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THEOEETICAL ELEMENTS
ELECTRICAL ENGINEERING
EPS PROTEUS STEINMETZ, A.M., PttD.
THIRD EDITION
THOROUGHLY REVISED AND GREATLY CORRECTED
The paper in this volume is brittle or the Inner margins are extremely narrow.
We have bound or rebound the volume
utilizing the best GENERAL BOOKBINDING Co,, CHESTER LAND, OHIO
COPYRIGHT, 1909, BY THE
MoGRAW-HILL BOOK COMPANY
NEW YORK
PREFACE.
THE first part of the following volume originated from a
series of University lectures which I once promised to deliver. This part can, to a certain extent, be considered as an intro-
duction to my work on "Theory and Calculation of Alternating
Current Phenomena/' leading up very gradually from the ordinary sine wave representation of the alternating current to the graphical representation by polar coordinates, from there to rectangular components of polar vectors, and ultimately to the symbolic representation by the complex quantity. The present work is, however, broader in its scope, in so far as it comprises the fundamental principles not only of alternating, but also of
direct currents.
The second part is a series of monographs of the more impor-
tant electrical apparatus, alternating as well as direct current. It is, in a certain respect, supplementary to "Alternating Current
Phenomena." While in the latter work I have presented the general principles of alternating current phenomena, in the present volume I intended to give a specific discussion of the particular features of individual apparatus. In consequence thereof, this part of the book is somewhat less theoretical, and
more descriptive, my intention being to present the most impor-
tant electrical apparatus in all their characteristic features as regard to external and internal structure, action under normal and abnormal conditions, individually and in connection with other apparatus, etc.
I have restricted the work to those apparatus which experience has shown as of practical importance, and give only those theories and methods which an extended experience in the design and operation has shown as of practical utility. I consider this the more desirable as, especially of late years, electri-
iii
iv
PREFACE.
cal literature has been haunted by so many theories (for instance
of the induction machine) which are incorrect, or too compli-
cated for use, or valueless in practical application. In the class
last mentioned are most of the graphical methods, which, while
they may give an approximate insight in the inter-relation of
phenomena, fail entirely in engineering practice owing to the
great difference in the magnitudes of the vectors in the same
diagram, and to the synthetic method of graphical representa-
tion, which generally require one to start with the quantity
which the diagram is intended to determine.
I originally intended to add a chapter on Rectifying Apparatus,
as arc light machines a-nd alternating current rectifiers, but had
to postpone this, due to the incomplete state of the theory of
these apparatus.
The same notation has been used as in the Third Edition of
"Alternating
Current
7
Phenomena/
that
is,
vector
quantities
denoted by dotted capitals. The same classification and nomen-
clature have been used as given by the report of the Standardiz-
ing Committee of the American Institute of Electrical Engineers.
PREFACE TO THE THIRD EDITION.
NEARLY eight years have elapsed since the appearance of the
second edition, during which time the book has been reprinted without change, and a revision, therefore, became greatly desired.
It was gratifying, however, to find that none of the contents of the former edition had to be dropped as superseded or anti-
quated. However, very much new material had to be added.
During these eight years the electrical industry has progressed at least as rapidly as in any previous period, and apparatus and phenomena which at the time of the second edition were of
theoretical interest only, or of no interest at all, have now
assumed great industrial importance, as for instance the singlephase commutator motor, the control of commutation by
commutating poles, etc. Besides rewriting and enlarging numerous paragraphs through-
out the text, a number of new sections and chapters have been added, on alternating-current railway motors, on the control of commutation by commutating poles ("interpoles"), on
converter heating and output, on converters with variable ratio of conversion ("split-polo converters")* on three-wire generators and converters, short-circuit currents of alternators, stability and regulation of induction motors, induction generators, etc.
In conformity with the arrangement used in my other books,
the paragraphs of the text have been numbered for easier reference and convenience.
When reading the book, or using it as text-book, it is recom-
mended :
After reading the first or general part of the present volume,
to
read
through
the
first
17
chapters
of
"
Theory
and
Calculation
of Alternating Current Phenomena," omitting, however, the
mathematical investigations as far as not absolutely required
vi
PREFACE TO THE THIRD EDITION.
for the understanding of the text, and then to take up the study of the second part of the present volume, which deals with
special apparatus. When reading this second part, it is recom-
mended to parallel its study with the reading of the chapter of "Alternating Current Phenomena" which deals with the same subject in a different manner. In this way a clear insight into the nature and behavior of apparatus will be imparted.
Where time is limited, a large part of the mathematical discussion may be skipped and in that way a general review of the
material gained.
Great thanks are due to the technical staff of the McGraw-
Hill Book Company, which has spared no effort to produce the third edition in as perfect and systematic a manner as possible, and to the numerous engineers who have greatly assisted me by pointing out typographical and other errors in the previous
edition.
CHARLES PROTEUS STEINMETZ.
SCHENECTADY, September, 1909.
CONTENTS.
PART I.
GENERAL SURVEY.
1. Magnetism and Electric Current. 2. Magnetism and E.M.F. 3. Generation of E.M.F. 4. Power and Effective Values. 5. Self-Inductance and Mutual Inductance. 6. Self-Inductance of Continuous-Current Circuits. 7. Inductance in Alternating-Current Circuits. 8. Power in Alternating-Current Circuits. 9. Polar Coordinates. 10. Hysteresis and Effective Resistance. 11. Capacity and Condensers. 12. Impedance of Transmission Lines. 13. Alternating-Current Transformer. 14. Rectangular Coordinates. 15. Load Characteristic of Transmission Line. ' 16. Phase Control of Transmission Lines, 17. Impedance and Admittance. 18. Equivalent Sine Waves.
PAGE 1 9
12 16 21 25 32 41 43 53 59 62 73 83 91 96 105 114
PART II.
SPECIAL APPARATUS.
INTRODUCTION.
120
A. SYNCHRONOUS MACHINES.
I. General.
125
II. Electromotive Forces.
127
III. Armature Reaction.
129
IV. Self-Inductance.
132
V. Synchronous Reactance.
136
VI. Characteristic Curves of Alternating-Current Generator.
138
viii
CONTENTS.
SYNCHRONOUS MACHINES (continued).
PAGE
VII. Synchronous Motor.
141
VIII. Characteristic Curves of Synchronous Motor.
143
IX. Magnetic Characteristic or Saturation Curve.
146
X. Efficiency and Losses.
149
XI. Unbalancing of Polyphase Synchronous Machines.
150
XII. Starting of Synchronous Motors.
151
XIII. Parallel Operation.
152
XIV. Division of Load in Parallel Operation.
154
XV. Fluctuating Cross-Currents in Parallel Operation.
155
XVI. High Frequency Cross-Currents between Synchronous
Machines.
159
XVII. Short-Circuit Currents of Alternators.
160
B. DIRECT-CURRENT COMMUTATING MACHINES.
I. General.
II. Armature Winding. III. Generated Electromotive Forces.
IV. V. VI. VII. VIII.
Distribution of Magnetic Flux. Effect of Saturation on Magnetic Distribution. Effect of Commutating Poles. Effect of Slots on Magnetic Flux. Armature Eeaction.
IX. Saturation Curves.
X. Compounding. XI. Characteristic Curves.
XII. Efficiency and Losses. XIII. Commutation.
XIV. Types of Commutating Machines. A. Generators. Separately excited and Magneto,
Shunt, Series, Compound. B, Motors. Shunt, Series, Compound.
166 168 178 179 183 185 190 192 194 196 197 198 198 206
208 215
C. ALTERNATING CURRENT COMMUTATING MACHINES.
I. General.
219
II. Power Factor.
220
III. Field Winding and Compensation Winding.
226
IV. Types of Varying Speed Single-Phase Commutator Motors. 230
V. Commutation.
236
VL Motor Characteristics.
250
VII. Efficiency and Losses.
259
VIII. Discussion of Motor Types.
260
IX. Other Commutator Motors.
266
CONTENTS.
D. SYNCHRONOUS CONVERTERS.
I. General.
II. Ratio of E.M.Fs. and of Currents.
III. Variation of the Ratio of E.M.Fs.
IV. Armature Current and Heating. V. Armature Reaction.
VI. VII. VIII. IX.
Reactive Currents and Compounding. Variable Ratio Converters (Split-Pole Converters).
Starting. Inverted Converters.
X. Frequency.
XL Double-Current Generators.
XII. Conclusion.
XIII. Direct-Current Converter.
XIV. Three-Wire Generator and Converter.
ix
PAGE
270 271 277 279 292 297 299 328 330 332 333 335 337 345
E. INDUCTION MACHINES.
I. General.
352
II. Polyphase Induction Motor.
1. Introduction.
356
2. Calculation.
357
3. Load and Speed Curves.
363
4. Effect of Armature Resistance and Starting.
368
III. Single-phase Induction Motor.
1. Introduction.
372
2. Load and Speed Curves. 3. Starting Devices of Single-phase Motors. 4. Acceleration with Starting Device. IV. Regulation and Stability. 1. Load and Stability. 2. Voltage Regulation and Output. V. Induction Generator.
376 380 385
387 392
1. Introduction.
407
2. Constant Speed Induction or Asynchronous Gen-
erator.
409
3. Power Factor of Induction Generator.
410
VI. Induction Booster.
417
VII. Phase Converter.
418
VIII. Frequency Converter or General Alternating-Current
Transformer.
421
IX. Concatenation of Induction Motors.
423
X. Synchronizing Induction Motors. XI. Self-exciting Induction Machines.
428 435
PAET L
GENERAL THEORY.
i. MAGNETISM AND ELECTRIC CURRENT.
A i.
magnet pole attracting (or repelling) another magnet
pole of equal strength at unit distance with unit force* is called
a unit magnet pole.
The space surrounding a magnet pole is called a magnetic
field of force , or magnetic field.
The magnetic field at unit distance from a unit magnet pole is called a unit magnetic field, and is represented by one line of magnetic force (or shortly "one line") per sq. cm., and from a unit magnet pole thus issue a total of 4 K lines of magnetic force.
The total number of lines of force issuing from a magnet
pole is called its magnetic flux.
m The magnetic flux $ of a magnet pole of strength is,
<> = 4 *m.
At the distance lr from a magnet pole of strength m, and
therefore of flux $ = 4 nm, assuming a uniform distribution in
all directions, the magnetic field has the intensity,
$m
4Tt7T7i/Lf3
2 I 6j
since the $ lines issuing from the pole distribute over the area
of a sphere of radius lr, that is the area 4 xlf.
A magnetic field of intensity 3C exerts upon a magnet pole m of strength the force,
m m Thus two magnet poles of strengths
and
l
v and distance
lr from each other, exert upon each other the force,
* That is, at one centimeter distance with such force as to give to the mass of one gram the acceleration of one centimeter per second,
1
2
ELEMENTS OF ELECTRICAL ENGINEERING.
2. Electric currents produce magnetic fields also; that is, the space surrounding the conductor carrying an electric current is a magnetic field, which appears and disappears and varies with the current producing it, and is indeed an essential part of the phenomenon called an electric
current.
Thus an electric current represents a magnetomotive force
(m.m.f.).
The magnetic field of a straight conductor, whose return
conductor is so far distant as not to affect the field, consists of
lines of force surrounding the conductor in concentric circles.
The intensity of this magnetic field is directly proportional to the current strength and inversely proportional to the dis-
tance from the conductor.
Since the lines of force of the magnetic field produced
by an electric current return into themselves, the magnetic field is a magnetic circuit. Since an electric current, at least a steady current, can exist only in a closed circuit, electricity flows in an electric circuit. The magnetic circuit produced by an electric current surrounds the electric circuit through which the electricity flows, and inversely. That is, the electric circuit and the magnetic circuit are interlinked with each
other.
Unit current in an electric circuit is the current which produces
in
a
magnetic
circuit
of
unit
length
the
field
intensity
4 TT,
that
is, produces as many lines of force per square centimeter as
issue from a unit magnet pole.
In unit distance from an electric conductor carrying unit
current, that is in a magnetic circuit of length 2 n, the field
= intensity is 2 7T 2, and in the distance 2 the field intensity
is unity; that is, unit current is the current which, in a straight conductor, whose return conductor is so far distant as not to affect its magnetic field, produces unit field intensity in distance 2 from the conductor.
One tenth of unit current is the practical unit, called one
ampere.
3. One ampere in an electric circuit or turn, that is, one
ampere-turn, thus produces in a magnetic circuit of unit length the field intensity, 0.4 TT, and in a magnetic circuit of length
MAGNETISM AND ELECTRIC CURRENT.
3
04"
I the field intensity '--'-, and & ampere-turns produce In a
6
magnetic circuit of length I the field intensity :
JC = ~~~~ lines of force per sq. cm.
6
regardless whether the 5 ampere-turns are due to ^ amperes
in a single
turn, or one ampere
in
SF
turns,
or
-F n
amperes
in
n
turns.
&, that is, the product of amperes and turns, is called magneto-
motive force (m.m.f.).
The m.m.f. per unit length of magnetic circuit, or ratio:
v ==
eft-
.
m.m.f.
length of magnetic circuit
is called the magnetizing force. Hence, m.m.f. is expressed in ampere-turns; magnetizing
force in ampere-turns per centimeter (or in practice frequently ampere-turns per inch), field intensity in lines of magnetic force per square centimeter.
At the distance lr from the conductor of a loop or circuit of 2F ampere-turns, whose return conductor is so far distant as not
= to affect the field, assuming the m.m.f. SF, since the length of the magnetic circuit = 2 nlTj we obtain as the magnetizing force,
and as the field intensity,
4. The magnetic field of an electric circuit consisting of two
parallel conductors (or any number of conductors, in a poly-
phase system), as the two wires of a transmission line, can be
considered as the superposition of the separate fields of the
conductors (consisting of concentric circles). Thus, if there are
/ amperes in a circuit consisting of two parallel conductors
(conductor
and
return
conductor),
at
the
distance
Z
t
from
the
4
ELEMENTS OF ELECTRICAL ENGINEERING.
first
and
1 2
from
the
second
conductor,
the
respective
field
intensities are,
^9 r
_!i-rl_. y
j
and
027
= and the resultant field intensity, if T angle between the direc-
ions of the two fields,
~ vV +2^ \/X + 3C -
+ 2
3C 2
3
2 CK^OCj cos T,
=
+2 Z2
cosr.
2
6^2
In the plane of the conductors, where the two fields are in the same, or opposite direction, the resultant field intensity is,,
where the plus sign applies to the space between, the minus sign the space outside of the conductors.
The resultant field of a circuit of parallel conductors consists of excentric circles, interlinked with the conductors, and crowded together in the space between the conductors.
The magnetic field in the interior of a spiral (solenoid, helix, coil) carrying an electric current, consists of straight
lines.
5. If a conductor is coiled in a spiral of I centimenter axial
N y = length of spiral, and turns, thus n
turns per centimeter
I
= length of spiral, and / current, in amperes, in the conductor,
the m.m.f. of the spiral is
$ =IN,
and the magnetizing force in the middle of the spiral, assuming the latter of very great length,
thus the field intensity in the middle of the spiral or solenoid,
X = 0.4 7i3C
= 0.4 nnL
MAGNETISM AND ELECTRIC CURRENT.
5
Strictly this is true only in the middle part of a spiral of such length that the m.m.f. consumed by the external or
magnetic return circuit of the spiral is negligible compared with the m.m.f. consumed by the magnetic circuit in the interior of the spiral, or in an endless spiral, that is a spiral whose axis curves back into itself, as a spiral whose axis is curved in a
circle.
Magnetomotive force & applies to the total magnetic circuit,
or part of the magnetic circuit. It is measured in ampere-
turns.
Magnetizing force <3C is the m.m.f. per unit length of magnetic circuit. It is measured in ampere-turns per centi-
meter.
Field intensity 3 is the number of lines of force per square
centimeter.
If
=
I length
of
the
magnetic circuit or a part of the
magnetic circuit;
= 0.4 TT^C = 1.257 3C
fjn
3C =
0.4 7T
= 3C 0.796 3G.
6. The preceding applies only to magnetic fields in air or
other unmagnetic materials.
If the medium in which the magnetic field is established is a
" magnetic
7
material/
the
number
of
lines
of
force
per
square
centimeter is different and usually many times greater. (Slightly
less in diamagnetic materials.)
The ratio of the number of lines of force in a medium, to the
number of lines of force which the same magnetizing force would
produce in air (or rather in a vacuum), is called the permeability or magnetic conductivity /* of the medium.
The number of lines of force per square centimeter in a magnetic medium is called the magnetic induction (B. The number of lines of force produced by the same magnetizing force in air
is called the field intensity 3C.
6
ELEMENTS OF ELECTRICAL ENGINEERING.
X In air, magnetic induction (B and field intensity
are
equal.
As a rule, the magnetizing force in a magnetic circuit is changed by the introduction of a magnetic material; due to the change of distribution of the magnetic flux.
= The permeability of air 1 and is constant.
The permeability of iron and other magnetic materials varies with the magnetizing force between a little above 1 and values
as high as 6000 in soft iron.
The magnetizing force 3C in a medium of permeability /JL
= produces the field intensity 3C 0.4 7t3& and the magnetic = induction (B 0.4 ^/z3C.
EXAMPLES.
A 7. (1.) pull of 2 grams at 4 cm. radius is required to hold a
horizontal bar magnet 12 cm. in length, pivoted at its center,
in a position at right angles to the magnetic meridian. What
is the intensity of the poles of the magnet, and the number of
lines of magnetic force issuing from each pole, if the horizontal
X = intensity of the terrestrial magnetic field
0.2, and the
= acceleration of gravity 980?
The distance between the poles of the bar magnet may be
assumed as five-sixths of its length.
m Let
= intensity of magnet pole. = lr 5 is the radius on
which the terrestrial magnetism acts.
m Thus 2 m3lr = 2 = torque exerted by the terrestrial mag-
netism.
2 grams weight = 2 X 980 = 1960 units of force. These at 4 cm. radius give the torque 4 X 1960 7840 g cm.
m Hence 2 = 7840. m 3920 is the strength of each magnet pole and
$ = 4 nm = 49,000, the number of lines of force issuing from
each pole.
A 8. (2.) conductor carrying 100 amperes runs in the direc-
tion of the magnetic meridian. What position will a compass needle assume, when held below the conductor at a distance of
50 cm., if the intensity of the terrestrial magnetic field is 0.2?
The intensity of the magnetic field of 100 amperes 50 cm.
= = = from the conductor, is 3C ~^~- 0.2 X ~
0.4, the direc-
IT
50
MAGNETISM AND ELECTRIC CURRENT.
7
tion is at right angles to the conductor; that is at right angles to
the terrestrial magnetic field.
If r = angle between compass needle and the north pole of
m the
magnetic
meridian,
=
I length
of
needle,
= intensity
of
its magnet pole, the torque of the terrestrial magnetism is JCml
= sin T 0.2 ml sin r, the torque of the current is
^ = 3Cm1 cos r = 0.2 Iml cos r
A 0.4
mi7
cos
r.
LT
= = = In equilibrium, 0.2 wZ sin T 0.4 #iZ cos r, or tan T 2, r
63.4.
9. (3.) What is the total magnetic = flux per I 1000 m. length,
passing between the conductors of a long distance transmission,
Fig. 1. Diagram of Transmission Line for Inductance Calculation.
= line carrying / amperes of current, if Id 0.82 cm. is the diameter of the conductors (No. B. & S.), k = 45 cm. the
spacing or distance between them? At distance lr from the center of one of the conductors (Fig. 1),
the length of the magnetic circuit surrounding this conductor
is 2 nlr, the m.m.f., / ampere turns; thus the magnetizing force
~- -~ 3C =
} and the field intensity 3C = 0.4 nK =
and the
9
2 nlr
Lr
= -^ flux in the zone dlr is d4>
-, and the total flux from the
Lr
surface of the conductor to the next conductor is,
= 0.2 Ildlr
0.2 II
e lr 1
["log
IT
~ = 0.2 II bge
-
8
ELEMENTS OF ELECTRICAL ENGINEERING.
The same flux is produced by the return conductor in the same direction, thus the total flux passing between the trans-
mission wires is,
or per 1000 m. = 105 cm. length,
~on
2 $ =0.4 X 105 / loge
- 0.4 X 105 X 4.70 / = 0.188 X 106 /,
U.Q-i
or 0.188 / megalines or millions of lines per line of 1000 m. of which 0.094 / megalines surround each of the two con-
ductors.
10. (4.) In an alternator each pole has to carry 6.4 millions
of lines, or 6.4 megalines magnetic flux. How many ampere-
turns per pole are required to produce this flux, if the magnetic
circuit in the armature of laminated iron has the cross section
of 930 sq. cm. and the length of 15 em., the air-gap between
stationary field poles and revolving armature is 0.95 cm. in
length and 1200 sq. cm. in section, the field-pole is 26.3 cm.
in length and 1075 sq. cm. in section, and is of laminated iron,
and the outside return circuit or yoke has a length per pole of
20 cm. and 2250 sq. cm. section, and is of cast iron?
= = The magnetic
densities
are;
(B 1
6880 in the armature, (B 2
= = 5340
in
the
air-gap,
(B 3
5950 in
the
field-pole,
and
(E 4
2850
^ in the yoke. The permeability of sheet iron is
= 2550 at
=
^ 6880,
/.3 =2380
at
& =5950. 3
The permeability of cast
iron is
//4 =280
at
(B 4
=2S50.
Thus the field intensity
= ~ = X X - = is,
3e x
=
2.7, 2 5340,
3
2.5,
OC 4
10.2.
\^\ The magnetizing force
X = = is,
1
2.15,
5C
2
= = - = = 3C 3
=1.99,
3t =8.13 4
ampere-turns
per cm.
(fr
3CZ)
is,
SF
X
32,
F 2
4040,
CF 3
52,
F 4
Thus the m.m.f.
163, or the total
m.m.f. per pole is
SF = 3^+
+ CF 2
^3+
$F = 4290 4
ampere-turns.
The permeability fi of magnetic materials varies with the
density <S, thus tables or curves have to be used for these quan-
tities. Such curves are usually made out for density <B and
MAGNETISM AND E.M.F.
9
magnetizing force 3C, so that the magnetizing force SC correspond-
ing to the density & can be derived directly from the curve.
Such a set of curves is given in Fig. 2.
50 100 130 200 2uO 300 350 400 4,10 500 530 COO 650 TOO 750 800
f/lagnjttzing ForceJjCAm jere-turnj per cm.Lengt !i of Magnetic Cfrcu it 5 J.O 15 20 25 30 35 40 45 DO 55 00 65 70 75 80
3Tig. 2. Magnetization Curves of Various Irons.
2. MAGNETISM AKD E.M.F.
ir. In an electric conductor moving relatively to a magnetic field, an e.m.f. is generated proportional to the rate of cutting of the lines of magnetic force by the conductor.
Unit e.m.f. is the e.m.f. generated in a conductor cutting one line of magnetic force per second.
10s times unit e.m.f. is the practical unit, called the volt.
10
ELEMENTS OF ELECTRICAL ENGINEERING.
Coiling the conductor n fold increases the e.m.f. n fold, by cutting each line of magnetic force n times.
In a closed electric circuit the e.m.f. produces an electric
current.
The ratio of e.m.f.to electric current produced thereby is called
the resistance of the electric circuit. Unit resistance is the resistance of a circuit in which unit
e.m.f. produces unit current. 109 times unit resistance is the practical unit, called the
ohm.
The ohm is the resistance of a circuit, in which one volt pro-
duces one ampere. The resistance per unit length and unit section of a conductor
is called its resistivity, p.
The resistivity p is a constant of the material, varying with
the temperature.
The resistance r of a conductor of length Z, area of section A,
= and resistivity p is r -
jHL
12. If the current in the electric circuit changes, starts, or stops, the corresponding change of the magnetic field of the
current generates an e.m.f. in the conductor carrying the current, which is called the e.m.f. of self-induction.
If the e.m.f. in an electric circuit moving relatively to a magnetic field produces a current in the circuit, the magnetic field produced by this current is called its magnetic
reaction.
The fundamental law of self-induction and magnetic reaction
is, that these effects take place in such a direction as to oppose their cause (Lentz's law).
Thus the e.m.f. of self-induction during an increase of current
is in the opposite direction, during a decrease of current in the
same direction as the e.m.f. producing the current. The magnetic reaction of the current produced in a circuit
moving out of a magnetic field is in the same direction, in a circuit moving into a magnetic field in opposite direction to the
magnetic field. Essentially, this law is nothing but a conclusion from the law
of conservation of energy.
MAGNETISM AND E.M.F.
11
EXAMPLES.
13. (L) An electromagnet is placed so that one pole sur-
rounds
the
other
pole
11
cylindrical!}
as
shown
in
section in
Fig.
3,
and a copper cylinder revolves between these poles at 3000 rev.
per min. What is the e.m.f. generated between the ends of this = cylinder, if the magnetic flux of the electromagnet is <J> 25
megalines? During each revolution the copper cylinder cuts 25 megalines.
= X It makes 50 rev. per sec. Thus it cuts 50 X 25 X 108 12.5 10s
Pig. 3. Unipolar Generator.
lines of magnetic flux per second. Hence the generated e.m.f.
E = is
12.5 volts.
Such a machine is called a "unipolar/' or more properly a
"nonpolar"
or
an
"
77
acyclic
generator.
14.
(2.)
The
field spools
of
the
20-pole
alternator in
Section
1 }
Example
4 ;
are
wound
each
with
616
turns
of
wire
No.
7
(B.
&
S.), 0.106 sq. cm. in cross section and 160 cm. mean length of
turn. The 20 spools are connected in series. How many
amperes and how many volts are required for the excitation of
X this alternator field, if the resistivity of copper is 1.8
6
10""
ohms per cm.3 *
Since 616 turns on each field spool are used, and 4280 ampere-
turns
required,
the
current
is
4280 -~-r
=
6.95
amperes.
* Cm. 3 refers to a cube whose side is one centimeter, and should not be confused with cu. cm.
12
ELEMENTS OF ELECTRICAL ENGINEERING.
The resistance of 20 spools of 616 turns of 160 cm. length,
^ ^ X 0.106 sq. cm. section, and 1.8
6
10"" resistivity is,
20
X
616
X
160
X
1.8
X
6
IP""
=
0.106
and the e.m.f. required 6.95 X 33.2 = 230 volts.
3. GEITORATION OF E.M.F,
A 15.
closed conductor, convolution or turn, revolving in a
magnetic field, passes during each revolution through two
positions of maximum inclosure of lines of magnetic force
A in Fig. 4, and two postions of zero inclosure of lines of mag-
B netic force in Fig. 4.
Fig. 4. Generation of e.m.f.
Thus it cuts during each revolution four times the lines of
force inclosed in the position of maximum inclosure. If $ = the maximum number of lines of flux inclosed by
/= the conductor,
the frequency in revolutions per second or
cycles, and n = number of convolutions or turns of the con-
ductor, the lines of force cut per second by the conductor, and
thus the average generated e.m.f. is,
E = 4 fn$ absolute units,
10"8 volts.
If /is given in hundreds of cycles, <& in megalines,
E = 4n$ volts.
If a coil revolves with uniform velocity through a uniform magnetic field, the magnetism inclosed by the coil at any instant
is,
< COS r
GENERATION OF E.M.F.
13
where $ = the maximum magnetism inclosed by the coil and r = angle between coil and its position of maximum inclosure
of magnetism.
The e.m.f. generated in the coil, which varies with the rate of cut-
$
p-^
ting or change of <> cos T, is thus,
E sin Q
I<lcos0
where E is the maximum value
of e.m.f., which takes place for
;
,
==:
;
..
=
T=90, or at the position of zero
inclosure of magnetic flux, since
Fig. 5. Generation of e.m.f. ^y Rotation.
in this position the rate of cutting is greatest.
= Since avg. (sin T) - , the average generated e.m.f. is,
71
E = -E9.
71
Since, however, we found above that,
E = 4:fn<& is the average generated e.m.f.,
it follows that
E Q
=2
7tfn$
is
the
maximum,
and
= e 2 7cfn& sin T the instantaneous generated e.m.f.
The interval between like poles forms 360 electrical-space
degrees, and in the two-pole model these are identical with the
mechanical-space degrees. With uniform rotation, the space
= angle, r, is proportional to time. Time angles are designated by
6, and with uniform rotation
r, r being measured in elec-
trical-space degrees.
The period of a complete cycle is 360 time degrees, or 2 TT or
- seconds. In the two-pole model the period of a cycle is that of
/
i
one complete revolution, and in a 2 np-pole machine,
of that
np
of one revolution.
Thus,
d~2xft
e=2 nfn$ sin 2 itft.
14
ELEMENTS OF ELECTRICAL ENGINEERING.
If the time is not counted from the moment of maximum
inclosure
of
magnetic
flux,
but
t = the l
time
at
this
moment,
we have
a2*/n*sin2*/(i -
or,
6=2 7:/7i* sin (0 - 0J.
Where
6 l
=27:ft 1
is
the
angle
at
which
the
position
of
max-
imum inclosure of magnetic flux takes place, and is called its
phase. These e.m.fs. are alternating.
If at the moment of reversal of the e.m.f. the connections
between the coil and the external circuit are reversed, the e.m.f.
E in the external circuit is pulsating between zero and
but
Q,
has the same average value E.
If a number of coils connected in series follow each other
successively in their rotation through the magnetic field, as the armature coils of a direct-current machine, and the connections
of each coil with the external circuit are reversed at the moment
of reversal of its e.m.f., their pulsating e.m.fs. superimposed in
the external circuit make a more or less steady or continuous
external e.m.f.
The average value of this e.m.f. is the sum of the average values
of the e.m.fs. of the individual coils.
Thus in a direct-current machine, if $ = maximum flux inclosed per turn, n = total number of turns in series from commutator brush to brush, and / = frequency of rotation through the
magnetic field.
E=* 4:fn$ generated e.m.f. (<E> in megalines, /in hundreds of cycles per second).
This is t"he formula of the direct-current generator.
EXAMPLES.
A 17. (1.)
circular wire coil of 200 turns and 40 cm. mean
diameter is revolved around a vertical axis. What is the
horizontal intensity of the magnetic field of the earth, if at a speed of 900 revolutions per minute the average e.m.f. generated in the coil is 0.028 volts?
The mean area of the coil is
- = 1255 sq. cm., thus the
GENERATION OF E.M.F.
15
terrestrial flux inclosed is 1255 X, and at 900 revolutions per
= 4X minute or 15 revolutions per second, this flux is cut
15
60
times per second by each turn, or 200 X 60 = 12,000 times by
the coil. Thus the total number of lines of magnetic force cut
X by the conductor per second is 12,000 X 1255
X == 0.151
10s
X, and the average generated e.m.f. is 0.151 5C volts. Since
= X = this is 0.028 volts,
0.1S6.
18. (2.) In a 550-volt direct-current machine of 8 poles and
drum armature, running at 500 rev. per min., the average volt-
age per commutator segment shall not exceed 11, each armature
coil shall contain one turn only, and the number of commutator
segments per pole shall be divisible by 3, so as to use the machine
as three-phase converter. What is the magnetic flux per field-
pole?
550 volts at 11 volts per commutator segment gives. 50, or
as next integer divisible by 3, n = 51 segments or turns per
pole.
8 poles give 4 cycles per revolution, 500 revs, per min. gives
/= 500/60 8.33 revs, per sec. Thus the frequency is,
4 X 8.33
= 33.3 cycles per sec.
= E The generated e.m.f. is
550 volts, thus by the formula of
direct-current generator,
or,
550 = 4 X 0.333 X 51 *,
= <i> 8.1 megalines per pole.
19. (3.) What is the e.m.f. generated in a single turn of a 20pole alternator running at 200 rev. per min., through a mag-
netic field of 6.4 megalines per pole?
2X
The frequency is /=
X ^
2t
oU
= e .E sin r,
E =2xfn$,
= 33,3 cycles.
*=6.4,
=1,
/= 0.333.
Thus,
$ = 2 ic X 0.333 X 6.4 = 13.4 volts maximum, or = e 13.4 sin 0.
16
ELEMENTS OF ELECTRICAL ENGINEERING
4. POWER AND EFFECTIVE VALUES.
E 20.
The
power
of
the
continuous
e.m.f. ;
producing con-
tinuous current / is P = EL
E = The e.m.f. consumed by resistance r is l Ir, thus the power consumed by resistance r is P = Pr.
E = E Either l
9 then the total power in the circuit is con-
E^ sumed by the resistance, or
E} then only a part of the
power is consumed by the resistance, the remainder by some
E E counter e.m.f.,
v
= If an alternating current i 7 sin 6 passes through a resist-
ance r, the power consumed by the resistance is,
= = fr
7
2r
2
sin
6
- (1 cos 2 6),
2i
thus varies with twice the frequency of the current, between
zero
and
7
2 r.
The average power consumed by resistance r is,
= since avg. (cos) 0. = Thus the alternating current i J sin 9 consumes in a resist-
ance r the same effect as a continuous current of intensity
v2 = The value /
Jb is called the effective value of the alter-
= nating current i 7 sin 6; since it gives the same effect.
E = v2 Analogously
JF
t is the effective value of the alternating
E e.m.f., e
sin #.
E =2 Since Q
nfn$, it follows that
= V2 jE?
Ttfn
= 4.44 /n$;
is the effective alternating e.m.f., generated in a coil of n turns
rotating at a frequency of / (in hundreds of cycles per second) through a magnetic field of $ megalines of force.
This is the formula of the alternating-current generator.
POWER AND EFFECTIVE VALUES.
21. The formula of the direct-current generator,
holds even if the e.m.fs. generated in the individual turns are not sine waves, since it is the average generated e.m.f.
The formula of the alternating current generator,
does not hold if the waves are not sine waves, since the ratios of
average to maximum and of maximum to effective e.m.f. are
changed. If the variation of magnetic flux is not sinusoidal, the effective
generated alternating e.m.f. is,
# = 7 V2 xfn$.
7 is called the form factor of the wave, and depends upon its shape, that is the distribution of the magnetic flux in the mag-
netic field.
Frequently form factor is defined as the ratio of the effective to the average value. This definition is undesirable since it gives for the sine wave, which is always considered the standard wave, a value differing from one.
EXAMPLES.
22. (1.) In a star connected 20-pole three-phase machine, revolving at 33.3 cycles or 200 rev. per min., the magnetic flux
per pole is 6.4 megalines. The armature contains one slot per pole and phase, and each slot contains 36 conductors. All these
conductors are connected in series. What is the effective e.m.f.
per circuit, and what the effective e.m.f. between the terminals of the machine?
Twenty slots of 36 conductors give 720 conductors, or 360 turns in series. Thus the effective e.m.f. is,
= V2 Ti
t
= 4.44 X 0.333 X 360 X 6.4 = 3400 volts per circuit.
The e.m.f. between the terminals of a star connected threephase machine is the resultant of the e.m.fs. of the two phases,
18
ELEMENTS OF ELECTRICAL ENGINEERING.
which differ by 60' degrees, and is thus 2 sin 60
that of one phase, thus,
E = E V3 l = 5900 volts effective.
= \/3 times
23. (2.) The conductor of the machine has a section of 22 sq. cm. and a mean length of 240 cm. per turn. At a resistivity (resistance per unit section and unit length) of copper
of p 1.8 X 10", what is the e.m.f. consumed in the machine
by the resistance, and what the power consumed at 450 kw.
output? 450 kw. output is 150,000 watts per phase or circuit, thus
- - ^ = = ,,
the
,
current
/7
150,000
4,4,.2
amperes
,.
effective.
O "iUU
The resistance of 360 turns of 240 cm. length, 0.22 sq. cm.
X section and 1.8
10"6 resistivity, is
r = 360
X
240
X
1.8
X
= ... 10~6
AO./l o,hms per circuit.
44.2
amp.
X
0.71 ohms
gives
31.5
volts
per
circuit
and
2
(44.2)
X 0.71 - 1400 watts per circuit, or a total of 3 X 1400= 4200
watts loss.
24. (3.) What is the self-inductance per wire of a three-
phase line of 14 miles length consisting of three wires No.
(Id =0.82 cm.), 45 cm. apart, transmitting the output of this
450 kw. 5900-volt three-phase machine?
450 kw. at 5900 volts gives 44.2 amp. per line. 44.2 amp.
effective gives 44.2 \/2 = 62.5 amp. maximum. 14 miles = 22,400 m. The magnetic flux produced by /
amperes in 1000 m. of a transmission line of 2 wires 45 cm.
apart and 0.82 cm. diameter was found in paragraph 1, example
- 3, as 2 $ 0.188 X 10 /, or * = 0.094 X 106 / for each wire.
Thus at 22,300 m. and 62.5 amp. maximum, the flux per
wire is
$ = 22.3 X 62.5 X 0.094 X 106 = 131 megalines.
Hence the generated e.m.f., effective value, at 33.3 cycles is,
= 4.44 X 0.333 X 131 = 193 volts per line;
POWER AXD EFFECTIVE VALUES.
19
the maximum value is,
VT= E - E X Q
273 volts per line;
and the instantaneous value,
e = E sin (0 -
= 273 sin (0 -
or, snce = xt =
have,
= 273 sin 210 (/
25. (4.) What is the form
factor (a) of the e.m.f. generated in a single conductor of
a direct-current machine hav-
ing 80 per cent pole arc and negligible spread of the mag-
netic flux at the pole corners,
and (6) what is the form factor of the voltage between two
collector rings connected to diametrical points of the armature of such a machine?
(a.) In a conductor during
the motion from position A,
shown in Fig. 6, to position
B, no e.m.f. is generated; Fig' 6 '
of Bipolar Generator.
B from position
to C a constant e.m.f. e is generated, from
E E F C to
again no e.m.f., from to
a constant e.m.f. e,
Fig. 7. E.m.f. of a Single Conductor, Direct-Current Machine 80 per cent Pole Arc.
A and from F to
again zero e.m.f.
as shown in Fig. 7.
The average e.m.f. is
= e l
0.8 e]
The e.m.f. wave thus is
20
ELEMENTS OF ELECTRICAL ENGINEERING.
hence, with this average e.m.f., if it were a sine wave, the maxi-
mum e.m.f. would be
and the effective e.m.f. would be
__ " "" e^ __ 0.4 xe
63 V2
\/2
The actual square of the e.m.f. is er for 80 per cent and zero for 20 per cent of the period, and the average or mean square thus is
0.8 e\
and therefore the actual effective value,
The form factor y, or the ratio of the actual effective value
e to 4
the
effective
value
e of 3
a
sine wave
of
the
same
mean
value and thus the same 'magnetic flux, then is
~= 4=
1.006;
that is, practically unity.
F (6.) While the collector leads a, b move from the position 9
C }
as
shown
in
Fig.
6,
to
5,
E,
constant
voltage
E
exists
between
them, the conductors which leave the field at C being replaced
by the conductors entering 'the field at 5. During the motion
E of the leads a, b from B, to C, F, the voltage steadily decreases,
Pig. 8. E.m.f. between two Collector Rings connected to Diametrical Points of the Armature of a Bipolar Machine having 80 per cent Pole Arc.
reverses, and rises again, to
E} as the conductors entering the
E field at
have an e.m.f. opposite to that of the conductors
leaving at C. Thus the voltage wave is, as shown by Fig. 8,
triangular, with the top cut off for 20 per cent of the half wave.
XELF-IXDVCTAXCE AND MUTUAL IXDUCTAXCE. 21
Then the average e.m.f. is
e, = 0.2 E + 2 X ^ -
0.6 E.
The maximum value of a sine wave of this average value is
and the effective value corresponding thereto is
=2
0.3 r.E
3 V2~ V2
E The actual voltage square is 2 for 20 per cent of the time, and rising on a parabolic curve from to E2 during 40 per cent of the
time, as shown in dotted lines in Fig. 8. The area of a parabolic curve is width times one-third of
height, or
'
3
hence, the mean square of voltage is
and the actual effective voltage is
..-
hence, the form factor is
or, 2.5 per cent higher than with a sine wave.
5. SELF-INDUCTANCE AND MUTUAL INDUCTANCE.
26. The number of interlinkages of an electric circuit with the lines of magnetic force of the flux produced by unit current
in the circuit is called the inductance of the circuit.
The number of interlinkages of an electric circuit with the lines of magnetic force of the flux produced by unit current in
a second electric circuit is called the mutual inductance of the second upon the first circuit. It is equal to the mutual indue-
22
ELEMENTS OF ELECTRICAL ENGINEERING
tance of the first upon the second circuit, as will be seen, and thus is called the mutual inductance between the two
circuits.
The number of interlinkages of an electric circuit with the
lines of magnetic flux produced by unit current in this circuit and not interlinked with a second circuit is called the self-
inductance of the circuit.
= If i current in a circuit of n turns, $ flux produced
thereby and interlinked with the circuit, n$ is the total number
of interlinkages, and L = ~- the inductance of the circuit. % If <3> is proportional to the current i and the number of
turns n,
=
<i>
and L =
',
the inductance.
(R
<R
m (R is called the reluctance and
the m.m.f. of the magnetic
circuit.
In magnetic circuits the reluctance (R has a position similar to that of resistance r in electric circuits.
The reluctance (R, and therefore the inductance, is not
constant in circuits containing magnetic materials, such as
iron, etc.
If (Rj is the reluctance of a magnetic circuit interlinked with
two
electric
circuits
of
n l
and
n 3
turns
respectively,
the
flux
produced by unit current in the first circuit and interlinked with
the second circuit is -1 and the mutual inductance of the first
M upon the second circuit is
=2
-~^- ,
that
is,
equal
to
the
mutual inductance of the second circuit upon the first circuit,
as stated above.
If no flux leaks between the two circuits, that is, if all flux is
M interlinked with
both
circuits, and
L 1
=
inductance
of
the
first,
L = inductance of the second circuit, and
= mutual induc-
2
tance, then
M2 = LtLr
M L^ If flux leaks between the two circuits, then 2 <
r
In this case the total flux produced by the first circuit consists of a part interlinked with the second circuit also, the mu-
SELF-IXDUCTAXCE AXD MUTUAL IXDUCTAXCE. 23
tual inductance, and a part passing between the two circuits,
that is, interlinked with the first circuit only, its self-inductance.
27.
Thus,
if
L t
and
L 2
are
the
inductances
of
the
two
circuits,
-1
n
and
2- is
n
the
total
flux
produced
bv
unit
current
in
the
first
l
2
and second circuit respectively.
Of the flux *- a part h is interlinked with the first circuit
n,
n,
only,
L Si
being
its
leakage inductance,
and
a
1
part
interlinked
/i
M with the second circuit also,
3
being the mutual inductance
M andi^tL = -^* +,
nnn
l
t
2
Thus, if
L and L
1
2
= inductance,
M = L and L
Sl
S2
self-inductance,
= mutual inductance of two circuits of n
n 2
turns
respectively,
we
have
and
or
,^
+
Sl
=+
2
S2
;
or
M - - 2 = (L,
L5l ) (L 2
L 5a ).
The practical unit of inductance is 109 times the absolute unit or 10s times the number of interlinkages per ampere (since
1 amp. = 0.1 unit current), and is called the henry (h); 0.001
of it is called the milhenry (mh.).
The number of interlinkages of i amperes in a circuit of
L henry inductance is iL 108 lines of force turns, and thus
the e.m.f. generated by a change of current di in time dt is
L ess
_. /7?* 2
io8 absolute units
dt
~L =
volts.
dt
A change of current of one ampere per second in the circuit
of one henry inductance generates one volt.
24
ELEMENTS OF ELECTRICAL ENGINEERING.
EXAMPLES.
28. (1.) What is the inductance of the field of a 20-pole
alternator, if the 20 field spools are connected in series, each spool contains 616 turns, and 6.95 amperes produces 6.4 mega-
lines per pole?
The total number of turns of all 20 spools is 20 X 616 = 12,320.
X Each is interlinked with 6.4 106 lines, thus the total number of interlinkages at 6.95 amperes is 12,320 X 6.4 X 106 = 78 X 109 .
6.95 amperes = 0.695 absolute units, hence the number of
interlinkages per unit current, or the inductance, is
7/05
v*
inu9
t
=
112
X
10 9
=
112 henrys.
0.69o
29. (2.) What is the mutual inductance between an alter-
nating transmission line and a telephone wire carried for 10 miles below and 1.20 m. distant from the one, 1.50 m. distant from the other conductor of the alternating line? and what is the e.m.f. generated in the telephone wire, if the alternating circuit carries 100 amperes at 60 cycles?
The mutual inductance between the telephone wire and the electric circuit is the magnetic flux produced by unit current in the telephone wire and interlinked with the alternating circuit, that is, that part of the magnetic flux produced by unit current in the telephone wire, which passes between the distances of 1.20 and 1.50 m.
At the distance lx from the telephone wire the length of mag-
= netic circuit is 2 xlx . The magnetizing force <3
- r- if / =
2 Tllx
current in telephone wire in amperes, and the field intensity
3C = 0.4 n$i = -0~27 and the flux in the zone dlx is ,
0.2 II dl
I = 10 miles = 1610 X 103 cm.
50-Q1211 77
X 322
103 / loge
= 72 / 103 ;
COXTLVUOUS-CURREXT CIRCUITS.
25
or, thus,
72 / 103 interlinkages, hence, for / = 10, or one absolute
unit,
M = 72 x 104 absolute
0.72 mh.
units, = 72 X 10~5
henrys =
100 amperes effective or 141.4 amperes maximum or 14.14
absolute units of current in the transmission line produces a
maximum flux interlinked with the telephone line of 14.14 X 0.72 X 10~3 X 10 - 10.2 megalines. Thus the e.m.f. generated
at 60 cycles is
E = 4.44 X 0.6 X 10.2 = 27.3 volts effective.
6. SELF-INDUCTANCE OF CONTINUOUS-CURRENT
CIRCUITS.
30. Self-inductance makes itself felt in continuous-current
circuits only, in starting and stopping or in general when the
current changes in value.
Starting of Current. = If r resistance,
E circuit,
= continuous e.m.f. impressed
L = inductance of upon circuit, i =
current in circuit at time t after impressing e.m.f. E, and di the
increase of current during time moment dt, then the increase of
magnetic interlinkages during time dt is
Ldi,
and the e.m.f. generated thereby is
'--
Lr
di
5"
By Lentz's law it is negative, since it is opposite to the impressed
e.m.f., its cause.
Thus the e.m.f. acting in this moment upon the circuit is
and the current is
or, transposing,
.
_
E
+
e i
E-L^-
=
dt^.
r
r
26
ELEMENTS OF ELECTRICAL ENGINEERING.
the integral of which is
- rt L
=
,
/.
log.U
-
E
\
r
where logec = integration constant.
This reduces to
E .
-42
= ^
C ,
[.
,
r
at = 0, i = 0, and thus
--E = c.
r
Substituting this value, the current is
and the e.m.f. of inductance is
At t = oo,
e = ir l
E =*
- = E t o
,
_tl
Es L
= 6i o.
Substituting these values,
and
= The expression tt
T
is called the "time constant of the cir-
cuit,"*
* The name time constant dates back to the early days of telegraphy, where it was applied to the ratio: - , that is, the reciprocal of what is above denoted
as time constant. This quantity which had gradually come into disuse, again became of importance when investigating transient electric phenomena, and
in this work it was found more convenient to denote the value =- as time conu
stant, since this value appears as one term of the more general time constant
+ of the electric circuit: =
|
7;]. (Theory and Calculation of Transient Elec-
tric Phenomena and Oscillations, Section IV).
CONTINUOUS-CURRENT CIRCUITS.
Substituted in the foregoing equation this gives
and
1
u
e
-!E!=: - 0.308 #.
r
31. Stopping of Current. In a circuit of inductance L and
resistance r, let a
current i Q
=
w
be
produced
by the
impressed
E e.m.f. E, and this e.m.f. be withdrawn and the circuit closed
through a resistance r r Let the current be i at the time t after withdrawal of the
e.m.f. E and the change of current during time moment dt be di.
di is negative, that is, the current decreases.
The decrease of magnetic interlinkages during moment dt is
Ldi.
Thus the e.m.f. generated thereby is
-r di
It
is
negative
since
di
is
negative,
and
e t
must
be
positive,
that
is, in the same direction as E, to maintain the current or oppose
the decrease of current, its cause.
Then the current is
.
e
,
_ L di.
r -f r l
+'
r
r^ dt
or, transposing,
_L + H^ = ^
L
i
the integral of which is
T -j- T
where = logec integration constant.
28
ELEMENTS OF ELECTRICAL ENGINEERING
This reduces to
for
r+rt
= i
ce
L
>
=
t
0,
i = -E = c. Q
Substituting this value, the current is
-^ E ~ (
i== e
}
r
and the generated e.m.f. is
= Substituting
i
Q
the current is
,
r+n
= i
ie
L'
and the generated e.m.f. is
/
-
\
At t = 0,
that is, the generated e.m.f. is increased over the previously
impressed e.m.f. in the same ratio as the resistance is increased.
When
r 1
= 0,
that
is,
when
in
withdrawing
the
impressed
E e.m.f. the circuit is short-circuited,
= E i
= - TJ-
_rL .
L
IB
Q
L the current, and
E e^^
= !l
eL
_ !l ijre L the generated e.m.f.
*
= In
this
case,
at t
==
0,
e 1
E, that is, the e.m.f. does not rise.
= In the case r t
oo
;
that
is,
if
in
withdrawing
the
e.m.f.
E,
= = the circuit is broken, we have t
and e l
oo
;
that
is,
the
e.m.f. rises infinitely.
The greater rv the higher is the generated e.m.f. e v the faster,
however,
do
e l
and
i
decrease.
= If r l
r, we have at t = 0,
CONTINUOUS-CURRENT CIRCUITS.
29
and
e n - ijr = E;
that
is,
if
the
external
resistance
r t
equals
the
internal
resistance
r, at the moment of withdrawal of the e.m.f. E the terminal
voltage is E.
The effect of the e.m.f. of inductance in stopping the current
at the time t is
= v + ie l
(r
L ...~o r-fr. f
r,} *
;
thus the total energy of the generated e.m.f.
W=
_
that
is,
the
energy
stored
as
magnetism
in
a
circuit
of
current i Q
and inductance L is
which is independent both of the resistance r of the circuit and
the
resistance
r 1
inserted
in
breaking
the
circuit.
This energy
has to be expended in stopping the current.
EXAMPLES-
32. (1.) In the alternator field in Section 1, Example 4, Sec-
tion 2, Example 2, and Section 5, Example 1, how long a time
E= after impressing the required e.m.f.
230 volts will it take for
\ the field to reach (a) J strength, (6) T strength?
(2.) If 500 volts are impressed upon the field of this alternator,
and a non-inductive resistance inserted in series so as to give
the required exciting current of 6.95 amperes, how long after
E = impressing the e.m.f.
500 volts will it take for the field to
& reach (a) J strength, (6)
strength, (c) and what is the resist-
ance required in the rheostat?
(3.) If 500 volts are impressed upon the field of this alternator without insertion of resistance, how long will it take for
the field to reach full strength?
(4.) With full field strength what is the energy stored as
magnetism?
30
ELEMENTS OF ELECTRICAL ENGINEERING.
(1.) The resistance of the alternator field is 33.2 ohms (Section
2, Example 2), the inductance 112 h. (Section 5, Example 1),
E the impressed e.m.f. is
= 230, the final value of current
=
rr
= 6.95 amperes. Thus the current at time t, is
r
= I
\
= G.95 (1 -
= - = (a.) i strength t -^-, hence (1
89
e-"- ")
0.5.
,J
-' = - = = 0.5, 0.296 1 log e log 0.5, t
g 5'
~^. .
log
2.34 seconds.
and
,
A = - = (6.)
strength: i
0.9
i ,
hence
(1
e- - 296 *)
0.9, and
t = 7.8 seconds
E (2.) To get i = 6.95 amperes, with
= 500 volts, a resist-
ance r = ~500- = 72 ohms, and thus a rheostat having a resist-
6.95
ance of 72 33.2 = 38.8 ohms is required.
We then have
- = 6.95 (1
e"- 643 0.
= = (a.) i
i
^^, after t
1.08 seconds.
= = (6.) i
0.9 i after t ,
3.6 seconds.
E (3.) Impressing
= 500 volts upon a circuit of = r 33.2,
L = 112, gives
. _ELH
2,
-st\ e L)
rv
7
- = 15.1 (1
-' 2mt
).
= i 6.95, or full field strength, gives
= - 6.95 15.1 (1
-- 296 0.
and t = 2.08 seconds.
CONTINUOUS-CURRENT CIRCUITS.
31
(4.) The stored energy is
L _ 2
^^' j_
6.952
X
112
_ 027702A0
,,
i
watt-seconds
or
i
joules
2i
2i
= 2000 foot-pounds.
= (1 joule 0.736 foot-pounds.)
Thus in case (3), where the field reaches full strength in 2.08
seconds ;
the
average
power
input
is
2000
j~
= 960
foot-pounds
J.OS
per second, = 1.75 hp.
In breaking the field circuit of this alternator, 2000 foot-
pounds of energy have to be dissipated in the spark, etc.
A 33. (5.)
coil of resistance r = 0.002 ohm and inductance
L = 0.005 milhenry, carrying current 7=90 amperes, is short-
circuited.
(a.) What is the equation of the current after short cir-
cuit?
(5.) In what time has the current decreased to 0.1, its initial
value?
(a.) i = 7e"i = 90s-400 '.
= = = (6.) i
0.1 7, <r400 '
0.1, after t
0.00576 second.
(6.) When short-circuiting the coil in Example 5, an e.m.f.
E =* 1 volt is inserted in the circuit of this coil, in opposite di-
rection to the current.
(a.) What is equation of the current?
(6.) After what time does the current become zero? (c.) After what time does the current reverse to its initial
value in opposite direction?
(d.) What impressed e.m.f. is required to make the current
die out in yoVo second?
E (e.) What impressed e.m.f. is required to reverse the current
in unnr second?
(a.) If e.m.f.
E is inserted, and at time t the current is
denoted by i, we have
_ e ^
L di the generated e.m.f.;
32
ELEMENTS OF ELECTRICAL ENGINEERING,
Thus,
-~ E + e = 1
and
^ == E +J_eL, =
r
Transposing,
E L , the total e.m.f.;
E
L di
.t, ht e
,
current:
r
r dt
T 7,
di
and integrating,
-y ~ rt
,
/E
\
^
log. (\ r7
t
+*) /
T
-log
C,
where loge c = integration constant.
"
At
t
-
0,
i
=
7,
thus
c
=
7
+
-
;
Substituting,
= - i
590 -*0()t
500.
= = = (6.) i
" - 40(
0,
0.85, after *
0.000405 second,
= - - (c.) i
= = = I
-400 *
90,
0.694, after t
0.00091 second
= (d) If i
at t = 0.0005, then
- = + (90
500 E) s- - 3
500 E,
E = -~^- = 0.81 volt.
- - = = = (e.) If i
/
90 at i
0.001, then
- 90 - (90 + 500 E) e- - 4 - 500 E,
4
O^+.f- E =
_ ) o.91 volt.
7. INDUCTANCE IN ALTERNATING-CURRENT CIRCUITS
= An 34,
= alternating current i 7 sin 2 rc/f or i 7 sin
can be represented graphically in rectangular coordinates by
curved line as shown in Fig. 9, with the instantaneous value
i as ordinates and the time t, or the arc of the angle corresponc
= ing to the time, 6 2 nfl, as abscissas, counting the time from th
zero value of the rising wave as zero point.
INDUCTANCE L\ ALTERNATING-CURRENT CIRCUITS. 33
If the zero value of current is not chosen as zero point of time,
the wave is represented by
or
- 7 sin(0 0'),
Fig. 9. Alternating Sine Wave.
where
f t
and
Of
are
respectively
the
time
and
the
corresponding
angle at which the current reaches its zero value in the ascen-
dant.
If such a sine wave of alternating current i = 7 sin 2 ~ft or i = 7 sin 8 passes through a circuit of resistance r and induc-
tance L, the magnetic flux produced by the current and thus its
interlinkages with the current, iL = 7 L sin 0, vary in a wave
line similar also to that of the current, as shown in Fig. 10 as <.
Fig. 10. Self-induction Effects produced by an Alternating Sine
Wave of Current.
The e.m.f. generated hereby is proportional to the change of
iL, and is thus a maximum where iL changes most rapidly, or at
its zero point, and zero where iL is a maximum, and according to Lentz's law it is positive during falling and negative during rising current. Thus this generated e.m.f. is a wave following
the wave of current by the time t =
~r where t ,
is time of one
complete period,
=
-
,
or
by
the
time
angle
= 90.
j
34
ELEMENTS OF ELECTRICAL ENGINEERING.
This e.ra.f. is called the counter e.m.f. of inductance. It is
,
T di
<?/=-Ly
dt
- - 2 7r/L/
cos 2 7T/55.
It is shown in dotted line in Fig. 10 as e 2'.
The quantity 2 TT/L is called the inductive reactance of the
circuit, and denoted by x. It is of the nature of a resistance,
and expressed in ohms. If L is given in 109 absolute units or
henrys, x appears in ohms.
The counter e.m.f. of inductance of the current, i = 7 sin
= 2 7T/35
7 sin 0, of effective value
is
,
e/
= - oJ
cos 2xft
- = xI Q
cos
0,
having
a
maximum
value
of
xI Q
and
an
effective
value
of
*>-%-*
that is, the effective value of the counter e.m.f. of inductance
equals the reactance, x, times the effective value of the current, I, and lags 90 time degrees, or a quarter period, behind the current.
35. By the counter e.m.f. of inductance,
= e/
x/ cos 6,
which is generated by
the current i = J sin
the change in d through the
flux due to the passage of circuit of reactance x, an
equal but opposite e.m.f.
= e 2
z/ cos 6
is consumed, and thus has to be impressed upon the circuit. This e.m.f. is called the e.m.f. consumed by inductance. It is 90 time degrees, or a quarter period, ahead of the current, and shown in Fig. 10 as a drawn line e T
Thus we have to distinguish between counter e.m.f. of induc-
tance 90 time degrees lagging, and e.m.f. consumed by inductance 90 time degrees leading.
INDUCTAXCE IX ALTERXATIXG-CTRREXT CIRCUITS. 35
These e.m.fs. stand in the same relation as action and reaction
in mechanics. They are shown in Fig. 10 as e./ and as <v The e.in.f. consumed by the resistance r of the circuit is pro-
portional to the current,
= = e l
ri
r/ sin 6,
and in phase therewith, that is, reaches its maximum and its
zero value at the same time as the current {, as shown by drawn
line
e l
in
Fig.
10.
E = Its effective value is
ri.
1
The resistance can also be represented by a (fictitious) counter
e.m.f.,
=
e^
rI sin Q
0,
opposite in phase to the current, shown as e^ in dotted line in
Fig. 10.
The counter e.m.f. of resistance and the e.m.f. consumed by resistance have the same relation to each other as the counter
e.m.f. of inductance and the e.m.f. consumed by inductance or
inductive reactance.
= 36. If an alternating current i / sin 6 of effective value
= /
*k exists in a circuit of resistance r and inductance L. that
V2
is, of reactance x = 2nfL, we have to distinguish:
E.m.f.
consumed
by
resistance,
= e 1
r/ sin d,
of
effective
value
E t
=
r/,
and
in
phase
with
the
current.
Counter e.m.f. of resistance, / =
r/ sin
of effective
<?,
E value l = ri, and in opposition or 180 time degrees displaced
from the current.
E.m.f.
consumed
by
reactance,
e = xI 2
cos #,
of
effective
E value
2
=
xl }
and
leading
the
current
by
90
time
degrees
or
a
quarter period.
Counter e.m.f. of reactance, e/ =
x/ cos 9, of effective
E value
' 2
=
#7,
and
lagging
90
time
degrees
or
a
quarter
period
behind the current.
The e.m.fs. consumed by resistance and by reactance are the
e.m.fs. which have to be impressed upon the circuit to overcome the counter e.m.fs. of resistance and of reactance.
Thus, the total counter e.m.f. of the circuit is
= + = + '
e
'
6i
ej
J (r sin 6
x cos #),
36
ELEMENTS OF ELECTRICAL ENGINEERING.
and the total impressed e.m.f., or e.m.f. consumed by the circuit,
is
= + = + e
e l
e 2
7 (r sin 6
x cos 6).
Substituting
and
it follows that
x = z sin QJ
and we have as the total impressed e.m.f.
e = s/ sin(0 + ),
shown by heavy drawn line e in Fig. 10, and total counter e.m.f.
shown
by
heavy
dotted
line
r
e
in
Fig.
10,
both
of
effective
value
e = zL
For 6 =
= # e 0?
0, that is, the zero value of e is ahead of
the zero value of current by the time angle , or the current lags
behind the impressed e.m.f. by the angle . is called the cm#Ze of time lag of the current, and
= Vr x 3
2
-f-
the impedance of the circuit. 6 is called the e.m.f. consumed by
impedance, er the counter e.m.f. of impedance.
E Since l == rl is the e.m.f. consumed by resistance, E 2 = xl is the e.m.f. consumed by reactance,
and
E = zl = Vr2 + x2 7 is the e.m.f. consumed by impe-
we have and
dance,
E - VWf+~E^y the total e.m.f.
E = E cos
l
,
E E% = sin # , its components.
The tangent of the angle of lag is
= = ,
*
tan 6
x-
27T/L *
,
r
r
and the time constant of the circuit is
^ = 2 7T/ COt
INDUCTANCE IN ALTERNATING-CURRENT CIRCUITS. 37
The total e.rn.f., e, impressed upon the circuit consists of two
components,
one,
e v in
phase
with
the
current,
the
other
one,
e
2?
in quadrature with the current.
Their effective values are
E E E,
cos ,
sin .
EXAMPLES. 37. (1.) What is the reactance per wire of a transmission
line of length I, if ld = diameter of the wire, 4 = spacing of the wires, and/ = frequency?
Fig. 11. Diagram for Calculation of Inductance "between two Parallel Conductors,
= If / current, in absolute units, in one wire of the trans-
mission line, the m.m.f. is I; thus the magnetizing force in a
zone dlx at distance lx from center of wire (Fig, 11) is 3 =
_
= and the field intensity in this zone is JC
- Thus
the magnetic flux in this zone is
d$ 3C ldlx
2 IldL
'
I*
hence, the total magnetic flux between the wire and the return
wire is
= 2/Z
sj
~2
"2
neglecting the flux inside the transmission wire.
S8
ELEMENTS OF ELECTRICAL ENGINEERING.
The inductance is
L
=
-
/
=
2 Hoge^^ absolute
units
^ = 2Hoge
10-9 henrys,
id
and
the
reactance
x
=
2
rJL
=
4
2I 2
rfl log, y- ,
in
absolute
units;
i^d
or
= y x
9 4 TT/Z loge
1 10
~9
,
in
ohms.
id
38. (2.) The voltage at the receiving end of a 33.3-cycle
threephase transmission line 14 miles in length shall be 5500
& between the lines. The line consists of three wires, No. B. = g. (id =0.82 cm.), 18 in. (45 cm.) apart, of resistivity p 1.8 X 10~8 .
(a.) What is the resistance, the reactance, and the impedance
per line, and the voltage consumed thereby at 44 amperes?
(&.) What is tHe generator voltage between lines at 44 amperes
to a non-inductive load?
(c.) What is the generator voltage between lines at 44 amperes
to a load circuit of 45 time degrees lag?
(d.) What is the generator voltage between lines at 44 amperes
to a load circuit of 45 time degrees lead?
Here I - 14 miles = 14 X 1.6 X 105 = 2.23 X 106 cm.
= ld 0.82cm.
A Hence the cross section, = 0.528 sq. cm.
= X -~~ = f N
(a.)
RT>esi.st, ance
per
r
line,
r
I
p
1.8
10-6 X 2.23 X 106
A u.52o
== 7.60 ohms.
Reactance
per
line,
x
=
4
2
nfllog,-
Z
X 10~9 == 4 K X 33.3 X
id
2.23 X 106 X log 110 X 10-9 - 4.35 ohms.
The impedance per line, z
if / = 44 amperes per line,
\/r2 + x2 = 8.76 ohms. Thus
E the e.m.f. consumed by resistance is l = rl = 334 volts,
the
e.m.f.
consumed
by
reactance
is
E 2
=
xl
=
192 volts,
and the e.m.f.
consumed
by
impedance
is
E 3
=
zl
=
385 volts.
INDUCTANCE IN ALTERNATING-CURRENT CIRCUITS. 39
5500
V3 (&.) 5500 volts between lines at receiving circuit give
3170 volts between line and neutral or zero point (Fig. 12),
V2 or per line, corresponding to a maximum voltage of 3170
=
4500 volts. 44 amperes effective per line gives a maximum
value of 44 \/2 = 62 amperes. Denoting the current by i = G2 sin 6, the voltage per line at
the receiving end with non-inductive
= load is e 4500 sin 0.
The e.m.f. consumed by resistance,
in phase with the current, of effective
value 334, and maximum value 334 - \/2 472, is
e = 472 sin 0. l
The e.m.f. consumed by reactance,
90 time degrees ahead of the current,
of effective value 192, and maximum
value 192 V2 = 272, is
D Fig. 12> Voltage ia^ram for
a Three-Phase Circuit,
= e 2
272 cos d.
Thus the total voltage required per line at the generator end
of the line is
e
Q
=
e
+
BI
+
e
2
=
(4500
+
472)
sin
+ 272 cos 6
= 4972 sin d + 272 cos 6.
. 272 Denoting
4972
tan we have ,
sin # =
tan fl fl
Vl + tan2 tf n
272 4980
cos<9 n
+ tan2 d n
4972 4980
Hence,
= + e Q
4980 (sin 9 cos #
cos 6 sin )
- 4980 sin (9 + 9 ).
Thus d is the lag of the current behind the e.m.f. at the
generator end of the line, = 3.2 time degrees, and 4980 the
40
ELEMENTS OF ELECTRICAL ENGINEERING.
E maximum voltage per line at the generator end; thus
=
4980
==
Q
V3 = 3520, the effective voltage per line, and 3520
= 6100, the
effective voltage between the lines at the generator.
(c.) If the current
i = 62 sin 6
lags in time 45 degrees behind the e.m.f. at the receiving end of
the line, this e.m.f. is expressed by
= - + e 4500 sin (6 + 45) 3170 (sin
cos ff) ;
that is, it leads the current by 45 time degrees, or is zero at =*
45 time degrees.
The e.m.f. consumed by resistance and by reactance being the
same as in (6), the generator voltage per line is
+ e o
=
e
+
&i
+
e 2
=
3642 sin
3442 cos 0.
Denoting
^ = tan
we have
,
= + e
5011 sin (0
).
Thus , the time angle of lag of the current behind the gen-
erator e.m.f., is 43 degrees, and 5011 the maximum voltage;
hence 3550 the effective voltage per line, and 3550 \/3 6160
the effective voltage between lines at the generator.
(d.) If the current { = 62 sin leads the e.m.f. by 45 time
degrees, the e.m.f. at the receiving end is
e
-
4500 sin
(0
45)
- = 3170 (sin 8
cos 0).
Thus at the generator end
+ + * - e o
*=
e
e i
e 2
3642 sin 6
2898 cos 0.
Denoting
= r-
tan , it is
OUttA
- e, -4654 sin (0
).
Thus , the time angle of lead at the generator, is 39 degrees,
and 4654 the maximum voltage; hence 3290 the effective voltage
per line and 5710 the effective voltage between lines at the
generator.
POWER IX ALTERNATING-CURRENT CIRCUITS. 41
8. POWER IN ALTERNATING-CURRENT CIRCUITS.
39. The power consumed by alternating current i = J sin 6,
of effective value / = = tance x 2 ;r/L, is
^, in a circuit of resistance r and reac-
V2
P = ei,
where e
+ / sin (0
the components
) is the impressed e.m.f., consisting of
and
e 1
= r/
sin 0, the e.m.f. consumed by resistance
e
2
=
xI cos Q
5. the
e.m.f.
consumed
by
reactance.
Vr z =
+ x2 is the impedance and tan = - the time-phase
angle of the circuit; thus the power is
= + p
2/
2
sin
6
sin
(6
)
=
^-(cos0
- cos (20 +
))
= zP (cos
- + cos (2
)).
Since the average cos (20 + ) = zero, the average power is
p = ZP cos
that is, the power in the circuit is that consumed by the resistance,
and independent of the reactance. Reactance or self-inductance consumes no power, and the
e.m.f. of self-inductance is a wattless e.m.f., while the e.m.f. of
resistance is a power e.m.f.
The wattless e.m.f. is in quadrature, the power e.m.f. in phase
with the current.
In general, if = angle of time-phase displacement between
the resultant e.m.f. and the resultant current of the circuit,
E / = current,
= impressed e.m.f., consisting of two com-
E ponents, one, t = E cos 0, in phase with the current, the other,
E E === 2
sin 0, in quadrature with the current, the power in the
= circuit is IE l
IE cos 0, and the e.m.f. in phase with the current
E E = t
cos is a power e.m.f., the e.m.f. in quadrature with
E E the current
= sin a wattless or reactive e.m.f.
42
ELEMENTS OF ELECTRICAL ENGINEERING.
40. Thus we have to distinguish power e.m.f. and wattless or reactive e.m.f., or power component of e.m.f., in phase with the current and wattless or reactive component of e.m.f., in quadra-
ture with the current.
Any e.m.f. can be considered as consisting of two components,
one, the power component, e v in phase with the current, and
the
other,
the
reactive
component,
e 2,
in
quadrature with the
current. The sum of instantaneous values of the two com-
ponents is the total e.m.f.
+ e == e l
er
E E If E, v 2 are the respective effective values, we have
E = VE* + E*, since
E E = 1
cos 6,
E = E 2
sin 0,
where d = time-phase angle between current and e.m.f. Analogously, a current / due to an impressed e.m.f. E with
a time-phase angle 6 can be considered as consisting of two
component currents,
7 X
=
7 cos
9j
the power component of
the
current,
and
7 2
==
7 sin
0,
the wattless
or reactive
component of
the
current.
The sum of instantaneous values of the power and reactive
components of the current equals the instantaneous value of the
total current,
+ = i l
i
a
i,
while their effective values have the relation
= +7 7
v/7 3 1
2 2.
Thus an alternating current can be resolved in two components, the power component, in phase with the e.m.f., and the wattless or reactive component, in quadrature with the e.m.f.
An alternating e.m.f. can be resolved in two components,
the power component, in phase with the current and the wattless or reactive component, in quadrature with the current.
The power in the circuit is the current times the e.m.f. times the cosine of the time phase angle, or is the power component of the current times the total e.m.f., or the power component of
the e.m.f. times the total current.
POLAR COORDINATES.
43
EXAMPLES.
41. (1.) What is the power received over the transmission line
in Section 7, Example 2, the power lost in the line, the power
put into the line, and the efficiency of transmission with non-
inductive load, with 45-time-degree lagging load and 45-clegree
leading load?
P The power received per line with non-inductive load is = El
= 3170 X 44 = 139 kw.
With a load of 45 time degrees phase displacement, P = El cos 45 = 98 kw.
P The power lost per line l = PR - 442 X 7.6 = 14.7 kw.
P P Thus the input into P the line =
+
= 151.7 kw. at
l
non-inductive load,
and
= 111.7 kw. at load of 45 degrees phase displacement.
The efficiency with non-inductive load is
P
151.7 = 90.3 per cent,
and with a load of 45 degrees phase displacement is
- = 1
p~
^ 86 - 8 P er cent -
jjjy
P The total output is 3 = 411 kw. and 291 kw., respectively.
P The total input 3
= 451.1 kw. and 335.1 kw., respectively.
9. POLAR COORDINATES.
42. In polar coordinates, alternating waves are represented, with the instantaneous values as radius vectors, and the time
as an angle, counting left-handed or counter clockwise, and one
revolution or 360 degrees representing one complete period.
The sine wave of alternating current i = I sin 9 is repre-
sented by a circle (Fig. 13) with the vertical axis as diameter,
equal in length 0/
to the maximum value J and shown as ,
heavy drawn circle.
The
e.m.f.
consumed
by
inductance,
e
2
=
x/
cos 0, is repre-
sented
by
a
circle
with
diameter
OE 2
in
horizontal
direction
to
the right, and equal in length to the maximum value, xIQ. Er
Analogously, the counter e.m.f. of self-inductance 2 is repre-
44
ELEMENTS OF ELECTRICAL ENGINEERING.
sented by a circle OE^in Fig. 13; the e.m.f. consumed by
^= OE resistance r by circle
of a diameter
l
r/ and the ,
counter e.m.f. of resistance $/ by circle OE^.
The counter e.m.f. of impedance
of a diameter equal in length to E',
aisndrelpargegsienngte1d80by-
circle OE'
# behind
the diameter of the current circle. This circle passes through
the points $/ and EJ, since at the moment 6 = 180 degrees,
Fig. 13. Sine waves represented in Polar Coordinates.
6/ 0, and thus the counter e.m.f. of impedance equals the
= counter e.m.f. of reactance ef
e
2 ',
and
at
6 = 270
degrees,
'
e
a
=
0,
and
the
counter
e.m.f.
of impedance
equals
the
counter
r
e.m.f. of resistance e
e/.
The e.m.f. consumed by impedance, or the impressed e.m.f.,
is represented by circle OE having a diameter equal in length
to E, and leading the diameter of the current circle by the angle
E E .
This circle passes through the points
and
1
r
An alternating wave is determined by the length and direction
of the diameter of its polar circle. The length is the maximum
value or intensity of the wave, the direction the phase of the
maximum value, generally called the phase of the wave.
POLAR COORDINATES.
45
43- Usually alternating waves are represented in polar co-
ordinates by mere vectors, the diameters of their polar circles, and the circles omitted, as in Fig. 14.
Fig. 14. Vector Diagram.
Fig. 15. Vector Diagram of two e.m.fs. Acting in the same Circuit.
Two e.m.fs., ^ and ev acting in the same circuit, give a result-
ant e.m.f. e equal to the sum of their instantaneous values. In
polar
coordinates
e l
and
e
2
are
represented
in
intensity
and
in
phase
by
two
vectors,
OE 1
and
OE^
Fig.
15.
The instantane-
OX ous values in any direction
are
the
projections
Oe v
Oe 2
of
OE 1
and
OE 2
upon
this
direction.
46
ELEMENTS OF ELECTRICAL ENGINEERING.
Since the sum of the projections of the sides of a parallelo-
gram is equal to the projection of the diagonal, thejsum of the
projections
Oe^
and
Oe 2
equals
thejprojection
Oe
of
OE,
the diago-
nal
of
the
parallelogram
with
OE t
and
OE 2
as
sides,
and
OE
is
thus the diameter of the circle of resultant e.m.f . ; that is, in polar
coordinates alternating sine waves of e.m.f., current, etc., are
combined and resolved by the parallelogram or polygon of sine
waves.
Since the effective values are proportional to the maximum
values, the former are generally used as the length of vector of the alternating wave. In this case the instantaneous values
are given by a circle with \/2 times the vector as diameter.
44. As phase of the first quantity considered, as in the above instance the current, any direction can be chosen. The further
quantities are determined thereby in direction or phase. In polar coordinates, as phase of the current, etc., is here and
in the following understood the time or the angle of its vector,
that is, the time of its maximum value, and a current of phase zero would thus be denoted analytically by i = 7 cos 6.
The zero vector OA is generally chosen for the most frequently
used quantity or reference quantity, as for the current, if a
number of e.m.fs. are considered in a circuit of the same cur-
rent, or for the e.m.f., if a number of currents are produced by the same e.m.f., or for the generated e.m.f. in apparatus such as transformers and induction motors, synchronous
apparatus, etc.
With the current as zero vector, all horizontal components
of e.m.f. are power components, all vertical components are
reactive components.
With the e.m.f. as zero vector, all horizpntal components of
current are power components, all vertical components of current
are reactive components.
By measurement from the polar diagram numerical values
can hardly ever be derived with sufficient accuracy, since the
magnitudes of the different quantities used in the same diagram are usually by far too different, and the polar diagram is there-
fore useful only as basis for trigonometrical or other calculation, and to give an insight into the mutual relation of the different quantities, and even then great care has to be taken to distin-
POLAR COORDINATES.
47
guish between the two equal but opposite vectors, counter e.ni.f. and e.m.f. consumed by the counter e.m.f., as explained
before.
45. In the polar coordinates described in the preceding, and used throughout this book, the angle represents the time, and is counted positive in left-handed or counter-clockwise rotation, with the instantaneous values of the periodic function as radii,
so that the periodic function is represented by a closed curve, and one revolution or 360 degrees as one period. This "time
diagram" is the polar coordinate system universally used In
other sciences to represent periodic phenomena, as the cosmic motions in astronomy, and even the choice of counter-clockwise
as positive rotation is retained from the custom of astronomy,
the rotation of the earth being such.
In the time diagram, the sine wrave is given by a circle, and
this circle of instantaneous values of the sine wave is represented,
in size or position, by its diameter. That is, the position of this
diameter denotes the time, t, or angle, = 2 nft, at which the
sine wave reaches its maximum value, and the length of this
diameter denotes the intensity of the maximum value.
OE If then, in polar coordinate representation, Fig. 16,
denotes
an e.m.f., 01 a current, this means that the maximum value of
e.m.f. equals OE, and is reached at the time, t v represented by
angle
AOE
=
O l
=
2 nft r
The current in this diagram then
has a maximum value equal to 01, and this maximum value is
reached at the time, t v represented by angle AOI =
= 2
2 xft y
If then
angle
6
2
>
6V
this
means
that
the
current
reaches its
maximum value later than the e.m.f., that is, the current in
- Fig.
16
lags
behind
the
e.m.f.,
by
the
angle
EOI
=
# 3
O= l
2 nf(t 2 ^), or by the time 3 t r
Frequently in electrical engineering another system of polar
coordinates is used, the so-called "crank diagram." In this,
sine waves of alternating currents and e.m.fs. are represented
as projections of a revolving vector upon the horizontal. That is,
a vector, equal in length to the maximum of the alternating
wave, revolves at uniform speed so as to make a complete
revolution per period, and the projections of this revolving
vector upon the horizontal then denote the instantaneous values
of the wave.
48 ELEMENTS OF ELECTRICAL ENGINEERING.
Obviously, by the crank diagram only sine waves can be
represented, while the time diagram permits the representation
of any wave shape, and therefore is preferable.
Let, for instance, 01 represent in length the maximum value
of current, i = / cos (0
2). Assume, then, a vector, OL to
Fig. 16. Representation of Current and e.m.f. by Polar Coordinates.
revolve, left-handed or in positive direction, so that it makes a
complete revolution during each cycle or period. If then at
a
certain
moment
of
time,
this
vector
stands
in
position
01 1
OA OA (Fig. 17), the projection,
of 01 on
lt
i
represents the instan-
taneous value of the current at this moment. At a later moment
01
has
moved
farther,
to
07 2,
and
the
projection,
OA 2J
of 07 on 2
OA is the instantaneous value. The diagram thus shows the
instantaneous condition of the sine waves. Each sine wave
reaches the maximum at the moment when its revolving vector,
01 j
passes
the
horizontal.
If Fig. JL8 represents the crank diagram of an e.m.f., OE, and a
APE current, 07, and if angle
> AOI, this means that the e.m.f.,
OE, is ahead of the current, 07, passes during the revolution the
zero line or line of maximum intensity, OA, earlier than the
current, or leads; that is, the current lags behind the e.m.f. The same Fig. 18 considered as polar diagram would mean that
the current leads the e.m.f.; that is, the maximum value of
A current, 07, occurs at a smaller angle, 07, that is, at an earlier time., than the maximum value of the e.m.f,, OE.
POLAR COORDINATES.
46. In the crank diagram, the first quantity therefore can be put in any position. For instance, the current, 07, in Fig. IS, could be drawn in position 07, Fig. 19. The e.m.f. then being
ITig. 17, Crank Diagram showing
instantaneous values.
Fig. 18. Crank Diagram of an e.m.f. and Current.
ahead of the current by angle EOI = 6 would come into the
position OE, Fig. 19.
A polar
diagram,
Fig.
16,
with
the
current,
0/ ?
lagging
behind
the e.m.f., OE, by the angle, 0, thus considered as crank diagram
would
represent
the
current
leading
the
e.m.f.
by
the
angle,
6 f
and a crank diagram, Fig, 18 or 19, with the current lagging
behind the e.m.f. by the angle, 6, would as polar diagram repre-
sent a current leading the e.m.f. by the angle, 6.
The main difference in appearance between the crank diagram
and the polar diagram therefore is that, with the same direction
of rotation, lag in the one diagram is represented in the same
manner as lead in the other diagram, and inversely. Or, a
50
ELEMENTS OF ELECTRICAL ENGINEERING.
representation by the crank diagram looks like a representation by the polar diagram, with reversed direction of rotation, and vice versa. Or, the one diagram is the image of the other and can be transformed into it by reversing right and left, or top and bottom. Therefore the crank diagram, Fig. 19, is the image
of the polar diagram, Fig. 16.
Fig. 19. Crank Diagram.
Since the time diagram, in which the position of the vector
represents its phase, that is, the moment of its maximum value,
is used in all other sciences, and also is preferable in electrical engineering, it will be exclusively used in the following, the positive direction being represented as counter-clockwise.
EXAMPLES.
47. In a three-phase long-distance transmission line, the volt-
age between lines at the receiving end shall be 5000 at no load,
5500 at full load of 44 amperes power component, and propor-
tional at intermediary values of the power component of the
current; that is, the voltage at the receiving end shall increase
proportional to the load. At three-quarters load the current
shall be in phase with the e.m.f. at the receiving end. The
generator excitation however and thus the (nominal) generated
e.m.f. of the generator shall be maintained constant at all loads,
and the voltage regulation effected by producing lagging or
leading currents with a synchronous motor in the receiving cir-
cuit. The line has a resistance r = 7.6 ohms and a reactance
l
x t
= 4.35
ohms
per
wire,
the
generator
is
star
connected,
the
resistance per circuit
being
= r 2
0.71,
and
the
(synchronous)
reactance is 3, = 25 ohms. What must be the wattless or
POLAR COORDINATES.
51
reactive component of the current, and therefore the total cur-
rent and its phase relation at no load, one-quarter load, one-
half load, three-quarters load, and full load, and what will be
the terminal voltage of the generator under these conditions?
The total resistance of
the line
and
generator
is r
=r l
+r 2
= 8.31
ohms;
the
total
reactance,
x
=
x l
+ -r = 29.35 ohms. 2
Fig. 20. Polar Diagram of e.m.f. aud Current in Transmission Line. Current Leading.
E OE Let, in the polar diagram, Fig. 20 or 21,
= represent
the
voltage
at
the
receiving
end
of
the
line,
01 1
=
7 t
the
power
component of the current corresponding to the load, in phase
with OE and 0/
}
2
7 = 2
the
reactive
component
of
the
current
Fig. 21. Polar Diagram of e.m.f. and Current in Transmission Line. Current Lagging.
in
quadrature
with
OE y
shown
leading
in
Fig.
20,
lagging
in
Fig. 21.
_
We then have total current 7 = 07.
Thus the e.m.f. consumed by resistance, OE = r7, is in phase
OE with 7, the e.m.f. consumed by reactance,
= 2 #7, is 90 degrees
52
ELEMENTS OF ELECTRICAL ENGINEERING.
OF ahead of /, and their resultant is
the e.m.f. consumed
37
by
impedance.
OEs combine^ with OE, the receiver voltage, gives the gener-
OE ator voltage
Q.
Resolving all e.m.fs. and currents into components in phase
and in quadrature with the received voltage E, we have
PHASE COMPONENT.
Current
/
t
E E.m.f. at receiving end of line,
E
E E.m.f. consumed by resistance,
=
l
rl t
E.m.f.
consumed
by
reactance,
E= 2
xI 2
Thus total e.m.f. or generator voltage,
E =E+E +E =
Q
l
2
E + rl + xI
t
2
QUADRATURE COMPONENT.
I 2
rI 2 xl^
r/ - xl
2
:
Herein the reactive lagging component of current is assumed
as positive, the leading as negative.
The generator e.m.f. thus consists of two components, which
give the resultant value
E,
=
V(E
+
rl,
+
xI2r
+
(rl t
xI.
substituting numerical values, this becomes
E - V(E + = Q
+ + 8.31 1,
29.35 7 2) 2
(8.31
/ 2
29.35
2
/J ;
at three-quarters load,
E=
5375 =~
=
3090 volts
per
circuit,
^=33,
J 2
=
0,
thus
EQ
=
V(3090
4-
8.31
X
2
33)
+
(29.35
X
2
33)
=
3520 volts
V3 per line or 3520 X
= 6100 volts between lines
as (nominal) generated e.m.f. of generator.
Substituting these values, we have
= + + - 3520
V(JS
8.31 /,
+ 29.35 /2) 2
(8.31 /,
29.35
2
J,) .
The voltage between the lines at the receiving end shall be:
No
i
}
|
FULL
LOAD. LOAD. LOAD. LOAD. LOAD.
Voltage between lines,
_
5000 5125 5250 5375 5500
W3, Thus, voltage per line,
#-= 2880 2950 3020 3090 3160
HYSTERESIS AND EFFECTIVE RESISTANCE
53
The power components of current
= per line, I,
11 22 33 44
Herefrom we get by substituting in the above equation
Reactive component of Lo^ D.
current,
7 - 21.6 2
*
Lo AD .
16.2
Loi D.
9.2
Lo| D.
j
-9.7
hence, the total current,
/ = VI* + / 2 = 21.6 2
19.6
23.9
33.0
45.05
and the power factor,
L cos g
o 56.0 92.0 100.0 97.7
the lag of the current,
0= 90
61
23
the generator terminal voltage per line is
-11.5
+ + + - 7.6 1,
4.35 / 2) 2
(7.6
/ 2
4.35 / t)
thus:
Per line,
E^ =
Between lines, E' V$ =
No
LOAD.
2980
5200
i
LOAD.
3106
5400
i
LOAD.
3228
5600
I LOAD.
3344
5800
LOAD
3463 6000
Therefore at constant excitation the generator voltage rises with the load, and is approximately proportional thereto.
10. HYSTERESIS AND EFFECTIVE RESISTANCE.
48. If an alternating current 01 = /, in Fig. 22, exists in a
circuit of reactance x 2 TtfL and of negligible resistance, the
magnetic flux produced by the current, 0$ = $, is in phase with
the current, and the e.m.f. generated by this flux, or counter
E e.m.f. of self-inductance, OE'" =
f" = xl,
lags
90
degrees
behind the current. The e.m.f. consumed by self-inductance
or impressed e.m.f. OE" = E" = xl is thus 90 degrees ahead of
the current.
Inversely, if the e.m.f. OE" = E" is impressed upon a circuit of reactance x = 2 nfL and of negligible resistance, the current
~c\n
i
01
=
I
=
lags 90 degrees behind the impressed e.m.f.
x
54
ELEMENTS OF ELECTRICAL ENGINEERING.
This current is called the exciting or magnetizing current of
the magnetic circuit, and is wattless. If the magnetic circuit contains iron or other magnetic material,
energy is consumed in the magnetic circuit by a frictional resistance of the material against a change of
|"
magnetism, which is called molecular mag-
netic friction.
I
If the alternating current is the only avail-
able source of energy in the magnetic circuit,
the expenditure of energy by molecular
magnetic friction appears as a lag of the
magnetism behind the m.m.f. of the current,
that is, as magnetic hysteresis, and can be
measured thereby.
Magnetic hysteresis is, however, a dis-
tinctly different phenomenon from molecular magnetic friction, and can be more or less
eliminated, as for instance by mechanical
vibration, or can be increased, without
changing the molecular magnetic friction.
E/'
Fig. 22. Phase Reiations of Magnetizing nd
""x ' f *
49. In consequence of magnetic hys-
= teresis, if an alternating e.m.f. OE" E" is
impressed upon a circuit of negligible resist-
ance, the exciting current, or current producing the magnetism, in this circuit is not
a wattless current, or current of 90 degrees lag, as in Fig. 22,
butJags less than 90 degrees, by an angle 90 a, as shown
_ by 01 = I in Fig. 23. Since the magnetism 0$ = <J> is in quadrature with the e.m.f.
E" due to it, angle a is the phase difference bet ween the magnetism
and the m.m.f., or the lead of the m.m.f., that is, the exciting
current, before the magnetism. It is called the angle ofhysteretic
lead.
In = this case the exciting current 01 / can be resolved in
two
components,
the magnetizing
current
01 2
=
7 2,
in
phase with
the magnetism 0$ = <J>, that is, in quadrature with the e.m.f.
W OE = /f
r ,
and
thus
wattless,
and
the
magnetic power
component
= of the
current
QY
the
hysteresis
current
01 1
Iv in phase with the
e.m.f. OE" = E", or in quadrature with the magnetism 0$ = $.
HYSTERESIS AND EFFECTIVE RESISTANCE.
55
Magnetizing current and hysteresis current are the two com-
ponents of the exciting current.
If the circuit contains^ besides the reactance x = 2 ~/L, a resistance r, the e.m.f. OE" = E" in the preceding Figs. 22 and
23 is not the impressed e.m.f., but the e.m.f. consumed by selfinductance or reactance, and has to be combined, Figs, 24 and
25, with the e.m.f. consumed__by the resistance, QE' = E' = Ir, to get the impressed e.m.f. OE = E.
Due to the hysteretic lead a, the lag of the current is less in
Figs. 23 and 25, a circuit expending energy in molecular magnetic friction, than in Figs. 22 and 24, a hysteresisless circuit.
Fig. 23. Angle of Hysteretic Lead.
Fig. 24. Effective Resistance on Phase Relation of Impressed e.m.f in a Hysteresisless Cir-
cuit.
As seen in Fig. 25, ir a circuit whose ohmic resistance is not negligible, the hysteresis current and the magnetizing current are not in phase and in quadrature respectively with the im-
pressed e.m.f., but with the counter e.m.f. of inductance or e.m.f.
consumed by inductance.
Obviously the magnetizing current is not quite wattless, since energy is consumed by this current in the ohmic resistance of
the circuit.
.
E OE Resolving, in Fig. 26, the impressed e.m.f.
=
into two
OW components,
E l
B
J1
in
phase,
and
OE 2
-
2 in quadrature
with
the
current
01
= /,
the
power
component
of
the
e.m.f.,
56
ELEMENTS OF ELECTRICAL ENGINEERING.
W E OE = l
v is greater than
Ir, and the reactive component
W OE = E is less than
= /x.
2
2
Fig. 25. Effective Resistance on Phase Relation of Impressed e.m.f. in a Circuit
having Hysteresis.
Fig. 26. Impressed e.m.f. Resolved into Components in Phase and in Quadrature with the Exciting
Current.
The value r'
___ power
e.m.f. .
is
called
the effective resist-
total current
ance and the value xf t /
wattless e.m.f. . is called the ap-
total current
parent or effective reactance of the circuit.
50. Due to the loss of energy by hysteresis (eddy currents, etc.), the effective resistance differs from, and is greater than, the ohmic resistance, and the apparent reactance is less than the
true or inductive reactance.
The loss of energy by molecular magnetic friction per cubic centimeter and cycle of magnetism is approximately
W=* 1 - 6 ^(B ,
where (B
W
the magnetic flux density, in lines per sq. cm.
energy,
in
absolute
units
or
ergs
per
cycle
(=
7
10~~
watt-seconds or joules), and TJ is called the coef-
ficient of hysteresis.
X In soft annealed sheet iron or sheet steel, >? varies from 0.75
10"3
to
2.5
X
10~3 ,
and
can
in
average,
for
good
material,
be
assumed as 2.00 X 10'3.
HYSTERESIS AND EFFECTIVE RESISTANCE.
57
r
The
loss
of
power
in
the
volume,
I ,
at
flux
density
(B
and
frequency /, is thus
p = X l - fl r/j<B
10"7 , in watts,
and, if I = the exciting current, the hysteretic effective resist-
ance is
If the flux density, (B, is proportional to the current, /, sub-
stituting for <B, and introducing the constant k, we have
that is, the effective hysteretic resistance is inversely proportional to the 0.4 power of the current, and directly proportional to the frequency.
51. Besides hysteresis, eddy or Foucault currents contribute to the effective resistance.
Since at constant frequency the Foucault currents are proportional to the magnetism producing them, and thus approximately proportional to the current, the loss of power by Foucault currents is proportional to the square of the current, the same as the ohmic loss, that is, the effective resistance due to Foucault currents is approximately constant at constant frequency, while that of hysteresis decreases slowly with the current.
Since the Foucault currents are proportional to the frequency, their effective resistance varies with the square of the frequency, while that of hysteresis varies only proportionally to the fre-
quency.
The total effective resistance of an alternating-current circuit
increases with the frequency, but is approximately constant, within a limited range, at constant frequency, decreasing somewhat with the increase of magnetism,
EXAMPLES.
A 52.
reactive coil shall give 100 volts e.m.f. of self-induc-
tance at 10 amperes and 60 cycles. The electric circuit consists
O of 200 turns (No. 8 B. & SO
0.013 sq. in.) of 16 in. mean
58
ELEMENTS OF ELECTRICAL ENGINEERING.
length of turn. The magnetic circuit has a section of 6 sq. in.
and a mean length of 18 in. of iron of hysteresis coefficient
= X TI
2.5
10~3 . An air gap is interposed in the magnetic cir-
cuit, of a section of 10 sq. in. (allowing for spread) , to get the
desired reactance.
How long must the air gap be, and what is the resistance, the
reactance, the effective resistance, the effective impedance, and
the power-factor of the reactive coil?
The coil contains 200 turns each 16 in. in length and 0.013
X sq. in. in cross section. Taking the resistivity of copper as 1.8
10~6 the resistance is ,
1.8 X 10- 6 X 200 X 16 0.013 X 2.54
0.175 ohm,
where 2.54 is the factor for converting inches to centimeters.
(1 inch = 2.54 cm.) E Writing = 100 volts generated, / = 60 cycles per second,
and n = 200 turns, the maximum magnetic flux is given by E - 4.44 fn$; or, 100 - 4.44 X 0.6 X 200*, and $ - 0.188
megaline.
This gives in an air gap of 10 sq. in, a maximum density
= (B
18,800 lines per sq. in., or 2920 lines per sq. cm.
Ten amperes in 200 turns give 2000 ampere-turns effective or
F = 2830 ampere-turns maximum.
Neglecting the ampere-turns required by the iron part of the
magnetic circuit as relatively very small, 2830 ampere-turns
have to be consumed by the air gap of density (B = 2920.
Since
the length of the air gap has to be
47rS 4 TT X 2830
With a cross section of 6 sq. in. and a mean length of 18 in.,
the volume of the iron is 108 cu. in., or 1770 cu. cm.
= The density in the iron, ffi,
= noo
'
31,330 lines per
sq. in., or 4850 lines per sq. cm.
HYSTERESIS AXD EFFECTIVE RESISTANCE.
59
With
an
hysteresis
coefficient
r = 2.5 X 10~ 3
y
,
and
= (B 1
4850, the loss of energy per cycle per cu. cm. is
W=
6
ryCB^-
= 2.5 X 10~3 X 48501 - 6
= 1980 ergs,
density
and the hysteresis loss at/ = 60 cycles and the volume F 1770
is thus
p = 60 X 1770 X 1980 ergs per sec. - 21.0 watts,
which at 10 amperes represent an effective hysteretic resistance,
W= *>1
r
"-
= ' 21 ohm -
Hence the total effective resistance of the reactive coil is
r
=
r t
+
r 2
=
0.175
+ 0.21
=
0.385
ohm;
the effective reactance is
x = ~ =10 ohms;
the impedance is
z = 10.01 ohms;
the power-factor is cos
- = 3.8 per cent;
the total apparent power of the reactive coil is Pz 1001 volt-amperes,
and the loss of power,
Pr = 38 watts.
ii. CAPACITY AND CONDENSERS.
53. The charge of an electric condenser is proportional to the
impressed voltage, that is, potential difference at its terminals, and to its capacity.
A condenser is said to have unit capacity if unit current exist-
ing for one second produces unit difference of potential at its
terminals.
The practical unit of capacity is that of a condenser in which
60
ELEMENTS OF ELECTRICAL ENGINEERING.
one ampere during one second produces one volt difference of
potential.
The
practical
unit
of
capacity
equals
9
10~~
absolute
units.
It
is called a farad.
One farad is an extremely large capacity, and therefore one
millionth of one farad, called microfarad, mf., is commonly
used.
If an alternating e.m.f. is impressed upon a condenser, the charge of the condenser varies proportionally to the e.m.f., and thus there is current to the condenser during rising and from the condenser during decreasing e.m.f., as shown in Fig. 27.
Fig. 27. Charging Current of a Condenser across which an Alternating e.m.f. is Impressed.
That is, the current consumed by the condenser leads the
impressed e.m.f. by 90 time degrees, or a quarter of a period.
E Denoting / as frequency and as effective alternating e.in.f . impressed upon a condenser of C mf. capacity, the condenser is
charged and discharged twice during each cycle, and the time
of one complete charge or discharge is therefore
Since E \/2 is the maximum voltage impressed upon the condenser, an average of CE \/2 10~6 amperes would have to
exist during one second to charge the condenser to this voltage,
_ and to charge it in r seconds an average current of 4 fCE
V2 10~6 amperes is required.
.
Since
effective current average current
n 2 \/2
the
effective
current
7
=
2
itfCE
10~6 ;
that
is,
at
an
impressed
CAPACITY AND CONDENSERS.
61
E C e.m.f. of effective volts and frequency/, a condenser of inf.
capacity consumes a current of
7 = 2 r.fCE ICT6 amperes effective,
which current leads the terminal voltage by 90 time degrees or a quarter period.
Transposing, the e.m.f. of the condenser is
The value
X Q
~-
2 7T/C
is
called
the condensive reactance of
the
condenser.
Due to the energy loss in the condenser by dielectric hysteresis, the current leads the e.m.f. by somewhat less than 90 time
degrees, and can be resolved into a wattless charging current and
a dielectric hysteresis current, which latter, however, is generally
- -- m so small as to be negligible. 54. The capacity of one wire of a transmission line is
Cn = X X 1.11
10~6
r-;
Z
.
,
,
mf.,
where Id = diameter of wire, cm.; ls = distance of wire from
X return wire, cm.; I = length of wire, cm., and 1.11
10~6 *
reduction coefficient from electrostatic units to mf.
O-M -- The logarithm is the natural logarithm; thus in common
logarithms, since log a = 2.303 Iog 10 a, the capacity is
nC =
x 10~6 x l i-n.mff.
,
The derivation of this equation must be omitted here.
The charging current of a line wire is thus
= 7
2
TtfCE
10-6 ,
E where/ = the frequency, in cycles per second, = the difference = of potential, effective, between the line and the neutral (E
i line voltage in a single-phase, or four-wire quarter-phase
V3 Y system, zrline voltage, or voltage, in a three-phase system).
62
ELEMENTS OF ELECTRICAL ENGINEERING.
EXAMPLES.
53. In the transmission line discussed in the examples in
37, 38, 41 and 47, what is the charging current of the line at 6000
volts between lines, at 33.3 cycles? How many volt-amperes
does it represent, and what percentage of the full load current of
44 amperes is it?
The length of the line is, per wire,
I = 2.23 X 106 cm.
The distance between wires,
ls = 45 cm.
The diameter of transmission wire, Id = 0.82 cm.
Thus the capacity, per wire, is
C = vX -9"41
*ift-e L/
7
logel
Id
The frequency is The voltage between lines,
= 0.26 mf.
/ = 33.3,
6000.
Thus per line, or between line and neutral point,
* -6^-3460 voter
V3
hence, the charging current per line is
7 = 2 xfCE 10 6 = 0.19 amperes,
or
0.43 per cent of full-load current;
that is, negligible in its influence on the transmission voltage. The volt-ampere input of the transmission is,
3 IJE = 2000 = 2.0 kv-amp.
12. IMPEDANCE OF TRANSMISSION LINES.
56. Let r = resistance; x = 2 *fL = the reactance of a trans-
E = mission line; Q the alternating e.m.f. impressed upon the line;
E / = the line current;
the e.m.f. at receiving end of the line,
and 6 = the angle of lag of current / behind e.m.f. E.
8 < thus denotes leading, 6 > lagging current, and 6 =
a non-inductive receiver circuit.
IMPEDANCE OP TRANSMISSION LINES.
63
The capacity of the transmission line shall be considered as
negligible.
Assuming the phase of the current 07 = 7 as zero in the polar
diagram, Fig. 28, the e.m.f. E is represented by vector OE, ahead
of 01 by angle 0. The e.m.f. consumed
OE by resistance r is
= E = 7r in
1
l
phase with the current, and the e.m.f.
consumed
by
reactance x is
OE = 2
E 2
=
Ix,
90 time degrees
ahead of the
current; thus the total e.m.f. con-
sumed by the line, or e.m.f. consumed
by impedance, is the resultant OE% of
OE and OE
E and is
= Iz.
{
2)
3
Combining
OE Z
and
OE
gives
,
the e.m.f. impressed upon the line.
Fig. 28. Polar Diagram of Current and e.m.fs. in a Transmission Line Assuming Zero
Capacity.
Denoting tan t = - the time angle of lag of the line impe-
dance, it is, trigonometrically,
OE*
- OE*
+
2
BE,
-
2
EE Q
cos
OEEQ .
Since
EE - OE, Iz,
OEE,
-
180
-
6 1
+
0,
we have
= E2 +
and
/) ^^ n
4 EIz sin3 L~ >
2t
and the drop of voltage in the line,
7/7 jG/
/ Cj \/ """""
TJ1
s==;
4/
/ Tr I/-/
7
I //w\2
"T" LZ)
E 57. That is, the voltage Q required at the sending end of a
line of resistance r and reactance or, delivering current 7 at voltage E, and the voltage drop in the line, do not depend upon current and line constants only, but depend also upon the angle of time-phase displacement of the current delivered over the line.
64
ELEMENTS OF ELECTRICAL ENGINEERING.
= If 8 0, that is, non-inductive receiving circuit,
(E
^ + that is, less than
Iz, and thus the line drop is less than Iz.
E E If Q = iy
is a maximum, =
+ Iz, and the line drop is the
impedance voltage.
E With decreasing 0, Q decreases, and becomes
E', that is,
no drop of voltage takes place in the line at a certain negative
Fig. 29. Locus of the Generator and Receiver e.m.fs. in a Transmission Line with Varying Load Phase Angle.
E value
of
6 which
depends
not
only on z and
O t
but
on
and /.
E Beyond this value of 0, Q becomes smaller than E; that is, a
rise of voltage takes place in the line, due to its reactance. This
can be seen best graphically;
Choosing the current vector 01 as the horizontal axis, for
E the sameje.m.f. received, but different phase angles 0, all
OE vectors
lie on a circle e with as center. Fig. 29. Vector
OE B is constant for a given line and given current I.
IMPEDANCE OF TRANSMISSION LINES.
65
E Since EJE^ = OE = constant, E Iks on a circle e with 3 as center and OE = E an radius.
OE To
const
the
ruct_
diagram
for
angle
7
is drawn at the
M" angle with 01, and
OE parallel to
3.
EE The distance
OE Q between the two circles on vector
Q is
the drop of voltage (or rise of voltage) in the line.
Fig. 30. Locus of the Generator and Receiver e.m.fs. in a Transmission Line with. Varying Load Phase Angle.
OE As seen in Fig. 30, J5? is maximum in the direction
3 as
= 0$ that is for
6 Q,
and
is
less
for
greater
as well,
", as smaller
E OE = angles 0. It is
in the direction
Q"', in which case 9 < 0,
and
minimum
in
the
direction
OE IV Q.
W S E The values of
corresponding to the generator voltages
/, EQ", EJ", E are shown by the points
E E" rn IV
66
ELEMENTS OF ELECTRICAL ENGINEERING.
E E respectively. The voltages
" and
Q
correspond to a watt-
E less receiver circuit E" and
JV
.
For non-inductive receiver
circuit ~OEv the generator voltage is OE*.
58. That is, in an inductive transmission line the drop of
voltage is maximum and equal to Iz if the phase angle of the
receiving circuit equals the phase angle
of the line.
Q
The
drop of voltage in the line decreases with increasing difference
between the phase angles of line and receiving circuit. It be-
comes zero if the phase angle of the receiving circuit reaches a
certain negative value (leading current). In this case no drop
of voltage takes place in the line. If the current in the receiving
circuit leads more than this value a rise of voltage takes place
in the line. Thus by varying phase angle of the receiving
circuit the drop of voltage in a transmission line with cur-
rent 7 can be made anything between Iz and a certain negative
value. Or inversely the same drop of voltage can be produced
for different values of the current / by varying the phase
angle.
Thus, if means are provided to vary the phase angle of the receiving circuit, by producing lagging and leading currents at will (as can be done by synchronous motors or converters) the
voltage at the receiving circuit can be maintained constant within a certain range irrespective of the load and generator
voltage.
E In Fig. 31 let OE = } the receiving_voltage; 7, the power
component
of
the
line
current;
thus
OE^
=
E^
= Iz,
the
e.m.f.
consumed by the power component of the current in the impe-
dance. This e.m.f. consists of the e.m.f. consumed by resist-
ance "OE^ and the e.m.f. consumed by reactance OE 2 .
Reactive components of the current are represented in the
diagram in the direction OA when lagging and OB when leading.
The e.m.f. consumed by these reactive components of the current
W OE in the impedance is thus in the direction <?/, perpendicular to 3 .
Combining
0$ and 3
gives
the
e.m.f.
OE 4
which
would
be
E required for non-inductive load . If JB is the generatorvoltage, Q
lies on a circle^ with OE as radius. Thus drawing EJ$ Q parallel
Ol 0$ E E to e,' gives
, the generator voltage;
'=
the e.m.f.
3
4
,
consumed in the impedance by the reactive component of the
F current; and as proportional thereto, OP =
the reactive
IMPEDANCE OF TRANSMISSION LINES.
67
current required to give at generator voltage J ami power
current / the receiver voltage E. This reactive current /', lags
E behind
f z
by less
than 90 and
more
than zero degree?.
59. In calculating numerical values, we can proceed either
trigonometrically as in the preceding, or algebraically by resolv-
Fig. 31. Regulation Diagram for Transmission Line.
ing all sine waves into two rectangular components; for instance,
a horizontal and a vertical component, in the same way as in mechanics when combining forces.
Let the horizontal components be counted positive towards the right, negative towards the left, and the vertical components positive upwards, negative downwards.
Assuming the receiving voltage as zero line or positive horizontal line, the power current / is the horizontal, the wattless
68
ELEMENTS OF ELECTRICAL ENGINEERING.
current /' the vertical component of the current. The e.m.f. con-
sumed in resistance by the power current / is a horizontal
component, and that consumed in resistance by the reactive
F current
a vertical component, and the inverse is true of the
e.m.f. consumed in reactance.
We have thus, as seen from Fig. 31.
Receiver voltage, E,
Power current, I,
HORIZONTAL VERTICAL COMPONENT, COMPONENT,
++ E10
Reactive current, I',
7'
E.m.f. consumed in resistance r by the
power current, 7r,
+ Ir
E.m.f. consumed in resistance r by the
reactive current, 7'r,
I'r
E.m.f. consumed in reactance x by the
power current, Ix,
Ix
E.m.f. consumed in reactance x by the
reactive current, I'x,
Fx
IrPx Thus, total e.m.f. required, or impressed
E e.m.f., w
E+
- I'r Ix]
hence, combined,
E V(E+ = Q
Ir
+ 2
I'x)
(
-2
I'r Ix)
or, expanded,
E VE E -
+ + + 2
2
(Ir
I'x)
22
(P 7' )z .
From this equation /' can be calculated; that is, the reactive
current
found
which
is
required
to
give
E Q
and
E
at
energy
current 7.
The" lag of the total current in the receiver circuit behind the
receiver voltage is
tan =
The lead of the generator voltage ahead of the receiver voltage
is
t
A_
vertical
component
of
E Q
E 1
horizontal component of Q
- I'r Ix
E+ ' Ir I'x
IMPEDANCE OF TRANSMISSION LINES.
69
and the lag of the total current behind the generator voltage is
=
0,
+ 0,.
As seen, by resolving into rectangular components the phase
angles are directly determined from these components.
The resistance voltage is the same component as the current to which it refers.
The reactance voltage is a component 90 time degrees ahead
of the current.
The same investigation as made here on long-distance trans-
mission applies also to distribution lines, reactive coils, trans-
formers, or any other apparatus containing resistance and reactance inserted in series into an alternating-current circuit.
EXAMPLES.
60. (1.) An induction motor has 2000 volts impressed upon
its terminals; the current and the power-factor, that is, the cosine of the angle of lag, are given as functions of the output
in Fig. 32.
ouurrppiur KW.
10 20 30 40 50 00 70 80 00 100 110 120 130 140 150 100 170 180 100 200
Fig. 32. Characteristics of Induction Motor and Variation of Generator e.m f. Necessary to Maintain Constant the e.m.f. Impressed upon the Motor.
The induction motor is supplied over a line of resistance
r = 2.0 and reactance x = 4.0.
(a.)
How
must
the
generator
voltage
e Q
be
varied
to
maintain
constant voltage e = 2000 at the motor terminals, and
70
ELEMENTS OF ELECTRICAL ENGINEERING.
(6.) At constant generator voltage eQ = 2300, how will the
voltage at the motor terminals vary?
We have
vA + e
izf
4
eiz
2
sin
2000.
4.472.
6 = 63.4. 1
cos 6 = power-factor.
Taking i from Fig. 32 and substituting, gives (a) the values
= of e for e
2000 ;
which
are
recorded
in
the
table,
and
plotted
in
Fig. 32;
(b.) The terminal voltage of the motor is e = 2000, the cur-
rent i, the output P, at generator voltage ey Thus at generator
'
voltage e
=
2300,
the
terminal
voltage
of
the
motor is
the current is
So
o
.,
2300 .
and the power is
p, _
P The values of e', i', T are recorded in the second part of the
table under (6) and plotted in Fig. 33.
IMPEDANCE OF TRANSMISSION LINES.
71
61. (2.) Over a line of resistance r
2.0 and reactance x =
6.0 power is supplied to a receiving circuit at a constant voltage
of e = 2000. How must the voltage at the beginning of the
line,
or
generator voltage,
e
Q,
be
varied
if
at
no
load
the
receiving
circuit
consumes
a
reactive
current
of
i = 20 2
amperes,
this
reactive current decreases with the increase of load, that is, of
power current iv becomes i2 =
at
i 1
=
50
amperes,
and
then
as leading current increases again at the same rate?
The reactive current,
= = i
20 at i
2
t
0,
i= z
at ^ = 50,
and can be represented by
--^)20
=
20
-0.4i t
72 ELEMENTS OF ELECTRICAL ENGINEERING.
the general equation of the transmission line is
=V (e
- (ijr
i
V =
- + 2^ (2000
+ -f 6? 2) 3
(2i 2
2
eij ;
hence,
substituting
the
value
of
i
2,
V - - e = Q
(2120
+ 2
0.4 ij
(40
6.8
3
ij
=
\/4 496 000
7
;
+46.4v
-2240ir
Substituting successive numerical
values
for i l
gives
the values
recorded in the following table and plotted in Fig. 34.
POWEf 3 ap K
Fig. 34. Variation of Generator e.m.f. Necessary to Maintain Constant Receiver Voltage if the Reactive Component of Receiver Current varies proportional to the Change of Power Component of the Current.
ALTERNATING-CURRENT TRANSFORMER.
73
13. ALTERNATING-CURRENT TRANSFORMER.
62. The alternating-current transformer consists of one mag-
netic circuit interlinked with two electric circuits, the primary circuit which receives energy, and the secondary circuit which
delivers energy.
Let
T\
= resistance,
x l
=
2 r/L S3
=
self-inductive or leakage
reactance of secondary circuit,
r = resistance, J = 2 z/L n = self-inductive or leakage
reactance of primary circuit,
where
L and S2
L 8l
refer
to
that
magnetic
flux
which
is
inter-
linked with the one but not with the other circuit.
= -Mums rLetj. a
^ -secondary,
ratio or
:
,
,*
<.
,
f
,.
v
(ratio of transformation).
primary
An alternating e.m.f. J impressed upon the primary electric
circuit causes a current, which produces a magnetic flux $ inter-
linked with primary and secondary circuits. This flux <I> gen-
E E erates e.m.fs.
and
l
i in secondary and in primary circuit,
^= which are to each other as the ratio of turns, thus
"P
- -
*
a
Let
E
=
secondary
terminal
voltage,
7 t
=
Secondary
current,
E 1 = lag of current 7j behind terminal voltage
(where
O l
<
denotes leading current).
Denoting then in Fig. 35 by a vector (IB = E the secondary
terminal
voltage,
01 i
=
/ x
is
the
secondary
current
lagging
by
the angle EOI - dr
The
e.m.f.
consumed
by
the
secondary
resistance r t
is
OE^
=
EI
=
Ir ll
in
phase
with
7r
The
e.m.f.
consumed
by
the
secondary
reactance x l
is
OE"
=
E" = I 1x v 90 time degrees ahead of 7 r Thus the e.m.f. con-
sumed
by
the
secondary
impedance
= Vr z l
x 2
2
-It
t
is
the
resultant
of
OF/
and
MS?,
or
77
OF/
-
E,"'
=_!&.
OE,"' combined with the terminal voltage OE = E gives the
secondary
e.m.f.
OE l
==
Er
Proportional thereto by the ratio of turns and in phase there-
with is the e.m.f. generated in the primary OEi = Ei where
*-.
74 ELEMENTS OF ELECTRICAL ENGINEERING.
To generate e.m.f. E^ and E^ the magnetic flux 0<f> = $ is
required,
90
time
degrees
ahead
of
OE l
and
OE^
To produce
flux < the m.m.f. of $ ampere-turns is required, as determined
from the dimensions of the magnetic circuit, and thus the
primary
current
7 00 ,
represented
by
vector
07 00 ,
leading
0$
by
the angle a.
Since the total m.m.f. of the transformer is given by the
primary
exciting
current
7 00 ,
there
must
be
a
component
of
Pig. 35. Vector Diagram of e.m.s. and Currents in a Transformer.
primary current
7',
corresponding
to
the
secondary
current
7 1?
which may be called the primary load current, and which is
opposite thereto and of the same m.m.f.; that is, of the intensity
I' =JiIv thus represented by vector OF 7' = alv
0/ oo;
the
primary
exciting
current,
and
the
primary
load
current OP, or component of primary current corresponding
to thejjecondary current, combined, give the total primary cur-
rent 07 = 7 .
The e.m.f. consumed by resistance in the primary is OEJ = EJ = 7 r in phase with 7 .
The
e.m.f.
consumed
by
the
primary
reactance
is
OE " Q
=
Ef
=7
x Q,
90
time
degrees
ahead
of
07
.
OE ' Q
and
OE"
combined
gives
OS/",
the
e.m.f.
consumed
by
the primary impedance.
ALTERNATING-CURRENT TRANSFORMER.
75
Equal and opposite to the primary counter-generated e.m.f.
OEl is the component of primary e.m.f., 0", consumed thereby.
OE' combined with OE"f gives QE"^ = E^ the primary
impressed
e.m.f. ,
and
angle #
= #/)/, the phase angle of the
primary circuit.
Figs. 36, 37, and 38 give the polar diagrams for
= 45 or
X
= lagging
current,
O l
zero or non-inductive circuit, and 6 =
45 or leading current.
Fig. 36. Vector Diagram of Transformer with Lagging Load Current.
E 63. As seen, the primary impressed e.m.f. required to proE duce the same secondary terminal voltage at the same current
7 1
is
larger
with
lagging
or
inductive
and
smaller
with
leading
current than on a non-inductive secondary circuit; or, inversely,
at
the
same
secondary
current
I I
the
secondary
terminal
voltage
E with lagging current is less and with leading current more,
than with non-inductive secondary circuit, at the same primary
E impressed e.m.f. .
The calculation of numerical values is not practicable by
measurement from the diagram, since the magnitudes of the
E different quantities are too different, Ef: Ef: E^. Q being
frequently in the proportion 1 : 10 : 100 : 2000.
76
ELEMENTS OF ELECTRICAL ENGINEERING.
Trigonometrieally, the calculation is thus :
OEE In triangle
V Fig. 35, writing
we have,
also,
hence,
OE?
=
OE2
+
EE\
-20EEE eos l
WE, - 7A
t = 180 - 0' + 0^
=
E*
+
2
J/^
+
2
EI.z, cos
(5'
Eig. 37. Vector Diagram of Transformer with Non-Inductive Loading.
This gives the secondary e.m.f., E v and therefrom the primary
counter-generated e.m.f.
In
triangle
EOE l
we
have
sin Ef>E
-r
sin E EO t
=
EE l
~
W^C
thus, writing
we have
- = /^ sin 6" -f- sin (0'
X)
-f- J?
1;
wherefrom we get
= = + 0*,
and
# jE 0/
1
1
/7
1
,
ALTERNATING-CURRENT TRANSFORMER
77
the phase displacement between secondary current and secondary e.m.L
In triangle 0/00/ we have
- 2
*
2 0/ 00/ 00/ cos
since
and
E><i> = 90, 0/00/ = 90 +
+ a,
= 07 00
7
00
==
exciting
current,
Fig. 38. Vector Diagram of Transformer with Leading Load Current.
calculated from the dimensions of the magnetic circuit.
the primary current is
V V ay =
+ + 2 a'/,
2
+ oo sin (0
a).
In
triangle
0/ 7 00
we have
sin
7 0/ 00
0/ -s- sin
007
=7 7 00
0/ -f;
writing
7 07 00
this becomes
sn
sn -T-
Q
+ <*
4- 7
t
;
therefrom we get 6", and thus
4 E'Ol*
=
^ 2
=
90
-a -
".
In triangle OE'E^ we have
QE* -
+ - E /2
2
S'jBo
2 OJS' JB'So cos OE' Q
Thus
78
ELEMENTS OF ELECTRICAL ENGINEERING.
writing
we have
^ = tan '
y
T Q
4 OE'E, - 180 - ff + Ov
OE'
E. Ei == -f- '
thus the impressed e.m.f. is
% 2 =
^o Ci
In
triangle
OE'E 9
sin S'OSo -^ sin
thus, writing
E'OE. =
we have
^ sin 0/' sin
herefrom we get #/', and
the phase displacement between primary current and impressed
e.m.f.
As
seen ;
the
trigonometric
method
of
transformer
calculation
is rather complicated.
64. Somewhat simpler is the algebraic method of resolving
into rectangular components.
Considering
first
the
secondary
circuit,
of
current
I t
lagging
behind the terminal voltage E by angle # r
E E The terminal voltage
has the components
cos
# x
in
phase,
E sin
9 i
in
quadrature
with
and
ahead
of
the
current
/r
The e.m.f. consumed by resistance r v I lr v is in phase.
The e.m.f. consumed by reactance x v I lx v is in quadrature
ahead of 7 r
Thus the secondary e.m.f. has the components
E + cos d l
Ir ll
in
phase,
/^ E sin + i
in quadrature ahead of the current Iv and
the total value,
cos + /)* + (S sin tf + /x) 2
ALTERNATING-CURRENT TRANSFORMER.
T9
and the tangent of the phase angle of the secondary circuit is
E ^ E + ,
-
= tan
sin 6-.1
Lx.
*
cos
O l
+
//i
Resolving all quantities into components in phase and in
quadrature with the secondary e.m.f. E v or in horizontal and
in vertical components, choosing the magnetism or mutual flux
as vertical axis, and denoting the direction to the right and upwards as positive, to the left and downwards as negative, we
have
HORIZONTAL COMPONENT.
VERTICAL COMPONENT.
Secondary current, I v
Secondary e.m.f., Ev
7 cos 6 l
E t
7 sin 9 t
Primary counter-generated e.m.f.,
T,7
pJfo i
-. -_&_,i_
;
a
"C1
&yL,
yn
a
Primary e.m.f. consumed thereby,
= jT&JV
77?
jCn,
Primary load current,- 7'
a/ 17
Magnetic flux, <,
Primary
exciting
current,
7 00 ,
con-
sisting of core loss current,
Magnetizing current,
~l
t
T"
a
+ a7 cos 6 t
700 sin a
AU
+ a7 sin 6 x
$
700 cos a
Hence, total primary current, 7 ,
HORIZONTAL COMPONENT.
+ a7 cos d
1
l
700 sin a
VERTICAL COMPONENT.
+ a/ sin O
1
l
700 cos a
E E.m.f. consumed by primary resistance r , '
with 7 ,
7 r in phase
HORIZONTAL COMPONENT,
VERTICAL COMPONENT.
+ r a7 cos 6 1
r 7 sin a 00
+ r al sin t
r
7
00
cos
a
E E.m.f.
consumed
by
primary
reactance
rc ,
= Q
7 x 90 ,
ahead of 7 ,
HORIZONTAL COMPONENT.
+ X a7 sin /? t
x
7 00
cos
a
VERTICAL COMPONENT.
x a7 cos 9 1
x /00 sin a
E E.m.f. consumed by primary generated e.m.f., f
L
a
horizontal.
80
ELEMENTS OF ELECTRICAL ENGINEERING.
E The total primary impressed e.m.f., Q ,
PI
+ a/ 1
(r
cos
ci
HORIZONTAL COMPONENT.
+X
sin 6)
+7 00
(>
sin a
+x
cos a).
a/ x
(r
sin
VERTICAL COMPONENT.
+ x cos 5)
/00 (r cos a
X Q
sin
a),
or writing since
tan
'= ? r o
_____________
Vr 2 + x 2 = 2 sin 07
= <^
' -^ and cos , #o
=^
'-
o
E Substituting this value, the horizontal component of is
^ O + a^J, cos ( -
+ ^oo sin (a +(?/);
tt
the vertical component of J5 is
og^ - + + sin (6
d,'}
z /00 cos (a
fl/),
and, the total primary impressed e.m.f. is
#0= v F'~+azo/iC
^ _ -_-
Combining the two components, the total primary current is
J
v + + + / (a/ 1 cos ^
/
00
sin
3
a)
(al sin <?
7 00
cos
a)
1+
+ sin (ff
.,
Since the tangent of the phase angle is the ratio of vertical com-
ponent to horizontal component, we have, primary e.m.f. phase,
tan ff =
- + + a*Ji sin (6
6 n')
zjnn cos (a
fl/)
| - + azj, cos ((?
+ - ')
/ g( 00 sin (a
')
primary current phase,
tan p
+ a/i sin fl
7 0fL cos5^
+ a/ cos 6 x
/ 00 sin a:
and lag of primary current behind impressed e.m.f.,
ALTERXATIXG-CURREXT TRANSFORMER.
81
EXAMPLES.
^ 65. (1.) In a 20-kw. transformer the ratio of turns is 20 1,
and 100 volts is produced at the secondary terminals at full
load. What is the primary current at full load, and the regu-
lation, that is, the rise of secondary voltage from full load to no load, at constant primary voltage, and what is this primary
voltage?
(a) at non-inductive secondary load, (6) with 60 degrees time lag in the external secondary circuit, (c) with (30 degrees time lead in the external secondary circuit.
The exciting current is 0.5 ampere, the core loss 600 watts,
the primary resistance 2 ohms, the primary reactance 5 ohms,
the secondary resistance 0.004 ohm, the secondary reactance 0.01 ohm.
Exciting current and core loss may be assumed as constant.
600 watts at 2000 volts gives 0.3 ampere core loss current,
hence Vo.52 3J = 0.4 ampere magnetizing current.
We have thus
X =5 Q
= 0.004 - 0.01
= /00 cos a
0.3
700 sin a
0.4
- /oo 0.5
a = 0.05
1. Secondary current as horizontal axis:
82
ELEMENTS OF ELECTRICAL ENGINEERING.
2. Magnetic flux as vertical axis :
Hence,
RECTANGULAR COORDINATES.
83
14. RECTANGULAR COORDINATES.
66. The polar diagram of sine waves gives the best insight into the mutual relations of alternating currents and e.m.fs.
For numerical calculation from the polar diagram either the trigonometric method or the method of rectangular components
is used.
The method of rectangular components, as explained in the above paragraphs, is usually simpler and more convenient than
the trigonometric method.
In the method of rectangular components it is desirable to
distinguish the two components from each other and from the
resultant or total value by their notation.
To distinguish the components from the resultant, small
letters are used for the components, capitals for the resultant. Thus in the transformer diagram of Section 13 the secondary
current
I t
has
the
horizontal
component
i
l
^ cos O v and
the vertical component i/ ==
I sin I
r
To distinguish horizontal and vertical components from each
other, either different types of letters can be used, or indices,
or a prefix or coefficient.
Different types of letters are inconvenient, indices distinguish-
ing the components undesirable, since indices are reserved for
distinguishing different e.m.fs., currents, etc., from each other.
Thus the most convenient way is the addition of a prefix or
coefficient to one of the components, and as such the letter j is
commonly used with the vertical component.
Thus the secondary current in the transformer diagram,
Section 13, can be written
+ A h
=
ft*
+ cos i
fli s* n r
(!)
This method offers the further advantage that the two com-
ponents can be written side by side, with the plus sign between
them, since the addition of the prefix j distinguishes the value
j{ 2
or
j7 t
sin
O l
as
vertical
component
from
the
horizontal
com-
ponent
i t
or
7 X
cos
r
/i =*i +Jh
(2)
thus
means
that
7 t
consists
of
a
horizontal
component
i t
and
a
vertical
component
i
2,
and
the
plus
sign
signifies
that
\
and
i
2
are
combined by the parallelogram of sine waves.
84
ELEMENTS OF ELECTRICAL ENGINEERING.
The secondary e.m.f. of the transformer in Section 13, Pig. 35,
E is written in this manner,
=
i
e v that is, it has the hori-
zontal component
e l
and
no
vertical
component.
The primary generated e.m.f. is
*
*
a
and the e.m.f. consumed thereby
F- + *.
(4)
The secondary current is
where
= = i l
7 X
cos
Ov
i 2
7 X
sin
6V
(6)
and the primary load current corresponding thereto is
The primary exciting current,
= where h
7
00
sin
a
is
the
hysteresis
current,
g
=*
7 00
cos
a
the
reactive magnetizing current.
Thus the total primary current is
K lo - /' + /oo - fa'i + >0 + J
+ 0).
(9)
The e.m.f. consumed by primary resistance r is
= + + + ro/o
r ai
o( i
A)
jV (ai a
0).
(10)
The horizontal component of primary current (ai\ 4- A) gives
as e.m.f. consumed by reactance x a negative vertical com-
ponent, denoted by
+ jx (ai l
A).
The vertical component
+ of primary current / (ai t gr) gives as e.m.f. consumed by reac-
tance
X Q
a
positive
horizontal
component,
denoted
by
x
(ai9 +g).
Thus the total e.m.f. consumed by primary reactance x is
x + (ai2
flf)
jx (ai t 4- A),
(11)
and the total e.m.f. consumed by primary impedance is
rfrtn x v^a at yj J_ o
ai/
\
i
JL,
i
rTif])\
'
/y (fin
ov 2
-JT_
/Tf\
J".T.
/|* Fy
j L"
^/yV
J- /i^ y/
^i y //Y1* JL /)M /*1 O^
^o
"* *vJ* \*^J
Thus, to get from the current the e.m.f. consumed in reac-
tance
X Q
by
the
horizontal
component
of
current;
the
coefficient
RECTANGULAR COORDINATES.
85
j has to be added; in the vertical component the coefficient
j omitted; or, we can say the reactance is denoted by jxQ for
X the horizontal and by -? for the vertical component of current.
1
= In other words, if /
i
-f
r
ji
is
a
current,
x
the reactance of its
circuit, the e.m.f. consumed by the reactance is
+ = jxi
xif
f
xi
jxi.
67. If instead of omitting j in deriving the reactance e.m.f.
for the vertical component of current, we would add j also (as done when deriving the reactance e.m.f. for the horizontal component of current), we get the reactance e.m.f.
jxi fxi'j
+ which gives the correct value
jxi
xi! if ',
f = -1;
(13)
that is, we can say, in deriving the e.m.f. consumed by reactance,
x, from the current, we multiply the
substitute f =
1.
By defining, and substituting, f =
current by jx, and 1, jx can thus be
called the reactance in the representation in rectangular coor-
dinates and r jx the impedance,
The primary impedance voltage of the transformer in the
preceding could thus be derived directly by multiplying the
current,
+ + = 7
(ai,
K) + j(ai2
g), (9)
by the impedance,
Z =r Q
jx
which gives
- [K = -
o' Z*! o fro Fo)
+ + + h) j (ai, g)]
- + + + - - + f (ai t
h)
j> (ai 2
g}
jx Q (ai t -f h)
fx, (at, g),
and substituting f
1,
Ed = + + + + - fro
ai
(
i
K)
X Q
(ai2 +g)]
j [r (ai a g)
x (at\ 4- A)],
and the total primary impressed e.m.f. is thus
^ Eo - + #o'
(14)
"
-^ + (ai2+9r) +? ro(ai2+9f)
(ai ^
J[
(15)
86 ELEMENTS OF ELECTRICAL ENGINEERING.
68* Such an expression in rectangular coordinates as
/ = i + jV
(16)
represents not only the current strength but also its phase. It means, in Fig. 39, that the
total current 01 has the two
rectangular components, the hori-
zontal component / cos
=i
and the vertical component /sin
=f i.
Thus,
f
i
Fig. 39. Magnitude and Phase in Rectangular Coordinates.
tantf=;r ?
(17)
that is, the tangent function of the
phase angle is the vertical component divided by the horizontal
component, or the term with prefix / divided by the term with-
out j. The total current intensity is obviously
VV / =
+2 i' -
(18)
The capital letter
/
in
the
symbolic expression
J
=i
+f ji
thus represents more than the / used in the preceding for total
current, etc., and gives not only the intensity but also the phase.
It is thus necessary to distinguish by the type of the latter the
capital letters denoting the resultant current in symbolic expres-
sion (that is, giving intensity and phase) from the capital letters
giving
merely
the
intensity
regardless
of
phase;
that
is
;
7
+ i
ji'
denotes a current of intensity
and phase
tan =
i
In the following, clotted italics will be used for the symbolic
expressions and plain italics for the absolute values of alternating
waves.
In the same way z = Vr2 + x2 is denoted in symbolic repre-
sentation of its rectangular components by
Z
-
r
jx.
(19)