/?), ..... (83)
where
(84)
Et' is a known quantity introduced for the sake of brevity. The
angle ^ in formula (83) is expressed through <j>' and ft because,
from Fig. 40,
^=<£'+/? ........ (85)
Another relation between Et and /? is obtained from the triangle
ODG, from which
Et = Esmp ........ (86)
A comparison of eqs. (83) and (86) gives that
Et' cos (<£' +fi) = E sin ft
Expanding and dividing throughout by cos /? we find the relation
(87)
from which the angle /? can be determined, and then En calculated
byeq. (82). -
The next step is to take from the no-load saturation curve the
value Mn of the net excitation necessary on the main poles in order
to induce the voltage En. The real excitation M / must be larger,
because part of it is neutralized by the direct armature raction Mj.
We thus have
...... (88)
where Md is calculated from eq. (79), the angle ([> being known
from eq. (85). When the load is thrown off, the only excitation
left is Mf, let it correspond to a voltage e0 on the no-load satura-
tion curve. From e0 and e the per cent voltage regulation of the
machine is determined from its definition as the ratio (e0—e)/e.
The same general method and the same equations apply in the
case of Fig. 41, when one is required to determine a point on one of
the phase characteristics of a synchronous motor. The beginner
must be careful with the sign minus in the case of the motor.
CHAP. VIII] REACTION IN SYNCHRONOUS MACHINES 157
Since <f>' > 90 degrees, the angle /? and the voltage Et are negative.
The angle </>2 also is usually negative. The cases of a leading
current in the generator and of a lagging current in the motor are
obtained by assigning the proper value and sign to the angle </>.
For the application of the Blondel diagram to the determination
of the overload capacity of a synchronous motor see the reference
given near the end of Art. 47.
A synchronous motor is sometimes operated at no load, and at
such a value of the field current that the machine draws reactive
leading kilovolt-amperes from the Ikie, thus improving the
power-factor of the system. In such a case the machine is called a
synchronous condenser, or better, a phase adjuster. The diagram in
Fig. 41 is greatly simplified in this case because the energy com-
ponent of the current can be neglected, as well as the drop ir,
and the e.m.f. Et. We then have i=i2 = id, and En=E=e+ix.
The direct armature reaction is determined from eq. (79) in which
^=90. When the motor is underexcited and draws a lagging
current from the line, i is to be considered negative, or ^=270
degrees. The same simplified diagram applies to a polyphase
rotary converter, operated from the alternating-current side, at no
load.
Prob. 14. It is required to calculate the field current and per cent
voltage regulation of a 12-pole, 150 kva., 2300- volt, 60-cycle, Y-connected
alternator, at a power factor of 85 per cent lagging. The machine has
two slots per pole per phase, and is provided with a full-pitch winding,
the number of turns per pole per phase being 18. The armature resist-
ance per phase of Y is 0.67 ohm, the reactance is 3.5 ohm. The number
of field turns per pole is 200. The no-load saturation curve is plotted
for the line voltage (not the phase voltage), and at first is a straight line
such that at 1800 volts the field current is 17.4 amp. The working part
of the no-load saturation curve is as follows :
Kilovolts 2.2 2.4 2.5 2.6 2.7 2.78
Field current, amp 22 25 27 30 34 40
Ans. 31 amp.; 14.3 per cent.
Prob. 15. Show that in the foregoing machine the short-circuit
current is equal to about two and a half times the rated current, at the
field excitation which gives the rated voltage at no-load. Hint: The
short-circuit curve is a straight line so that one can first calculate the
field current for any assumed value of the armature current and e=0.
Prob. 16. From the results of the calculations of the preceding
problem show that the cross-magnetizing effect and the ohmic drop are
negligible under short-circuit, in the machine under consideration.
158 THE MAGNETIC CIRCUIT [ART. 50
Assuming that r is usually small as compared to x, describe a simple
method for calculating the short-circuit curve, using only the reactance
of the machine and the demagnetizing ampere-turns of the armature.
In practice, the influence of the neglected factors is accounted for in short-
circuit calculations by taking sin (p in formula (79) as equal to between
0.95 and 0.98 instead of unity.
Prob. 17. Plot the no-load phase characteristic of the machine
specified in problem 14, when it is used as a motor. The iron loss and
friction amount to 8.5 kw.
Ans. Field amperes 14.9 23.4 32.6
Armature amperes 30 2.13 30
Prob. 18. The machine specified in problem 14 is to be used as a
motor, at a constant input of 150 kw. Plot its phase characteristics,
i.e., the curves of the armature current and of power-factor against the
field current as abscissae.
Ans. Field amperes 32.6 24.3 16.65
Armature amperes . . 47 . 00 37 . 65 47 . 00
Power-factor 0 . 80 1 . 00 0 . 80
Prob. 19. Write complete instructions for the predetermination of
the regulation of alternators and of the phase characteristics of synchro-
nous motors, by BlondePs method. The instructions must give only
the successive steps in the calculations, without any theory or explana-
tions. Write directions and formula? on the left-hand side of the sheet,
and a numerical illustration on the right-hand side opposite it.
Prob. 20. Calculate the overload capacities of the foregoing motor
at field currents of 25 amp. and 35 amp., by the two methods described
in the articles refered to near the end of Art. 47.
Prob. 21. Show that for a machine with non-salient poles BlondePs
and Potier's diagrams are identical.
50. The Calculation of the Value of the Coefficient of Direct
Reaction in Eq. (79) ^ The average value 0.83 of the ratio of the
effective armature m.m.f. over a pole-face to the maximum m.m.f .
at the center of the pole is given in Art. 48 without proof. The
following computations show the reasonable theoretical limits of
this ratio. If the armature m.m.f. (direct reaction) at the center
of the, N pole (Fig. 39) is M, its value at some other point along
the air-gap is M cos x, where x is measured in electrical radians.
Let the permeance of the active layer of the machine per electrical
radian be (P at the center of the pole, and let this permeance vary
along the periphery of the armature according to a law f(x) , so
that at a point determined by the abscissa x the permeance per
1 This and the next article can be omitted, if desired, without impairing
the continuity of treatment.
CHAP. VIII] REACTION IN SYNCHRONOUS MACHINES 159
electrical radian is (Pf(x). The function/Or) must be periodic and
such that /(O) = 1, and fQn) = 0, /(TT) = 1, etc., because the perme-
ance reaches its maximum value under the centers of the poles
and is practically nil midway between the poles.
The direct armature m.m.f., acting alone, without any excita-
tion on the poles, would produce in each half of a pole a flux
0= C+**.Mcosx(Pf(x)dx.
Jo
The magnetomotive force Md placed on the real poles, acting
alone, must produce the same total flux, so that
Equating the two preceding expressions we get
MC**cosxf(x)dx~MdC**f(x)dx. . . . (89)
«/0 »/0
The ratio of Md to M can be calculated from this equation, by
assuming a proper law f(x) according to which the permeance of
the active layer varies with x, in poles of the usual shapes. Hav-
ing a drawing of the armature and of a pole, the magnetic field can
be mapped out by the judgment of the eye, assisted if necessary
by Lehmann's method (Art. 41 above). A curve can then be
plotted, giving the relative permeances per unit peripheral length,
against x as abscissae. Thus, the function f(x) is given graphically,
and the two integrals which enter into eq. (89) can be determined
graphically or be calculated by Simpson's Rule. Or else,
f(x) can be expanded into a Fourier series and the integration
performed analytically. Such calculations performed on poles of
the usual proportions give values of Md/M of between 0.81 and
0.85.
It is also possible to assume for f(x) a few simple analytical
expressions, and integrate eq. (89) directly. Take for instance
f(x) =cos2 x. By plotting this function against x as absciss® the
reader will see that the function becomes zero midway between
the poles, is equal to unity opposite the centers of the poles, and
has a reasonable general shape at intermediate points. Substi-
tuting cos2 x for f(x) into eq. (89) and integrating, gives f M =
, from which Md/M =0.85.
160 THE MAGNETIC CIRCUIT [ART. 51
Another extreme assumption is that of poles without chamfer,
with a constant air-gap. Neglecting the fringe at the pole-tips,
f(x) = 1 from x =0 to x =6, and f(x) =0 from x =6 to x =%n. Inte-
grating eq. (89) between the limits 0 and 6 we obtain
(90)
The poles usually cover between 60 and 70 per cent of the periph-
ery. For 0=0.6(j7r) the preceding equation gives Md/M=0.86,
and for 6=0.7(%n), Md/M=0.81.
Prob. 22. Let the permeance of the active layer decrease from the
center of the poles according to the straight-line law, so that
/(*)-! -(2/*)s.
What is the ratio of Md/Mf Ans. 0.81 1 .
Prob. 23. The permeance of the active layer decreases according
to a parabolic law, that is, as the square of the distance from the center
of the poles. What is the ratio of Md/M? Ans. 0.774.
Prob. 24. The law f(x} =cos2 x assumed in the text above presup-
poses that the permeance varies according to a sine law of double
frequency with a constant term, because cos2 z=£ +£ cos 2x. In reality,
the permeance varies more slowly under the poles and more rapidly
between the poles than this law presupposes (Fig. 39). A correction can
be brought in by adding another harmonic of twice the frequency to the
foregoing expression, thus making it un symmetrical, and of the form
f(x) =a + b cos 2x+c cos 4x. Show that f(x) =2 cos2 x — cos4 x contains
the largest relative amount of the fourth harmonic, consistent with the
physical conditions of the problem, and compare graphically this curve
with/(z) =cos2 x.
Prob. 25. What is the value of Md/M for the form of f(x) given in
the preceding problem? Ans. 0.815.
Prob. 26. Plot the curve /(x) for a given machine, estimating the
permeances by Lehmann's method, and determine the value of the coef-
ficient in formula (79).
51. The Calculation of the Value of the Coefficient of Transverse
Reaction in Eq. (81). The average value 0.33 of the ratio of the
maximum distorting armature m.m.f . to the equivalent number of
ampere-turns, Mt, on the fictitious poles is given in Art. 48 without
proof. The following computations show the reasonable theoret-
ical limits of this ratio. The problem is more complicated than
that of finding the ratio of Md/M, because there the field ampere-
turns, the actual demagnetizing armature-m.m.f., and the equiva-
lent ampere-turns Md are all acting on the same permeance of the
CHAP. VIII] REACTION IN SYNCHRONOUS MACHINES 161
active layer, and the wave form of the flux is very little affected by
the direct armature reaction. In the case of the transverse reac-
tion, however, the wave form of the flux produced by the actual
cross-magnetizing ampere-turns of the armature is entirely differ-
ent from that produced by the coil Mt acting on the fictitious pole
(Fig. 39). Namely, the actual curve of the transverse flux has a
large " saddle " in the middle, due to the large reluctance of the
space between the real poles. The flux distribution produced by
the fictitious poles is practically the same as that under the main
poles, the two sets of poles being of the same shape.
The addition of the vectors Et and En in Figs. 40 and 41 is legit-
imate only when Et is induced by a flux of the same density dis-
tribution as En, and this is the reason for representing the trans-
verse reaction as due to fictitious poles of the same shape as the real
poles. Therefore, for the purposes of computation, the flux dis-
tribution, produced by the actual distorting ampere-turns on the
armature, is resolved into a distribution of the same form as that
produced by the main poles and into higher harmonics. The m.m.f .
Mt is calculated so as to produce the first distribution only. This
fundamental curve is not sinusoidal, but will have a shape depend-
ing on the shape of the pole shoes. The effect of the sinusoidal
higher harmonics on the value of Et is disregarded, or it can be
taken into account by correcting the value of the coefficient in
formula (81) from the results of tests.
The first harmonic of the armature distortion m.m.f. is M sin x,
because this m.m.f. reaches its maximum between the real poles;
x is measured as before from the centers of the real poles. The
permeance of the active layer, with reference to the real poles, can
be represented as before by (Pf(x) . The flux density produced by
the transverse reaction of the armature at a point denned by the
abscissa x is therefore proportional to M sin x (Pf(x) . The per-
meance of the active layer with reference to the fictitious poles is
<Pf(x+fa)- The flux density under the fictitious poles follows
therefore the law Mt(Pf(x+%7c). As is explained before, the two
distributions of the flux density differ widely from one another,
and the real distribution is resolved into the fictitious distribution,
and higher sinusoidal harmonics ; the prominent third harmonic is
clearly seen in Fig. 39. Thus, we have, omitting (P,
M sin xf(x) =Mtf(x+fa) +A3 sin 3x+A5 sin 5z+etc. (91)
162 THE MAGNETIC CIRCUIT [ABT. 51
In order to determine Mt, the usual method is to mulitply both
sides of this equation by sin x and integrate between 0 and xf
because then all upper harmonics give terms equal to zero. In
this particular case the limits of integration can be narrowed down
to 0 and \K, because the symmetry of the curve is such that the
segment between 0 and \n is similar to all the rest. Thus, we
get
M f^sin2 xf(x)dx =Mt f^sin xf(x+fr)dx. . (92)
«/o a/o
From this equation the ratio Mt/M can be calculated by the
methods shown in Art. 50, i.e., by assuming reasonable forms of
the function f(x). Taking again f(x) =cos2 x and integrating eq.
(92) we get &nM =%Mt, from which Mt/M =0.295. Taking the
other extreme case, viz., f(x) =1 from x=0 to x=6, and f(x) =0
from x =6 to x =%n, gives, after integration
(93)
For 0=0.6(4*:), M,/M=0.29; for 0=0.7(4*), M</M=0.39.1 It
will be noted that the cross-magnetizing action of the armature
increases considerably with the increasing ratio of pole-arc to pole-
pitch, while the direct reaction slowly diminishes with the increase
of this ratio. In machines intended primarily for lighting pur-
poses it is advisable to use a rather small ratio of pole-arc to pole-
pitch, in order to reduce transverse reaction which affects the volt-
age regulation at high values of power-factor in particular.
Prob. 27. What is the value of Mt/M for the form of f(x) given in
problem 24; namely, for/(z) = 2 cos2 z-cos4 x? Ans. 0.368.
Prob. 28. Determine the numerical value of the coefficient in formula
(81) for the machine used in problem 26.
1 These values are higher than those derived by E. Arnold. The fact
that Arnold's values for the coefficient of transversal reaction are low has
been pointed out by Sumec in Elektrotechnik und Maschinenbau, 1906, p.
67; also by J. A. Schouten, in his article " Ueber den Spannungsabfall mehr-
phasiger synchroner Maschinen, " Elektrotechnische Zeitschrift, Vol. 31 (1910),
p. 877.
CHAPTER IX
ARMATURE REACTION IN DIRECT-CURRENT
MACHINES
52. The Direct and Transverse Armature Reactions. Let
Fig. 42 represent the developed cross-section of a part of a direct-
current machine, either a generator or a motor. For the sake of
simplicity the brushes are shown making contact directly with the
armature conductors, omitting the commutator. Electrically
this is equvialent to the actual conditions, because the commutator
segments are soldered to the end-connections of the same conduc-
tors. The brushes are shifted by a distance d from the geometrical
neutral, to insure a satisfactory commutation; d being expressed
hi centimeters, measured along the armature periphery, the same
as the pole-pitch T.
The actual armature conductors and currents are replaced, for
each pole-pitch, by a current sheet, or belt, of the same strength.
Let, for instance, the pole-pitch be 40 cm., and let the machine
have 120 armature conductors per pole. If the current per con-
ductor is 100 amp., the total number of ampere-conductors per
pole is 12,000 ; the total current of the equivalent belt, which con-
sists of one wide conductor, must be 12,000 amp., or 300 amp. per
linear cm. of the pole-pitch. The latter value, or the number of
armature ampere-conductors per centimeter of periphery, is some-
times called the specific electric loading of the machine. The mag-
netic action of the equvialent current sheet on the magnetic flux of
the machine is practically the same as that of the actual armature
conductors, because in a direct-current machine the slots are com-
paratively numerous and small. The current in the cross-hatched
belts is supposed to flow from the reader into the paper, and the cur-
rent in the belts marked with dots — toward the reader. With the
directions of the flux and of the current shown in the figure, the
directions of rotation of the machine when working as a generator
and as a motor are those shown by the arrow-heads (see Art. 24).
163
164
THE MAGNETIC CIRCUIT
[ART. 52
The polarity of the brushes cannot be indicated without knowing
the actual connections in the winding. It is preferable, therefore,
for our purposes to designate the brushes as E and W (east and
west), according to their position with respect to the poles of the
machine, the observer looking from the commutator side. The
whole interpolar regions to the right of the north poles can be called
the eastern regions, those to the left the western regions; the same
notation can be also applied to the commutating poles.
The armature currents exert a two-fold action upon the main
field of the machine: they partly distort it, and partly weaken it.
For the purposes of theory and calculation it is convenient to
separate these two actions, the same as in the case of the synchro-
cj
0,1
FIG. 42. — The direct and transverse armature reactions in a
direct-current machine.
nous machine in the preceding chapter. Let the sheets of current
be divided into parts denoted by the letters D and T with sub-
scripts corresponding to their location with reference to the poles
and brushes. The belts denoted by D are comprised within the
space d, one each side of the geometrical neutrals; those denoted
by T are (Jr— <J) centimeters wide.
The belts D exert a direct demagnetizing action upon the poles.
Namely, the belts Dn Dn can be considered as two sides of a coil the
axis of which is along the center line CnCn'. The m.m.f. of this
coil opposes that of the field coil on the north pole. In the same
way, the m.m.f. of the coil DSDS opposes' the action of the field coil
on the south pole. The foregoing is true no matter what the
actual connections of the armature conductors are, provided that
the winding-pitch is nearly 100 per cent. With a fractional-pitch
CHAP. IX] ARMATURE REACTION IN D.C. MACHINES 165
winding the currents within each D belt flow partly in the opposite
directions and neutralize each other's action.
Let the specific electric loading of the machine, as defined
above, be (AC) . Then, with a full-pitch winding, the demagnetiz-
ing ampere-turns per pole are1
M,=(AC)d (94)
The belts TeTe constitute together a coil the center of which is
along the axis OeOer] the adjacent belts TwTwiono. a coil with its
axis along 0W0W'. The m.m.f. distribution of these coils is indi-
cated by the broken line ABC, which shows that the T belts pro-
duce a transverse armature reaction. The line ABC is also the
curve of the flux density distribution which would be produced by
the transversal reaction alone, if the active layer of the machine
were the same throughout (non-salient poles). On account of a
much higher reluctance of the paths in the interpolar regions the
flux density there is much lower, and is shown by the dotted lines.
The actual distribution of the field in a loaded machine is obtained
considering from point to point the field and armature m.m.fs.
acting upon the individual magnetic paths.
The transverse reaction opposes the field m.m.f. under one-half
of each pole and assists it under the other half, so that the main
field is distorted. In a generator the field is shifted in the direc-
tion of rotation, in a motor it is crowded against the direction of
rotation of the armature. This is the same as what happens in
synchronous machines, when the armature is revolving and the
poles stationary (see Fig. 36).
The brushes must be shifted in the same direction in which the
flux is shifted, because the magnetic neutral is displaced with
respect to the geometric neutral. Usually, the brushes are shifted
beyond the magnetic neutral, in order to obtain a proper flux den-
sity for commutation. Namely, to assist the reversal of the cur-
rent in the conductors which are short-circuited by the brushes,
these conductors must be brought into the fringe of a field of such
a direction as assists the commutation. In the case of a gener-
ator this means the field under the influence of which the conduc-
1 The effect of the coils short-circuited under the brushes is not considered
separately, for the sake of simplicity. For an analysis of the reaction of the
short-circuited coils upon the field see E. Arnold, Die Gleichstrommaschine
Vol. 1 (1906), Chap. 23.
166 THE MAGNETIC CIRCUIT [AET. 53
tors come after the commutation. In a motor the armature cur-
rent flows against the induced e.m.f.; it is therefore the field which
cuts the conductors before the commutation that assists the rever-
sal of the current. This explains the direction of the shift of the
brushes in the two cases. The student should make this clear to
himself by considering in detail the directions of the currents and
of the induced voltages in a particular case.
The maximum m.m.f. per pole produced by the distorting
belts is equal to (AC) (%T—d), but since this m.m.f. acts along the
interpolar space of high reluctance its effect is not large (except in
machines with commutating poles). Of much more importance
is the action of the distorting belts under the main poles. At each
pole-tip the armature m.m.f. is
w, (95)
where- w is the width of the pole shoe. This m.m.f. decreases
according to the straight line law to the center of each pole and
is of opposite signs at the two tips of the same pole.
Prob. 1. Determine the polarity of the brushes in Fig. 42 for a
progressive and a retrogressive winding, in the case of a generator and
of a motor.
Prob. 2. Indicate the D and the T belts in a fractional-pitch winding
(a) with the brushes in the geometric neutral, and (b) with the brushes
shifted by d.
Prob. 3. A 500 kw., 230 volt, 10 pole, direct-current machine has
a full-pitch multiple winding placed in 165 slots. There are 8 con-
ductors per slot, and two turns per commutator segment. What are
the demagnetizing ampere-turns per pole when the brushes are shifted
by 4 commutator segments? Ans. 3470.
Prob. 4. What is the amplitude of the broken line ABC in the
preceding machine? Ans. 10850 amp-turns.
Prob. 5. For a given machine, draw the curves of the flux density
distribution under a pole, at no-load and at full load, by considering
the m.m.fs. acting upon the individual paths, and the reluctance of the
paths.1
53. The Calculation of the Field Ampere-turns in a Direct-
current Machine under Load. The net ampere-turns per pole are
determined from the no-load saturation curve of the machine for
1 For details and examples of such curves see Pichelmayer, Dynamobau
(1908), p. 176; Parshalland Hobart, Electric Machine Design (1906), p. 159;
Arnold, Die Gleichstrommachine, Vol. 1 (1906), p. 324.
CHAP. IX] ARMATURE REACTION IN D.C. MACHINES 167
the necessary induced voltage. In a generator the induced voltage
is equal to the specified terminal voltage plus the internal ir drop
in the machine. In a motor the induced voltage is less than the
line voltage by the amount of the internal voltage drop. The
actual ampere-turns to be provided on the field poles are larger
than the net excitation by the amount necessary for the compensa-
tion of the armature reaction.
The direct reaction is compensated for by adding to each field
coil the ampere-turns given by eq. (94). For instance, in
prob. 3 above, 3470 ampere-turns per pole must be added to the
required net excitation, in order to compensate for the effect of the
direct armature reaction. The necessary shift of the brushes is
only roughly estimated from one's experience with previously
built machines, though it could be determined more accurately
from the distribution of flux density in the pole-tip fringe. The
poles usually cover not over 70 per cent of the armature periphery,
so that the distance between the geometric neutral and the pole-
tip is about 15 per cent of the pole pitch. In preliminary esti-
mates, the brush shift may be taken to be about 10 per cent of the
pole pitch; this brings the brushes not quite to the pole-tips
though well within their fringe. In actual operation a smaller
shift may be expected. In machines provided with commutating
poles, and in motors intended for rotation in both directions, the
brushes are usually in the geometric neutral, so that the demag-
netizing action is zero.
In a machine with a low saturation in the teeth and in the pole-
tips, the cross-magnetizing m.m.f. of the armature does not affect
the magnitude of the total flux per pole, because the flux is
increased on one-half of the pole as much as it is reduced on the
other half. It is shown in Art. 31 that the induced e.m.f. of a
direct-current machine is independent of the flux distribution,
provided that the total flux is the same, so that no extra field
ampere-turns are necessary in such a machine to compensate for
the distortion of the flux.
However, the teeth and the pole-tips are usually saturated to
a considerable extent, so that the flux is increased on one side of
the pole less than it is reduced on the other side. Thus, the useful
flux is not only distorted by the transverse armature reaction, but
is also weakened. This latter effect has to be counterbalanced by
additional ampere-turns on the field poles. In most cases these
168
THE MAGNETIC CIRCUIT
[ART. 53
additional ampere-turns are estimated empirically, on the basis
of one's previous experience, because the amount is not large, and
a correct computation is rather tedious.1
The theoretical relation between the distorting ampere-turns
and the field ampere-turns required for their compensation is shown
in Fig. 43. Let OX represent the no-load saturation curve of the
machine for its active layer only, that is, for the air-gap, the teeth,
and the pole shoe, if the latter is sufficiently saturated. The
ordinates represent the induced e.m.f . between the brushes, or, to
another scale, the useful flux per pole ; the abscissae give the corre-
sponding values of the field ampere-turns per pole, disregarding
the reluctance of the field poles, field yoke, and armature core.
a a' d d1
Ampere-turns and pole face areas
FIG. 43.— A construction for determining the field m.m.f. needed for the
compensation of the transverse reaction.
In other words, the abscissae give the values of the difference of
magnetic potential across the active layer of the machine, at no
load. It is proper to consider here the m.m.fs. across the active
layer only, because the distorting action of the armature extends
only over this layer. No matter how irregular the flux distribu-
tion in the air-gap and in the teeth may be, the flux density in the
pole cores and in the yoke is practically uniform (compare Fig. 36) .
1 For Hobart's empirical curves for estimating the field excitation required
for overcoming the armature distortion see the Standard Handbook, under
' Generators, direct-current, ampere-turns, estimate."
CHAP. IX] ARMATURE REACTION IN D.C. MACHINES 169
For the sake of simplicity, we replace the actual pole shoe by
an equivalent one, without chamfer, of rectangular shape, and
with a negligible pole-tip fringing. The ordinates of the curve OX
represent to another scale the average flux density on the surface
of the pole shoe, because this density is proportional to the total
useful flux, or to the induced voltage. Let ab be the flux density
corresponding to the required induced e.m.f., and let Oa=M be
the necessary net m.m.f ., without the transverse reaction. Let ac
and ad represent the distorting ampere-turns, M%, at the pole-tips,
as given by eq. (95) ; then Oc and Od are the resultant m.m.f s. at
the pole-tips. The flux density at the pole-tips of the loaded
machine is then equal to eg and dh respectively.
Since the distorting ampere-turns vary directly as the distance
from the center of the pole, and the area of strips of equal width
is the same, the abscissae from a represent to scale either distances
along the pole face, or areas on the pole face, measured from the
center of the pole. Thus, the part gh of the curve OX represents
also the distribution of the flux density under the pole, in the
loaded machine. Likewise, the line ef represents the distribution
of the flux density under the pole at no-load.
Since the ordinates represent to scale the flux densities and the
abscissae the areas of the different parts under the poles, the area
under the flux density distribution curve also represents to scale
the total flux per pole. The total flux at no load is represented by
the area of the reactangle cefd, and to the same scale, the flux in the
loaded machine is represented by the area cghd. If the saturation
curve were a straight line, the two areas would be equal, so that
the distortion would not modify the value of the total flux per pole.
In reality, the area geb is larger than the area bhf, and the differ-
ence between the two represents the reduction in the flux, due to
the transverse armature reaction.
Let now the field excitation be increased by an unknown amount
aaf to the value Oa'=M', in order to compensate for the above
explained decrease in the flux. All the points in Fig. 43 are shifted
by the same amount, and are denoted by the same letters with the
sign " prime." The new flux in the loaded machine is represented
by the area c'g'h'd', the point a' being the center of the line c'd' =cd.
If the point a! has been properly selected we must have the con-
dition that the
area cefd= area c'g'h'd', (96)
170 THE MAGNETIC CIRCUIT [ART. 53
that is, the flux corresponding to the excitation Oa at no-load is
the same as the flux corresponding to the excitation Oa' when the
machine is loaded. The problem is then, knowing the point a and
the distance cd=c'd' to find the point a' such that equation (96)
is satisfied. The two areas have the common part c'g'bfd, and
the parts cee'c' and dff'd' are equal to each other. Therefore,
eq. (96) is satisfied if the
area be'g' =- area bfh' (97)
The position of the point a' is found either by trials, or as the
intersection of the curves rrf and ss'; the ordinates of the curve rrf
represent the area be'g' for various assumed positions of the point
a', and the ordinates of the curve ssf represent the corresponding
values of the area bf'h'.
Thus, the total field ampere-turns required for a direct-current
generator under load are found as follows: (a) To the specified
terminal voltage add the voltage drop in the armature, under the
brushes, and in the windings (if any) which are in series with the
armature, viz.; the series field winding, the compensating winding,
and the interpole winding. This will give the induced voltage E.
(b) From the no-load saturation curve find the excitation corre-
sponding to E. (c) Estimate the amount of the brush shift (if
any) and calculate the corresponding demagnetizing ampere-turns
according to eq. (94) . (d) Determine the ampere-turns aar (Fig. 43)
required for the compensation of the armature distortion, (e)
Add the ampere-turns calculated under (b), (c) and (d). In the
case of a motor subtract the voltage drop in the machine from
the terminal voltage, to find the induced e.m.f. E, but otherwise
proceed as before.
If the machine is shunt-wound or series-wound, the field wind-
ing is designed so as to provide the necessary maximum number of
ampere-turns at a required margin in the field rheostat for the
specified voltage or speed variations. When the machine is com-
pound wound, the shunt winding alone must supply the required
number of ampere-turns at no-load. The series ampere-turns is,
then, the difference between the total m.m.f. required at full-load
and that supplied by the shunt-field. When a generator is over-
compounded, the shunt excitation is larger at full load than it is at
no load, and allowance must be made for this fact.
CHAP. IX] ARMATURE REACTION IN D.C. MACHINES 171
In the case of a variable-speed motor, the problem of predict-
ing the exact speed at a given load can be solved by successive
approximations only. First, this speed is determined neglecting
the transverse armature reaction altogether, or assigning to it a
reasonable value. Then, the construction shown in Fig. 43 is
performed for a few speeds near the approximate value, and thus
a more correct value of the speed is found by trials. Or else
different values of speed are assumed first, and the corresponding
values of the armature current are found for each speed. These
armature currents must be such that, considering the total arma-
ture' reaction, the field m.m.f. is just sufficient to produce the
required counter-e.m.f. in the armature. The details of the solu-
tion are left to the student to investigate. He must clearly
understand that the problem can be solved only for a given or an
assumed speed, because of the necessity of using the active layer
characteristic OX (Fig. 43).
Prob. 6. In a direct-current machine it is desired to have a full
load a flux density of not less than 3500 maxwells per sq.cm. at the
pole-tip at which the commutation takes place; the m.m.f. across the
active layer is 7500 amp .-turns, the air-gap is 11 mm., the air-gap
factor 1.25. The ratio of the ideal pole width to the pole pitch is 0.7.
What are the permissible ampere-turns on the armature, per pole?
Solution: The net m.m.f. across the active layers at the pole-tip under
consideration is 3500X0.8X1.1X1.25 = 3850 amp .-turns. Hence
M2 = %(AC) X0.7r = 7500 -3850 =3650. Ans. i(A<7)r=5200.
Prob. 7. The specific electric loading in a direct-current machine
is 250 amp.-conductors per centimeter; the average flux density on
the pole-face is 8.5 kl. per sq. cm., at no load and at the rated full load
terminal voltage; the width of the equivalent ideal pole is 32 cm. The
estimated internal voltage drop at full load is about 5 per cent of the
terminal voltage. Calculate the ampere-turns per pole required to
compensate for the transverse armature reaction; the active layer
characteristic (Fig. 43) is as follows :
M= 4 5 6 7 8 9 11 13 kilo ampere-turns;
5 = 5.40 6.75 7.75 8.50 8.90 9.20 9.55 9.78 kl. per sq. cm.
Ans. 950.
Prob. 8. For the preceding machine, calculate the total required
excitation at full load, per pole; the brush shift is 8 per cent of the
pole-pitch; the pole-pitch r = 46 cm.; 2000 ampere-turns are required
for the parts of the machine outside the active layer.
Ans. 12,000 amp .-turns.
Prob. 9. A shunt-wound motor is designed so as to operate at a
certain speed at full load. Show how to predict its speed at no-load.
172 THE MAGNETIC CIRCUIT [ART. 54
Prob. 10. Show how, in a compound-wound motor, the total
required field excitation must be divided between the shunt and series
windings in order to obtain a prescribed speed regulation between no-
load and full load.
Prob. 11. Assume the curve OX in Fig. 43 to be given in the form
of an analytic equation, B=f(M). Show that the unknown excitation
Oa'=M' is determined by the equation 2M2f(M) = F(M'+M2) -
F(M'-M2), where the function F is such that dF(M)/dM=f(M};
MQ = Oa is the excitation corresponding to the given value of e or B = ab.
Prob. 12. Apply the formula of the preceding problem to the case
when the active layer characteristic can be represented by (a) the log-
arithmic curve y = a log (1 +bx) ; (6) the hyperbola y=gx/(h+x) ; (c) a
part of the parabola (y*-y02) =2p(x-Xo), continued as a tangent straight
line passing through the origin.
54. Commutating Poles and Compensating Windings. The
two limiting factors in proportioning a direct-current machine are,
first, the sparking under the brushes, and secondly, the armature
reaction. In order to reverse a considerable armature current
in a coil during the short interval of time that the coil is under a
brush, an external field of a proper direction and magnitude is
necessary. In ordinary machines (Fig. 42) this field is obtained
by shifting the brushes so as to bring the short-circuited armature
conductors under a pole fringe. However, with this method the
specific electric loading and the armature ampere-turns must be
kept below a certain limit with reference to the ampere-turns on
the field; otherwise the armature reaction would weaken the field
to such an extent as to reduce the flux density in the fringe below
the required value. Therefore, in many modern machines, instead
of moving the brushes to the poles, part of the poles, so to say, are
brought to the brushes (Fig. 44). These additional poles are
called commutating poles or interpoles. Their polarity is under-
stood with reference to Fig. 42 : Since the E brush had to be shifted
toward the north pole, now a north interpole is placed over each E
brush.
The armature belts Te, Te, create a strong m.m.f. along the axis
of the commutating pole Nc, in the wrong direction. Therefore,
the winding on each interpole must be provided first with a num-
ber of ampere-turns equal and opposite to that of the armature,
and secondly with enough additional ampere-turns to establish
the required commutating flux. These additional ampere-turns
are calculated only for the air-gap, armature teeth, and the pole
body itself. The m.m.f. required for the armature core and the
CHAP. IX] ARMATURE REACTION IN B.C. MACHINES
173
yoke of the machine is negligible, because the commutating flux is
small as compared with the main flux and is displaced with respect
to it by ninety electrical degrees. The winding on the interpoles
is connected in series with the main circuit of the machine, because
the armature m.m.f. is proportional to the armature current, and
also because the density of the reversing field must be proportional
to the current undergoing commutation.
The flux density under the interpoles is determined from the
condition that the e.m.f. induced in the armature conductors by
the commutating flux be equal and opposite to the e.m.f. due to
the inductance of the armature coils undergoing commutation.
For practical purposes, the inductance can be only roughly esti-
mated (see Art. 68 below), but on the other hand an accurate cal-
Compensating Commutating
Main pole Winding /pole or interpole
Generator
t
Motor
FIG. 44. — Interpoles and a compensating winding in a direct-current machine.
culation is not necessary, because the number of ampere-turns on
the interpole is easily adjusted by a shunt around its winding, as
in the case of a series winding on the main poles. Or else the
commutating flux can be increased by " shimming up " the
interpoles. The m.m.f. required for establishing a required
commutating flux is calculated in the same manner as in the case
of the main flux, viz., the saturation curves (Figs. 2 and 3) are
used for the pole body and the teeth, while the reluctance of the
air-gap is estimated as is explained in Arts. 36 and 37. The
commutating poles must be of such a width that all the coils
undergoing commutation are under their influence.
A comparatively large leakage factor, say between 1.4 and 1.5,
or over, is usually assumed for the commutating poles, on account
of the proximity of the main poles. It is advisable to concentrate
the winding near the tip of the interpole, in order to reduce the
174 THE MAGNETIC CIRCUIT [ART. 54
magnetic leakage. The leakage factor of the main poles is also
somewhat increased by the presence of the interpoles; this is one
of their disadvantages. Some other disadvantages are : the ven-
tilation of the field coils is more difficult, and a smaller ratio of
pole arc to pole pitch must be used. However, the advantages
gained by the use of commutating poles are such that their use
is rapidly becoming universal.
The interpole winding removes the effect of the transverse
belts T, T, in the commutating zone, but does not neutralize their
distorting effect under the main poles. Hence, the distortion and
its accompanying reduction of the main flux are practically the
same as without the. interpoles. To remove this distortion a com-
pensating winding (Fig. 44), connected in series with the main cir-
cuit of the machine, is sometimes placed on the main poles. The
connections are such that the compensating winding opposes the
magnetizing action of the armature winding. By properly select-
ing the specific electric loading of the compensating winding the
transverse armature reaction under the poles can be removed,
either completely or in part. This winding was invented inde-
pendently by Deri in Europe and by Professor H. J. Ryan in this
country; on account of its expense, it is used in rare cases
only.
When a compensating winding is used in addition to the inter-
poles, the number of ampere-turns on the interpoles is consider-
ably reduced, because the compensating winding can be made to
neutralize the larger portion or all of the armature reaction. In
such a machine a much higher specific loading can be allowed than
in an ordinary machine of the same dimensions. Therefore, such
compensated machines are particularly well adapted for rapidly
fluctuating loads, and for sudden overloads or reversals of rotation
in the case of a motor.
Prob. 13. From the following data determine the ampere-turns
required on each commutating pole of a turbo-generator: The com-
mutating poles are made of cast steel; the average flux density on
the face of the interpole is 6000 maxwells per sq.cm.; the pole-face
area 250 sq.cm.; the pole cross-section 160 sq.cm.; the radial length
of the interpole 27 cm.; the leakage factor, 1.5. The air-gap reluctance
is 2.7 millirels, the true tooth density 20 kilolines per sq.cm., the height
of the tooth 4.5 cm., and the armature ampere-turns per pole 9500.
Ans. About 15,300.
Prob. 14. The rated current of the machine in the preceding problem
CHAP. IX] ARMATURE REACTION IN D.C. MACHINES 175
is 1200 amp., and in addition to the interpoles the machine is to be
provided with a compensating winding. Show that each interpole
should have at least 8 turns, and each main pole be provided with 10
bars for the compensating winding, in order to neutralize the armature
distortion under the main poles and provide the proper commutating
field.
Prob. 15. Machines provided with interpoles are very sensitive
as to their brush position. By shifting the brushes even slightly from
the geometrical neutral the terminal voltage of such a generator can be
varied to a considerable extent. In the case of a motor the speed can
be regulated by this method, without adjusting the field rheostat. Give
an explanation of this " compounding" effect of the brush shift.
55. Armature Reaction in a Rotary Converter. The actual
currents of irregular form which flow in the armature winding of a
rotary converter may be considered as the resultants of the direct
current taken from the machine and of the alternating currents
taken in by the machine. The resultant magnetizing action upon
the field is the same as if these two kinds of currents were flowing
through two separate windings. Therefore, the armature reaction
in a rotary converter can be calculated by properly combining the
armature reactions of a synchronous motor and of a direct-current
generator.
The alternating-current input into a rotary converter may be
either at a power factor of unity, if the field excitation is properly
adjusted, or the input may have a lagging or a leading component,
the same as in the case of a synchronous motor. The armature
reaction due to the energy component of the input consists chiefly
in the distortion of the field, against the direction of rotation of the
armature. But the action of the direct current is to distort the
field in the direction of rotation, and since the two m.m.fs. are not
much different from one another, the resultant transverse arma-
ture reaction is very small. The direct reaction of the direct cur-
rent depends upon the position of the brushes, and the direct
reaction due to the alternating currents is determined by the
reactive component of the input, which component may vary
within wide limits. Thus, the resultant direct reaction of a rotary
converter may be adjusted to almost any desired value.
The ohmic drop in the armature of a rotary converter has a
different expression than in either a direct-current or a synchro-
nous machine, because the i2r loss must be calculated for the actual
shape of the superimposed currents. Rotary converters are some-
176 THE MAGNETIC CIRCUIT [ART. 55
times provided with interpoles, in order to improve the commu-
tation and to use a higher specific electric loading.1
1 For further details in regard to the armature reaction in a rotary con-
verter see Arnold, Wechselstromtechnik, Vol. 4 (1904), Chap. 28; Pichel-
mayer, Dynamobau (1908), p. 276; Standard Handbook, index under " Con-
verter, synchronous, effective armature resistance"; Parshall & Hobart,
Electric Machine Design (1906) p. 377; Lamme and Newbury, Interpoles in
Synchronous Converters, Trans. Amer. Inst. Electr. Engrs., Vol. 29 (1910),
p. 1625.
CHAPTER X
ELECTROMAGNETIC ENERGY AND INDUCTANCE
56. The Energy Stored in an Electromagnetic Field. Experi-
ment shows that no supply of energy is required to maintain a
constant magnetic field. The power input into the exciting coil or
coils is in this case exactly equal to that converted into the i2r heat
in the conductors, and this power is the same whether a magnetic
field is produced or not. Another familiar example is that of a
permanent magnet, which maintains a magnetic field without any
supply of energy from the outside, and apparently without any
decrease in its internal energy. Nevertheless, every magnetic field
has a certain amount of energy stored within it, though in a form
yet unknown. This is proved by the fact that an expenditure of
energy is required to increase the field, and, on the other hand,
when the flux is reduced, some energy is returned into the
exciting electric circuit.
The conversion of electric into magnetic energy and vice
versa is accomplished through the e.m.f. induced by the vary-
ing flux. Let a magnetic flux be excited by a coil CC (Fig. 45)
supplied with current from a source of constant voltage E, and let
there be a rheostat r in series with the coil. Let part of the resist-
ance in the rheostat be suddenly cut out in order to increase the
current in the coil. It will be found that the current rises to its
final value not instantly ; namely, when the current increases, the
flux also increases, and in so doing it induces in the electric circuit
an e.m.f., say e, which tends to oppose the current. This et
together with the iR drop, balances the voltage E, so that for a
time the power Ei supplied by the source is larger than the power
i2R lost in the total resistance of the circuit. The difference,
Ei —i2R, is stored in the magnetic field created by the coil and by
the other parts of the circuit. The energy eidt supplied by the
electric circuit during the element of time dt is converted into the
magnetic energy of the field, by the law of the conservation of
177
178
THE MAGNETIC CIRCUIT
[ART. 56
energy, while the amount (E —e)idt=i2Rdt is converted into
heat.
If now the same resistance is suddenly introduced into
the circuit, the current gradually returns to its former value, and
during this variable period the i2R loss is larger than the power Ei
supplied by the source. The applied voltage is in this case assisted
by the voltage e induced by the decreasing field, the e.m.f . e sup-
plying part of the i2R loss.
To make the matter more concrete let the source of electric
energy be a constant-voltage, direct-current generator, driven by
FIG. 45. — The magnetic field produced by a coil, showing complete and
partial linkages.
a steam turbine. Let the load consist of coils of variable resist-
ance R, and let the coils produce a considerable magnetic field.
As long as the load current is constant, the rate of the steam con-
sumption is determined by the i2R loss in the circuit. When the
current increases, the steam is consumed at each instant at a
higher rate than it would be with a constant current of the same
instantaneous value. The energy of steam is thus partly stored
in the magnetic field of the coils. When the current is returned
to its former value, the steam consumption during the transitional
CHAP. X] ENERGY AND INDUCTANCE 179
period is less than that which corresponds to the i2R loss in the
circuit, so that practically the same amount of steam is saved
which was expended before in increasing the magnetic field.
These phenomena may be explained, or at least expressed in
more familiar terms, by assuming the magnetic field to be due to
some kind of motion in a medium possessed of inertia (Art. 3).
When the field strength is increased it becomes necessary to accel-
erate the parts in motion, overcoming their inertia. When the
field is reduced, the kinetic energy of motion is returned to the
electric circuit. One can also conceive of the energy of the mag-
netic field to be static and in the form of some elastic stress.
Under this hypothesis, when a current increases, the magnetic
stress also increases at the expense of the electric energy. In
either case, when the current is constant no energy is required
to maintain the field, any more than to maintain a constant rota-
tion in a fly-wheel or a constant stress in an elastic body.
It seems the more probable that the magnetic energy of a
circuit is stored in some kinetic form, because the current which
accompanies the flux is itself a kinetic phenomenon. On the other
hand, it appears more likely that electrostatic energy is due to some
elastic stresses and displacements in the medium, and thus it
may be said to be potential energy. Electric oscillations and
waves consist, then, in periodic transformations of electrostatic into
electromagnetic energy, or potential into kinetic energy, and vice
versa, similar to the mechanical oscillations and waves in an
elastic body. In the familiar case of current or voltage resonance
the total energy of the circuit at a certain instant is stored in the
form of electrostatic energy in the condenser (permittor) con-
nected into the circuit, or in the natural permittance of the circuit ;
the current and the magnetic energy at this instant are equal
to zero. At another instant, when the current and the magnetic
field are at their maximum, the energy stored is all in the form
of magnetic energy, and the voltage across the condenser and
the stress in the dielectric are equal to zero. An oscillating
pendulum offers a close analogy to such a system. The resistance
of the electric circuit, and the magnetic and dielectric hysteresis,
are analogous to the friction and windage which accompanies
the motion of the pendulum.
As it is of importance hi mechanical machine design to know
the inertia of the moving parts of a machine, so it is often necessary
180 THE MAGNETIC CIRCUIT [ART. 57
in the design and operation of electrical apparatus and circuits to
know the amount of energy stored in the magnetic field, or the
electro-magnetic inertia. This inertia modifies the current and
voltage relations in the electric circuit in somewhat the same way
in which the inertia of the reciprocating parts in an engine modi-
fies the useful effort. In mechanical design a revolving part is
characterized by its moment of inertia -from which the stored
energy can be calculated for any desired speed. So in electrical
engineering a circuit or an apparatus is characterized by its electro-
magnetic inertia or inductance, from which it is possible to calculate
the magnetic energy stored in it at any desired value of the cur-
rent. In this and in the two next chapters expressions are deduced
for the inductance of some of the principal types of apparatus
used in electrical engineering.
Prob. 1. A stationary electromagnet attracts and lifts its armature
with a weight attached to it. Explain how the energy necessary for
the lifting of the weight is supplied by the electric circuit.
Prob. 2. A direct-current generator driven by a water-wheel is
subjected to very large and sudden fluctuations of the load, which
the governor and the gate mechanism are unable to follow properly.
A heavy fly-wheel on the generator shaft would improve the operating
conditions. Would a reactance coil in series with the main circuit
achieve the same result, provided that it could be made large enough
to store the required excess of energy ?
Prob. 3. What experimental evidence could be offered to support
the contention that the energy of an electric circuit is contained in the
magnetic field linked with the current, and not in the current itself ?
The flow of current is usually compared to that of water in a pipe;
is not all the kinetic energy stored in the moving water itself and not
in the surrounding medium, and if so, is a current of electricity really
like a current of water ?
Prob. 4. Describe in detail current and voltage resonance 1 and
free electrical oscillations, from the point of view of the periodic conversion
of electromagnetic into electrostatic energy and vice versa, taking account
of the dissipation of energy in the resistance of the circuit.
57. Electromagnetic Energy Expressed through the Linkages
of Current and Flux. In order to obtain a general expression for
the energy stored in the magnetic field of an electric circuit, con-
sider first a single loop of wireaa (Fig. 11) through which a steady
electric current i is flowing. Let the cross-section of the wire be
small as compared to the dimensions of the loop, so that the flux
1 See the author's Experimental Electrical Engineering, Vol. 2, pp. 17 to 25.
CHAP. X] ENERGY AND INDUCTANCE 181
inside the wire may be disregarded. The electromagnetic energy
possessed by the loop is equal to the electrical energy spent in
establishing the current i in the loop against the induced e.m.f.
Let it and $t be the instantaneous values of the current in
amperes and the flux in webers at a moment t during the period of
building up the flux, and let et be the instantaneous applied
voltage. Let the flux increase by dOt during an infinitesimal
interval dt; then the electrical energy (in joules) supplied from
the source of power to the magnetic field is
dW =itetdt =it(d$t/dt)dt =itd®t,
where— et= —dtf>t/dt is the instantaneous e.m.f. induced in the
loop by the changing flux. The total energy supplied from the
electrical source during the period of building up the field to its
final value 4> is
TF= I itd$t (98)
In a medium of constant permeability the integration can be
easily performed, because the flux is proportional to the current,
or, according to eq. (2) in Art. 5, $t =(Pit, where (P is the permeance
of the magnetic circuit, in henrys. Eliminating by means of this
relation either it or <Dt from eq. (98) we can obtain any one of the
following three expressions for the electromagnetic energy stored
in the loop :
(99)
In the first form, eq. (99) expresses the fact that the magnetic
energy stored in a loop is equal to one-half the product of the cur-
rent by the flux; in the second form, it shows that the stored
energy is proportional to the square of the current and to the per-
meance of the magnetic circuit. Both forms are of importance in
practical applications.
Take now the more general case of a coil of n turns (Fig. 45) ;
the flux which links with a part of the turns is now of a magnitude
comparable with that of the flux which links with all the turns of
the coil. We shall consider the complete linkages and the partial
linkages separately. Consider first the energy due to the flux
182 THE MAGNETIC CIRCUIT [ART. 57
which links with all the turns of the coil. The e.m.f. induced in
the coil by this flux, when the current changes, is — et = —n(d0t/df),
and the relation between the current and the flux is 0t=(Pcnit}
where (Pc is the permeance of the path of the complete linkages.
By repeating the reasoning given above in the case of a single loop
we find that
Wc=%ni#c, ..... . . (100)
or
...... (lOOa)
where the subcript c signifies that the quantities refer to the com-
plete linkages only (Fig. 45) . Two forms only are retained, being
those that are of the most practical importance.
The energy of the partial linkages is calculated in a similar
manner. Let A$P be a small annular flux (Fig. 45) which links
with Up turns of the coil, where np may be an integer or a fraction.
For these turns the linkage with A$P is a complete linkage, while
for the external (n—np) turns it is no linkage at all and represents
no energy, because no e.m.f. is induced by A®? in the turns external
to it. Thus, the energy due to the flux A$P, according to eqs.
(100) and (lOOa), is equal to \nviA^P, or to %nP2i24(PP. The
total energy of the partial linkages is the sum of such expressions,
over the whole flux $P, or
WP=%iI>npA$p, ........ (101)
or
WP=%i22nP24(PP. . . ..... (lOla)
The total energy of the coil is
A$Pl ..... (102)
or
W=^i2[n2(Pc + I,nP2A(PPl .... (102a)
where the first term on the right-hand side refers to the complete
linkages and the second to the partial linkages of the flux and the
current. In these expressions the current is in amperes, the fluxes
in webers, the permeances in henrys, and the energy in joules
(watt-seconds) . If other units are used the corresponding numeri-
cal conversion factors must be introduced.
CHAP. X] ENERGY AND INDUCTANCE 183
Some new light is thrown upon these relations by using the
m.m.f . M instead of the ampere-turns ni. Namely, eqs. (102) and
(102a) become:
1fHEtfA*-2iffJ*4 .... (103)
or
. . . . (103a)
These expressions are analogous to those for the energy stored in
an electrostatic circuit, viz., %EQ, and %E2C (see The Electric Cir-
cuit). The m.m.f. Mc is analogous to the e.m.f. E; the magnetic
flux 0C is analogous to the electrostatic flux Q, and the permeance
(Pc is analogous to the permittance C.
We can assume as a fundamental law of nature the fact that
with a given steady current the magnetic field is distributed in such
a way that the total electromagnetic energy of the system is a
maximum. All known fields obey this law, and, in addition, it can
be proved by the higher mathematics. Eq. (102a) shows that this
law is fulfilled when the sum nc2(Pc + 2nP2(Ppisa, maximum. When
the partial linkages are comparatively small, the energy stored is a
maximum when the permeance (Pc of the paths of the total linkages
is a maximum. This fact is made use of in the graphical method
of mapping out a magnetic field, in Art. 41 above.
Prob. 5. The no-load saturation curve of an 8-pole electric gen-
erator is a straight line such that when the useful flux is 10 megalines
per pole the excitation is 7200 amp .-turns per pole; the leakage factor
is 1.2. Show that at this excitation there is enough energy stored
in the field to supply a small incandescent lamp with power for a few
minutes.
Prob. 6. Explain the function and the diagram of connections of
a field-discharge switch.
Prob. 7. Prove that the magnetic energy stored in an apparatus
containing iron is proportional to the area between the saturation curve
and the axis of ordinates. The saturation curve is understood to give
the total flux plotted against the exciting ampere-turns as abscissae.
Hint: See Art. 16.
Prob. 8. Deduce expression (102a) directly, by writing down an
expression for the total instantaneous e.m.f. induced in a coil (Fig. 45).
Prob. 9. Explain the reason for which, in the formulae deduced above,
it is permissible to consider n to be a fractional number.
58. Inductance as the Coefficient of Stored Energy, or the
Electrical Inertia of a Circuit. Eq. (102a) shows that in a mag-
netic circuit of constant permeability the stored energy is proper-
184 THE MAGNETIC CIRCUIT [ART. 58
tional to the square of the current which excites the field. The
coefficient of proportionality, which depends only upon the form
of the circuit and the position of the exciting m.m.f., is defined
as the inductance of the electric circuit. The older name for
inductance (is the coefficient of self-induction. It is assumed
here that the magnetic circuit is excited by only one electric cir-
cuit, so that there is no mutual inductance. Thus, by definition
.... (104)
where the inductance is
. . . . . (105)
or, replacing the summation by an integration,
L =n*(Pc + CnnP2d(PP. (106)
JQ
Since the permeances in eq. (102a) are expressed in henrys, and
the numbers of turns are numerics, the inductance L in the defin-
ing eqs. (105), or (106), is also in henrys. If the permeances are
measured in millihenrys or in perms, the inductance L is measured
in the same units. As a matter of fact, the henry was originally
adopted as a unit of inductance, and only later on was applied to
permeance.1
In some cases it is convenient to replace the actual coil (Fig. 45)
by a fictitious coil of an equal inductance, and of the same number of
turns, but without partial linkages. Let (Peq be the permeance of
the complete linkages of this fictitious coil; then, by definition,
eqs. (105) and (106) become
L=n2(Peq. ...... (106a)
This expression is used when the permeance of the paths is calcu-
lated from the results of experimental measurements of inductance,
because in this case it is not possible to separate the partial
linkages. Use is made of formula (106a) in chapter XII, in cal-
culating the inductance of armature windings.
1 The use of the henry as a unit of permeance was proposed by Professor
Giorgi. See Trans. Intern. Elec. Congress at St. Louis (1904), Vol. 1, p. 136.
The connection between inductance and permeance seems to have been first
established by Oliver Heaviside; see his Electromagnetic Theory (1894), Vol.
1, p. 31.
CHAP. X] ENERGY AND INDUCTANCE 185
The inductance L is related in a simple manner to the electro-
motive force induced in the exciting electric circuit when the cur-
rent varies in it. Namely, the electric power supplied to the cir-
cuit or returned from the circuit to the source is equal to the rate
of change of the stored energy, so that we have from eq. (104)
dW/dt=i(-e)=Li(di/dt),
or, canceling i,
e = -L(di/d£), . . ,' . (107)
The sign minus is used because e is understood to be the induced
e.m.f. and not that applied at the terminals of the circuit. There-
fore, when dW/dt is positive, that is, when the stored energy
increases with the time, e is induced in the direction opposite to that
of the flow of the current, and hence by convention is considered
negative. Inductance is sometimes defined by eq. (107), and then
eqs. (104) and (105) are deduced from it. The definition of L by
the expression for the electromagnetic energy seems to be a more
logical one for the purpose of this treatise, while the other defini-
tion in terms of the induced e.m.f. is proper from the point of
view of the electric circuit.
Looking upon the stored magnetic energy as due to some kind
of a motion in the medium, eq. (104) suggests the familiar expres-
sions %mv2 and %Kaj2 for the kinetic energy of a mechanical system.
Taking the current to be analogous to the velocity of motion, the
inductance becomes analogous to mechanical mass and moment of
inertia. The larger the electromagnetic inertia L the more energy
is stored with the same current. Equation (107) also has its
analogue in mechanics, namely in the familiar expressions mdv/dt
and Kdco/dt for the accelerating force and torque respectively.
The e.m.f. e represents the reaction of the circuit upon the source
of power when the latter tends to increase i the rate of flow of elec-
tricity. While these analogies should not be carried too far, they
are helpful in forming a clearer picture of the electromagnetic
phenomena.
The role of inductance, L, in the current and voltage relations of
alternating-current circuits is treated in detail in the author's
Electric Circuit. In this book inductance is considered from the
point of view of the magnetic circuit, i.e., as expressed by eqs. (104)
to (106) . In the next two chapters the values of inductance are
186 THE MAGNETIC CIRCUIT [ART. 58
calculated for some important practical cases, from the forms and
the dimensions of the magnetic circuits, using the fundamental
equations (104), (105) and (106). The reader will see that the
problem is reduced to the determination of various permeances and
fluxes; hence, it presents the same difficulties with which he is
already familiar from the study of Chapters V and VI.
Inductance of electric circuits in the presence of iron. When
iron is present in the magnetic circuit, three cases may be distin-
guished :
(1) The reluctance of the iron parts is negligible as compared
to that of the rest of the circuit ;
(2) The reluctance of the iron parts is constant within the
range of the flux densities used ;
(3) The reluctance of the iron parts is considerable, and is
variable.
In the first two cases, eqs. (104), (105) and (106) hold true, and
the inductance can be calculated from the constant permeances of
the magnetic circuit. In the third case, inductance, if used at all,
must be separately defined, because eq. (1026) does not hold when
the permeance of the circuit varies with the current. The equa-
tion of energy is in this case
W=n (\d0,c + 2 rnpitd(4$tp). (108)
Jo Jo
This equation is deduced by the same reasoning as eq. (98) .
The following three definitions of inductance are used by differ-
ent authors when the reluctance of a magnetic circuit is variable :
(a) the expressions (104) and (108) are equated to each other, and
L is calculated separately for each final value i of the current.
Thus L is variable, and neither eq. (105) nor (107) hold true, (b)
L is defined from eq. (107) ; in this case neither eq. (104) nor (105)
are fulfilled, (c) L is defined at a given current by eq. (105) so
that Li represents the sum of the linkages of the flux and the cur-
rent. Therefore eq. (107) becomes e = —d(Li)/dt, and dW =id(Li) .
With each of the three definitions L is variable, and therefore is not
very useful in applications. The author's opinion is that when the
permeance of the circuit is variable, L should not be introduced at
all, but the original equation of energy (108) be used directly. Or
else in approximate calculations, a constant value of L can be
used, calculated for some average value of i or 0.
CHAP. X] ENERGY AND INDUCTANCE 187
Prob. 10. It is desired to make a standard of inductance of one
millihenry by winding uniformly one layer of thin flat conductor upon
a toroidal wooden ring of circular cross-section. How many turns are
needed if the diameter of the cross-section of the ring is 10 cm. and
the mean diameter of the ring itself is 50 cm.? Ans. 400.
Prob. 11. An iron ring of circular cross-section is uniformly wound
with n turns of wire, the total thickness of the winding being t; the
mean diameter of the ring is D and the radius of its cross-section r.
What is the inductance of the apparatus, assuming the permeability
of the iron to be constant and equal to 1500 times that of the air?
Hint: d6>p = tt2n(r+x)dx/nD', np=n(x/t).
Ans. 1.25(n2/jD) [1500r2 + t($r + #)] X 10~8 henry.
Prob. 12. A ring is made of non-magnetic material, and has a
rectangular cross-section of dimensions b and h] the mean diameter
of the ring is D. It is uniformly wound with n turns of wire, the total
thickness of the winding being t. What is the inductance of the winding?
Ans. 0.4(n2/#) \bh + $t(b +h +3f)]lQ-* millihenry.
Prob. 13. The ring in the preceding problem has the following
dimensions: D = 50 cm.; /i = b = 10 cm.; it is to be wound with a con-
ductor 3 mm. thick (insulated). How many turns are required in order
to get an inductance of 0.43 of a henry? Hint : Solve by trials, assum-
ing reasonable values for t; the number of turns per layer decreases as
the thickness of the winding increases. Ans. About 5300.
Prob. 14. It is desired to design a choke coil which will cause a
reactive drop of 250 volts, at 10 amp. and 50 cycles. The cross-section
of the core (Fig. 12) is 120 sq.cm., and the mean length of the path
130 cm.; the maximum flux density in the iron must be not over 7
kl. per sq. cm. What is the required number of turns and the length
of the air-gap in the core?1 Ans. 150; 3.7 mm.
Prob. 15. An electrical circuit, which consists of a Leyden jar
battery of 0.01 mf. capacity and of a coil having an inductance of 10
millihenrys, undergoes free electrical oscillations in such a way that the
maximum instantaneous voltage across the condenser is 10,000 volts.
What is the current through the inductance one-quarter of a cycle
later, neglecting any loss of energy during the interval ?
Ans. 10 amp.
Prob. 16. Suggest a practical experiment which would prove directly
that the stored electromagnetic energy is proportional to the square
of the current.
Prob. 17. Prove that the inductance of a coil of given external
dimensions is proportional to the square of the number of turns, taking
into account the complete and the partial linkages. Show that the
ohmic resistance of the coil is also proportional to the square of the
number of turns, provided that the space factor is constant.
NOTE 1. The theoretical calculation of the ~ inductance of short
1 For a complete design of a reactive coil see G. Kapp, Transformers (1908),
p. 105.
188 THE MAGNETIC CIRCUIT [ART. 58
straight coils and loops of wire in the air is rather complicated, because
of the mathematical difficulties in expressing the permeances of the
paths. Those interested in the subject will find ample information
in Rosa and Cohen's Formula; and Tables for the Calculation of Mutual
and Self-Inductance, in the Bulletins of the Bureau of Standards, Vol.
5 (1908), No. 1. The article contains also quite a- complete bibliography
on the subject. See also Orlich, Kapazit'dt und Inductivit'dt (1909),
p. 74 et seq.
Note 2. In the formulae deduced in this and in the two following
chapters, it is presupposed that the current is distributed uniformly
over the cross-section of the conductors. Such is the case in conductors
of moderate size and at ordinary commercial frequencies, unless perchance
the material is itself a magnetic substance. With high frequencies,
or with conductors of unusually large transverse dimensions, as also with
conductors of a magnetic material, the current is not distributed uni-
formly over the cross-section of the conductors, the current density
being higher near the periphery. The result is that, as the frequency
increases, the inductance becomes lower and the ohmic resistance higher.
This is known as the skin effect. For an explanation, for a mathematical
treatment in a simple case, and for references see Heinke, Handbuch
der Elektrotechnik, Vol. 1 (1904) part 2, pp. 120 to 129. Tables and for-
mulas will be found in the Standard Handbook, and in Foster's Pocket-
Book. See also C. P. Steinmetz, Alternating-Current Phenomena (1908),
pp. 206-208, and his Transient Electric Phenomena (1909), Section III,
Chapter VII; Arnold, Die Wechselstromtechnik, Vol. 1 (1910), p. 564; A.
B. Field, Eddy Currents in Large Slot-wound Conductors, Trans. Amer.
Inst. Electr. Engrs., Vol. 24 (1905), p. 761.
CHAPTER XI.
Let a direct current of i
THE INDUCTANCE OF CABLES AND OF TRANS-
MISSION LINES.
59. The Inductance of a Single-phase Concentric Cable. Let
Fig. 46 represent the cross-section of a concentric cable, which
consists of an inner core A and of an external annular conductor
D, with some insulation C between them. Let the radii of the
conductors be a, b, and c, respectively. The insulation outside of
D and the sheathing are not shown,
amperes flow through the inner
conductor away from the reader
and return through the outer
conductor. The magnetic field
produced by this current links
with the current, and for reasons
of symmetry the lines of force
are concentric circles. The field
is confined within the cable,
because outside the external con-
ductor D the m.m.f. isi — i=0.
In the space between the two
conductors the lines of force are
linked with the whole current,
and since there is but one turn,
the m.m.f. is equal to i. The length of a line of force of a radius x
cm. is 27rzsothat the magnetic intensity isH =i/2nx, amp. turns per
cm., the corresponding flux density B=/j.i/2nx maxwells per sq.cm.
Thus, the flux density decreases inversely as the distance from the
center; it is represented by the ordinates of the part qr of a hyper-
bola.
In the space within the inner conductor A, a line of force of
radius x is linked with a current ix=i(nx2/na?) =i(x/a)2, provided
189
Fig. 46— The magnetic field within
a concentric cable.
190 THE MAGNETIC CIRCUIT [ART. 59
that the current is distributed uniformly over the cross-section of
the conductor. The length of the line of force is 2xx, so that
H=ix/2nx=xi/(2na2), and B=fix i/(2na2). Thus, in this part
of the field the flux density increases as the distance from the
center and is represented by the straight line Oq.
In the space inside the conductor D, a line of force of a
radius x is linked with the current — i(x2 — 62)/(c2— b2) of the
external conductor and with the current +i of the internal con-
ductor, or altogether with the current ix=i(c2—x2)/(c2—b2).
Consequently, here, the flux density is represented by the
hyperbola rs, the equation of which is
B=fjLix/27tx = fjii(c2/x -x)/[2n(c2 -62)].
The curve Oqrs gives a clear physical picture of the field dis-
tribution in the cable, and helps one to understand the linkages
which enter into the calculation of the inductance of the cable.
The inductance of the cable is calculated according to the
fundamental formula (106), the complete linkages being in the
space between the two conductors, and the partial linkages being
within the space occupied by the conductors themselves. Con-
sider a piece of the cable one centimeter long. The permeance of a
tube of force of a radius x and of a thickness dx is fj.dx/2nx, so that
the permeance of the complete linkages is,
LC'=(PC'= CbfjLdx/2nx = (fjL/2n)Ln(b/a) perm/cm., (109)
where Ln is the symbol for natural logarithms. In this case the
permeance is equal to the inductance because the number of turns
n = l. The sign "prime" indicates that the quantities Lcr and
(Pef refer to a unit length of the cable.
For the space inside the inner conductor np = (x/a)2, this being
the fraction of the current with which the line of force of radius x
is linked. Hence, the part of the inductance of the cable due to
the field inside the conductor A is
LA' = (/£/27r) C*(x/a)*(dx/x) =n/8n =0.05 perm/cm. (110)
^o
This formula shows that the part of the inductance due to the
field within the inner conductor is independent of the radius of the
CHAP. XI] INDUCTANCE OF TRANSMISSION LINES 191
conductor, and is always equal to 0.05 perm, per cm., or 0.05 milli-
henry per kilometer length of the cable.
The exact expression for the part of the inductance of the cable
LD' due to the linkage within the outer conductor is given in
problem 10 below. The formula is rather complicated for prac-
tical use, especially in view of the fact that this part of the induct-
ance is comparatively small, because the flux density on the part
rs of the curve is small. It is more convenient, therefore, to make
simplifying assumptions, when the thickness t of the outer con-
ductor is small as compared to b. Namely, the length of all the
paths within the outer conductor may be assumed to be equal
to 2nb, so that the permeance of an infinitesimal path of a radius x
and thickness dx nearly equals fjidx/2xb. Furthermore, the volume
of the current in the outer conductor, between the radii b and x>
may be assumed to be proportional to the distance z— &, and
hence equal to i (x —b}/(c —b). A line of force of a radius x is linked
therefore with the whole current i in the inner conductor and with
the above-stated part of the current in the outer conductor, and,
since the currents flow in opposite directions, this line of
force is linked altogether with the current i(c—x)/(c—b). Hence,
it is linked with np = (c —x)/(c —b) turns. Thus, the inductance of
the cable, due to the outer partial linkages, is, in the first approxi-
mation,
LD'=n/(2nbt2) C\c-x)2dx=^t/b perm/cm. . (Ill)
"b
If a closer approximation is desired, it is convenient to expand
eq. (114) in ascending powers of t/b, as is explained in problem 11.
The result is
LD' =rV/^[^ ~ro(^/^)2^~ A(V^)3 • • •] perm/cm. . (112)
It will thus be seen that eq. (Ill) is an accurate approximation,
because eq. (112) contains in the parentheses no term with the
first power of the ratio t/b.
Thus, the total inductance of a concentric cable, I kilometers
long, is
L =[0.46 Log10 (b/a) +0.05+LD']Z millihenrys, . (113)
where LD' is given by eqs. (Ill), (112), or (114), according to the
accuracy desired. Expressions (110) and (114) are correct only at
low frequencies, such as are used for power transmission. With
192 THE MAGNETIC CIRCUIT [ART. 59
very high frequencies, the skin effect becomes noticeable, that is,
the current in the inner conductor is forced outward and that in
the outer conductor inward. In the limit, when the frequency
is infinite, the currents are concentrated on the opposing surfaces
of the conductors, and the partial linkages are equal to zero.
Thus, each of the expressions in question must be multiplied by a
variable coefficient k which, for a given cable, is a function of
the frequency. At ordinary frequencies k = 1, and gradually
approaches zero as the frequency increases to infinity.1
Prob. 1. A concentric cable is to be designed for 750 amperes, the
current density to be about 2.2 amp. per gross sq.mm., and the thickness
of the insulation between the conductors to be 6 mm. What are the
dimensions of the conductors assuming them to be solid, that is, not
stranded? The fact that they are in reality stranded is taken care of
in the permissible current density.
Ans. a = 10.5 ; b = 16.5 ; c = 19.6 mm.
Prob. 2. Plot the curve Oqrs of distribution of the flux density
in the cable given in the preceding problem.
Ans. At x= a, B= 143; at x = b, B= 91 maxwells per sq.cm.
Prob. 3. What is the total flux in megalines per kilometer of the
cable specified in the two preceding problems? Ans. 15.1.
Prob. 4. Show how to plot the cilrve of the distribution of flux density
in a three-phase concentric cable, at some given instantaneous values
of the three currents.
Prob. 5. What is the inductance of a 25-km. cable in which the
diameter of the inner conductor is 12 mm., the thickness of insulation
is 3 mm., and the dimension c is such that the current density in the
outer conductor is 10 per cent higher than in the inner one ?
Ans. 25 [0.0810 + 0.050 + 0.0122] = 3.58 millihenry.
Prob. 6. A cable consists of three concentric cylinders of negligible
thickness; the radii of the cylinders are rv, r2, and r3, beginning with
the inner one. What is the inductance in millihenrys per kilometer,
when a current flows through the inner cylinder and returns equally
divided through the two others? ,
Ans. 0.46 [log (r2/rO +0.25 log (r3/r2) .]
Prob. 7. In the cable given in the preceding problem the total
current i flows through the middle cylinder, the part mi returns through
the inner cylinder, and the rest, ni, returns through the outer one. What
is the total inductance per kilometer of length?
Ans. 0.46 [m2 log (r,/rO +n2 log (r./r,)].
1 For the field distribution in and the inductance of non-concentric cables
see Alex. Russell, Alternating Currents, Vol. 1 (1904), Chap. XV; for the
reactance of armored cables see J. B. Whitehead, " The Resistance and React-
ance of Armored Cables, Trans. Amer. Inst. Electr. Engrs., Vol. 28 (1909),
p. 737.
CHAP. XI] INDUCTANCE OF TRANSMISSION LINES 193
Prob. 8. At what ratio of b to a in Fig. 46 is the magnetic energy
stored within the inner conductor equal to that stored between the
two conductors ?• Ans. 1.28.
Prob. 9. It is required to replace the solid inner conductor A in
Fig. 46 by an infinitely thin shell of such a radius a' that the total
inductance of the cable shall remain the same. What is the radius of
the shell ? Hint : ( ft /2n) Ln (a /a') = /I/STT.
Ans. a'/a = £-°'25=0.779.
Prob. 10. Prove that the part of the inductance due to the linkages
within the outer conductor in Fig. 46 is expressed by
;r_i(3^+^-4c^)]. . . (114)
Hint: dVp^ndx/lnx; np = l-n(x2-b*)/7t(c*/b2).
Prob. 11. Deduce eq. (112) from formula (114), assuming the
ratio t/b to be small as compared to unity. Hint: Put c = b(l+y)
where y = t/b is a small fraction. Expand Ln(l-y) into an infinite
series, and omit in the numerator of eq. (114) all the terms above f/5;
expand (c2 — 62) in the denominator in ascending powers of y, and divide
the numerator by this polynomial.
60. The Magnetic Field Created by a Loop of Two Parallel
Wires. Let Fig. 47 represent the cross-section of a single-phase
or direct-current transmission line, the wires being denoted by
A and B. With the directions of the currents in the wires shown
by the dot and the cross, the magnetic field has the directions
shown by the arrow-heads, one-half of the flux linking with each
wire. Before calculating the inductance of the loop it is instruct-
ive to get a clear picture, quantitative as well as qualitative, of the
field itself.
The field distribution is symmetrical with respect to the line
AB and the axis 00' . The whole flux passes in the space between
the wires, so as to be linked with the m.m.f . which produces it, and
then extends to infinity on all sides. The flux density is at its
maximum near the wires and gradually decreases toward 00'
and toward ± oo, as is shown by the curve pqsts'q'p'. The ordi-
nates of this curve represent the flux densities at the various points
of the line passing through the centers of the wires. The reason
for which the flux density is larger near the wires is that the path
there is shorter, although the m.m.f. acting along all the paths
is the same. This m.m.f. is numerically equal to the current i in
the wires, the number of turns being equal to one.
It is proved below that the magnetic paths outside the conduc-
194
THE MAGNETIC CIRCUIT
[ART. 60
tors themselves are eccentric circles, with their centers on the
line AB extended. The equipotential surfaces are circular
cylinders, which are shown in Fig. 47 as circles passing through the
centers A and B of the conductors. Within the conductors them-
selves there are no equipotential surfaces.
For purposes of analysis it is convenient to regard the field in
Fig. 47 as the result of the superposition of two simpler fields, simi-
lar to those of the concentric cable of the preceding article. Con-
FIG. 47. — The magnetic field produced by a single-phase transmission line.
sider the conductor A, together with a concentric cylinder of an
infinitely large radius, as one conducting system. Let the current
flow through A toward the reader, and return through the infinite
cylinder. Let the conductor B with a similar concentric cylinder
form another independent system. The currents in the conduc-
tors A and B are to be the same as the actual currents flowing
through them, but each infinite cylinder is to serve as a return for
the corresponding conductor, as if there were no electrical connec-
tion between A and B. The currents in the two cylinders are
flowing in opposite directions and the cylinders themselves
CHAP. XI] INDUCTANCE OF TRANSMISSION LINES 195
coincide at infinity, because the distance AB between their axes is
infinitely small as compared to their radii. Hence, the two cur-
rents in the cylinders cancel each other, and the combination of
the two component systems is magnetically identical with the
given loop.
In a medium of constant permeability, the resultant magnetic
intensity H, produced at a point by the combined action of several
independent m.m.fs., is equal to the geometric sum of the intensi-
ties produced at the same point by the separate m.m.fs.1 This
being true of the intensities, the component flux densities at any
point are also combined according to the parallelogram law
because they are proportional to the intensities. Hence, the
resultant flux can be regarded as the result of the superposition
of the fluxes created by the component systems.
The field produced by the system A consists of concentric cir-
cles, the flux density outside the conductor A being inversely pro-
portional to the distance from the center of A (curve qr in Fig. 46).
The field created by the system B consists of similar circles around
B, and the field shown in Fig. 47 is a superposition of these two
fields. Thus the resultant field intensity H at a point P is a
geometric sum of
Hi=i/2*ri, (115)
and
H2=i/2xr2, (116)
HI and H2 being perpendicular to the corresponding radii vectors
TI and r2 from the centers of the conductors to the point P. The
directions of HI and H2 are determined by the right-hand screw
rule. Since HI and H2 are known in magnitude and direction at
each point of the field, the resultant intensity H may also be deter-
mined.
To deduce the equation of the lines of force in the resultant
field, we shall express analytically the condition that the total flux
which crosses the surface CP is equal to zero, provided that C and
P lie on the same line of force. This total flux may be considered
1 This principle of superposition can be considered (a) as an experimental
fact ; (6) as an immediate consequence of the fact that in a medium of constant
permeability the effects are proportional to the causes*; (c) as a consequence
of Laplace's law dH = Const. Xids sin 0/10r2, according to which the total
field intensity is regarded as the sum of those produced by the infinitesimal
elements of the current, or currents.
196 THE MAGNETIC CIRCUIT [ART. 60
as the resultant of the fluxes due to the systems A and B. Accord-
ing to eq. (115), the flux density due to A, at a distance x from A,
is BI = jj.i/2xx, so that the flux due to A which crosses CP is
= C
t/A
(117)
This flux is directed to the left, looking from C to P. By analogy,
the flux due to the system B is
02f = (fjLi/2n)'Ln(r2/BC) maxwells/cm., . . . (118)
and is directed to the right, looking from C to P. The condition
that no flux crosses CP is, that </Y is equal to $2', or
Ln(r!/AC)=Ln(r2/£C),
or
ri/r2=AC/BC=Const ..... (119)
This is the equation of a line of force in " bipolar " co-ordinates;
the curve is such that the ratio of r*i to r2 remains constant, How-
ever, this constant is different for each line of force, because each
line has its own point C.
Eq. (119) may be proved to represent a circle, by selecting an
origin, say at A, and substituting for rx and r2 their values in terms
of the rectangular co-ordinates x and y. The following proof by
elementary geometry leads to the same result. Produce AP and
lay off PD=PB=r2. According to eq. (119), BD is parallel to
CP, and consequently PC bisects the angle APB =to. Let the
point C' lie on the same line of force with C; then no flux passes
through PC', and by analogy with eq. (119) we have
n/r2=AC'/BC'=C<mat. .... (120)
By plotting PD' =r2 (not shown in figure) along PA, in the oppo-
site direction from PD, and connecting D' to B, one can show as
before that PC' bisects the angle BPD = l8Q0-w. But the
bisectors of two supplementary angles are perpendicular to each
other; consequently, CPC' is a right angle, and the point P lies
on a semicircle erected on the diameter CC'. This semicircle is the
line of force itself, because all the points, such as P, which are deter-
mined by C and C' must lie on it. Another semicircle below the
line AB closes the line of force.
CHAP. XI] INDUCTANCE OF TRANSMISSION LINES 197
From eqs. (119) and (120) the following expressions are obtained
for the radius R of the line of force :
BC
or
BC'
• •••• • • • (122)
so that the line of force can be easily drawn for a given C or C".
To prove that the equipotential lines are also circles we proceed
as follows. The line AB is evidently an equipotential line, because
it is perpendicular to all the lines of force. The difference of mag-
netic potent'al or the m.m.f. between AB and P, contributed by
the system A, is equal to i(d\/2n) ampere-turns, where the angle
BAP is denoted by 0\. This is because the m.m.f. due to the sys-
tem A, taken around a complete circle concentric with A, is equal
to i, and is distributed uniformly along the circle, for reasons of
symmetry. Or else it follows directly from eq. (115). By
analogy, the difference of magnetic potential between AB and P,
due to the system B, is i(6i/lTi). Thus the total difference of
potential between AB and P is
Mcp = (i/^)(dl+d2)=(i/2n)(n-^. . . (123)
This shows that the m.m.f. between any two points in the field is
proportional to the difference in the angles CD at which the line AB
is visible from these points. For any two points on the same
equipotential line Mcp is the same, so that the equation of such a
line is
&>=const ........ (124)
This represents the arc of a circle passing through A and B, and
corresponding to the inscribed angle a).
Prob. 12. A single-phase transmission line consists of two conductors
1 cm. in diameter, and spaced 100 cm. between the centers. Draw
the curve of flux density distribution (pqst in Fig. 47) for an instantaneous
value of the current equal to 100 amp.
Ans. z = 50.0 25.0 0.5 -0.5 -50.0 -oo cm.;
B= 1.60 2.1380.40 -79.60 -0.53 0 maxw./sq. cm.
Prob. 13. For the transmission line in problem 12 draw the lines
198 THE MAGNETIC CIRCUIT [ART. 61
of force which divide the total flux between the wires into ten equal
parts (not counting the flux within the wires).
Ans. The circles nearest to 00' cross AB at a distance of 48.4
cm. from each other.
Prob. 14. Referring to the two preceding problems, draw ten
equipotential circles which divide the whole m.m.f. of 100 ampere-turns
into ten equal parts.
Ans. The arcs nearest to AB intersect 00' at a distance of
32.5 cm. from each other.
Prob. 16. A telephone line runs parallel to a single-phase power
transmission line. The position of one of the telephone wires is fixed;
show how to determine the position of the other wire so as to have a
minimum of inductive disturbance in the telephone circuit. Hint:
The center lines of the two telephone wires must intersect the same line
of force due to the power line.
Prob. 16. A telephone line runs parallel to a 25-cycle, single-phase
transmission line. The distances from one of the telephone wires to
the power wires are 3.5 m. and 2.7 m.; the distances from the other
telephone wire to the power wires are 3.6 and 2.5 m. (in the same order).
What voltage is induced in the telephone line per kilometer of its length,
when the current in the power line is 100 effective amperes?
NOTE: In practice, this voltage is neutralized by transposing
either line after a certain number of spans. Ans. 0.33 volt.
61. The Inductance of a Single-phase Line. The inductance of
a single-phase line (Fig. 47) can be calculated according to the fun-
damental formulae (105) or (106), provided that the permeances
of the elementary paths be expressed analytically. However, in
this case it is much simpler to use the principle of superposition
employed in the preceding article, and to consider the actual flux
as the resultant of two fluxes each surrounding concentrically one
of the wires and extending to infinity. The fluxes produced by
the two component systems are equal and symmetrical with
respect to the wires. It is therefore sufficient to calculate the
linkages of the loop AB with the flux produced by one of the sys-
tems, say that corresponding to A, and to multiply the result by
two.
The flux produced by A and having A as a center link, with
the current in the loop AB. These linkages may be divided into
the following:
(a) Linkages within the wire A ; that is, from x =0 to x =a;
(6) Linkages between the wires A and B, that is, from
x=a to x =b— a;
CHAP. XIJ INDUCTANCE OF TRANSMISSION LINES 199
(c) Linkages outside the loop, that is, from x=b+a to x =
infinity,
(d) Linkages within the wire B, that is, from
^.
x=b— a to x=b+a.
The linkages (a), (b) and (c) are the same as in a concentric
cable (Fig. 46), because the shape of the lines of force and the
number of turns with which they are linked are the same. The
partial linkages (d) are somewhat difficult to express analytically.
When the distance b between the wires is large as compared to
their diameters, the whole current in B may be assumed to be con-
centrated along the axis of the wire B, instead of being spread over
the cross-section. With this assumption, the partial linkages (d)
are done away with, the linkages (6) are extended to x =b, and the
linkages (c) begin from x =b. The expressions for the linkages (a)
and (b) are given by eqs. (110) and (109) respectively. The link-
ages (c) are equal to zero, because in this region the lines of force
produced by A are linked with both A and B, and therefore
with i— i=Q ampere -turns. Thus, the inductance in question is
L' =0.46 logio(6/o) +0.05 (125)
This gives the inductance of a single-phase line in perms per centi-
meter length, or in millihenry s per kilometer length of the wire.1
To obtain the inductance per unit length of the line this expression
must be multiplied by two, because the linkages due to the flux
produced by the system B are not taken into account in eq. (125).
However, for the purposes of the next two articles it is more con-
venient to use expression (125), and to consider separately the
inductance of each wire, remembering that the two wires of a loop
are in series, and that therefore their inductances are added.2
Prob. 17. Check by means of formula (125) some of the values for
the inductance and reactance of transmission lines tabulated in the
various pocketbooks and handbooks.
1 It is of interest to note that the exact integration over the partial linkages
(d) leads to the same Eq. (125), so that this formula is correct even when
the wires are close to each other. See A. Russell, Alternating Currents, Vol.
1 (1904), pp. 59-60.
2 The inductance of two or more parallel cylinders oY any cross-section can
be expressed through the so-called " geometric mean distance," introduced
by Maxwell. For details see Orlich, Kapazitdt und Induktivitat (1909),
pp. 63-74.
200 THE MAGNETIC CIRCUIT [ART. 61
Prob. 18. Show by means of tables or curves that the inductance
of a transmission line varies much more slowly than (a) the spacing
with a given size of wire, (6) the size of wire with a given spacing.
Prob. 19. When the diameters of the two wires A and B are different,
prove that the inductance of the complete loop is the same as if the
diameter of each wire was equal to the geometric mean of the actual
diameters.
Prob. 20. Show that the inductance of a single-phase line with a
ground return can be calculated from eq. (125) by putting b=2h where
h is the elevation of the wire above the ground. Hint: In Fig. 47 the
plane 00' may be considered to be the surface of the earth, assumed
to be a perfect conductor. If the earth be removed, an "image" con-
ductor B must be added in order to provide a return path for the
current, such that the field surrounding A would remain the same.
Prob. 21. Two single-phase lines are placed on two cross-arms of
the same tower, one directly above the other, at a vertical distance
of c cm apart. What is the total inductance of the combination, when
the two lines are connected in parallel and each line carries one-half
of the total current?
Solution: Consider the four wires as forming four fictitious systems,
with cylinders at infinity as returns. Let b be the spacing in each
loop, and let b be larger than c. Denote the wires in one loop by 1
and 2, in the other by 1' and 2', and let d be the diagonal distance
between 1 and 2'. Assume all the wires except 1 to be of an infinitesimal
cross-section. Then, the linkages of the flux produced by the system
1 with the currents in the four wires are
Ln(c/o) +0.2i
+0.2(ii)2 Ln(d/6) milli joules /km.
Thus, allowing the same amount for the linkages due to the cur-
rent in the wire 1', we get that the inductance of the split line, each way,
is
(bd/ca) +0.05] millihenrys/km.,
instead of the expression (125) for the single line. The same result
is obtained when b is smaller than c. Hence, by splitting a line in
two the inductance is considerably reduced, because partial linkages are
substituted for some of the complete linkages. If d were equal to c
the reduction would be 50 per cent; but since d is always larger than
c the gain is less than 50 per cent. However, when the two lines are
very far apart the saving is very nearly 50 per cent.
Prob. 22. A certain single-phase transmission line has been designed
to consist of No. 000 B. & S. conductor with a spacing of 180 cm. It
is desired to reduce the reactive drop by about 20 per cent, without
increasing the weight of copper, by using two lines in parallel, with
the same spacing. What is the size of the conductor and the distance
between the loops? Ans. No. 1 B. & S. ; about 8 cm.
Prob. 23. Solve problem 21 when the load is divided unequally
CHAP. XI] INDUCTANCE OF TRANSMISSION LINES 201
between the two loops, the currents being mi and ni respectively, where
Ans. L' = 0.46[(ra2 + tt2) log (c/a) +log (6/c) +2mn log
0.05(ra2+n2), when 6>c, and L'=0.46[(ra2+n2) log (6 /a) +
2ranlog (d/c)]+0.05(w2+n2), when b<c.
Prob. 24. A single-phase line consists of three conductors, the
total current flowing through conductor 1, and returning through con-
ductors 2 and 3 in parallel. If the current in one return conductor is
mi, and in the other ni, where m+n = l, prove that the inductance of
the line per kilometer of its length is, in millihenrys,
L' = 0.46 [log (ftu/oO +w2 log (623/<z2) +n2 log (623/a3) +m log (bia/618)
(&12/623) +nlog (&18/b28)] +0.05(1 +m2+n2), when bn>bl3>b23.
In the particular case when bl2 = b23 = bn, al=a2 = a3, and m=n = %, the
inductance is reduced by 25 per cent as compared to that of a single loop.1
62. The Inductance of a Three-phase Line with Symmetrical
and Semi-symmetrical Spacing. The magnetic field which sur-
rounds a single-phase line varies in its intensity from instant to
instant, as the current changes, but the direction of the magnetic
intensity and of the flux density at each point remains the same.
In other words, the flux is a pulsating one. The field created by
three-phase currents in a transmission line varies at each point in
both its magnitude and direction. At the end of each cycle, the
field assumes its original magnitude and direction. If the spacing
of the wires is symmetrical, the field at the end of each third of a
cycle has the same magnitude and position with respect the next
wire. The field may therefore be said to be revolving in space.
This revolving flux, like that in an induction motor, induces
e.m.fs. in the three phases. The problem is to determine these
counter-e.m.fs. in the transmission line, knowing the size of the
wires, the spacing, and the load. In transmission line calculations,
especially in determining the voltage drop and regulation, it is
convenient to consider each wire separately, and to determine the
voltage drop in phase and in quadrature with the current. Thus,
having expressed the e.m.fs. induced by the revolving flux in terms
of the constants of the line, each wire is then considered as if it
were brought outside the inductive action of the two other wires.
We shall consider first the case of an equidistant spacing of the
three wires, because in most practical calculations of voltage drop
1The splitting of conductors discussed in problems 21 to 24 has been
proposed for extremely long transmission lines, in order to reduce their induct-
ance and at the same time increase their electrostatic permittance (capacity).
202 THE MAGNETIC CIRCUIT [ART. 62
an unsymmetrical spacing is replaced by an equivalent equidistant
spacing. The exact solution for an unsymmetrical spacing is given
in the next article. Let the instantaneous values of the three cur-
rents in the wires A, B and C of a three-phase transmission line be
ii, 12 and 13. The sum of the three currents at each instant is
zero, or
t'i +12+^3=0 (126)
Let 4>eq be the equivalent flux which links at any instant with the
wire A. The instantaneous e.m.f. induced in this wire is
e1 = -d#eq/dt. . (127)
The equivalent flux consists of the actual flux outside the wire plus
the sum of the fluxes inside the wire, each infinitesimal tube of flues
being reduced in the proper ratio, according to the fraction of the
cross-section of the wire with which it is linked. Or, what is the
same thing, each wire is replaced by an equivalent hollow cylinder
of infinitesimal thickness, without partial linkages, as in problem
9 in Art. 59 (consult also the definition of equivalent permeance
given in Art. 58).
In order to determine 4>eq we replace the three-phase system
by two superimposed single-phase systems. The current i\ in the
wire A may be thought of as the sum of the currents —12 and —13,
each flowing in a separate fictitious wire, and both of these wires
coinciding with A. The currents +12 in B and — 12 in A form
one single-phase loop, while the currents +t'3 in C and — i3 in A
form the other loop. The flux ®eq which surrounds A is the sum
of the fluxes produced by these two loops. The flux per unit
length of the line, due to the first loop, is equal to —(Peqi2, since
the number of turns is equal to one. For the same reason (P'eq =
U where U is determined by eq. (125). Hence, the flux per
unit length of the line, due to the first loop, is — L'i2. Similarly,
the flux due to the second loop and linked with A is equal to
—L'iz, the same value of U being used because the spacing and
the sizes of all of the wires are the same. Thus,
'eq
(128)
the last result being obtained by substituting the value of 12+^3
from eq. (126). Thus, eq. (127) becomes simply
ei = -L'dii/dt, (129)
CHAP. XI] INDUCTANCE OF TRANSMISSION LINES 203
that is, the induced e.m.f. is the same as in a single-phase line carry-
ing the current i\. Thus, the inductance of a three-phase line with
symmetrical spacing, per wire, is the same as the inductance of a
single-phase line, per wire, with the same size of wire and the same
spacing. The total e.m.f. induced in each wire is in quadrature
with the current in the wire.
In reaching this conclusion the following facts were made use
of : (a) The current in each wire at any instant is equal to the sum
of the currents in the two other wires; (6) the fluxes due to sepa-
rate m.m.fs. can be superimposed in a medium of constant perme-
ability; (c) The inductance of the loop A-B is equal to that of A-C
because of the same spacing. No other suppositions in regard to
the character of the load or the voltages between the wires were
made. Therefore, the conclusion arrived at holds true:
(a) For balanced as well as unbalanced loads;
(6) For balanced or unbalanced line voltages;
(c) For a three-wire two-phase system, three-wire single-
phase system, monocyclic system, etc.
(d) For sinusoidal voltages as well as for those departing from
this form.
Semi-symmetrical Spacing. When two out of the three distances
between the wires in a three-phase line are equal to each other, the
arrangement is called semi-symmetrical. Two common cases of
this kind are : (a) When the wires are placed at the vertices of an
isosceles triangle; (6) when they are placed at equal distances in
the same plane, for instance on the same cross-arm, or are fastened
to suspension insulators, one above the other. In such cases the
inductive drop in the symmetrically situated wire is the same as if
the wire belonged to a single-phase loop, carrying the same current,
and with a spacing equal to the distance of this wire to either of
the other two wires. Let, for instance, the distance A-B be equal
to B-C, and let the distance A-C be different from the two. The
proof given above can be repeated for the wire B, and the same
conclusion will be reached because the spacing A-C is not used
in the deduction. But, of course, the proof does not hold true for
either wire A or C.
When the three wires are in the same plane,^ the inductance of
each of the outside wires is larger than that of the middle wire.
This can be shown as follows : Let the three wires be in a horizontal
plane, and let them be denoted from left to right by A, B, C. Let
204 THE MAGNETIC CIRCUIT [ART. 62
the distance between A and B be equal to b and the distance
between A and C be equal to 26. If the wire B were moved to
coincide with C, the inductance of A would be the same as if it
belonged to a single-phase loop with a spacing 26. If C were
moved to coincide with B, the inductance of A would be that of a
wire in a single-phase loop with a spacing b. Thus, the inductance
of A corresponds in reality to a spacing intermediate between b and
26. The inductance of the middle wire B is the same as that of a
wire in a single-phase loop with the spacing 6, as is proved above.
Thus, the inductance of either A or C is larger than that of B.
An inspection of a table of the inductances or reactances of
transmission lines will show that the inductance increases much
more slowly than the spacing. For instance, according to the
Standard Handbook, the reactance per mile of No. 0000 wire, at
25 cycles, is 0.303 ohm with a spacing of 72 inch, and is 0.340 ohm
with a spacing of 150 inch. Therefore, in practical calculations,
when the spacing is semi-symmetrical, the values of inductance are
taken the same for all the three wires, for an average spacing
between the three, or, in order to be on the safe side, for the maxi-
mum spacing. In the most unfavorable case, even if an error of
say 10 per cent be made in the estimated value of the inductance,
and if the inductive drop is say 20 per cent of the load voltage, the
error in the calculated value of voltage drop is only 2 per cent of
the load voltage, and that at zero power factor. At power factors
nearer unity, when the vector of the inductive drop is added at
an angle to the line voltage, the error is much smaller.
Prob. 25. Show that the instantaneous electromagnetic energy
stored per kilometer of a three-phase line with symmetrical spacing is
equal to iZ/fo'+^'+tV) milli joules per kilometer, where U has the
value given by eq. (125). If this is true, then each wire may be con-
sidered as if it were subjected to no inductive action from the other
wires and had an inductance L' expressed by eq. (125) . This is another
way of proving eq. (129), and the statement printed in italics above.
Solution: Consider each wire, with a concentric cylinder at infinity,
as a component system. Determine the linkages of the field created
by the system A with the currents in A, B, and C, as in Art. 61. The
result is equal to \L'i?. Similar expressions are then written by analogy
for the fluxes due to the systems B and C.
Prob. 26. Show graphically that, when the distances A-B and
A-C are equal, the equivalent flux linking with A is independent of
the spacing B-C, and is the same as if B and C coincided. That is,
prove that the inductance of A is the same as if it belonged to a single-
CHAP. XI] INDUCTANCE OF TRANSMISSION LINES
205
phase loop. Solution: Let 7t, 72, and 73 (Fig. 48), represent the vectors
of the three currents at an unbalanced load. The current /x in A is
replaced by — 12 and — 73, and the system is split into two single-phase
loops, A-B and A-C. The fluxes due to these systems and linked with
A are denoted by fl>n and 013. They are in phase with the corresponding
currents, and are proportional to the magnitudes of these currents,
because of the equal spacing. Hence, the triangles of the currents and
of the fluxes are similar, and the resultant flux ^l linking with A is
in phase with 7t. If 0i2 = £(P72, and i03 = i(P73, where (P is the equivalent
permeance of each single-phase loop, then the result shows that (^i = ^(P71.
If the wires B and C coincided the equivalent permeance (P would be
the same, and hence the proposition is proved. The voltage drop, Eit
due to the flux 01 is shown by a vector leading /i by 90 degrees.
63. The Equivalent Reactance and Resistance of a Three-
phase Line with an Unequal Spacing of the Wires. In the case
of an unequal spacing of the wires eqs. (126) and (127) still hold
true, because they do not depend upon the spacing; but eq. (128)
becomes
./.. (130)
-i
where Z/i2 is the value of the inductance per unit length, calcu-
lated by eq. (125) for the spacing between A and B, and L'i3 is
the value of the inductance per unit
length, for the spacing A-C. Substi- Ti
tuting the valueof 4>eq from eq. (130)
into eq. (127) we get
This shows that with an unequal
spacing the effect of the mutual
induction of the phases cannot be
replaced by an equivalent inductance
in each phase, because, generally
speaking, the currents 12 and i3 cannot
be eliminated from this equation by
means of eq. (126).
Let us apply now eq. (131) to
the case of sinusoidal currents and
voltages. Let the current in the wire B be 12 = v2l 2 sin Inft,
where 72 is the effective value of the current; then the first term
on the right-hand side of eq. (131) becomes 27r/7/i2V272 cos 2-ft.
FIG. 48.— The currents and
fluxes in a three-phase line
with a symmetrical spacing.
206 THE MAGNETIC CIRCUIT [ART. 63
In the symbolic notation this is represented as jx^'l, where #12'
=27r/Li2/ is the reactance corresponding to 1/12', 1 2 is the vector
of the effective value of the current in the wire B, and j signifies
that the vector x\z\2 is in leading quadrature with the vector 12.
Consequently, the voltage drop E\ in the wire A, equal and
opposite to the induced e.m.f ., is
Ei = -3Xi2'l2-jxi3'Iz ...... (132)
When the currents are given, /2 and 73 can be expressed in the
usual way through their components, and the drop EI is then
expressed through its components as e\+jef\. The reactances
x\2 and £13' are taken from the available tables, for the specified
frequency and the appropriate spacings, or else they can be
calculated using the value of L' from eq. (125).
The voltage drop E± in eq. (132) can be represented as if it were
due to an equivalent reactance x\ and an equivalent resistance r\
in the phase A (the latter in addition to the actual ohmic resistance
of the wire) . This is possible when 1 2 and 1 3 can be expressed in
terms of /i, and is especially convenient whenever the phase differ-
ence between these currents and their ratio is constant. Namely,
let the current I2 lead the current l\ in phase by <£i2 electrical
degrees (Fig. 49)*. Then
. . . (133)
where (/2//i) is the ratio of the effective values of the currents,
apart from their phase relation. Multiplying the vector /i by
(/2//i) changes its magnitude to that of 72, while multiplying it by
(cos (£12 + 7 sin $12) turns it counter-clockwise by </>i2 degrees. By
analogy we also have that
/3 = (VW{i(cos<£i3+/sin<£i3). . . . (134)
Both (£12 and <£i3 are measured counter-clockwise. Substituting
these values into eq. (132) and separating the real from the
imaginary part we get
EI =Ii[(l2/h)xi2 sin fa + Vs/Idxis' sin <£i3]
-H\[(h/h)x\* cos <f>i2 + (h/h)xiz' cos 018]. . . (135)
Thus, the drop El is the same as if it were caused by a fictitious
reactance
Xi = - (l2/Il)Xi2 COS $12 - (/3//l)Zl3' COS <£i3, . (136)
CHAP. XI] INDUCTANCE OF TRANSMISSION LINES
207
and a fictitious resistance
sn
sn
(137)
-i
Both x\ and r\ may be either positive or negative, depending
upon the constants of the circuit and of the load. The resistance
TI does not involve any loss of
power, converted into heat; it
merely shows that energy is trans-
ferred inductively from phase A
into B or C, at a rate I-pr\, due to
a lack of symmetry in the resultant
field.
Prob. 27. Show that with a
balanced three-phase load
0.866 (z12' -
Prob. 28. When the three wires
are in the same plane, .the spacings
being equal, and the three-phase load
balanced, show that the equivalent reactance of each outside wire
FIG. 49. — The currents and fluxes
m a three-phase line with an
unsymmetriccd spacing.
s</=*m'+0.435/XlO-3 ohm/km.,
(139)
where Xm is the reactance of the middle wire per kilometer, in ohms.
The equivalent resistance of the middle wire is zero, and that of the
two outside wires is
r0' = ±0.753/X10-3 ohm/km.,
(140)
where the sign plus refers to the wire in which the current leads that in
the middle wire.
Prob. 29. Compare the vector diagram in Fig. 49 with that in Fig.
48, and shown that with an unsymmetrical spacing the induced voltage
E! is not in quadrature with the corresponding current Ilt so that the
action of the other two wires cannot be replaced by an equivalent
inductance alone, but only by an inductance and a resistance. Show
graphically that the latter may be either positive or negative.
CHAPTER XII
THE INDUCTANCE OF THE WINDINGS OF
ELECTRICAL MACHINERY.
64. The Inductance of Transformer Windings. When a
transformer is operated at no load, i.e., with its secondary cir-
cuit open, practically the whole flux is concentrated within
the iron core. When, however, the transformer is loaded, so that
considerable currents flow in both windings, appreciable leakage
fluxes are formed (Fig. 50), which are linked partly with the
primary winding, and partly with the secondary winding. When
the load current is considerable, the primary and the secondary
ampere-turns are large as compared to the exciting ampere-
turns, so that at each instant the secondary ampere-turns are
practically equal and opposite to the primary ampere-turns.
An inspection of Fig. 50 will show that the m.m.f. acting upon
the useful path in the iron is equal to the difference between the
m.m.fs. of the primary and secondary windings, while the m.m.f.
acting upon leakage paths is equal to the sum of the m.m.fs.
of both windings.
Take, for instance, the line of force fghk ; with respect to the
part fg of its path, the secondary coil Si and the adjacent half
PI of the primary coil form together a fictitious annular coil
(leakage coil). The m.m.f. of this coil is equal to in^, where
ii is the primary current, and n^ is the number of turns in the
whole primary coil P. Similarly, the coil S2 and the part P%
of the primary coil may be said to form another fictitious coil
linking with the part hk of the path of the lines of force.
It will be seen from the dots and crosses that the m.m.fs.
of the two fictitious coils assist each other, and that the paths
of the leakage flux are as indicated by the arrow-heads. Some
lines of force are linked with the total m.m.f. of the fictitious
coils, others are linked with only part of the turns. Although
208
CHAP. XII]
INDUCTANCE OF WINDINGS
209
the reluctance of the leakage paths is very high as compared
to that of the useful path in iron, yet the m.m.f. acting upon
.Actual Field
Simplified Fiel
Flux Density Distribution
X
X
Jr
<e£G-»j
?v2
""ir
X
^
FIG. 50. — The leakage field in a transformer with cylindrical coils.
the leakage paths is also many times greater than that acting
upon the path in iron. As a consequence, the leakage fluxes
reach appreciable magnitudes.
210 THE MAGNETIC CIRCUIT [ART. 64
The leakage fluxes induce e.m.fs. in the windings in lagging
quadrature with the respective currents, and thus affect the
voltage regulation of the transformer. That part of the applied
voltage which balances these e.m.fs. is known as. the reactance
drop in the transformer. It is customary to speak about the
primary reactance and the secondary reactance, also about the
primary and the secondary leakage fluxes, as if they had a
separate and independent existence. However, it must be under-
stood that each leakage flux is produced by the combined action
of both windings, as is explained above. Moreover, where the
leakage fluxes enter the iron they combine with the main flux
in the proper direction, so that they form there only a com-
ponent of the resultant flux.
In reality, the primary ampere-turns are not exactly equal
and opposite to the secondary ampere-turns, so that, in addition
to the leakage fluxes shown in Fig. 50, there is a leakage flux
due to the magnetizing ampere-turns. However, this correction
is negligible, when the load is considerable, and the calculation
of the leakage flux is greatly simplified by assuming the primary
ampere-turns to be exactly equal and opposite to the secondary
ampere-turns.
The effect of the leakage reactance upon the performance
of a transformer is treated in The Electric Circuit; there the
value of the reactance is supposed to be given. Here the problem
is to show how to calculate the leakage inductance from the given
dimensions of a transformer. The two types of winding to be
considered are the one with cylindrical coils (Fig. 50) and the
one with flat coils (Fig.51). Both types of winding can be used
with any of the three kinds of magnetic circuit which are used
with transformers (Figs. 12, 13, and 14).
The problem of calculating the leakage inductance, according
to the fundamental formula (106), is reduced to that of finding
the permeances of the individual paths of the leakage flux. It
would be out of the question here to determine the actual paths
and to express their permeances mathematically. Therefore, in
accordance with Dr. Kapp's proposal,1 simplified paths are
assumed, shown in Fig. 50 to the right. The inductance so
calculated is corrected by an empirical factor, obtained from
experiments on transformers of similar type and proportions.
1 G. Kapp, Transformers (1908), p. 177.
CHAP. XII] INDUCTANCE OF WINDINGS 211
The simplifying assumptions are (a) that the paths within and
between the coils are straight lines, and (6) that the reluctance
of the paths in the space outside the coils is negligible, because
the cross-section of these paths is practically unlimited.
(1) Cylindrical Coils. We shall calculate first the primary
inductance of a transformer having cylindrical coils, i.e., the
inductance due to the linkages of the leakage flux with the
primary winding. The permeance of the path of the complete
linkages is (Pci = J*.a<i0m/2l perms, where Om is the mean length
of a turn in the coil P, and 0,1 is the radial thickness of the flux.
The notation is shown in the detail drawing, at the bottom of Fig.
50. In this expression a,iOm is the mean cross-section of the
path, being an average between the cross-sections within the
spaces PI~SI and P2-S2. The length of the paths within the
coils is 21, and the reluctance of the paths outside the coils is
neglected. This path is linked with n^ turns.
Similarly, the permeance of an infinitesimal annular path
within the primary coil, at a distance x from its center, and
of a width dx, is d(PPi = jj.0mdx/2l perms. This path is linked
with npi = ni(2x/bi) turns. Substituting these values into eq.
(106) we obtain
a,
or
L1 = (fJLU12Om/2l)(a1 + ib1) perms. . * «' . (141)
By a somewhat similar reasoning we should find for the com-
bination of the two secondary coils, assuming them to be connected
electrically in series,
L2=(/m22Ow/2Z)(a2 + i&2) perms. .... (142)
In the operation of a transformer it is the total equivalent
inductance of the two windings reduced to one of the circuits
that is of importance. Since resistances and reactances can be
transferred from, the primary circuit to the secondary or vice
versa, when multiplied by the square of the ratio of the numbers
of turns (Art. 446), the equivalent inductance, reduced to the
primary circuit, and per leg of the core, is
henrys. . (143)
212 THE MAGNETIC CIRCUIT [ART. 64
All the lengths here are expressed in centimeters, and /*= 1.257,
also a=di +a2 is the spacing between the coils, which is a known
quantity. Thus, the unknown distances ax and 0% which enter
into the expressions for LI and L2 are eliminated from the
formula for the equivalent inductance.
The coefficient k corrects for the difference between the
actual linkages shown in Fig. 50 at the left, and the assumed
linkages shown at the right. The values of k, found from experi-
ments, vary within quite wide limits, depending upon the pro-
portions of the coils. For good modern transformers Arnold
gives the limits of k between 0.95 and 1.05.1 See also eq. (147)
below. Formula (143) gives the inductance of one leg only;
the equivalent inductance of the whole transformer depends
upon the electrical connections between the coils.
In designing a transformer the coils are usually arranged
in such a way as to reduce the leakage reactance to the least
possible amount.2 Eq. (143) shows that in order to achieve
this result, a comparatively small number of turns must be used,
and the coils must be thin and long. The space a between the
coils must be kept as small as is compatible with the require-
ments for insulation and cooling.
The usual arrangement of coils shown in Fig. 50 gives a
considerably smaller leakage inductance than the simpler arrange-
ment shown in Fig. 12. Namely, with the arrangement shown
in Fig. 12, the permeance of the path of the complete linkages
in the space between the coils is fj.aiOm/l. This expression
differs from that used before in that I stands in the denominator
in place of 21. For the partial linkages np = nl(x/b1), where
x is measured now from the edge of the primary coil, furthest
from the secondary coil. Thus, the primary inductance is in
this case
n a,
or
By symmetry we can write the expression for L2, and hence,
!E. Arnold Wechselstromtechnik (2d edition), Vol. 1., p. 561.
2 In some cases a considerable leakage reactance is specified as a protection
against violent short circuits.
CHAP. XII] INDUCTANCE OF WINDINGS 213
after combining,
. . . (144)
This value is between two and four times as large as the
value given by eq. (143). For this reason, in most transformers,
the low-tension coil is split into two sections; compare also with
Fig. 14.
Formula (143) and the values of k given above have been
deduced for the core-type transformer. It is clear, however,
that the same formulae will apply to the shell-type and the cruci-
form type transformers with cylindrical coils, though the coeffi-
cient k may have different values in each case. Until more
reliable and numerous experimental data are available the same
values of k will have to be used for these types as for the core-
type.1
(2) Flat Coils. With flat coils (Fig. 51) the inductance of
a part of the winding between AB and CD can be calculated
in precisely the same way as in Fig. 50. If the primary winding
is split into q sections, the inductance per section, by analogy
with eq. (143), is
Leq/q=k[fjL(n1/q)2Om/2l][a + i(b1+b2)]lO-^henTys} . (145)
where the dimension I is again measured in the direction of
the lines of force and Om is the mean length of a turn. The
dimensions a, &x, and 62 are indicated in Fig. 51. The inductance
of the whole winding is
Leq=k(/jLnl2Om/2ql)[a+±(b1 + b2)]W-*henrys, . . (146)
where all the lengths are expressed in centimeters, and /*= 1.257.
This formula presupposes that the m.m.fs. are balanced, or
in other words, that there are two half-sections of the same
winding at the ends; such is usually the case in order to reduce
the leakage reactance. (See also Fig. 13.)
Eq. (146) shows that the leakage reactance is considerably
reduced, and consequently the voltage regulation improved, by
subdividing the windings and placing the ' primary and the
1 See also the Standard Handbook for Electrical Engineers under Trans-
former, leakage reactance.
214
THE MAGNETIC CIRCUIT
[ART. 64
secondary windings on the core alternately. At a given voltage,
and with a given type of construction, the spacing a between
the coils may be considered as constant and independent of the
number of sections. In transformers for extra-high voltages a
is large as compared to J(61+62), so that the leakage reactance
is almost inversely proportional to the number of sections q.
In low-voltage transformers a is small as compared to 61 and
62; hence, Leq is almost inversely proportional to q2, because
bi and 62 are themselves inversely proportional to q. Thus,
FIG. 51. — The leakage field in a transformer with flat coils.
in general, the inductance of a transformer is inversely pro-
portional to qn, where n has a value between 1 and 2.1
Dr. W. Rogowski has given an exact mathematical solution
for the flux distribution in the case of flat transformer coils,
assuming the coils and the core to be indefinitely long in the
direction perpendicular to the cross-section shown in Fig. 5 1.2
1 For experimental data in regard to the effect of the subdivision and
arrangement of transformer windings upon the leakage reactance see Dr. W.
Rogowski, Mitteilungen Ueber Forschungsarbeiten, Heft 71 (Springer, 1909),
p. 18, also his article Ueber die Streuung des Transformators, Elektrotechnische
Zeitschrift, Vol. 31 (1910), pp. 1035 and 1069; also Faccioli, Reactance of
Shell-type Transformers, Electrical World, Vol. 55 (1910), p. 941.
2 Dr. W. Rogowski, loc. cit.
CHAP. XII] INDUCTANCE OF WINDINGS 215
With certain simplifying assumptions he arrived at the same
formula for inductance as eq. (146) in which approximately
Because of the simplifying assumptions made in the deduc-
tion of this formula, the values of k calculated from the results
of tests on actual transformers differ slightly from those given
by the formula. Let k' be an empirical correction coefficient,
then
] ..... (147)
In Dr. Rogowski's experiments the actually measured induct-
ance was on the average 6 per cent higher than the calculated
one. Until more experimental data are available it is therefore
advisable to use in eq. (147) the value of A;7 =1.06. Eq. (147)
is applicable to transformers of all the three types (Figs. 12 to
14), though in the case of a shell-type or cruciform transformer,
the presence of iron outside the coils tends to increase the value
of &'. However, Dr. Rogowski states, that with the space usually
allowed for insulation between the coils and the iron, the influence
of the iron in increasing the leakage reactance is negligible. Eq.
(147) holds approximately true for cylindrical coils also, though
there are asx yet no conclusive tests for the value of the cor-
rection factor to be used with such coils.
The general similarity between the equations for leakage induct-
ance given above raises the question, as to what element they
possess in common. This is found in the conception of a leakage
coil, which is the " fictitious coil " spoken of above. An inspec-
tion of Figs. 50 and 51 will show that the successive lines of
force converge upon lines which may be called the " hearts "
of the flux system. These hearts are located in places where
the net m.m.f. is zero. This is at the edge of the half-coils and
at the center of the whole-coils, in the two figures mentioned.
In deriving eq. (144) for the case where the coils are not split,
the heart is assumed to be at the edge of both coils. If we
define a leakage coil as that part of the winding between two
successive hearts, then eq. (144) will always apply to it. In
eq. (143) 6X and 62 refer to the width of the double leakage coil,
hence if we substitute in eq. (144) %bi and J62 for &i and b2,
216 THE MAGNETIC CIRCUIT [ART. 64
we get the coefficient J, which appears in eq. (143). In eq.
(143) ni and Leg refer to the double leakage coil. Substituting
in eq. (144) \n^ for n± and %Leq for Leq, we get eq. (143). Thus,
the differences between the two equations is readily explained.
In case the coils are divided in any unusual manner, we must
first locate the hearts by noticing where the m.m.fs. are balanced.
Then we should figure out the inductance by eq. (144) for each
leakage coil separately. The only precaution to be observed is
that the various quantities refer to the leakage coil. Finally
(if the coils are in series) we should add the various induct-
ances together. The arrangement with half coils on the ends
gives the minimum of inductance for a given number of coils.
Prob. 1. The approximate assumed dimensions of a 1 5-kw., 2200/1 1 0-
v., 60-cycle, cruciform-type transformer with cylindrical coils (Fig. 14)
are: Om =140 cm.; ^=4.5 cm.; b2 = 3 cm.; a = l cm. The maximum
useful flux is 1.03 megalines. Show that the relationship between the
height I of the winding and the percentage reactive voltage drop is
a^ = 166. Assume fc = 1.10.
Prob. 2. Referring to the preceding problem, what is the permeance
of the space between the outside low-tension coil and the high-tension
coil, per centimeter of the height of the coils, and what is the effective
value of the flux density in this space, at full load ?
Ans. 197.5 perms per cm., 3420 /I lines per sq.cm.
Prob. 3. Each leg of a core-type transformer is provided with six
flat high-tension coils of 530 turns each, interposed with the same
number of low-tension coils of 40 turns each, one of the low-tension
coils being split in two and placed at the ends. The high-tension coils
are wound of 3 mm. round wire, 53 layers, 10 turns per layer (61 = 3 cm.) ;
the low-tension coils are wound of 8 mm. square wire, in 20 layers, 2 turns
per layer (62 = 1.6 cm.). The distance between the coils is 20 mm.
Taking the inductance of this transformer to be unity, calculate the
relative inductances of the transformer when the high-tension winding
is divided into three coils and also into two coils, assuming fc, I, and
a to be the same in all cases. Ans. 2.55; 4.66.
Prob. 4. Solve the preceding problem, taking into account the
change in k. Ans. 2.42; 4.14.
Prob. 6. The following results were obtained from a short-circuit
test on a 22/2-kv., 2500-kva., 60-cycle, shell-type transformer, with
flat coils: With the high-tension winding short-circuited, and full rated
current flowing through the low-tension winding the voltage across
the secondary terminals was 73.5 v., and the wattmeter reading was
27 kw. The transformer winding consists of 12 high-tension coils of
100 turns each, and of 11 low-tension coils interposed between the
high-tension coils, together with 2 half-coils at the ends. The dimen-
sions of the coils are : Om=2.6m.; Z = 18cm.; 61 = 16mm.; 62 = 1
CHAP. XII] INDUCTANCE OF WINDINGS 217
a = 12 mm. Calculate the correction factor k' in formula (147). Hint:
Eliminate the ir drop, using the wattmeter reading. Ans. About 1.05.
Prob. 6. What is the greatest permissible thickness of the coils
of a 60-cycle transformer, if the reactive drop must not be larger than
three times the resistance drop? The ducts a are 1 cm. The space
factor of the copper in each coil is 0.55. The primary coils and the
whole coils of the secondary are of the same thickness. fc = 0.98.
Hint : Assume Om, I, and nt and show that they cancel out.
Ans. b = 3.66 cm.
Prob. 7. In order to provide a better cooling, and at the same
time save on insulation, two flat high-tension coils are frequently placed
side by side, with a small air-space in between, and in the same way
the low-tension coils may be subdivided. Show that no leakage flux
passes in the space between the two adjacent coils which belong to
the same winding, so that the inductance of the winding is not appreciably
increased by these spaces, and can be calculated as if these spaces did
not exist.1
Prob. 8. Compare the formulae given above and the numerical values
of transformer leakage reactance obtained therefrom with the formulae
and data given in the Standard Handbook for Electrical Engineers. Do
this for any transformer, the dimensions of which are available.
Prob. 9. The equivalent reactance of a transformer is Zj reduced
to the primary circuit, and is x2 reduced to the secondary circuit. All
the primary coils are connected in series, and all the secondary coils
are also in series. Show that these equivalent reactances become
XI/GI and x2/c22 respectively, when the primary winding is divided
into cx branches in parallel, and the secondary winding is divided into
c2 branches in parallel. The division is supposed to be made in such
a way as to keep the m.m.fs. in the adjacent coils balanced. Hint:
If the equivalent reactance of one primary branch is z/, that of c-^
branches in series is x1 = x1c/, and that of cx branches in parallel is a^'/Ci
no matter whether the secondary coils are connected in series or in
parallel.
Prob. 10. Prove that the equivalent reactance of a transformer
is the same, whether the coils are in series or in parallel, provided that
the total number of turns in series is the same.
Note: Sometimes, because of the difficulty in using heavy con-
ductors, it is desirable to multiple the coils. In such case, the parallel
coils must have the same number of turns, and they should be sym-
metrically arranged, so as to prevent exchange currents.
65. The Equivalent Leakage Permeance of Armature Windings.
In certain problems relating to the design and operation of
electrical machinery it is necessary to calculate the inductance
1 This fact has been verified experimentally; see Arnold, Wechselstrom-
technik, Vol. 2 (1910), p. 29.
218
THE MAGNETIC CIRCUIT
[ART. 65
of the armature windings. This inductance affects the per-
formance of the machine, because the leakage fluxes created
by the armature currents induce e.m.fs. in the machine. Such
leakage fluxes are shown in Figs. 23 and 36, in an induction
machine and a synchronous machine respectively. For purposes
of theory and computation these leakage fluxes are usually
subdivided into three parts, namely:
(a) Leakage fluxes linked with the parts of the windings
embedded in the armature iron (Figs. 36 and 54). These paths
are closed partly through the slots, and partly through the tooth-
tips (slot leakage and tooth-tip leakage).
(b) Leakage fluxes linked with the parts of the armature
windings in the air-ducts.
>
•N
CEP
S
> N
'S
FIG. 52. — Undivided end-connections. FIG. 53. — Divided end-connections.
(c) Leakage fluxes linked with the end-connections of the
armature windings (Figs. 52 and 53).
Usually the fluxes (a) and (b) are merely distortional com-
ponents of the main flux of the machine, and only the fluxes
(c) have a real existence.
It will be readily seen that the paths of the tooth-tip leakage
and of that around the end-connections are too complicated
to allow the corresponding permeances to be calculated theo-
retically. For this reason, various empirical and semi-empirical
formula are used in practice for estimating the leakage inductance
of armature windings, the coefficients in these formulae being
determined from tests on similar machines.
The most rational procedure is to express the inductance
through the equivalent permeance of the paths, as defined by
eq. (106a) in Art. 58. Let there be CPP conductors per pole
CHAP. XII] INDUCTANCE OF WINDINGS 219
per phase, such as are indicated for instance in Fig. 15. Then
the inductance of a machine, per pole per phase, is given by the
equation
LPP=CPP2(Peg, . . .... (148)
where (Peq is an empirical value of the equivalent permeance
per pole per phase. This formula presupposes that all the
partial linkages are replaced by the equivalent complete linkages
embracing all the Cpp conductors. Moreover, the value of (Peq
is such as to take into account the inductive action of the other
phases upon the phase under consideration. The total inductance
of the machine, per phase, depends upon the electrical connec-
tions in the armature winding. If all the p poles are connected
in series, the foregoing expression for LPP must be multiplied
by p; if there are two branches in parallel, the inductance
of each branch is %pLPP, and the combined inductance of the
whole machine is %(%pLpp) = \pLPP.
The leakage inductance of a machine is usually determined
by sending through it an alternating current of a known frequency,
under conditions which depend upon the kind of the machine
(the field to be removed in a synchronous machine, and the
armature to be locked in an induction machine). From the
readings of the current of the applied voltage and the watts
input, the reactance x of the machine is calculated (after elim-
inating the ohmic drop) . Then, knowing the frequency / of the
supply and the number of poles of the machine, the inductance
LPP=x/(2xfp) per pole is calculated. Substituting into eq.
(148) this value of Lpp and the known number of conductors
Cpp, the equivalent permeance (Peq per pole per phase is deter-
mined. By performing such tests on machines built on the
same punching, but of different embedded lengths, the permeance
due to the embedded parts of the winding is separated from
that due to the end-connections; the values so obtained are then
used in new designs.
The leakage permeances in the embedded parts are pro-
portional to the widths of these parts in the direction parallel
to the shaft, in other words, to the length of conductors which
are surrounded by the leakage lines. Experiment shows that
the permeance of the paths in the air-ducts and around the
end-connections is also approximately proportional to the lengths
220 THE MAGNETIC CIRCUIT [ART. 65
of the corresponding parts of the armature coils. Since all these
permeances are in parallel, (Peq is equal to their sum, or
e .... (149)
Here the letter I denotes the lengths in cm. of the coil, or, what
is the same thing, the width of the paths of flux. The subscripts
i, a, and e refer to the iron, the air-ducts, and the end-connections
respectively. Thus, ^ is the semi-net length of the core, that is,
the length exclusive of ducts but inclusive of the space between
the laminations.1 The corresponding permeance per centimeter is
(Pj. The sign " prime " signifies that the corresponding <P's refer
to one centimeter width of path.
The coefficient a. is equal to 1 when the end-connections
are arranged as in Fig. 52, and a=J when they are arranged
according to Fig. 53. Namely, in the first case the number of
conductors Ce per group of the end-connections is equal to the
number CPP in the embedded part. In the second case Ce is
equal to JCPP. In the first case there are as many groups of
end-connections per phase as there are poles; in the second
case there are twice as many groups per phase as there are poles,
so that two groups (one pointing to the right and one to the left)
must be counted per pole. Thus, with undivided end-con-
nections, the inductance is Cpp2(pe'le, while with divided end-
connections it is (%CPP)2(Pef.2le=Cpp2. \(Pele- This accounts for
the value of a = J, in formula (149), and shows that the inductance
due to the end-connections is reduced twice by subdividing
them into two groups. When estimating the leakage reactance
it is therefore necessary to know the exact arrangement of the
end-connections.
In preliminary calculations, before the armature coils are
drawn to scale, the length le of the end-connections in a full-
pitch winding is usually assumed to be equal to about 1.5r, where
T is the pole pitch. For fractional-pitch windings, le varies
roughly as the winding-pitch, or Ze=1.5^r (see Art. 29).
1 The semi-net length is used in getting the leakage permeance in the
slot, because the flux spreads as it comes out of the iron almost immediately.
The spaces between the laminations do not affect the density in the air
because they are so small as compared to the dimensions of the slot.
CHAP. XII] INDUCTANCE OF WINDINGS 221
Substituting the value of (Peq from eq. (149) into eq. (148)
we obtain
Lpp=CPp2((Pifli+(Pafla + ^efle^~8 he*ryS. . (150)
In the following three articles formula (150) is applied to
the calculation of the leakage reactance of
(a) Induction machines;
(6) Synchronous machines;
(c) Coils undergoing commutation in a direct-current machine.
In each case somewhat different values of the unit permeances
(Pf are used, because of the diversity of the magnetic paths.
66. The Leakage Reactance in Induction Machines. It is
explained in Art. 35 and shown in Fig. 23 that the actual flux
in a loaded induction machine is the resultant of three fluxes,
of which the useful flux $ is linked with both the primary and
the secondary windings. The component fluxes 4>l and $2,
linked with the primary and secondary windings respectively,
are called the leakage fluxes. They induce in the windings e.m.fs.
in quadrature with the corresponding currents, and these e.m.fs.
have to be balanced by a part of the terminal voltage. Con-
sequently, that part of the applied voltage which is balanced by
the useful flux is reduced; in other words, the useful flux and
the useful torque are reduced with a given current input. As
a matter of fact, the maximum torque and the overload capacity
of an induction machine are essentially determined by its leakage
fluxes, or what amounts to the same thing, its leakage inductances.
Knowing the primary and secondary leakage reactances, the
actual induction machine is replaced by an equivalent electric
circuit (or a circle diagram is drawn for it), after which its
performance can be predicted at any desired load. The problem
here is to determine the values of these leakage reactances and
inductances, from the given dimensions of a machine. The
rest of the problem is treated in the Electric Circuit.
The leakages fluxes, which are indicated schematically in
Fig. 23, are shown more in detail in Fig. 54. The primary
conductors in one of the phases and under one pole are marked
with dots, and the adjacent secondary conductors are marked
with crosses, to indicate the currents which are flowing in them.
Assuming the rotor to be provided with a squirrel-cage
222
THE MAGNETIC CIRCUIT
[ART. 66
winding, the distribution of the secondary currents is prac-
tically an image of the primary currents. Neglecting the mag-
netizing ampere-turns necessary for establishing the main or
useful flux, the secondary ampere-turns per pole per phase are
equal and opposite to the primary ampere-turns. A similar
assumption is also made in the preceding article, in the case of
the transformer. This assumption is not as accurate in the
case of an induction machine, because here the magnetizing
current is proportionately much larger, due to the air-gap;
nevertheless, the assumption is sufficiently accurate for most
FIG. 54. — The slot and zig-zag leakage fluxes in an induction machine.
practical purposes. Even if the magnetizing current is equal
to say 25 per cent of the full-load current, the difference between
the primary and the secondary ampere-turns should be less
than 10 per cent, because the magnetizing current is considerably
out of phase with the secondary current.1
With this assumption, the primary and the secondary current
belts shown in Fig. 54 may be considered as two sides of a narrow
fictitious coil which excites the leakage flux, causing it to pass
circumferentially along the active layer.2 Neglecting the mutual
1 See the circle diagram of an induction motor, for instance, in the author's
Experimental Electrical Engineering, Vol. 2, p. 167.
2 Although the secondary frequency is different from the primary, with
respect to the revolving rotor it is the same as the primary frequency with
respect to the stator. Let s be the slip expressed as a fraction of the primary
frequency. Then the speed of the rotor is (1 — s), and the frequency of the
secondary currents with respect to a fixed point on the stator is s + (1 — s) = 1.
CHAP. XII] INDUCTANCE OF WINDINGS 223
action of the consecutive phases, the length of this flux is approx-
imately r/m, where r is the pole pitch and m is the number
of the stator phases. Part of the flux is linked with the primary
current belt, and part with the secondary belt. The conditions
are essentially the same as between the transformer windings
P1 and $! in Fig. 50. Knowing the equivalent permeances
of the individual paths the inductance can be calculated from
eq. (150).
In an induction machine with a squirrel-cage rotor the total
leakage in the embedded part may be resolved into three com-
ponents shown in Fig. 54, namely :
(1) The primary slot leakage, 4>8l;
(2) The secondary slot leakage, tf>s2;
(3) The tooth-tip or zigzag leakage, ®z.
The fluxes tf>sl and 4>s2 are alternating fluxes of the frequency
of the corresponding currents. The zigzag flux ®z varies according
to a much more complicated law, because the permeance of its
path changes from instant to instant in accordance with the
relative position of the stator and rotor teeth; compare posi-
tions (1) and (2) in Fig. 23. Moreover, Fig. 54 shows only the
simplest case, which never occurs in practice, namely; when
the stator tooth pitch is equal to that in the rotor. In reality,
the two pitches are always selected so as to be different, in order
to avoid the motor sticking at sub-synchronous speeds (due to the
higher harmonics in the fluxes and in the currents). Therefore,
the paths of the zigzag leakage flux are much more complicated
than is shown in Fig. 54, and in calculations the average per-
meance of the zigzag path is used.
In a machine with a phase-wound rotor the main flux is
further distorted, due to the fact that the primary and secondary
phase-belts are not exactly in space opposition at all moments.
While the total m.m.fs. of the primary and secondary are
balanced, there is a local unbalancing which changes from
instant to instant. This distortion is the same as if it were due
to an additional leakage, which was named by Professor C. A.
Adams the belt leakage.1 This part of the leakage usually
constitutes but a small part of the total leakage, and will not
be considered here separately. Those interested are referred to
1 C. A. Adams, The Leakage Reactance of Induction Motors, Trans.
Intern. Electr. Congress, St. Louis, 1904, Vol. 1, p. 711.
224 THE MAGNETIC CIRCUIT [ART. 66
the original paper and to the works mentioned at the end of
this article.
Let there be CPP1 conductors per pole per phase in the stator
winding; then the fictitious coil (Fig. 54) made up of the
primary and secondary conductors has CPP1 turns, when reduced
to the primary circuit. This is because the secondary winding
can be replaced by an equivalent winding with a " one to one "
ratio, that is, with the same number of conductors as the
primary winding. In this case, the secondary current is equal
to the primary current (Art. 446). Therefore, eq. (150) can
be made to give the equivalent inductance, including the pri-
mary inductance and the secondary inductance reduced to the
primary circuit, provided that the permeances of the paths
linking with the secondary conductors are included in the values
of (P"s. Such is naturally the case when the values are deter-
mined from a test with the rotor locked.
Extended tests have shown that in a given line of machines
the permeance (Pj is inversely proportional to the peripheral
length of the equivalent leakage flux, that is, inversely proportional
to (r/w), where T is the pole pitch and ra is the number of
primary phases.1 This shows that the permeance per centi-
meter of peripheral length of the active layer is fairly constant,
in spite of different dimensions and proportions, as long as
these are varied within reasonable limits. Thus
where G>" is the leakage permeance of the active layer in the
embedded part per one centimeter of axial length and per centi-
meter of the peripheral length of the path. Thus, the final
formula for the equivalent leakage inductance of an induction
machine, per pole per phase, reduced to the primary circuit, is
Z^PP= CPP12[CP/V(V™) + (Pa'la + a(Pe'le]W-* henrys. (151)
In this formula the following average values of unit per-
meance may be used for machines of usual proportions, unless
more accurate data are available.
1 H. M. Hobart, Electric Motors (1910), table on p. 397. The values
for (P/' given below have been computed from this table, and the results
multiplied by 2, because the table gives the values of the primary per-
meances only.
CHAP. XII] INDUCTANCE OF WINDINGS 225
The equivalent permeance (Pi" of the embedded part in perms
per centimeter of the semi-net axial length of the machine, and
per centimeter of peripheral length of the air gap, is for
Open slots Half-open slots Completely closed slots
11.5 14.5 18.
The equivalent permeance around the end-connections, and
around the parts of the conductors in the air-ducts, decreases
with the increasing number of slots per pole per phase, for the
same reason that the slot permeance decreased. In induction
motors usually at least three slots are used per pole per phase,
and under these conditions Mr. H. M. Hobart uses <P/ = 0.8
perm per centimeter, with phase-wound rotors, and <P/ = 0.6
perm per centimeter with squirrel-cage rotors. (Pa' may be
taken in all cases equal to 0.8 perm per cm. The lengths le
and la are always understood to refer to the stator winding.
The foregoing data refer to machines with full-pitch windings
in the stator and in the rotor. With a fractional pitch winding
the equivalent leakage permeances are somewhat smaller, due to
longer phase belts and to the mutual induction of the over-
lapping phases. Let the winding-pitch factor (Art. 29) of the
primary winding be kwl and that of the secondary winding
kw2. Then the leakage inductance of the machine, calculated
for a full-pitch winding (but for Ze=1.5£r), is multiplied by
kwi-kW2- This is an empirical correction, which is accurate
enough for ordinary practical purposes. In reality, of two
machines designed for a given duty, one with a fractional pitch
winding and the other not (but otherwise both alike) the first
often has a higher inductance than the second. This is because
more turns are required with the fractional pitch winding, if the
flux densities in the iron and in the air-gap are to be the same
in both cases.1
With the data given above the calculation of the leakage
reactance of a given induction motor is quite simple, and one
engaged regularly in the commercial design of induction-motors
can obtain sufficiently accurate data for their design or for the
1 For a theoretical and experimental investigation of the effect of a
fractional pitch upon the leakage reactance in induction machines see
C. A. Adams, Fractional-pitch Windings for Induction Motors, Trans. Amer.
Inst. Elec. Engrs., Vol. 26 (1907), p. 1488.
226 THE MAGNETIC CIRCUIT [ART. 66
computation of their performance, provided that he adapts the
numerical values of the unit permeances to each individual case,
on the basis of his previous experience. Many authors have
given theoretical formulae for the leakage inductance of induction
machines.1 These formulae, curves, and methods, while useful
in accurate work, are too elaborate to be quoted here ; at any rate
they are of interest only to a specialist in design. Two examples
of the theoretical calculation of leakage inductance are given
below, in order to show the student the general method used,
and thus introduce him into the literature on the subject.
(a) A Theoretical Calculation of the Slot Leakage Permeance.
We shall calculate the equivalent permeance for the half closed
slot (Fig. 55), an open slot being a special case of it. With
the notation shown in sketch, and with SPP slots per pole per
phase, we have:
PC' = Ab2/s2 + 63/^3 + bt'/8t]/SPP; d(? P' = fidx/ (s4SPP),
and nx=CPPx/b4. Hence
and
. . (152)
The equivalent permeance of one slot, per unit of the semi-
net axial length of the machine is
^.'=M&2/*2 + &3/«3+(J&4 + &40/*4]. - - - (153)
This shows that the slot permeance depends only upon the
proportions of the slot and not upon its absolute dimensions.
.(b) A Theoretical Calculation of the Zigzag Leakage Permeance.
The calculation of the zigzag leakage permeance is simple only
1 C. A. Adams, loc. cit.; also Trans. Amer. Inst. Elec. Engrs., Vol. 24
(1905), p. 338; ibid., Vol. 26 (1907), p. 1488. Hobart, Electric Motors,
(1910), Chap, xxi; Arnold, Wechselstromtechnik, Vol. 5, part 1 (1909), pp.
49-54; R. Goldschmidt, Appendix to his book on The Alternating Current
Commutator Motor (1909); R. E. Hellmund, Practische Berechnung des
Streuungskoeffizienten in Induktionsmotoren, \Elektrotechnische Zeitschrift,
Vol. 31 (1910), p. 1111 and 1140; W. Rogowski, Zur Streuung des Dreh-
strommotors, ibid., pp. 356, 1292, and 1316. See also an extensive series
of articles by J. Rezelman in La Lumiere Electrique, 1909-1911, and in
the (London) Electrician.
CHAP. XII]
INDUCTANCE OF WINDINGS
227
when the stator tooth-pitch and the rotor tooth-pitch are equal
to each other; otherwise the paths become too complicated for
mathematical analysis. Since the position of the secondary teeth
varies with respect to the primary teeth, the permeance of the
zigzag leakage also varies, and it is necessary to take its average
value over one-half of the tooth-pitch X, that is, between the posi-
tions (1) and (2) in Fig. 23. In some intermediate position (Fig.
55), determined by the overlap y, the reluctance of the path /,
per unit of semi-net axial length of the iron, is a/ (//?/) rels.
FIG. 55. — The notation used in the calculation of the slot and zigzag leakage
The reluctance of the path g is a/fi(t2 —s1 —y) rels. The com-
bined reluctance of / and g is equal to the sum of the foregoing
expressions. The permeance of / and g in series, being the
reciprocal of the combined reluctance, is
The permeance in question varies according to this law, for the
positions of the secondary tooth between ?/ = i02~si) and 2/ = 0,
the tooth moving to the left. From y=0 to* 2/=i(^2~si~~"^) the
permeance is practically equal to zero, because the secondary tooth
bridges over the primary slot no more. The student is advised
228
THE MAGNETIC CIRCUIT
[ART. 66
to mark the positions of the rotor teeth on a strip of paper and
to place them in different positions with respect to the primary
slot in order to see the variations in the overlap. Thus, the
average value of the zigzag permeance per tooth pitch is
i(fc-
(Pv'dy=p(t2-81)2/(6aX) perms/cm. (154)
Prob. 11. Draw a sketch similar to Fig. 54, but with the rotor
tooth-pitch different from that in the stator, and indicate roughly the
general character of the paths of the zigzag leakage.
Prob. 12. Show that increasing the number of slots per pole in a
given machine does not alter materially the slot leakage, but reduces
considerably the zigzag leakage.
Prob. 13. Calculate the equivalent leakage inductance per phase
of a three-phase, 10-pole, induction machine, with 15 slots per pole in
the stator, and a phase-wound rotor. Both windings have 100 per
cent pitch, the slots are semi-closed on both punchings, the individual
stator coils in each phase are connected in series, and there are 20 con-
ductors in each stator slot. The bore of the machine is 110 cm., the gross
length of the core is 30 cm. ; there are three vents of 8 mm. each. The
end-connections are arranged according to Fig. 53. Ans. 57.5 mh.
Prob. 14. The design of the machine in the preceding problem
has been modified in the following respects: The rotor is provided
with a squirrel-cage winding, the winding pitch in the stator is shortened
to 80 per cent, the stator slots are made open, and the end connections
are divided, as in Fig. 54. What is the inductance of the machine?
Ans. 43.3 mh.
Prob. 15. Check the values of <P/' given in the text above with
those in Hobart's table.
Prob. 16. For the usual limits of proportions of slots, teeth, and
air-gap calculate the values of (P/'
from eqs. (153) and (154) and com-
pare the results with the average
experimental values given in this
text.
Prob. 17. Calculate the equi-
valent leakage permeance of a round
"N slot (Fig. 56), assuming the con-
ductors to completely fill it, and
{ } the lines of force to be straight lines.
^ •' Hint: Select the angle a. as the inde-
FIG. 56. — A semi-closed round slot, pendent variable, and integrate eq.
(106) between a = 0 and a = it.
See Arnold, Wechselstromtechnik, Vol. 4 (1904), p. 44.
Ans. (5Y = //(0.623 +b/s) perms per cm.
CHAP. XII] INDUCTANCE OF WINDINGS 229
67. The Leakage Reactance in Synchronous Machines. The
physical nature of the armature reactance in a synchronous
machine is explained in Art. 46; the influence of this reactance
upon the performance of a machine is shown in Figs. 37, 38, 40,
and 41. The problem here is to calculate the numerical value
of this reactance for a given machine, using eq. (150) with
empirical coefficients (P'.1
It may also be stated here that for standard machines, partic-
ularly in preliminary estimates, the ix drop is sometimes taken
as a certain percentage of the rated voltage of the machine
instead of estimating the inductance from formula (150). In
synchronous generators the ix drop at the rated volt-ampere
load varies from 5 to 10 per cent of the rated terminal voltage.
In synchronous motors, where some inductance is useful, the
ix drop ranges from 8 to 15 per cent of the rated voltage. For
60-cycle machines, and for machines with a comparatively large
number of armature ampere-turns, values must be taken nearer
the higher limit. For 25-cycle machines, and for machines
with a comparatively small number of armature ampere-turns,
values must be taken nearer the lower limit. A considerable
error in estimating the value of ix has but little effect upon the
calculated performance at unity power factor, because the vector
ix is then perpendicular to e (Figs. 37, 38, 40 and 41). However, a
considerable error may be introduced at lower values of the
power factor if the reactive drop ix has not been estimated
with a sufficient accuracy.
The values of (P' for synchronous machines are different
from those given above for induction machines, because of the
absence of any secondary current-belts. Parshall and Hobart 2
give the following values for (P{ :
1 For a theoretical calculation of the coefficient (Pf, see Arnold, Wechsel-
stromtechnik, Vol. 4 (1904), pp. 41-52; Hawkins and Wallis, The Dynamo,
Vol. 2 (1909), pp. 901-904. For a comparison between the calculated and
actually measured values see an extended series of articles by J. Rezelman
in La Lumiere Electrique, 1909-1911, and in The (London) Electrician.
2 Electric Machine Design (1906), p. 478. These values are corroborated
by those obtained by Pichelmayer; see his Dynamobau, 1908, pp. 208 and
504. Pichelmayer's values for (Pj (which he denotes by £) are somewhat
high because the end-connection leakage is not considered separately.
Arnold's values, given in his Wechselstromtechnik, Vol. 4, p. 280, should be
used with discretion, because they apply to a different formula; namely,
230 THE MAGNETIC CIRCUIT [ART. 67
Uni-coil windings in open slots .... 3 to 6 perms per cm.
Thoroughly distributed windings in
open slots 1.5 to 3 " "
Uni-coil windings in completely
closed slots 7 to 14 " "
Thoroughly distributed windings in
completely closed slots 3 to 6 " "
The much larger values of #>/ for closed slots, as compared
to. those with open slots, were to be expected because the bridge
which closes the slot offers a path of high permeance. The
lower values for windings distributed in several slots per pole
per phase, as compared to uni-slot windings, are due to the
fact that the partial linkages become more and more pronounced
as the winding is distributed into a larger number of separate
coils, and also because the length of the paths is greater. This
is somewhat analogous to splitting a transmission line into two
or more lines; see prob. 21 in Art. 61. The greatest reduction
in the value of the inductance results when the number of slots
is increased from one to two; a further subdivision is of much
less importance. For instance, if the permeance <P{ with a
uni-slot coil is 7, then dividing the same coil into two slots
reduces the permeance to less than 5. On the other hand, a
change from four to five slots per phase per pole would hardly
reduce the equivalent permeance more than from say 3.5 to 3.4.
The data in the table above give rather a wide range from which
to select a value of (Pj for a particular machine, and the designer
must exercise his judgment as to whether his machine will have
a permeance nearer the upper or lower limit. This judgment
comes with experience, by comparing the predicted performance
of machines with that actually observed.
The values of (Pa' and (Pe' depend upon the number of coils
per group, in other words, upon the number of slots per pole
per phase. Until more accurate and detailed data are available^
he considers separately the equivalent permeance (P8 of each slot, instead
of the group of slots per pole per phase. Thus, his formula for the leakage
inductance per pole, with our notation, is LPP = SPPCS2(PS, where Spp is
the number of slots per pole per phase, and Cs is the number of armature
conductors per slot. The values of unit permeance, which he gives and denotes
by A, refer to this formula.
CHAP. XII] INDUCTANCE OF WINDINGS 231
we shall assume (Pe' = (Pa', and use the following values, based
upon Mr. Hobart's experiments: J
Number of slots per pole per phase .1 2 3 more than 3
(Pef = <pa', in perms per centimeter ..0.8 0.7 0.6 0.5
With a fractional-pitch winding the inductance is some-
what reduced (see the end of the preceding article). As an
empirical correction, the inductance calculated for a full-pitch
winding may be multiplied by the winding pitch factor kw.
The leakage reactance of the armature cannot be calculated
from a short-circuit test, because the short-circiut current is
essentially determined by the direct armature reaction. A
difference of 50 or even 100 per cent in the armature reactance
would change the short-circuit current by only a few per cent.
A much closer approximation is obtained from the so-called
air-characteristic.2 Namely, it has been found by numerous experi-
ments that the armature inductance, when the field is revol-
ving synchronously, is nearly equal to the armature inductance
with the field completely removed, and the armature supplied
with alternating currents from an external source. The air-
characteristic is the relation between the current and the voltage
under these conditions. Eliminating the ohmic drop, the induct-
ance of the machine is easily calculated, and the value so found
can be used in the prediction of the performance of the machine,
Such an air-characteristic is easily taken in the shop or in the
power house before the machine is completely assembled. From
the three observed curves, namely, the no-load saturation curve,
the short-circuit curve, and the air-characteristic, the perform-
ance of a synchronous machine at any load can be predicted
to a considerable degree of accuracy.
Prob. 18. What is the inductance per phase of a 6-pole, 3-phase,
turbo-alternator, the armature of which has the following dimensions:
Bore, 1.2 m., gross length of core, 1.2 m., 20 air-ducts of 1 cm. each,
90 open slots? There are 8 conductors per slot, the winding is of the two-
layer type, the winding pitch is 11/15. Assume (P/ = 2. 5 perms /cm.
Ans. 24.3 mh.
Prob. 19. The inductance of the machine specified in the preceding
problem was determined experimentally (from - an air-characteristic),
1 Journ. Inst. Electr. Eng. (British), Vol. 31, pp. 192 ff.
2 Pichelmayer, Dynamobau (1908), p. 207.
232 THE MAGNETIC CIRCUIT [ART. 68
and compared to that measured on a similar machine, the gross length
of the core of which was 80 cm. and which was provided with 12 ducts
of 1 cm. each. The equivalent leakage permeance of the shorter machine
was found to be 30 per cent less than that of the other machine. What
are the actual values of (P/ and (Pe'(=(P</) for both machines?
Ans. (P/ = 2.46, OY=(P«' = 0.56 perm/cm.
Prob. 20. An alternator has 3 slots per phase per pole, and the
equivalent permeance is (P/ = 1.8; (Pe' = 0.6. What would be the value
of the same constants per slot? Ans. 5.4 and 1.8.
Prob. 21. A 3-phase alternator has 4 slots per phase per pole.
If the coils were connected up for a 2-phase machine without change
what would be the ratio of the new L to the old? Ans. 3:2.
68. The Reactance Voltage of Coils undergoing Commuta-
tion. Let Fig. 57 represent a part of the armature winding
and commutator of a direct-current machine, with two adjacent
sets of brushes. During the interval of time when an arma-
ture coil, such as CD, is short-circuited by a set of brushes,
the current in the coil is reversed from its full value in one
direction to an equal value in the opposite direction. The coil
is then said to undergo commutation.
Under unfavorable conditions this reversal of current is
accompanied by sparking between one of the edges of the brushes
and the commutator. Unless a machine is provided with inter-
poles, its output is usually limited by this sparking at the com-
mutator. It is of importance, therefore, to have a practical
criterion for judging the quality of commutation to be expected
of a given machine. Numerous formulae and methods have
been proposed for the purpose; all rational formulae contain,
as a factor, the inductance of the coils undergoing commutation,
because this inductance determines essentially the law according
to which the current is reversed with the time. For this reason,
the subject of commutation is treated in this chapter, under the
general topic of the inductance of windings. The method of
calculation of the inductance and the criterion of commutation
given below are due to Mr. H. M. Hobart.1
A description of the phenomenon of commutation. The phenom-
enon of commutation may be briefly described as follows:
Let, for the sake of explanation, the armature and the commu-
1 See Hobart, Elementary Principles of Continuous-Current Dynamo Design
(1906), Chap. 4; also Parshall and Hobart, Electric Machine Design (1906),
pp. 171-194.
CHAP. XII]
INDUCTANCE OF WINDINGS
233
tator be assumed to be stationary, and the brushes revolving in
the direction of the horizontal arrow, shown in Fig. 57. Let
the machine be provided with a multiple winding (lap winding),
so that there are as many armature circuits and sets of brushes
as there are poles. The current through each armature branch
is, therefore
(155)
where / is the total armature current and p the number of poles.
Commutator
mim
I i ' I I 0
nun
Ml
Toe^*
1
^HeelJ
J
Negative
brush
Direction of motion
of the brushes
FIG. 57. — Part of the armature winding, commutator, and brushes in a
direct current machine.
At each set of brushes two branches of the armature winding
are connected in parallel, so that the current through each set
of brushes is equal to 2/x With reference to Fig. 57, it will
be seen that the two armature branches, X and Y, which begin
at each set of brushes, may be called, with respect to this set,
the left-hand branch and the right-hand branch.
234 THE MAGNETIC CIRCUIT [ART. 68
In the position of the positive brushes just preceding that
marked 1, the coil CD is not short-circuited, and carries the full
current /i, being the first coil of the right-hand branch. The
lead d is idle, and the total current 2/! is delivered to the brushes
through the leads c, m, and n. In the position of the positive
brushes just after that marked 2, the coil CD is again not short-
circuited, but is carrying the full current /i in the opposite direc-
tion, being the first coil of the left-hand branch.
In the positions of the brushes between 1 and 2 the coil CD
is short-circuited by the brushes through the leads c and d,
and the current in the coil changes gradually from + /! to — /i.
If the coil possessed no inductance, the variation in the current
would be practically determined by the contact resistance
between the brush and the commutator, the resistance of the
coil itself and of the leads being negligible (with carbon brushes).
Under these conditions the current in the short-circuited coil
would vary with the time according to the straight-line law,
and the current density under the heels and the toes of the
brushes would be the same. This is called the " pure resistance "
commutation, or the perfect commutaton, because it is not accom-
panied by sparking. Such a commutation is approached in
machines with inteupoles, when the effect of the inductance is
correctly compensated for by the commutating flux (Art. 54).
In reality, the short-circuited coil possesses a considerable
inductance, which has the effect of electromagnetic inertia,
retarding the reversal of the current. Consequently, at the
beginning of the commutation period the lead d and the corre-
sponding commutator segment do not carry their proper share
of the current, which they would carry with a perfect com-
mutation. At the end of the commutation period the current
must then be reversed quickly, because the whole current must
be transferred from the lead c to the other leads. If the inductance
is considerable, the current in the lead c is still of a considerable
magnitude when the toe of the brush is about to leave the
corresponding commutator segment. Therefore, the last period
of the reversal is accomplished through the air between the
brush and the segment, in the form of an electric arc. This
is known as the sparking at the brushes. Besides, during the
last moments of reversal, the current density under the toe
is much higher than the average density under the brush, and
CHAP. XII] INDUCTANCE OF WINDINGS 235
this high density causes a glowing at the edge of the brushes,
making the commutation still less satisfactory.
The average reactance e.m.f. in the coils of a full-pitch lap
winding. For an empirical criterion of the quality of commutation
Mr. Hobart takes the average reactance voltage induced in the coil.
This is a reasonable criterion, because the ratio of the maximum
voltage occurring when the brush leaves a segment to the average
voltage, will be more or less the same in machines of usual
design constants. Of course, the average reactance voltage
is only a relative criterion, to be used with great discretion,
and applied only for comparison with machines which proved
in actual operation to commutate satisfactorily.
Let the inductance of an armature coil between two adjacent
commutator segments be Leq. The subscript eq (meaning equiva-
lent) is added to indicate that the value of L includes not only the
true inductance of the coil itself, but also the average inductive
action of the coils which are undergoing commutation simultane-
ously with it. Let the frequency of the current in the coil under-
going commutation be / cycles per second. Then the current is
reversed in a time J/. The flux changes during this time
from +LeQIi to —Leqli: Hence, according to the fundamental
eq. (26) Art. 24, the average reactance voltage, which is taken
as the criterion of commutation, is
In order to obtain a satisfactory commutation, the voltage
eave must not exceed a certain value, determined from actual
experience with machines in regular operation. Mr. Hobart
recommends values for eave not to exceed 3 to 4 volts, provided
that one does not depend upon the fringe flux of the main poles
to facilitate commutation.
The inductance Leq, which enters in the foregoing formula,
is calculated according to the general formula (150), as follows:
Assume first that there is no common flux or mutual induction
between the coil under consideration and the other coils which
are simultaneously short-circuited. Then, if q is the number
of turns per commutator segment (in Fig. 57 #=1) we must put
CPP=q. This will give the inductance of one side of the coil,
say C. To obtain the inductance of both sides, C and D, the
result must be multiplied by 2, or Leq=2LPP.
236 THE MAGNETIC CIRCUIT [ART. 68
In reality there is a common flux which links with the coil
under consideration and with the other coils undergoing com-
mutation at the same time. This flux is changing with the
time, and consequently it induces additional voltages in the
coil CD. The induced e.m.f. depends upon the relative
position of CD and the other coils (whether in the same slot
or in the adjacent slots) and upon the rate of change of the cur-
rent in each coil of the group. It would be too complicated for
practical purposes to take all these factors into account with
any degree of accuracy. Therefore, Hobart makes a further
assumption, namely, that the current in all the coils, which are
short-circuited at the same time, varies at the same rate and that
the whole leakage flux is linked with all the coils of the group (Fig.
57).
Let s be the average number of coils simultaneously short-
circuited under a set of brushes (the actual number varies from
instant to instant). Consider a group of mutually influencing
conductors, such as are shown at C or at D. One-half of the
conductors in the same group are short-circuited by the positive
brushes, the other half by the adjacent negative brushes. The
total number of coils in each group is 2s, and since by assumption
the current in all of them varies at the same rate, and all of
the flux is linked with all of the coils, the equivalent inductance
of the coil A B is 2s times larger than if this coil were alone.
Thus for a multiple-wound armature
L^ = 2LPP . 2s = 4sq2 ((Pfli + (Pa'la + %(P >e'le) X 10~8 henrys. (157)
On the basis of Mr. Hobart 's tests and until more accurate data
are available, the following average values of the unit permeances
may be used: (Pj = 4 and (Pa' = <Pe' = 0.8 perms per centimeter.
The frequency / which enters into formula (156) is calculated
as follows : The time between the positions 1 and 2 of the brushes
corresponds to one-half of one cycle, because during this interval
the current changes from +/i to —I\. Let v be the peripheral
velocity of the commutator, in meters per sec., let b be the thick-
ness of the brushes, and b' the thickness of the mica insulation
between the commutator segments, both in millimeters. The
time between the positions 1 and 2 of the brushes is (6 — bf)/WOOv
seconds. Hence
f=5QOv/(b-V) cy./sec (158)
CHAP. XII] INDUCTANCE OF WINDINGS 237
The number of simultaneously short-circuited coils varies
periodically with the position of the brushes. Thus, in Fig. 57
sometimes two and sometimes three coils are short-circuited
by one set of brushes. On the average
s=(b-b')/a, . . . . . . (159)
where a is the width of one commutator segment including the
mica insulation. Thus, all the values which enter into the
formula (156) are determined, and the reactance voltage for a
given machine can be easily calculated.
Formula (156) is used not only as a criterion of the commuta-
tion, but also for the calculation of the flux density, under the
interpoles where such are required. Namely, this flux density must
be such that the average voltage induced by the commutating
flux is approximately equal and opposite to the average reactance
voltage; see Art. 24. prob. 6. A still closer compensation for the
influence of the inductance is achieved by properly grading the
commutating flux, so as to compensate not only for the average
reactance voltage, but also to some extent for the instantaneous
induced e.m.fs.
The average reactance e.m.f. induced in some other direct cur-
rent windings. With two-circuit wave windings two cases must
be considered, namely, (a) when the machine is provided with
only two sets of brushes, (b) when there are more than two sets
of brushes. In the first case eq. (156) is used, where
and Leq is understood to comprise the short-circuited conductors
under all the poles, per commutator segment. Let there be
again q turns per coil, that is, per unit of winding per pair of
poles. Since the corresponding conductors under all the poles
are in series, we have that Leq=pLPP. The influence of the
other simultaneously short-circuitqd coils is expressed as before
by the factor 2s, where s is given by formula (159). Thus, for
a two-circuit winding with two sets of brushes
Leg=2psq2((Pi'li+(Pa'la + i(Pe'le)XlO-1 henrys. . (161)
The frequency /is given as before by eq. (158).
When more than two sets of brushes are used, the sets of
equal polarity are connected in parallel by the stud connections
238 THE MAGNETIC CIRCUIT [ART. 68
outside the armature windings, and beside there are two short-
circuiting paths through the coils undergoing commutation: a
long path and a short path. Thus, the problem becomes
indefinite, because it is not possible to tell the relative amounts of
the current through these different paths. Disregarding the short
path, the criterion becomes the same as in the case of a machine
with two sets of brushes only.1 This is on the safe side, and
the commutation may be expected to be better than that cal-
culated, or at the worst, as good.
With multiplex windings, the expression for eave is the
same as that given above, provided that proper values are selected
for /!, s, and /. With regard to the- latter quantity it must
be remembered that bf is much larger than the actual thickness
of mica. • Namely, with respect to the component winding under
consideration the metal of the commutator segments belonging
to the other component windings is equivalent to insulation.
This fact must not be lost sight of in choosing the correct value
for bf to be used in the expression (158).
With fractional-pitch windings the reactance voltage is smaller
than with the corresponding full-pitch winding, because the
conductors short-circuited under the adjacent sets of brushes
(Fig. 57) are situated in part or totally in different slots, and
have a smaller common magnetic flux, or none at all. When
the winding pitch is reduced considerably, s instead of 2s must
be used in the preceding formulae; otherwise a value between
s and 2s must be chosen, according to one's judgment.2
Prob. 22. The armature of a 6-pole, 600-r.p.m., multiple-wound,
direct-current machine has the following dimensions: Diameter, 85 cm.;
gross length, 22 cm.; three air-ducts, 1 cm. each; 1008 face conductors.
1 For an analysis of this case see C. A. Adams, Reactance E.M.F. and
the Design of Commutating Machines, Electrical World and Engineer, Vol.
46 (1905), p. 346.
2 For an advanced and more scientific theory of commutation, see Arnold,
Die Gleichstromaschine, Vol. 1 (1906), pp. 354 to 513; in particular the
approximate formula (170) on bottom of p. 498; also Vol. 2 (1907), chapter
14. A simpler and more concise treatment will be found in Tomalen's
Electrical Engineering. A good practical treatment will also be found in
Pichelmayer's Dynamobau, pp. 86-118; it is considerably simplified as
compared to Arnold's treatment, and is accurate enough for practical pur-
poses, because the numerical values of unit permeances are known only
approximately.
CHAP. XII] INDUCTANCE OF WINDINGS 239
The commutator diameter is 52 cm.; the number of segments, 252;
the mica insulation is 1 mm. thick; the brushes are 15 mm. thick.
What is the average reactance voltage when the total armature current
is 320 amp.? Ans. 4.52 volts.
Prob. 23. Show that the answer to the preceding problem would
be nine times larger if by mistake the winding were assumed to be of
the two-circuit type.
Prob. 24. The peripheral velocity of a commutator is 18 meters
per sec., the width of each segment (without mica) is 4.5 mm.; the
thickness of the mica is 0.9 mm. The commutator is to be used in con-
nection with a duplex winding. What is the smallest permissible
thickness of the brushes if the frequency of commutation must not
exceed 800 cycles per sec.? Ans. 17.5 mm.
Prob. 25. For a perfect commutation and for an imperfect one
draw the following curves to time as abscissae: (a) the current in the
short-circuited coil; (6) the currents in the leads c and d] (c) the
current densities under the heel and toe of the brush. Take the width
of the brushes to be equal to that of one commutator segment, and
assume the mica insulation to be of a negligible thickness.
Prob. 26. Show that the width of the brushes has comparatively
little net effect upon the commutation of a machine.
Prob. 27. What flux density is needed in the interpolar zone in
prob. 22 to secure perfect commutation? Ans. 2.23 kl./sq. cm.
CHAPTER XIII
THE MECHANICAL FORCE AND TORQUE DUE TO
ELECTROMAGNETIC ENERGY.
69. The Density of Energy in a Magnetic Field. The reader
is already familiar with the fact that a certain amount of energy
is required to establish the flux within a magnetic circuit, and
that this energy remains stored in the field. This stored energy
may be conveniently thought of as the kinetic energy of vor-
tices around the lines of force (Art. 3). Various expressions
for the total stored electro-magnetic energy are given in Arts.
57 and 58; the problem here is to find a relation between the
distribution of the flux density and that of the energy in the
field.
Consider first the simplest magnetic circuit (Fig. 1) con-
sisting of a non-magnetic material. According to the last eq.
(99), the total energy stored in such a circuit is
joules, . . . . :. (162)
if 0 is in webers, I and A in cm., and /*= 1.257X10"8 henrys
per cm. cube. The volume of the field is V=IA cubic cm.
Since the flux density is uniform, the energy is also uniformly
distributed, and the density of the energy is
Denoting the density of the energy W/V by W, and introducing
the flux density B= 4>/A, we get
TF' = fB2//* joules per cu.cm. . . . (163)
Either B or // can be eliminated from this expression by means
of the relation B = pH, so that we have two other expressions
for the density of the energy:
....... (164)
(165)
240
CHAP. XIII] TORQUE AND TRACTIVE EFFORT 241
Two more expressions for the density of the energy can be written,
using the reluctivity v instead of the permeability /*.
In a uniform field the preceding expressions represent the
actual amounts of energy stored per cubic centimeter. In a
non-uniform field W is the density of energy at a point, or the
limit of the expression AW/AV. This is analogous to what we
have in the case of a non-uniform distribution of matter, where
the density of matter at a point is the limit of the ratio of the
mass to the volume. Thus, the total energy stored in a non-
uniform field is
(166)
where the integration is to be extended over the volume of the
whole magnetic circuit. Similarly, from eqs. (164) and (165)
we get
flr-i/il H2dV, (167)
W=%C HBdV (168)
These expressions are consistent with eqs. (102) and (102a)
as is shown in pfob. 6 below.
When fj. is variable, the preceding formulae do not hold true,
and the density of energy is represented by eq. (19), Art. 16.
Prob. 1. Deduce an expression for the magnetic energy stored in
the insulation of a concentric cable (Fig. 46), between the radii a and b,
the length of the cable being I cm. and the current i. Hint: For an
infinitesimal shell of a radius x and thicknesss dx we have : H = i/2nx,
and dV=2nxl dx. Ans. W = 0.23ft2 log (6/o)10-8 joules.
Prob. 2. Check the answer to the preceding problem by means
of eqs. (104) and (109).
Prob. 3. In a concentric cable (Fig. 46) a = 7 mm. and b is 20 mm.
What is the density of the energy at the inner and outer conductors,
when i = 120 amp.?
Ans. 4.68 and 0.57 microjoules per cu.cm.
Prob. 4. Deduce expression (110) from eq. (167).
Prob. 5. Taking the data from the various problems given in this
book as typical, show that ordinarily in generators and motors a large
proportion of the total energy of the field is stored in the air-gap.
Prob. 6. Show that eqs. (166) to (168) are consistent with eqs.
(102) to (103u). Solution : Take an infinitesimal tube of partial linkages
242 THE MAGNETIC CIRCUIT [ART. 70
(Fig. 45). The energy contained in this tube is dW = %MPd(I>p; but
MP= I Hdl, and d$=BdA. Since d@ is the same through all cross-
sections of the tube, d@ can be introduced under the integral sign, and
we have dW ' = i \ HdlBdA =J f HBdV, the integration being extended
over the volume of the tube. The total energy of the circuit is found
by extending the integration over the volume of all the tubes of the
field. The other equations are proved in a similar manner.
70. The Longitudinal Tension and the Lateral Compression
in a Magnetic Field. The existence of mechanical forces in a
magnetic field is well known to the student. He needs only
to be reminded of the supporting force of an electromagnet, of
the attraction and repulsion between parallel conductors carrying
electric currents, of the torque of an electric motor, etc. These
mechanical forces must necessarily exist, if the magnetic field
is the seat of stored energy. This is because, if we deform the
circuit, we must in general change the stored energy and hence
do mechanical work. The lines of force tend to shorten them-
selves and to spread laterally, so as to make the permeance of
the field a maximum, with the complete linkages. Where there
are partial linkages, it is the total stored energy that tends toward
a maximum (Art. 57). This fact is entirely consistent with
the hypothesis of whirling tubes of force, because the centrifugal
force of rotation produces exactly the same effect, that is, a lateral
spreading and a tension along the axis of rotation. A good
analogy is afforded by a short piece of rubber tube filled with
water and rotated about its longitudinal axis.
(a) The Longitudinal Tension. Consider again the simple
magnetic circuit (Fig. 1), and let it be allowed to shrink, due
to the longitudinal tension of the lines of force, so as to reduce
its average length by Al, without changing the cross-section A.
Let at the same time the current be slightly decreased so as to
keep the same total flux as before. Let Ftf be the mechanical
tension along the lines of force, per square cm. of cross-section
A; then the mechanical work done against the external forces
which hold the winding stretched is (Ft'.A) Al. The density
of energy W remains the same because B is the same, but the
total stored energy is decreased by W'(AM), because the volume
of the field is decreased by AM. Since the change was made
CHAP. XIII] TORQUE AND TRACTIVE EFFORT
243
in such a way as to keep the total flux constant, no e.m.f. was
induced in the winding during the deformation, and consequently
there was no interchange of energy between the electric and
the magnetic circuit. Thus, the decrease in the stored energy
Coil
FIG. 58. — A lifting electromagnet.
is due entirely to the mechanical work performed.
the two preceding expressions, we have that
Equating
. .,{. . (169)
If W is in joules per cu.cm., F't is in joulecens per sq.cm. (see
Appendix I) , so that in a rational system of units the mechanical
stress per unit area is numerically equal to the density of the stored
244 |;| THE MAGNETIC CIRCUIT [ART. 70
energy. The physical dimensions of F' and W are also the
same.
If Ft is in kg. per sq.cm., B in kilolines per sq.cm., and
H in kiloampere-turns per cm., the preceding formula becomes,
when applied to air,
(170)
These formulae apply directly to the lifting magnet (Fig.
58), and give the carrying weight per unit area of the contact
between the core and its armature. The total weight which
the magnet is able to support is
where A is the sum of the areas denoted by Si and 82. Of course,
H is taken for the air-gap, which is the only part of the circuit
that is changing its dimensions when the armature is moved.
(b) The Lateral Compression. Let now the simple magnetic
circuit be allowed to expand laterally by a small length Js in
directions perpendicular to the surface of the toroid. Let Fc'
be the pressure (compression) exerted by the lines of force upon
the winding, per sq. cm. of the surface of the toroid. Then
the mechanical work done by the magnetic forces in expanding
the ring against the external forces which hold the winding,
is SFC'AS, where S is the surface of the toroid. Let again the
current be slightly decreased during the deformation, so as to
keep the flux constant. No voltage is induced in the winding,
and hence there is no interchange of energy between the electric
and the magnetic circuit. Thus we can find Fc', as we found
the stress in the case of the tension, by equating the work done
to the decrease in the stored energy. The stored energy is
expressed by eq. (162), in which A is the only variable; hence
by differentiating W with respect to A we get :
This is a negative quantity, because the stored energy decreases.
But IAA represents the increase in the volume of the ring, so
that IAA = SJs, and consequently
Fe' = l&/n=ti,H* = W'~Ft' ..... (172)
In other words, the lateral compression is nummerically equal to
CHAP. XIII] TORQUE AND TRACTIVE EFFORT
245
the longitudinal tension, and both are numerically equal to the
density of the stored energy.
As an application of the lateral action, consider a constant-
current or floating-coil transformer (Fig. 59), used in series
arc-lighting. The leakage flux is similar in its character to that
shown in Figs. 50 and 51. The lateral pressure of the leakage
lines between the coils tends to separate them, acting against
the weight of the floating coil. A part of this weight has to be
balanced by a counter-weight Q because the electro-magnetic
forces under normal operation are comparatively small.
Since the currents are alternating, the force is pulsating,
_T I/J /Core
Hi
j|
>
1
1
I
1
. Tn Constant*
/ Current
'/— — Series A.C.
s Lamps
Counter-
^ weight
^
Secondary
Goil
i5
H-
v^
|t* * 'I
From a Constant
Potential A.C.
Supply
T
Primary Goil/
FIG. 59. — A floating-coil constant-current transformer.
but is always in the same direction, tending to separate the coils.
The average force depends upon the average value of H2t in
other words, upon the effective value of the current. According
to eqs. (170) and (172), we have
(Fc)ave=(Fc')ave.S=HeffS/15.Qkg., . .. (173)
where S is the area of the floating coil in contact with the flux.
With the assumed paths for the lines of force, and neglecting
the reluctance of the iron core, we have that
Heff=nIeff/WQOl kiloamp.-turns/cm., . . (174)
where I is the length of the lines of force in * the air, in cm., and
nleff is the m.m.f. of either coil. The force of repulsion is pro-
portional to the square of the current, and is independent of the
246 THE MAGNETIC CIRCUIT [ART. 70
distance h between the coils. Hence, a constant weight Q
regulates for a constant current. When the coils are further
from each other, the induced secondary voltage is less, on account
of a much higher leakage flux. When the current increases
momentarily, due to a decreasing line resistance, the coil is
overbalanced and rises till the induced voltage and current fall
to the proper value. Thus, the coil always floats at the proper
height to induce the voltage needed on the line.
Formulae (173) and (174) apply also to the mechanical forces
between the primary and the secondary coils of a constant-
potential transformer (Figs. 13 and 51). Under normal con-
ditions these forces are negligible, but in a violent short-circuit
the end-coils are sometimes bent away and damaged, unless
they are properly secured to the rest of the winding. Such
short-circuits are particularly detrimental in large transformers,
having a close regulation, that is, having a very small internal
impedance drop, and which are connected to systems of practically
unlimited power and constant potential. As Dr. Steinmetz puts
it, the closest approach to the appearance of such a transformer
after a short-circuit is the way two express trains must look
after a head-on collision at high speed.
Another interesting example of the effect of the mechanical
forces produced by a magnetic field is the so-called pinch phe-
nomenon.1 The lines of force which surround a cylindrical con-
ductor may be compared to rubber bands, which tend to com-
press it. With a liquid conductor and large currents, such
for instance as are carried by a molten metal in some electro-
metallurgical processes, the pressure of the magnetic field is
sufficient to modify and to reduce the cross-section of the liquid
conductor. This was first observed by Mr. Carl Hering and
called by him the pinch phenomenon. In passing a relatively
large alternating current through a non-electrolytic liquid con-
ductor contained in a trough, he found that the liquid contracted
in cross-section and flowed up-hill lengthwise in the trough,
climbing up on the electrodes. With a further increase of
1 E. F. Northrup, Some Newly Observed Manifestations of Forces in the
Interior of an Electric Conductor, Physical Review, Vol. 24 (1907), p.
474. This article contains some cleverly devised experiments illustrating
the pinch phenomenon, and also a mathematical theory of the forces which
come into play.
CHAP. XIII] TORQUE AND TRACTIVE EFFORT
247
current, this contraction of cross-section became so great at
one point that a deep depression was formed in the liquid, with
steeply inclined sides, like the letter V.
In most cases of mechanical forces in a magnetic field, these
forces and the resulting movements are due to the combined
Stop
Coil
Core
FIG. 60. — A tractive electromagnet. FIG. 61.-
-Two bus-bars and their
support.
action of longitudinal tensions and transverse compressions and
not to one of these actions alone. For- instance, a loop of flexible
wire, through which a large current is flowing, tends to stretch
itself so as to assume a maximum opening, that is, a maximum
permeance of the magnetic field linked with it. This action
is due to both the longitudinal tension and the lateral pressure.
In such cases the mechanical forces are best computed by the
principle of virtual displacements explained in the next article.
248
THE MAGNETIC CIRCUIT
[ART. 70
Prob. 7. Show that the required flux density in the air-gap of a
lifting electromagnet (Fig. 58) can be calculated from the expression
B = 15.7VoF/A, in kl/sq.cm., where F is the rated supporting force,
in metric tons, A is the area of contact in sq. dm., and a is the factor
of safety.
Prob. 8. Show that in an armored tractive magnet (Fig. 60) the
tractive effort F varies with the air-gap s according to the law Fsz = 3.G8
kg-cm., when the excitation is 2000 amp.-turns and the cross-section
of the plunger and of the stop is 12 sq.cm. Assume the leakage and the
reluctance of the steel parts to be negligible.
Prob. 9. Referring to the preceding problem, what is the true
average pull between the values s = l and s=4 cm., and what is the
arithmetical mean pull? Ans. 0.77 and 1.63 kg.
Prob. 10. Indicate roughly the principal paths of magnetic leakage
in Fig. 60, and explain the influence of the leakage upon the tractive
effort, with a small and a large air-gap.
Prob. 11. The flux between two thin and high bus-bars, placed at
a short distance from each other, has the general character shown in
Fig. 61. Calculate the force per meter length that pushes the bus-bars
apart when, during a short-circuit, the estimated current is 50 kilo-
amperes. Ans. About 800 kg. per meter.
Prob. 12. Deduce an expression for the magnetic pull due to the
eccentric position of the armature in an electric machine (Fig. 62).
A certain allowance is usually made for this pull in addition to the
weight of the revolving part, in determining the safe size of the shaft.
Solution : Since the pull is proportional
to the square of the flux density, we
replace the actual variable air-gap
density by a constant radial density
acting upon the whole periphery of the
armature and equal to the quadratic
average (the effective value) of the
actual flux density distribution. Let
this value be Beff kl.per sq. cm. when
the armature is properly centered.
Let the original uniform air-gap be a,
and the eccentricity be s. Since a and
e are small as compared to the diameter
of the armature, the actual air-gap
at an angle a from the vertical is
approximately equal to a — ecosa. Neg-
lecting the reluctance of the iron parts
of the machine, the flux density is inversely as the length of the air-
gap, so that we have
Ba=Beffa/(a-£ cos a) =Beff/[l - (e/o) cos a].
a— e
FIG. 62. — An eccentric armature-
Let A be the total air-gap area to which B refers; then, according
CHAP. XIII] TORQUE AND TRACTIVE EFFORT 249
to eq. (171), the vertical component of the pull upon the strip of the
width dais
a>cos a.
The horizontal component of the pull is balanced by the corresponding
component on the other half of the armature. The total pull downward is
/•*
Fex=[2A/(27tX24.7)] I Ba2 cos a da.
Jo
Putting tan a =z and integrating we get the so-called Sumec formula
for the eccentric pull, in kg. :
Fex = (ABeff*/24.7)(e/a)[l-(s/ay]-™. . . . (175)
The integration is simplified; if, before integrating, the difference is
taken between the vertical forces at the points corresponding to
a and to n — a. The limits of integration are then 0 and %x.1
Prob. 13. The average flux density under the poles of a direct-
current machine is 6.5 kl/sq.cm.; the poles cover 68 per cent of the
periphery. The diameter of the armature is 1.52 m.; the effective
length is 56 cm. What is the magnetic pull when the eccentricity is
10 per cent and when it is 50 per cent of the original air-gap?
Ans. 3.2 and 24 metric tons.
Prob. 14. The 22/2-kv., 2500-kva., transformer specified in prob.
5, Art. 64, had a total impedance drop of 73.5 volts at full load current
on the low-tension side. What average force is exerted on each coil-
face during a short-circuit, provided that the line voltage remains
constant? The transformer winding consists of 12 high-tension coils of
100 turns each, and of 11 low-tension coils interposed between the high-
tension coils, together with 2 half-coils at the ends. Two of the dimen-
sions of the coils are repeated here: Om=2.6 m.; J=18 cm. Hint: The
short-circuit current is equal to 2000/73.5 = 27.2 times the rated current.
Ans. About 22. metric tons.
Prob. 16. What is the mechanical pressure on the surface of the
conductors in prob. 3?
Ans. 47.6 and 5.8 milligrams per sq.cm.
71. The Determination of the Mechanical Forces by Means
of the Principle of Virtual Displacements. In order to determine
the mechanical force or torque between two parts of a mag-
netic circuit the general method consists in giving these parts
an infinitesimal relative displacement and applying the law of
1 J. K. Sumec, Berechnung des einseitigen magnetischen Zuges bei Excen-
trizitat, Zeitschrift fur Elektrotechnik (Vienna), Vol. 22 (1904), p. 727. This
periodical is continued now under the name of Electrotechnik und Maschin-
enbau.
250 THE MAGNETIC CIRCUIT [ART. 71
the conservation of energy to this displacement. From the
equation so obtained the component of the force in the direction
of the displacement can be calculated. Taking other displace-
ments in different directions, a sufficient number of the com-
ponents of the forces are determined to enable one to calculate
the forces themselves. Since the forces in a given position of
the system are perfectly definite, the result is the same no matter
what displacements are assumed, provided that these displace-
ments are possible, that is, consistent with the given conditions
of the problem. Therefore displacements are selected which
give the simplest formula? for the energies involved. We have
had two applications of this principle in the preceding article,
in deriving the expressions for the tension and the compression
in the field, by giving the simple magnetic circuit the proper
" virtual " displacements. In applying this method, not only
the mechanical displacement has to be specified, but also the
electric and the magnetic conditions of the circuit, in order to
make the energy relations entirely definite. Thus, in the pre-
ceding article, the electromagnetic condition was 0 = const.1
First let us take the case when the partial linkages are neg-
ligible; then according to the third eq. (99), the stored energy is
, ..... . . (176)
where (ft is the reluctance of the circuit. Let F be the unknown
mechanical force between two parts of the magnetic circuit at
a distance s, and let one part of the system be given an infini-
tesimal displacement ds. Let F be considered positive in the
direction in which the displacement ds is positive. The mechan-
ical work done is then equal to Fds. As in the preceding article,
let this displacement take place with a constant flux, so that
there is no interchange of energy between the magnetic circuit
under consideration and the electric circuit by which it is excited.
Then the work is done entirely at the expense of the stored energy
of the magnetic circuit, and we have :
Fds=dWm=-dW8, ..... (177)
where dWm is the mechanical work done. The sign minus before
1 The principle of virtual displacements is much used nowadays in the
theory of elasticity and in the calculation of the mechanical stresses in the
so-called statically-indeterminate engineering structures.
CHAP. XIII] TORQUE AND TRACTIVE EFFORT 251
dW8 is necessary because the stored energy decreases. From
eqs. (176) and (177) we get
F= -%<P2.d(R/ds ....... (178)
In some cases it is more convenient to express F through M
and (P. We have
or
F=+%M2d(P/ds. ...... (179)
In the preceding formulae F is in joulecens, M is in ampere-
turns, ^ is in webers, (P is in henrys, and (ft in yrnehs. With
other units the formulae contain an additional numerical factor.
It is to be noted that the mechanical forces are in such a direc-
tion that they tend to increase the permeance and decrease the
reluctance of the circuit. This agrees with previous statements.
See Arts. 41 and 57.
If the partial linkages are of importance, it is convenient
to express the stored energy in the form TFs=Ji2!/, because the
inductance L takes account of the partial linkages; see eqs.
(105) and (106), Art. 58. The energy equation, according to
eq. (177), is then
-dW8 = -%d(v*L) ..... (180)
and the condition that there is no interchange of energy with
the line is
d(iL) = Q. . ...... (181)
The latter equation becomes clear by reference to eq. (106a),
because Li=ntf>eq, where ^ is the equivalent flux under the
supposition of no partial linkages. The condition that there
shall be no e.m.f. induced in the winding during the displace-
ment, is d(n<t>eg)/dt=Q, whence eq. (181) follows directly.
Performing the differentiations in eqs. (180) and (181), and
substituting the value of Ldi from the second equation into the
first, we get that
(182)
When there are no partial linkages, L = n2(P, and eq. (182)
becomes identical with (179).
252 THE MAGNETIC CIRCUIT [ART. 71
Formulae for the average force of direct current electromagnets.
In a tractive magnet (Fig. 60) or a rotary magnet (Fig. 63)
it is often required to know the average pull over a finite travel
of the moving part. For the average force, the equation
FavedS=4Wm= — 4W8 holds, which is analogous to eq. (177) ;
the only difference being that finite instead of infinitesimal incre-
ments are used. If the motion takes place at a constant flux,
or at least the values of the flux are the same in the initial and
the end-positions, we get from the preceding formulae :
• V . (183)
-Sl); . . (184)
-s1). .... (185)
The finite travel of the plunger often takes place at a con-
stant current, for instance, in the regulating mechanism of a
FIG. 63. — A rotary electromagnet.
series arc-lamp, also approximately in a direct-current electro-
magnet connected across a constant-potential line. Under such
conditions the foregoing formulae are not directly appliable,
because they have been deduced under the assumption of no
interchange of energy between the electric and the magnetic
circuits, so that this case has to be considered separately.
Let the motion be in the direction of the magnetic attraction,
and let the current remain constant during the motion. The
stored energy is larger in the end-position than in the initial
position, because the flux is larger and the current is the same.
Therefore, the energy supplied during the motion from the line
must be sufficient to perform the mechanical work, and to
CHAP. XIII] TORQUE AND TRACTIVE EFFORT 253
increase the energy stored in the magnetic circuit. The increase
in the stored energy is
and the energy supplied from the line is calculated as follows:
The average voltage induced during the motion of the plunger
is eave=i(L2—Li)/t, where t is the duration of the motion. The
energy supplied from the line is therefore
Thus, the energy supplied from the line is twice as large as the
work performed, and we have the following important law
(due to Lord Kelvin) : When in a singly excited magnetic circuit,
without saturation, a deformation takes place, at a constant current,
the energy supplied from the line is divided into two equal parts,
one half increasing the stored energy of the circuit, the other half
being converted into mechanical work.
According to this law we have, for a constant current electro-
magnet, that the mechanical work done is equal to the increase
in the energy stored in the magnetic field. Hence
Thus,
Fave=^i2(L2-Ll)/(s2-s1)'} ...... (186)
or, if the partial linkages are negligible,
s1)') . . . (187)
(188)
When a magnet performs a rotary motion (Eig. 63), the
preceding formulae are modified by substituting TdO in place of
Fds, or Tave(62— 61) in place of Fave(s2 — si). Here T is the torque
in joules and (62—61) or dO is the angular displacement of the
armature in radians. Or else, in the foregoing formulas F may
be understood to stand for the tangential force, and the dis-
placement to be ds=r dd, where r is the radius upon which
the force F is acting. Then the torque is T= Fr. If, for instance,
we apply eq. (179) to a rotary motion, it becomes
T=Fr=+%M2d(P/dO. ' ..... (189)
The other equations may be written by analogy with this one.
254 THE MAGNETIC CIRCUIT [ART. 71
Alternating-Current Electromagnets. The preceding formulae are
deduced under the supposition that the magnetic field is excited
by a direct current. They are, however, applicable also to alter-
nating-current electromagnets, because the pulsations in the
current merely cause the energy to surge to and from the magnetic
circuit, without any net effect, so far as the average stored energy
and the mechanical work are concerned. The average stored
energy corresponds to the effective values of the current and the
flux.
In practice two types of alternating-current electromagnets
are of importance, namely, those operating at a constant voltage
and those operating at a constant current. As an example of
the first class may be mentioned the electromagnets used for
the operation of large switches at a distance (remote control);
the windings of such electromagnets are usually connected directly
across the line. Constant-current magnets are used in the
operating mechanism of alternating-current series arc-lamps. In
an A. C. electromagnet practically all of the voltage drop is
reactive and hence proportional to the flux.
In a constant-potential electromagnet the effective value of
the equivalent flux is the same for all positions of the plunger
(neglecting the ohmic drop in the winding). Therefore, formula
(185) holds true. Let e be the effective value of the constant
voltage, or more accurately the reactive component alone, and
let/ be the frequency of the supply. Then, e=2xfLiii = 2nfL2i2,
so that the formula for the pull becomes
Fave = e(il-i2)/l^f(s2-s1)] ..... (190)
For a constant-current A.C. electromagnet eq. (186) applies;
introducing again the reactive volts ei = 2/r/L1t and e2 = 2nfL2i,
we get
-si)]. .... (191)
In both cases the mechanical work performed is proportional
to, the difference in the reactive volt-amperes consumed in the
two extreme positions of the moving part.1
1 For further details in regard to electromagnets consult C. P. Steinmetz,
Mechanical Forces in Magnetic Fields, Trans. Amer. Inst. Elec. Engs., Vol.
30 (1911), and the discussion following this paper; also C. R. Underbill
Solenoids, Electromagnets, and Electromagnetic Windings (1910), chapters 6
CHAP. XIII] TORQUE AND TRACTIVE EFFORT 255
Prob. 16. Derive formula (171) for the lifting magnet by means
of the principle of virtual displacements. Solution: The reluctance
of the air-gap is (R = s/(/*A), so that d(R/ds = l/(/*A), hence, according
to eq. (178) F= -J^VCM) = - AB2/2p. The minus sign indicates that
the stress is one of tension.
Prob. 17. Derive expression (172) from formula (179).
Prob. 18. Derive the formula for the repulsion between the wingings
in a transformer from eq. (182).
Prob. 19. Derive from eq. (182) the force of repulsion between two
infinitely long, parallel, cylindrical conductors placed at a distance
of b meters apart, and forming an electric circuit (Fig. 47).
Ans. 2Mi2(l/fy X10~8 kg. for I meters of the loop.
Prob. 20. What deformation of the windings may be expected
during a severe short-circuit of a core-type or a cruciform type trans-
former (Figs. 12 and 14) with cylindrical coils, (a) when the centers
of the coils are on the same horizontal line, and (6) when one of the
windings is mounted somewhat higher than the other ?
Prob. 21. Show that in a constant-current rotary magnet (Fig.
63) Ta= Const. — that is, the torque in the different positions of the
armature is inversely proportional to the air-gap at the entering pole-
tip. Hint: cKP = pwrdO/a, where w is the dimension parallel to the
shaft.
Prob. 22. State Kelvin's law when mechanical work is done against
the forces of the magnetic field.
Prob. 23. A 60-cycle, 8-amp., series arc-lamp magnet has a stroke
of 32 mm.; the reactive voltage consumed in the initial position is 9
v., and in the final position 20 v. What is the average pull ?
Ans. 372 grams.
72. The Torque in Generators and Motors. The magnetic
circuits considered in the preceding articles of this chapter are
singly excited, that is, they have but one exciting electric circuit.
From the point of view of mechanical forces this also applies
to each air-gap in a transformer, because, neglecting the mag-
netizing current, the primary and the secondary coils may be
combined into equivalent leakage coils (Art. 64). On the other
hand, a generator or a motor under load has a doubly-excited
magnetic circuit, the useful field being linked with both the
field and the armature windings. The two m.m.fs. not being
in direct opposition in space, the flux is deflected from the shortest
to 9 incl.; S. P. Thompson, On the Predetermination of Plunger Electro-
magnets, Intern. Elect. Congress, St. Louis, 1904, Vol. 1, p. 542; E. Jasse,
Ueber Elektromagnete, Elecktrotechnik und Maschinenbau, Vol. 28 (1910),
p. 833; R. Wikander, The Economical Design of Direct-current Electro-
magnets; Trans. Amer. Inst. Elect. Engs., Vol. 30 (1911).
256
THE MAGNETIC CIRCUIT
[ART. 72
o
path, and the torque is due to the tendency of the tubes of force
to shorten themselves longitudinally, and to spread laterally.
Consider the simplest generator or motor, consisting of a
very long straight conductor which can move at right angles
to a uniform magnetic field of a density B (Fig. 64). Let the
ends of the conductor slide upon two stationary bars through
which the current is conducted into a load circuit in the case of
a generator action, and through which the power is supplied
in the case of a motor action. If the magnetic circuit contains
no iron, the resultant field is a
superposition of the original uniform
field and of the circular field created
by the current in the conductor.
With the direction of the current
indicated in Fig. 64, the resultant
field is stronger on the left-hand side
of the conductor than it is on the
right-hand side, and there is a
resultant lateral pressure exerted
upon the conductor to the right.
In the case of a motor action the
direction of the motion is in the
same direction as the pressure from
the stronger field. In the case of a
generator action the conductor is
moved by an external force against this pressure. Compare
also the rule given in Art. 24.
To find the mechanical force between the conductor and the
field, we will apply again the principle of virtual displacements.
Let the current through the conductor be i, and let the con-
ductor be moved against the magnetic pressure by a small amount
ds. Assume that the stored magnetic energy of the electric
circuit to which the conductor belongs is the same in the various
positions of the conductor. (Such is the case in actual machines.)
Then the work done by the external force is entirely converted
into electrical energy, and we have
Gen
Mdtor
FIG. 64. — A straight conductor
in a uniform magnetic field.
(192)
where e is the induced e.m.f. Let both F and e refer to a length
CHAP. XIII] TORQUE AND TRACTIVE EFFORT 257
I of the conductor. Substituting the value of e from eq. (27),
Art. 24, we have
Fds/dt=iBlv,
or, since ds/dt=v,
F=iBl. . . \ . . . . (193)
In this expression i is in amperes, B is in webers per sq.cm.,
I is in cm., and F is in joulecens. With other units the formula
contains a numerical factor.
Formula (193) maybe used also with anon-uniform field, and
also when the direction of the conductor is not at right angles
to that of the line of force. In such cases the formula becomes
dF=iBdl, where dF is the force acting upon an infinitesimal
length dl of the conductor, and B is understood to be the com-
ponent of the actual flux density perpendicular to dl.
As an application of formula (193), consider the attraction
or the repulsion between two straight parallel conductors carrying
currents i\ and 1*2, and placed at a distance b from each other.
The circuit of each conductor may be considered closed through
a concentric cylindrical shell of infinite radius, as in Art. 60.
It is apparent from symmetry that the field produced by each
system gives no resultant force with the current in the same
system. Thus, the mechanical force is due to the action of the
field 1 upon the current 2, and vice versa.
The flux density due to the system 2 at a distance b from
the conductor 2 is B= /jL.i2/2xb, so that, according to eq. (193),
F=fiiii2l/ 2nb joulecens, (194)
where n= 1.257 X 10~8. In kilograms the same formula is
F = 2Mili2(l/b)W-8, (194a)
provided that I and b are measured in the same units. The
force is an attraction or a repulsion according to whether the
two currents are flowing in the same or in the opposite directions.
When ii = i2, this formula checks with that given in prob. 19
above.
Formula (193) applies also to the tangential force between
the field and armature conductor in any ordinary generator or
motor, provided that (a) the conductors are placed upon a
smooth-body armature, and (6) the conductors are distributed
258 THE MAGNETIC CIRCUIT [ART. 72
uniformly over the armature periphery, so that the stored mag-
netic energy is the same in all positions of the armature. It
would be entirely wrong, however, to apply this formula to
a slotted armature, using for B the actual small flux density
in the slot within which the conductor lies. This would give
the force acting upon the conductor itself, and tending to press
it against the adjacent conductor or against the side of the slot;
but the actual tangential force exerted upon the armature as
a whole is many times greater, and practically all of it is exerted
directly upon the steel laminations of the teeth.
At no-load, the flux distribution in the active layer of the
machine is symmetrical with respect to the center line of each
pole (Fig. 24), so that the resultant pull along the lines of force
is directed radially. The armature currents distort the field
as a whole, and also distort it locally around each tooth, the
general character of distortion being shown in Fig. 36. The
unbalanced pull along the lines of force has a tangential com-
ponent which produces the armature torque. This torque,
although caused by the current in the armature conductors,
is largely exerted directly upon the teeth, because the flux density
there is much higher.
Thus, in order to determine the total electromagnetic torque
in a slotted armature, it is again necessary to apply the principle
of virtual displacements. The reluctance of the active layer
per pole varies somewhat with the position of the armature,
so that the energy stored in the field is also slightly fluctuating.
It is convenient, therefore, to take a displacement which s a
multiple of the tooth pitch, in order to have the same stored
energy in the two extreme positions. This gives the average
electromagnetic torque.
(a) The Torque in a Direct-current Machine. Let the virtual
displacement be equal to 6 geometric degrees and be accomplished
in t seconds. Then we have
TaveO=iEt, . (195)
where T is the torque, i is the total armature current, and E is
the total induced e.m.f. Eq. (195) states the equality of the
mechanical work done and of the corresponding electrical energy
supplied. The average induced e.m.f. is independent of the
flux distribution, or of the presence or absence of teeth (see
CHAP. XIII] /TORQUE AND TRACTIVE EFFORT 259
Art. 24 and prob. 18 in Art. 26). Take 6 to correspond to
two pole pitches, or 6=2n/(%p); then t=l/f, where / is the
frequency of the magnetic cycles. Substituting these values and
using the value of E from eq. (37), Art. 31, we get, after
reduction,
Tave=iNp$/n joules, (196)
where $ is in webers; or
Tave = 0.0325iNp& X 10~2 kg-meters, . . (196o)
^ being in megalines.
This formula does not contain the speed of the machine,
the torque depending only upon the armature ampere-turns iN
and the total flux pti>. Consequently, the formula can be used
for calculating the starting torque or the starting current of
a motor. Eqs. (196) and (196a) give the total electromagnetic
torque, part of which serves to Overcome the hysteresis, eddy
currents, friction and windage. The remainder is available on
the shaft. When calculating the starting torque, it is necessary
to take into account the effort required for accelerating the
revolving masses.
(b) The Torque in a Synchronous Machine. The equation of
energy is
Taved=miEeos<j>'.t, (197)
where m is the number of phases, i and E are the effective values
of the armature current and the induced voltage per phase, and
$ is the internal phase angle (Fig. 37). Taking again a dis-
placement over two poles and using the value of E from eq. (3)
Art. 26, we get
Tave= Q.Q36lkbmiN cos <£'p01(r2 kg-meters. . (198)
(c) The Torque in an Induction Machine. The torque being
exerted between the primary and the secondary members of
an induction machine, it may be considered from the point of
view of either member. This is because the torque of reaction
upon the stator is equal and opposite to the direct torque upon
the rotor. For purposes of computation it is more convenient
to consider the torque from the point of view of the primary
winding, in order to be able to use the primary frequency and
260 THE MAGNETIC CIRCUIT [ART. 72
the synchronous speed. Therefore eq. (198) gives the torque of
an induction machine (including, as before, friction and hysteresis),
where the various quantities refer to the stator. However, these
quantities may equally well be taken in the secondary, but in
this case, since hysteresis occurs mainly in the stator, the torque
to overcome hysteresis is not included.
Formula (198) is hardly ever used in practice, especially
for the computation of the starting torque, because it is difficult
to eliminate the large leakage flux which gives no torque. It
is much more convenient to determine the torque from the circle
diagram, or from the equivalent electric circuit.
In case the torque is determined from the equivalent electric cir-
cuit, we can write from eqs. (195) and (197) the expressions for the
torque directly, by substituting for 6 its value 2-Tu- 1- (R.P.M.)/60.
For a direct-current machine,
Tave = 0.0325—^-— Kg.-meters. . . . (199)
TU' (xi.i .M..)
Here the induced e.m.f. E = Et±iRa, according to whether the
machine is a generator or a motor; Et is the terminal voltage
and Ra the resistance of the armature, brushes, and series field.
For an alternating-current machine,
. . . (200)
In a synchronous machine E is the induced e.m.f. and <ft is
the internal phase angle. In an induction machine, iE cos <£'
is the power per phase delivered to the rotor, that is, the input
minus hysteresis and primary PR loss. The term (R.P.M.) is
in all cases the synchronous speed of the machine.
Prob. 24. Two single-conductor cables from a direct-current machine
are installed parallel to each other at a distance of 16 cm. between their
centers, on transverse supports spaced 80 cm. apart. The rated current
through the cables is 850 amp. What is the force acting upon each
support under the normal conditions, and when the current rises to
twenty times its rated value during a short-circuit?
Ans. 0.0737 and 29.5 kg.
Prob. 25. A 4-pole, series direct-current motor must develop a
starting torque of 74 kg.-m. (including the losses). The largest possible
flux per pole is about 2.5 ml. ; there are 240 turns in series between the
brushes. Calculate the starting current. Ans. 95 amp.
CHAP. XIII] TORQUE AND TRACTIVE EFFORT 261
Prob. 26. Explain the reason for which the compensating winding
(Art. 54), while removing the armature reaction, does not affect appre-
ciably the useful torque of the motor. Hint: this can be shown by
applying the method of virtual displacements; also from the fact that
the local distortion of the flux in the teeth is not removed.
Prob. 27. On the basis of Art. 15, explain the mechanism by which
the phenomenon of hysteresis in an armature core causes an opposing
torque. Explain the same for eddy currents.
Prob. 28. Demonstrate that formula (193) may be used for slotted
armatures, provided that B stands for the average flux density per
tooth pitch. Hint: take a virtual displacement of one tooth pitch,
and express the induced voltage through the flux A® =Bl\ per tooth
pitch.
Prob. 29. Show that in a single-phase synchronous motor the
torque at the synchronous speed pulsates at double the frequency of
the supply; also that the torque is zero at any but the synchronous
speed.
Prob. 30. Prove that in an induction machine the torque near syn-
chronism is approximately proportional to the square of the voltage
and to the per cent slip. Hint: i= Const. X $ Xslip.
Prob. 31. Describe in detail how to calculate the maximum starting
torque of a given induction motor, and how to calculate the amount
of secondary resistance necessary for a prescribed torque.
APPENDIX I
THE AMPERE-OHM SYSTEM OF UNITS
THE ampere and the ohm can be now considered as two
arbitrary fundamental units established by an international
agreement. Their values can be reproduced to a fraction of a
per cent according to detailed specifications adopted by practically
all civilized nations. These two 'units, together with the centi-
meter and the second, permit the determination of the values of
all other electric and magnetic quantities. The units of mass and
of temperature do not enter explicitly into the formulae, but are
contained in the legal definition of the ampere and of the ohm.
The dimension of resistance can be expressed through those of
power and current, according to the equation P=I2R, but it
is more convenient to consider the dimension of R as fundamental
in order to avoid the explicit use of the dimension of mass [M].
For the engineer there is no more a need of using the electro-
static or the electromagnetic units; for him there is but one
ampere-ohm system, which is neither electrostatic nor electro-
magnetic. The ampere has npt only a magnitude, but a physical
dimension as well, a dimension which with our present knowledge
is fundamental, that is, it cannot be reduced to a combination
of the dimensions of length, time, and mass (or energy). Let
the dimensions of current be denoted by [I] and that of resistance
by [R] ; let the dimensions of length and time be denoted by the
commonly recognized symbols [L] and [T]- The magnitudes and
the dimensions of the important magnetic units are expressed
through these four, as is shown in the following table. For the
expressions of the electric and the electrostatic quantities in the
ampere-ohm system see the author's " Electric Circuit."
Other units of more convenient magnitude are easily created
by multiplying the above-tabulated units by powers of 10, or
by adding prefixes milli-, micro-, kilo-, mega-, etc.
A study of the physical dimensions of the magnetic quantities
262
AMPERE-OHM SYSTEM
263
is interesting in itself, and gives a better insight into the nature
of these quantities. Moreover, formulae can sometimes be checked
by comparing the physical dimensions on both sides of the equa-
tion. Let, for instance, a formula for energy be given
where k is a numerical coefficient. Substituting the physical
dimensions of all the quantities on the right-hand side of the
equation from the table below, the result will be found to have
the dimension of energy. This fact adds to one's assurance
that the given formula is theoretically correct.
TABLE OF MAGNETIC UNITS AND THEIR DIMENSIONS IN
THE AMPERE-OHM SYSTEM
Symbol and Formula
Quantity.
Dimension.
Name of the Unit.
M=nl
Magnetomotive force
'[I]
Ampere-turn.
H = M/l
Field intensity, or
[IL-1]
Ampere-turn per
m.m.f. gradient
centimeter.
@=ET
Magnetic flux
[IRT]*
Weber (maxwell).
B = #/A
Magnetic flux density
[IRTL-2]
Webers (maxwells)
per square centi-
meter.
(P = 0/M
Permeance
[RT]
Henry (perm).
(R=M/$ = 1/(P
Reluctance
[R-'T-1!
Yrneh (rel).
Henries (perms) per
v=B/H
Permeability
[RTL-1]
centimeter cube.
v=H/B = l/fi
Reluctivity
[R-iT-iL]
Yrnehs (rels) per
centimeter cube.
W = \M®
Magnetic energy or
[I2RT]
Joule or watt-sec-
work
ond.
W/V=±BH
Density of magnetic
[PRTL-3]
Joules per cubic
energy
centimeter.
F = W/l
Force
PRTL-1]
Joulecen.
* This is also the dimension of the magnetic pole strength. The concept
of pole strength is of no use in electrical engineering, and, in the author's
opinion, its usefulness in physics is more than doubtful. The whole theory
of electromagnetic phenomena can and ought to be built up on the two
laws of circuitation, as has been done by Oliver Heaviside in his Electro-
magnetic Theory.
A small irregularity is due to the use of the maxwell and of
its multiples instead of the weber. As long as this usage persists
264 THE MAGNETIC CIRCUIT
it is convenient to use the corresponding units for reluctance
and permeance, to which the author has ventured to give the
names of rel and perm. Since one maxwell is equal to 1/108 of a
weber, one perm is equal to 1/108 of one henry, and one rel is
108 yrnehs. Accordingly, permeabilities and reluctivities are
measured in perms per centimeter cube and in rels per centimeter
cube respectively.
In order not to break with the established usage, the maxwell,
the perm, and their multiples are employed in numerical compu-
tations in this book, while the weber and the henry are used in
the deduction of the formula, being the natural fundamental
units of flux and permeance in the ampere-ohm system. It is pos-
sible that the constant necessity for multiplying or dividing results
by 10~8, due to the use of the maxwell, may prove to be more and
more of an inconvenience in proportion as magnetic computations
come into common engineering practice. Then the weber, the
henry, and their submultiples will be found ready for use, and
the system of magnetic units will be completely coordinated.
Another irregularity in the system as outlined above is caused
by the use of the kilogram as the unit of force, because it leads to
two units for energy and torque, viz., the kilogram-meter and
the joule; 1 kg.-meter= 9.806 joules. Force ought to be measured
in joules per centimeter length, to avoid the odd multiplier. Such
a unit is equal to about 10.2 kg., and could be properly called
the joulecen (=107 dynes). There is not much prospect in sight
of introducing this unit of force into practice, because the kilo-
gram is too well established in common use. The next best
thing to do is to derive formulae and perform calculations, whenever
convenient, in joulecens, and to convert the result into kilo-
grams by multiplying it by g= 9.806. This is done in some
places in this book.
Thus, leaving aside all historical precedents and justifications,
the whole system of electric and magnetic units is reduced to
this simple scheme : In addition to the centimeter, the gram, the
second and the degree Centigrade, two other fundamental units
are recognized, the ohm and the ampere. All other electric and
magnetic units have dimensions and values which are con-
nected with those of the fundamental six in a simple and almost
self-evident manner (see the table above).
To appreciate fully the advantages of the practical ampere-
AMPERE-OHM SYSTEM
265
ohm system over the C.G.S. electrostatic and electromagnetic
systems, one has only to compare the dimensions, for instance,
of magnetomotive force and of flux in these three systems,
as shown below.
The Ampere-
Ohm System.
C.G.S. Electro-
magnetic System.
C.G.S. Electro-
static System.
Dimension of m.m.f.
Dimension of flux.
[11
[IRT1
Li/feT*r-y-*
LiAfir-y
L!M*T-2/cl
L*M*/c-i
APPENDIX II
AMPERE-TURN vs. GILBERT
THE reader has probably been taught before that the per-
meability of air is equal to unity in the electromagnetic C.G.S.
system; silent assumption was then probably made that /*=!
also in the practical ampere-ohm system. The true situation
is, however, as follows: In any system of units whatsoever, the
fundamental equation $ = (fiA/l) . M holds true, being a mathe-
matical expression of an observed fact. Now let the quantities
be expressed in the ampere-ohm system, and assume the centi-
meter to be the unit of length. The flux is then expressed
either in maxwells or in webers, both of which are connected
with the ampere-ohm system through the volt. The natural
(though not the only possible) unit for the magnetomotive force
is one ampere-turn. Therefore, all the quantities in the fore-
going equation are determinate, and the value of /* cannot be
prescribed or assumed, but must be determined from an actual
experiment, the same as the electric conductivity of a metal,
or the permittivity of a dielectric have to be determined.
Experiment shows that /*= 1.257 when the maxwell is used as the
unit of flux, and hence ju= 1.257X 10~8 if the flux is measured in
webers.
It is possible to assume /£= 1, provided that the unit of mag-
netomotive force is not prescribed in advance. In this case,
the unit of magnetomotive force, as determined from experiment,
comes out equal to 1/1.257 of an ampere-turn. This unit is
called the gilbert, and it must be understood that the permeability
of non-magnetic materials is equal to unity only if the magneto-
motive force is measured in gilberts. To the author the advan-
tages of such a system for practical use are more than doubtful.
In the first place, the gilbert is a superfluous unit, because the
results of calculations must after all for practical purposes be
converted into ampere-turns in order to specify the number of
266
AMPERE-TURN vs. GILBERT 267
turns and the exciting current of windings. Thus, one would
have to deal with two units of magnetomotive force, the gilbert
and the ampere-turn, one being about 0.8 of the other. In the
second place, with the assumption /*= 1 for non-magnetic materials
B becomes numerically equal to H, which is a grave inconvenience,
because B and H are different physical quantities. B and H
have different physical dimensions, because JJL has a definite
physical dimension, even though the numerical value of it is
assumed to be equal to unity for air. Therefore, to be sure
that proper physical dimensions are preserved, one has to remem-
ber where /* is omitted in formula, and for a physical interpretation
of results it is much more convenient to have it there, explicitly.
Still another objection to using the gilbert and to putting
jj. equal to unity for air is that the ratio of the ampere-turn
to the gilbert is equal to a quasi-scientific constant 47T/10.
To the author's knowledge, there is no simple, elementary way
of deducing the value of this constant, without going over
the whole mathematical theory of electricity and magnetism.
Thus, a constant is retained in practical formulae, the significance
of which remains a puzzle to the engineer all his life. It is true
that the value of /*= 1.257 is equal to the same 4^/10 after all;
but in this case there is nothing "absolute," mysterious, or
sacred about the value of 4^/10. The student is simply told
that 1.257 happens to be eaual to 4^/10 because the value of
the ampere was unfortunately so selected. It is not necessary
to go into further details, because the historical reasons which
led to the selection of the values of unit pole and unit current
hardly hold at present. All calculations would be just as con-
venient if p were equal to 2.257, or any other value, instead of
1.257.
For these reasons the author unhesitatingly discards the
gilbert in teaching as well as in practice and uses the ampere-
turn as the natural unit of magnetomotive force. The value
of permeability becomes then an experimental quantity which
depends upon the units selected for flux and length.
A,
INDEX
Active layer characteristic, definition of 168
Aging of laminations 49
Air-gap ampere-turns 89
factor, definition of 90
factor for induction machines 98
flux, nature of its distribution 88
permeance and m.m.f., accurate method 92-97
. simplified, permeance of 89
Alternating current electromagnets, tractive effort of 254
machines, torques of 259-260
machines. See also Synchronous and Induction.
Alternator. See Synchronous machine.
calculation of regulation 155
definition of load characteristic 141
definition of regulation 156
wave-form 72
Ampere-Ohm system 262
Ampere-turns, the cause of magnetism. See also M.m.f 4
Ampere-turn,
unit of m.m.f 5
vs. Gilbert 266
Analogue, mechanical, to hysteresis 36
to inductance 185
Annealing of laminations 50
Apparent flux density in teeth 101
Armature, eccentric, force upon 248
reactance in synchronous machine, nature and definition of .... 140
See also Inductance.
reaction, definition of 140
in a direct current machine 163
in an induction machine 131
in a rotary converter 175
in a synchronous machine 139
Asynchronous. See Induction machines.
Auxiliary poles. See Commutating poles.
Axial length of armature, effective 94
269
270 INDEX
Back ampere turns. See Demagnetizing, Direct.
Belt leakage, description of 223
Belts of current in a D.C. machine 163
B-H curves for iron 20-24
relation for air 15
Blondel diagram 154-155
Breadth factor, definition of 65
formula for 68
Brush shift in D.C. machines 165
Bus-bars, repulsion between 248
Cable, concentric, flux distribution in 189
induction of a single phase 191
Carter's curve for fringe permeance 95
Castings 21
Characteristic, active layer, definition of 168
air 231
load, definition of in alternator 141
open circuit. See Saturation curves, and Exciting cur-
rent.
phase, synchronous motor 142
Circuit, a simple magnetic. (See also Magnetic circuit) .> 1
magnetic, containing iron, series-parallel 29
definition of 3
Coercive force 32
Commutating poles, calculations for. (See also Interpoles) 173
definition of and location 172
Commutation, criterion of 235
description of 233
frequency of 236
of a fractional pitch winding 238
of a lap winding 235
of a multiplex winding 238
of a two circuit winding 237
vs. inductance 232
Compensating winding for D.C. machine 174
Complete linkages, definition of 181
Compounding in a D.C. machine 170
Compression, lateral, in a magnetic field 244
Computations for interpoles 173
Condenser, synchronous 157
Conductor, force on, in a magnetic field • 256
Cores of iron wire 41
of revolving machinery, m.m.f . for 105
of transformers, m.m.f for 81
Core loss current, in an induction motor. See Core loss in revolving
machinery.
INDEX 271
PAGE
Core loss current, in a transformer 82
curves 45
extrapolation of 54
in transformers 82
in revolving machinery 46
measurement of 44
or iron loss, definition of 42
Cross ampere-turns. See Reaction.
Current. See also Exciting current.
belts in a D.C. machine 103
secondary, in induction machine 132
Demagnetization and distortion. See Reaction.
Demagnetizing reaction, in direct current machines 164
in synchronous machines. See Direct reaction.
Density of flux, definition of 14
in teeth, apparent 101
Deri winding 174
Dimensions of units, table of 263
Direct current electromagnet, average tractive effort of 252
Direct current machine, brush shift 165, 167
compensating windings for 174
compounding in 170
current belts in 163
demagnetizing reaction in 164
e.m.f . induced in 75
interpoles in 172
inductance of coils in 236
loaded, field m.m.f 170
loaded, flux distribution 168
loaded, m.mf . to overcome distortion 169
torque , 258-260
transverse reaction in 165
Direct current machines. See also Saturation curves.
windings, vs. induced e.m.f 76
vs. inductance 237
Direct current motor, speed under load 171
Direct reaction. See also Demagnetizing reaction.
calculation of coefficient of 158
definition of coefficient of 153
nature of '. 150
Displacements, virtual, principle of 249
Distortion and demagnetization. See Reaction.
of field in a direct current machine, m.m.f. -needed to over-
come 169
Distributed windings, advantages and disadvantages of 66
definition of . . .121
272 INDEX
PAGE
Distributed windings, for alternator fields 121
e.m.f. of. See E.m.f.
, m.m.f. of D.C 165
m.m.f . of single-phase 123
m.m.f. of polyphase 128
Eccentric armature, force upon 248
Eddy currents, nature of 40
prevention of 41
separation of, from hysteresis 53
Electric circuit, equivalent, torque from 260
Electric loading, specific, definition of 163
Electromagnet, A.C., average tractive effort of 254
average tractive effort of D.C 252
lifting 243
rotary, torque of 253
tractive, force of 248
Electromagnetic inertia 180
E.m.f., average reactance, as criterion for commutation 235
formula for induced 57
induced in A.C. machines 65
induced in D.C. machines 75
induced in an unsymmetrically spaced transmission line 205
methods of inducing 55
ratio. (See Ratio of transformation.)
ratio in a rotary converter 78
regulation in an alternator 156
regulation in a rotary converter 78
vs. inductance 185
Energy, density of, in magnetic field 240
lost in hysteresis cycle 38
of magnetic field, none consumed in 177
stored in magnetic field, description of 177
stored in magnetic field, formulae for 181
Equivalent electric circuit, torque from 260
permeance, definition of 184
secondary winding, reduced to primary 133
Exciting current in an induction motor . . .> 130
in a transformer vs. magnetizing current 80
of a transformer, exciting volt-amperes 82
of a transformer, saturated core 84
unsaturated core 81
of machinery. See Saturation curves.
Factor, air-gap, definition of 90
in induction motor 98
amplitude 84
INDEX 273
Factor, breadth factor 65
leakage, calculation of 108
leakage, definition of 107
slot factor 69
space factor in iron 41
winding pitch factor 71
Faraday's law of induction 57
Fictitious poles as assumed in synchronous machines 151
Field frames, jn.m.f . for 106
Field, magnetic (See also Magnetic field) 1
m.m.f. at no load. See Saturation curves.
m.m.f . in loaded D.C. machine 170
loaded synchronous machines 148, 156
Field pole leakage, calculation of 110
effect of 108
effect of load upon 113
effect of saturation upon 112
Field poles, m.m.f. for 107
Figure of loss , 47
Fleming's rule 59
Flux. See also Leakage flux.
air-gap, nature of its distribution 88
density, definition of 14
in teeth, apparent 101
description of .' . 5
distribution of, in a concentric cable 189
in a loaded D.C. machine 169
in a loaded synchronous machine 139
distortion. See Reaction .
fringing, definition of 88
permeance of ; 93
gliding or revolving 126
leakage, definition of 16
in induction machine, description of 221
in revolving machinery, nature of 86
refraction 119
units of 6
Force, lines of magnetic 2
mechanical, in magnetic field, average tractive effort in D.C.
electromagnets 252
formulae for the actual force .... 251
lateral compression 244
longitudinal tension 242
of A.C. magnets 254
of a lifting magnet 243
of a tractive magnet 248
on an eccentric armature 248
274 INDEX
PAGE
Force, mechanical, in magnetic field, on conductors carrying current . . 256
on transformer coils 245
pinch phenomenon 246
reason for 242
repulsion between bus-bars 248
torque in a rotary magnet 253
Fourier, method for analyzing waves for their harmonics, used 124, 161
Fractional pitch windings, advantages and disadvantages of 67
effect on inductance in induction machines . . 225
effect on inductance in synchronous machines 231
vs. commutation in direct current machines. . . 238
Frequency of commutation 236
Fringing, definition of 88
Full load m.m.f . in a D.C. machine 170
in an induction machine 131
in a synchronous machine 148, 156
Gauss 14
Generator action 56
Gilbert 12, 266
Gliding and pulsating m.m.f 126
Harmonics, of e.m.f ., in alternating current machines 72
vs. voltage ratio in a rotary coverter 79
of rectangular m.m.f. wave 123
upper, of m.m.f. in an induction machine 136
Heart, definition of 215
Heating due to hysteresis 38
Henry as unit of inductance 184
permeance, definition of 9
Herring's experiment ." 56
Hysteresis and saturation, explanation of 34
cycle, energy lost in 34
irreversible 34
description of 32
empirical equation for 48
loop 33
vs. heating 34
separated from the eddy current loss 53
measurement of 40
mechanical analogue to 36
Induced e.m.f., formulae for 57
in a D.C. machine 75
in an alternator and in an induction machine 65
in a transformer 62
Inductance and e.m.f., relation between 185
INDEX 275
Inductance as electromagnetic inertia 180
definition of, and formula} for 184
henry and perm as units of 184
mechanical analogue of 185
of a concentric cable, single phase 191
of circuits in the presence of iron 186
of coils and loops 188
of coils in a D.C. machine 236
of synchronous machines, measurement of, in A.C. machines
219, 231
of transformers, constants for 215
formulae for 211-214
vs. the end coils 213
vs. the number of coils 213
vs. the shape of the coils 212
of transmission lines, single phase 199
three- wire symmetrical spacing 201
of windings, formula for 219
fractional pitch 225
how measured 219
vs. commutation 232
vs. leakage permeance 184, 219
Induction, law of 6, 57
machines, armature reaction 131
e.m.f . induced in 65
higher harmonics in the m.m.f. of a 136
inductance vs. winding pitch -. 225
leakage flux, description of the 221
leakage permeance, values of the '. 225
motor, ratio of transformation in a 134
secondary current in a 132
torque in 259-260
of e.m.f., methods of 55
Inertia, electromagnetic 180
Insulator, magnetic 17
Intensity, magnetic, definition of 13
Interference. See Armature reaction, and Reaction.
Interpoles, definition of (See also Computating poles) 172
Irregular paths, permeance of; how to map field in 116
Iron, grades of and their use 22
laminations, preparation of 50
reason for 41
loss, definition of 42
effects of '. 43
figure of loss ". 47
magnetization curves 20-24
properties of, general 20, 32
276 INDEX
PAGE
Iron, saturation curves 20-24
silicon steel 49
space factor in laminations 41
used in permanent magnets 50
virgin state of 32
vs. inductance 186
wire cores 41
Joints in transformers 81
Joulecen, definition of 264
Kelvin's law 253
Knee of saturation curve 25
Laminations, grades of 44
preparation of 50
reason for 41
Lateral compression in the magnetic field 244
Law of induction 6, 57
Law, Ohm's, for magnetic circuit 7
Laws of circulation 7, 57
Leakage coil, definition of 215
factor, calculation of 108
definition of 107
vs. leakage flux and permeance 109
Leakage flux 16
about armature windings 218
belt, description of 223
in field poles, as affected by load 113
as affected by saturation 112
effect of 108
in induction motors, description of 221
in transformers 208
nature of, in machinery 86
zig-zag, description of 223
Leakage inductance. See Inductance.
See also Leakage permeance.
Leakage permeance between field poles 108
in induction machines, values of 225
in synchronous machines, values of 230
of coils in a D.C. machine 236
of slots, calculation of 226
of windings, formula for . 220
of zig-zag or tooth tip leakage, calculation of 227
Lehmann, Dr. Th., method of finding permeance of irregular field 18
Lifting magnet 243
INDEX 277
Lines of force 2
Linkages, fact of 3
as a measure of energy 180
partial and complete, definition of 181
Load characteristics of alternator, definition of 141
Loaded D.C. machine, flux distribution in 169
field m.m.f. in 170
D.C. motor, speed of a 171
induction machine, m.m.f. relations in 132
synchronous machine, flux distribution in ; 139
Loading, specific electric, definition of 163
Longitudinal tension in magnetic field 242
Loop, hysteresis 33
Machinery, core loss in revolving 46
leakage flux in, nature of 86
Machines, revolving, how torque is produced in 255, 258
torque in 259, 260
types of synchronous 63
Magnets, A.C., average tractive effort in 254
average tractive effort in D.C 252
lifting 243
molecular 34
permanent, iron used in 50
torque of rotary magnets 253
tractive, force of a 248
Magnetic circuit, definition of 3
Ohm's law for the 7
types of, in revolving machinery 85
simple 1
with iron, in series and parallel 29
Magnetic field, description of 1
density of energy in 240
energy stored in, description of 177
formulae for 181
formulae for the actual force in a 251
formulas for average force in a 252
lateral compression in 244
longitudinal tension in 243
of transmission line, description of 193
shape of 196
reason for mechanical forces in 242
Magnetic insulation 17
intensity, definition of 13
relation to m.m.f .- 13
potential 16
state. . 1
278 INDEX
PAGE
Magnetization curves. (See also Saturation curves.) 20-24
Magnetizing current vs. exciting current 80
Magnetism, cause of 1,4
M.m.f . or magneto motive force, definition of 5
for air-gap 89
for armature cores 105
for a D.C. machine to compensate for distortion 168
for field frames 106
for field poles 107
for teeth, saturated 101
for teeth, tapered 100
gliding 126
in induction machines, higher harmonics of 136
of field, in a loaded D. C. machine 170
of windings, concentrated 121
distributed single phase 123
distributed polyphase 128
relations in an induction machine 132
a synchronous machine 143
wave, harmonics of 123, 136
Maxwell, definition of 6
Measurement of core loss 44
of hysteresis : 40
of inductance 219
Mechanical analogue of hysteresis 36
of inductance 185
force. See Force, mechanical.
Methods of inducing e.m.f 55
Molecular magnets 34
Multiplex windings, commutation of 238
Mutual induction, in transmission lines 205
in the windings of machines 219
See also Transformer action 55
Neutral zone, in D.C. machine 163
Ohm's law for the magnetic circuit 7
Open circuit characteristic. See Saturation curves.
current. See Exciting current.
Overload capacity of a synchronous motor 148
Parallel combination of permeances 16
-series circuits containing iron 29
Partial linkages, definition of 181
Perm, definition of 9
Permeability curves 24
definition of . . 11
INDEX 279
PAGE
Permeability, equation 24
of air, discussed 266
of non-magnetic materials 11
relative 24
vs. reluctivity 11
vs. saturation 24
Permeance, combinations of, in parallel and series 16
definition of 9
equivalent, definition of 184
leakage, in D.C. machines 236
in induction machines, values of 225
in synchronous machines, values of 230
measurement of 219, 231
of slot, formula for 226
of windings, formula for 220
zigzag, formula for 228
of air-gap, accurate method 92-97
simplified 89
of irregular paths 116
of pole fringe 94
of tooth fringe 93
units of * 9
vs. dimensions 11
Phase characteristics of a synchronous motor 142
relation of m.m.fs. in an induction machine 132
in a synchronous machine 145
Pinch phenomenon 246
Pole face windings, Ryan 174
fringe permeance 94
Poles, fictitious, assumed in synchronous machines 151
non-salient in a synchronous machine 121, 143
salient, definition of 142
Potential, definition of 16
Potier diagram 146
Pulsating and gliding m.m.fs 126
Ratio of transformation 134
of voltages in a rotary converter 78
Rayleigh, Lord, method for finding permeance of a field 116
Reactance, equivalent, of an unsymmetrical transmission line 206
in synchronous machines, nature of 140
voltage, average, as criterion of commutation 235
Reaction. See also Armature reaction.
demagnetizing, in a D.C. machine 164
direct and transverse, nature of 150
calculation of the coefficient of 158
definition of the coefficient of . 153
280 INDEX
PAGE
Reaction, distorting. See Transverse reaction.
transverse, calculation of the coefficient of 160
definition of the coefficient of 153
in a direct current machine 165
in a synchronous machine 150
Rectangular m.m.f . wave, harmonics of 123
Regulation of alternators, calculation of 146-8, 155-6
definition of 156
Rel, definition of 8
Relative permeability 24
Reluctance, definition of 7
in series and in parallel 16
of irregular paths 116
of various magnetic circuits. See Permeance.
unit of 8
vs. dimensions 11
Reluctivity, definition of 11
of air 11
vs. permeability 11
Remanent magnetism. See Residual.
Repulsion between bus-bars 248
between transformer coils 245
Residual magnetism 32
Resistance, equivalent, of unsymmetrical transmission line 207
Revolving field 126
machinery, core loss in 46
types of magnetic circuit in 85
See also particular kind of machinery.
Right-hand screw rule 2
Rotary converter, armature reaction in a 175
voltage ratio in a 78
voltage regulation in a 78
Ryan, pole face winding for D.C. machines 174
Salient poles, definition of 142
in synchronous machines, Blondel diagram for 154
Saturation and hysteresis, an explanation of 34
and permeability 24
curve, knee of 25
curves of iron 20-24
of machinery. See also Transformers, exciting
current : 85-87
effect of, upon pole leakage 112
per cent of 26
Secondary current, calculation of, in an induction motor 132
winding, equivalent, reduced to the primary 133
Semi-net length 220
INDEX 281
PAGE
Semi-symmetrical spacing 203
Series-parallel circuits with iron 29
Series-reluctances 16
Sheet metal. See Laminations.
Short chord or short pitch windings. See Fractional pitch windings.
Silicon steel 49
Simple magnetic circuit 1
Skin effect 188
Slot factor, definition of 68
formulae for " 69
leakage permeance, calculation of 226
Slotted armature, how torque is produced in 258
Solenoidal, term applied to magnetic field 6
Space factor in iron 41
Sparking. See Commutation.
Speed of a variable-speed D.C. motor under load 171
Squirel-cage winding 133
Steinmetz's law for hysteresis 48
Steel. See Iron.
Stray flux or leakage 16
Superposition, principle of 194
Synchronous condensers 157
Synchronous machines, armature reaction in 140
e.m.f . induced in 65
fictitious poles assumed in 151
loaded, flux distribution in 139
measurement of leakage inductance 219, 231
overload capacity of motor 148
phase relation of m.m.fs. in a 143-145
reactive drop in a (limits of) 229
torque in 259, 260
types of 63
values of leakage permeance of the windings of. . 230
vector diagrams of, with non-salient poles . . 145, 147
with salient poles 154
Synchronous motor, definition of phase characteristices of 142
overload capacity of 148
Teeth, apparent flux density in 101
permeance of tooth fringe 93
saturated, m.m.f . for 101
tapered, m.m.f. for 100
Tension. See also Force, mechanical
longitudinal, in magnetic field 242
Torque. See also Force, mechanical.
from equivalent electric circuit. 260
in revolving machines, how produced 258
282 INDEX
PAGE
Torque in slotted armatures, how produced . . 258
of rotary magnets . 253
of revolving machinery 258 260
Tractive effort, actual 251
average, in a D.C. magnet 252
in an A.C. magnet, average 254
magnet, force of 248
Transformer action 56
Transformers, core loss current in a 82
exciting current in a .' 80, 83
induced e.m.f . in a 62
inductance of 211-214
as affected by end coils 213
as affected by number of coils 213
as affected by the shape of the coils 212
values of the constants 215
joints in the magnetic circuit of a 81
leakage flux in a 208
magnetizing current in a 80
repulsion between the coils of a 245
types of 60
Transmission line, description of the field of a 193
shape of the field calculated 196
single phase, inductance of 199
three-wire, symmetrical spacing, inductance of 201
unsymmetrical spacing, e.m.f. induced in 205
unsymmetrical spacing, equivalent reactance of 206
unsymmetrical spacing, equivalent resistance of 207
Transverse and direct reaction, nature of 150
reaction, calculation of the coefficient of 160
definition of the coefficient of 153
in a D.C. machine 165
in a synchronous machine 150
Turning moment in machinery 258-260
Unit of flux 6
of inductance .N 184
of permeance 9
of reluctance 8
Units, table of the dimensions of 263
V curve or phase characteristics '. 142
Variable-speed D.C. motor under load 171
Vector relations in a synchronous machine 145, 147, 154, 155
with non-salient poles 145
with salient poles 154
Virgin state of iron 32
r
INDEX 283
Virtual displacements, principle of 249
Voltage. See E.m.f.
Wave form of e.m.f., effect of type of windings upon 73
how to get sine wave 72
Wave-wound armatures, commutation of 237
Weber, definition of 6
Wire, iron, for cores 41
Winding-pitch factor, values of 71
Windings, compensating, for D.C. machines, 174
concentrated, m.m.f . of 121
for D.C. machines, effect of type on voltage 76
distributed, definition of 121
in A.C. machines, advantages and disadvantages
of 66
formula for the inductance of 219
fractional pitch, advantages and disadvantages of 67
commutation of 238
effect of, on induced e.m.f 67, 76
on inductance 225, 231
inductance and reactance, how measured 219
leakage permeance of, formula for 220
multiplex, commutation of 238
pole-face for D.C. machines 174
polyphase, m.m.f. of 128
wave or two-circuit, commutation of 237
Yrneh, definition of 10
Zig-zag leakage, description of 223
permeance, calculation of 227
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EN
DEC 1 3 1947
APR 6 1948
MAR' 11 M949 0q •*-
NOV 12 ][
rx i o
MAR 1 1 1952
APR 16 1952'
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