Electrical
Drives,
Machines,
and Power Systems
/
Theodore Wildi
Electrical Machines,
and Power
Systems
Drives,
Fifth Edition
Theodore Wildi
Professor Emeritus, Laval University
Prentice
Hall
Upper Saddle
River,
New Jersey
Columbus, Ohio
Library of Congress Cataloging-in-Publication Data
page 136 by Weston Instilments; pages 204, 239, 251, 312,
by ABB; page 207
by Hammond; pages 209, 232, 777, 778, 796, 797 by
339, 344, 370, 583, 584, 626, 700, 701, 782
Theodore.
Wildi,
Electrical
machines,
Theodore Wildi.— 5th
p.
and
drives,
power systems
/
ed.
cm.
Includes bibliographical references and index.
ISBN
1.
3.
0-1 3-093083-0 (alk.
Electric
paper)
machinery.
Electric driving.
I.
2.
Electric
power systems.
Title.
TK2182.W53 2002
621.31'
042— dc21
2001051338
Editor in Chief: Stephen Helba
Assistant Vice President and Publisher: Charles
E. Stewart, Jr.
Production Editor: Alexandrina Benedicto Wolf
Design Coordinator: Diane Ernsberger
Cover Designer: Thomas Borah
Cover Art: Corbis Stock Market
Production Manager: Matthew Ottenweller
set in Times Roman and Helvetica by
Communications, Ltd. It was printed and
bound by R.R. Donnelley & Sons Company. The
cover was printed by Phoenix Color Corp.
This
book was
Carlisle
Pearson
Pearson
Pearson
Pearson
Pearson
Pearson
Pearson
Pearson
Pearson
Photo
Education Ltd., London
Education Australia Pty. Limited, Sydney
Education Singapore, Pte. Ltd.
Education North Asia Ltd., Hong Kong
Education Canada, Ltd., Toronto
Education de Mexico, SA. de C.V.
Education-Japan, Tokyo
Education Malaysia, Pte. Ltd.
Education, Upper Saddle R iver, New Jersey
credits:
Pages
21, 86, 87, 88, 107, 115, 371, 643, 640,
651, 652, 691, 701, 702, 708, 713, 731, 741, 743, 762, 773
General Electric; pages
1 1 7,
Westinghouse; pages 232, 250 by Ferranti-Paekard; pages
704 by Montel, Sprecher and Schnh; page 235 by
American Superior Electric; pages 252, 386, 644, 668, 669,
670, 689, 691, 703, 706, 708, 764, 765 by Hydro-Quebec; pages
265, 354, 392, 628, 737, 839, 840 by Lab Volt; pages 2(57, 301
by Brook Crompton-Parkinson Ltd.; page 290 by ElectroMecanik; pages 294, 295, 315, 354, 441, 442, 452, 547, 744 by
Siemens; pages 300, 301, 391, 404 by Gould; page 304 by
Reliance Electric; pages 337, 338, 340, 645 by Marine
Industrie; pages 342, 344, by Allis-Chalniers Power Systems,
Inc.; page 345 by Air France; pages 424, 425, 433 by Pacific
Scientific, Motor and Control Division, Rockford, IL; pages
425, 426 by AIRPAX Corporation; pages 440, 442, 447, 450,
452, 458, 785 by Square D; pages 440, 441, 448, 449 by
Klockner-Moeller; page 441 by Potter and Brumfield; pages
443, 451 by Telemeeanique, Group Schneider; page 454 by
Hubbel; pages 477, 493, 500 by International Rectifier; page
595 by Robicon Corporation; page 627 by Carnival Cruise
Lines; page 645 by Les Ateliers d'Ingeniere Dominion; page
647 by Tennessee Valley Authority; page 649 by Novenco,
Inc.; page 694 by Pirelli Cables Limited; page 650 by FosterWheeler Energy Corporation; page 651 by Portland General
Electric; pages 653, 654 by Electricity Commission of New
South Wales; page 658 by Connecticut Yankee Atomic
Power Company Georges Betancotirt; page 659 by Atomic
Energy of Canada; page 667 by Canadian Ohio Brass Co.,
Ltd.; page 674 by IREQ; page 700 by Canadian General
Electric; pages 703, 704, 714 by Dominion Cutout; pages 703,
704, 705, 707 by Kearney; page 731 by Sangamo; page 735 by
Gentec Inc.; page 738 by Service de la C.I.D.E.M., Ville de
Montreal; page 758 by GEC Power Engineering Limited,
England; page 759 by Manitoba Hydro; page 762 by New
Brunswick Electric Power Commission; pages 763, 764 by
United Power Association; page 772 by EPRI; page 289 by
Services Electromecaniques Roberge; page 830 by Fluke
Electronics Canada, Inc.; pages 842, 843, 844, 849 by Oinron
Canada Inc.; pages 851, 852, 853, 855 by St. Lawrence
Stevedoring; pages 854, 855, 856, 857 by Schneider Electric;
page 133 by Leroy Somer and Emerson Electric; page 436 by
233,
09, 100,
264, 300, 402, 59 1, 605, 619
by
289 by H. Roberge; pages
by Baldor Electric Company;
—
Emerson
Electric.
Copyright © 2002, 2000, 1997, 1991, 1981 by Sperika Enterprises Ltd. and published by Pearson
Education, Inc., Upper Saddle River, New Jersey 07458. All rights reserved. Printed in the United States
of America. This publication is protected by Copyright and permission should be obtained from the publisher
prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any
means, electronic, mechanical, photocopying, recording, or likewise. For information regarding
permission(s), write to: Rights and Permissions Department.
Prentice
Hall
10
98765432
ISBN 0-13-093083-0
Preface
This
fifth
edition
was prompted
in part
by the great
in-
is
crease of computers in industrial controls and au-
in
tomation,
which has produced computer programs
can simulate relays and relay contacts. These
that
no longer pertinent
isolation
to discuss
dc and ac machines
because wherever they are being
age. Consequently, the term drive
motor alone but the
now
on/off discrete controls have eliminated the wiring
the
and installation of hardware components
torque and speed of the machine. This
virtual relays
a keyboard.
and contacts
The devices
that
that
in
favor of
can be programmed on
perform these operations
are called
Programmable Logic Controllers (PLCs),
or simply
programmable
were
initially
controllers.
These devices
stand-alone computers that controlled a
specific robot or
manufacturing operation. However,
with the advent of the Internet, they have
now been
integrated with the overall manufacturing process,
leading seamlessly to integration with
sales,
procurement, and consumer satisfaction.
The
20
is
management,
relay control of
machines covered
now supplemented by coverage
Chapter 3
in
of PLC controls
Chapter 3
of
PLCs and shows, by way of example, how they
used
.
1
covers the basic principles
are
running the activities of a large service enter-
in
prise.
1
This
new
chapter illustrates
computer-based
setting
how
these trend-
activities involving controls
and automation are being integrated with other business activities, including e-commerce.
As
I
mentioned
in the last edition, similar up-
heavals have occurred in
ply
amazing
influence on the
involves not
entire unit that directs the
way
power technology. It is simof power elec-
to witness the entrance
tronics into every facet of industrial drives. Thus,
it
having a
is
machinery
electrical
courses are being taught.
How
has this dramatic change
come about?
mainly due to the high-power solid
state
It is
switching
devices, such as insulated gate bipolar transistors
(IGBTs), which can operate
at
frequencies of up to
20 kHz. The change has also been driven by
tors
and gate turn-off thyristors (GTOs)
is
thyris-
that
handle currents of several thousand amperes
ages of up to 5 kV. Another key element
Chapter
in
direct
in-
an electronic control forms part of the pack-
stalled,
can
at volt-
the
com-
puting power of microprocessors that can process
signal data in real time with incredible speed.
The high switching frequencies of IGBTs permit
the use of pulse-width-modulation techniques in
power
converters. This, in turn, enables torque and
speed control of induction motors
down
to zero
was not feasible in rectangular-wave
converters that were employed only a few years ago.
Most industrial drives are in the fractional horsepower to the 500 hp range. That is precisely the range
now available for control by IGBTs. The result has
speed. This
been an explosion
in the retrofitting
Lower maintenance
costs,
of existing drives.
higher efficiency,
and
PREFACE
economically
attractive.
pedagogical quality. As a
made such changeovers
greater productivity have
Thus, dc drives are being
placed by induction motor drives, which require less
•
Every sector of
is
•
industrial
this
revolutionary con-
verter technology. Electric elevators, electric
tives, electric transit vehicles,
and
ing, ventilating
air
locomo-
•
is
— an
power
distribution of electric
Most
I
distorted wave.
are
wave
in
Power Research
(EPR1)
Institute
in
Palo
Alto, California, in collaboration with several electrical
manufacturers, has also resulted
in
the cre-
ation of high-power static switches, thyristor-eon-
and converters
trolled series capacitors,
that
electric power.
method
that enables
into
its
Once they know how
harmonic components,
harmonics quickly
in a
unravel a
to
their interest
rises.
at all.
All the important changes first introduced in
Important development work, carried out by the
Electric
they affect the behav-
students to calculate the harmonic content
seeing large rotating machines, such as synchronous
have no moving pails
how
also devised a simple
condensers and frequency changers, being replaced by
solid-state converters that
have added a new chapter
I
generated and
and the quality of
and
we
importantly,
ics are
industry that has
been relatively stable for over 50 years. Here,
im-
this
ior of capacitors, inductors, cables, transformers,
new technology.
utilize this
also affecting the transmission
make
on harmonics. Chapter 30 reveals how harmon-
servomechanisms, heat-
conditioning systems, fans,
being modified to
The change
to
portant topic easier to understand.
compressors, and innumerable industrial production
lines are
their solutions
Chapter 7 on Active, Reactive, and Apparent
Power was completely revised
and commercial activity
by
therefore being affected
The end-of-chapter problems and
were revised and double-checked.
maintenance while offering equal and often superior
dynamic performance.
more than 20
result,
percent of the pages were altered.
re-
can
fill
previous editions have been kept
tion.
in this fifth edi-
Thus, the writing of circuit equations, the
discussion of higher frequency transformers, and
the equivalent circuit
diagram of the single-phase
induction motor have
all
been retained.
the role of phase-shift transformers.
These new methods of power flow
FACTS
by the acronym
(Flexible
control,
AC
known
Systems) will permit existing transmission and
bution lines to carry more
demand
creasing
extremely
bilize a
power
for electricity.
to
On
that
It is
sta-
is
may suddenly be menaced by an
in electric
Most students
in
many
formulating them.
I
particularly easy to follow. Readers will be
re-
minder of the circuit-solving procedure.
on
all rest
•
motor drives
is
Chapter
panded
similar to
ers.
employed to control the flow of power in elecAs a result, everything falls neatly and
coherently into place. The teaching and learning of
electric machines, drives, and power systems are
1
1
on Special Transformers was ex-
to include higher
The reader
is
frequency transform-
guided through the reasoning
that
behind the design of such transformers, and
tric utilities.
they
thereby
made much
ex-
dis-
glad to refer to this section as a convenient
base. In other words, the converter tech-
nology used
2.
such equations, but
close an ac/de circuit-solving methodology that
account of their
remarkable that these innovations
common
to solve
perience difficulty
unexpected disturbance.
a
section covering the writing of circuit
know how
distri-
can also
A new
equations was added to Chapter
meet the ever-in-
fast response, the converters
network
•
Transmission
become smaller
why
as the frequency increases.
High-frequency transformers are directly related
to the
higher frequencies encountered
in
switch-
ing converters.
easier.
The following changes have been made
in
the
•
Chapter 16 on Synchronous Generators has been
expanded
fourth and fifth editions:
to
show why an
increase
in size in-
evitably leads to higher efficiencies and greater
•
Every page of the original work was examined
for clarity of expression
and reviewed as
to its
outputs per kilogram. This fundamental aspect of
machine design
will interest
many
readers.
PREFACE
•
A new
section
was added
Chapter
to
1
phase induction motor.
many worked-out problems,
to solve the circuit,
presents a rigorous, yet
It
which permits
a better under-
Chapter 2
1
,
Electronics,
Fundamental Elements of Power
was revised and expanded to in-
modulation
(PWM)
techniques.
how
made
they can be
to
converters and
generate almost any
motors operating
major addition
Electric Utility
to Part
It
power
sags, swells,
It
power becomes
become
visit
students
made
may
in
to establish
how the
lines.
book requires
user-friendly treatment of even
topics, this
book
broad range of readers.
will
First,
meet
it
is
the needs of a
appropriate for
students following a two-year electrical
community
versities.
in
colleges, technical institutes, and uni-
Owing
to its
very broad coverage, the text
can also be incorporated
program.
program
Many
for their electric
in
a 4-year technology
universities have adopted the
book
power service courses.
wealth of practical information that can be di-
rectly applied to that greatest laboratory
electrical industry itself.
actual use.
at close
hand the
The photographs help convey
the
1
chapters, a conscious effort
was
coherence, so that the reader can see
various concepts
fit
together. For
example, the
to those
found
lines, in turn,
And
reactive
transmission
in
bring up the question
power
is
an important
aspect in electronic converters. Therefore, knowledge
it
in
is
one sector
applied
is
strengthened and broadened
in another.
As
a result, the learning
of electrical machines, drives, and HPwer sy^ems be-
comes a challenging, thought-prov^^f^experience.
In order to convey the real-w|ftd aspects df machinery and power systems, particular. attentio|i has
been paid
to the inertia
of revolving masses, the
physical limitations of materials, ai^l .the problems
created by heat. This
appro^kfjal^^^
multidisciplinary programs of
many
fftfe
and
eo^lle^es
technical institutes.
Instructors responsible for industrial training will
find a
in
the transmission and distribution of
Transmission
when
complex
the impor-
not have had the opportunity to
machines are similar
increasingly important.
gained
its
book shows
terminology and power equations for synchronous
and some trigonometry.
to
the
an industrial plant or to see
only a background in basic circuit theory, algebra,
Owing
and Websites
articles,
by diagrams and pictures, showing
illustrated
Throughout the 3
also
as regards
a reality, these
subject matter covered in this
will also
invited to con-
is
various stages of construction or
in
of reactive power.
The
also avail-
magnificent size of these devices and machines.
methods of controlling the quality of
electricity will
is
end of most chapters
at the
electrical energy.
harmonics, and brownouts. As dereg-
ulation of electric
electronic
power
Manual
Industrial Application prob-
of books, technical
equipment used
explains the
electronically.
discusses the quality of electric
appear
The
Reference section toward the end of the book.
Some
vector control.
IV dealing with
Power Systems.
practical, inter-
tance given to photographs. All equipment and sys-
them
technologies that are being developed to control
the flow of electric
that
tems are
at
Chapter 29, Transmission and Distribution represents a
lems
A Solutions
A quick glance through
special section explains the
PWM drives and flux
basics of
—
appeal to hands-on users. The reader
in the
Chapter 23, Electronic Control of Alternating
A
the end of each chapter are di-
at
end of the book.
sult the list
/^Current Motors, was greatly expanded to cover
variable speeds.
exercises
mediate, and advanced. Furthermore, to encourage
the
waveshape and frequency.
the properties of induction
The
vided into three levels of learning
able for instructors.
illustrates the
It
oflGBT
its
particularly suitable
the reader to solve the problems, answers are given at
clude switching converters and pulse width
extraordinary versatility
is
for self-study.
programmed
standing of this ubiquitous single-phase machine.
•
being de-
is
voted to continuing education, this book, with
motor. Hand-held computers can be
•
effort
velop the equivalent circuit diagram of a single-
simple approach, based on the 3-phase induction
•
when much
Finally, at a time
8 to de-
v
of
all
—
the
In
summary,
I
employ
a tlicM>Bk&$ ipfefeAal,
multidisciplinary approach to give a broad under-
standing of modern electric power. Clearly,
longer the staid subject
it
was considered
it
is
to
no
be
PREFACE
vi
some years ago. There
this
is
dynamic, expanding
good reason to believe that
field will open career op-
and Bernard Oegema of Schneider Canada; Carl
Tobie of Edison Electric
Institute;
Damiano Esposito
portunities for everyone.
and Vance
make a final remark concerning the
As mentioned previously, power
technology has made a quantum jump in the past
Scott Lindsay of Daiya Control
I
would
like to
use of this book.
Belisle,
machines, drives, and power systems, there will
now
be a long period of consolidation during which existing
machines and devices
will
be replaced by newer
Systems; Louis
and Jean Lamontagne of Lumen; Benoit
Arsenault and Les Halmos of Allen Bradley.
eight years, mainly on account of the availability of
fast-acting semiconductors. In the field of electrical
E. Gulliksen of Carnival Cruise Lines;
extend a special note of thanks to Professor
I
Thomas
Young
of
Rochester
the
Technology, to Dr. Robert
University,
College, and to Jean Anderson of Lab- Volt Ltd. for
models. But the basic technology covered herein will
having extensively reviewed and commented
not change significantly in the foreseeable future.
on various aspects of this book and
Consequently, the reader will find that
this
book can
of
Peros of Seneca
Professor Martin
to
Institute
H. Alden of McMaster
T.
their valued viewpoints.
also
I
in
depth
for having offered
want
to
acknowledge
also be used as a valuable long-term reference.
the contribution of Professor Stephane Montreuil of
Acknowledgments
end-of-chapter problems and the solutions manual.
CEGEP Levis-Lauzon for having gone over all the
the
want
I
In preparing this edition
and previous editions of
my
acknowledge
book,
1
would
like to
the impor-
tant contribution of the following persons.
Consultant; David Krispinsky, Rochester Institute of
Technology; Athula Kulatunga, Southeast Missourri
State University; Rick Miller, Ferris State University;
Nehir,
Montana
State University; Martin
University;
Chandra
James
E,
Sekhai;
University;
Gerald Sevigny, Southern Maine Technical College;
Philippe Viarouge, Laval University; Stacy Wilson,
Western
Kentucky
Rochester
Institute
A& M
and
University;
Sri R. Kolla,
University;
Thomas
of Technology; Dr.
P
Texas
Ted James, Pasadena City College;
Bowling Green
State University.
Commercial, industrial and institutional contributors:
Andre Dupont, Raj Kapila, G. Linhofer,
Katherine Sahapoglu of ABB; Roger Bullock, Gerry
McCormick, James Nanney, Darryl J.
Van Son, and Roddy Yates of Baldor Electric
Company; Jacques Bedard, Guy Goupil, and Michel
Lessard of Lab-Volt Ltd.; Richard B. Dube of
General Electric Company; Abdel-Aty Edric and
Ashock Sundaram of Electric Power Research
Institute; Neil H. Woodley of Westinghouse Electric
Rene Poulin of
Inc. in
of
Lawrence
in the application
and
to
of pro-
photographer Hughes
also
also hereby acknowledged.
is
want
E. Stewart,
Editor;
the Centre de Robotique Industrielle
reviewing and describing the essential features
PLCs
and
Jr.,
to
to express
my
appreciation to Charles
Publisher; to Delia Uherec, Associate
Alexandrina B. Wolf, Senior Production
Editor, of Prentice Hall, for planning, coordinating,
and administrating
As
to
in
this text.
previous editions,
my
provide his valuable help
art,
in
son Karl continued
preparing the line
photographs, and word processing of this
latest
edition.
My
thanks also go to
ing supported
me
thor, consultant,
Goyette, Jim
Corporation; Maurice Larabie, Jean-Louis Marin,
controllers,
St.
providing industrial ex-
in
Chicoine for his work. The important contribution of
Young,
Enjeti,
know-how
perience and
grammable
M.
Roach, Bob Jones
Purdue
appreciation to Jean- Serge
and Giles Campagna of
Electric,
I
Peros, Seneca College;
my
Stevedoring for their help
Professors and reviewers: Robert T. H. Alden,
McMaster University; Ramon E. Ariza, Delgado
Community College; Fred E. Eberlin, Educational
M. H.
to express
Lamirande of Omron, Pierre Juteau of Schneider
I
tors
in
my
my
wife, Rachel, for hav-
continuing vocation as au-
and teacher.
also wish to voice
my
gratitude to the instruc-
and students, practicing engineers, and techni-
cians
who
asked questions and made suggestions by
e-mailing their messages to wildi@wildi-theo.com.
You
are cordially invited to
do the same.
Theodore Wildi
PART
I.
FUNDAMENTALS
Distinction between sources and
2.2
loads
1.
UNITS 3
l.O
Introduction 3
LI
Systems of units 3
1.2
Getting used to SI 4
1.3
Base and derived units of the SI 4
\A
Definitions of base units 5
1.
Definitions of derived units 5
1.6
Double-subscript notation for
units 7
Conversion charts and
their use 8
Per-unit system with one base
l.ll
Per-unit system with
Sign notation for voltages
2.6
Graph of an
2.7
Positive and negative currents
18
19
Sinusoidal voltage
2.9
Converting cosine functions into sine
2.
10
2.
1
two bases
10
1
19
Effective value of an ac voltage 20
Phasor representation 2
2.12
Harmonics 23
2.13
Energy
in
14
Energy
in a
2.15
Some
an inductor 25
capacitor 25
useful equations 26
1
12
ELECTROMAGNETISM
2.
MAGNETISM, AND CIRCUITS 15
2.0
Introduction
2.
Conventional and electron current
flow 15
17
alternating voltage
2.8
1
6
FUNDAMENTALS OF ELECTRICITY,
1
17
2.5
2.
l.IO
Questions and Problems
17
functions 20
The per-unit system of
measurement 9
1.
2.4
Multiples and submultiples
Commonly used
1.
Sign notation
voltages
of SI units 7
1.
16
2.3
15
Magnetic
density
2.17
2.
1
8
2.19
field intensity
H and
B 27
B-H curve
B-H curve
of
f ACULTAD
vacuum 27
DE
q r ~n
of a magnetic material" 27
Determining the relative
permeability 28
*
WIN
A
_ -
-
BlBi-iO I fcw*
CONTENTS
i
2.20
Faraday's law of electromagnetic
3.9
induction 29
a conductor 30
2.21
Voltage induced
2.22
Lorentz force on a conductor 3
2.23
2.24
2.25
Kinetic energy of rotation,
inertia
in
3.10
Torque,
inertia,
and change
Direction of the force acting on a
3.11
Speed of a motor/load system 57
straight conductor 3
3.12
Power flow
Residual flux density and coercive
in a
mechanically coupled
system 58
force 32
3.13
Motor driving a load having
Hysteresis loop 33
3.14
Electric motors driving linear motion
Hysteresis loss 33
Hysteresis losses caused by
3.15
Heat and temperature 60
rotation 33
3.16
Temperature scales 61
Eddy
Eddy
3.17
currents 34
3.18
Eddy-current losses
in a revolving
core 35
2.31
Current
in
an inductor 36
Transmission of heat 62
Heat transfer by conduction 62
3.20
Calculating the losses by
convection 63
Kirchhoffs voltage law 40
2.33
Kirchhoffs voltage law and double-
3.22
Heat transfer by radiation 64
3.23
Calculating radiation losses 64
Questions and Problems 65
40
2.34
Kirchhoffs current law 41
2.35
Currents, impedances, and associated
PART
II.
voltages 41
2.36
Kirchhoffs laws and ac
2.37
KVL and
2.38
Solving ac and dc circuits with sign
2.39
circuits
4.
44
Circuits and hybrid notation
ELECTRICAL MACHINES AND
TRANSFORMERS
43
sign notation 43
45
Questions and Problems 46
DIRECT-CURRENT GENERATORS
4.0
Introduction 71
4.1
Generating an ac voltage 71
4.2
Direct-current generator 72
4.3
Difference between ac and dc
generators 73
FUNDAMENTALS OF MECHANICS
AND HEAT 50
4.4
Improving the waveshape 73
4.5
Induced voltage 75
Neutral zones 76
3.0
Introduction 50
4.6
3.1
Force 50
4.7
Value of the induced voltage 76
3.2
Torque 51
4.8
Generator under load: the energy
3.3
4.9
3.5
Mechanical work 5
Power 52
Power of a motor 52
3.6
Transformation of energy 53
3.7
Efficiency of a machine 53
4.11
commutation 78
Commutating poles 79
3.8
Kinetic energy of linear motion 54
4.12
Separately excited generator 79
3.4
58
temperature
Heat transfer by convection 63
3.21
2.32
notation
to raise the
3.19
CIRCUITS AND EQUATIONS
subscript notation
Heat required
of a body 6
currents in a stationary iron
core 35
2.30
inertia
loads 59
2.26
2.28
in
speed 57
2.27
2.29
moment of
54
conversion process 77
4.
10
Armature reaction 77
Shifting the brushes to improve
71
CONTENTS
4. 13
No-load operation and saturation
5.18
curve 79
4.14
Shunt generator 80
5.
4.15
Controlling the voltage of a shunt
5.20
1
9
generator 81
4.16
Dynamic braking and mechanical
constant
1
Armature reaction
5.21
1 1
Flux distortion due to armature
reaction
Equivalent circuit 82
1
1
Commutating poles 113
Compensating winding 114
Separately excited generator under
5.22
load 82
5.23
Basics of variable speed control
4. 18
Shunt generator under load 83
5.24
Permanent magnet motors
4.19
Compound
4.20
Differential
4.
1
7
Questions and Problems
generator 83
compound
1
1
Load
4.22
Generator specifications 84
84
6.
EFFICIENCY AND HEATING OF
ELECTRICAL MACHINES 120
120
Introduction
6.1
Mechanical losses
Field 84
6.2
Electrical losses
4.24
Armature 85
6.3
Losses as a function of load
4.25
Commutator and brushes 86
6.4
Efficiency curve
123
4.26
Details of a multipole generator 88
6.5
Temperature
125
4.27
The
The
6.6
Life expectancy of electric
6.7
Thermal
4.28
ideal
commutation process 91
practical
commutation process 92
1
20
20
1
rise
DIRECT-CURRENT MOTORS 96
Introduction 96
classification of
126
6.8
Maximum
ambient temperature and
6.9
Temperature
6.10
Relationship between the speed and
Counter-electromotive force
5.2
Acceleration of the motor 97
Mechanical power and torque 98
5.4
Speed of rotation 100
5.5
Armature speed control 101
rise
size of a
machine
1
Shunt motor under load 103
ACTIVE, REACTIVE,
POWER 134
7.0
Introduction
7.1
Instantaneous power
7.2
Active power 136
7.3
Reactive power 137
7.
102
Series motor
Series
5.10
Applications of the series motor
104
motor speed control 105
1
5.11
Compound motor
5.12
Reversing the direction of rotation
5.13
Starting a shunt
5.14
Face-plate starter 108
5.15
Stopping a motor 109
06
106
1
07
7.4
motor 108
5.16
Dynamic braking 109
5.17
Plugging 110
7.6
3
134
134
Definition of reactive load and
reactive source
7.5
!
AND APPARENT
Field speed control
5.7
5.8
27
30
Questions and Problems
5.6
5.9
1
by the resistance
method 129
(cemf) 96
5.3
23
insulators
hot-spot temperature rise
5.1
1
equipment 126
Questions and Problems 93
5.0
1
1
6.0
CONSTRUCTION OF DIRECT-CURRENT GENERATORS
4.23
1
17
generator 84
4.21
characteristics
tin
1
138
The capacitor and
power 139
reactive
Distinction between active and
reactive
power 140
CONTENTS
Combined active and
apparent power 141
7.7
reactive loads:
Relationship between
7.9
7.10
Power
Power
7.11
Further aspects of sources and
loads
7.12
7.
3
1
1
4
Q, and S
141
8.20
Varmeter 177
8.2
A remarkable
1
144
7.
1
6
17
Systems comprising several loads 146
9.
148
Solving
AC
circuits using the
method 148
Power and vector notation
power
Rules on sources and loads (sign
154
Rules on sources and loads (double
subscript notation)
Introduction
9.
Voltage induced
1
Introduction
83
Elementary transformer
9.4
Polarity of a transformer
9.5
Properties of polarity marks
9.6
Ideal transformer at no-load; voltage
1
84
1
85
1
86
186
187
ratio
9.7
Ideal transformer under load; current
9.8
Circuit
188
symbol
for an ideal
transformer 191
158
Polyphase systems 158
9.9
Single-phase generator 159
9.
Impedance
10
Power output of a single-phase
ratio
191
Shifting impedances from secondary
to
primary and vice versa 192
Questions and Problems
60
Two-phase generator 160
Power output of a 2-phase
8.5
1
9.3
8.1
1
a coil
Applied voltage and induced
8.2
generator
in
9.2
ratio
8.4
80
1
183
9.0
154
THREE-PHASE CIRCUITS 158
8.3
178
THE IDEAL TRANSFORMER 183
voltage
151
Questions and Problems 155
8.0
single-phase to 3-phase
144
notation)
7.
3-phase,
Questions and Problems
triangle
7.15
in
177
transformation
Reactive power without magnetic
fields
7.
P,
143
triangle
Power measurement
4-wire circuits
7.8
factor
19
8.
10.
195
PRACTICAL TRANSFORMERS 197
generator 16
10.0
Introduction
8.6
Three-phase generator 162
10.
Ideal transformer with an imperfect
8.7
Power output of
generator
1
a 3 -phase
10.2
8.8
Wye
Voltage relationships
connection
164
8.10
Delta connection
Power transmitted by
line
8.12
8.13
97
199
Primary and secondary leakage
10.3
167
reactance 200
a 3-phase
1
0.4
168
Equivalent circuit of a practical
transformer 202
Active, reactive and apparent
3-phase circuits
1
197
Ideal transformer with loose
coupling
165
8.
1
core
62
8.9
1
1
power
in
1
Construction of a power
0.5
169
transformer 203
Solving 3-phase circuits
170
10.6
Standard terminal markings 204
8.14
Industrial loads
171
10.7
Polarity tests
8.15
Phase sequence 174
10.8
Transformer taps 205
10.9
Losses and transformer rating 206
8.
1
6
Determining the phase sequence
8.
1
7
8.
1
8
Power measurement
Power measurement
3-wire circuits
176
in
ac circuits
in
3-phase,
1
75
1
76
204
10.10
No-load saturation curve 206
10.11
Cooling methods 207
10.
1
2
Simplifying the equivalent circuit 209
CONTENTS
Voltage regulation 211
10.14
Measuring transformer
transformers 260
impedances 212
Questions and Problems 260
10.15
Introducing the per unit method 215
1
11.
0.
6
1
Impedance of a transformer 2
Typical per-unit impedances 216
10.18
Transformers
1
THREE-PHASE INDUCTION
MOTORS
13.0
Introduction 263
13.1
Principal
13.2
Principle of operation
13.3
The
TRANSFORMERS
225
Introduction 225
1
1
1
.2
Autotransformer 226
1
.3
Conventional transformer connected
1
1
.5
1
1
.6
Number
3.5
1
1
cage motor 273
228
Acceleration of the rotor-slip 274
13.7
Motor under load 274
and slip speed 274
Voltage transformers 230
13.8
Current transformers 23
13.9
Slip
13.10
Voltage and frequency induced
Opening
the secondary of a
CT can
be
rotor
Toroidal current transformers 234
.7
11.8
Variable autotransformer 235
11.9
High-impedance transformers 236
1
1
.
1
0
1
1
.
1
1
1
2.
1
Active power flow 278
Torque versus speed curve 28
13.15
Effect of rotor re s stance 282
13.16
Wound-rotor motor 284
Introduction 243
13.17
Three-phase windings 285
Basic properties of 3-phase
13.18
1
2.4
Wye-delta connection 247
12.6
Open-delta connection 248
Three-phase transformers 249
12.8
Step-up and step-down
1
2.
1
1
0
1
12. 12
19
3.2
1
i
Sector motor 288
Linear induction motor 289
Traveling waves 291
Properties of a linear induction
motor 291
13.22
Wye-wye connection 248
12.7
3.
13.20
244
1
1
an induction
13.14
Delta-wye connection 246
2.
in
13.13
2.3
1
Estimating the currents
High-frequency transformers 238
1
12.9
13.12
Questions and Problems 24
Delta-delta connection
2.5
Characteristics of squirrel-cage
motor 277
transformer banks 243
1
13.11
Induction heating transformers 237
2.2
1
in the
275
induction motors 276
THREE-PHASE TRANSFORMERS 243
12.0
of poles-synchronous
Starting characteristics of a squirrel-
3.6
dangerous 233
1
264
265
rotating field
speed 271
1
as an autotransformer
components 263
Direction of rotation 270
13.4
transformer 225
11. 4
263
Questions and Problems 221
in parallel
Dual-voltage distribution
1
.
Polarity
219
SPECIAL
1
13.
1
10.17
11.0
12.
marking of 3-phase
10.13
12.13
Magnetic
levitation
293
Questions and Problems 295
autotransformer 25
SELECTION AND APPLICATION OF
THREE-PHASE INDUCTION
Phase-shift principle 253
MOTORS
14.
Three-phase to 2-phase
14.0
transformation 254
1
4.
1
4.2
1
Calculations involving 3-phase trans-
299
Introduction 299
Standardization and classification of
induction motors 299
Phase-shift transformer 256
formers 258
xi
Classification according to environ-
ment and cooling methods 299
CONTENTS
Classification according to electrical
14.3
16.
16.0
Introduction 335
16.
Commercial synchronous
335
14.4
Choice of motor speed 303
14.5
Two-speed motors 303
14.6
Induction motor characteristics under
16.2
Number
various load conditions 305
16.3
16.4
Main
Main
Field excitation and exciters 342
14.7
Starting an induction
1
generators 335
motor 308
of poles 335
336
features of the stator
features of the rotor
340
14.8
Plugging an induction motor 308
16.5
14.9
Braking with direct current 309
16.6
Brushless excitation 343
14.10
Abnormal conditions 310
Mechanical overload 310
16.7
Factors affecting the size of
16.8
No-load saturation curve 345
14.11
1
4.
1
2
Line voltage changes 3
Single-phasing 310
14.14
Frequency variation 311
4.
1
5
synchronous generators 344
1
14.13
1
circuit
of an ac generator 346
Induction motor operating as a
16.10
Determining the value of
generator 3
16.11
Base impedance, per-unit
1
Complete torque-speed characteristic
of an induction machine 314
14.17
Features of a wound-rotor induction
Start-up of high-inertia loads 315
14.19
Variable-speed drives 315
Frequency converter 3
s
348
s
349
350
1
6.
1
2
Short-circuit ratio
1
6.
1
3
Synchronous generator under
16.14
14.18
X
X
load 350
motor 315
4.20
Synchronous reactance-equivalent
16.9
14.16
1
SYNCHRONOUS GENERATORS
and mechanical properties 301
Regulation curves 352
1
6.
1
5
Synchronization of a generator 353
1
6.
1
6
Synchronous generator on an
infinite
bus 355
1
Questions and Problems 3
16.17
1
Infinite bus-effect of varying the
exciting current 355
15.
EQUIVALENT CIRCUIT OF THE
INDUCTION MOTOR 322
15.0
15.1
16.18
Infinite bus-effect
of varying the
mechanical torque 355
Introduction 322
The wound-rotor induction motor 322
Power relationships 325
16.19
Physical interpretation of alternator
behavior 357
1
6.20
Active power delivered by the
motor 326
1
6.2
Control of active power 359
15.4
Breakdown torque and speed 327
16.22
5.5
Equivalent circuit of two practical
1
15.2
1
1
5.3
Phasor diagram of the
nfl,
i
generator 358
iction
1
6.23
motors 327
15.6
15.7
5.8
15.9
Power
transfer
between two
sources 361
Calculation of the breakdown
16.24
Efficiency, power, and size of
torque 328
electrical
Torque-speed curve and other
Questions and Problems 364
characteristics
1
Transient reactance 359
machines 362
329
Properties of an asynchronous
17.
SYNCHRONOUS MOTORS
generator 330
17.0
Introduction 369
Tests to determine the equivalent
17.1
Construction 370
circuit
33
l
Questions and Problems 333
1
7.2
17.3
Starting a synchronous
Pull-in torque
372
369
motor 372
CONTENTS
]
1
Motor under load-general
7.4
7.5
18.19
Deducing
xiii
the circuit diagram of a
description 372
single-phase motor 411
Motor under load-simple
Questions and Problems 4
1
calculations 373
Power and torque 376
17.7
Mechanical and
17.8
Reluctance torque 378
19.1
Elementary stepper motor 417
Losses and efficiency of a
19.2
Effect of inertia
1
7.9
19.
377
electrical angles
1
synchronous motor 379
17.10
1
7.
1
1
power 380
Excitation and reactive
Power
1
9.0
418
Effect of a mechanical load 4
9.3
19.5
Start-stop stepping rate
19.6
Slew speed 421
420
V-curves 382
17.13
Stopping synchronous motors 383
19.7
Ramping 422
The synchronous motor versus
induction motor 385
Synchronous capacitor 385
19.8
Types of stepper motors 422
14
7.
17.15
the
Motor windings and associated
424
19.10 High-speed operation 427
Modifying the time constant 428
19.11
19.12 Bilevel drive 428
19.13 Instability and resonance 434
19.14 Stepper motors and linear drives 434
Questions and Problems 434
19.9
drives
Questions and Problems 388
SINGLE-PHASE
1
8.
MOTORS
391
Introduction 39
18.0
Construction of a single-phase
1
induction motor 39
18.2
Synchronous speed 393
18.3
Torque-speed characteristic 394
18.4
Principle of operation
PART
III.
394
ELECTRICAL AND ELECTRONIC
DRIVES
Locked-rotor torque 396
1
8.5
1
8.6
Resistance split-phase motor 396
1
8.7
Capacitor-start motor 398
1
8.8
Efficiency and
1
1
Torque versus current 420
19.4
factor rating 38
Introduction 41
17.12
1
18.
STEPPER MOTORS 417
17.6
8.9
18.10
1
8.
1
1
8.
12
1
power
20.
factor of single-
BASICS OF INDUSTRIAL
CONTROL
MOTOR
439
phase induction motors 399
20.0
Introduction 439
Vibration of single-phase motors 40!
20.1
Control devices 439
Capacitor-run motor 402
20.2
Normally-open and normally-closed
contacts 443
Reversing the direction of
403
20.3
Relay
Shaded-pole motor 403
20.4
Control diagrams 445
rotation
coil exciting current
443
18.13
Universal motor 404
20.5
Starting
methods 446
18.14
Hysteresis motor 405
20.6
Manual
across-the-line starters
18.15
Synchronous reluctance motor 407
20.7
Magnetic across-the-line
18.16
Synchro drive 408
20.8
Inching and jogging 450
EQUIVALENT CIRCUIT OF A SINGLE-PHASE
MOTOR
Reversing the direction of
20.9
rotation 451
1
8.
1
7
18.18
Magnetomotive force
Revolving
motor 410
mmfs
in
distribution
a single-phase
409
20.10
20.
1
1
20.12
447
starters
Plugging 453
Reduced-voltage starting 454
Primary resistance starting 454
448
CONTENTS
xiv
20.
1
3
20.14
Autotransformer starting 458
21.17
Power gain of
Other starting methods 460
21.18
Current interruption and forced
20.15
Cam
20.16
Computers and controls 462
a thy ristor
494
commutation 495
switches 461
21.19
Basic thyristor power circuits 496
2 .20
Controlled rectifier supplying a
1
ELECTRIC DRIVES
passive load (Circuit
7
Fundamentals of electric drives 462
1
8
Typical torque-speed curves 463
1
9
Shape of the torque-speed
20.
1
20.
20.
curve 464
20.20
Current-speed curves 466
20.21
Regenerative braking 467
21 D)
Controlled rectifier supplying an ac-
21 .22
Line-commutated inverter (Circuit
tive load (Circuit 2,
21.0
2 .23
1
472
Potential level
Voltage across some circuit
.2
21 .25
2
1
.4
2
1
.5
Battery charger with series
.7
6,
Table 2 D) 502
1
Delayed triggering-rectifier
21.29
Delayed triggering-inverter mode 507
21.30
Triggering range 508
21.31
Equivalent circuit of a
2 1 .32
Currents in a 3-phase, 6-pulse
converter 5
Single-phase bridge rectifier 480
481
Filters
2 .9
Three-phase, 3-pulse diode
21.10
Three-phase, 6-pulse rectifier 485
21.11
Effective line current, fundamental
1
Power factor 511
21.34 Commutation overlap 514
21.35 Extinction angle 514
21.33
483
489
Distortion power
line current
21.12
Three-phase, 6-pulse controllable
mode 505
476
Battery charger with series
rectifier
Table
converter 509
21.8
1
5,
Three-phase, 6-pulse rectifier feeding
1
inductor 478
21
Cycloconverter (Circuit
an active load 504
The diode 475
Main characteristics of a diode 476
resistor
Table
2 .27
THE DIODE AND DIODE CIRCUITS
.6
4,
Basic principle of operation 503
1
1
switch (Circuit
500
21 .26
2 .28
2
3,
21D) 501
elements 474
21.3
static
converter (Circuit
21.1
1
AC
2 ID)
Introduction 472
2
Table 2 ID) 497
Table 2 ID) 498
2 1 .24
FUNDAMENTAL ELEMENTS OF
POWER ELECTRONICS 472
Table
21.21
Questions and Problems 468
21.
1,
496
factor
490
Displacement power factor, total
power factor 490
21.14 Harmonic content, THD 491
21.13
DC-TO-DC SWITCHING CONVERTERS
21.36
DC-to-DC switching converter
21.38
Rapid switching 519
21.39
Impedance transformation 522
2 .40
Basic 2-quadrant dc-to-dc
21.41
Two-quadrant electronic
2 1 .42
Four-quadrant dc-to-dc
1
converter 522
THE THYRISTOR
converter 525
AND THYRISTOR CIRCUITS
21.15
Thethyristor 492
21.16
Principles of gate firing
Semiconductor switches 515
21.37
converter 526
492
21.43
Switching losses 528
5
1
CONTENTS
Dc-to-ac rectangular wave
ELECTRONIC CONTROL OF
ALTERNATING CURRENT MOTORS 575
converter 529
23.0
Introduction 575
Dc-to-ac converter with pulse-width
23.1
Types of ac drives 575
modulation 530
23.2
DC-TO-AC SWITCHING CONVERTERS
2 .44
1
2 .45
1
23.
21.46
Dc-to-ac sine wave converter 532
21.47
Generating a sine wave 533
PWM pulse train
Synchronous motor drive using
current-source dc link 577
Synchronous motor and
23.3
cycloconverter 580
534
21.48
Creating the
21.49
Dc-to-ac 3-phase converter 535
21.50
xv
Cycloconverter voltage and frequency
23.4
control
Conclusion 537
580
Squirrel-cage induction motor with
Questions and Problems 537
23.5
ELECTRONIC CONTROL OF DIRECT-
23.6
cycloconverter 582
22.
CURRENT MOTORS
Squirrel-cage motor and static voltage
controller
541
quadrant speed control 541
22.1
First
22.2
Two-quadrant control-field
reversal
544
SELF-COMMUTATED INVERTERS
23.8
Self-commutated inverters for cage
23.9
Current-source self-commutated
motors 592
Two-quadrant control-armature
22.3
reversal
545
frequency converter (rectangular
Two-quadrant control-two
22.4
wave) 593
converters 545
Four-quadrant control-two converters
22.5
23.10
Two-quadrant control with positive
22.6
23.
torque 549
22.7
Four-quadrant drive 549
22.8
Six-pulse converter with freewheeling
23.
1
1
1
2
diode 551
Half-bridge converter 556
22.10
Detraction 558
Motor drive using
a dc-to-dc
22. 12
Introduction to brushless dc
22.
Commutator replaced by reversing
Recovering power
in a
wound-rotor
23.14
Pulse-width modulation and ind
modulation 602
motors 604
Synchronous motor
22.
1
1
5
6
U^'COi"
TORQUE AND SPEED CONTROL
as a brushless dc
OF INDUCTION MOTORS
machine 568
22.
wound-
Review of pulse-width
switches 566
22. 14
motor 597
23.13
motors 565
3
control of a
rotor induction
PULSE-WIDTH MODULATION DRIVES
switching converter 560
1
wave) 594
Chopper speed
induction motor 599
22.9
1
Voltage-source self-commutated
frequency converter (rectangular
with circulating current 546
1
590
Introduction 54
22.0
22.
589
Soft-starting cage motors
23.7
BGi'l. VAO Q[ ^!3 L fTRM
Standard synchronous motor and
23.15
Dc motor and
brushless dc machine 569
23.16
Slip speed, flux orientation,
Questions and Problems 571
604
23.17
Features of variable-speed controlconstant torque
mode 607
A
aBdBwiOiECA
torque 605
Practical application of a brushless dc
motor 569
flux orientation
CONTENTS
xvi
23. 8
1
Features of variable-speed controlconstant horsepower
23.19
24.
Features of variable-speed control-
1
Induction motor and
its
24.
1
2
Equivalent circuit of a practical
Volts per hertz of a practical
Speed and torque control of induction
motors 614
Carrier frequencies 615
23.25
Dynamic
Condenser 650
24.15
Cooling towers 650
24. 6
Boiler- feed
24. 7
Energy flow diagram for a steam
24.
Thermal
1
8
stations
and the
environment 652
616
Principle of flux vector control
23.27
Variable-speed drive and electric
NUCLEAR GENERATING STATIONS
618
Principal
23.29
Operating
mode
mode
Composition of an atomic nucleus;
24.20
The source of uranium 655
isotopes 655
of the 3-phase
converter 622
Operating
24. 9
1
components 621
23.28
of the single-phase
converter 624
Conclusion 629
Questions and Problems 629
IV.
pump 65
plant 651
control of induction
23.26
PART
thermal generating
648
Turbines 650
motors 615
23.31
646
24.14
1
23.24
23.30
installations
24.13
1
traction
of a hydropower plant 644
Makeup of a
station
motor 613
23.23
Makeup
Pumped-storage
THERMAL GENERATING STATIONS
equivalent
motor 612
23.22
0
24.11
circuit 61
23.2
1
mode 610
generator
23.20
Types of hydropower stations 643
24.9
mode 610
ELECTRIC UTILITY POWER
24.21
Energy released by atomic fission 656
24.22
Chain reaction 656
24.23
Types of nuclear reactors 657
24.24
24.25
Example of a light-water reactor 658
Example of a heavy-water reactor 659
24.26
Principle of the fast breeder
reactor
SYSTEMS
24.27
660
Nuclear fusion 661
Questions and Problems 661
24.
GENERATION OF ELECTRICAL
ENERGY
635
25.
24.0
Introduction 635
24.
Demand
1
TRANSMISSION OF ELECTRICAL
ENERGY
of an electrical system 635
24.2
Location of the generating station 637
24.3
Types of generating
24.4
Controlling the power balance
24.5
between generator and load 638
Advantage of interconnected
stations
25.0
25.
1
637
Conditions during an outage 641
Frequency and
electric clocks
642
components of a power
664
25.2
Types of power
25.3
Standard voltages 667
25.4
Components of a
line
24.7
lines
HV
665
transmission
667
25.5
Construction of a line 668
25.6
Galloping lines 669
Corona effect-radio interference 669
Pollution 669
Lightning strokes 670
HYDROPOWER GENERATING STATIONS
25.7
Available hydro power 642
25.9
25.8
24.8
Principal
distribution system
systems 639
24.6
664
Introduction 664
CONTENTS
25.10
Lightning arresters on buildings 671
25.11
Lightning and transmission lines 671
25.12
Basic impulse insulation level
26 A]
Low-voltage distribution 709
PROTECTION OF MEDIUM-VOLTAGE
DISTRIBUTION SYSTEMS
(BIL) 672
25.14
Ground wires 673
Tower grounding 673
25.15
Fundamental objectives of a
25.13
26. 2
1
Coordination of the protective
devices 714
26.
transmission line 675
1
3
26.14
25.16
Equivalent circuit of a line 676
25.17
Typical impedance values 676
25.18
Simplifying the equivalent circuit 678
25.19
xvii
Fused cutouts 7
1
Reclosers 716
26.15
Sectionalizers 716
26.16
Review of
Voltage regulation and power-
MV protection
717
LOW-VOLTAGE DISTRIBUTION
transmission capability of
26.17
transmission lines 680
25.20
Resistive line
680
25.21
Inductive line 681
25.22
Compensated inductive
25.23
25.24
25.25
25.26
25.27
25.28
25.29
26.
line
683
1
8
LV
distribution system
Grounding
717
electrical installations 7
Electric shock
26.20
Grounding of 120
V
240V/120V
and
systems 720
Inductive line connecting two
systems 685
26.21
Equipment grounding 721
Review of power transmission 686
line voltage 687
Methods of increasing the power
capacity 689
Extra-high-voltage lines 689
Power exchange between power
centers 692
Practical example of power
exchange 693
26.22
Ground-fault circuit breaker 723
26.23
Choosing the
Rapid conductor heating:
2
I
t
factor
724
26.24
The
26.25
Electrical installation in
role of fuses
725
buildings 725
26.26
Principal
components of an
electrical
725
installation
Questions and Problems 727
Questions and Problems 695
THE COST OF ELECTRICITY 729
26.
DISTRIBUTION OF ELECTRICAL
Introduction 729
ENERGY 698
Tariff based
upon energy 730
Tariff based
upon demand 730
26.0
Introduction 698
Demand
meter 730
SUBSTATIONS
Tariff based
26.1
Substation equipment 698
upon power
factor
732
Typical rate structures 733
Demand
698
26.2
Circuit breakers
26.3
Air-break switches 702
26.4
Disconnecting switches 702
26.5
Grounding switches 702
26.6
Surge arresters 702
1
719
26.19
Power
^giiTAD Of
ILF
controllers 733
factor correction
Measuring
\
737
electrical energy, the
watthourmeter 740
27.9
Operation of the watthourmeter 741
26.7
Current-limiting reactors 705
27.10
Meter readout 742
26.8
Grounding transformer 706
27.11
Measuring three-phase energy and
26.9
Example of a substation 707
power 743
26.10
Medium-voltage distribution 709
Questions and Problems 743
CONTENTS
xviii
28.
DIRECT-CURRENT TRANSMISSION 746
Why
Distribution system 785
Compensators and
28.0
Introduction 746
28.
Features of dc transmission 746
29. 10
1
28.2
Basic dc transmission system 747
28.3
Voltage, current, and
Power
fluctuations
29.
power
1
1
characteristic
28.6
Power
on a dc
line
29.13
752
28.9
Power reversal 755
Components of a dc transmission
line
30.
755
Inductors and harmonic
28.12
Converter transformers 756
28. 13
Reactive power source 757
compensator: principle of
Conclusion 796
Harmonic filters on the ac side 757
28.15 Communications link 757
28.16 Ground electrode 757
28. 7 Example of a monopolar converter
station 757
28. 8 Thyristor converter station 758
28.19 Typical installations 760
30.0
Introduction 799
Harmonics and phasor diagrams 799
Effective value of a distorted
wave 800
Crest factor and total harmonic
30.3
distortion
30.5
Displacement power factor and
power
Non-linear loads 804
Generating harmonics 805
30.8
Correcting the power factor 807
30.9
Generation of reactive power 808
EFFECT OF HARMONICS
30. 10
30.
29.
Thyristor-controlled series capacitor
1
1
30.12
Harmonic current in a capacitor 809
Harmonic currents in a
conductor 810
Distorted voltage and flux in a
810
Harmonic currents
coil
30.13
in
a 3-phase,
4-wire distribution system 812
(TCSC) 769
29.2
Vernier control 771
29.3
Static
29.4
Eliminating the harmonics 776
29.5
Unified power flow controller
Harmonics and resonance 813
Harmonic filters 818
30.16 Harmonics in the supply
network 819
(UPEC) 776
30.17
29.6
Static
synchronous compensator 773
total
804
30.6
SOLID-STATE CONTROLLERS 768
Introduction 768
factor
circuits
30.7
TRANSMISSION AND DISTRIBUTION
29.0
802
Harmonics and
Questions and Problems 765
TRANSMISSION POWER FLOW CONTROLLERS
(THD) 801
30.4
1
1
799
30.1
on the
28. 14
1
series
HARMONICS
30.2
filters
dc side (6-pulse converter) 756
29.
The
754
Bipolar transmission line 754
1
principle of
Questions and Problems 797
control 753
Effect of voltage fluctuations
1
The shunt compensator:
operation 793
28.7
28.
29. 12
75
28.8
28. 10
circuit
operation 787
Typical rectifier and inverter
28.5
784
analysis 787
relationships 748
28.4
PWM converters?
29.8
29.9
30.14
30.15
Transformers and the
K
factor 821
frequency changer 780
HARMONIC ANALYSIS
DISTRIBUTION
CUSTOM POWER PRODUCTS
30.
29.7
1
8
Procedure of analyzing a periodic
Disturbances on distribution
wave 823
systems 782
Questions and Problems 827
CONTENTS
31.
PROGRAMMABLE LOGIC
31.18
Getting to
CONTROLLERS
831
31.19
Linking the PLCs 853
Introduction 83
31.20
Capacity of industrial
PLCs 83
31.21
Programming the PLCs 853
The transparent enterprise 855
Elements of
system 832
31.0
3
1
.
3
1
.2
3
1
.3
3
l
.4
1
a control
31.10
31.11
Conventional control circuits and
31.6
3
1
.7
3
1
.8
3
1
.9
circuits
Appendixes 865
AXO
Conversion Charts 865
AX1
Properties of Insulating
Materials 869
AX2
PLC
Mechanical and
Thermal Properties of Some
Electrical,
Common
Conductors (and
Insulators) 870
Security rule 847
31.13
Programming the PLC 847
Programming languages 847
31.14
References 859
844
31.12
31.15
851
Questions and Problems 856
Examples of the use of a PLC 835
The central processing unit
(CPU) 838
Programming unit 838
The I/O modules 839
Structure of the input modules 839
Structure of the output modules 840
Modular construction of PLCs 84
Remote inputs and outputs 84
31. 5
know PLCs
AX3
Properties of
Round Copper
Conductors 871
Advantages of PLCs over relay
Answers
cabinets 848
MODERNIZATION OF AN INDUSTRY
PLCs 850
31.16
Industrial application of
31.17
Planning the change 850
to
Problems 873
Answers to Industrial Application
Problems 877
Index 879
xix
1
To Rachel
Part One
Fundamentals
Chapter
1
Units
example,
Introduction
1.0
in
measuring length some people use the
inch and yard, while others use the millimeter and
an important role
Units play
everything
effect,
we buy and
thing
we
familiar that
sell is
Some
means of units.
we
our daily
how
lives. In
meter. Astronomers
them
become
for granted,
they started, or
deal with the rod and chain. But these units of length
so
can be compared with great accuracy because the
seldom
why
is based upon the speed of light.
Such standards of reference make it possible to
compare the units of measure in one country, or in
standard of length
they
were given the sizes they have.
was defined
as the length
of 36 barleycorns strung
end
and the yard
was the distance from the
tip
Centuries ago the foot
to the
end of
we have come
measure more
now based upon
This
way
precisely.
in
improvement
in
in
in
defining
Most
units are
which
terms of the speed of
our standards of measure has
hand with the advances
in
1.1
Systems
Over
the years systems of units
of units
have been devised
to
A
may be
described as one
in
which
the
units bear a direct numerical relationship to each
technology,
other, usually expressed as a
whole number. Thus
the English system of units, the inch, foot,
the other.
are related to each other by the
Although the basic standards of reference are recall
any
meet the needs of commerce, industry, and science.
system of units
and the one could not have been achieved without
ognized by
in
the anchors that tie together the units used in the
and time by the duration of atomic vibrations.
gone hand
measure
Standard units of length, mass, and time are
world today.
and reproducible. Thus the me-
and yard are measured
light,
a long
specialty, with the units of
other.
of King Edgar's nose
the physical laws of nature,
are both invariable
ter
one
his outstretched hand.
Since then
our units of
to end,
employ the parsec, physicists
some surveyors still have to
use the angstrom, and
measured and compared by
of these units have
often take
stopping to think
in
see and feel and every-
The same
countries of the world, the units of
numbers
1
in
and yard
2, 3,
and
36.
correlation exists in metric systems,
except that the units are related to each other by
everyday measure are far from being universal. For
multiples of ten.
3
Thus
the centimeter, meter,
and
FUNDAMENTALS
4
kilometer are related by the numbers
1
00 000.
1
00,
1
000, and
4.
It
centimeters than to convert yards into feet, and this
decimal approach
scientist, the
layman, thereby blending the theoretical and
one of the advantages of the
is
can be used by the research
technician, the practicing engineer, and by the
therefore easier to convert meters into
It is
the practical worlds.
metric system of units.*
Despite these advantages the SI
Today
the officially recognized metric system
the International
System of Units, for which the
is SI. The SI was formally
universal abbreviation
introduced
1960,
in
the
at
"Systeme international d
to everything. In specialized areas of
not the answer
and even
in
atomic physics,
The
used to
official introduction
of Units, and
its
be
measure
will continue to
plane angles in degrees, even though the SI unit
radian. Furthermore,
1
day and hour
will
is
the
be used,
still
unites.
despite the fact that the SI unit of time
1.2 Getting
may
day-to-day work, other units
more convenient. Thus we
General
Eleventh
Conference of Weights and Measures, under the
official title
is
is
SI
Base and derived units
1.3
of the International System
adoption by most countries of the
is
the second.
of the SI
The foundation of the International System of Units
rests upon the seven base units listed in Table A.
l
world, did not, however, eliminate the systems that
were previously employed.
become
habits, units
cannot readily
from yards
to
let
go.
lates to the
It is
because long familiarity with a
an idea of
magnitude and how
its
(particularly in the electrical
it
it
re-
growing importance of SI
the
and mechanical
know
necessary to
TABLE 1A
BASE UNITS
Quantity
Symbol
Unit
And
physical world.
Nevertheless,
makes
we
not easy to switch overnight
meters and from ounces to grams.
this is quite natural,
unit gives us
Just like well-established
a part of ourselves, which
Length
meter
m
Mass
Time
kilogram
kg
second
s
Electric current
ampere
Temperature
kclvin
A
K
Luminous
candela
cd
mole
mol
fields)
the essentials of this
Amount
intensity
of substance
measurement system. Consequently, one must be
able to convert
ple,
from one system
unambiguous way.
to
another
in a
sim-
From
In this regard the reader will
these base units
we
derive other units to
discover that the conversion charts listed in the
express quantities such as area, power, force, mag-
Appendix are particularly
netic flux,
and so on. There
number of
units
The
SI possesses a
tures shared
helpful.
number of remarkable
by no other system of
fea-
we can
is
really
derive, but
no
limit to the
some occur so
frequently that they have been given special names.
units:
Thus, instead of saying that the unit of pressure
1
.
2.
It is
It
a decimal system.
employs many
dustry and
pere, kilogram,
3.
It is
units
commerce;
commonly used
some of the most
ships in electricity, mechanics,
The metric
Canada
am-
that
have special names are
listed in
Table
l
B.
and watt.
a coherent system that expresses with star-
tling simplicity
*
in in-
for example, volt,
unit of length
is
the official spelling
heat.
spelled either meter or metre. In
is
metre.
TABLE 1B
DERIVED UNITS
basic relation-
and
is
newton per square meter, we use a less cumbersome name, the pascal. Some of the derived units
the
Quantity
Unit
Symbol
Electric capacitance
farad
F
Electric charge
coulomb
C
Electric conductance
Siemens
S
UNITS
A
TABLE 1 B
5
tuned to the resonant fre-
epiartz oscillator,
(continued)
quency of cesium atoms, produces a highly accuQuantity
Symbol
Unit
and stable frequency.
The ampere (A) is that constant current which, if
maintained in two straight parallel conductors of inrate
Electric potential
volt
V
Electric resistance
ohm
tt
Energy
joule
J
finite length,
Force
newton
N
placed
Frequency
hertz
Hz
tween these conductors a force equal
newton per meter of
Illumination
lux
lx
iicni y
n
Luminous flux
lumen
lm
Magnetic flux
weber
Wb
Magnetic flux density
tesla
T
1
1 1
LI LI<_
LuIlLC
Plane angle
radian
The kelvin
ature,
watt
W
pascal
Pa
steradian
in
units illustrate the
ated with this
water,
ice,
italics is
sr
and water vapor
definition.
The
The candela
definitions of the SI base
extraordinary precision associunits.
The
text in
explanatory and does not form part of the
by light
in
the length of the path travelled
vacuum during
a time interval of 1/299
until
A
coexist
equal
point
is
to
called the
is
273.16
equal
kelvins,
to 0.01 de-
temperature ofO °C
is
therefore
(cd)
1983 the speed of light was defined
792 458 m/s exactly.
to
he 299
the luminous intensity, in a
is
given direction, of a source that emits monochromatic radiation of frequency 540
X
10
12
hertz and
that has a radiant intensity in that direction of 1/683
watt per steradian.
is
the
amount of substance of
system that contains as many elementary
a
entities as
there are atoms in 0.012 kilogram of carbon 12.
Note:
792 458 of a second.
///
is
triple
The mole (mol)
is
cooled
cell is
equal to 273.15 kelvins, exactly
definition:
The meter (m)
of water.
triple point
an evacuated
point of water and
triple
base units
modern system of
thermodynamic temperthermodynamic
begins to form. The resulting temperature where
ice
gree Celsius (°C).
official
length.
the fraction 1/273.16 of the
is
temperature of the
by
The following
10~ 7
X
to 2
rad
Power
1.4 Definitions of
vacuum, would produce be-
in
(K), unit of
Pure water
Pressure
Solid angle
of negligible circular cross-section, and
meter apart
1
tities
When
the
mole
is
used, the elementary en-
must be specified and may be atoms, mole-
cules, ions, electrons, other particles, or specified
The kilogram (kg) is the unit of mass; it is
equal to the mass of the international prototype of
groups of such particles.
the kilogram.
The international prototype of the kilogram
is
a
1.5 Definitions of derived units
particular cylinder of platinum-iridium alloy that
is
preserved
in
International
a vault at Sevres, France, by the
Bureau of Weights and Measures.
Duplicates of the prototype exist in all important
standards laboratories
in the
world. The platinum-
iridium cylinder (90 percent platinum, 10 percent
iridium)
is
about 4 cm high and 4 cm
The second
(s) is the
in
diameter.
duration of 9 192 63
1
770
periods of the radiation corresponding to the transition
state
between the two hyperfine levels of the ground
of the cesium- 33 atom.
1
Some
of the more important derived units are de-
fined as follows:
The coulomb (C)
transported in
(Hence
1
l
is
the quantity of electricity
second by a current of
coulomb =
/
The degree Celsius (°C)
and
is
used
in
l
ampere.
ampere second.)
is
equal to the kelvin
place of the kelvin for expressing
Celsius temperature (symbol
t)
defined by the equa-
= T — Ta where T is the thermodynamic
perature and Tn = 273.15 K, by definition.
tion
t
tem-
FUNDAMENTALS
6
The farad
(F)
is
the capacitance of a capacitor
between the plates of which there appears a
ence of potential of
volt
I
when
quantity of electricity equal to
force
coulomb.
farad
coulomb per volt)
The henry (H) is the inductance of a closed circuit in which an electromotive force of
volt is produced when the electric current in the circuit varies
uniformly at a rate of ampere per second. (Hence
=
1
(I
/
=
I
volt
tromotive force. (Hence
second per ampere.)
angle with
that force
is
which gives
kilogram an acceleration of
second per second. (Hence
I
newton
=
to
I
a
angle with
kilogram
mass and an acceleration,
defined
in
The
terms of a
Exponent form
24
000 000 000 000 000 000 000 000
1()
000 000 000 000 000 000 000
10
000 000 000 000 000 000
000 000 000 000 000
10
l
I
I
I
in length to the radius.
is
the unit of electric conduc-
vertex at the center of a sphere and en-
tesla (T)
000 000
10
I
000
00
1
l()
0.1
0.01
2
IN
0.000 000 001
0.000 000 000 001
6
3
1()
10
2
10'
10-'
fi
l()"
9
10
10"
0.000 000 000 000 000 001
10
i2
15
Symbol
yotta
Y
zetta
Z
E
peta
P
tera
T
tr't-
mana
G
mega
M
kilo
k
hecto
h
dec a
da
deci
d
centi
c
milli
m
micro
nano
n
pico
P
fern to
f
alto
a
~ lx
10" 21
10
SI
2
10
10
0 000 000 000 000 000 000 000 001
Prefix
12
uf
0.000 000 000 000 001
0.000 000 000 000 000 000 001
UNITS
exa
0.001
0.000 001
SI
'
l()
10
l
in length to the radius.
the unit of magnetic flux density
,s
000 000 000 000
000 000 000
I
is
equal to one weber per square meter.
also applies to sta-
it
Multiplier
I
its
PREFIXES TO CREATE MULTIPLES AND SUBMULTIPLES OF
TABLE 1C
(S)
of a square with sides equal
is
ampere. )
closing an area of the spherical surface equal to that
meter per second squared.
Although the newton
J volt per
ohm. (The Siemens
was formerly named the mho. )
The steradian (sr) is the unit of measure of a solid
meter per
I
ohm =
tance equal to one reciprocal
newton meter)
1
I
vertex at the center of a circle and sub-
The Siemens
1
The newton (N)
its
tended by an arc equal
1
mass of
am-
The pascal (Pa) is the unit of pressure or stress
equal to one newton per square meter.
The radian (rad) is the unit of measure of a plane
I
I
I
pere, this conductor not being the source of any elec-
The hertz (Hz) is the frequency of a periodic
phenomenon of which the period is second.
The joule (J) is the work done when the point of
application of
newton is displaced a distance of
meter in the direction of the force. (Hence J joule
=
every application where a
to
1
1
henry
and
involved.
points, produces in this conductor a current of
1
I
is
The ohm (il) is the electric resistance between
two points of a conductor when a constant difference
of potential of
volt, applied between these two
charged by a
is
it
tionary objects
differ-
24
zepto
z
yocto
y
UNITS
the difference of electric poten-
electricity.
They contain notes
between two points of a conducting wire carry-
the reader
who
The volt (V)
tial
is
ing a constant current of
(Hence
I
=
volt
is
power
the
is
(Hence
I
=
watt
The weber (Wb)
ing a circuit of
one
motive force of
uniform rate in
J joule
watt.
1
in
Quantity
second. {Hence
I
an electro-
it
reduced
is
it
to zero at a
weber
=
I
volt
second.
Multiples
1.6
and submultiples
of SI units
by adding appropriate prefixes
Thus prefixes such as
to the units.
mega, nano, and centi
kilo,
multiply the value of the unit
by factors
radian
rad
square meter
nr
Energy (or work)
joule
J
Force
newton
N
Length
meter
m
Mass
Power
kilogram
kg
watt
W
Pressure
pascal
Pa
Speed
meter per second
m/s
1
rad/s
Nm
Volume
Volume
cubic meter
m
liter
L
rotation
.
Although the radian
ID, IE, and IF
2.
Most
countries, including
COMMON
UNITS
3.
units
some common
IN
A
The newton
is
4.
The pascal
5.
In this
is
units
6.
in
spelled liter or
litre.
(
1
rad/s
= 9.55
The
W
joule per (kilogram kelvin)
J/kg-Kor J/kg °C
Temperature
kelvin
K
Temperature difference
kelvin or degree Celsius
Thermal conductivity
watt per (meter-kelvin)
Kor °C
W/m-Kor W/m°C
is
organi-
1
N/nr.
Canada
It
is litre.
Note
Specific heat
K
some
r/min).
official spelling in
J
I
we
57.3°).
mainly used for liquids and gases.
is
Symbol
SI unit
exactly equal to a temperature difference of
I
°C.
I
2
The °C
is
l
I
a recognized SI unit and, in practical
often used instead of the kelvin.
Thermodynamic, or absolute, temperature
°C.
is
This unit of volume
watt
is
~
the revolution per minute (r/min) to
joule
it
(as well as
a very small pressure equal to
book we use
Heat
temperature difference of
rad
(I
a very small force, roughly equal to the
Thermal power
calculations,
2.
Canada
book
THERMODYNAMICS
Quantity
1.
in this
force needed to press a doorbell.
encountered in mechanics, thermodynamics, and
TABLE 1E
6
meter.
10 watts.
list
5
the SI unit of angular measure,
is
designate rotational speed
Tables
4
radian per second
6
Commonly used
1.7
3
zations in the United States), use the spelling metre instead of
seconds,
10
megawatt —
1
newton meter
amperes,
-9
=
Note
2
Torque
use the degree almost exclusively
/^ampere = 1000
nanosecond
Symbol
SI unit
Area
J
1
MECHANICS
listed in
Table 1C. For example,
1
IN
Angle
Speed of
Multiples and submultiples of SI units are generated
UNITS
per second. )
produces
turn,
COMMON
TABLE 1D
joule per sec-
the magnetic flux that, link-
is
volt as
1
1
1
that gives rise to the
production of energy at the rate of
ond.
equal to
watt per ampere.)
I
The watt (W)
particularly useful to
not yet familiar with the SI.
ampere, when the power
1
between these points
dissipated
is
1
The absolute temperature
T is
is
expressed
in kelvins.
On
related to the Celsius temperature
t
the other hand, the temperature of objects
by the equation
T=
t
+
273.
1
5.
is
usually expressed
FUNDAMENTALS
8
COMMON
TABLE 1F
UNITS
ELECTRICITY AND
IN
MAGNETISM
Quantity
Symbol
SI unit
Capacitance
farad
F
Conductance
Siemens
c
Electric charge
c
Electric current
coulomb
ampere
F nprtrv
joule
j
Erecjuencv
hertz
Hz
Inductance
henry
Potential difference
volt
i
i
A
Figure
2
watt
Resistivity
ohm
ohm
Magnetic
ampere
Drawn from
and Conversion Charts" by Theodore
Enterprises Ltd. All rights reserved.
n
"Metric Units
Om
meter
for units of length.
Conversion chart adapted and reproduced with permission. Copyright © 1991, 1995 by Sperika
W
Power
1.1
Conversion chart
H
V
Resistance
field strength
Note
Wildi.
A/m
IEEE
Press, Piscataway NJ, 08855-1331.
3
per meter
Magnetic flux
weber
Wb
Magnetic flux density
tesla
T
4
Magnetomotive force
ampere
A
5
1
.
2.
3.
4.
5.
listed in
descending order of size, and the
Formerly called mho.
of the connected units: the yard
Hz = cycle per second.
A/m = ampere turn per meter.
T - Wb/nr.
What was formerly called an ampere
than the inch, the inch
I
I
ampere:
I
A
I
ampere
turn
is
now simply
simple method.
Conversion charts and their use
Unfamiliar units can be converted to units
is
an arithmetic process that often leaves us
wondering
if
from yard
move downward
in the
Appendix eliminate
this problem because they show the relative size of
a unit by the position it occupies on the page. The
charts in the
in
units are
we reach
we want to
to
two arrows
millimeter.
convert from millime-
we
move
up-
ward against
the direction of the arrows until
we
start at
millimeter and
we
apply
the following rules:
1
If, in
.
traveling
move
in the
from one
unit to another,
direction of the arrow,
we
we
multi-
ply by the associated number.
between.
Conversely,
2.
The
if
to mil-
we have
ters to yards,
largest unit is at the top, the smallest at the bottom,
and intermediate units are ranked
direction of the
(36 and 25.4) until
Conversely,
from yards
in Fig. 1.1,
reach yard. In making such conversions
our calculations are correct.
The conversion
to convert
limeters. Starting
we know
well by using standard conversion tables. But this
strictly
we can
turn.
Suppose we wish
1.8
36 times larger
25.4 times larger than the
convert from one unit to any other by the following
I
called
is
is
millimeter, and so on. With this arrangement
1
1
1
lines join-
them bear an arrow that always points toward
the smaller unit. The numbers show the relative size
ing
connected by arrows, each of which
if
we move
against the arrow,
we
divide.
bears a number.
to the smaller
hence,
its
The number
is
the ratio of the larger
of the units that are connected and,
value
is
always greater than
row always points toward the smaller
In Fig.
I.I, for
unity.
The
ar-
unit.
example, five units of length
the mile, meter, yard, inch,
and millimeter
—
are
Because the arrows point downward,
this means
when moving down the chart we multiply, and
when moving up, we divide. Note that in moving
from one unit to another, we can follow any path we
that
please; the conversion result
is
always the same.
UNITS
The rectangles bearing SI units extend
toward the
Each rectangle bears
from other units.
for the unit as
ENERGY
slightly
of the chart to distinguish them
left
name
well as the
the
9
TNT
kilotonne of
J
symbol
1.167 x 10 6
of the unit written
.
out in full.
|
I
kilowatt hour
.
kW-h
|
| 3.6
mega joule
MJ|
|
Example
I- 1
1000
Convert 2.5 yards to millimeters.
British thermal unit
Btu
|
|
| 1.055
Solution
kJ
fkilojoule
|
Starting
(Fig.
from yard and moving toward millimeter
we move downward
1.1),
the arrows.
1000
in the
calorie
direction of
|
1
| 4.184
We must therefore
multiply the numbers
joule
|
associated with each arrow:
N-m
|n ewton-meter
2.5
-
yd
2.5
(
= 2286
X
36)
X
(
25.4) millimeters
van second
mm
I
I
6.24 x 10 18
Example 1-2
eV
[electronvolt
|
Convert 2000 meters into miles.
Figure 1.2
Solution
Starting
move
See Example
from meter and moving toward mile, we
and then against, the direction of
with,
first
the arrows.
Consequently,
2000 meters
= 2000 X
1
(
= 2000 X
we
1 995 by Sperika Enterprises
Drawn from "Metric Units and
Conversion Charts" by Theodore Wildi. IEEE Press,
sion.
Copyright© 1991,
Ltd. All rights reserved.
obtain
.0936) (- 1760) miles
Piscataway, NJ, 08855-1331.
1.0936
1760
=
1-3.
Conversion chart adapted and reproduced with permis-
The per-unit system
of measurement
1-9
1.24 mi
The
SI units just described enable us to specify the
Example 1-3
magnitude of any quantity. Thus mass
Convert 777 calories to kilowatt-hours.
kilograms,
volts.
Solution
Referring to the chart on
moving from calorie
travel
downward
to
ENERGY
(Fig. 1.2)
kilowatt-hour,
we
and
first
(with the arrow 4.184) and then
upward (against the arrows 1000, 1000, and
Applying the conversion rule,
we
power
in watts,
and
However, we can often
size of
expressed
in
get a better idea of the
something by comparing
thing similar. In effect,
is
electric potential in
it
to the size of
some-
we can create our own unit and
specify the size of similar quantities
compared
to this
arbitrary unit. This concept gives rise to the per-unit
3.6).
find
method of expressing
the
magnitude of a quantity.
For example, suppose the average weight of
777 calories
= 777 (X 4.184) (- 1000)
adults in
1000) (- 3.6)
New
any individual
=
9.03
X
10"
4
kW
h
York
is
130
lb.
Using
this arbitrary
weight as a base, we can compare the weight of
a
in
terms of
person weighing 160
lb
this
base weight. Thus
would have
a per-unit
FUNDAMENTALS
10
weight of 160
weighing
mb/ 130
I
The
lb/
lb
=
130 lb
-
Another person
1.23.
would have
15 lb
1
n
3500
weight of
a per-unit
a
4800
0.88.
measurement has
per-unit system of
vantage of giving the size of a quantity
±
the ad*2
terms of
in
450
xv
3000
n
n
a particularly convenient unit, called the per-unit
in
reference to our previ-
a football
player has a per-unit
base of the system. Thus,
ous example,
weight of
1.7
if
we immediately know
above average. Furthermore,
far
1.7
X
130
Note
=
that
221
1
.7
per-unit,
To
where
it
is
Figure 1.3
weight
is
Conventional
are given, they
would be absurd
weighs
state that the football player
is
weight
1
.7 lb.
to
His weight
the selected base unit
generalize, a per-unit system of
consists of selecting one or
is
130
lb.
Fig.
1
.3,
composed of several
we
them. In this book
decide to use an impedance of
base, the per-unit
measuring
*,(pu)
and impedance.
The base may be
power
is
40
ratings of 25 hp,
hp,
The
1
the base
=
3.
is
of 15 hp. In
40 hp/50 hp
-
0.8 and
world
in this per-unit
and 150 hp/ 15 hp
-
^50
n
4800 n
if
expressed
_
"
=
1.67, 40 hp/ 15 hp
3.2
3000
a
1500
ft
in
vector notation
shown
per-unit values.
in Fig.
1
is
same
impedances are
We can
solve this
other circuit. For example,
used, the per-unit circuit
is
that
.5.
power
-
2.33(pu)
3.2(pu)
-nrrv
2.67,
10.
therefore important to
value, the actual values of
per-unit
_
~
real circuit, but the
we would any
R 2 (pu)
know the magnitude
If we do not know
the quantities we are
0.30
i
dealing with cannot be calculated.
The
_ 0.30
~
150011
of the base of the per-unit system.
its
12
1500
and 3 pu, respectively.
well have selected a base
case the respective per-unit rating
this
2.33
50 hp, the three motors have
would be 25 hp/ 15 hp
It is
0.5;
ratings of 0.5, 0.8,
We could equally
=
per-unit circuit (Fig. 1.4) contains the
now
Thus,
=
elements as the
circuit as
of 50 hp. The corresponding per-unit ratings
hp
XL (pu)
said to have
PH
where
power
a
3500
a power, a voltage,
and 150 hp. Let us select an arbitrary base power
=
=
Xc (pu)
a current, or a velocity. For example, suppose that
50 hp/50 hp
=
R 2 (pu)
one quantity as our
system
stick, the per-unit
are then 25 hp/50
we
150011
system with one base
three motors have
If
as the
impedances are as follows:
more convenient mea-
select the size of only
a single base.
500 ohms
are particularly interested in
current, power, torque,
we
1
measurement
selecting convenient measuring sticks for voltage,
1.10 Per-unit
resistors, capacitors,
and inductors having the impedances shown.
suring sticks and comparing similar things against
If
circuit.
lb.
whenever per-unit values
always pure numbers. Thus
are
his
his actual
method can
also be applied to im-
pedances. Consider, for example, the circuit
in
Figure 1.4
Per-unit circuit.
2(pu)
UNITS
2.33
3.2
1
1
In order to understand the significance of this re-
j
sult, the
reader should study the two following ex-
amples. The bases are the same as before, namely
4kV
0.30
500
Figure 1.5
=
/B
kW
ZB =
A
125
32 ft
Example 1-4
Per-unit circuit with
notation.
i
A 400 II resistor carries a current of 60 A.
above base values,
a.
system
with two bases
Per-unit
1.11
b.
c.
when two bases
particularly useful
are
bases are usually a base voltage £" B
P B Thus
power
4
becomes
used. The
electrotechnology the per-unit system
In
.
kV and
d.
e.
The per-unit resistance
The per-unit current
The per-unit voltage across the resistor
The per-unit power dissipated in the resistor
The actual E and P of the resistor
and a base
the selected base voltage
may
Solution
be
a.
the selected base
power 500 kW.
The
per-unit resistance
The two base values can be selected quite indeb.
unit
The
per-unit current
system
is
that
it
automatically establishes a cor-
/>w
For example,
is
base power
PB
base voltage
EH
impedance Z B
and the base
if
—
is
d.
base voltage
The
the base voltage
/
4
is
500 kW, the base current
ZB = £ b /'b =
In effect,
system
impedance.
per-unit
is
kV
and the base
is
1
The
0.48
is
V/125
A =
32
ft
also get a base current and a base
Consequently,
the
so-called
=
0.48
=
6
=
X
X
tf(pu)
12.5
is
E(pu)
=
6
-
2.88
X
X
/(pu)
0.48
is
E — E B X E(pu)
= 4 kV X 6
= 24 kV
is
4( )00
/(pu)
actual voltage across the resistor
A
25
=
power
P(pu)
by selecting the voltage/power per-
we
The
n
= ^b/^b = 500 000/4000 =
The base impedance
system.
12.5
per-unit voltage across the resistor
EH
—
base current
e.
'b
per-unit
=
/ B is
£(pu)
unit
ft/32 ft
= 60 A/ 125 A =
/(pu)
Thus
c.
base current
power
is
interesting feature of the voltage/power per-
responding base current and base impedance.
the
= 400
tf(pu)
pendently of each other.
One
Using the
calculate:
2-base
system really gives us a 4-base per-unit
The
actual
power
dissipated in the resistor i$^;^'as''
P = PB x
^(pu)
= 500 kW X
= 1440kW
BCOLTAD
2.88
Df.
RU,]?\
vlii TA
A
©IBUCUfcCA
FUNDAMENTALS
12
Example 1-5
A
7.2
/|.(Pu)
.
= 2.844
kV
source delivers power to a 24 II resistor
and a 400
kW
electric boiler (Fig. 1.6).
Draw
equivalent per-unit circuit diagram. Use the
base values as
Example
in
the
same
/i(pu)
/,(pu)
= 0.444
= 2.4
1-4.
R{pu)
boiler
[
0.75
j
Calculate
The
The
The
The
The
The
a.
b.
c.
d.
e.
f.
per-unit E(pu), /?(pu), PCpu)
per-unit current
P(pu)
0.8
/ 2 (pu)
per-unit line current
(pu)
l
{
power absorbed by the resistor
power absorbed by the resistor
per-unit
actual
Figure 1.7
Per-unit version of Figure
1
.6.
actual line current
The
per-unit line current
/t
(pu)
ft
24
n
r
400 kW
The
d.
per-unit
is
=
/,(pu)
+
/ 2 (pu)
=
0.444
+
2.4
=
2.844
power
P(pu)
Figure 1 .6
See Example
/L
in the resistor is
=
=
E(pu)
=
4.32
X
X
1.8
/ 2 (pu)
2.4
1-5.
The
e.
actual
power
in the resistor is
Solution
a.
The
per-unit line voltage
£,(pu)
P 2 = P a X «pu)
= 500 kW X 4.32
is
- 7.2kV/4kV =
1.8
= 2160 kW
The
per-unit resistance
is
The
f.
R(pu)
= 24
n=
fi/32
actual line current
=
12
The
per-unit
power of the
P(pu)
We
b.
can
The
boiler
=
is
= 400 kW/500 kW =
now draw
c.
The
=
£(pu)//?(pu)
=
1.8
-
2.4
-
-
1
/,
125
Name
/,
X
(pu)
2.844
=
355.5
A
the seven base units of the
International
0.75
1
-2
1-3
per-unit current
X
Questions and Problems
is
1
/ 2 (pu)
IB
0.8
the per-unit circuit (Fig. 1.7)
per-unit current f2
is
0.75
Name
System of
Units.
five derived units of the SI.
Give the symbols of seven base
units,
paying
is
particular attention to capitalization.
/,(pu)
=
P(pu)/E(pu)
-
0.444
=
0.8/1.8
1-4
Why
are
some derived
units given special
UNITS
1-5
What
ergy,
1-6
1-58
revolution
1
-60
oersted
1-59
degree
1
-6
ampere
are the SI units of force, pressure, en-
power, and frequency?
Give the appropriate prefix for the following
multipliers: 100, 1000, 10
1/1000,
10" 6
,
Make
the following conversions using the conver-
1/10, 1/100,
,
io'-\
1-62
10 square meters to square yards
1-63
250
1-64
1645 square millimeters
Express the following SI units in symbol form:
1-7
megawatt
1-21
millitesla
1-8
terajoule
1-22
millimeter
1-9
millipascal
turn
sion charts:
10" 9
,
6
1
1
1-23
1
revolution
1-10
kilohertz
1-24
megohm
1-11
gigajoule
1-25
megapascal
1-12
milliampere
1-26
millisecond
-65
1
3
MCM to square millimeters
000
circular mils to square millimeters
1-66
640 acres
1-67
81
1-68
to square inches
square kilometers
to
000 watts
to
Btu per second
33 000 foot pound-force per minute to kilowatts
1-13
microweber
1-27
picofarad
1-14
centimeter
1-28
kilovolt
1-15
liter
1-29
megampere
1-16
milligram
1-30
kiloampere
1-17
microsecond
1-31
kilometer
1-69
1
-70
1-71
1-18
millikelvin
1-19
milliradian
1-20
terawatthour
1-32
1-33
1
-34
flow
I
-38
density
1-39
power
1-36
plane angle
1-40
temperature
1-41
microjoules
10 pound-force to kilogram-force
60 000
-72
1
-73
1
-74
50 ounces
1
-75
76 oersteds
1
-76
5
1
-77
1
lines per square inch to teslas
kilogauss
.2 teslas to
to
kilograms
to
000 meters
amperes per meter
to miles
and
frequency
magnetic flux
to
meters
milliliter
1-35
1-37
0 foot pound-force
1
symbol:
rate of
1
feet to cubic
nanometer
State the SI unit for the following quantities
write the
250 cubic
mass
80 ampere hours
1-78
25 pound-force
1-79
25 pounds
1-80
3 tonnes to
1-81
1-82
Give the names of the SI units that eorresponcl to
1
00 000
0.3
to
to
coulombs
newtons
to
kilograms
pounds
lines of force to
webers
pounds per cubic inch
kilograms per
to
cubic meter
the following units:
1-83
1-42
Btu
1-51
bar
1-43
horsepower
1-52
pound-mass
1-44
line
of flux
1-53
1
-84
1
-85
pound-force
2 inches of mercury to millibars
200 pounds per square inch
to pascals
70 pounds-force per square inch
to
newtons
per square meter
1-45
inch
1-54
kilowatt-hour
1-46
angstrom
1-55
gallon per
1-47
cycle per second
1-48
gauss
1-49
line per
1-50
°F
1
1
square inch 1-57
mho
-87
1-88
pound-force per
1
square inch
1
5 revolutions per minute to radians per
second
minute
1-56
-86
-89
A temperature of 20 °C to kelvins
A temperature of 200 °F to kelvins
1
A temperature
kelvins
difference of
1
20° Celsius to
14
I
-90
FUNDAMENTALS
A resistance of 60 ft
is
Industrial application
selected as the base
resistance in a circuit. If the circuit contains
1
-94
A
motor has an efficiency of 92.6%. What
is
the efficiency in per-unit?
A
variable-speed motor having a nameplate
three resistors having actual values of 100 fl,
3000
II,
and 20 O, calculate the per-unit
1-95
value of each resistor.
1-91
A power of 25 kW
rating of 15 hp,
and a voltage of 2400
are selected as the base
voltage of a
V
power and base
power system. Calculate
890 r/min develops a
torque of 25 newton meters
at
1260 r/min.
Calculate the per-unit values of the torque,
the
speed, and power.
value of the base impedance and the base
1-96
Three
resistors
have the following
ratings:
current.
1
-92
A resistor
has a per-unit value of 5.3.
base power
12
is
470
is
250
If the
resistor
resistance
A
10012
kW and the base voltage
ohmic value of
V, calculate the
the resistor.
1-93
A
length of 4
m
is
selected as a base unit.
b.
the per-unit length of
1
mile
C
300
II
the per-unit length of
1
foot
of resistors
1
d.
e.
the per-unit value of a
the magnitude of the base area (in
m
f.
ft
W
W
40 W
24
75
and voltage
values of the resistance, power,
m2
3
the magnitude of the base volume (in m
c.
50
Using resistor A as a base, determine the per-unit
Calculate
a.
B
power
3
the per-unit value of an area of 2 square
A
rating
C, respectively.
30 hp cage motor has the following cur-
rent ratings:
)
volume of 6000
-97
B and
)
FLA:
full-load current
LRA: locked
NLA:
36
A
rotor current 2 8
1
A
no-load current 14 A.
miles
Calculate the per-unit values of LRA and NLA.
Chapter 2
Fundamentals of Electricity,
Magnetism, and Circuits
Introduction
2.0
positive
^
negative
(
terminal (+) -fli-ftr
This
chapter briefly reviews
some of
t
ermina
the funda-
mentals of electricity, magnetism, and circuits.
We assume
the reader
already familiar with the
is
dry
cell
including the solution of electric circuits.
basics,
However, a review
useful because
is
it
focuses on
those items that are particularly important in
technology. Furthermore,
used throughout this
Some
currents.
it
book
power
establishes the notation
and
to designate voltages
Figure 2.1
of the topics treated here will also
Dry
cell,
provide the reader with a reference for subjects covered in later chapters.
electron current flow
Conventional and electron
2.1
current flow
Consider the dry cell
positive
(
shown
+ and one
difference of potential
volts) is
tive
terminal
If
(
—
)
1
,
having one
terminal.
The
between them (measured
in
to an excess of electrons at the nega-
compared
we connect
tential
in
due
in Fig. 2.
negative
)
to the positive terminal.
a wire across the terminals, the po-
difference causes an electric current to flow
is composed of a steady
come out of the negative
the circuit. This current
stream of electrons that
terminal,
move along
the wire, and reenter the cell
Figure 2.2
by the positive terminal (Fig. 2.2),
Electron flow.
15
'
<-)
FUNDAMENTALS
16
Before the electron theory of current flow was
fully understood, scientists of the 17th century arbi-
decided
trarily
that current in a
from the positive terminal
to the negative terminal
This so-called conventional current flow
(Fig. 2.3).
used today and
is still
conductor flows
is
the accepted direction of
power technology.
book we use the conventional current
current flow in electric
In this
flow, but
flow
is
it is
worth recalling
that the actual electron
opposite to the conventional current flow.
In
order to establish a general rule, consider two
black boxes
A and B
ally
changing
between sources
/ that is
in direction (Fig. 2.4).
drop along the wires
assumed
is
that are
connected
to be zero.
some way
in
A2
to the external ter-
.
Under such highly
are also continually changing.
is
how can we
tell
To answer
sometimes important
and loads
in
an electric
to identify the sources
circuit.
By
definition, a
source delivers electrical power whereas a load absorbs
it.
Every
electrical device (motor, resistor,
thermocouple, battery, capacity, generator,
How
can
we
tell
the
A or B
the question,
suppose we have appro-
instantaneous polarity
the terminals
(
+ )( —
)
determine the
of the voltage across
and the instantaneous direction of
conventional current flow. The following rule then
applies:
etc.) that
carries a current can be classified as either a source
or a load.
whether
a source or a load?
priate instruments that enable us to
is
Each
and B h B 2 A variable voltage exists
across the terminals, and its magnitude and polarity
minals A],
and loads
It
continu-
The voltage
box contains unknown devices and components
variable conditions,
2.2 Distinction
connected by a pair of
that are
wires carrying a variable current
•
A device
is
a source
whenever current flows out
of the positive terminal.
one from the other?
•
A device
is
a load
whenever current flows
into a
positive terminal.
conventional current flow
If the
instantaneous polarities and instantaneous
current flow are as
from the
load.
rule that
However,
if
shown
box
A
is
in
and box
The above
is
A the
follows
B
is
a
box B would become
load.
rule for establishing
a source or load
it
the current should reverse while
the polarity remains the same,
the source
Fig. 2.4,
a source and box
is
whether a device
very simple, but
it
has impor-
tant applications, particularly in alternating current
circuits.
Figure 2.3
Some
devices, such as resistors, can behave only
as loads.
Other devices, such as photocells, can act
only as sources. However,
many
Conventional current flow.
either as sources or as loads.
/
Figure 2.4
between a source and a
Distinction
load.
devices can behave
Thus when a
battery
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
power,
delivers electric
flows out of the
recharged,
(4
)
(
+
)
it
when
being
is
it
acts as a load (current flows into the
it
The
acts as a source (current
terminal);
ties
•
on a system, but they can briefly behave
generators
When
is
thing
when
is
like
£AB = +
tween
100 V, which reads: The voltage be-
A and B
is
100 V, and
A
is
(
the capacitor
it
+
is
acts as a source
On
terminal.
)
charging up,
and current flows into the
+
(
)
it
•
positive with
the
£ BA = 100 V, which reads: The voltage between A and B is 100 V, and B is negative with
respect to A.
acts
As another example,
terminal.
if
we know
ator voltage in Fig. 2.6 has a value
that the gener-
E2 = —
\
100 V,
we know that the voltage between the terminals is
00 V and that terminal 2 is negative with respect to
Sign notation
2.3
relative polari-
true of capacitors.
discharging
and current flows out of the
as a load
and the
can be designated by the
respect to B.
The same
a capacitor
other hand,
B
and
the electromechanical conditions are
if
appropriate.
A
double-subscript notation, as follows:
terminal). Similarly, electric motors usually act
as loads
potential difference
of terminals
1
1
In
arithmetic
we
scribe addition
mechanics,
direction
)
to de-
etc.,
to indicate the
compared
+
that the direction
This interpretation of
(
+
to an ar-
the speed
if
—400 r/min,
100 r/min to
of rotation has reversed.
)
and
—
(
terminal
2.5 Sign notation for voltages
Although we can represent the value and the polarity
of voltages by the double-subscript notation (£, 2
£ AB
in the
signs
is
fre-
we
etc.),
£2 V
etc.)
(
+
sign.
)
in
marked with
to
G
having a positive terminal
terminal B. Terminal
A is positive
minal B. Similarly, terminal
spect to terminal A.
tive
a positive
(
+
)
by
sign.
is
itself:
it
is
Note
B
is
A and
The other terminal
2
shows
a
a negative
with respect to ter-
negative with re-
that terminal
A is
not posi-
only positive with respect to B.
Figure 2.6
= 100
If E
21
terminal
V,
terminal 2
is
a
arbitrarily
describe a system of notation that enables us
indicate the polarity of voltages. Fig. 2.5
source
(£,,
For example. Fig. 2.7 shows
which one of the terminals
Double-subscript notation
for voltages
We now
symbol
,
It
and identifying one of the terminals by a
chapters that follow.
source £,
2.4
often prefer to use the sign notation.
consists of designating the voltage by a
,
)
,
positive
met
quently
1.
and
of an electric current, of a mechanical
motor changes from
means
)
In electricity
meaning
the
(
chosen direction. For example,
bitrary
it
we broaden
+ and —
(
and subtraction.
of a rotational speed,
force,
of a
use the symbols
negative with respect
1.
Figure 2.5
Figure 2.7
Double-subscript notation to designate a voltage.
Sign notation to designate a voltage.
to
FUNDAMENTALS
18
unmarked, but
is
is
automatically assumed to be neg-
ative with respect to the
With
•
If
(
we
this notation the
+
= +
state that £,
)
indicated in the diagram.
+
(
)
sign
that the
The terminal bearing
then actually positive and the
is
other terminal
is
negative. Furthermore, the
magnitude of the voltage across the terminals
Figure 2.9
islOV.
Solution of
Conversely,
•
means
10 V, this
of the terminals corresponds to that
real polarity
the
terminal
following rules apply:
if
the terminals
£,
= -
shown on
diagram. The terminal bearing the
(
+
)
sign
actually negative, and the other terminal
is
The magnitude of the voltage across
tive.
terminals
is
the
is
the
10 V.
circuit of Fig. 2.8 consists of three sources
7
V
V2 and V?
,
,,
with a positive
in series to
— each
(
+
)
—
having a terminal marked
The sources are connected
using jumper wires A, B, C,
sign.
a resistor R,
Determine the actual value and polarity of the
voltage across each source,
V,
spectively. In effect, point
V2 = +
10 V, and
knowing
V? = -40
that
V
=
ues and polarities are as
in
—
That
is
a negative sign.
2.6
Graph
of
an alternating voltage
In the chapters that follow,
whose voltages change
alternating voltages
voltage
at
2.
1
may be
0).
we encounter
sources
polarity periodically.
The
Such
represented by means
vertical axis indicates the
each instant, while the horizontal axis
in-
dicates the corresponding time. Voltages are positive
when they
when
are
E2
above the horizontal axis and nega-
they are below. Figure 2.10 shows the
\
produced by the generator of Fig,
2.6.
find that the true val-
shown in Fig. 2.9. However,
jumper A, it seems im-
2
directing our attention to
possible that
(
re-
negative with respect
to point C.
voltage
we
only
{
V.
Solution
stated,
is
B and positive with respect
why A carries both a positive and
tive
Using the rules just
A
It
jumpers B and C,
to point
of a graph (Fig.
and D.
-4
inherently positive nor inherently negative.
has a polarity with respect to
posi-
Example 2-1
The
2-1
10 V, the real polarity of
the reverse of that
is
Example
).
it
can be both positive
However, we must remember
Figure 2.8
Example
Circuit of
2-1.
(
+
)
and negative
that
A
is
neither
Figure 2.10
Graph of an alternating voltage having a peak
of
1
00
V.
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
from zero,
Starting
taining
+ 100 V
falls to
zero
E2]
gradually increases,
after 0.5 second.
spect to terminal
because
1
E2
During the interval from
1
spect to terminal
polarities
onds are
1
.
2.7 Positive
is
insets
at 0.5,
I, IT,
E2X
is
negative with re-
The instantaneous
of the generator
shown by
positive.
to 2 seconds,
negative; therefore, terminal 2
0
this
positive with re-
is
is
\
at-
then gradually
It
end of one second. During
at the
one-second interval, terminal 2
1
and
voltages and
.5,
and
TIT
of Fig. 2.10.
2.
1
7 sec-
and negative currents
0
We
also
indicate
1
make use of positive and negative signs to
the direction of current flow. The signs are
*-
time
>v
X
|
allocated with respect to a reference direction given
on the circuit diagram. For example, the current
a resistor (Fig. 2.11)
Y to
X.
One
be positive
may flow from X
of these two directions
+
(
is
in
Y or from
considered to
and the other negative
)
to
(
—
Figure 2.13
and the corresponding graph
Electric circuit
The arrow
).
of current.
indicates the positive direction of current flow.
Solution
According
zero to
Because
may
is
the interval
flow from
X
to
Y
or from
Y
to X.
interval
from
from
+2 A to
in the resistor.
to
l
it
to
l
second.
from B
to
A
of the arrow). During the
2 seconds, the current decreases
zero, but
circulates
it still
Between
increases from zero to
ative,
from 0
positive, the current flows
in the resistor (direction
Figure 2.11
Current
it
from
to the graph, the current increases
+2 A during
from B
to
A
2 and 3 seconds, the current
—2 A and,
because
it
is
neg-
really flows in a direction opposite to that
of the arrow; that
is,
from
A
to
B
in the resistor.
Figure 2.12
Circuit
element showing positive direction
of current flow.
2.8 Sinusoidal voltage
The
The positive direction
means of an arrow (Fig.
2
A flows from X to Y,
and
if
is
it
is
2.
1
shown
2).
flows
Thus,
of the arrow),
from
is
it
if
by
a current of
+2
Y to X (direction opposite to that
designated by the symbol
graph
a resistor
in
shown
of this graph.
in
Figure
2.
R
1
—2
A.
varies according to the
3.
It
may
therefore
be expressed by the equation
e
= Em
cos
{lirft
+
6)
(2.1)
A. Conversely,
Example 2-2
The current
ac voltage generated by commercial alternators
very nearly a perfect sine wave.
in the positive direction
designated by the symbol
current flows
arbitrarily
is
Interpret the
meaning
where
e
—
Em —
/ =
instantaneous voltage [V]
peak value of the sinusoidal voltage [V]
frequency [Hz
t
=
time
6
=
a fixed angle [radj
[s]
FUNDAMENTALS
20
The expression lirft and 6 are angles, expressed
However, it is often more convenient to
The voltage
in radians.
express the angle
in
e ab
degrees, as follows:
e
= Em
cos (360 ft
*
= £ ni
cos
+
In these
(2.3)
0)
equations the symbols have the same sig-
s is
100 cos 488 622°
X 50 X 27.144 +
30°)
V
moment
at this
terminal a
Note
21 AAA-
100 cos (360
-20.8
Thus,
+
=
=
or
<<|>
t
=
(2.2)
6)
at
the voltage
is
V
—20.8
and
negative with respect to terminal b.
is
that an angle of
488 622° corresponds
to
488
622/360 = 357 complete cycles plus 0.2833 cycle.
The latter corresponds to 0.2833 X 360° = 102°,
and 100 cos 102° = -20.8v
1
nificance as before, and the time-dependent angle
(= 360
ft) is
Example 2-3
The sine wave
_
in Fig.
2.14 represents the voltage
£. lb across the terminals a
operates at 50 Hz.
and b of an ac motor
Knowing
that 6
100 V, calculate the voltage
27 .1 44
<j)
also expressed in degrees.
at
t
2.9 Converting cosine functions
into sine functions
that
= 30°, and E m =
= 0 and at t =
We can convert a cosine function of voltage or current
into a sine function
Em
Solution
The voltage
at
t
e, xh
=
At
this
is
moment
0
Em
is
= E m cos (360 ft + 9)
- 100 cos (360 X 50 X
= 100 cos 30°
-
a
by adding 90°
to the angle 6.
Thus,
s.
86.6
Similarly,
0
+
30°)
+86.6
V
and terminal
B
+
=
90°)
we can convert a sine function
/m
is
0)
(360 ft
cos (360 ft
+
6)
(2.4)
into a cosine
function by subtracting 90° from the angle
V
the voltage
+
(360 ft
sin
/ m sin
therefore positive with respect to terminal b.
+
cos (360 ft
6.
Thus,
=
+ 6-90)
(2.5)
2.10 Effective value of an ac voltage
Although the properties of an ac voltage are known
when its frequency and peak value Em are specified, it
is much more common to use the effective value £e(j
.
For a voltage
E
between
cfi
-
that varies sinusoidally, the relationship
and
£ni
is
given by the expression
Ecn = EJ,2
(2.6)
-
The
effective value of an ac voltage
is
some-
times called the
RMS
the voltage.
a measure of the heating effect of
It is
(root
compared
the ac voltage as
mean square) value of
to that
of an equivalent
dc voltage. For example, an ac voltage having an
fective value of
1
ing effect in a resistor as does a dc voltage of
Figure 2.14
Sinusoidal voltage having a peak value of 100
expressed by e ab =
Em
cos (360
ft
+
30°).
V and
ef-
35 volts produces the same heat1
35 V.
The same remarks apply to the effective value of
an ac current. Thus a current that varies sinusoidally
and whose peak value is / m possesses an effective
value
/C
given by
|
t
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
U
n
2
(2.7)
Most alternating current instruments are
brated to
rent
show
c.
e
cali-
the effective value of voltage or cur-
and not the peak value (Fig. 2.15).
value of an alternating voltage or current
When
given
is
339
it
understood
that
it
the
is
effective
E^-and
Furthermore, the subscript in
Owing
E
given by
sin 21
600
t
t
to the phase lag of 30°, the current
is
value.
given by
/eir is
dropped
/
and the effective values of voltage and current are
simply represented by the symbols
is
= E m sin 360 ft
= 339 sin 360 X 60
the
d.
is
Let us assume the voltage
21
and
/.
= /m sin (360 ft - 30)
= 14.1 sin (21 600 f
=
14.1 sin
(<J>
-
30)
30)
Example 2-4
A 60 Hz source having an effective voltage of
240 V delivers an effective current of 10 A to a circuit. The current lags the voltage by 30°. Draw the
waveshape for
E and
Em = E
b.
/
In
The peak voltage
x
2
m =
/
1
Phasor representation
most power studies the frequency
we simply
is
= 240
x
2
val-
1
take
it
is
fixed,
for granted. Furthermore,
and so
we
n
2
is
magnitudes and phase angles.
= 10x2=
14.1
are
not particularly concerned with the instantaneous
= 339 V
voltages and currents but more with their
The peak current
/
The waveshapes giving the instantaneous
ues of e and are shown in Figure 2. 6.
2.1
/.
Solution
a.
e.
A
ages are measured
in
And because the
RMS
volt-
terms of the effective values
E
Figure 2.15
Commercial voltmeters and ammeters are graduated
ing
up
to
2500 A and 9000
V.
(Courtesy of General Electric.)
in
effective values. This
range
of instruments
has scales rang-
FUNDAMENTALS
22
\
16.67
\
30°
14
•
A
\
\
(a)
ms
A
1
'*
9'
<b)
1
t
E
h-
1.39
E
/
V
339
I
ms
Figure 2.17
current phasor
The
/
and voltage phasor
E are
in
phase.
Figure 2.16
Graph showing the instantaneous values of voltage and
current. The current lags 30° behind the voltage. The effective voltage is 240 V and the effective current is 10 A.
E and
interested in
E m we
peak values
rather than the
are really only
,
Figure 2.18
Phasor / lags behind phasor
0.
E by an
angle of
0
degrees.
This line of reasoning has given rise to the phasor method of representing voltages and currents.
The
basic purpose of phasor diagrams
sense that
it
A
phasor
is
phasor
show
represents.
the
same angle
If a
phasor
bears an arrow, and
its
length
is
The angle between two phasors
to the electrical phase angle
propor-
3.
is
it
equal
between the quantities.
1
.
Two
phasor
E is
/ is
phasors are said to be
are parallel to each other
them
in
phase when they
and point
in the
same
Two
one
The phase angle between
direction as phasor
said to lead phasor
/.
same
Fig. 2.
is
1
8,
it
/ lags
behind
make
then
Conversely, a
E if phasor
make
has to be rotated counterclockwise to
point in the
/,
it
direction. Thus, referring to
clear that phasor
£ by
we could
E
leads phasor /
equally well say that
0 degrees.
then zero.
is
phasors are said to be out of phase
they point in different directions.
gle
to be rotated clockwise to
same
by 6 degrees. But
4.
2.
rotate
line up.
said to lag behind phasor
rules apply to phasors:
direction (Fig. 2.17).
we
sweep through
to
make them
to
E has
point in the
phasor
/
The following
we have
similar to a vector in the
tional to the effective value of the voltage or current
it
Consequently, whether
/.
phasor or the other,
between voltages
the magnitudes and phase angles
and currents.
to
is
between them
is
point in the
same
The phase
the angle through
one of the phasors has
when
to be rotated to
Referring
phasor
/
now
to Fig. 2.
1
9
we could rotate
(3 to make it
clockwise by an angle
an-
which
make
it
direction as the other. Thus,
referring to Fig. 2. 1 8, phasor / has to be rotated
counterclockwise by an angle 6 to
in the
same
phasor
E has
gle 6 to
make
it
point
direction as phasor E. Conversely,
make
to be rotated
it
clockwise by an an-
point in the
same
direction as
Figure 2.19
Phasor / leads
£ by
B
E by
degrees.
(3
degrees. But phasor
/
also lags
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
same
point in the
direction as phasor E.
could then say that phasor
23
We
E by
leads phasor
/
P degrees. But this is the same as saying that
phasor / lags phasor E by 0 degrees. In practice,
we always
select the smaller phase angle
between the two phasors
to designate the lag
or lead situation.
5.
common
Phasors do not have to have a
may be
shown
that
in Fig. 2.20.
£|
same
from each
entirely separate
By
phase with
in
is
applying rule
3,
we can
because they point
l
see
in the
x
direction. Furthermore,
sor £,
origin but
other, as
by 90°, and
E2
phasor
/ 2 leads
lags behind /2 by
1
pha-
35°.
Figure 2.21
Different ways of showing the phase relationships
between three voltages that are mutually displaced
at 120°.
Solution
To draw
trary
equivalent to 240 V. Phasor
Figure 2.20
Phasors do not have to start from a
show
their
the phasor diagram,
common
origin to
magnitudes and phase relationships.
it
we
direction for phasor E,
E with a length
lags 30° behind
(Fig. 2.22).
/ is
select any arbimaking its length
then drawn so that
Knowing
equivalent to
frequency
that the
is
1
0A
60 Hz,
between the positive peaks
the time interval
is
given by
In the
in Fig.
same way,
£bc
the three phasors
2.2 1 a can be rearranged as
without affecting the
shown
,
and
£ca
in Fig. 2.2
1
= 360 ft
30 = 360 X 60
= 1.39 ms
9
b
phase relationship between
t
£ab in Fig. 2.21b still points in the
as Eab in Fig. 2.21a, and the same is
them. Note that
same direction
true for the
Fig. 2.2
three
240 V
other phasors.
1
c
shows
still
t
another arrangement of the
phasors that does not
in
any way
alter their
magnitude or phase relationship.
The angle
0
between two phasors is a measure
peak positive val-
of the time that separates their
ues.
Knowing
the frequency,
we can
calculate the
of the voltage
and current given
in
Figure 2.16.
time.
Example 2-5
Draw the phasor diagram of the voltage and current
in Fig. 2. 16. Calculate the time interval between the
positive
Figure 2.22
Phasor diagram
peaks of
E
and
/.
2.12
Harmonics
The voltages and
currents in a
power
The
quently not pure sine waves.
circuit are freline
voltages
FUNDAMENTALS
24
usually have a satisfactory
waveshape but
Fig. 2.23. This distortion
distorted, as
netic saturation in the cores of transformers or
switching action of thyristors or
IGBTs
In order to understand the distorting effect of a
the cur-
shown in
can be produced by mag-
sometimes badly
rents are
by the
in electronic
drives.
harmonic,
us consider two sinusoidal sources
in series (Fig. 2.24a).
quencies are respectively 60
Hz and
1
Their
c,
fre-
80 Hz. The
V and 20 V.
The fundamental (60 Hz) and the third harmonic
80 Hz) voltages are assumed to pass through zero
at the same time, and both are perfect sine waves.
Because the sources are in series, the terminal
corresponding peak amplitudes are 100
(
1
voltage
—4
let
and e 2 connected
is
equal to the
sum of
the instantaneous
The resulting
wave (Fig. 2.24b).
voltages produced by each source.
—
A
0
4- V
terminal voltage
is
a flat-topped
-
Thus, the sum of a fundamental voltage and a har-
monic voltage yields a nonsinusoidal waveform
whose degree of distortion depends upon the mag-
—
nitude of the harmonic (or harmonics)
60
1
20
1
300
240
80
it
contains.
420
360
Figure 2.23
This severely distorted 60
Hz
current obtained on an
electronic drive contains the following harmonics: funda-
mental (60 Hz) = 59 A; fifth harmonic (300 Hz) - 15.6 A;
seventh harmonic (420 Hz) = 10.3 A. Higher harmonics
are also present, but their amplitudes are small.
(Courtesy of Electro-Mecanik.)
The
to the
distortion of a voltage or current can be traced
harmonics
it
contains.
A harmonic
age or current whose frequency
is
is
any
volt-
an integral multi-
ple of (2. 3, 4. etc., times) the line frequency.
Consider a
set
of sine waves in which the lowest
frequency is/ and
multiples of /
the lowest
the other
By
all
frequency
waves
other frequencies are integral
definition, the sine
is
wave having
called the fundamental and
are called harmonics. For example,
waves whose frequencies are 20, 40,
and 380 Hz is said to possess the following
a set of sine
1
00.
components:
(b)
fundamental frequency: 20
Hz
(the lowest
frequency)
Figure 2.24
second harmonic: 40 Hz (2
fifth
harmonic: l()0Hz(5
X 20 Hz)
a.
X 20 Hz)
Hz
(
19
X 20 Hz)
sinusoidal sources having different frequen-
cies connected
b.
nineteenth harmonic: 380
Two
in
series.
A fundamental and
third
harmonic voltage can
gether produce a flat-topped wave.
to-
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
We
can produce a periodic voltage or current of
any conceivable shape. All
gether a fundamental
we have
to
do
component and an
of harmonic components. For example,
to
is
add
to-
power
we can
distorted
gen-
wave having an amplitude of 00 V and
a frequency of 50 Hz by connecting the following
sine wave sources in series, as shown in Table 2A.
In
1
100 V
SQUARE WAVE
waveshapes
Freq.
[VI
[Hz|
that are rich in
mental
voltage
harmonics.
produce
together
power. This fundamental power
motor
a
is
fundamental
power
the useful
and an arc furnace
to rotate
The product of
heat up.
Amplitude
in
ac circuits the fundamental current and funda-
the corresponding
Harmonic
produced whenever voltages
electronic circuits. All these circuits produce
that causes a
TABLE 2A
are also
and currents are periodically switched, such as
arbitrary set
square
erate a
They
circuits.
25
to
harmonic voltage times
harmonic current also produces a
Relative
amplitude
harmonic power. The
latter is
usually dissipated as
heat in the ac circuit and, consequently, does no
127.3
50
third
42.44
150
1/3
fifth
25.46
250
1/5
fundamental
useful
1
work.
Harmonic
and
currents
voltages
should therefore be kept as small as possible.
It
should be noted that the product of a funda-
mental voltage and a harmonic current yields zero
seventh
18.46
350
1/7
ninth
14.15
450
1/9
net power.
Harmonics are covered
in
greater detail
in
Chapter 30.
2.13 Energy in an inductor
lh
I27
1.00
6350
1/127
A
energy
coil stores
carries a current
/.
W
127.3/n
A
tal
square
wave
wave and an
is
thus
infinite
50 n
1/n
W=
composed of a fundamennumber of harmonics. The
and they are consequently
less
wave.
and pointy corners of the square
In practice,
square waves are not produced by
adding sine waves, but the example does
l
-
field
when
it
given by
,
LI
(2.8)
energy stored
L = inductance of
/
=
in the coil [J]
the coil [H|
current [A]
important.
However, these high-frequency harmonics produce
the steep sides
=
is
where
higher harmonics have smaller and smaller amplitudes,
magnetic
in its
The energy
show
that
any waveshape can be built up from a fundamental
wave and an appropriate number of harmonics.
we can decompose a distorted periwave into its fundamental and harmonic components. The procedure for decomposing a distorted wave is given in Chapter 30.
Harmonic voltages and currents are usually undesirable, but in some ac circuits they are also unavoidable. Harmonics are created by nonlinear
If
the current varies, the stored energy rises and falls
in step
with the current. Thus, whenever the current
increases, the coil absorbs energy and
current
The
cussed
falls,
energy
is
properties of an inductor are
in
Section 2.3
whenever
the
released.
more
fully dis-
1
Conversely,
odic
loads,
such as electric arcs and saturated magnetic
2.14 Energy
in
A capacitor stores
ever a voltage
energy
is
E
a capacitor
energy
in its electric field
appears across
its
when-
terminals.
The
given by
W='c£
2
(2.9)
FUNDAMENTALS
26
W=
where
W=
energy stored
C—
E=
capacitance of the capacitor [F]
in the
capacitor
fJ]
8
The energy
Example 2-6
having an inductance of 10
in series
with a 100
capacitor.
jjlF
current in the circuit
is
40
A
voltage across the capacitor
energy stored
this
X
1/2
10
10" 3
X
X 40 2
J
stored in the capacitor
is
voltage [V)
W=
A coil
1
Ml LI =
in
the electric
mH
=
is
connected
2
1/2
32
CE =
1/2
X
100
X 10" 6 X 800 2
J
The instantaneous
Some
2.15
and the instantaneous
800 V. Calculate the
and magnetic fields at
is
useful equations
We terminate this
tions (Table
moment.
2B)
section with a
list
of useful equa-
that are frequently required
solving ac circuits.
The equations
when
are given without
proof on the assumption that the reader already pos-
Solution
The energy
stored in the coil
sesses a
is
knowledge of ac
circuits in general.
IMPEDANCE OF SOME COMMON AC CIRCUITS
TABLE 2B
Impedance
Circuit diagram
XL =
Equation
(2-I0)
2TTJL
(2-ll)
2irfC
Z = \R 2 + X
—
o
o
1|
o-m—it——
Ac
H
Z=
\
R~
2
(2-12)
+ Xc r
(2-
1
3)
XL
Z = VR- +
(X}
- XC Y
(2-14)
RX,
a*
(2-15)
\R 2 + X
:
(
RX C
(2-16)
Vr 2 + Xc 2
Xc \ R +_XL 2
1
2
VR +
(X L
- Xcf
(2-17)
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
ELECTROMAGNETISM
X
by def10~ 7 or
approximately 1/800 000. This enables us
to write
In the SI, the
inition.
Magnetic
2.16
and
Whenever
a
component,
H
field intensity
B
magnetic flux
exists in a
it
field intensity
is
due
H, given by
H=U/I
(2.18)
in the
approximate form:
//= 800 000 3
magnetic
field intensity
a flux density of
H—
U=
/
=
magnetic
field intensity
f
rubber, and air have
length of the
that
component [m]
The resulting magnetic flux density
is
1)
A/m
of 800
produces
millitesla.
I
Nonmagnetic materials such
A/m]
magnetomotive force acting on the
component [A] (or ampere turn)
(2.2
The B-H curve of vacuum is a straight line. A
vacuum never saturates, no matter how great the
flux density may be (Fig. 2.25). The curve shows
that a
where
fixed,
is
has a numerical value of 4tt
body or
presence of a magnetic
to the
It
Eq. 2-20
flux density
cf>
magnetic constant
27
B-H
as copper, paper,
curves almost identical to
of vacuum.
mT
2.0,
given by
B = $IA
r
.
—
1
,
1
(2.19)
where
B =
flux density [T]
§ =
A =
flux in the
There
component WbJ
[
cross section of the
is
density (B)
component [m
2.17
In
]
a definite relationship between the flux
and the magnetic
field intensity
any material. This relationship
graphically
2
by the
B-H curve
B-H curve
of
is
(H) of
usually expressed
Figure 2.25
of the material.
B- /-/curve of
of
nonmagnetic materials.
vacuum
vacuum, the magnetic flux density
B
is
directly
2.18
expressed by the equation
B =
The
il 0
H
(2.20)
B-H curve
of a magnetic
material
proportional to the magnetic field intensity H, and
is
vacuum and
flux density in a magnetic material also de-
pends upon the magnetic
is
subjected.
value
Its
is
field intensity to
which
it
given by
where
B =
B =
flux density [T]
H=
magnetic
|x 0
—
field intensity
where
[A/m
magnetic constant [= 4tt
X
7
I0~ l*
B,
|jl
before, and
0
,
and
(jl,.
is
H have
ijL
oK /y
the
(2.22)
same significance
as
the relative permeability of the
material.
The value of
ijl,.
is
not constant but varies with
the flux density in the material. Consequently, the
Also called the permeability of vacuum. The complete expression for
(jl
0 is
4
tt
X
K)
7
henry/meter.
relationship between
this
makes Eq. 2.22
B and
H
is
not linear, and
rather impractical to use.
We
FUNDAMENTALS
28
prefer to
show
the relationship by
saturation curve. Thus, Fig. 2.26
means of a B-H
shows typical
saturation curves of three materials
used
in electrical
and cast
steel.
field intensity
of 1.4
sity
T
commonly
machines: silicon iron, cast
iron,
ial, it is
The curves show that a magnetic
of 2000 A/m produces a flux den-
in cast steel
but only 0.5
T
jjl,-
~
800 000
(2.23)
fl///
in cast
where
2.19 Determining the relative
permeability
tio
easy to calculate the relative permeability
using the approximate equation
iron.
The
would be produced in vacuum, under the
same magnetic field intensity H.
Given the saturation curve of a magnetic matersity that
relative permeability
of the flux density
in
B =
flux density in the magnetic material [TJ
H
corresponding magnetic field intensity [A/ml
—
Example 2-7
jjl,.
of a material
is
Determine the permeability of
the ra-
flux density of
the material to the flux den-
1
silicon iron
.4 T.
teslas
0
1000
2000
3000
4000
*H
Figure 2.26
B-H saturation curves
of three
magnetic materials.
5000
6000
A/m
( 1
%)
at
a
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
the magnetic materials saturate
Solution
Referring to the saturation carve (Fig. 2.26),
that
a flux density of
intensity
jlil..
1
.4
see
a magnetic field
eventually
all
the
B-H
more and more and
B-H curve
curves follow the
of vacuum.
of 1000 A/m. Consequently,
=
2.20 Faraday's law of
800 000 B/H
= 800 000 X
At
T requires
we
29
1.4/1000
this flux density, silicon iron is
more permeable than vacuum (or
electromagnetic induction
=1120
1120 times
air).
shows the saturation curves of a broad
from vacuum to Permalloy®, one
most permeable magnetic materials known.
Fig. 2.27
In
1831, while pursuing his experiments, Michael
Faraday made one of the most important discoveries in
electromagnetism.
Now known
as
Faraday's
it
revealed a
range of materials
law of electromagnetic induction,
of the
fundamental relationship between the voltage and
Note that as the magnetic field intensity increases,
flux in a circuit. Faraday's law states:
Figure 2.27
Saturation curves of magnetic
curve of
vacuum where
H is
and nonmagnetic materials. Note
high.
that
all
curves
become asymptotic
to the
B-H
FUNDAMENTALS
30
function of time, a voltage
tween
its
of the induced voltage
tional to the rate of
By
1
definition,
when
units,
its
E =
propor-
is
and according
terminals. Consequently,
V
if
induced be-
is
is
The induced voltage
(2.24)
in
A/
where
In
=
—
At
coil
[Wb]
fixed in space.
is
common)
(although
case,
it
is
induction
with
relative
is
induced ac-
in this special
easier to calculate the
induced voltage with reference
electromagnetic
move
The
in the flux linking the
cording to Faraday's law. However,
[s]
of
generators, the coils
coils and, consequently, a voltage
time interval during which the flux
law
many motors and
motion produces a change
in the coil
change of flux inside the
Faraday's
induced
a conductor
respect to a flux that
induced voltage [V]
changes
zero as soon as the flux
falls to
ceases to change.
2.21 Voltage
A<I>
A<J>
X1/10
1000
given by
N~—
= number of turns
3
the flux varies in-
side a coil of /V turns, the voltage induced
/V
in
is
= 60 V
system of
the flux inside a loop varies at the rate of
E=
AO
= 2000 X
N—
At
to the SI
1
change takes place uniformly
this
1/10 of a second (A/), the induced voltage
change of flux.
weber per second, a voltage of
tween
induced be-
is
terminals.
The value
2.
Because
linking a loop (or turn) varies as a
If the flux
1.
to the conductors,
rather than with reference to the coil
itself. In effect,
established the basis of operation of transformers,
whenever a conductor cuts a magnetic field, a voltage is induced across its terminals. The value of the
generators, and alternating current motors.
induced voltage
opened the door to a host of practical applications and
is
given by
E=
Example 2-8
A coil of 2000 turns surrounds a flux of 5 mWb produced by a permanent magnet
net
is
(Fig. 2.28).
The mag-
suddenly withdrawn causing the flux inside
the coil to drop uniformly to 2
second.
What
is
mWb
in
1/10 of a
the voltage induced?
flux
E—
B=
=
/
induced voltage [VJ
flux density [Tl
active length of the conductor in the
magnetic
change
AO
(5
v
is
mWb -
N=
2
mWb)
3
(2.25)
where
Solution
The
Blv
—
field
[m]
relative speed of the conductor fm/s|
mWb
Example 2-9
The stationary conductors of a
2000
an active length of 2
N
S
\
0.6 teslas.
,>
moving
at
m
large generator
and are cut by a
have
field
Calculate the voltage induced in each conductor.
5
4> 2
mWb
=2mWb
Solution
A/ = 1/10
s
According to Eq. 2-25. we find
Figure 2.28
Voltage induced by a moving magnet.
See Example
2-8.
E=
Blv
=
0.6
X
of
a speed of 100 m/s (Fig. 2.29).
2
X
100
-
120
V
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
120
3
V
Figure 2.30
Force on a conductor.
Figure 2.29
Voltage induced
Example
in
a stationary conductor.
See
2-9.
-
on a conductor
2.22 Lorentz force
When
conductor
a current-carrying
magnetic field,
it
is
is
placed
force
is
of fundamental importance because
stitutes the
and of
tors,
it
con-
basis of operation of motors, of genera-
many
nitude of the force
electrical instruments.
The force
is
parallel to
it
=
0.
The mag-
when
greatest
the conductor
perpendicular to the field (Fig. 2.30) and zero
is
Figure 2.31
Force
depends upon the orientation of
conductor with respect to the direction of the
field.
it
we
electromagnetic force, or Lorentz force. This
call
the
1
N
in a
subjected to a force which
—
(Fig. 2.3
l
).
is
when
Calculate the force on the conductor
BII
tremes, the force has intermediate values.
tor is
straight
perpen-
it is
Solution
Between these two ex-
The maximum force acting on a
if
dicular to the lines of force (Fig. 2.30).
X
0.5
X 200 = 300 N
3
conduc-
given by
2.23 Direction of the force acting
F=
BII
on a straight conductor
(2.26)
where
Whenever
F=
force acting on the conductor [N]
B =
flux density of the field [T]
a conductor carries a current,
rounded by a magnetic
into the
field.
For
page of this book, the circular
have the direction shown
in
it
a current
lines
Figure 2.32a.
of force
=
active length of the conductor [m]
figure
/
-
current in the conductor A]
N, S poles of a powerful permanent magnet.
[
the magnetic field created
The magnetic
shape shown
Example 2-10
A conductor
3
m long carrying a current of 200 A is
placed in a magnetic field
whose density
is
0.5 T.
in
field
sur-
The same
/
shows
is
flowing
between
does not, of course, have
the
the
the figure because lines of force
never cross each other. What, then,
the resulting field?
is
the shape of
FUNDAMENTALS
32
N
//
H
cross-section
.
Xi
-
- ct>
I
.
I
(a)
Y turns
length
-
/
Figure 2.33a
Method of determining the B-H properties
of
a mag-
netic material.
curve oa
(b)
a value
Figure 2.32
a. Magnetic field due to magnet and conductor.
b. Resulting magnetic field pushes the conductor
downward.
If
in
Bm
Figure 2.33b. The flux density reaches
for a magnetic field intensity
the current
the flux density
curve, but
To answer
the question,
we observe
that the lines
of force created respectively by the conductor and
permanent magnet
the
act in the
above the conductor and
low
in
same
direction
opposite directions be-
Consequently, the number of lines above the
it.
B
we
the magnetic
H nv
gradually reduced to zero,
does not follow the original
moves along
oa. In effect, as
sity,
now
is
a curve
ab
situated
above
reduce the magnetic field inten-
domains
that
were lined up under
Hm tend to retain their origiphenomenon is called hystereConsequently, when H is reduced to zero, a sub-
the influence of field
nal orientation. This
sis.
stantial flux density remains.
It
is
called residual
flux density or residual induction (B r ).
conductor must be greater than the number below.
The
resulting magnetic field therefore has the shape
given
in
Figure 2.32b.
Recalling that lines of flux act like stretched
elastic bands,
upon
it
is
easy to visualize that a force acts
the conductor, tending to
push
it
downward.
2.24 Residual flux density
and coercive force
Consider the
coil
of Figure 2.33a, which surrounds
a magnetic material
formed
in the
shape of a
ring.
A
magnetic
field intensity
current source, connected to the coil, produces a
current
whose value and direction can be changed
from zero, we gradually increase /,
Figure 2.33b
H
Residual induction and coercive force.
at will. Starting
so that
and B increase. This increase traces out
//
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
we wish
If
to eliminate this residual flux,
33
we
have to reverse the current in the coil and gradually
H
increase
in the
opposite direction.
we move along curve
As we do
so,
be. The magnetic domains
gradually change their previous orientation until the
becomes zero
flux density
point
at
The magnetic
c.
required to reduce the flux to zero
field intensity
is
called coercive force
In
also
B
reducing the flux density from
have
to furnish
overcome the
The energy supplied
A
to zero,
magnetic
in orientation.
dissipated as heat
is
we
used to
is
frictional resistance of the
domains as they oppose the change
material.
r
energy. This energy
the
in
very sensitive thermometer would in-
dicate a slight
temperature
rise as the ring is
being
Figure 2.34
Hysteresis loop.
If
B
expressed
is
amperes per meter, the area
demagnetized.
dissipated per cycle,
in
in
and
teslas
of the loop
is
H in
the energy
joules per kilogram.
2.25 Hysteresis loop
Transformers and most electric motors operate on
such devices the flux
alternating current. In
iron
tion.
changes continuously both
The magnetic domains
in
depends upon the frequency. Thus,
pressed
if
at
a rate that
the flux has a
in
J/nr
)
is
equal to the area
(in
T-A/m) of
the hysteresis loop.
value and direc-
are therefore oriented
one direction, then the other,
first in
in the
To reduce
hysteresis losses,
we
select
magnetic
materials that have a narrow hysteresis loop, such
as the grain-oriented silicon steel used in the cores
of alternating-current transformers.
frequency of 60 Hz, the domains describe a com-
every 1/60 of a second, passing succes-
plete cycle
through peak flux densities +Z? m and —B m
the peak magnetic field intensity alternates be-
sively
as
tween
+ // m and — // m
as a function
of H,
.
we
we
If
v
B
obtain a closed curve called
hysteresis loop (Fig. 2.34).
B and coercive force
plot the flux density
H
c
The
residual induction
have the same
signifi-
cance as before.
2.27 Hysteresis losses caused
by rotation
Hysteresis losses are also produced
for example, an armature
volves
in
AB, made of
S (Fig. 2.35). The magnetic domains
armature rotates, the
describing a hysteresis loop, the flux
and
+# m
.
+ 5 m +B n
,
0,
and
a, of
moves
— B m —B n
,
corresponding respectively
b, c, d, e, f,
energy
that the
is
0,
to points a,
Figure 2.34. The magnetic
material absorbs energy during each cycle
this
iron, that re-
in the
armature
field, irrespective
of the position of the armature. Consequently, as the
2.26 Hysteresis loss
successively from
a piece of
a field produced by permanent magnets N,
tend to line up with the magnetic
In
when
iron rotates in a constant magnetic field. Consider,
dissipated as heat.
We
and
can prove
amount of heat released per cycle
(ex-
first
sal
toward
A and
N
poles of the domains point
then toward B.
A complete
rever-
occurs therefore every half-revolution, as can be
seen
in
Fig.
2.35a and 2.35b. Consequently, the
magnetic domains
cally,
in
the armature reverse periodi-
even though the magnetic
field is
constant.
Hysteresis losses are produced just as they are in an
ac magnetic field.
FUNDAMENTALS
34
Figure 2.37
Concentric conductors carry ac currents due to ac
flux
<1>.
currents are progressively smaller as the area of the
(b)
loops surrounding the flux decreases.
In Fig. 2.38 the ac flux passes
Figure 2.35
Hysteresis losses due to rotation.
through a solid
metal plate.
It is
packed
of rectangular conductors touching
set
basically equivalent to a densely
each other. Currents swirl back and forth inside
Eddy currents
2.28
Consider an ac flux
that links a rectangular-shaped
<E»
conductor (Fig. 2.36). According to Faraday's law,
an ac voltage
If the
E
is
{
tor to heat up. If a
the
first,
induced across
conductor
alternating current
is
/,
current
U
short-circuited, a substantial
second conductor
is
is
conduc-
placed inside
induced because
centric
it
is the power disshows four such conloops carrying currents I 7 2 7 3 and / 4 The
is
less than
penetrated by an ac flux can
this regard, special care
/,
and
so, too,
x
,
,
,
The
flux
c|)
in Fig.
be increasing. As a
the
flux
<1>
in trans-
2.37 and 2.38
result,
eddy currents flow
in
is
assumed
to
due to the Lenz's law,
such a
way
as to
oppose
the increasing flux.
.
conductor
Figure 2.36
has to be taken
is
hot. In
sections of the enclosing tanks to overheat.
eddy currents
An ac
become very
formers so that stray leakage fluxes do not cause
Consequently, the short-circuit
sipated in this loop. Fig. 2.37
in the figure.
These so-called eddy currents (or Foucault currents) can be very large, due to the low resistance
of the plate. Consequently, a metal plate that
terminals.
its
will flow, causing the
a smaller voltage
links a smaller flux.
shown
the plate, following the paths
V
—
metal plate
Figure 2.38
induces voltage
Ev
Large ac eddy currents are induced
in
a
solid
metal plate.
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
Eddy currents
2.29
iron
35
a stationary
in
core
insulation
The eddy -current problem becomes particularly
important
when
iron has to carry an ac flux. This
the case in all ac
2.39a
shows a
coil carrying
A large
of the core.
ally
become red hot (even
due
to these
at a
its
Figure 2.39b
Eddy currents are reduced by
splitting the
core
in half.
in
two
The voltage
of what it was
length, taking care to insulate the
from each other
induced
in
each section
(Fig. 2.39b).
is
one half
before, with the result that the
eddy currents, and
corresponding losses, are considerably reduced.
we
If
continue to subdivide the core,
we
composed of stacked laminations, usually
of a millimeter thick. Furthermore, a small
silicon is
find that
decrease progressively. In practice, the core
the losses
ity,
/'•-•''
core could eventu-
frequency of 60 Hz)
can reduce the losses by splitting the core
two along
is
currents
eddy-current losses.
sections
the
Eddy
up as shown and they flow throughout the
entire length
We
i.-A
an ac current that pro-
duces an ac flux in a solid iron core.
are set
is
motors and transformers. Figure
alloyed with the steel to increase
thereby reducing the losses
still
more
a fraction
amount of
its
(Fig. 2.39c).
The cores of ac motors and generators are
fore
always laminated.
A
eddy current in
one lamination
resistiv-
there-
Figure 2.39c
Core built up of
thin,
insulated laminations.
thin coating of insulation
covers each lamination to prevent electrical contact
between them. The stacked laminations are tightly
held in place
by
bolts
and appropriate end-pieces.
For a given iron core, the eddy-current losses decrease in proportion to the square of the
laminations.
number of
2.30 Eddy-current losses
in a revolving core
The
stationary field in direct-current motors and gen-
erators produces a constant dc flux. This constant
flux induces
eddy currents
To understand how they
in the
revolving armature.
are induced, consider a solid
cylindrical iron core that revolves between the poles
of a magnet (Fig. 2.40a). As
lines and,
duced along
Owing
it
its
its
is in-
length having the polarities shown.
to this voltage, large
core because
resistance
eddy currents flow
is
These eddy currents produce
are immediately converted
is
turns, the core cuts flux
according to Faraday's law, a voltage
in the
very low (Fig. 2.40b).
large
FR
into heat.
losses
which
The power
loss
proportional to the square of the speed and the
square of the flux density.
To reduce
Figure 2.39a
Solid iron core carrying
an ac
flux.
the eddy-current losses,
we
laminate
the armature using thin circular laminations that are
FUNDAMENTALS
36
rotation
(b)
(b)
Figure 2.41
Armature
a.
Figure 2.40
Much
b.
up
of thin laminations.
a revolving armature.
a.
Voltage induced
b.
Large eddy currents are induced.
in
built
smaller eddy currents are induced.
of current. However,
known and we want
insulated from each
other.
The laminations
tightly stacked with the flat side
are
running parallel to
the flux lines (Fig. 2.41).
rent
/.
We
To
often happens that e
knowledge of advanced mathemat-
get around this problem,
we can
use a
graphical solution, called the volt-second method.
2,31 Current in
It is
well
known
an inductor
yields the
that in an inductive circuit the volt-
known
Ai
= L~~
how
=
instantaneous voltage induced in
the circuit
|
V]
L = inductance of the
A/7 A/
=
E
/,
time
.
e,
later, after
carries a current
an interval At.
circuit [H]
/, at
a time
From Eq. 2-27 we
the rate of
change
=
1
eAt
L
of change of current [A/s]
when we know
which a
We want to determine the current a very short
Ai
rate
in
appears across an inductance L.
can write
This equation enables us to calculate the instanta-
neous voltage
=
response to
in
applied voltage.
Suppose the inductance
/
e
the current in an induc-
and decreases with time,
variable voltage
where
It
and has the advantage of en-
Consider, for example, Fig. 2.42,
(2.27)
At
results
abling us to visualize
a
e
same
tor increases
age and current are related by the equation
is
can use the same equation, but the solu-
tion requires a
ics.
it
to calculate the resulting cur-
which means
that the
short interval At
is
change
given by
in
current Ai during a
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
Volt-seconds are gained (and
it
-«
voltage
1
is
when a
lost)
37
variable
applied across an inductor.
Figure 2.42
Variable voltage applied across
sulting
change
in
current.
Initial
an inductor and
current
re-
is A,.
mental changes
—
(f 2
f|).
As
U =
initial
A =
/,
in
current A/ during the long period
we
a result
current
find current
/,
+
+
e 2 At 2
(A/,
+
A/ 2
I 2 at
+
time
+
A/ 3
.
t
.
2
.)
average voltage e during the interval
A/
-
X
duration Af of the interval
I
L
2
=
+ |
+
/,
(^Af,
-
=
<?
3
I
curve during the interval
.)
.
.
.)
.
/ algebraic sum of
all
+
between
and
t
A under
t
2
the voltage
-seconds across the induccurve between
tance during the interval At
The values of
is
tive
(
+
+
Af)
=
initial
current
+
A/
e2
or negative
)
eas AAj,
(f,
and
t
f
{
Therefore, the current in the inductance after the in-
/at time
little
areas under the voltage curve
net area
volt
the
I
4 =
Am
x
Af
.
2
\
terval
+
Af 3
+ AA 2 + AA 3 +
(A/4,
/ area A/\ under the voltage
A/
+
AA 2
The sum of
,
A/\ 3
these
(
and so on may be posi-
,
)
and, therefore, the
and so on may be
,
(
—
+
)
2
and
(
—
)
(
+
)
little ar-
or
(
—
).
values of the A/\s
gives the net area under the voltage curve between
/,
+
AA
fj
are usually
current at a time
ter
f,
f2
(Fig. 2.43).
more
,
2
.
in Fig.
2.44 the net area/\ after time inter-
val Tis equal to (A
interested in calculating the
when
We
f
Thus,
L
We
and
f
2 is
many Af
ize, the
,
— A2
)
volt-seconds.
To general-
current after an interval Tis always given by
intervals af-
then have to add the incre-
/
=
/,
+ AIL
(2.28)
FUNDAMENTALS
38
volts
Figure 2.44
The
T is
A and A 2
net volt-seconds during interval
algebraic
sum
areas
of the
equal
the
to
.
A
where
/(
=
current at start of interval
T
/
=
current after lime interval
T\A]
A —
under the volt-time curve dur-
net area
T
ing lime
/.
—
[
V-s]
inductance [HI
Consider, for example an inductor
L,
having negli-
gible resistance, connected to a source
whose
volt-
age varies according to the curve of Fig. 2.45a.
the initial current
is
zero, the value at instant
=A
F
t\
If
Figure 2.45
is
/L
l
As time goes
However, the current reaches
time
/->
because
at this
its
moment
maximum
value
at
the area under the
any more. Beyond
becomes negative and, consequently,
vol age curve ceases to increase
l
t,
the voltage
the net area begins to diminish.
ample, the net area
equal to
is
the corresponding current
f=(A +
i
At instant
/4 ,
At instant ^, for ex+ A 2 — A$) and
(/\,
stant
is
/
the negative area 04 3
words,
it
the current
is
to
is
A
is
ex-
+ A 2 The
net
4-
4)
).
l
the inductor.
onds during the
interval
from 0
to
t
2
.
As
it
becomes
direct proportion to the volt-seconds received.
during the discharge period from
t
2
in
Then
to /4 the inductor
loses volt- seconds and the current decreases accordingly.
An
inductor, therefore, behaves very
a capacitor.
in
known
much
like
However, instead of storing ampere-secstores volt-seconds.
a capacitor having a capacitance
that the voltage
E across
its
C
For
it
is
terminals
is
given by
also zero. After inin
other
where
changes direction.
at
in
charged up with volt-seconds, the current increases
well
becomes negative;
Another way of looking
inductor stores volt-seconds.
example,
A 2 - A y )IL
zero and so the current
4,
Current
onds (coulombs), an inductor
is
actly equal to the positive area (A
area
An
b.
in-
and so does the current.
progressively
creases
under the curve
by, the area
a.
the situation (Fig. 2.45),
consider that the inductor accumulates volt-sec-
in
E
is
}
the initial voltage and
Q
.
L
is
the charge
coulombs (ampere seconds, positive or negative)
the capacitor received during a given interval.
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
In the
same way,
for an inductor having an in-
ductance L, the current
/ it
carries
Example 2-11
The voltage across
given by
is
39
the terminal s of an inductor of 2
H
varies according to the curve given in Fig. 2.46.
a.
Calculate the instantaneous current
cuit,
where
in
/] is
the initial current and
knowing
the cir-
/ in
that the initial current
is
zero.
1
QL
is
the "charge'
b.
Repeat the calculations for an
current
initial
of 7 A.
volt-seconds (positive or negative) that the in-
ductor received during a given interval.
It
1
interesting to note that
is
volt-second.
turns
coil of
1
weber-turn
600 turns
of 20 milliwebers has stored
a flux
netic
Thus a
12V-s/3H
tained
equal to
mag-
a total
12
000mWb
12 volt-seconds. If the inductor has an induc-
tance of 3 henries,
Fig.
it
X 20mWb =
charge of 600 turns
=
in
is
that surrounds
it
is
carrying a current of
Q
{
/L
=
=4 A.
the voltage of Fig. 2.45a
an inductance of 100 H.
The
and the current rises to a
maximum
initial
is
applied to
current
of 6.9
zero,
is
A
before
again dropping to zero after a time interval of
Important Note:
an interval
T
is
value to all the
If the
27
s.
current at the beginning of
not zero,
we
simply add the
initial
ampere values calculated by the
volt-second method.
Figure 2.46
See Example 2-11.
a.
Interval
from zero
to 3
s:
During
this interval
the area in volt-seconds increases uniformly
and progressively. Thus,
area
A
and so
is
4
V
forth.
s;
after
after
one second, the
two seconds
8
it is
Using the expression
=
/
V
s;
AIL, the
current builds up to the following respective
2.45b shows the instantaneous current ob-
when
Solution
values: 2 A, 4 A, and so on, attaining a final
value of 6
Interval
A after three
from 3
s to
5
seconds.
s:
The area continues
to
increase but at a slower rate, because voltage
is
smaller than before.
When =
t
5
surface starting from the beginning
therefore the current
Interval
from 5
s to
by 4 squares, which
is
7
is
16 V-s/2
s:
s,
is
H =
The surface
equivalent to 8
E
the total
16
V
s;
8 A.
increases
V
s.
FUNDAMENTALS
40
means that the sum of the voltage rises is equal to
sum of the voltage drops. In our methodology it
Consequently, the current increases by 4 A,
thus reaching 12 A. Note that the current no
the
is
age rise" or a "voltage drop/'
is
not constant during this interval.
from 7
Interval
s to
8
We
The voltage sud-
s:
denly changes polarity with the result that the
V
8
during this interval subtract from the
s
ously.
The
from the beginning
net area
therefore 24
V
-
s
8
V
=
s
Consequently, the current
interval
7=16
is
Interval
voltage
from 8
is
s
at the
=
V-s/2 77
to 10
s:
V
16
ence.
from
end of
2.33 Kirchhoff's voltage law and
double-subscript notation
we assumed
is
may
elements
zero
in
which
six circuit
The
s:
—
at t
14
is
Beyond
zero.
current of
+7
s,
this point
be sources or loads, and the conI
to 4.
we have
A,
to
add
we can start with any node
cw or ccw direction until we
starting point. In so doing, we
in either a
come back
to the
This ordered
voltage
The new current wave is simply 7 A
above the curve shown in Fig. 2.46. Thus, the
=
1
1
s is
CIRCUITS
When
+
6
7
2.7.
We
1
set
of labels
subscripts.
We
is
used to establish the
write the
voltage sub-
scripts in sequential fashion, following the
it is
order as the nodes
cw around
in
presume the reader
same
meet.
loop
ABCD, we
successively encounter
nodes 2-4-3-1-2. The resulting
KVL
equation
is
therefore written
essential to fol-
E24 + £43 + E31 + E l2 =
based upon the voltage
rules that are
we
For example, starting with node 2 and moving
3 A.
AND EQUATIONS
and current notations covered
and
=
writing circuit equations,
low certain
going
elements A, E, and D,
and move
ously.
at t
In
encounter the labeled nodes one after the other.
each of the currents calculated previ-
current
elements
around a circuit loop, such as the loop involving
negative volt-
equal to the positive area,
and so the net current
A to
Consider Fig. 2.47
A, B, C, D, E, and F are connected together. The
the current changes direction,
7
by the sign notation.
later
nections (nodes) are labeled
the negative area
initial
followed
this
8 A.
seconds continue to accumulate and
With an
a matter of individual prefer-
will begin with the double-subscript nota-
Because the terminal
that
10 s to 14
in
The choice
tion,
coil resistance).
Interval
can be expressed
that voltages
is
a "volt-
is
second area does not change and neither does
(remember
We
is
s.
zero during this interval, the net volt-
the current
have seen
either double-subscript or sign notation.
of one or the other
volt-seconds that were accumulated previ-
b.
not necessary to specify whether there
longer follows a straight line because the volt-
age
0
Sections 2.4, 2.5,
familiar with the
is
solution of such equations, using linear and vector
method
will
review only
the writing of these equations, using
Kirchhoff s
algebra. Consequently, our
voltage law
(KVL) and
By following
a
current law (KCL).
few simple
rules,
solve any circuit, ac ordc, no matter
it
is
possible to
how complex.
We
begin our explanation of the rules regarding voltages.
2.32 Kirchhoff's voltage law
Kirchhofrs voltage law
sum
states that the algebraic
of the voltages around a closed loop
Thus,
in a
3
is
zero.
closed circuit, Kirchhoff's voltage law
Figure 2.47
Rule
for writing
notation.
KVL
equations using double-subscript
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
If we select loop CEF and start with node 4 and
move ccw, we successively encounter nodes 4-2-3-
4.
KVL equation
The resulting
+
£42
The
set
usually
0
ac or dc. If they are ac, the voltages will
to
it is
essential to equate
all
we have done so far and
We do not recommend attempts
equations to zero as
continue to do.
equate voltage rises to voltage drops.
In
£,
set
even represent instantaneous values.
order to prevent errors,
will
and so
be expressed as phasors, having certain
of voltages can
KVL
= -10 V
KVL equa-
magnitudes and phase angles. In some cases the
In
= ~E ]2 ~~ E 2 $
= -E X2 + E yl
= -40 + 30
£31
of voltages designated by the
may be
tions
Transposing terms,
is
+ em
£23
= +10 V
3
indicating that terminal
is
1
positive with respect to
terminal 3 and that the voltage between the two
is
10 V.
2.34 Kirchhoff s current law
finding the solution to such double-subscript
equations,
pressed as
it
useful to
is
EXY
remember that
a voltage ex-
can always be expressed as
— E YX
Kirchhoff s current law
sum
means
zero. This
rents that leave
Example 2-12
shows two sources connected in series,
having terminals (nodes)
2, and 3. The magnitude
2.48
1
+40 V and £ 32
Fig. 2.49
and
is
equal to the
sum of
the cur-
shows
five currents arriving at a
The sum of
common
the currents flowing
,
E l2 and E^ 2 are specified as E i2 =
= +30 V. We wish to determine the
I
equal to
is
the currents that
it.
terminal (or node).
into the
is (I 2
+
node
I4
+
is (/,
7 S ).
magnitude and polarity of the voltage between the
open terminals
point
at a
sum of
that the
flow into a terminal
and polarity of
states that the algebraic
of the currents that arrive
,
and vice versa.
Fig.
41
/,
+
/ 3 ),
while the
sum
that leaves
it
Applying KCL, we can write
+
/3
=
I2
+
I4
+
/,
3.
Solution
In
writing the loop equation, let us start at terminal
and
move ccw
The resulting
we
till
again
come back
KVL equation
£,
2
to terminal
1
1
is
+ £ 23 + E M =
0
Figure 2.49
?3
1t
Rule
for writing
KCL
equations.
2.35 Currents, impedances,
and associated voltages
Consider an impedance
2
Z
carrying a current A
connected between two terminals marked
Figure 2.48
(Fig. 2.50).
See Example 2.12.
will
A
1
and 2
voltage E, 2 having a magnitude
,
/Z,
appear across the impedance. The question of
FUNDAMENTALS
42
Let us write the circuit equations for Fig. 2.5
Loop 2312,
starting with
node 2 and moving cw:
=0
+ /4 Z4 + E 3I -/,Z,
Figure 2.50
E12 = + /Z.
Voltage
/4
Z4
is
preceded by a
+
(
)
sign,
going around the loop we are moving
now
polarity
The
arises: Is
question
is
across an impedance
direction as the current flow
voltage IZ
or
— IZ?
/,
Z
in
write £, 2
= +/Z
the
the associated
Conversely,
moving across an impedance against
ative sign.
Thus,
or dc, and the
ductive
In
(
jX,
most
),
E2]
the direction
resistive (R), in-
(—jXc ).
impossible to predict the
or capacitive
circuits
it
is
On
the other hand, voltage I{Z
Loop 3423,
starting with
+ /3Z3 Voltages /3Z3 and
because
in
Fig. 2.5
and
Zj,
1
in
,
I2
node 3 and moving ccw:
Z2 +
/4
Z4 =
0
/4
Z4
are preceded by a
(
+
sign,
)
the direction of the respective currents. Voltage
is
preceded by a negative sign because
ing against current
Loop 242,
I2
are
Z2
mov-
.
starting with
£24
we
in
/2
node 2 and moving cw:
~ h^i =
0
Consider for example the circuit of
which two known voltage sources E, 3
E24 are connected to four known impedances
Z 2 Z3 and Z4 Because the actual directions of
,
pre-
going around the loop we are moving
actual direction of current flow in the various circuit elements.
is
X
against the direction of/,.
in
~IZ. The current can be ac
impedance can be
.
in
when
of current flow, the voltage /Zis preceded by a neg-
=
tion of /4
because
in the direc-
ceded by a negative sign because we are moving
preceded by a positive sign. Thus,
is
we
Fig. 2.50,
+ /Z
equal to
resolved by the following rule:
When moving
same
£ l2
1
KCL
node
at
2:
/s
=
U +h
/.
.
unknown, we simply
assume arbitrary directions as shown in the figure.
It is a remarkable fact that no matter what directions are assumed, the final outcome after solving
current flows are presently
the equations (voltages, currents, polarities, phase
angles, power, etc.)
is
always correct.
KCL
node
at
3:
h+
/.
= h
Example 2-13
Write the circuit equations and calculate the currents flowing in the circuit of Fig. 2.52,
that
£AD - + 108 V
and
ECD = + 48
knowing
V.
Solution
We first
and
/3
select arbitrary directions for currents
and write the
6
108
Figure 2.51
Writing
KVL and KCL
V-—-
See Example
a
Ail
12
Figure 2.52
equations.
2.13.
/,,
circuit equations as follows:
Q
^
48 V
/2
,
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
For the loop
DABCD composed of the two sources
we
and the 6 fi and 4 fl resistors,
For the loop
DCBD
Applying
1
composed of
the
(cw)
48
V
source
2 II resistors:
E DC +
47 2
+
-48 +
4/ 2
+
KCL at
12/ 3
=
12/,
=
0
(ccw)
Figure 2.53
See Example
Solution
a.
We conclude
and
73
72
To solve
= -3 A
we
73
that the directions
flows from
To
= +5 A
establish an arbi-
first
we assumed
assumed because
Thus, suppose
between points a and
write the circuit equation,
let
/
b.
move cw
us
c.
This
yields
for
I\
2 is
bears a nega-
/2
left to right
around the loop, starting from terminal
However, the actual direction of
opposite to that
we
find
ECd +
were correct because both currents carry a
positive sign.
the circuit,
trary direction of current flow.
*3
Solving these simultaneous equations,
= +8 A
2.14.
0
node B, we get
h + h =
/,
763
obtain
£da + 6'i - 4/ 2 + ECD = 0
-108 + 67, - 47 2 + 48 - 0
and the 4 il and
16
43
7(16
+
j
Substituting the values of
tion
+ £bc =
63)
Eac and Ehi
and combining the terms
in
.
we
/,
0
in this
equa-
get
tive sign.
-200
2.36 Kirchhoff s
laws and ac circuits
The same basic rules of writing double-subscript
equations can be applied to ac circuits, including
3-phase circuits.
sistive
sistive,
elements
The only difference
in
all
Solving
b.
765
A.
+
75.8°
we
100
find that
I
=
L
1
150°
.9 Z.
=
0
20.5°.
To determine £. lb we write the following equation, moving cw around the loop:
,
that the re-
£ua + £
;ib
+ £bc =
0
Transposing terms,
three. Furthermore, the volt-
magnitudes and phase angles.
is
+
this equation,
ages and currents are expressed as phasors, having
sor equations
120°
dc circuits are replaced by reor capacitive elements, or a
inductive,
combination of
is
A.
The
more time-consuming, but
equations themselves can be written
by inspection. Let us consider
= £ac — £ bc
= 200 L 120° -
solution of phathe
down almost
100
l_
150°
Using vector algebra, we find
two examples.
E
ilb
=
123.9
L
96.2°
Example 2-14
In the circuit
ate the
of Fig. 2.53, two sources A,
B
gener-
2.37
KVL and
Voltages
E, c
sign notation
following voltages:
= 200 L
E bc =
120°
100
L
150°
in
ac and dc circuits are frequently indi-
cated with sign notation and designated by symbols
such as
E. v e m
for such circuits,
Calculate
a.
The value of the current
b.
the value of
£ ab
and
its
/ in
the circuit
phase angle
,
and so on. To write the equations
we employ
As we cruise around a
larity (+ or — ) of the first
the following rule:
loop,
we observe
the po-
terminal of every voltage
FUNDAMENTALS
44
(£,, E. v e m
etc.)
,
we
meet.
of the voltage source
minal
If
only the
(
+
terminal
)
marked, the unmarked
is
ter-
2.38 Solving ac and dc circuits
with sign notation
taken to be negative. This observed polar-
is
(+ or — precedes the respective voltages as we
write them down in the KVL equation. The following example illustrates the application of this rule.
ity
In circuits using sign notation,
wc
treat the
IZ
volt-
)
ages
in
same way
the
subscript notation. In other words, the IZ voltage
across an impedance
sign
Example 2-15
In Fig. 2.54,
it is
known
wish
age
Z
preceded by a positive
is
whenever we move across
the
impedance
in
the direction of current flow. Conversely, the IZ
,
1
determine the value and polarity of the volt-
to
Ec
EA and £ B
E A = + 37 V and E R - - 5 V. We
given the polarity marks of
that
as in circuits using double-
voltage
move
is
preceded by a negative sign whenever
The following example
across the open terminals.
to
we
against the direction of current flow.
illustrates the
procedure
be followed.
Solution
First,
we
assign an arbitrary polarity
minal voltage
Ec We
.
then proceed
+ to the tercw around the
)
(
loop in Fig. 2.54, starting with voltage
EA
.
This
yields the following equation:
Example 2-16
The circuit of Fig.
E =
1600
/L
60°.
2.55 is powered by an ac source
The values of the respective im-
pedances are indicated
-EA + Ec -E B =
(cw)
0
Ec
Calculate
a.
b.
T2
T1
in the figure.
The current flowing in each element
The voltage Ex across the 72 ohm capacitive
reactance.
Solution
a.
To solve
to
flow
this
problem, the currents are assumed
in the arbitrary directions
shown.
We
then write the following equations.
Moving cw around
-
E-
40)
Figure 2.54
Rule
Note
for writing
KVL
that the sign
sponds
equations using sign notations.
preceding each voltage correof the terminal that was
to the polarity
encountered
in
going
cw around
first
the loop.
Transposing terms.
Ec ~ E A + E H
= + 37 = +22 V
Thus, the magnitude of
of terminal Tl
is
to
is
22 V, and the polarity
indeed positive with respect to
minal T2. The polarity
happens
Ec
5
we assumed
have been the correct one.
at
ter-
the outset
Figure 2.55
See Example
2.15.
the loop
-
7,(30)
BDAB, we
+
I2
(-
j
obtain
37)
=
0 (cw)
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
moving ccw around
Then,
we
the second loop
45
ABCA,
obtain
/2 (-j
Finally,
-
37)
(-
/3
KCL at
applying
+
/,
/2
j
-
72)
=
21 / 3
0
(ccw)
N
node A, we have
+ h =
o
Upon solving these equations, we obtain the
fol-
lowing results:
=
/,
44.9
zl
215°
=
/3
b.
I2
=
30.3
Figure 2.56
L
40°
See Example
2.16.
14.9/1 24°
We can think of E x as being a voltmeter
connected across the capacitor. As a result,
Solution
the 'Voltmeter"
and capacitor together
form a closed loop for which
we can
To meet
write
ing
a circuit equation, as for
Thus, in traveling
we
any other loop.
cw around
requirement,
this
KVL
we
which
equations,
write the followthe
should
reader
verify:
the loop
write
-/,(-j72) + E x =
£ l2 + Eb - E, =
£23 + Ec - Eh =
E M + £a - E c =
0
Thus
0
0
0
Transposing terms, we obtain
Ex =
=
/3
(-j72)
14.9 zl 24°
(-
j
E 12 = Ea - Eb = 26
72)
45
and so
£23 = Eb
Ex =
1073
zl
Ec =
26
£ 31 - Ec ~ E =
iX
45
and hybrid notation
2.39 Circuits
In
tation
We
employ both sign noand double-subscript notation as shown in the
some
circuits
it
useful to
is
Ll
I
notation).
We
wish
to
determine the values of
£ 23 and £ 31 (double-subscript
,
notation).
E ]2
*
20°
-26/1
240°
- 26
0°
ing the loop created by
E
Therefore
E, N
=
x
=
240°
Z_
=
210°
can even express the sign notation
in
and terminals
N
terms of
in
follow-
and
1
,
we
KVL equation
£ N! +
Example 2-17
Fig. 2.56 shows a 3-phase system in which E —
26 L 0°, Eb = 26 zl 20°, and Ec = 26/1 240° (sign
1
L
L
-
90°
double-subscript notation. For example,
can write the
following example.
zl
26
120°
-30°
zl
45
-66°
-26/1
0°
zl
zl
E,
E N = — £,,, which
,
0
can be expressed as
E. v
This completes our review on the writing of dc
and ac
circuit equations.
46
FUNDAMENTALS
Questions and Problems
2-6
What
2-1
Three dc sources G h G 2 and
generate voltages as follows:
,
£34
= -100 V
= -40 V
£56
= +60 V
£,2
Show
the actual polarity
minals
2-2
In
in
G3
(Fig. 2.57)
c.
d.
Magnetomotive force
b.
2-7
Problem
(
+ )( — )
of the
ter-
2-8
2-3
if
2-9
the
show
the voltage
and the actual polarity of the generator
minals
2-4
at instants
A conductor 2 m
1
,
2, 3,
long
and
moves
60 km/h through a magnetic
2-10
is
moved, and
the coil falls to 1.2
Figure 2.57
2-3.
to
A.
Draw
the force on the
moving N
pole.
N pole act in the
direction as the direction of rotation?
the
waveshape of a sinusoidal
V
and a frequency of 5 Hz.
at a
b.
speed of
If the
voltage
is
=
5
voltage at
field
t
having a
the flux linking
mWb in 0.2
the average voltage induced.
Figure 2.58
Calculate the force on the
Does
same
a.
carries a cur-
2-11
A sinusoidal
zero
at
t
=
0,
what
is
the
ms? 75 ms? 150 ms?
current has an effective value
of 50 A. Calculate the peak value of current.
The magnet
See Problem
T
mm.
Calculate the force on the conductor.
b.
4.
A coil having 200 turns links a flux of
3 mWb, produced by a permanent magnet.
2-1
produce a flux density of 0.6
voltage having a peak value of 200
voltage.
See Problems
to
gap having a length of 8
c.
ter-
flux density of 0.6 T. Calculate the induced
2-5
air
mmf required.
Conductor AB in Figure 2.29
rent of 800 A flowing from B
a.
Terminals 1-4 and 3-6
Terminals 1-3 and 4-6
Referring to Figure 2.58,
want
an
at 0.2 T, 0.6 T,
T.
Calculate the
determine the voltage
following terminals are connected together.
a. Terminals 2-3 and 4-5
c.
We
in
and polarity across the open terminals
b.
of cast iron
tive permeability
2-1, if the three sources are
in series,
Referring to Figure 2.26, calculate the rela-
and 0.7
each case.
connected
the SI unit of
is
Magnetic flux
Magnetic flux density
Magnetic field intensity
a.
and
2-2.
s.
Calculate
2-12
A sinusoidal
voltage of 120
V
is
applied to
a resistor of 10 O.
Calculate
a.
the effective current in the resistor
b.
the peak voltage across the resistor
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
2-13
c.
the
d.
the
power dissipated by the resistor
peak power dissipated by the resistor
A distorted
voltage contains an
th
1
2- 8
1
fre-
2-
1
9
quency of the fundamental.
2-14
The current
lags
in
a 60
Hz
(
—
)
,
and curthe
is
each case, indicate which phasor
is
(
)
)
{
36 degrees behind the voltage.
2-20
The
alternating voltage e 2 in Fig. 2.24a
is
given by the expression
angle between the following phasors and,
c.
is
B 2 which box
The resistance of the conductors joining the
two boxes in Figure 2.4 is zero. If A, is +
with respect to A 2 can B be — with respect to B 2 ?
Referring to Fig. 2.59, determine the phase
b.
terminal A,
A 2 to
,
positive peaks of voltage and current.
a.
if
from
(
single-phase motor
Calculate the time interval between the
2-15
Figure 2.4,
source?
har-
monic of 20 V, 253 Hz. Calculate the
In
rent flows
47
e2
= 20
cos (360 ./f- 9)
in
If 6
lagging.
=
1
In d it stria I
2-2
1
and/ =
50°
value of e 2
/, and / 3
li and / 3
E and /,
at
t
=
0,
1
80 Hz, calculate the
and
at
/
=
262.37
s.
Appl icatio n
In Fig. 2.60 write the
KVL circuit equa-
tions for parts (a), (b), (c),
and
(d).
(Go cw
around the loops.)
6
A
Figure 2.59
See Problem 2-15.
2-
1
6
The voltage applied
to an ac
the current
is /
= 20
magnet
is
E — 60 sin
- 60°), all
sin
given by the expression
cj>
1
and
an-
(c|>
gles being expressed in degrees.
a.
Draw
the phasor
diagram for
E and
/,
Draw
the
function of
c.
1
a.
/ as
In Fig. 2.6
1
write the
for parts (a), (b),
<\>.
and
KCL circuit equations
(c),
and determine the
true direction of current flow.
peak negative power
in the circuit.
2-23
leads of the third
if
cw around
the
harmonic source are
In Fig. 2.62 write the
equations for parts
Referring to Fig. 2.24, draw the wave-
shape of the distorted sine wave,
b.
a
Calculate the peak positive power and
the
2- 7
2-22
waveshape of E and
2.21.
us-
ing effective values.
b.
Figure 2.60
See Problem
2-24
An
KVL and KCL circuit
(a), (b), (c),
and
(d).
(Go
the loops.)
electronic generator produces the out-
shown
reversed.
put voltage pulses
Calculate the peak voltage of the result-
this voltage is applied across a 10 fl resis-
ing waveshape.
tor,
calculate
in Fig. 2.63. If
1
FUNDAMENTALS
48
a.
the fundamental frequency of the
2-25
current
Repeat the calculations of Problem 2-24 for
the
b.
the peak power, in watts
c.
the energy dissipated per cycle, in
2-26
In Fig. 2.65 write the
power per cycle
the average
e.
the value of the dc voltage that
in the
—
the effective value of the voltage in the
figure
o
1
0
the average voltage
8
4
V
+ 100
resistor
7A
y*4A
2
in parts
the loops.)
would
produce the same average power
g.
KVL and KCL circuit
(Go cw around
(a) to (g).
d.
in Fig. 2.64.
equations for the ac circuits shown
joules
f.
waveshape shown
A
A
3
A
/
^
2
4
6
8
s
Figure 2.63
See Problem
2.24.
^4_A
A
1'
+ 100
(b)
(a)
v
(c)
0
4
-
Figure 2.61
See Problem 2.22.
2
-100 V
Figure 2.64
See Problem
'l
2
5 LI
R
11
/1
|
(b)
(a)
6"
I
—
T
4 LI
1
7Q I—
',1
T
(c)
Figure 2.62
See Problem
2.23.
CSJ
6
1
0
2.25.
seconds
FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS
E l2 = 100
20
0°
>50U
12
tf
(a)
(b)
(c)
£-13
£ ab =
49
£ B = 50 150_°
101.30°
E A = 20
= 30 1-3 0°
Eba =100L0 c
[45_°
7
£ 3 = 100(0°
12
J
40
U
'l
A
60
12
) i
30
*
(^O
3'
1
24
12
^
40
12
^
"1
(d)
Figure 2.65
See Problem 2.26.
(e)
(g)
30
hr
12
Chapter 3
Fundamentals of Mechanics
and Heat
3.0 Introduction
In
order to get a thorough grasp of electrical
technology,
is
it
mechanics and
large motors
is
essential to
power
know something about
For example, the starting of
heat.
determined not only by the magnitude
of the torque, but also by the inertia of the revolving
And
parts.
the overload capacity of an alternator
determined not only by the size of
also by the temperature that
its
its
the
—
—
is
we know
exert a muscular effort to
is
overcome
tional force that continually pulls
the force of
lift
it
a stone,
we
the gravita-
downward.
There are other kinds of forces, such as the force
by exploding dynamite. All these forces are ex-
windings can safely
And we could mention many more cases where
comprehensive approach
familiar force
For example, whenever we
exerted by a stretched spring or the forces created
of the conductors as by the currents they
chanical/thermal approach
gravity.
is
And the stringing of a transmission line is
determined as much by the ice-loading and mechancarry.
The most
conductors, but
withstand.
ical strength
Force
3.1
pressed
in
terms of the newton (N), which
is
the SI
unit of force.
The magnitude of
the force of gravity depends
upon the mass of a body, and
is
given by the ap-
proximate equation
the electrical/me-
F = 9.8w
essential to a thor-
(3. 1)
ough understanding of power technology.
For
this reason, this introductory
chapter covers
certain
fundamentals of mechanics and heat. The top-
ics are
not immediately essential to an understanding
of the chapters which follow, but they constitute a
may wish
Consequently, we rec-
valuable reference source, which the reader
to consult
ommend
from time
a quick
to time.
first
reading, followed by a closer
study of each section, as the need arises.
where
F=
m =
9.8
=
force of gravity acting on the body [N]
mass of the body
[kg]
an approximate constant that
applies
when
objects are relatively
close to the surface of the earth
(within 30 km).
FUNDAMENTALS OF MECHANICS AND HEAT
Example 3-1
51
F
Calculate the approximate value of the force of
on a mass of
gravity that acts
1
2 kg.
Solution
The force of gravity
m =
= 17.6 N
F=
9.8
is
X
9.8
=
12
Figure 3.1
Torque T=
17.6 newtons
1
Fr.
1
When using the
make
to
the
equal to 0.453
hand, a
between the pound
a distinction
pound-force
A pound
(lbf).
592 37 kg,
pound-force
is
On
exactly.
and
(lb)
T=
F=
r =
the other
X
equal to (9.806 65
0.453
592 37) newtons exactly, or about 4.448 N.
If the
pulley
around
Example 3-2
(3.2)
where
mass
a unit of
is
T = Fr
we have
English system of units,
is
torque |N m|
force [N]
radius [mj
move,
free to
it
will begin to rotate
axis.
its
Calculate the approximate value of the force of
gravity that acts
on a mass of 140
newtons and
sult in
Express the
lb.
re-
Example 3-3
A
in pounds-force.
motor develops
a starting torque of 150
pulley on the shaft has a diameter of
Solution
the braking force needed to prevent the
Using the conversion charts in Appendix
massif 140
Eq.
3.
1
-
lb
9.8
=
140 (- 2.205)
the force of gravity
F=
m =
9.8
X
AXO,
= 622.3(^4.448) =
1
=
63.5
622.3
39.9 pound-force
=
39.9 lbf
1
mass of 140
entirely different
The
N
F =
lb.
139.9
is
N
lbf.
radius
T/r
action
is
from a mass of 140
al-
lb.
3.3 Mechanical
it
rotate.
Mechanical work
pose a string
/•
is
(Fig. 3.
1
Torque
If
we
pull
on the
cl
is
done whenever a force
in the direction
the pulley will tend to rotate.
The torque exerted
is
F moves
of the force. The work
Fd
(3.3)
where
W=
F=
cl =
work
(
J]
force [N]
distance the force
moves [m]
a ra-
string with a force
on the pulley by the tangential force
be
is
of the force. For example, sup-
wrapped around a pulley having
).
N would
given by
between the axis of rotation and the
point of application
required. If the ra-
work
W=
on a body, tending to make
is
sufficient to prevent rotation.
equal to the product of the force times the perpen-
F
= 300 N
almost
However,
produced when a force exerts a twisting
dicular distance
motor from
m; consequently, a braking force
50/0.5
1
dius were 2 m, a braking force of 75
is
dius
0.5
is
=
Torque
Torque
If a
turning.
a distance
3.2
m.
63.5 kg. Using
though the numbers are nearly the same, a force of
is
N
m, calculate
Solution
Note that the force of gravity of
exactly equal to the
a
is
Again using the conversion charts, a force of 622.3
140 lbf
l
given by
Example 3-4
A
mass of 50 kg
3.2).
is
lifted to a height
Calculate the work done.
of 10
m
(Fig.
1
FUNDAMENTALS
52
500 kg
Figure 3.2
W=
Work
Fd
Figure 3.3
Power P - Wit
Solution
The force of
gravity acting on the 50 kg
F=
m =
9.8
The work done
X
9.8
50
mass
is
F=
= 490 N
The work done
X
490
10
W=
= 4900 J
The power
Power
Power
is
9.8
X 500 = 4900 N
is
is
W= Fd =
3.4
m=
9.8
P =
the rate of doing work.
It is
given by the
=
Fd = 4900 X 30 = 147 000
J
is
Wit
147 000/12
=
12
250
W=
12.25
kW
equation:
Expressed
P=
Wit
(3.4)
in
horsepower,
P=
12 250/746
=
16.4
hp
where
P = power [W]
W = work done
=
t
The
unit of
power
is
the watt (W).
corresponds
is
use the
sometimes expressed
to the
is
motor
equal
average power output
height of 30
m
a mass of
lifts
in
power developed by
12
s
500 kg through a
the motor, in kilowatts and in
Solution
ity
in the
acting on the
cable
mass
is
The mechanical power output of a motor depends
upon its rotational speed and the torque it develops.
The power is given by
where
P =
T—
n =
9.55 =
(Fig. 3.3). Calculate the
horsepower.
The tension
equal to the force of grav-
that
is
being
motor
in
Example 3-5
electric
of a
to
of a dray horse.
An
Power
1000 W. The
One horsepower
units.
[s]
We often
equal to
is
power output of motors
horsepower (hp)
It
[J]
time taken to do the work
kilowatt (kW), which
746 W.
3.5
lifted:
mechanical power [WJ
torque [N-m]
speed of rotation [r/min]
a constant to take care of units
(exact value
=
30/tt)
We can measure the power output of a motor by
means of a prony brake. It consists of a stationary
flat belt that presses against a pulley mounted on the
motor shaft. The ends of the belt are connected to
two spring
scales,
and the
belt pressure
is
adjusted
FUNDAMENTALS OF MECHANICS AND HEAT
V (Fig. 3.4). As the motor turns,
by tightening screw
53
3.6 Transformation of energy
we can increase or decrease the power output by adpower developed by the motor
into heat
When
register
The mechanical
tension of the belt.
justing the
by the
the
entirely converted
is
it
not running, the spring scales
equal pulls and so the resulting torque
(as
P 2 The
re-
does
in Fig. 3.4), pull
Thermal energy (heat released by
ley
is
therefore
a radius
ters.
late
f>,
exceeds pull
— P 2 newtons. If the pulley has
T = (P — P 2 )r newton me-
(/>,
Knowing
a stove, by
by the sun)
Chemical energy (energy contained
3.
in
dyna-
mite, in coal, or in an electric storage battery)
)
the net torque
r,
.
friction,
on the circumference of the pul-
in
mov-
ing car)
is
However, when the motor turns clockwise
sultant force acting
one of the following forms:
a coiled spring or the kinetic energy of a
2.
zero.
exist in
Mechanical energy (potential energy stored
1.
rubbing against the pulley.
belt
motor
is
Energy can
produced by
Electrical energy (energy
4.
a gen-
x
the speed of rotation,
we can
erator, or
calcu-
the power, using Eq. 3.5.
by lightning)
Atomic energy (energy released when
5.
cleus of an atom
is
the nu-
modified)
Although energy can be neither created nor destroyed,
it
can be converted from one form to an-
other by means of appropriate devices or machines.
For example, the chemical energy contained
in
coal
can be transformed into thermal energy by burning
The thermal energy contained
the coal in a furnace.
in
steam can then be transformed into mechanical
energy by using a turbine. Finally, the mechanical
energy can be transformed into electrical energy by
means of a
In the
generator.
above example, the furnace, the turbine,
Figure 3.4
and the generator are the machines
Prony brake.
ergy transformation.
that
Unfortunately, whenever energy
the output
Example 3-6
During a prony brake test on an electric motor, the
25
spring scales indicate
(Fig. 3.4).
turns at
0.1
N
and 5 N, respectively
Calculate the power output
if
the
motor
1700 r/min and the radius of the pulley
is
always
transformed,
because
all
ma-
chines have losses. These losses appear in the form of
heat, causing the temperature
of the machine to
rise.
Thus, the electrical energy supplied to a motor
is
partly dissipated as heat in the windings. Furthermore,
is
some of its mechanical energy
m.
is
less than the input
do the en-
is
also
lost,
due
to bear-
ing friction and air turbulence created by the cooling
Solution
fan.
The torque
T=
=
The power
The mechanical
losses are also transformed into
put of a motor
Fr
(25
-
5)
X
0.1
=
2
Nm
is
less than the electrical input.
3.7 Efficiency of a
machine
is
The efficiency of a machine
P =
ai
mechanical power out-
heat. Consequently, the useful
is
779.55
=
1700
X
2/9.55
= 356
W
The motor develops 356 W, or about 0.5 hp.
is
given by the equation
P
t\
= p X
100
(3.6)
FUNDAMENTALS
54
where
Kinetic energy
—
Tl
P0 =
The
energy
power
input
\/2mv
power of the machine fW]
output
efficiency
is
is
machine
to the
particularly
[
W]
low when thermal
from 25
the efficiency of steam turbines ranges
that of internal
gines (automobile engines, diesel motors)
we must remember
it
lies
how low
that a
having an efficiency of 20 percent loses,
of heat, 80 percent of the energy
Ek = kinetic energy [JJ
m = mass of the body [kg]
v = speed of the body fm/s]
to
be-
A bus having a mass of 6000 kg moves at a speed of
machine
100 km/h. If it carries 40 passengers having a total
mass of 2400 kg, calculate the total kinetic energy
of the loaded vehicle. What happens to this energy
in the
form
receives.
efficiently.
Example 3-8
these
Electric motors transform electrical energy into
mechanical energy much more
when
the bus
Solution
pending on the size of the motor.
Total
150
kW
percent
when
mass of
it
motor has an efficiency of 92
operates
at full-load.
The speed
+ 2400 = 8400
=
100 km/h
v
100
refers to the
The mechanical output power
=
163
kW
To stop the
is
is
\/2mv
3
2
1/2
245 928
J
X 8400 X
3.25 MJ
27.
2
bus, the brakes are applied and the re-
sulting frictional heat
is
entirely
produced
at the
ex-
pense of the kinetic energy. The bus will finally
come
losses are
150
27.8 m/s
kinetic energy
Ek =
=
P 0 = 150kW
163
m
s
mechanical
The
150/0.92
kg
X 1000
3600
The 150 kW rating always
power output of the motor.
The input power is
The
6000
is
is
Solution
{
to a stop?
Calculate the
losses in the machine.
P = PJi] =
braked
the loaded bus
m =
_
electric
is
Their ef-
ficiency ranges between 75 and 98 percent, de-
Example 3-7
(3.7)
combustion en-
tween 15 and 30 percent. To realize
efficiencies are,
2
where
converted into mechanical energy. Thus,
40 percent, while
A
a form of mechanical energy
efficiency [percent]
—
Pi
is
given by the equation
13
to rest
when
all
the kinetic energy (3.25
MJ)
has been dissipated as heat.
kW
Considering the high efficiency of the motor,
the losses are quite moderate, but they
be enough to heat a large
home
in the
would
still
winter.
3.8 Kinetic
3.9 Kinetic
A
energy of linear motion
A falling stone or a swiftly
sess kinetic energy,
which
moving automobile posis
energy due to motion.
energy of
moment
middle of
Its
rotation,
of inertia
revolving body also possesses kinetic energy.
magnitude depends upon the speed of rotation
and upon (he mass and shape of the body. The
netic energy of rotation
on page 56.
is
ki-
given by the equation
FUNDAMENTALS OF MECHANICS AND HEAT
TABLE 3A
MOMENT OF
INERTIA J AROUND AN AXIS
OF ROTATION
0
Figure 3.5
Mass
m revolving at a distance
J
r
around axis
0.
= mr
(3.9)
Figure 3.6
Solid disc of
mass
m and
radius
r.
(3.10)
Figure 3.7
Annular
ring of
mass
J
m having a
rectangular cross-section.
= -(Rf + R 2 I
(3.11)
)
Figure 3.8
Straight bar of
mass
m pivoted on
its
center.
J
(3.12)
12
Figure 3.9
Rectangular bar
of
mass
J
m revolving around
= -(Rf + /V
axis 0.
(3.13)
55
1
FUNDAMENTALS
56
5.48
X
10"'V/7-
(3.8)
The
b.
Ek =
=
where
E k = kinetic energy [J]
J = moment of inertia [kg-rrrl
= rotational speed [r/minj
X 1()~ 3 = constant to take care of units
2
Lexact value = tt /1800]
//
5.48
The moment of inertia./ (sometimes simply called
depends upon the mass and shape of the body.
inertia )
may be
value
Its
calculated for a
shapes by using Eqs. 3.9 to
3.
1
3 given in Table 3 A. If
body has a complex shape,
the
number of simple
can always be bro-
it
ken up into two or more of the simpler shapes given
in the table.
are then
The
individual Js of these simple shapes
added to give the
kinetic energy
=
Note
as
5.48
X
5.48
X
3.
1
is
l0"-V/r
X
(3.8)
175
X
1800
2
MJ
that this relatively small flywheel possesses
much
kinetic energy as the loaded bus
Example
tioned in
men-
3-8.
Example 3-10
A
flywheel having the shape given
in Fig. 3.1
1
is
composed of a ring supported by a rectangular hub.
The ring and hub respectively have a mass of 80 kg
and 20 kg. Calculate the moment of
inertia of the
flywheel.
J of the body.
total
Inertia plays a very important part in rotating
machines; consequently,
it
is
worth our while to
solve a few problems.
80 kg
Example 3-9
A
1
1400 kg
solid
m
flywheel has a diameter of
steel
and a thickness of 225
mm (Fie.
225
3. 10).
mm
1400 kg
V
1800 r/m
Figure 3.11
Flywheel in Example 3.10.
Figure 3.10
Flywheel in Example
3.9.
Calculate
Its
b.
The
at
Solution
moment
a.
of inertia
kinetic energy
For the
when
ring,
the flywheel revolves
J
I800r/min
]
Solution
a.
Referring to Table 3 A,
inertia
we
find the
moment
of
= m (Rr + R 2 2 )/2
= 80 (0.4 2 + 0.3 2 )/2 =
(3.1 1)
10 kg
m
2
For the hub,
is
J2
J
(3.10)
1400
X
0.5
The
2
total
= mL 2 /\2
= 20 X (0.6) 2 /12 =
moment
= 175ks-m 2
J
=
J
(3. 12)
0.6 kg
m
2
of inertia of the flywheel
X
+
J2
=
10.6
kg
m
is
2
k,
FUNDAMENTALS OF MECHANICS AND HEAT
3.10 Torque, inertia,
volving body, and that
a
change the speed of a
to
is
to subject
The
given period of time.
depends upon the
Speed
1
In electric
way
only one
is
3-1
speed
in
There
and change
it
re-
to a torque for
of change of speed
rate
inertia, as well as
on the torque.
A
=
A/7
9.55
power technology,
a system there are three
erted by the load, and the speed.
(3.14)
T=
The load
a shaft (Fig. 3. 2).
interval of time during
torque
J
9.55
is
— moment
= constant
applied
creasing
or
Suppose
the
T
[sj
.
Y
to take care
=
in a
We now
explain
exerts a constant torque
counterclockwise direction.
TM developed by
the
can be varied by
in-
the torque
acts clockwise,
and
decreasing
system
it
the
current
electric
rest
is initially at
and
that
/.
7M =
Because the torques are equal and opposite, the
net torque acting on the shaft
2
zero, and so
is
it
has
]
no tendency
to rotate.
of units
Load
mm
30/ttJ
the torque acts in the direction of rotation, the
speed rises. Conversely,
tion
which the
of inertia fkg-m
[exact value
If
motor
[Nm]
torque
=
Ar
speed [r/min]
in
to consider:
they interreact.
1
change
main factors
the torque developed by the motor, the torque ex-
Tu that always acts
On the other hand,
=
often happens that
Consider a load coupled to a motor by means of
TMJ
where
A/?
it
system
an electric motor drives a mechanical load. In such
how
simple equation relates these factors:
of a motor/load
57
if
of rotation, the speed
acts against the direc-
it
The term An may
falls.
therefore represent either an increase or a decrease
in
speed.
'
M
Example 3-11
The
to
fly wheel
increase
torque of
20
of Fig. 3.11 turns
Nm.
at
60 r/min.
We wish
Motor
600 r/min by applying a
For how long must the torque be
speed
its
to
Figure 3.12
applied?
Shaft
is
stationary
TM
71-
Solution
The change
in
A/?
speed
-
The moment of
(600
-
inertia
J
As
is
=
60)
= 540
a result of the
twists and
r/min
becomes
shaft
deformed, but other-
wise nothing happens.
Suppose we want
is
10.6 kg-
opposing torques, the
slightly
/z,
To do so, we increase the motor current so
TM exceeds T The net torque on the shaft acts
speed
2
that
.
.
x
clockwise, and so
Substituting these values in Eq. 3.14
the load to turn clockwise at a
it
begins to rotate clockwise. The
speed increases progressively with time but as soon
9.55 7Af/./
A/7
(3.14)
as the desired
540
=
9.55
x 20
A/7 10.6
TL The
.
yielding
speed
motor current so
30
s
is
TM
reached,
is
let
us reduce the
again exactly equal to
net torque acting on the system
and the speed
Ar
/z,
that
any more (Fig.
/z,
is
now
zero
will neither increase or decrease
3. 13).
FUNDAMENTALS
58
Load
Load
Figure 3.13
Shaft turns
Figure 3.14
cw 7M
Shaft turns
71-
Whenever
This brings us to a very important conclusion.
The speed of a mechanical load remains constant
when the torque TM developed by the motor is
equal and opposite to the torque T L exerted by the
load.
At
first, this
we
accept, because
when TM =
7^
rather difficult to
is
the
opposite
to
a motor remains constant
motor torque is exactly equal and
load
motor/load system
is
torque.
In
the
effect,
then in a state of dynamic
equilibrium.
With the load now running clockwise
suppose we reduce
TM
so that
it
is
at
a speed
TL
less than
.
on the shaft now acts counterclock-
net torque
wise. Consequently, the speed decreases and will
continue to decrease as long as
T
imbalance between
the
enough, the speed
we
then reverse. If
TM — TL when
motor torque 7M and load torque
and opposite, the speed will
change. The rate of change depends upon the inerof the rotating parts, and this aspect
tia
more
is
covered
in
detail in Section 3. 13.
Power flow
3.12
our reasoning (and reality) shows.
the
the
are not exactly equal
{
T,
will eventually
exceeds
TM
and
TM
.
If
long
lasts
become zero and
control the motor torque so that
the reverse speed reaches a value n 2
,
in
a mechanically
coupled system
The speed of
repeat:
whenever
The
TL
.
are inclined to believe that
the system should simply stop. But
,
this is not so, as
We
conclusion
ccw TM = TL
Returning again to Fig.
3. 13,
we
torque
TM
acts in the
torque
T
acts opposite to speed n
see that motor
same direction (clockwise) as
speed n lt This means that the motor delivers mechanical power to the shaft. On the other hand, load
{
x
the load receives mechanical
.
Consequently,
power from
the shaft.
We can therefore state the following general rule:
When the torque developed by a motor acts in
same direction as the speed, the motor delivpower to the load. For all other conditions,
the motor receives power from the load.
the
ers
In Fig. 3.14, for example, the
power from
the load because
7M
motor receives
acts opposite to n 2
.
the system will continue to run indefinitely at this
Although
new speed
brief periods in electric trains and electric hoists.
(Fig. 3. 14).
In conclusion, torques
Figs. 3.12, 3.13,
TM and T
{
are identical in
and 3.14, and yet the shaft may be
turning clockwise, counterclockwise, or not
The
actual
steady-state
whether TM was greater or
speed
less than
depends
7L
this is
an unusual condition,
it
occurs for
The behavior of the motor under these conditions
will be examined in later chapters.
at all.
upon
3.13 Motor driving a load
for a certain
having inertia
period of time before the actual steady-state condi-
was reached. The reader should ponder
moments over this statement.
tion
a
few
When
a motor drives a mechanical load, the speed
is
usually constant. In this state of dynamic equilibrium.
FUNDAMENTALS OF MECHANICS AND HEAT
the torque
TM developed by the motor is exactly equal
T imposed by the load.
and opposite to the torque
The
inertia
of the revolving parts does not
However,
play under these conditions.
torque
speed
is
raised so that
it
increase,
will
come
if
we have
already
when
less than that
of
speed drops. The increase or decrease
in
speed (An)
motor torque
is
TM
torque Tis now replaced by the net torque (TM — 7L
=
A/7
- T )AtIJ
9.55 (rM
x
stays con-
paper remains
must be greater than the load torque
It
.
We
in
have
given by the Eq. 3.14, except that
is still
Nm)
in the
order for the speed to increase.
seen.
Conversely,
because the tension
unchanged. Let the required motor torque be
exceeds the load torque, the
as
the speed increases from 120 r/min to 160
stant
motor
the
As
r/min, the load torque (5400
into
the load, the
the
b.
x
59
-
A/7
):
J
(3.15)
At
where
160
-
= 4500
= 5 s
120
m
kg
r/min
- TL )At
9.55 (7\,
An =
= 40
2
J
= change in speed [r/minj
TM = motor torque [N-m]
7L = load torque [N m]
At = time interval during which
TM and TL are acting [s]
J = moment of inertia of all
A/7
_
40
-
9.55
4500
Thus,
TM - 5400 = 3770
rM = 9170
revolving parts [kg-irrj
The motor must
Example 3-12
A
large reel
torque of 9170
of paper installed
machine has
a
diameter of
moment of
and a
1
inertia of
at the
end of
a
paper
kgm 2
.
It is
therefore develop a constant
Nm
during the acceleration
period.
m, a length of 5.6 m,
.8
4500
5400) 5
The mechanical power of the
driven
reel
motor
160 r/min
at
accelerating
is
by a directly coupled variable-speed dc motor turn-
The paper
ing at 120 r/min.
tension of
a.
160 r/min
c.
in 5
c.
from 120 r/min
to
seconds, calculate the torque that
power of
the
this interval.
motor
after
it
As soon
7 = Fr = 6000 X
1
reel
.8/2
The power developed by the
P =
nT
120
P =
is
is
N
m). The power
nT _
~
160
X 5400
9.55
9.55
90.5
kW (equivalent to
121 hp)
is
= 5400 N m
reel
motor
3.14 Electric motors driving linear
motion loads
is
Rotating loads such as fans, pumps, and machine
X 5400
9.55
kW
160 r/min)
therefore reduced to
tools are well suited for direct mechanical coupling
to electric motors.
67.85
(
equal to the load torque (5400
has
(3.5)
9.55
kW (equivalent to 206 hp)
as the desired speed
of the motor
Solution
The torque exerted on the
9.55
reached, the motor only has to develop a torque
reached the desired speed of 160 r/min.
a.
160X9170-
-
153.6
the reel
speed of 120 r/min.
motor must develop during
Calculate the
-
power of the motor when
the speed has to be raised
the
nT
P =
6000 N.
Calculate the
If
kept under a constant
9.55
turns at a constant
b.
is
(about 91 hp)
move
in
On
the other hand, loads that
a straight line, such as hoists, trains, wire-
FUNDAMENTALS
60
drawing machines,
must be equipped with a
etc.,
where
motion converter before they can be connected to a
rotational speed [r/min]
T=
torque [N-m]
F=
v =
9.55 =
we seldom
converters are so utterly simple that
=
n
The motion converter may be a
rope-pulley arrangement, a rack and pinion mechanism, or simply a wheel moving over a track. These
rotating machine.
force [N]
linear speed [m/s|
a constant [exact value
=
30/tt]
think of the important part they play.
Straight-line motion involves a linear speed v
and a force
while rotary motion involves a rota-
speed n and a torque
tional
tities
F,
related
when
T.
How
are these quan-
a motion converter
is
used?
Consider a jack driven by a motor that rotates
T (Fig.
a linear speed
plied in raising the load
is
v.
motive turns
the other hand, the
needed
is
at
1
to pull an electric train
The motor on board
the loco-
200 r/min. Calculate the torque de-
at
F
Solution
The power sup-
given by
power
kN
This
nT=
9.55fY
1200 T =
X
P a = Fv
On
of 25
a speed of 90 km/h.
veloped by the motor.
3. 15).
causes a vertical ram to exert a powerful force
at
A force
at
a speed n while exerting a torque
while moving
Example 3.13
9.55
(3.16)
25 000
j = 4974 N-m =
input to the jack
5
X
(90 000/3600)
kN
m
is
given by
= nT
P:
3.15 Heat and temperature
(3.5)
1
9.55
Assuming
verter,
Whenever
there are no losses in the motion con-
we have
the SI unit
p.
1
1
heat
applied to a body,
is
mal energy. Heat
is
'
energy?
Consequently,
nT=
9.55Fv
a
body receives
this type
of
atoms of the body vibrate more
First, the
intensely. Second,
we can
receives ther-
the joule.
What happens when
= po
it
therefore a form of energy and
is
its
temperature increases, a fact
verify by touching
it
or by observing the
(3.16)
reading of a thermometer.
For a given amount of heat, the increase
tem-
in
perature depends upon the mass of the body and the
material of which
100 kJ of heat to
is
it
made. For example,
if
by 24°C. The same amount of heat supplied
of copper raises
we add
kg of water, the temperature
1
rises
to
temperature by 263°C.
its
therefore obvious that heat and temperature are
kg
1
It
is
two
quite different things.
If
we remove
heat from a body,
its
temperature
drops. However, the temperature cannot
low a lower
zero.
or
It
limit.
corresponds
— 273.1 5 °C. At
This limit
to a
fall
be-
called absolute
is
temperature of 0 kelvin
absolute zero
all
and the only motion
Figure 3.15
tions cease
Converting rotary motion into linear motion.
that of the orbiting electrons.
atomic vibra-
that subsists
is
FUNDAMENTALS OF MECHANICS AND HEAT
1806
T
iron melts
T
450 K
i
1083
aluminum melts
933
660
1220
lead melts
600
327
621
water boils
water freezes
373
273
100
212
32
0
-273.15
®
®
1981
-459.67
Fahrenheit scale
Celsius scale
Kelvin scale
2791
810°F
450 °C
1, 356
copper melts —
T
1533
61
Figure 3.16
Temperature scales.
3.16
The
Temperature scales
given
specific heat capacity of several materials
Table
in
AX2
in
the
is
Appendix.
The kelvin and the degree Celsius are the SI units of
temperature. Fig. 3. 6
1
tionships
between
shows some
interesting rela-
Kelvin,
the
Celsius,
and
Fahrenheit temperature scales. For example, iron
melts at
1
806
K
or
1
533°C or 279 °F
1
Example 3-14
Calculate the heat required to raise the temperature
of 200
L of water from 10°C
tank
perfectly insulated (Fig. 3.17).
is
heat capacity of water
3.17
Heat required to raise
the temperature of a body
The temperature
heat
it
terial.
receives,
rise
its
weighs
1
kg.
of a body depends upon the
mass, and the nature of the ma-
The relationship between these
quantities
is
given by the equation
Q = mcAt
(3. 17)
where
Q =
quantity of heat added to
(or
m =
c
=
removed from)
mass of the body
=
body
[JJ
specific heat capacity of the
material
At
a
|kg]
making up
the
body
[J/(kg-°C)l
Figure 3.17
change
Electric
in
temperature [°C]
water heater.
is
to
70°C, assuming the
The
specific
41 80 J/kg °C, and one
liter
n
FUNDAMENTALS
62
Solution
quired
convection
convection
The mass of water
is
200 kg, and so the heat
re-
is
Q ~ mcAt
= 200 X 4180 X
= 502 MJ
(70
-
10)
Referring to the conversion table for Energy (see
Appendix),
we
find that 50.2
MJ
is
equal to 13.9
kW-h.
3.18 Transmission of heat
Many problems
in electric
power technology
lated to the adequate cooling of devices
are re-
and ma-
knowledge of the
mechanism by which heat is transferred from one
body to another. In the sections that follow, we
Figure 3.18
Heat transmission by convection, conduction, and
ra-
diation.
chines. This, in turn, requires a
briefly
review the elementary physics of heat trans-
We
mission.
also include
some simple
but useful
Referring to Fig. 3.19,
we can
equation
equations, enabling us to determine, with reason-
P =
able accuracy, the heat loss, temperature rise, and so
on of
electrical
equipment.
its
we
-
—
atoms
(Fig. 3. 18). This
A =
—
atomic vibration
]
t
2)
—
end of the
bar.
atom
is
to the next, to the other
at
d = thickness of the body [m]
Consequently, the end opposite the
flame also warms up, an observation
one time or another.
ferred along the bar
we have
In effect, heat
is
all
trans-
by a process called conduction.
The rate of heat transfer depends upon the thermal conductivity of the material. Thus, copper is a
better thermal
conductor than
steel
is,
and plastics
and other nonmetallic materials are especially poor
conductors of heat.
The SI
unit of thermal conductivity
per meter degree Celsius
conductivity of several
in
Tables
AX
1
and
AX2
[W/(m
common
in the
°C)J.
is
the watt
The thermal
materials
Appendix.
is
given
2
body [m
difference of temperature between
surface area of the
opposite faces [°C]
transmitted from one
made
(3.18)
[W/(m-°C)J
bring a hot flame near one end of an iron bar,
temperature rises due to the increased vibration
its
]
P = power (heat) transmitted [W]
X = thermal conductivity of the body
(t
of
-
where
3.19 Heat transfer by conduction
If
calculate the ther-
mal power transmitted through a body by using the
Figure 3.19
Heat transmission by conduction.
]
FUNDAMENTALS OF MECHANICS AND HEAT
Example 3-15
63
tank
The temperature difference between two sides of a
mica
sheet of
50°C
is
cm" and thickness
ing
3
is
through the sheet,
(Fig. 3.20). If
mm,
its
area
is
200
calculate the heat flow-
in watts.
Sol ui ion
AX1,
According to Table
of mica
ducted
is,
W/m
0.36
is
the thermal conductivity
The thermal power con-
°C.
therefore.
-
\A(t
P
t
l
2)
(3.18)
0.36
d
Figure 3.21
X
Convection currents
0.02 (120
70)
120
in oil.
W
0.003
The warm
0.02
m
2
it
comes
in
contact with the cooler tank,
X = 0.36
chills,
nal tank.
120'
oil
becomes heavier, sinks to the bottom, and
moves upward again to replace the warmer oil now
moving away. The heat dissipated by the body is,
therefore, carried away by convection to the exter-
mica
The
tank, in turn, loses
convection to the surrounding
x
its
heat by natural
air.
70°C
3
3.21
mm
Calculating the losses
by convection
Figure 3.20
Mica sheet,
The
Example 3-15.
heat loss by natural convection in air
is
given
by the approximate equation
P =
3.20
Heat transfer by convection
In Fig.
3.18 the air in contact with the hot steel bar
3/Uf,
-
t
2
y
(3.19)
where
warms up and, becoming
a
chimney.
As
the bar,
smoke
moves upward, it is
which, in turn, also warms
the hot air
placed by cooler air
A continual
lighter, rises like
current of air
removing
its
is
P —
A —
f, =
t2 =
in
re-
up.
therefore set up around
heat by a process called nat-
heat loss by natural convection [W|
surface of the body [m]
surface temperature of the body [°C]
ambient temperature of the surrounding
air l°CJ
ural convection.
The convection process can be accelerated by
employing a fan
to create a rapid circulation
In the
case of forced convection, such as that
produced by a blower, the heat carried away
is
of
given approximately by
fresh
in
air.
Heat transfer by forced convection
is
used
P = 1280 Va {t 2
most electric motors to obtain efficient cooling.
when
oil. The
Natural convection also takes place
body
is
immersed
contact with the
currents
in a liquid,
such as
(3.20)
where
oil in
body heats up, creating convection
which follow the path shown
/,)
a hot
in Fig. 3.21.
P — heat loss by forced convection [W]
— volume of cooling air |m Vs]
V.
x
FUNDAMENTALS
64
=
i\
temperature of the incoming (cool) air
3.22 Heat transfer by radiation
[°C|
f
—
2
We
temperature of the outgoing (warm)
have
air|°C]
same
when hydrogen,
Surprisingly, Eq. 3.20 also applies
a
much
lighter gas,
basked
all
is
used as the cooling medium.
properties
Example 3-16
area of 1.2 nr.
When
operates
it
at full-load,
60°C
surface temperature rises to
20°C
in
the
an ambient of
(Fig. 3.22). Calculate the heat loss
by natural
as
energy
light,
P =
=
diate
when
only converted to heat
is
have discovered
even those
heat,
as the physi-
that
that all bodies ra-
very cold. The
are
amount of energy given off depends upon
the tem-
perature of the body.
On
L25
-
3A{t
t2
}
the other hand,
all
bodies absorb radiant en-
3
X
(60
1.2
ergy absorbed depends upon the temperature of
)
-
20)
L25
W
= 362
the surrounding objects. There
tinual
ial
362
i
f
)
\
W
H
bodies,
as
convection
body
the
is
same
when
in
as that of
body then radiates
a
body
as
the temperature of a
surroundings. The
its
much energy
is
hotter than
is
tinually lose heat
in
therefore, a con-
each were a miniature sun.
if
Equilibrium sets
\
is,
exchange of radiant energy between mater-
and the net radiation
zero.
its
On
as
it
receives
the other hand,
environment,
by radiation, even
if
it
it
will
is
if
con-
located
vacuum.
3.23 Calculating radiation losses
Figure 3.22
Convection and radiation losses
a
in
totally
enclosed
motor.
The heat
that a
body
loses by radiation
is
given by
the equation
P = kA
Example 3-17
kW
fan rated at 3.75
through a 750
kW
motor
the inlet temperature
is
3
1
is
blows 240 rrrVmin of
to carry
22°C and
away
air
away by
P =
A =
T, =
T2 =
the outlet temper-
the circulating
air.
lx
=
1280
(t 2
-
k
=
X 240/60
heat radiated
(31
-
22)
(3.21)
= 46 kW
IW]
surface area of the body \m'\
absolute temperature of the body [K]
absolute temperature of the surround-
a constant,
ture of the
f,)
(approximate)
)
ing objects IK]
Consequently, the losses are
P = 1280 V
- T2 4
where
°C. estimate the losses in the motor.
losses are carried
4
(7,
the heat. If
Solution
The
passes
readily
it
ergy from the objects that surround them. The en-
Solution
ature
and
meet a solid body, such
earth. Scientists
convection.
A
the
and living things on the surface of the
cal objects
enclosed motor has an external surface
totally
warmth produced by
through the empty space between the sun and the
earth. Solar
the sun's rays
A
in the
sun's rays. This radiant heat energy possesses the
Table
3B
which depends upon the na-
body surface
gives the values of k for surfaces
monly encountered
in electrical
equipment.
com-
FUN DA MENTA LS OF MFC HA NfCS A NI) HFA I
3-3
RADIATION CONSTANTS
TABLE 3B
Give
the SI unit
symbol
Type of surface
polished silver
bright
oxidized copper
3
aluminum paint
3
Nichrome
2
tungsten
2
oxidi/ed iron
4
insulating materials
paint or nonmetallic
perfect emitter
5
enamel
5
5.669
(blackbody)
)
X l()"
X 10 s
X 10~ s
X 10 s
X 10" s
X 10 8
X 10'"*
X 10" s
X 10 s
X 10'"*
1
force
work
pressure
area
mass
temperature
thermal energy
thermal power
3-4
200
force of
An
tion,
coated with a non-
knowing
surrounding objects are
that all
mechanic exerts a
end of a wrench hav-
automobile engine develops a torque of
at
an
at
a speed of
power output
4000
r/min. Calculate
watts and in horse-
in
power.
by radia-
lost
at the
ing a length of 0.3 m. Calculate the torque
The motor
is
N
power
he exerts.
the
Example 3-16
electrical
In tightening a bolt, a
Example 3-18
in
energy
electrical
600 N-m
enamel. Calculate the heat
mechanical power
mechanical energy
3-5
metallic
and the corresponding SI
for the following quantities:
4
s
0.2
copper
oxidized
W/(m 2 -K
Constant k
65
3-6
ambient temperature of 20°C.
A crane
lifts
200
ft in
and
in
15
a mass of 600 lb to a height of
s.
Calculate the power
watts
in
horsepower.
Solution
An
3-7
=
= 60°C
or (273. 15 + 60) = 333 K
T2 = surrounding temperature = 20°C
or (273. 15 4- 20) - 293 K
7,
surface temperature
From Table 3B. k
The power
=
5
X
10
by radiation
lost
s
W/(m 2 -K 4
is,
line
4
=
5
-
7\
10" 8
X
(7,
X
3-8
(3.21)
)
4
-
293
20 kW.
c.
A
large flywheel has a
it
(hp]
J
lb-fr. Calculate
rotates at
its
moment
of inertia of
kinetic energy
when
60 r/min.
4
The rotor of an induction motor has a moment of inertia of 5 kg-nr. Calculate the
3-9
)
W (approximate)
= 296
kW from the
20
The power output of the motor [kW] and
The efficiency of the motor
The amount of heat released Btu/h]
b.
therefore,
(333
1
to
).
4
1.2
motor draws
and has losses equal
Calculate
a.
500
P = kA
electric
energy needed to bring the speed
It
is
interesting to note that the
almost as
much
motor dissipates
heat by radiation (296
W)
as
it
does
by convection (362 W).
3-
1
0
Questions and Problems
3-11
3-
1
A cement
is
from zero
b.
from 200 r/min
c.
from 3000 r/min
Name
ried
Practical level
What
What
block has a mass of 40 kg.
the force of gravity acting on it?
3-2
is
needed
How much
to
energy
to
the three
to
400 r/min
to
400 r/min
ways whereby heat
from one body
A motor develops a cw
50 N-m.
If this situation persists for
some
time, will the
direction of rotation eventually be
needed
to
lift
a sack
of flour weighing 75 kg to a height of 4
m?
car-
torque of 60 N-m,
lift it?
is
is
to another.
and the load develops a ccw torque of
a.
force
200 r/min
a.
b.
What
value of motor torque
the speed constant?
is
cw
needed
or
to
ccw?
keep
FUNDAMENTALS
66
3-12
A
motor drives
r/min.
Nm, and
15 Nm.
12
of
cw speed of 1000
a cw torque of
exerts a ccw torque
a load at
3- 9
1
The motor develops
the load
a.
Will the speed increase or decrease?
b.
If this situation persists for
some
1
negligible, calculate the following:
what
time, in
a.
direction will the shaft eventually rotate'?
3-13
Referring to Fig. 3.12,
what
3-14
is
the
if
Referring to Fig. 3.13,
=50
TM = 40 N
power delivered by
if
the
b.
m,
power
3-20
motor?
TM = 40 N m
r/min, calculate the
= 50
Calculate the heat [MJJ required to raise the
temperature of 100 kg of copper from 20°C
deliv-
3-2
1
Repeat Problem 3-20 for
1
00 kg of
alu-
minum.
Referring to Fig. 3.14,
n2
The torque developed by the motor [N-m]
The force opposing the motion of the bus [N]
to 100°C.
and
ered by the motor.
3-15
The electric motor in a trolley bus develops
a power output of 80 hp at 200 r/min as the
bus moves up a hill at a speed of 30 miles
per hour. Assuming that the gear losses are
if
TM = 40 N m and
r/min, calculate the
3-22
power received
by the motor.
The motor in Fig. 3.23 drives a hoist, raising a mass m of 800 kg at a uniform rate of
5 m/s. The winch has a radius of 20 cm.
Calculate the torque [N-m] and speed
Intermediate level
[r/min] of the motor.
3- 6
1
During a prony brake
test
on a motor (see
and speed
Fig. 3.4), the following scale
readings were noted:
P2 = 5
n =
1
If
P,
lbf
=28
lbf
160 r/min
the diameter of the pulley
calculate the
12 inches,
is
power output of
the
motor
in
kilowatts and in horsepower.
3- 7
1
A
motor drives a flywheel having a moment
of inertia of 5 kg-nr.
The speed
increases
ij
from 1600 r/min
to
1800 r/min
in 8
Motor:
s.
Calculate
a.
The torque developed by
b.
The energy
c.
The motor power [W]
d.
The power
in the
input
the
flywheel
[W]
at
at
motor [N-m]
1800 r/min
Figure 3.23
|kJ|
to the flywheel at
A
3-23
dc motor coupled
velops 120 hp
at
a constant speed of 700
The moment of
2
parts is 2500 lb-ft
ing
a.
|
inertia of the revolv-
1
Problem 3-22
m/s, calculate the
r/min] and torque
[ft-
lbf]
is
re-
new speed
of the motor.
i
Calculate the torque [N-m] developed by the
3-24
How many
Btus are required to raise the
temperature of a 50 gallon (U.S.) reservoir
of water from 55°F to
Calculate the motor torque IN-m] needed so
that the
{Note;
to
Industrial appl cat ion
.
motor.
b.
Tf the hoisting rate in
duced
to a large grinder de-
r/min.
Problem 3-22.
1750
r/min
3-18
Electric hoist,
1600 r/min
speed will increase
to
The torque exerted by
mains the same.)
750 r/min
in 5
the grinder re-
s.
the tank
will
it
is
take
1
80°F, assuming that
perfectly insulated.
if
the tank
electric heater?
is
How
heated by
long
a 2
kW
FUNDAMENTALS OF MECHANICS AND HEA T
3-25
A
large indoor transformer
using an
is
painted a non-
It is
proposed to refurbish
aluminum
paint. Will this affect
metallic black.
it
the temperature of the transformer? If so,
will
3-26
An
it
electrically heated
perature
is
m X
cement floor covers
30 m. The surface tem-
25 °C and the ambient tempera-
ture
is
23 °C. Approximately
heat
is
given
the point of
is
off, in
The cable and other
electrical
how much
kilowatts? Note: from
view of heat radiation, cement
considered to be an insulator.
components
inside a sheet metal panel dissipate a total
of 2 kW.
A blower
inside the panel keeps
the inside temperature at a uniform level
throughout. The panel
run hotter or cooler?
an area of 100
3-27
67
high, and 2
Assuming
tion
ft
is
4
ft
wide, 8
ft
deep, and totally closed.
that heat
is
radiated by convec-
and radiation from
all
sides except the
bottom, estimate the temperature inside the
panel
if
the ambient temperature
The panel
enamel.
is
is
30 °C.
painted with a nonmetallic
Part Two
Electrical
Machines and
Transformers
Chapter 4
Direct-Current Generators
This
Introduction
4.0
We
begin our study of rotating machinery with
the direct-current generator.
generators are not as
common
because direct current,
when
are discussed.
as they used to be,
required,
is
mainly
without
current
any
using
moving
important because
tion to the
and
it
is
4.1
motor and vice versa. Owing
similar construction, the
reason
fundamental properties of
erator
after
we
We show
when
operates
meant by the neutral
fine
what
how
the induced voltage
termines
it
at
is
its
is
point.
it
is
(ac) generator.
The
that the voltage generated in
any dc gen-
inherently alternating and only
becomes dc
is
has been rectified by the commutator.
1
begin with the basic principles
the importance of brush position
the study of a direct-
Fig. 4.
shows an elementary ac generator
composed of a coil that revolves at 60 r/min between the N, S poles of a permanent magnet. The
applied to a dc motor.
chapter
may seem,
edge of the alternating-current
to their
iearn about a dc generator can be di-
of a 2-pole generator
it
current (dc) generator has to begin with a knowl-
generators and motors are identical. Consequently,
In this
Generating an ac voltage
Irrelevant as
built
same way; consequently, any dc generator can
we
actual
including multipole designs.
many dc
Commercial dc generators and motors are
rectly
their voltage-regulation characteristics.
physical construction of direct-current machines,
brief periods.
anything
next.
of dc generators
represents a logical introduc-
behavior of dc motors. Indeed,
operate as a
commutating poles and
The chapter ends with a description of the
industry actually operate as generators for
in
for
We then discuss the major types
parts.
Nevertheless, an understanding of dc generators
The need
problem of pole-tip saturation are covered
the
produced by electronic rectifiers. These rectifiers
the
followed by a study of the behavior of the
current flow, and the importance of armature reaction
Direct-current
can convert the current of an ac system into direct
motors
is
generator under load. Mechanical torque, direction of
is due to an external driving force, such
motor (not shown). The coil is connected to
two slip rings mounted on the shaft. The slip rings
are connected to an external load by means of two
no-load.
rotation
and de-
as a
We show
generated and what de-
stationary brushes x and
magnitude.
71
y.
1
ELECTRICA L MA CHINES AND TRANSFORMERS
72
rotation (Fig. 4.2).
60 r/min
The waveshape depends upon
We
shape of the N, S poles.
designed
The
assume
the
were
the poles
generate the sinusoidal wave shown.
to
coil
in
our example revolves
uniform
at
speed, therefore each angle of rotation corresponds
to
makes
a specific interval of time. Because the coil
one
turn per second, the angle of
360°
4.2 cor-
in Fig.
responds to an interval of one second. Consequently,
we can
also represent the induced voltage as a func-
tion of time (Fig. 4.3).
v
1
+20
cycle -
N
\
F Al>
c
an elementary ac generator
of
1
revolution per second.
\
V
y_
\
\
\
/
\
/
\
f
\
/
\
turning at
\
/
\
1
\
/
/
\
Figure 4.1
Schematic diagram
\
/
\
\
/
\
\
/
\
s
1
As
the coil rotates, a voltage
A
is
induced (Eq. 2-25)
between
its
between
the brushes and, therefore, across the load.
terminals
The voltage
is
time
1.25
/
/
and D. This voltage appears
/
-
1
cycle
generated because the conductors of
the coil cut across the flux produced by the N, S
poles.
The induced voltage
(20 V, say)
when
the coil
is
zontal position, as shown.
coil
is
momentarily
quently the voltage
is
No
maximum
therefore
momentarily
flux
is
in the hori-
cut
when
in the vertical position;
at
these instants
feature of the voltage
is
that
its
is
zero.
Figure 4.3
Voltage induced as a function of time.
the
conse-
Another
polarity changes
4.2 Direct-current generator
If
the brushes in Fig. 4.1 could be switched from
every time the coil makes half a turn. The voltage can
one
therefore be represented as a function of the angle of
about to change,
slip ring to the other
every time the polarity was
we would
obtain a voltage of con-
stant polarity across the load.
V
Brush x would always
be positive and brush y negative. We can obtain this
result by using a commutator (Fig. 4.4), A commu-
+ 20
tator in
that
is
its
simplest form
composed of
is
cut in half, with each
a slip ring
segment insulated from
t-'.A I)
the other as well as
\
180
-
360 degrees 450
angle
connected
from
to coil-end
the shaft.
The commutator revolves with
age between the segments
tionary brushes x and
One segment
A and the other to coil-end
is
the coil and the volt-
picked up by two
The voltage between brushes
nating voltage in the coil
Figure 4.2
in
the ac generator as a function of
the angle of rotation.
sta-
y.
x
and y pulsates
but never changes polarity (Fig. 4.5).
Voltage induced
is
D.
is
rectified
The
alter-
by the com-
mutator, which acts as a mechanical reversing
switch.
DIRECT-CURRENT GENERATORS
Due
60 r/min
to the constant polarity
73
between the brushes,
Hows in
The machine represented in Fig.
the current in the external load always
the
same
4.4
is
direction.
called a direct-current generator, or
4.3 Difference
dynamo.
between ac
and dc generators
The elementary
ac and dc generators in Figs. 4.1
and 4.4 are essentially
and an ac voltage
chines only differ
same way.
built the
between the poles of
case, a coil rotates
is
induced
in the
way
We
commutator
Elementary dc generator
machines which carry both
is
simply an ac generator
rectifier called
a
commu-
(Fig. 4.6a).
tator (Fig. 4.6c).
generators carry
while dc generators require a
Figure 4.4
equipped with a mechanical
each
magnet
The ma-
the coil.
in
In
the coils are connected
to the external circuit (Fig. 4.6): ac
slip rings (Fig 4.6b)
a
sometimes build small
slip rings
and a
commu-
Such machines can function
si-
multaneously as ac and dc generators.
tator.
4.4 Improving the
Returning
+20
to the
waveshape
dc generator,
we can improve
the
pulsating dc voltage by using four coils and four
segments, as shown
shape
but
it
is
given
never
The resulting waveThe voltage still pulsates
in Fig. 4,7.
in Fig. 4.8.
falls to zero;
it
is
much
closer to a steady
dc voltage.
By increasing the number of coils and segments,
we can obtain a dc voltage that is very smooth.
0
0
90
180
270
360 degrees
Modern dc generators produce
The
angle 9
ripple of less than 5 percent.
Figure 4.5
voltages having a
coils are lodged in
the slots of a laminated iron cylinder.
The elementary dc generator produces a pulsating dc
the cylinder constitute the
voltage.
The
coils
The percent ripple is the ratio of the RMS value of
the ac component of voltage to the dc component,
expressed
(b)
(a)
in percent.
(
C
)
Figure 4.6
The three armatures
rings or
and
armature of the machine.
(a), (b),
and
(c)
have
identical windings.
a commutator), an ac or dc voltage
is
obtained.
Depending upon how they are connected
(to slip
ELECTRICAL MACHINES AND TRANSFORMERS
74
rotation
rotation
A
Figure 4.7
Schematic diagram of a dc generator having 4
and 4 commutator bars. See Fig. 4.9.
Figure 4.9
coils
The actual physical construction of the generator
shown in Fig. 4.7. The armature has 4 slots, 4 coils,
and 4 commutator bars.
v
A
schematic diagram such as Fig. 4.7
where the
tips,
and so on. But we must remember
coil sides (a h a 2
;
b,,
b2
ally located at 180° to
side as Figure 4.7
ol
0
I
I
90
180
I
I
|_
I
360
270
degrees
*0
Figure 4.8
The voltage between the brushes
than
in Fig.
that
4.5.
one
of Fig. 4.7, because
we
in
will
be using similar drawings
of dc machines. The four coils
the figure are identical to the coil
At the instant shown,
coil
A
is
shown
in Fig. 4.
1
not cutting any flux
The reason is that the coil sides
of these two coils are midway between the poles. On
the other hand, coils B and D are cutting flux coming
and neither
is
coil C.
from the center of the
N
and S poles. Consequently,
the voltage induced in these coils
possible value (20 V, say). That
is at its
is
maximum
also the voltage
across the brushes at this particular instant.
each other and not side by
seems
to indicate.
coil side is at the
is at
a, is in
important to understand the physical meaning
to explain the behavior
that the
of each coil are actu-
etc.)
The actual construction of this armature is
shown in Fig. 4.9. The four coils are placed in four
slots. Each coil has two coil sides, and so there are
two coil sides per slot. Thus, each slot contains the
conductors of two coils.
other
It is
;
For reasons of symmetry, the coils are wound so
more uniform
is
us
between the poles, under the poles, near the
cated:
pole
tells
coil sides of the individual coils are lo-
the top of slot
bottom of
bottom of a
slot 3.
The
1,
while coil side a 2
coil
the
is in
connections to the com-
mutator segments are easy to follow
armature.
and the
slot
the top. For example, in Fig. 4.7 coil side
The reader should compare
in this
simple
these connec-
tions with those in Fig. 4.9 to verify that they are the
same. Note also the actual position and schematic
position of the brushes with respect to the poles.
the position of the coils when
moved
the armature has
through 45°. The sides a,,
A
are now sweeping past pole tip
and
a 2 of coil
Fig. 4.10
shows
1
pole
same
tip 4.
The
sides of coil
C
are experiencing the
flux because they are in the
A. Consequently, the voltage
e.A
same
slots as coil
induced
in coil
A is
DIRECT-CURRENT GENERATORS
rotation
15
rotation
Figure 4.10
when
Position of the coils
rotated
the armature of Fig. 4.9 has
through 45°.
Figure 4.11a
Physical construction of a dc generator having 12
exactly the
same
as the voltage e c induced in coil C.
A
C
while coil
coils,
12
slots,
and 12 commutator
moving downward,
moving upward. The polarities of e a
however, that coil
Note,
is
bars.
is
rotation
and ec are, therefore, opposite as shown.
The same reasoning leads us
and? are equal and opposite
Ll
+
+
+
=
conclude that e h
to
in polarity.
This means
at all times.
Consequently,
no current will flow in the closed loop
formed by the
that e.d
eh
ec
This
four coils.
is
circulating current
ed
0
most fortunate, because any such
would produce
2
I
R
The voltage between the brushes
c c (or e a
+
c\\)
at the instant
minimum
to the
shown.
shown
voltage
losses.
is
equal to e b
It
in Fig. 4.8.
The armature winding we have just discussed
called a
lap winding.
the
It is
+
corresponds
most
common
is
type of
winding used in direct-current generators and motors.
4.5
Figure 4.11b
Schematic diagram
Induced voltage
induced
Figures 4.
1
and
la
4.
1
show
lb
a
more
in
When
the armature rotates, the voltage
E induced
each conductor depends upon the flux density
which
it
cuts.
This fact
is
E=
Because the density
point to point,
coil
depends
based upon the equation
Blv
in the air
(2.25)
gap varies from
the value of the induced voltage per
upon
its
of the
armature and the voltages
the 12 coils.
realistic ar-
mature having 12 coils and 12 slots instead of only
4.
in
instantaneous
position.
Consider, for example, the voltages induced
armature when
Fig. 4.11.
it
in
the
occupies the position shown
The conductors
in slots
in
and 7 are ex-
1
actly
between the poles, where the
flux density
is
zero.
The voltage induced
coils
lodged
in
slots
1
and 7
is,
in the
therefore, zero.
two
On
the other hand,
the conductors in slots 4 and 10 are directly under
the center of the poles,
greatest.
where the
The voltage induced
in
flux density
the
two
is
coils
ELECTRICAL MACHINES AND TRANSFORMERS
76
lodged
in
Finally,
due
duced
in
as that
slots
induced
in the coils
lodged
and
in slots 5
0V
same
the
is
rotation
in-
1
coil
A
/
1
shows the instantaneous voltage ineach of the 2 coils of the armature. They
18, and 20 V, respectively. Note that the
in
4.
1
lb
1
are 0, 7,
brushes short-circuit the coils
is
maximum.
therefore,
is,
magnetic symmetry, the voltage
to
the coils lodged in slots 3 and 9
Figure
duced
these
which the voltage
in
momentarily zero.
Taking polarities into account, we can see that
between the brushes
the voltage
18
+
to
brush
7)
=
70
and brush x
V,
is
+
(7
is
18
+ 20 +
positive with respect
y.
This voltage remains essentially con-
stant as the
armature rotates, because the number of
coils
between the brushes
always the same,
is
irre-
spective of armature position.
Note
4.11b straddles two
that brush x in Fig.
commutator segments
Consequently,
the
that are
brush
connected
However, since the induced voltage
momentarily zero, no current
brush.
sitioned on the
to
brush
If
That
we were
the case in Figs. 4.
is
(0
+
7
+
1
1
is
a
momen-
and
4.
1
1
b.
18
+ 20 +
18)
=
63
continually short-circuit coils that generate 7 V.
Large currents will flow
in the short-circuited coils
and brushes, and sparking
with coils that are momentarily
age between the brushes and
the
at
same time
sparking occurs, there
is
poor commutation.
a neutral zone.
4.7 Value of the induced voltage
The voltage induced
winding
is
in a
dc generator having a
lap
given by the equation
E0 =
E0 =
Z=
n =
(J> =
Z/?*/60
(4J)
voltage between the brushes [V]
total
number of conductors on
the armature
speed of rotation [r/min]
tlux per pole
|Wb]
This important equation shows that for a given generator the voltage
only holds true
is
zones
Neutral zones are those places on the surface of the
armature where the tlux density
is
zero.
When
the
generator operates at no-load, the neutral zones are
located exactly between the poles.
No
voltage
is
in-
is
directly proportional to the flux
per pole and to the speed of rotation.
tion. If the
4.6 Neutral
in
We
in contact
will result. Thus, shifting
the brushes off the neutral position reduces the volt-
When
through the neutral zone.
brushes so they are
where
V.
decreases. Furthermore, in this position, the brushes
said to be
try to set the
between the brushes would be-
Thus, by shifting the brushes the output voltage
causes sparking.
in a coil that cuts
always
brush yoke by 30° (Fig.
to shift the
4.12), the voltage
come
which
y,
The brushes are
neutral position when they are pocommutator so as to short-circuit
duced
coil B.
those coils in which the induced voltage
tarily zero.
A.
coil
in this coil is
flow through the
will
The same remarks apply
momentarily short-circuits
said to be in the
to coil A.
short-circuits
Figure 4.12
Moving the brushes off the neutral point reduces the
output voltage and produces sparking.
if
the brushes are
Example 4-1
The armature of a
slots.
Each
Wb.
equation
brushes are shifted off neutral, the
equivalent to reducing the
0.04
The
on the neutral
posieffect
number of conductors Z
_
6-pole,
coil has
600 r/min generator, has 90
4 turns and the tlux per pole
is
Calculate the value of the induced voltage.
DIRECT-CURRENT GENERATORS
The current delivered by
Solution
Each turn corresponds to two conductors on the armature, and
The
coils arc required to
fill
the
number of armature conductors
total
Z = 90
90
coils
X
4 turns/coil
X
90
slots.
flows through
2 conductors/turn
The speed
we
in the
same
discover
direction in
The same
is
N
mo-
true for conductors that are
mentarily under a S pole. However, the currents unis
n
= 600
der the
r/min
N
pole flow
the opposite direction to
in
those under a S pole. Referring to Fig.
Consequently,
En = Z//4V60 = 720 X 600 X
= 288 V
The voltage between the brushes
fore
always flows
we would
those conductors that are momentarily under a
pole.
= 720
generator also
the
the armature conductors. If
could look inside the machine,
that current
is
all
77
at
4. 13, the ar-
mature conductors under the S pole carry currents
0.04/60
that
flow into the page, away from the reader.
Conversely, the armature currents under the
no-load
is
there-
N
pole
flow out of the page, toward the reader.
Because the conductors
288 V, provided the brushes are on neutral.
lie in
a magnetic field,
they are subjected to a force, according to Lorentz's
4.8
Generator under load: the energy
conversion process
When
a direct-current generator is
under load, some
fundamental flux and current relationships take place
that are directly related to
the mechanical-electrical
energy conversion process. Consider for example, a
2-pole generator
that
is
while delivering current
/
driven counterclockwise
to a load (Fig. 4.
1
3).
law (sections 2.22 and 2.23).
find that the individual forces
all
we examine
F on
the di-
we
act clockwise. In effect, they
the conductors
produce a torque
which the gen-
that acts opposite to the direction in
erator
is
we must
being driven. To keep the generator going,
exert a torque on the shaft to
overcome
this
opposing electromagnetic torque. The resulting mechanical
which
rotation
If
rection of current flow and the direction of flux,
how
is
power
is
converted into electrical power,
delivered to the generator load. That
is
the energy conversion process takes place.
torque due to F
4.9
Armature reaction
Until now,
we have assumed
motive force (mmf) acting
to the field.
in a
that the only
dc generator
However, the current flowing
ture coils also creates a powerful
that distorts
we
we
and
field
weakening takes place
reaction.
the impact of the armature
mmf.
return to the generator under load (Fig. 4.13). If
consider the armature alone,
magnetic
chanical torque.
magnetomotive force
mmf is called armature
To understand
Figure 4.13
The energy conversion process. The electromagnetic
torque due to Fmust be balanced by the applied me-
due
arma-
both motors and generators. The effect produced by
the armature
load
that
in the
and weakens the flux coming from the
poles. This distortion
in
magnetois
at right
field as
shown
it
in Fig. 4.14.
will
produce a
This
field acts
angles to the field produced by the N, S poles.
The intensity of the armature flux depends upon its
mmf, which in turn depends upon the current carried
by the armature. Thus, contrary
armature flux
is
to the field flux, the
not constant but varies with the load.
ELECTRICAL MACHINES AND TRANSFORMERS
78
armature flux
rotation
neutra zone
,
neutral zone
Figure 4.14
Magnetic field produced by the current flowing
armature conductors.
We can
in
the
immediately foresee a problem which the
armature flux will produce. Fig.
zone
flux in the neutral
is
shows
4. 14
that the
Figure 4.15
Armature reaction
S poles.
may
sparking
by the brushes. As
will
The
occur.
depend upon
a result, severe
intensity of the sparking
the armature flux
the load current delivered
that
is
it
distorts the flux
mmf
and
field
whose shape
is
mmf
in the direction
The
all
The
neutral
still
another effect:
the higher flux density in pole tips 2. 3 causes saturation to set
in.
Consequently, the increase
in flux
decrease
in flux
under pole
tips 2, 3 is less than the
under pole
tips
1,
4.
As
a result, the total flux pro-
duced by the N, S poles
less than
may be
1
as
.
was when
induced voltage given
as 10 percent.
in flux
under load,
we
could
zone when the gener-
move
the brushes to re-
For generators, the brushes are shifted
to the
new
zone by moving them
tation.
For motors, the brushes are shifted against
in
the direction of ro-
the direction of rotation.
As soon
However,
We
as the brushes are
improves,
tion
forth
it
For large machines, the decrease
much
ro-
neutral
no-load. This causes a cor-
at
in the
does not
duce the sparking.
rises
responding reduction
4.
to the shift in the neutral
is
the
is
generator was running
by Eq.
Due
ator
of rotation of the
dc generators.
flux distortion produces
it
improve commutation
to
produced by the
illustrated in Fig. 4.15.
zones have shifted
in space;
4.10 Shifting the brushes
produces a magnetic field
armature. This occurs in
important to note that the orientation of the
with the armature.
tate
the armature
combination of the armature
poles. In effect, the
It is
armature flux remains fixed
and hence upon
by the generator.
The second problem created by
mmf
N,
no longer zero and, con-
sequently, a voltage will be induced in the coils that
are short-circuited
produced by the
distorts the field
and
if
meaning
moved, the commuta-
there
is
less
sparking.
the load fluctuates, the armature
falls
and so the neutral zone
shifts
mmf
back and
between the no-load and full-load positions.
would therefore have to move the brushes back
and forth
to
cedure
not practical and other
is
obtain sparkless commutation. This pro-
means
are used to
DIRECT- CURRENT GENERA TORS
resolve the problem.
ever, the
to
For small dc machines, how-
brushes are set
ensure reasonably
in
an intermediate position
good commutation
at all loads.
mmf of the
the
commutating poles
is
greater than the armature
mmf. This
flux in the neutral zone,
which aids
made
19
slightly
creates a small
the
commuta-
tion process (see Section 4.28).
Commutating poles
4.11
Fig. 4. 16
shows how
the
commutating poles of
a 2-pole machine are connected. Clearly, the direc-
To counter the
of armature
effect
reaction
in
medium- and large-power dc machines, we always
place a set of
commutating poles* between the main
These narrow poles carry wind-
poles (Fig. 4. 16).
ings that are
connected
with the armature.
in series
The number of turns on the windings
that the
equal
and opposite
mmfa
of the armature.
the
designed so
As
exactly
bucking each other
ing the
armature
mmf
flowing through the windings
in-
opposite to the
mmf of the commutating poles acts
mmf of the armature and, therefore,
neutralizes
effect.
is
its
restricted to the
However,
the neutralization
narrow brush zone where com-
mutation takes place. The distorted flux distribution
under the main poles, unfortunately, remains
the same.
the load current varies,
rise
at all
in this
between the main poles
we no longer have
mmfc
magnetomotive force
the
to
two magnetomotive forces
space
is
poles develop a magnetomotive force
tion of the current
dicates that the
and
fall
times.
together,
By
way, the flux
is
4.12 Separately excited generator
nullifyin the
always zero and so
to shift the brushes. In practice,
Now
that
we have learned some basic facts about
we can study the various types and
dc generators,
their properties.
magnets
Thus, instead of using permanent
to create the
magnetic
field,
we can use a
shown
pair of electromagnets, called field poles, as
O (+)
in Fig. 4. 17.
When
generator
supplied by an independent source
is
the dc field current in such a
(such as a storage battery or another generator,
called an exciter), the generator
is
said to be sepa-
rately excited. Thus, in Fig. 4. 7 the dc source con1
nected to terminals a and b causes an exciting current / x to flow. If the armature
is
or a diesel engine, a voltage
£0
brush terminals x and
driven by a motor
appears between
y.
O (-)
Figure 4.16
Commutating poles produce an
the
mmf,
of the
mmf c
that
opposes
armature.
Commutating poles are sometimes called mierpoles.
Figure 4.17
Separately excited 2-pole generator. The N, S
field
poles are created by the current flowing
field
windings.
in
the
1
ELECTRICAL MACHINES AND TRANSFORMERS
80
How does the saturation curve relate to the induced
£n ? If we drive the generator at constant speed,
4.13 No-load operation
voltage
and saturation curve
E0 is directly proportional to the flux
When
a separately excited dc generator runs at no-
load (armature circuit open), a change
the excit-
in
ing current causes a corresponding change in the in-
We now examine
duced voltage.
whose shape
4.
8a.
1
which increases the
If
we
f|>
plot
as a function of I X1
per pole.
is
1
is
mmf of the
obtain the satu-
ration curve of Fig. 4. 8a. This curve
whether or not the venerator
obtained
turning.
identical to the saturation curve of Fig.
result
is
shown
in Fig. 4.
1
8b;
rated voltage of a dc generator
above the knee of the curve.
usually a
is
In Fig. 4.
example, the rated (or nominal) voltage
varying the exciting current,
called the
it is
we can
is
1
1
8b. for
20
V.
By
vary the
in-
duced voltage as we please. Furthermore, by
re-
versing the current, the flux will reverse and so, too,
will the polarity of the induced voltage.
Induced voltage
vs speed.
For a given exciting
current, the induced voltage increases in direct pro-
rated flux
portion to the speed, a result that follows from Eq. 4.
A
.
obtain a curve
little
flux
we
Consequently,
we
Let us gradually
so that the
,
is
,
no-load saturation curve of the generator.
The
raise the exciting current / x
field increases,
The
as a function of 7 X
the relationship
between the two.
Field flux vs exciting current.
Ea
by plotting
If
we
of the induced voltage also reverses. However,
ity
we
1
reverse the direction of rotation, the polar-
reverse both the exciting current
and
if
the direc-
tion of rotation, the polarity of the induced voltage
remains the same.
Figure 4.18a
Flux per pole versus exciting current.
When
the flux
the exciting current
relatively small,
is
small and the iron in the machine
is
saturated. Very
little
mmf
is
needed
un-
is
to establish
the flux through the iron, with the result that the
mmf developed
by the
field coils
is
available to drive the flux through the air gap.
Because the permeability of
air
3
A
almost entirely
is
constant, the
Figure 4.18b
Saturation curve of a dc generator.
flux increases in direct proportion to the exciting
current, as
shown by
the linear portion
0a of the
4.14 Shunt generator
saturation curve.
However, as we continue
current, the iron in the field
to saturate.
A
to raise the exciting
and the armature begins
large increase in the
quired to produce a small increase
mmf is now
in flux, as
by portion be of the curve. The machine
to
is
re-
shown
now
said
be saturated. Saturation of the iron begins to be
important
when we
reach the so-called "knee" ab of
the saturation curve.
A
shunt-excited generator
shunt-field winding
is
is
connected
a
machine whose
in parallel
with the
armature terminals, so that the generator can be
self-excited (Fig. 4. 19).
this
connection
is
that
it
The
principal advantage of
eliminates the need for an
external source of excitation.
How
is
generator
self-excitation achieved?
is
When
started up, a small voltage
is
a shunt
induced
in
i
DIRECT-CURRENT GENERATORS
81
field rheostat
y
Figure 4.20
Controlling the generator voltage with a field rheostat.
rheostat
is
a resistor with an adjustable
To understand how
suppose that
p
is in
tact
Ea
is
1
the output voltage varies,
V when
20
movable contact
the
we move the conresistance R between
the center of the rheostat. If
toward extremity m, the
points
A
sliding contact.
{
p and b diminishes, which causes
the excit-
ing current to increase. This increases the flux and,
£tv On
consequently, the induced voltage
hand,
if
(b)
we move
the other
R
the contact toward extremity n,
l
increases, the exciting current diminishes, the flux
Figure 4.19
b.
diminishes, and so
Self-excited shunt generator.
a.
Schematic diagram
field is
of
one designed
a shunt generator.
to
be connected
in
A
We
shunt
shunt
(alter-
know
points
the
armature, due to the remanent flux in the poles.
produces a small exciting current
This voltage
the
shunt field.
The
resulting small
mmf acts
/ x in
in the
same direction as the remanent flux, causing the flux
per pole to increase.
which increases
/x ,
The increased
flux increases
which increases
more, which increases
E0
the flux
even more, and so
This progressive buildup continues until
a
maximum
E0
forth.
reaches
value determined by the field resistance
and the degree of saturation. See next section.
p and
R
E0
if
we
of the shunt field circuit between
{
We draw
b.
ing to the slope of
a straight line correspond-
R and superimpose
on the
it
{
uration curve (Fig. 4.2
sat-
This dotted line passes
1).
through the origin, and the point where
it
intersects
the curve yields the induced voltage.
Ea
still
will fall.
the saturation curve of the generator and the
total resistance
nate term for parallel) with the armature winding.
£0
can determine the no-load value of
For example,
50
12
if
the shunt field has a resistance of
and the rheostat
R = 50
12.
{
The
line
is
m,
then
R must
E = 50 V, / =
pass
set at
extremity
corresponding
to
{
through the coordinate point
I
A.
This line intersects the saturation curve where the
voltage
is
150
V
(Fig. 4.21).
That
is
the
maximum
voltage the shunt generator can produce.
By changing
4.15 Controlling the voltage
of a
shunt generator
to
It
is
easy to control the induced voltage of a shunt-
excited generator.
rent
We
simply vary the exciting cur-
by means of a rheostat connected in series with
the shunt field (Fig. 4.20).
the setting of the rheostat, the total
resistance of the field circuit increases, causing
decrease progressively. For example,
creased to 120
12,
A\
is
E0
in-
the resistance line cuts the satu-
ration curve at a voltage
we continue
if
Ea
of 120 V.
a critical value will
be
reached where the slope of the resistance line
is
If
to raise
A*,,
ELECTRICAL MACHINES AND TRANSFORMERS
82
Figure 4.22
Equivalent circuit of a dc generator.
revolving conductors. Terminals
l ,
2 are the external
armature terminals of the machine, and F,, F 2 are the
field
now
winding terminals. Using
study the more
common
we
this circuit,
will
types of direct-current
generators and their behavior under load.
4.17 Separately excited generator
under load
Figure 4.21
The no-load voltage depends upon the resistance
the shunt-field
of
Let us consider a separately excited generator that
circuit.
driven
equal to that of the saturation curve
region.
When
unsaturated
in its
this resistance is attained, the
induced
voltage suddenly drops to zero and will remain so
for any
R
v
greater than this critical value. In Fig. 4.2
the critical resistance corresponds to
200
at
a battery (Fig. 4.23).
and so
is
The
When
load, terminal voltage
E0
resistance
whose field
the
E {2
zero.
is
constant
machine operates
is
However,
is
excited by
The induced voltage £0
is
at no-
equal to the induced
because the voltage drop
is
is
exciting current
the resultant flux.
therefore fixed.
voltage
il.
constant speed and
if
in the
armature
we connect
a load
4.16 Equivalent circuit
across the armature (Fig. 4.23), the resulting load
We
R Kr Terminal voltage E l2 is now less than the induced
voltage £
As we increase the load, the terminal
current / produces a voltage drop across resistance
set
have seen
that the
of identical coils,
all
armature winding contains a
of which possess a certain re-
()
sistance.
which
The
exists
machine
the
armature resistance R^
total
between the armature terminals when
is
stationary.
It is
measured on the com-
mutator surface between those segments that
der the
ally
(
+
)
that
is
and
(
—
)
The
brushes.
resistance
lie
is
.
voltage diminishes progressively, as
4.24.
shown
as a function of
cun e
of the generator.
load current
is
called the load
y
un-
usu-
very small, often less than one-hundredth of an
ohm.
Its
value depends mainly upon the
voltage of the generator.
circuit,
we can
represent
one of the brushes.
If
To simplify
machine has
resistance of these windings
The equivalent
circuit
posed of a resistance R„
(Fig. 4.22).
The
is
in series
latter is the
with
interpoles, the
included
of a generator
in series
power and
the generator
R 0 as if it were
the
in
R ir
is
thus
com-
with a voltage
voltage induced
Ea
in the
in Fig.
The graph of terminal voltage
Figure 4.23
Separately excited generator under load.
DIRECT-CURRENT GENERATORS
V
tions
100
produce corresponding changes
in
83
the genera-
tor terminal voltage, causing the lights to flicker.
Compound generators eliminate this problem.
A compound generator (Fig. 4.25a) is similar
to a shunt generator,
field coils
connected
except that
in series
has additional
it
with the armature.
These series field coils are composed of a few turns
of heavy wire, big enough to carry the armature cur0
10
5
rent.
A
The
4.25b
showing the shunt and
Figure 4.24
of the series coils
total resistance
fore, small. Figure
When
In practice,
ration
the induced voltage
Ev also decreases
with increasing load, because pole-tip satu-
slightly
tends to decrease the field flux. Consequently,
voltage
the terminal
E l2
falls off
more
rapidly than
can be attributed to armature resistance alone.
the series coils
series field connections.
As
the generator
off more sharply with increasing load than that of
a separately excited generator.
field
The reason
current in a separately excited
constant,
whereas
citing current falls
in a self-excited
load to full-load is
drop
coils acts in the
is
said to
15% and 10%,
Compound
its
raises the value of
original
Ea
.
If
practically constant
from no-load
it
respectively.
generator
prevent
voltage of a dc generator from de-
we
can usually tolerate a reasonable drop in terminal
voltage as the load increases, this has a serious effect
on lighting circuits. For example, the distribu-
tion
system of a ship supplies power to both dc ma-
chinery
and
incandescent
lamps.
The
current
mmf
no-load value,
properly designed, the terminal voltage remains
regulation
creasing with increasing load. Thus, although
flows
developed
the series coils are
voltage from no-
to
now
direction as the
machine remains
The voltage
however,
Consequently, the field flux un-
generator the ex-
The compound generator was developed
the terminal
same
/c
mmf
The
that the
whereas for a separately excited generator
usually less than 10 percent.
be
field.
coils.
about 15 percent of the full-load
is
4.19
in
coils,
loaded, the terminal volt-
as the terminal voltage drops. For a
self-excited generator, the
voltage,
is
is
through the series field
der load rises above
The terminal voltage of a self-excited shunt generator
The shunt
zero.
age tends to drop, but load current
which
falls
is
flux, just as in a standard self-excited shunt generator.
by these
Shunt generator under load
there-
diagram
carry exciting current / x which produces the field
of the shunt
4.18
is,
the generator runs at no-load, the current
Load characteristic of a separately excited generator.
in
a schematic
is
Figure 4.25
delivered by the generator fluctuates continually, in
a.
Compound
response to the varying loads. These current varia-
b.
Schematic diagram.
generator under load.
to full-load.
The
ELECTRICAL MACHINES AND TRANSFORMERS
84
rise in the
induced voltage compensates for the
ar-
Load characteristics
4.21
mature IR drop.
In
the
some cases we have
to
compensate not only for
armature voltage drop, but also for the IR drop
in
between the generator and the load.
the feeder line
The generator manufacturer then adds one or two extra turns
on
winding so
the series
that the terminal
ma-
voltage increases as the load current rises. Such
chines are called over-compound generators.
compounding
placed
is
too strong, a low resistance can be
with the series
in parallel
the current in the series field
as reducing the
If the
field.
This reduces
The load characteristics of some shunt and compound generators are given in Fig. 4.26. The voltage of an over-compound generator increases by
1
value of the diverter resistance
is
if
the
equal to that of the
series field, the current in the latter
is
reduced by
half.
compound
flat-compound generator remains constant.
tor
15 percent
is
compound generator the
mmf of the
As
a re-
terminal voltage falls drastically with in-
We
can make such a generator by
simply reversing the series field of a standard com-
pound
generator. Differential
were formerly used
in
below
its
the
is
30 percent
lower.
4.22 Generator specifications
The nameplate of
a generator indicates the power,
voltage, speed, and other details about the machine.
ratings, or
nominal characteristics, are the
following information
compound
nameplate of a 100
Power
100
Voltage
250
20
generators
kW
V
A
50° C
Exciting current
Temperature
rise
These specifications
tell
dc arc welders, because they
tended to limit the short-circuit current and to stabi-
punched on
is
the
kW generator:
Speed
1200r/min
Type
Compound
B
Class
us that the machine can
power of 100
deliver, continuously, a
lize the arc
On
no-load value, while that
of a differential-compound generator
ple, the
series field acts opposite to the shunt field.
creasing load.
applied, whereas that of a
values guaranteed by the manufacturer. For exam-
generator
sult, the
is
other hand, the full-load voltage of a shunt genera-
These
In a differential
full-load
and has the same effect
number of turns. For example,
4.20 Differential
when
percent
kW at a volt-
age of 250 V, without exceeding a temperature
rise
during the welding process.
The voltage regulation of the differential compound generator in Fig. 4.26 is (no-load — full00-70)770 = 42.9%.
load )/full- load =
(
of 50°C.
can therefore supply a load current of
It
100 000/250
= 400 A.
and the current
in the
It
possesses a series winding,
shunt field
is
20 A.
In practice,
1
the terminal voltage
its
%
100
rating of
overcom pound
power from
compound
ceed
1
separate excitation
class
B
used
in the
80
shunt
60
differential
00
250
V.
is
adjusted to a value close to
We may draw
any amount of
the generator, as long as
kW and the current
is
it
does not ex-
400 A. The
less than
designation refers to the class of insulation
machine.
compound
40
CONSTRUCTION OF
20
DIRECT-CURRENT GENERATORS
We have
0-
50
100
described the basic features and properties
%
of direct-current generators.
We now
look
at
the
Load current
mechanical construction of these machines,
direct-
Figure 4.26
ing our attention to the field, the armature, the
Typical load characteristics of dc generators.
mutator, and the brushes.
com-
DIRECT-CURRENT GENERATORS
85
4.23 Field
The
field
chine.
It
produces the magnetic flux
is
composed of
side
in the
ma-
basically a stationary electromagnet
a set
of salient poles bolted to the
in-
of a circular frame (Figs. 4.27 and 4.28). Field
mounted on the poles, carry the dc exciting
coils,
current.
The frame
is
usually
stacked iron laminations. In
solid cast
our discussions so far
2-pole generators.
ator or
poles.
some generators
the
created by permanent magnets.
flux is
In
made of
whereas the pole pieces are composed of
steel,
we have considered only
However,
motor may have
in practice a
2, 4, 6,
or as
many
The number of poles depends upon
flux
dc generas
24
Figure 4.29
Adjacent poles
of multipole
generators have opposite
magnetic
polarities.
ical size
of the machine; the bigger
the phys-
poles
frame
will have.
it
By using
is,
it
the
more
a multipole design,
we
can reduce the dimensions and cost of large mafield
chines, and also
The
improve
field coils
their
performance.
of a multipole machine are con-
nected together so that adjacent poles have oppo-
commutator
site
magnetic polarities (Fig. 4.29). The shunt
composed of
coils are
field
several hundred turns of
wire carrying a relatively small current. The coils
armature
are insulated
from the pole pieces
to
prevent short-
circuits.
Figure 4.27
mmf
The
Cross section of a 2-pole generator.
developed by the coils produces a
magnetic flux that passes through the pole pieces,
the frame, the armature,
is
the short space
pieces.
It
and the
air gap.
The
air
gap
between the armature and the pole
ranges from about
1
erator rating increases from
1
Because the armature and
.5 to
kW
mm as the gen-
5
to
100 kW.
field are
composed of
magnetic materials having excellent permeability,
most of the
mmf
produced by the
field
is
used to
drive the flux across the air gap. Consequently, by
reducing
its
shunt field
made
fect
length,
coils.
we can
becomes too
must be
enough so
Figure 4.28
Cutaway view
does not overheat when
It
has 3
field, the coils are
top of the shunt-field coils.
tor size
a 4-pole shunt generator.
gap cannot be
great.
generator has a series
wound on
of
air
too short otherwise the armature reaction ef-
If the
brushes per brush set.
diminish the size of the
However, the
large
rent of the generator.
it
The conduc-
that the
winding
carries the full-load cur-
ELECTRICAL MACHINES AND TRANSFORMERS
86
4.24 Armature
The armature
the rotating part of a dc generator.
is
It
consists of a commutator, an iron core, and a set of
The armature
coils (Fig. 4.30).
keyed
is
revolves between the field poles.
composed of slotted,
to
form a
to a shaft
The
and
iron core
is
iron laminations that are stacked
solid cylindrical core.
The laminations
are
individually coated with an insulating film so that
they do not
As
come
in electrical
contact with each other.
The
a result, eddy-current losses are reduced.
are lined
up
to provide the space
needed
slots
to insert the
armature conductors.
The armature conductors carry the load current
They are insulated from
the iron core by several layers of paper or mica and
delivered by the generator.
are firmly held in place by fiber slot sticks. If the ar-
mature current
is
below 10 A, round wire
is
Figure 4.31
Armature lamination with tapered
but for currents exceeding 20 A, rectangular con-
ductors are preferred because they
make
slots.
used;
iron teeth
fiber slot stick
better use
of the available slot space. The lamination of a
small armature
tion
view of the
is
shown
slot
in Fig.
4.31.
A
of a large armature
cross sec-
is
shown
in
Fig. 4.32.
Figure 4.32
Cross-section of a slot containing 4 conductors.
4.25
Commutator and brushes
The commutator
is
composed of an assembly of
ta-
pered copper segments insulated from each other by
mica
sheets,
and mounted on the shaft of the ma-
chine (Fig. 4.33). The armature conductors are con-
nected to the commutator
in
a
manner we
will ex-
plain in Section 4.26.
Great care
Figure 4.30
Armature of a dc generator showing the commutator,
stacked laminations, slots, and shaft.
(Courtesy of General Electric Company, USA)
is
taken
in
building the commutator
because any eccentricity will cause the brushes
bounce,
producing
unacceptable
sparking.
to
The
sparks burn the brushes and overheat and carbonize
the commutator.
DIRECT-CURRENT GENERATORS
yoke
that permits the entire brush
tated through an angle
position. In going
87
assembly to be
and then locked
ro-
in the neutral
around the commutator, the suc-
cessive brush sets have positive and negative polarities.
Brushes having the same polarity are connected
together and the leads are brought out to one positive
and one negative terminal
(Fig. 4.34b).
The brushes are made of carbon because it has
good electrical conductivity and its softness does
not score the commutator. To improve the conductivity, a small amount of copper is sometimes mixed
with the carbon. The brush pressure is set by means
of adjustable springs.
friction
tator
A
of
is
too great, the
and brushes; on the other hand,
if
commuit
is
too
a dc machine.
2-pole generator has two brushes fixed dia-
metrically opposite to
slide
pressure
weak, the imperfect contact may produce sparking.
Figure 4.33
Commutator
If the
produces excessive heating of the
each other (Fig. 4.34a). They
on the commutator and ensure good electrical
contact
between the revolving armature and the
sta-
tionary external load.
Multipole machines possess as
they
have poles. The brush
of one or
that
many brush
sets, in turn, are
more brushes, depending upon
sets as
composed
the current
has to be earned. In Fig. 4.35c, for example,
brushes
mounted side-by-side make up the brush
The brush sets are spaced
at
two
set.
equal intervals around the
commutator. They are supported by a movable brush
+
(a)
9
6
(b)
Figure 4.34
a.
Brushes of a 2-pole generator.
b.
Brushes and connections of a 6-pole generator.
Figure 4.35
Carbon brush and ultraflexible copper lead.
b. Brush holder and spring to exert pressure.
c. Brush set composed of two brushes, mounted on
a.
rocker arm.
(Courtesy of General Electric Company, USA)
88
ELECTRICAL MACHINES AND TRANSFORMERS
The pressure
is
usually about 15
kPa
2 lb/in"),
and the permissible current density
mately 10
A/cm
2
65 A/in
is
In order to get a better
2
).
Thus, atypical brush
cm X cm
of 4.5 N
having a cross section of 3
1.2 in
1
X
lb)
and can
the construction of a
modern
0.4 in) exerts a pressure
1
carry a current of about 30 A.
Fig. 4.36
shows
4-pole dc generator.
In
4.26 Details of a multipole generator
approxigenerators,
was
1
2-
the schematic diagram
is
of such a machine having 72 slots on the armature,
72 segments on the commutator, and 72
coils.
The
armature has a lap winding, and the reader should
note
erator that
understanding of multipole
us examine the construction of a
pole machine. Fig. 4.38a
order to appreciate the
progress that has been made. Fig. 4.37 shows a gen-
let
how
similar
it is
2-pole machine (Fig.
built in 1889.
Figure 4.36
Sectional view of a 100 kW, 250 V, 1750 r/min 4-pole dc generator.
(Courtesy of General Electric Company, USA)
to the
4.
1
1
schematic diagram of a
b).
Coils
A and C
are
mo-
DIRECT-CURRENT GENERATORS
89
The voltage generated between brushes x and y
sum of the
equal to the
connected
coils
commutator segments
to
brush sets are similarly generated by five
(
form the
+
+
-2, 2-3,
)
(
coils.
brush sets are connected together
The
terminal.
)
(
—
connected to form the
larly
I
and 5-6. The voltages between the other
3-4, 4-5,
The
is
voltages generated by the five
to
brush sets are simi-
)
—
(
terminal. These
)
connections are not shown on the diagram. For similar
reasons of clarity,
that are placed
Fig.
diagram. Coil
is
B
stalled in
1889
250 A
at
a voltage of
1 1
0
It
deliv-
Other prop-
V.
pioneering machine include the following:
Speed
Total
first in-
to light the streets of Montreal.
ered a current of
erties of this
Thompson generator was
1
300 r/min
2390 kg
weight
mm
330 mm
292
Armature diameter
Stator internal diameter
Number of commutator
Only
A has its coil sides in slots
B
are in slots
is
connected
the three
4 and
10.
I
and
7,
while
Furthermore, coil
is
poles.
segments 3 and
B
The voltage
I,
4.
A
are
between the poles. Consequently,
induced
the coil sides of
to
shown, the coil-sides of coil
the neutral zone
no voltage
Figure 4.37
y.
C are shown so as not to complicate the
In the position
This direct-current
the interpoles
connected to commutator segments 72 and
while coil
in
show
between brushes x and
A, B, and
those of coil
A
not
4.38b gives a detailed view of the armature
coils lying
coils
we do
between the N, S poles.
A.
in coil
On
the other hand,
are directly under the
in coil
B
is
maximum
N
and S
at this
mo-
ment. Consequently, the voltage between adjacent
commutator segments 3 and 4 is maximum.
The voltage in coil C is also zero because its coil
sides are sweeping across the neutral zone. Note
that the positive
circuit coils
and negative brushes each short-
having zero induced voltage.
76
bars
#4
Armature conductor size
#
Shunt field conductor size
14
Example
4-2
The generator
in Fig.
4.38 generates 240
V
between
adjacent brushes and delivers a current of 2400
A modern generator having the same power and
speed weighs 7 times less and occupies only 1/3 the
floor
Calculate
space.
a.
b.
mentarily in the neutral zone, while coil
coming from
the flux
The
coil
width (known as coil pitch)
the coil sides
tween poles
l.
B
and the center of pole
of coil
is
cutting
A
is
such that
coming from adjacent N,
Thus, the coil sides of coil
center of pole 2
B
c.
The current delivered per brush set
The current flowing in each coil
The average voltage induced per coil
the center of the poles.
the coil sides cut the flux
S poles.
A
to the load.
lie
under the
3.
Similarly,
Solution
a.
A current of 2400 A flows out of the +
(
and back
into the
(
There are 12 brush
The
—
)
sets,
2. 3.
/
terminal
6 positive and 6 negative.
current per brush set
is
are in the neutral zones be-
2 and poles
)
terminal of the generator.
2400/6
= 400 A
90
ELECTRICAL MACHINES AND TRANSFORMERS
Figure 4.38b
Closeup view of the armature
coils
between adjacent brushes.
DIRECT-CURRENT GENERA TORS
b.
Each positive brush
coils to the right
/
c.
-
400/2
each
in
200
coil
is
A
71
There are six coils between adjacent brush
The average voltage per
Eavge
The
4.27
coil
240/6
a generator
|
{
The
commutation
72
I
currents flowing
the
If
80 A, the
shown
carry
all
40
versal takes place
other.
left. If
In Fig.
1
,
the current
is
how commutation
4.39a the brush
and the 40
short distance,
now
is
A from
brush unite
is
The
to
U
re-
|
(c)
80
di-
we
right
A
and
output.
1
40
40
contact with segment
2,
1
.
Owing
the total current,
72
|
1
|
only one-fourth of
namely 0.25 X 80
V
|
3
|
4040
7
(e)
80
between the
only one-fourth of the total contact area, and so
is
I
to the
commutator is proportional to the conThe area in contact with segment 2 is
from segment 2
40
©| |©
71
while 75 per-
contact resistance, the conductivity
40
40
V.
and 25 percent of the brush surface
contact with segment
the current
3
|
middle of seg-
on the
give the 80
in
tact area.
©
,2
|
to the
changes
takes place,
in the
the coils
in
brush and
N
40 40
4.39b the commutator has moved a
Fig.
1
called commutation.
brush produces a voltage drop of about
is
|
j
The contact resistance between the segment and
cent
72
4.39a to 4.39e.
the left of the
In
40
40
during the millisecond interval that
rection in this brief interval
ment
I
means
move from one end of the brush
To understand
|
(b)
©
71
on the right-hand side of the
The process whereby
refer to Figs.
3
I
80
0
40 A.
current in these coils must reverse.
a coil takes to
©
2
1/
T
moving from
are
40
40
in Fig.
brush will soon be on the left-hand side. This
that the
|
flow toward
and the
the right
coils
commutator segments
right to left, the coils
3
|
6020
the armature
in
that the currents in the coils
the load current is
2
I
©
V
|
under load, the individual coils
coming both from
the brush,
/
20
40
40
Note
1
= 40 V
windings next to a positive brush are
4.39a.
®
(a)
on the armature carry one-half the load current carried
by one brush.
©
40
sets.
71
is
40
is
process
When
72
|
40
ideal
40
40
40
to the left of the brush.
Consequently, the current
=
from the
set gathers current
and
9
=
20 A. By
Figure 4.39
Commutation
of the current
in coil 1.
are neglected and current reversal
brush contact resistance.
is
Inductive effects
caused by the
ELECTRICA L MA CHINES A ND TRA NS FORMERS
92
same token,
the
brush
is
we now
If
the current
X 80 - 60
0.75
from segment
cover that the current flowing
and 2
now
are the
ities
means
dis-
must be 20 A.
1
A to
dropped from 40
and the brush area
is
in coil
X
1/10
a
little
in
L =
the currents are equal. This
is
zero
Segment 2
left.
is
now
still
far-
contact with 75
in
A from
from segment 2 and 20
If coil
I
again 20 A, but
is
the opposite direction to
now understand how
what
it
segments
In Fig.
plete
slide
e
and the current
in coil 2 is
1
is
com-
about to be reversed.
it
is
important
(amperes per square
centimeter) remains the same
at
the brush face. Thus, the heat
produced by the con-
resistance
is
every point across
spread uniformally across the
brush surface. Unfortunately, such ideal commutanot possible in practical machines, and
tion
is
now
investigate the reason why.
induced voltage [V]
change of current [A/s]
rate of
LA//A/
=
It
we
The
practical
X
6
10
X
5.75
+ 40 -
[
X
V
the presence of this induced voltage (attribut-
is
that
flow
in coil
1
coil is considered.
when
We
ues for these currents
1,
and the currents
creasing or decreasing.
tance does not
come
that
that
it
takes place
very short time; consequently, the current can-
The reason
is
the
armature coils have inductance and
it
it
should.
strongly opposes a rapid change in current.
in
Fig.
4.39 has 72 bars and that the armature turns
at
600
r/min.
in 1/10
One
revolution
that the
is,
therefore,
The
currents
in Fig. 4.39.
the middle of seg-
is in
in the coils are neither in-
As
a result, the coil induc-
into play.
4.40b the current
in coil
1
is
changing due
However, the
in-
value of 20 A. Suppose the coil current
35 A. From Kirchhoff s current law, the currents
flowing from segments
then respectively 75
20 A. Note
completed
of a second and during this short period 72
form over the brush
touches segment
into the brush are
5 A, instead of
face.
2,
60
A and
no longer uniis
low where
and high where
it
1
In Fig. 4.40c the brush
1
is
The density
segment
cally placed as regards
rent in coil
and 2
1
A and
that the current density
the brush touches
commutator
Suppose, for example,
val-
order to determine the re-
4.40a the brush
In Fig.
ment
currents
have assumed plausible
in
should be compared with those
is
not reverse as quickly as
new
the self-inductance of the
sulting current flows in the brush.
to its ideal
is
40)]
duced voltage e prevents the current from dropping
process
in a
-
(
3
10
able to L), that opposes the change in current.
In Fig.
commutation
The problem with commutation
in-
is
to the contact resistance effect.
4.28
given by
(4.2)
Figs. 4.40a to 4.40e illustrate the
in coil
commutation process,
is
LM&t
1.39
over the brush.
to note that the current density
tact
=
100
We
the brush contact resis-
4.39e the current reversal
In this ideal
self-induction
=
seg-
tance forces a progressive reversal of the current as
the
only
inductance of the coil [H|
Hows
did before!
it
is
has an inductance of, say, 100 fxH. the
1
Applying Kirchhoff" s current law, we find
1.
that the current in coil
can
=
duced voltage
percent of the brush, and so the currents divide ac-
A
1
at this instant.
moved
4.39d the commutator has
cordingly: 60
in
=
e
1
in coil
which
1
the same. Consequently, the conductiv-
same and so
past the brush. Thus, the
or 1.39 ms!
s
The voltage induced by
A//A/
In Fig.
1/720
e
contact with segments
in
that the current in coil
ther to the
ment
-
1/72
20 A.
moved
4.39c the commutator has
In Fig.
further,
we
contact with the brush, the cur-
in
rent in this coil has
commutator bars sweep
time available to reverse the current
apply Kirchhoff s current law,
Thus, by coming
to the
1
A.
is
momentarily symmetri-
segments
1
and
2.
has not fallen to zero, and
But the curis still,
say.
DIRECT-CURRENT GENERATORS
30 A. As a
40
40
40
40
40
while that
sity
©
71
72
|
segment 2
2
3
|
f
still
35
40
2
/
|
3
I
I
has
left-
moved beyond
it
in coil
1
the
has
has a value of 20 A,
from segment
to the brush
is
is
The
1
resulting high current den-
sity
causes the brush to overheat
720
coils are being
<t>)
at the tip.
Because
commutated every second,
overheating raises the brush
75 5
71
therefore, 7
40
©
V
|
A
despite the fact that the contact area
getting very small.
72
1
Assuming
not reversed.
now 60 A,
|
is,
midpoint of the brush and the current
the current flowing
71
70
is
1
0 A. The current den-
will tend to overheat.
4.40d segment
In Fig.
(a)
80
40
1
on the left-hand side of the brush
hand side of the brush
4040
40
only
is
times greater than on the right-hand side. The
vl /l
|
segment
result, the current in
in
93
tip to the
this
incandescent
point and serious sparking will result.
In
80
designing dc motors and generators, every
is made to reduce the self-inductance of the
One of the most effective ways is to reduce
the number of turns per coil. But for a given output voltage, this means that the number of coils
must be increased. And more coils implies more
effort
coils.
40
40
40
30
40
©
71
|
72
1^1/2
|
3
I
1
commutator
7010
(c)
80
bars. Thus, in practice, direct-current
generators have a large
number of
mutator bars — not so much
the output voltage but to
coils
and com-
to reduce the ripple in
overcome
problem of
the
commutation.
20
40
40
40
40
Another important factor
©
71
72
|
J
1
|
2
7
is
made
3
|
that the
(d)
80
zone.
As
©
|
72
1
|
x
2
,
|
the brush
3
I
4
40 4C
(e)
80
tance
opposes the reversal
always
is
commutation
induced
is
1
.5 V.
the brush
.
The
of current.
in the
to the self-
measures, the composition of
carefully chosen.
It
affects the brush
V
to as
much
This drop occurs between the surface of
and the commutator surface.
A large brush
drop helps commutation, but unfortunately
1
mmf.
created in the neutral
coil.
creases the losses.
of the current in coil
is
voltage drop, which can vary from 0.2
as
Figure 4.40
Commutation
is
armature
the
which opposes the voltage due
In addition to these
71
than
this tlux, a voltage
inductance of the
40
40
aiding commutation
the coil side undergoing
sweeps through
coil
40
greater
Therefore, a small tlux
60 20
40
slightly
in
commutating poles
4
|
tht\
40
mmf of the
As
a result, the
it
in-
commutator and
brushes become hotter and the efficiency of the
coil
induc-
generator
is
slightly reduced.
ELECTRICAL MACHINES AND TRANSFORMERS
94
Questions and Problems
resistance
the
Practical level
Sketch the main components of a dc gener-
4-1
ator.
4- 2
1
4-2
Why
are the brushes of a de
ways placed
at
machine
al-
100
is
<},
calculate the
machine operates
a.
At no-load
b.
At full-load
Fig. 4.
1
mmf when
rated voltage
at
8b shows the no-load saturation
curve of a separately excited dc generator
the neutral points?
when
it
revolves
1500 r/min. Calculate
at
4-3
Describe the construction of a commutator.
the exciting current needed to generate
4-4
How
120
is
the induced voltage of a separately
excited dc generator affected
a.
the speed increases?
b.
the exciting current
How
4-5
do we adjust
4- 3
if
1
is
reduced?
tion
the voltage of a shunt
is
4. 10, the
induced voltage
momentarily 18
V, in the posi-
shown. Calculate the voltages induced
C
A, B, and
same
at the
Referring to Fig. 4.
1
age induced
A when
decreases with increasing load. Explain.
has rotated by 90°; by 120°.
over-
4- 5
1
load
and
differential
compound generators
as to construction
b.
as to electrical properties
1
instant.
lb, calculate the volt-
the armature
positive with respect to brush y
1
b.
Show
Does
the polarity of each of
the polarity reverse
when
a coil turns through 180°?
4-16
The generator of
Fig. 4.38 revolves at
r/min and the flux per pole
is
20
960
mWb.
Calculate the no-load armature voltage
if
each armature coil has 6 turns.
Intermediate level
separate excited dc generator turning at
4-17
How many
a.
127 V. The armature resistance
machine delivers
is
brush sets are needed for the gen-
erator in Fig. 4.38?
1400 r/min produces an induced voltage of
the
is
in coil
the 12 coils.
a.
A
Brush x
in Fig. 4,
Explain the difference between shunt, compound,
4-9
330 r/min.
The terminal voltage of a shunt generator
increases.
4-8
1
D
in coils
4-14
Explain why the output voltage of an
compound generator increases as the
4-7
at
Referring to Fig.
in coil
generator?
4-6
V
If the
b.
2 12 and
machine delivers
a total load current of
1800 A. calculate the current flowing
a current of 12 A.
armature
in
each
coil.
Calculate
Advanced
a.
The terminal voltage |V]
b.
c.
The heat dissipated in the armature [W]
The braking torque exerted by the armature
A
separately excited dc generator pro-
4-18
[Nm|
4-10
duces a no-load voltage of 115
V.
segments 3 and 4 must be greater than
40 V?
What
happens if
a. The speed is increased by 20 percent?
b. The direction of rotation is reversed'.'
c. The exciting current is increased by 10 percent?
d. The polarity of the field is reversed?
level
The voltage between brushes x and y is
240 V in the generator shown in Fig. 4.38.
Why can we say that the voltage between
4- 9
1
Referring to Fig.
ity
of
£ xv when
4.
1
0,
determine the polar-
the armature turns counter-
clockwise.
4-20
a.
In Fig. 4.38
determine the polarity of
tween commutator segments 3 and
4-
1
1
Each pole of a
1
00 kW, 250
V flat-compound
generator has a shunt field of 2000 turns and
a series field of 7 turns.
If
the total shunt-field
ing that the armature
b.
At the same
instant,
is
4.
£ 34
be-
know-
turning clockwise,
what
segment 35 with respect
is
to
the polarity of
segment 34?
DIRECT-CURRENT GENERATORS
4-2]
The armature shown
5.4 (Chapter 5)
in Fig.
has 81 slots, and the commutator has 243
segments.
lap
It
will be
winding having
flux per field pole
is
wound
/ n du stria I
4-24
to give a 6-pole
30
mWb,
The induced voltage
at a
calculate the
is 60 ohms and
2
The I R loss in
are 15
mm
wide and
commutator
4-22
A 200 W,
1
20
is
V,
that the
diameter of the
1
4-23
dc generator has a
winding on the armature.
The
rated armature current
The
The
I
total losses in the
2
R
effi-
is
is
5 A.
0.023 pu.
machine
losses in the armature
The generator in Problem 4-24 weighs
2600 lb. Calculate the output in watts per
4-26
In
Problem 4-24 calculate
The full-load current of the generator
The current carried by the armature coils
the torque re-
quired to drive the generator
(The shunt
field
is
at
1750 r/min.
powered by a separate
source.)
4-27
A 4-pole
2 8 A.
dc generator delivers a current of
The average brush voltage drop on
each of the four brush sets
Calculate
b.
the rated current
the armature
a.
1
a.
r/min separately
kilogram.
equal to 1.33 ms.
A 4-pole 250 kW, 750V
lap
4-25
800 r/min dc generator
The brush width is
such as to cover 3 commutator segments.
Show that the duration of the commutation
is
1
b.
c.
450 mm.
has 75 commutator bars.
process
V 750
Calculate
The average flux density per pole
The time needed to reverse the current in
each armature coil, knowing that the brushes
c.
500
speed of
1200 r/min
b.
A 240 kW,
ciency of 94%. The shunt field resistance
following:
a.
Appl tea t ion
excited dc generator has an overall
turn per coil. If the
1
95
is
found
to
be
0.6 V. Calculate the total brush loss in the
machine, neglecting friction
loss.
1
Chapter 5
Direct-Current Motors
Today,
5.0 Introduction
this
general statement can be challenged
because the availability of sophisticated electronic
Now
thai
erators,
we have
we can
a
good understanding of dc gen-
motors transform
Direct-current
drives has
begin our study of de motors.
energy
electrical
hoists, fans,
cars.
vari-
ment has given
it
service and
force (cemf)
has to drive, and this require-
rise to three basic types
Direct-current motors are built the
of motors:
erate either as a
trate,
Series motors
3.
Compound motors
motor or
means of a switch
all
electric utility sys-
electric trains,
it
is
in steel mills,
As soon
form the alternating current
der to use dc motors.
flows
mines, and
sometimes advantageous
low.
to trans-
is
To
connected to a dc source
(Fig. 5.
1
).
illus-
which the armature,
E
The armature has
is
s
by
a re-
created by a set
in
The
as the switch
is
closed, a large current
the armature because
its
resistance
individual armature conductors are
is
very
imme-
diately subjected to a force because they are im-
into direct current in or-
The reason
as
of permanent magnets.
tems furnish alternating current. However, for speapplications such as
in
sistance R, and the magnetic field
Direct -current motors are seldom used in ordinary
because
as a generator.
consider a dc generator
initially at rest, is
industrial applications
same way
generators are; consequently, a dc machine can op-
Shunt motors
2.
mersed
that the torque-
in
the magnetic field created by the perma-
nent magnets. These forces add up to produce a
speed characteristics of dc motors can be varied over
a
in
Counter-electromotive
5.1
The torque-
speed characteristic of the motor must be adapted to
the type of the load
cial
still
thousands more are being produced every year.
definite torque-speed
pump or fan) or a highly
able one (such as a hoist or automobile).
.
possible to use alternating current
pumps, calendars, punch-presses, and
These devices may have a
characteristic (such as a
1
it
there are millions of dc motors
mechanical energy. They drive devices such as
into
made
motors for variable speed applications. Nevertheless,
wide range while retaining high efficiency.
powerful torque, causing the armature
96
to rotate.
DIRECT-CURRENT MOTORS
motor
5.2 Acceleration of the
The
net voltage acting in the armature circuit in Fig.
(£ s
—£
The
resulting armature current
5.2
is
/ is
limited only by the armature resistance R, and so
()
volts.
)
=
/
(£ s
- EJ/R
(5.1)
When the motor is at rest, the induced
£0 = 0, and so the starting current is
Figure 5.1
Starting a
line.
The
On
the other hand, as
gins to turn, a
cut a
in
soon as the armature be-
second phenomenon takes place: the
We know
generator effect.
duced
that a voltage
Ea
is in-
the armature conductors as soon as they
magnetic field (Fig.
5.2).
This
is
always
true,
The value and
the same as those
no matter what causes the rotation.
polarity of the
obtained
induced voltage are
when
the
machine operates
The induced voltage
the
E0
is
as a generator.
therefore proportional to
speed of rotation n of the motor and to the flux
per pole, as previously given by Eq. 4.
tf>
£0 in
or the circuit-breakers to
the type
the
(4.
polarity
.
is
equal to
It
£
(1
is
force (cemf) because
always acts against the source voltage
acts against the voltage in the sense that the
net voltage acting in the series circuit
equal to (£ s
— £
()
)
volts
and not (£ s
of Fig. 5.2
+ £n
)
they are
consequent rapid acceleration of the armature.
As
the speed increases, the counter-emf
creases, with the result that the value of (£ s
diminishes.
It
follows from Eq.
5.
)
armature
that the
1
£0 in— £u
current / drops progressively as the speed increases.
maximum
volts.
is
to accelerate until
()
voltage
£
s
.
voltage (£s
to act
ical
In effect,
if
£0 were
reaches a def-
less than the source
equal to
— E0 would become
)
the current
The driving
/.
£
s
,
the net
zero and so, too,
forces
would cease
on the armature conductors, and the mechan-
drag imposed by the fan and the bearings would
immediately cause the motor
to
slow down. As the
speed decreases the net voltage (Es
and so does the current
fall
it
speed. At no-load this speed pro-
duces a counter-emf £ slightly
would
case of a motor, the induced voltage
called counter-electromotive
£s
the armature and
Z
if
ductors produce a powerful starting torque and a
1
a constant that de-
of winding. For lap windings
However,
trip.
mo-
blow
Although the armature current decreases, the
number of armature conductors.
In the
its
Z is
the fuses to
absent, the large forces acting on the armature con-
motor continues
number of turns on
be 20 to 30 times
would cause
tor. In practice, this
1
Z//4V60
the case of a generator.
pends upon the
may
current
starting
greater than the nominal full-load current of the
inite,
As
voltage
= EJR
/
dc motor across the
97
as soon as the torque
current
is
/.
— £0
The speed
)
increases
will cease to
developed by the armature
equal to the load torque. Thus,
motor runs
when
a
no-load, the counter-emf must be
at
£
slightly less than
s?
to flow, sufficient to
so as to enable a small current
produce the required torque.
Example 5-7
The armature of a permanent-magnet dc generator has
a resistance of
when
1
H
is
and generates a voltage of 50
500
r/min. If the armature
is
V
con-
nected to a source of 150 V, calculate the following:
Figure 5.2
Counter-electromotive force (cemf)
the speed
in
a dc motor.
a.
The
starting current
"1
ELECTRICAL MACHINES AND TRANSFORMERS
98
b.
The counter-emf when
the
motor runs
at
1000
Mechanical power and torque
5.3
r/min. At 1460 r/min.
c.
The armature current
at
1000 r/min. At 1460
The power and torque of
r/min.
motor are two of
a dc
most important properties.
its
We now derive two sim-
ple equations that enable us to calculate them.
Solution
a.
moment of start-up,
At the
tionary, so
rent
E0 = 0 V
1
the armature
(Fig. 5.3a).
The
is
.
sta-
According
to Eq. 4.
wound armature
starting cur-
£ =
limited only by the armature resistance:
is
(t
- EJR =
/
150 V/l
n=
A
150
the
1
cemf induced
in a lap-
given by
is
Z/i$/60
(4.1)
Referring to Fig. 5.2, the electrical power P. supx
b.
Because
the generator voltage
r/min, the
cemf of
the
1000 r/min and 146
c.
The
V
motor
at
is
50
V
will be
at
100
V
at
E
s
P*
However,
drop
s
150
-
100
E
in the
= EJ
s
is
equal to the
sum of E0
E = Ea +
s
=
(E S
=
50/1
When
the
A
(Fig. 5.3b)
= (E0 +
= EJ +
motor speed reaches 1460 r/min, the
cemf will be 146 V, almost equal to the source voltage. Under these conditions, the armature current
is
The
2
I
R term
IR)I
2
I
R
only
electrical
=
=
(Es
4
- EJ/R =
(150
-
146)/1
(5.4)
represents heat dissipated in the ar-
mature, but the very important term
/
(5.3)
EJ
P*
50
IR
follows that
- EJ/R
-
plus the IR
is
It
I
(5.2)
armature:
= 50 V
The corresponding armature current
I:
1000
is
E - Eu =
equal to the supply voltage
is
multiplied by the armature current
1460 r/min.
net voltage in the armature circuit at
r/min
500
plied to the armature
power
that is
EJ
is
the
converted into mechanical
power. The mechanical power of the motor
therefore exactly equal to the product of the
A
is
cemf
multiplied by the armature current
and
the
corresponding
motor torque
smaller than before (Fig. 5.3c).
is
much
P = EJ
(5.5)
DIRECT- CURRENT MOTORS
where
where
P —
E0 -
mechanical power developed by the
T
motor [W]
Z
induced voltage
in
=
total
torque (N-mJ
number of conductors on the armature
[Wb|*
-
effective flux per pole
<&
=
/
=
armature current A|
6.28
=
constant, to take care of units
the armature (cemf)
IV]
/
[
current supplied to the armature [A]
[exact value
2.
99
Turning our attention to torque
that the
mechanical power
P
is
T,
=
2tt
1
we know
given by the
Eq. 5.6 shows that
motor
expression
either
we can
raise the torque of a
by raising the armature current or by
raising the flux created by the poles.
P = n 7/9.55
where n
is
(3.5)
Example 5-2
The following
the speed of rotation.
Combining Eqs.
/?779.55
3.5, 4.1,
and
5.5,
we
obtain
- EJ
= Z/i3>//6()
hp),
250
V,
details are given
on a 225 k\V (~ 300
1200 r/min dc motor (see Figs. 5.4 and
5.5):
243
armature coils
turns per coil
and so
1
type of winding
T = Z*//6.28
The torque developed by a lap-wound motor
therefore given
armature slots
81
commutator segments
243
is
field poles
by the expression
T = Z$//6.28
lap
(5.6)
6
diameter of armature
559
axial length of armature
235
The
effective flux
is
niven by
<I>
mm
mm
- 60 EJZn.
Figure 5.4
225 kW, 250 V, 1200 r/min. The armature core has a diameter
and an axial length of 235 mm. It is composed of 400 stacked laminations 0.56 mm thick. The armature
slots and the commutator has 243 bars. (H. Roberge)
Bare armature and commutator of a dc motor rated
of
559
has 81
mm
ELECTRICAL MACHINES AND TRANSFORMERS
00
1
Figure 5.5
Armature
a.
of Fig. 5.4 in the
process
b.
One
c.
Connecting the
d.
Commutator connections ready
of the 81 coils
coil
of
ready to be placed
ends
to the
being wound; coil-forming machine gives the coils the desired shape.
in
the slots.
commutator
bars.
for brazing. (H.
Roberge)
T=
Calculate
a.
The
b.
The number of conductors per
c.
The
rated armature current
slot
9.55 Pin
=
9.55
X 225 000/1200
=
1791
N-m
flux per pole
The
flux per pole
is
Solution
a.
=
6.28 TiZI
nearly equal to the applied voltage (250 V).
=
(6.28
The
=
25.7
We
can assume
that the
induced voltage
rated armature current
/
- F/E0 =
E0
d>
is
is
225 000/250
5.4
= 900 A
Speed
When
b.
Each
coil
is
made up of 2 conductors,
gether there are 243
X
2
= 486
so alto-
conductors on
Conductors per
slot
Coil sides per slot
The motor torque
1790)/(486
is
=
=
486/81
6
=
IR drop due
always small compared
On
900)
a dc motor drives a load between no-load and
full-load, the
equal to
6
X
of rotation
to
armature resistance
to the supply voltage
This means that the counter-emf
the armature.
c.
X
mWb
E
s
£0
is
is
E
s
.
very nearly
.
the other hand,
may be expressed by
we have
already seen that
E0
the equation
En =
Z/7*/60
(4.1)
DIRECT-CURRENT MOTORS
101
(variable)
!
motor armature
motor
field
(fixed)
O
(
j
O
3-phase
motor
Figure 5.6
Ward-Leonard speed control system.
Replacing
E0
by
£ we
s
3-phase
obtain
,
as
Zn*/60,
the
line. This method of speed control, known
Ward-Leonard system, is found in steel
and paper
'A mills, high-rise elevators, mines,
That
RKTR
is
60£
v
(approx)
A
'SiaJU3$6^A
modern
In
installations the generator
placed by a high-power electronic converter that
changes the ac power of the
electrical utility to dc,
The Ward-Leonard system
= speed of rotation [r/min]
E = armature voltage fVJ
Z = total number of armature
armature of a dc motor.
s
shows
This important equation
is
conductors
speed of the
that the
directly proportional to the armature supply
and inversely proportional
voltage
We will now
study
more than
is
simple way of applying a variable dc voltage
//
pole.
mills.
often re-
by electronic means.
where
motor
is
to the flux per
how this equation is
applied.
It
just a
to the
can actually force the mo-
tor to
develop the torque and speed required by the
load.
For example, suppose
slightly
higher than the
Current will then flow
5.6,
E
is
s
E
cemf
i}
adjusted to be
of the motor.
in the direction
and the motor develops
shown
in Fig.
a positive torque.
The
armature of the motor absorbs power because
Armature speed control
5.5
According to Eq. 5.7,
constant
with
Now, suppose we reduce Es by reducing the gen<1>
G As soon as E s becomes less than
£u current / reverses. As a result,
the motor torque
speed depends only upon the
reverses and (2) the armature of the motor delivers
if
the flux per pole
(permanent magnet
fixed excitation), the
E By
armature voltage
s
.
In
practice,
we can
tion
or field
fall in
vary
E
s
of the motor
s,
the
proportion.
by connecting the
is
G
(Fig. 5.6).
The
field excita-
kept constant, but the generator
can be varied from zero
and even reversed.
to
maximum
The generator output voltage
f can therefore be varied from zero
s
mum, with
£
kept
M to a separately excited variable-
voltage dc generator
excitation / x
field
<I> is
raising or lowering
motor speed will rise and
motor armature
to
either positive or negative
maxi-
polarity.
Consequently, the motor speed can be varied from
zero to
maximum
generator
is
/
flows into the positive terminal.
in either direction.
Note
erator excitation
.
(
,
1
)
power to generator G. In effect, the dc motor suddenly
becomes a generator and generator G suddenly becomes a motor. The electric power that the dc motor
now
netic
its
the
delivers to
G
is
derived
at the
expense of the
connected mechanical load. Thus, by reducing
motor
is
£
s,
suddenly forced to slow down.
What happens to the dc power received by genG? When G receives electric power, it operates as a motor, driving its own ac motor as an asynchronous generator!* As a result, ac power is fed
erator
that the
driven by an ac motor connected to a
ki-
energy of the rapidly decelerating armature and
The asynchronous generator
is
explained
in
Chapter
14.
1
ELECTRICAL MACHINES AND TRANSFORMERS
02
back into the
P = EJ = 380 X 3000 =
normally feeds the ac motor.
line that
The tact that power can be recovered this way
makes the Ward-Leonard system very efficient, and
constitutes another of
Example 5-3
A 2000 kW, 500
by a 2500
kW
control system
generator, using a
shown
T = 9.55/Vw
motor
V, variable-speed
in Fig. 5.6.
is
driven
Ward-Leonard
The total resis-
tance of the motor and generator armature circuit
The motor turns
when E0 is 500 V.
10 mi}.
r/min,
at a
is
nominal speed of 300
The motor torque and speed when
The speed of
of
this
(400
=
n
with the armature (Fig. 5.7). The current
It is
E
s
,
in the
which subtracts
yielding a smaller
lot
below
its
nominal
recommended for small motors beof power and heat is wasted in the rheoonly
and the overall efficiency
is
is
low. Furthermore,
poor, even for a fixed setting
increases as the armature current increases. This
produces a substantial drop
motor armature
speed with increasing
in
mechanical load.
is
P = EJ = 380 X 2000 - 760
The motor speed
to control
to place a rheostat in se-
is
380)/0.01
A
to the
connected me-
of the rheostat. In effect, the /R drop across the rheo-
-
stat
The power
motor
the speed regulation
is
s
its
electromechanical braking torque.
cause a
stat,
= (E - EJ/R =
motor and
Rheostat Speed Control Another way
speed.
Solution
2000
the
ables us to reduce the speed
s
The armature current
kN-m
supply voltage across the armature. This method en-
E = 350 V and £0 = 380 V
-=
47.8
140 000)/228
1
from the fixed source voltage
The motor torque and speed when
/
=
X
chanical load will rapidly drop under the influence
ries
s
a.
(9.55
rheostat produces a voltage drop
E = 400 V and E0 = 380 V
b.
=
the speed of a dc
Calculate
a.
kW
140
Braking torque developed by the motor:
advantages.
its
1
kW
is
X 300 = 228
(380 V/500 V)
The motor torque
r/min
is
T = 9.55P/R
b.
Because
-
(9.55
=
31.8
X 760
000)/228
kN-m
E a — 380
V, the
motor speed
is still
228 r/min.
The armature current
/
is
= (E
EJ/R =
= - 3000 A
s
The current
is
(350
-
Power returned by
38())/0.01
it
flows
in re-
motor torque also reverses.
the
and the 10 mi} resistance:
5,6 Field
According
negative and so
verse; consequently, the
Figure 5.7
Armature speed control using a rheostat.
motor
to the
generator
a dc
speed control
to Eq. 5.7
we
can also vary the speed of
motor by varying the
keep the armature voltage
numerator
in
Eq. 5.7
is
field flux
E
s
<t>.
Let us
now
constant so that the
constant. Consequently, the
motor speed now changes
in
inverse proportion to
DIRECT-CURRENT MOTORS
the flux
drop,
cr>:
if
we
increase the flux the speed will
This
method of speed control
frequently used
is
when
the
called
base speed. To control the flux (and hence,
motor has
the speed),
to run
we connect
above
a rheostat
its
R
rated speed,
in series
To understand
at
constant speed.
less
method of speed
this
motor
in Fig.
5.8a
is
The counter-emf
running
slightly
ZTn
is
£
due
s,
we suddenly
to the
increase the
resistance of the rheostat, both the exciting current
/x
and the flux
<J>
reduces the
cemf
jump
a
to
to
will diminish. This
immediately
causing the armature current
much
is
again almost equal to
E
/
higher value. The current
.
£
will accelerate until
It
same E0 with
Clearly, to develop the
motor must turn
raise the
motor speed above
shunt-wound motors,
£0
and
s
()
s.
flux, the
troducing a resistance
E
motor develops a
the
field,
greater torque than before.
with
control, sup-
initially
than the armature supply voltage
IR drop in the armature. If
Despite the weaker
value depends
its
between
the very small difference
{
the field (Fig. 5.8a).
pose that the
changes dramatically because
upon
and vice versa.
103
its
weaker
a
can therefore
nominal value by
with the
in series
this
We
faster.
in-
For
field.
method of speed control
enables high-speed/base-speed ratios as high as 3 to
1.
Broader speed ranges tend
produce
to
instability
and poor commutation.
Under
abnormal conditions, the
certain
exciting current of a shunt motor
dentally, the only flux
manent magnetism
that the
motor has
remaining
in the poles.*
is
is
may
flux
drop to dangerously low values. For example,
if
the
interrupted acci-
due
that
This flux
to the re-
is
so small
dangerously high
to rotate at a
speed to induce the required cemf. Safety devices are
introduced to prevent such runaway conditions.
(a)
5-7
Shunt motor under load
Consider a dc motor running
chanical load
at
no-load. If a
suddenly applied
is
me-
to the shaft, the
small no-load current does not produce enough
torque to carry the load and the motor begins to
slow down. This causes the cemf to diminish,
p.u.
—/
Tvsn
2
—
"-7
/
—
/
/
/
/
higher torque.
*
Tvs
motor
I
is
When
the torque developed by the
exactly equal to the torque imposed by the
mechanical load, then, and only then, will the speed
/
remain constant (see Section
t
3.1 1).
To sum
up, as
the mechanical load increases, the armature current
•rated load
(b)
re-
sulting in a higher current and a corresponding
rises
and the speed drops.
"—7
The speed of
/
stant
<L
2
1
^ speed
a shunt
from no-load
motor stays
only drops by 10 to 15 percent
when
it
is
p. u.
/
The term residual magnetism
Figure 5.8
also used. However, the
is
IEEE Standard Dictionary of Electrical and
a shunt motor including the
a.
Schematic diagram
b.
Torque-speed and torque-current characteristic of
induction;
a shunt motor.
will be less than the residual induction."
of
full-load
n
armature current
field
relatively con-
to full-load. In small motors,
rheostat.
Terms
states.
.
netie eireuit, the
if
.
If
Electronics
there are no air gaps ... in the inag-
remanent induction
there are air gaps
.
.
.
will equal the residual
the remanent induction
"1
1
ELECTRICAL MACHINES AND TRANSFORMERS
04
applied. In big machines, the drop
in part, to
is
even
less,
due
By
the very low armature resistance.
A
51
ad-
justing the field rheostat, the speed can, of course,
be kept absolutely constant as the load changes.
Typical torque-speed and torque-current charac-
of a shunt motor are shown in Fig. 5.8b.
The speed, torque and current are given in per-unit
values. The torque is directly proportional to the arteristics
mature current. Furthermore, the speed changes
only from
from 0 pu
pu
1.1
pu as the torque increases
to 0.9
to 2 pu.
Example 5-4
A
shunt motor rotating
120
V
1500 r/min
at
and the shunt-field resistance
mature resistance
a.
b.
c.
is
fed by a
source (Fig. 5.9a). The line current
is 0.
is
II. If
The
=
Figure 5.9
See Example
120 V/120
P = 6000 - 250 - 5750
= 7.7
motor
(equivalent to 5750/746
=
/
b.
5
The voltage across
W
O= A
1
= 50 A
I
the armature
E=
and
in
armature iron losses.
120
5.8 Series
is
A
series
motor
motor
identical in construction to a
is
shunt motor except for the
V
Voltage drop due to armature resistance
nected
= 50X0.1 =
The cemf generated by
5
V
field.
The
field
con-
is
This series field
is
armature current (Fig. 5.10a).
composed of a few
turns of wire
having a cross section sufficiently large
the armature
is
with the armature and must, there-
in series
fore, carry the full
IR
hp)
is
-
I
is
The actual mechanical output is slightly less than
5750
because some of the mechanical power is
dissipated in bearing friction losses, in windage
losses,
The armature current
5.4.
W
the
field current (Fig. 5.9b) is
/x
A
Mechanical power developed by the armature
The current in the armature
The counter-emf
The mechanical power developed by
Solution:
a.
51
the ar-
calculate the following:
ii,
1
120
is
to carry
is
the current.
E0 =
-
120
5
=
V
115
Although the construction
ties
c.
The
total
power supplied
to the
motor
is
similar, the proper-
of a series motor are completely different from
is
those of a shunt motor. In a shunt motor, the flux F
P,
=
El
=
120
Power absorbed by
X
= 6120
51
W
per pole
field
the armature
is
is
constant
P n = El = 120 X 50 = 6000
W
armature
is
IR
2
=
50
2
X
0.
1
in
a series motor
When
the current
is
large and vice versa. Despite these
is
differences, the
P =
But
depends upon the armature current
and. hence, upon the load.
large, the flux
in the
loads because the shunt
line.
is
the flux per pole
Power dissipated
at all
connected to the
- 250
W
same
basic principles and equations
apply to both machines.
DIRECT-CURRENT MOTORS
105
series field
(b)
Figure 5.10
a.
Series motor connection diagram.
b.
Schematic diagram of a series motor.
When
a series motor operates at full-load, the
per pole
flux
identical
same
as that of a shunt
starts up, the
armature current
3 p.u.
motor of
power and speed. However, when
motor
series
the
is
speed
/;
armature current
the
/
higher
is
Figure 5.11
than normal, with the result that the flux per pole
also greater than
normal,
torque of a series
motor
ft
is
is
follows that the starting
of
ing the
On
T versus
/
curves of Figs. 5.8 and
the other hand, if the
than full-load, the
pole are
5.
1
Conversely, the speed
motor operates
necting an external resistor
at less
ture
The weaker field
same way as it would
smaller than normal.
shunt motor with a
weak shunt
the load current of a series
For ex-
field.
motor drops
ample,
if
half
normal value, the flux diminishes by half and
its
so the
speed doubles. Obviously,
if
the load
is
to
ate at
no-load,
ft
tends to run away, and the resulting
could tear the windings out of the
centrifugal forces
and the
sistor
and
field.
The
field
reduces
be lowered by con-
in series
total
with the arma-
IR drop across the
armature
the
voltage, and so the speed must
re-
supply
fall.
Typical torque-speed and torque-current characteristics are
ferent
shown
in Fig. 5.
1
1
.
They
are quite dif-
from the shunt motor characteristics given
in
Fig. 5.8b.
small,
may rise to dangerously high values. For
reason we never permit a series motor to oper-
speed
this
may
1
armature current and the flux per
causes the speed to rise in the
for a
and current-torque characteristic
a series motor.
considerably greater than
of a shunt motor. This can be seen by compar-
that
the
Typical speed-torque
Example 5-5
A
15 hp,
240
V,
1780 r/min dc series motor has
full-load rated current of
teristics are
54 A.
Its
a
operating charac-
given by the per-unit curves of Fig. 5.11.
armature and destroy the machine.
Calculate
5.9
When
Series motor speed control
a series
motor carries
a load,
its
speed
a.
The current and speed when
24 N m
b.
The efficiency under
may
the load torque
is
these conditions
have to be adjusted slightly. Thus, the speed can be
low resistance
increased by placing a
with the
series
field.
smaller than before,
and an increase
in
The
field
in parallel
current
which produces a drop
speed.
is
then
in flux
Solution
a.
We
first
establish the base power, base speed,
and base current of the motor. They correspond
to the full-load ratings as follows:
ELECTRICAL MACHINES AND TRANSFORMERS
106
PH =
//
=
hp
15
X 746 =
15
11
190
W
are also used in electric cranes and hoists: light loads
are lifted quickly
= 1780r/min
H
= 54A
/B
Compound motor
5.11
The base torque
therefore,
is,
A compound dc
P vh
9.55
Tv =
=
X
9.55
11
780
190/1
mmf of the
Fig. 5.
N-m
torque of 24
corresponds to a per-
two
runs
=
T(pu)
=
24/60
2
1
at
Referring to Fig.
5.
tained at a speed of
From
the
X
/j(pu)
T vs
/
1
.4
1
winding
0.4
1
,
a torque of 0.4
pu
pu. Thus, the speed
nB
=
1.4
shows
low and the
is
However,
is at-
by current
machine:
is
As
X 1780
/x
it
curve, a torque of 0.4 pu re-
is
The
total
current
b.
X
/(pu)
-
/B
0.6
To calculate the
and
X 54 =
away
like a shunt
no-load.
at
mmf of the
series field
remains
flux per
falls
with increasing load and the
to full-load is
generally
between 10 percent and 30 percent.
A
we have
efficiency,
to run
to
know P
32.4
x
P0 =
/2779.55
= 7776
= 2492 X
()
W
24/9.55
= 6263 W
= PJP, = 6263/7776 =
0.805 or 80.5%
5.10 Applications of the series
motor
Series motors are used on equipment requiring a high
starting torque.
They
which must run
series
motor
is
at
are also used to drive devices
high speed
at light
loads.
The
particularly well adapted for traction
purposes, such as in electric trains. Acceleration
rapid
because the torque
is
high
at
is
low speeds.
Furthermore, the series motor automatically slows
down
as the train goes up a grade yet turns at top
speed on
flat
ground. The power of a series motor
tends to be constant, because high torque
is
accom-
panied by low speed and vice versa. Series motors
Figure 5.12
Connection diagram of a dc compound motor.
b. Schematic diagram of the motor.
a.
is
fully excited
Pr
P = EI = 240 X
T|
32.4
is
therefore greater under load than at no-load.
speed drop from no-load
=
the series
of the series field
mmf of the shunt field
mmf (and the resulting
The motor speed
is
the motor
/ in
and so the motor behaves
the load increases, the
constant.
pole)
When
the shunt field
does not tend
increases but the
r/min
mmf
quires a current of 0.6 pu. Consequently, the load
/
and
the connection and schematic
no-load, the armature current
negligible.
- 2492
fields add.
diagrams of a compound motor.
unit torque of
=
a series field
compound motor, the
The shunt field is always
stronger than the series field.
= 60 N-m
//
motor carries both
a shunt field. In a cumulative
"is
A load
and heavy loads more slowly.
DIRECT-CURRENT MOTORS
If
1.6
the series field
the shunt field,
we
is
connected so
increasing load.
1.2
creases, and this
it
The speed
may
opposes
compound
obtain a differential
motor. In such a motor, the total
1.4
that
107
mmf decreases with
rises as the
lead to instability.
load in-
The
differ-
-r-+ential
1.0
j
a
Fig. 5.
a
x>
compound motor
differential
0.8
compound
compound
a>
0)
a
of shunt,
1
3
shows
the typical torque-speed curves
compound and
basis. Fig. 5. 14
w
has very few applications.
shows
series
motors on a per-unit
a typical application of dc
0.6
motors
in a steel mill.
0.4
5.12 Reversing the direction
0.2
of rotation
0
0
0.2
0.4
0.6
0.8
Torque
1.2
1.0
1.4
1.6
(per-unit)
the direction of rotation of a dc motor,
must reverse either
(2) both the shunt
Figure 5.13
Typical
To reverse
speed versus torque characteristics
of various
and
l
)
in
we
the armature connections or
series field connections.
terpoles are considered to
The change
dc motors.
(
connections
form
is
The in-
part of the armature.
shown
in Fig. 5. 15.
Figure 5.14
Hot
strip finishing mill
to the
runout table
composed
(left
of
6 stands, each driven by a 2500
kW
dc motor. The wide steel
foreground) driven by 161 dc motors, each rated 3 kW.
(Courtesy of General Electric)
strip is
delivered
"1
1
ELECTRICAL MACHINES AND TRANSFORMERS
OS
(-)
(+)
(->
(+)
<+>
(0
(b)
(a)
|
I
Figure 5.15
compound
a.
Original connections of a
b.
Reversing the armature connections to reverse the direction of rotation.
c.
Reversing the
field
motor.
connections to reverse the direction of
5.13 Starting a shunt motor
[f
we apply
full
rotation.
5.14 Face-plate starter
voltage to a stationary shunt motor,
Fig. 5. 16
shows
the schematic diagram of a manual
the starting current in the armature will be very high
face-plate starter for a shunt motor. Bare copper
and we run the
contacts are connected to current-limiting resistors
risk of
/?,,
Burning out the armature;
a.
Damaging
b.
the
commutator and brushes, due
to
*
and
R4 Conducting arm
.
when
it
of insulated handle
is
1
sweeps
across
pulled to the right by means
2. In the position
shown,
the
M
heavy sparking;
c.
R2
the contacts
Overloading the feeder;
d.
Snapping off the shaft due
e.
Damaging
the driven
to
mechanical shock;
equipment because of the
sudden mechanical hammerblow.
arm touches dead copper contact
and the motor
circuit is open. As we draw the handle to the right
the conducting arm first touches fixed contact N.
The sii; y voltage Es immediately causes full
field current / x to flow, but the
armature current
/ is
limited by the four resistors in the starter box. The
All dc motors must, therefore, be provided with
a
means
to limit the starting current to
values, usually between
rent.
One
solution
is
to
1
.5
reasonable
and twice full-load cur-
connect a rheostat
in series
motor begins
to turn and, as the
cemf E0
the armature current gradually falls.
speed ceases
builds up,
When
any more, the arm
the motor
is
pulled to
the
next contact, thereby removing resistor
R from
the
to rise
}
with the armature. The resistance
is
duced as the motor accelerates and
eliminated entirely,
full
when
the
gradually reis
eventually
machine has attained
Today, electronic methods are often used to
control.
and
to
provide speed
circuit.
The
current immediately jumps to
higher value and the motor quickly accelerates to
next higher speed.
we move
speed.
limit the starting current
armature
arm
When
the speed again levels off
to the next contact,
and so
finally touches the last contact.
netically held in this position
net 4,
which
is in
a
the
forth, until the
The arm
is
mag-
by a small electromag-
series with the shunt field.
DIRECT-CURRENT MOTORS
109
Figure 5.16
Manual face-plate starter for a shunt motor.
If the
supply voltage
the field excitation
is
suddenly interrupted, or
electromagnet releases the arm, allowing
to
it
to return
dead position, under the pull of spring
its
3.
This
prevents the motor from restarting un-
safety feature
expectedly
if
should accidentally be cut, the
when
the supply voltage
is
reestablished.
When
the
tion of the
cemf Ea
motor
are as
shown
armature IR drop,
If
is
running normally, the direc-
armature current
£
()
is
we suddenly open
motor continues
I
x
and the polarity of the
Neglecting the
in Fig. 5.17a.
equal to
E
s.
the switch (Fig. 5.17b), the
to turn, but
its
speed will gradually
drop due to friction and windage losses.
5.15
Stopping a motor
One
inclined to believe that stopping a dc
other hand, because the shunt field
induced voltage
is
almost
a simple,
always
trivial,
not
a heavy inertia load,
the
true.
system to
come
is
same
it
a generator
a large dc motor
may
to a halt. For
many
coupled
is
take an hour or
more
Ea
whose armature
is
motor
under these circumstances,
is
motor
is
by simple mechanical
armature
is
suddenly connected to
reasons such
the external resistor (Fig. 5. 7c). Voltage
way we stop a
car.
this current
motor
original current
same
A more elegant method consists of
electrically.
flows
in the
brake the
circulating a reverse current in the armature, so as to
brake the
mediately produce an armature current
a braking
friction, in the
Two methods
are
/,. It
(
nected to a source
nected to the
throw switch.
£
s,
whose
field is directly
and whose armature
is
either the line or to
an external resistor
R (Fig.
very smooth stop.
concon-
5. 17).
im-
However,
/2
.
is
The
reverse torque brings the machine to a rapid, but
l
same source by means of a doubleThe switch connects the armature to
.
developed whose magnitude depends upon
Dynamic braking
Consider a shunt motor
E0 will
opposite direction to the
dynamic braking and (2) plugging.
5.16
/2
follows that a reverse torque
em-
ployed to create such an electromechanical brake:
now
Let us close the switch on the second set of contacts so that the
we must apply
One way to
at the
is
for
often unacceptable and,
torque to ensure a rapid stop.
the
on open-circuit.
1
a lengthy deceleration time
On
excited,
continues to exist, falling
rate as the speed. In essence, the
operation. Unfortunately, this
When
is
to
motor
is still
Figure 5.17a
Armature connected
to a
dc source
Es
.
1
1
ELECTRICAL MACHINES AND TRANSFORMERS
0
Figure 5.17b
Armature on open
circuit
generating a voltage
E0
\x
Time
.
Figure 5.18
Speed versus time curves
for
various braking methods.
reversing the armature current by reversing the
ter-
minals of the source (Fig. 5.19a).
Under normal motor conditions, armature
rent
is
/,
/,
Dynamic braking.
where
In practice, resistor
R
R0
cur-
given by
is
=
(E s
- £0 )//? 0
the armature resistance. If
we
suddenly
reverse the terminals of the source, the net voltage
is
chosen so that the
initial
acting on the armature circuit becomes (E 0 + E s
The so-called counter-emf E0 of the armature is no
).
braking current
rent.
The
initial
is
about twice the rated motor cur-
braking torque
is
then twice the nor-
longer counter to anything but actually adds to the
mal torque of the motor.
As
E0
in
E
supply voltage
the
motor slows down, the gradual decrease
produces a corresponding decrease
in
/2
becoming zero when
finally
the arma-
The speed drops quickly
This net voltage would produce
greater than the full-load armature current. This
current
tor,
ture ceases to turn.
.
.
Consequently, the braking torque becomes smaller
and smaller,
s
an enormous reverse current, perhaps 50 times
would
initiate
an arc around the commuta-
destroying segments, brushes, and supports,
at first
and then more slowly, as the armature comes
even before the
line circuit breakers
to a
halt.
The speed decreases exponentially, somewhat
like
the
voltage across a discharging capacitor.
Consequently, the speed decreases by half
intervals of time
Ta To
dynamic braking,
.
in
equal
illustrate the usefulness
Fig. 5. 18
of
compares the speed-
time curves for a motor equipped with dynamic
braking and one that simply coasts to a stop.
5.17 Plugging
We
a
can stop the motor even more rapidly by using
method called plugging.
It
consists of suddenly
Figure 5.19a
Armature connected
to
dc source
Es
.
could open.
DIRECT-CURRENT MOTORS
value.
However,
much
is
it
draw the
easier to
speed-time curves by defining a new time con-
T0 which
stant
the time for the speed to de-
is
crease to 50 percent of
its
original value. There
is
a direct mathematical relationship between the
conventional time constant
T0
constant
T and
the half-time
given by
It is
.
Ta = 0.6937
We
Figure 5.19b
is
Plugging.
can prove that
this
given by
Jnr
Tn =
(5.9)
°
To prevent such a catastrophe, we must
by introducing a
reverse current
resistor
R
in series
As
in
dynamic
with the reversing circuit (Fig. 5. 9b).
1
braking, the resistor
U
braking current
With
oped even
effect, at
when
Ta =
the armature has
E0 =
= moment
J
devel-
=
is
braking
motor
P =
cuit,
otherwise
interruption
it
the armature cir-
will begin to run in reverse. Circuit
mounted on
the
motor
The curves of Fig. 5.18 enable us
dynamic braking
plugging and
for the
to
completely after an interval
if
dynamic braking
the
its
is
2T0 On
.
it
same
initial
is still
25 per-
original value at this time. Nevertheless,
more popular
in
motor
to the braking resistor
a constant (exact value
=
a constant
This equation
is
braking effect
pated
is
in the
exact value
=
log c 2]
due
that the
energy
dissi-
resistor. In general, the
motor
entirely
braking
to the
subjected to an extra braking torque due to
windage and
friction,
and so the braking time
be less than that given by Eq.
hp),
250
V,
1
280 r/min dc motor
mechanical time constant
drives a large flywheel and the total
We
can therefore speak of a mechanical
much
same way we speak of
time constant
T
the electrical
time constant of a capacitor that dis-
in
the
charges into a resistor.
essence,
motor
T
is
it
is
speed
is
its
initial
connected
to a
210
V
is
kW.
moment
177 kg
of
m2
.
It
in-
The
dc source, and
1280 r/min just before the armature
switched across a braking resistor of 0.2
its
is
il.
Calculate
The mechanical time constant T0 of
the braking
system
takes for the speed
36.8 percent of
of the flywheel and armature
motor
a.
the time
to fall to
will
5.9.
has windage, friction, and iron losses of 8
ertia
of the
|
Dynamic braking and
braking.
[W]
=
most applications.
We mentioned that the speed decreases exponentially
with time when a dc motor is stopped by dynamic
In
the
based upon the assumption
is
Example 5-6
A 225 kW (- 300
5.18
starts [r/min]
power delivered by
2
0.693
comparative simplicity of dynamic braking ren-
ders
shaft
compare
the other hand,
used, the speed
motor
(30/77) /log c 2]
shaft.
braking current. Note that plugging stops the motor
cent of
=
131.5
usually controlled by an automatic
is
null-speed device
of inertia of the rotating
initial
l
we must immediately open
stops,
fall to
|sj
speed of the motor when
initial
i
as the
previous value
lkg-nr]
/?
which
its
parts, referred to the
to a stop. In
= EJR,
As soon
value.
its initial
come
0, but I2
is
time for the motor speed to
one-half
to limit the initial
reverse torque
circuit, a
zero speed,
about one-half
designed
131.5 7,
where
about twice full-load current.
to
plugging
this
is
limit the
(5.8)
mechanical time constant
b.
The time
for the
motor speed
to
drop
to
20 r/min
ELECTRICAL MACHINES AND TRANSFORMERS
c.
The time
for the speed to drop to
only braking force
friction,
due
that
is
20 r/min
to the
if
The stopping time increases
the
windage,
20 r/min
and iron losses
Solution
a.
We
;
note that the armature voltage
the speed
When
the armature
210V. The
the resistor
P,
210
V
is
(276/10)
=
28 min
This braking time
switched to the brak-
initial
dynamic braking
very
is still
power delivered
to
X 60 =
1656
2
/R
2
= 210
/0.2
= 220 500
The time constant Ta
s
Ta =
7//,
177
131.5
-
10
28 times longer than when
used.
to a
is
dynamically
complete stop.
In practice,
however, we can assume that the machine stops
W
is
2
is
is
a motor which
Theoretically,
braked never comes
ter
an interval equal to 5
If the
/(131.5 Py)
(5.9)
motor
af-
seconds.
plugged, the stopping time has a
is
2
X 1280
X 220 500
Tn
definite value given by
ts
= 2T0
(5.10)
where
s
—
ts
b.
=
and
is
= E
approximately
is
1280 r/min.
is
ing resistor, the induced voltage
close to
is
proportion to the
in
time constant. Consequently, the time to reach
The motor speed drops by 50 percent every
The speed versus time curve follows the se-
10
s.
stopping time using plugging
Tn — time
constant as given
in
[
sj
Eq. 5.9
[sj
quence given below:
speed (r/min)
time
Example 5-7
The motor in Example 5-6
(s)
1280
0
640
320
10
ing resistor
30
Calculate
80
40
a.
40
20
50
b.
an interval of 60
60
to
20 r/min
The initial braking
The stopping time
after
The
We
ro -
=
.///,
/>,
/(131.5 P,
X
= 276 s =
(177
The
initial
/,
The
,
initial
power
is
s
braking current
- EiR =
initial
is
420/0.4
braking power
=
1050
= 8000
According
to Eq. 5.9,
Tn
A
is
X
{
220.5
kW
has the same value as
before:
is
T0 =
10
s
)
2
1280 )/(131.5
4.6
current and braking
P = EJ = 210 X 1050 =
have
1280
The new time constant
2
so that the
fl,
as before.
E = E + E = 210+ 210 - 420 V
s.
of the braking time.
=
same
net voltage acting across the resistor
(
x
the
Solution
The initial windage, friction, and iron losses are
8 kW. These losses do not vary with speed in
exactly the same way as do the losses in a braking resistor. However, the behavior is comparable, which enables us to make a rough estimate
n
is
20
160
The speed of the motor drops
c.
plugged, and the brak-
increased to 0.4
is
braking current
.
.
is
min
X
8000)
The time
to
come
ts
to a
complete stop
= 2T0 = 20
s
is
DIRECT-CURRENT MOTORS
Armature reaction
5.19
now we have assumed
Until
ing in a dc
motor
current flowing
is
that
due
the flux
that the only
However,
to the field.
N
mmf actthe
armature conductors also cre-
in the
magnetomotive force
ates a
113
that distorts
and weakens
coining from the poles. This distortion and
zone
field
weakening takes place
in
motors as well as
We recall that the magnetic action
mmf is called armature reaction.
generators.
armature
in
of the
Figure 5.20
Flux distribution
a motor running
in
at no-load.
due
5.20 Flux distortion
to armature reaction
When
motor runs
a
at no-load, the
small current
flowing in the armature does not appreciably affect
the flux
when
coming from
<2>,
the
the poles (Fig. 5.20). But
armature carries
its
normal current,
it
duces a strong magnetomotive force which,
would create a
acted alone,
superimposing
and
<1>|
increases
under the
we
2,
(Fig. 5.2
if
it
By
1).
obtain the resulting
Figure 5.21
Flux created by the full-load armature current.
our example the flux density
(Fig. 5.22). In
flux
<J>
02
flux
pro-
half of the pole and
left
de-
it
ceases under the right half. This unequal distribution
produces two important effects. First the neutral
zone shifts toward the
rotation).
The
left
result
is
(against the direction of
poor commutation with
sparking at the brushes. Second,
flux
density
in
pole
tip
A,
due
to the
saturation
higher
sets
Consequently, the increase of flux under the
hand side of the pole
the right-hand side.
is
Flux 4> 3
than flux
slightly less
less than the
<F, at
chines the decrease in flux
percent and
therefore
no-load. For large
may
left-
decrease under
at full- load is
be as
much
zone
in.
Figure 5.22
Resulting flux distribution
a motor running at
full-
main the
as 10
case of a dc generator, these narrow poles de-
velop a magnetomotive force equal and opposite
causes the speed to increase with load.
it
in
load.
the
mmf of the armature
to
so that the respective mag-
Such a condition tends to be unstable; to eliminate
the
problem,
we sometimes add
netomotive forces
a series field of
rise
and
fall
one
current varies. In practice, the
or
two turns to increase the flux under load. Such
tating poles
motors are said to have a stabilized-shunt winding.
Commutating poles
5.21
set
of
improve commutation, we always place a
commutating poles between the main poles of
medium- and large-power dc motors
made
(Fig. 5.23).
As
slightly greater than that of the
armature. Consequently, a small flux subsists
in the
region of the commutating poles. The flux
is
signed to induce
To counter the effect of armature reaction and
thereby
is
together as the load
mmf of the commu-
in the coil
tion a voltage that
is
equal and opposite to the self-
induction voltage mentioned in Section 4.28.
result,
commutation
de-
undergoing commuta-
is
As
a
greatly improved and takes
place roughly as described in Section 4.27.
DIRECT-CURRENT MOTORS
Figure 5.24
Six-pole
dc motor having a compensating winding distributed
in
slots in the
main poles. The machine also has 6
commutating poles.
(Courtesy of General Electric
Company)
base speed. In so doing, the rated values of armature
current,
armature voltage, and field flux must not be
may be
we assume an
exceeded, although lesser values
In
rately
making our
E. v
is
negligible (Fig. 5.25).
the
armature current
exciting current /f
in
per-unit values.
£a happens
/a is
that
it
1
.
The advantage of the
Thus, the per-unit torque
used.
ideal sepa-
unit flux
<I>
f
to
,
The armature
/a ,
the flux
and the speed n are
Thus,
be 240
if
all
<t>
is
T
is
given by the per-
times the per-unit armature current
T=$> r I
ll
/.,
(5.11)
voltf,
the
expressed
the rated armature voltage
By
the
voltage
Ea
same reasoning,
is
1
shunt field flux O, has a per-unit
the per-unit armature
equal to the per-unit speed n times the
per-unit flux O,
=
V and the rated armature current
600 A, they are both given a per-unit value of
Similarly, the rated
per-unit approach
renders the torque-speed curve universal.
excited shunt motor in which the armature re-
sistance
age
analysis,
value of
The
(5.12)
logical starting point of the torque-speed
curve (Fig. 5.26),
is
the condition
where the motor
1
1
6
ELECTRICAL MACHINES AND TRANSFORMERS
(T=
develops rated torque
'a
The
rated speed
)
—
speed (n
at rated
I
).
order to reduce the speed below base speed,
In
we
l
often called base speed.
is
gradually reduce the armature voltage to zero,
and
while keeping the rated values of
at their per-unit
value of
l
.
3> r constant
Applying Eq.
(5.
7=1
corresponding per-unit torque
X
1
1
the
),
=
1
1.
Furthermore, according to Eq. (5.12), the per-unit
Figure 5.25
£a = n X
£a /a
voltage
Per-unit circuit diagram
the state of
operation,
,
= n. Figures 5.27 and 5.28 show
]
and
known
<£>,
during this phase of motor
mode.
as the constant torque
Next, to raise the speed above base speed,
alize that the armature voltage
anymore because
The only solution
T
already at
it is
is
to
keep
means
=
base speed, the per-unit flux
OH
1
0
'
2.0
1.0
^
speed
n
is
Eq. (5.12),
1.
1
this
Thus, above
\/n.
equal to the recipro-
of the per-unit speed. During this operating
cal
mode, the armature current can be kept
of
level
T=
Figure 5.26
=
and so
1,
rated level of
its
flux. Referring to
{
re-
E,d at its rated level of
and reduce the
that n<$
we
cannot be increased
ct»
1.
=
/
r a
Recalling Eq. (5.11),
(\/n)
X
it
at its rated
follows that
-\ln. Consequently, above base
1
speed, the per-unit torque decreases as the reciprocal of the per-unit speed.
motor
1
is
clear that since the per-
and armature voltage are both
unit armature current
equal to
It is
during this phase, the power input to the
equal to
1
.
Having assumed an
chine, the per-unit mechanical
equal to
is
why
1,
which corresponds
the region
ideal
power output
is
maalso
to rated power. That
above base speed
is
named
the
constant horsepower mode.
We
conclude that the ideal dc shunt motor can
operate anywhere within the limits of the torque-
speed curve depicted
in Fig. 5.26.
In practice, the actual torque-speed
fer considerably
from
that
shown
curve
may
in Fig. 5.26.
dif-
The
curve indicates an upper speed limit of 2 but some
machines can be pushed
to limits
of 3 and even 4, by
reducing the flux accordingly. However,
speed
is
raised
lems develop and centrifugal forces
<H
'
0
1.0 1.25
>-
speed
n
L2.0
dangerous.
When
the ventilation
when
the
above base speed, commutation probthe motor runs
becomes poorer and
tends to rise above
its
may become
below base speed,
the temperature
rated value. Consequently, the
armature current must be reduced, which reduces the
Figure 5.28
torque. Eventually,
when
the speed
is
zero,
all
forced
DIRECT-CURRENT MOTORS
ventilation ceases
and even the
field current
must be
117
magnet motors
5-24 Permanent
reduced to prevent overheating of the shunt field
coils.
As a
result, the
permissible stalled torque
only have a per-unit value of 0.25.
tical
torque-speed curve
The
ishes
The resulting pracspeed dimin-
can be largely overcome by using an external
of air, no matter
what
It
delivers a constant stream
the speed of the
Under these conditions,
approaches that
shown
We
have seen
and a
that shunt-field
field current to
motors require coils
produce the
flux.
consumed, the heat produced, and
in Fig. 5.29.
drastic fall-off in torque as the
blower to cool the motor.
to be.
shown
is
may
motor happens
the torque-speed curve
large space taken
up by the
vantages of a dc motor.
By
The energy
the relatively
field poles are disad-
using permanent mag-
nets instead of field coils, these disadvantages are
overcome. The
result
is
a smaller
motor having a
higher efficiency with the added benefit of never
risking run-away due to field failure.
in Fig. 5.26.
A further advantage
is
that the effective air
The reason
that
that the
is
many
increased
times.
magnets have a permeability
nearly equal to that of
is
mature
sible
is
of using permanent magnets
gap
air.
As
a result, the ar-
mmf cannot create the intense field that is pos-
when
soft-iron
pole
pieces
are
employed.
Consequently, the field created by the magnets does
not
become distorted,
armature reaction
is
as
shown
in Fig. 5.22.
Thus, the
reduced and commutation
is
im-
proved, as well as the overload capacity of the motor.
0
0.2 0.4 0.6 0.8 1.0
—
A further advantage is that the long air gap reduces the
2.0
speed
inductance of the armature and hence
n
much more
quickly to changes
in
it
responds
armature current.
Permanent magnet motors are particularly advan-
Figure 5.29
Torque-speed curve of a typical dc motor.
tageous
in capacities
below about
5 hp.
The magnets
Figure 5.30
Permanent magnet motor rated
slots 20;
1
.5 hp,
commutator bars: 40; turns per
0.34 a.
(Courtesy of Baldor Electric
Company)
90
V,
2900
coil: 5;
r/min, 14.5 A.
conductor
Armature diameter: 73 mm; armature length: 115 mm;
17 AWG, lap winding. Armature resistance at 20°C:
size: No.
ELECTRICAL MACHINES AND TRANSFORMERS
are ceramic or rare-earth/cobalt alloys.
shows
PM
motor.
and
tia
fast
Its
the series winding, per pole.
5.30
Fig.
2900 r/min
elongated armature ensures low iner-
the construction of a
response
hp,
.5
when used
The only drawback of
tively high cost
1
has a
V,
motors
motor
of the magnets and the inability to
a.
b.
obtain higher speeds by field weakening.
5-12
Questions and Problems
Name
1
make
Explain what
is
a
5- 3
1
meant by the generator
What determines
ef-
ity
in
If the
line, calcu-
following:
full-load
when
V
the armature
is
connected
200
1
to a
source. Calculate the armature voltat
1500
The following details are known about a
250 hp, 230 V, 435 r/min dc shunt motor:
A
nominal full-load current: 862
the magnitude and polar-
of the counter-emf
V
r/min. At 100 r/min.
fect in a motor.
5-3
23 A.
separately excited dc motor turns at
115
three types of dc motors and
is
field
and the
age required so that the motor runs
sketch of the connections.
5-2
A
The shunt
15 (1,
1
connected to a 230
mmf per pole at
mmf at no-load
The
The
r/min
Practical Level
5-
is
late the
the rela-
is
of
total resistance
nominal armature current
servo applications.
in
PM
90
H
insulation class:
a dc motor?
weight: 3400 kg
The counter-emf of a motor
5-4
is
always
mm
external diameter of the frame: 915
slightly less than the applied armature volt-
length of frame: 1260
mm
age. Explain.
and efficiency
a. Calculate the total losses
Name two methods
5-5
that are
used to vary
the speed of a dc motor.
5-6
why
Explain
ing current
a starting resistor
motor up
needed
to bring a
percent of the total losses
Show one way to reverse the direction
rotation of a compound motor.
5-8
A 230 V
5-9
20 per-
Calculate the value of the armature resistance
knowing
as well as the counter-emf.
speed?
to
field excit-
the shunt field causes
if
cent of the total losses.
c.
is
approximate shunt
b. Calculate the
the armature current of a shunt
motor decreases as the motor accelerates.
Why
5-7
at
full-load.
of
that
at full-load are
50
due
to armature resistance.
d.
If
we wish
to attain a
what should be
shunt motor has a nominal arma-
the
speed of
1
100 r/min,
approximate exciting
current?
ture current of
tance
0.15
11,
If the
armature
c.
The mechanical power developed by
[kW] and
the armature
if
the
motor
across the 230
b.
V
the
mo-
inal
is
initial starting
shown
The value of
20 hp, 240
V,
400
nom400 A, calculate
in Fig. 5. 17. If the
is
the braking resistor
to limit the
125 percent of
maximum
its
R
if
we
braking current
to
nominal value
line,
to limit the initial current to
1
b.
The braking power [kW] when the motor has
decelerated to 200 r/min, 50 r/min, 0 r/min.
a.
The motor
15 A.
5-15
in
Problem
5-
1
4
is
now
stopped
by using the plugging circuit of Fig. 5.19.
The compound motor of Fig. 5.12 has 1200
turns on the shunt
1
armature current
want
directly connected
Intermediate level
1
wish to stop a
the following:
a.
Calculate the value of the starting resistor
needed
5-1
We
ing circuit
[WJ
[hp]
a. In Problem 5-9 calculate the
current
5-14
r/min motor by using the dynamic brak-
The counter-emf [V|
The power supplied to
tor,
resis-
calculate the following:
b.
a.
5-10
is
60 A.
winding and 25 turns on
Calculate the
the
maximum
new braking
resistor
braking current
is
R
so that
500 A.
DIRECT-CURRENT MOTORS
Calculate the braking power [kW]
b.
motor has decelerated
to
when
the
5-20
200 r/min, 50 r/min,
ing:
0 r/min.
Compare
c.
the braking
power developed
at
200
r/min to the instantaneous power dissipated
in resistor R.
Advanced
5-16
5-2
1
tor has a
225 kW,
a
diameter of 559
length of 235
mm.
1
The
The
c.
at a
The value of
c.
The
the armature
the counter-cmf at
full
flux per pole, in milliwebers
A standard
20 hp, 240
V.
1
load
[mWb]
500 r/min
self-
it
200 r/min
Calculate the following:
ing.
total kinetic
ings and
commutator
is
the
decided to cool the machine by
the air by
tor
equal to the J calcu-
in-
blower and channeling
means of an
The highest
30°C and
exits the mo-
air duct.
the temperature of the air that
J of the wind-
from
1500 r/min without overheat-
expected ambient temperature
energy of the revolving parts
if
Ft is
to
stalling an external
turns at 1200 r/min
speed of 600 r/min,
requirement has arisen whereby the mo-
tor should run at speeds ranging
kinetic energy of the armature alone
when
The number of conductors on
b.
A
200 r/min mo-
mm and an axial
The approximate moment of inertia, knowing
3
that iron has a density of 7900 kg/m
b.
a.
cooled dc motor has an efficiency of 88%.
level
The armature of
a.
Referring to Fig. 5.30, calculate the follow-
is
should not exceed 35°C. Calculate the
capacity of the blower required,
in
cubic
lated in (a)
feet per minute. (Hint: see Section 3.21.)
5-17
If
we reduce
the normal exciting current of
a practical shunt
motor by 50 percent,
speed increases, but
it
5-22
the
A
250
hp,
nominal
never doubles.
500 V dc shunt motor draws
field current of 5
A
a
under rated
The field resistance is 90 12. Calculate
ohmic value and power of the series re-
load.
Explain why, bearing
in
mind
the saturation
the
of the iron under normal excitation.
5- 8
1
The speed of
a series
motor drops with
ing temperature, while that of a shunt
ris-
mo-
tor increases. Explain.
Industrial
5-19
5-23
magnetism per 100°C increase
The motor runs
at
3%
in
of
its
tempera-
a no-load speed of
2500 r/min when connected
to a 150
A5
500
hp dc motor draws a
In each case, calculate the
V
field as a
What
source in an ambient temperature of 22°C.
room where
that the field current drops
the shunt field and resistor
V
source.
field current
of
A when the field is connected to a
150 V source. On the other hand, a 500 hp
motor draws a field current of 4.3 A when
the field is connected to a 300 V dc source.
magnet motor equipped with
Estimate the speed
A when
0.68
cobalt-samarium magnets loses
ture.
needed so
are connected to the
A pplication
A permanent
sistor
to 4.5
if
the
motor
is
placed
the ambient temperature
is
in
a
40°C.
power required
for the
percentage of the rated power of the motor.
conclusions can you draw from these results?
Chapter 6
Efficiency and Heating
of Electrical Machines
6.0 Introduction
Whenever
a
one form
loss.
The
6.1
machine transforms energy from
to another, there
is
loss takes place in the
causing
(
duction
in efficiency.
1
)
an increase
in
Mechanical losses are due
machine
bearing friction,
to
brush friction, and windage. The friction losses
always a certain
depend upon
itself,
temperature and (2) a
Mechanical losses
the speed of the
machine and upon
the design of the bearings, brushes, commutator,
re-
and
Windage
depend on
the
speed and design of the cooling fan and on the
tur-
slip
rings.
losses
From the standpoint of losses, electrical machines may be divided into two groups: those that
bulence produced by the revolving
have revolving parts (motors, generators,
sence of prior information,
those that do
Electrical
(transformers,
not
etc.)
reactors,
and mechanical losses are produced
and
tests
of these mechanical losses.
In this
chapter
chines, but the
we analyze
same
internal fan
the losses in dc
losses are also found in
mamost
in
us a clue as to
We
how
is
they
important because
may be
it
determine the value
mounted on
the
motor
shaft.
It
cool air from the surroundings, blows
the windings, and expels
machines operating on alternating current. The
study of power losses
parts. In the ab-
usually conduct
Rotating machines are usually cooled by an
machines.
in stationary
itself to
etc.).
in ro-
tating machines, while only electrical losses are
produced
on the machine
we
it
draws
it
over
again through suitable
vents. In hostile environments, special cooling
methods are sometimes used,
gives
reduced.
as
illustrated in
Fig. 6.1.
also cover the important topics of tempera-
ture rise
and the service
life
of electrical equipment.
We show
that both are related to the class
of insula-
tion used
and
have been
that these insulation classes
6.2 Electrical losses
standardized.
Electrical losses are
120
composed of the following:
EFFICIENCY AND HEATING OF ELECTRICAL MACHINES
1.
Conductor
/
R
losses (sometimes called
copper
R =
losses)
2.
Brush losses
3.
Iron losses
1.
Conductor Losses The
in
resistance, in turn,
depends upon
the length, cross section, resistivity,
and tempera-
A =
rent
ture
it
carries.
of the conductor.
The following equations
determine the resistance
able us to
ture
resistance and the square of the cur-
The
at
at)
(6.2)
which
losses in a conductor de-
R =
L =
its
(6.1)
P A
= pod +
p
pend upon
121
en-
p
resistance of conductor [il]
cross section of conductor
length of conductor [m]
=
resistivity
[
m2
]
of conductor
at
temperature
conductor
at
0°C
/
any tempera-
and for any material:
=
a =
resistivity of
Po
m]
temperature coefficient of resistance
0°C
t
[il
=
at
l/°C]
[
temperature of conductor [°C]
The values of p and a for different materials are
listed in Appendix AX2. In dc motors and generacopper losses occur
tors,
shunt
field, the
armature, the series
in the
field, the
commutating
poles,
and
compensating winding. These I~R losses show
the
up as
rise
conductor temperatures
heat, causing the
to
above ambient temperature.
Instead of using the
2
I
R
we sometimes
equation,
number
prefer to express the losses in terms of the
of watts per kilogram of conductor material. The
losses are then given by the equation
P c = l00Oy 2 p/£
(6.3)
where
Pc —
J =
Figure 6.1
Totally
tor for
enclosed, water-cooled, 450 kW, 3600 r/min mo-
use
machine
is
in
Warm
a hostile environment.
air inside
Westinghouse nameplate. After releasing
its
water-cooled pipes, the cool
air
reenters the
chine by
pipes located diagonally
as cooling-water inlet
on the heat exchanger serve
and
outlet respectively.
(Courtesy of Westinghouse)
(W/kg|
]
p
£
=
density of the conductor [kg/m
1000
=
constant, to take care of units
conductor [nll-m)
1
]
heat to a
maway of two rectangular pipes leading into the
end bells. The cooling air therefore moves in a closed
circuit, and the surrounding contaminated atmosphere
never reaches the motor windings. The circular capped
set of
[A/mm
loss
2
resistivity of the
the
above the
current density
power
=
blown upward and through a water-cooled
heat exchanger, situated immediately
specific conductor
According
mass
is
density. For
tween
1
to this equation, the
loss per unit
proportional to the square of the current
.5
copper conductors, we use densities be-
A/mm 2
losses vary
and 6
A/mm 2
from 5 W/kg
to
.
The corresponding
90 W/kg
(Fig. 6.2).
The
higher densities require an efficient cooling system
to prevent
an excessive temperature
rise.
ELECTRICAL MACHINES AND TRANSFORMERS
122
9.6
W/kg
cm
1
carbon
brush
2
200
80° C
copper conductor
commutator
Figure 6.2
Copper losses may be expressed
in
watts per
Figure 6.3
Brush contact voltage drop occurs between the brush
face and commutator.
kilo-
gram.
2.
Brush Losses The
2
I
R
losses in the brushes are
negligible because the current density
A/mm", which
0.1
per.
is
only about
used
and eddy currents, as previously explained
in
Sections 2.27 and 2.30. Iron losses depend upon the
cop-
magnetic flux density, the speed of rotation, the
However, the contact voltage drop between the
quality of the steel, and the size of the armature.
is
far less than that
brushes and commutator
may produce
in
significant
pending on the type of brush, the applied pressure,
They typically range from 0.5 W/kg to 20 W/kg.
The higher values occur in the armature teeth,
where the flux density may be as high as .7 T. The
and the brush current (Fig.
losses in the armature core are usually
losses.
3.
The drop
varies from 0.8
V
to
1.3 V,
de-
6.3).
Iron Losses Iron losses are produced
in the ar-
mature of a dc machine. They are due to hysteresis
1
The
losses can be
much
lower.
minimized by annealing the
steel
(Fig. 6.4).
Figure 6.4
kW electric oven is used to anneal punched steel laminations. This industrial process, carried out in a conatmosphere of 800°C, significantly reduces the iron losses. The laminations are seen as they leave the oven.
(Courtesy of General Electric)
This 150
trolled
EFFICIENCY AND HEATING OF ELECTRICAL MACHINES
Some
iron losses are also
They
faces.
are
due
produced
in
the pole
to flux pulsations created as suc-
and
cessive armature teeth
sweep across
slots
the
Strange as
chanical drag
it
may seem,
on the armature, producing the same
mechanical
effect as
impose a me-
iron losses
power
the line to continue to rotate. This no-load
overcomes
the friction, windage, and iron losses,
and provides for the copper losses
2
pole face.
123
The I R losses in
commutating field
load current
in the
shunt
field.
the armature, series field,
seldom more than 5 percent of
is
and
are negligible because the nothe
nominal full-load current.
friction.
As we load
machine the current increases
the
PR
in
Example 6-1
the armature circuit. Consequently, the
Adc machine turning at 875 r/min carries an armawinding whose total weight is 40 kg. The cur-
the armature circuit (consisting of the armature and
ture
A/mm 2
rent density is 5
ture
is
80°C. The
amount
to
1
1
and the operating tempera-
total iron losses in the
armature
00 W.
all
the other windings in series with
the other hand, the no-load losses
losses in
will rise.
it)
On
mentioned above
remain essentially constant as the load increases,
unless the speed of the machine changes appreciably.
It
follows that the total losses increase with
Calculate
load.
a.
b.
The copper losses
Because they are converted
into heat, the tem-
perature of the machine rises progressively as the
The mechanical drag [N
inJ
due
to the iron losses
load increases.
However, the temperature must not exceed
Solution
a.
Referring to Table
sistivity
p
of copper
=
p0
=
15.88
=
21.3
(
at
in the
80°C
Appendix, the
re-
maximum
used
is
the
+
1
AX2
at)
machine. Consequently, there
in the
power
that the
+ 0.00427 X
80)
The specific power
loaded beyond
is
loss
8890 kg/m
deliver.
is
a limit to
This temper-
inal or rated
nllm
The density of copper
machine can
power enables us to establish the nompower of the machine. A machine
ature-limited
(1
the
allowable temperature of the insulation
heat.
3
The
its
nominal rating
inevitably shortens the service
will usually over-
more
insulation deteriorates
life
rapidly,
which
of the machine.
is
If
a
machine runs
intermittently,
it
can carry heavy
2
P L = 100O/ p/£
= 1000 X 5 2 X
(6.1)
21.3/8890
overloads without overheating, provided that the operating time
rating of
= 60 W/kg
1
is
copper loss
is
is
P = 60x 40 = 2400
W
The braking torque due
to iron losses
is
it
can be
for
P = «779.55
= 875
T=
12
A
(
3.5)
779.55
N-m
or approximately 8.85 ft-lbf
Losses as a function of load
physically impossible for a generator
kW to deliver an output of
100 kW, even
6.4 Efficiency curve
The efficiency of
ful
output
power
it
must absorb some power from
machine
is
the ratio of the use-
to the input
=
P
-
-
X
power
P- (Section
x
plus the losses p.
ti
However,
a
power PG
Furthermore, input power
dc motor running at no-load develops no useful
power.
kW for
one millisecond.
3.7).
6.3
2
from
calculated
1100
1
for higher loads the capacity
limited by other factors, usually electrical. For in-
stance,
rated at 10
b.
Thus, a motor having a nominal
However,
short periods.
Total
short.
kW can readily carry a load of
0
100
We
=
is
equal to useful
can therefore write
P
°
X
100
(6.4)
ELECTRICAL MACHINES AND TRANSFORMERS
124
The
where
=
T|
ful
output power [W]
P =
input
power [W]
zero
is
losses
the
25 percent of
rating, the
nominal
its
copper losses
we have
V,
in the
2
=
50 r/min, 230
loaded
is
to
armature curof
1/4)
its full-
the following:
to calculate the
efficiency of a dc machine.
Example 6-2
A dc compound motor having
motor
approximately 25 percent (or
is
square of the current,
1
no-load because no use-
load value. Because the copper losses vary as the
|W1
The following example shows how
1
at
developed by the motor.
25 percent load When
rent
x
=
is
efficiency [%]
P0 —
p
efficiency
power
armature circuit
W
X 595 = 37
(1/4)
no-load losses
a rating of 10
kW,
50 A, has the following losses
W
= 830
at
total losses
full
load:
=
bearing friction loss
brush friction loss
windage
loss
mechanical losses
(1)
total
(2)
iron losses
(3)
copper
loss in the shunt field
copper losses at
in the
b.
in the series field
c.
in
the
total
(4)
commutating winding
copper
W
50 W
200 W
290 W
420 W
120 W
=
=
=
500
load
at full
—
load
at 25, 50, 75, 100,
595
Pa =
at
no-load
of the machine.
Draw
a graph
showing
10
kW X
power supplied
P,
830
In the
(1/4)
to the
=
867
W
= 2500
at
motor
W
25 percent
same way, we
At 50 percent load
2
(6.2)
100
= 74%
find the losses at 50, 75,
100, and 150 percent of the
(1/2)
W
is
= (PJPJ X 100
= (2500/3367) X
efficiency as a function of mechanical load (neglect
3.35 hp)
is
= 2500 + 867 - 3367
and the efficiency
W
and 150 percent of the nom-
+
is
T!
Calculate the losses and efficiency
inal rating
W
25 W
70 W
37
Useful power developed by the motor
loss in the
armature circuit
and
40
full load:
armature
a.
=
=
=
=
=
=
nominal load:
the losses are
X 595 + 830 = 979
W
the losses due to brush contact drop).
At 75 percent load
Solution
2
No-load The copper losses
(3/4)
in the
At 100 percent load
sum of the mechanical
losses (1), the iron losses (2),
595
W
the losses are
+ 830 =
1425
W
and the shunt-field
At 150 percent load
losses (3):
2
(1.5)
no-load losses
1165
armature circuit
are negligible at no-load. Consequently, the no-
load losses are equal to the
the losses are
X 595 + 830 =
= 290 + 420 +
120
- 830
the losses are
X 595 + 830 = 2169
W
W
The
efficiency calculations for the various loads
6A and
These losses remain essentially constant as the load
are listed in Table
varies.
graphically in Fig. 6.5.
the results are
shown
EFFICIENCY AND HEATING OF ELECTRICAL MACHINES
125
Figure 6.5
Losses and efficiency as a function of mechanical power.
See Example
It is
LOSSES AND EFFICIENCY OF A DC
TABLE 6A
when
Total
Output
Load
losses
power P
m
[Wl
[W]
75
1
100
1
l
r
/r]
830
0
74
979
5
000
5
980
83.6
165
7
500
8 665
86.5
10
15
000
1
000
1
selecting a motor to
425
87.5
17 170
roughly equal to the load
We
3 367
425
The efficiency curve
mature
rises sharply as the load in-
fall.
This
is
typical of the
efficiency curves of all electric motors, both ac
dc. Electric
motor designers usually
and
try to attain the
above calculation of efficiency we could
have included the losses due to brush voltage drop.
Assuming a constant drop,
brush loss at full-load
brushes
=
say,
amounts
losses,
V per brush, the
to 0.8 V X 50 A X 2
of 0.8
80 W. At 50 percent load, the brush loss
would be 40 W. These losses,
do
loads the
Consequently,
a particular job,
we
W
(
1
6.5
it
has to drive.
maximum
our example
(830
+
at that
load where the ar-
copper losses are equal to the no-load
this
corresponds to a
= 660 W,
830)
1
an output of
1
total
8
1
may wish
to
check these
Temperature
The temperature
rise
1
The
5.8 hp) and an efficiency of 87.68 percent.
results.
rise
of a machine or device
difference between the temperature of
its
is
the
warmest
accessible part and the ambient temperature.
It
may
be measured by simply using two thermometers.
peak efficiency at full-load.
In the
circuit
losses. In
loss of
87.4
over a broad range of power,
and then slowly begins to
at light
poor.
can prove that the efficiency of any dc ma-
chine reaches a
reader
creases, flattens off
is
Efficiency
P-
[Wj
0
2 169
150
power
t)
motor
should always choose one having a power rating
500
830
867
50
Input
2
0
important to remember that
efficiency of any
MOTOR
25
6.2.
when added
modify the efficiency curve only
to the other
slightly.
However, due
to the practical difficulty of placing a
to the really warmest spot inside
method is seldom used. We usuupon more sophisticated methods, de-
thermometer close
the machine, this
ally
rely
scribed in the following sections.
Temperature
rise
has a direct bearing on the
power rating of a machine or device.
It
also has a di-
ELECTRICAL MACHINES AND TRANSFORMERS
126
rect bearing
on
its
temperature rise
useful service
is
life.
Consequently,
In crystallizing, organic insulators
and
a very important quantity.
brittle.
ical vibration will
expectancy of
equipment
6.6 Life
electric
a
Apart from accidental electrical and mechanical
is
its
insulation:
higher the temperature, the shorter
made on many
its
insulating materials have
cause them
to break.
life.
The
Low
1
200°C
for the
same length of time.
temperatures are just as harmful as high
temperatures are, because the insulation tends
that
tors
their flexibility at temperatures as
life
means
ture of 105°C,
will
it
have a service
years at a temperature of 115°C, of
125°C, and of only one year
The
at
factors that contribute
at
life
a tempera-
6.7
of only four
two years
6.6).
Because of these
lation
tallize
it
upon
their ability to withstand heat.
correspond to the
(6) time (Fig.
maximum
slowly begins to crys*
rapidly as the temperature rises.
Such
as
IEEE. Underwriters Laboratories. Canadian
Standards Association.
humidity
high temperature
chemicals
fungus
(UU so
dust
noxious gases
rodents
Figure 6.6
Factors that
may
These classes
temperature levels
shorten the service
life
of
an
of:
105°C, 130°C, 155°C, 180°C. and 220°C (formerly
and the transformation takes place more
time
that set standards*
have grouped insulators into five classes, depending
factors, the state of the insu-
changes gradually;
Thermal classification
Committees and organizations
to the deteriora-
and
retain
— 60°C.
of insulators
tion of insulators are (1) heat, (2) humidity, (3) vi-
bration, (4) acidity, (5) oxidation,
low as
at
135°C!
most
have been developed, however, which
motor has a
that if a
expectancy of eight years
to
freeze and crack. Special synthetic organic insula-
approximately by half every time the temperature
increases by 10°C. This
that
On the other
synthetic polymers can withstand temper-
atures as high as
Tests
shown
some
does not exceed 00 °C.
the service life of electrical apparatus diminishes
normal
Under normal
expectancy of eight to ten years provided
life
hand,
expectancy of electrical apparatus
limited by the temperature of
hard
conditions of operation, most organic insulators have
their temperature
failures, the life
become
Eventually, the slightest shock or mechan-
insulator.
vibration
EFFICIENCY AND HEATING OF ELECTRICAL MACHINES
represented
by the
letters
A, B,
thermal classification (Table
6B)
F,
is
H, and R). This
40°C, This standardized temperature was estab-
a cornerstone in
lished for the following reasons:
design and manufacture of electrical apparatus.
the
1
6.8
Maximum ambient temperature
and hot-spot temperature rise
maximum ambient
TABLE 6B
Class
l()5°C
A
temperature, which
Illustrative
B
F
when immersed
H
be included
life at
220°C
R
240°C
S
above
240°C
C
in this class if
such as
and paper when suitably impregnated or
silk,
oil.
by experience or accepted
Other materials or combinations of materials
tests
shown
they can be
shown
to
may
be included
have comparable thermal
life at
to
have comparable
substances. Other materials or combinations of materials
shown
to
have comparable
may
life at
with suitable bonding
etc.,
in this class if
by experience
130°C.
Materials or combinations of materials such as mica, glass fiber, asbestos,
with suitable bonding
etc.,
be included in this class
if
by experience
155°C.
Materials or combinations of materials such as silicone elastomer, mica, glass fiber, asbestos,
etc.,
with suitable bonding substances such as appropriate silicone resins. Other materials or combinations of
may
they can be
shown
to
have
Materials or combinations of materials which by experience or accepted tests can be
shown
to
have
shown
to
have
shown
to
have
be included
life at
in this class if
by experience or accepted
tests
180°C.
the required thermal life at 200°C.
Materials or combinations of materials which by experience or accepted tests can be
the required thermal life at 22()°C.
Materials or combinations of materials which by experience or accepted tests can be
the required thermal life at 240°C.
Materials consisting entirely of mica, porcelain, glass, quartz, and similar inorganic materials. Other
materials or combinations of materials
they can be
implies that electrical
sive,
performance guarantees.
I05°C.
shown
to
continuously
at
may
be included
have the required thermal
The above insulation classes indicate a normal
ated
to give
Materials or combinations of materials such as mica, glass fiber, asbestos,
materials
N
likely to encounter.
enables them to standardize the size of their
substances. Other materials or combinations of materials
comparable
200°C
It
machines and
in a dielectric liquid
or accepted tests they can be
80°C
machines are
examples and definitions
or accepted tests they can be
I55°C
2.
Materials or combinations of materials such as cotton,
coated or
may
130°C
enables electrical manufacturers to foresee
their
usually
is
It
CLASSES OF INSULATION SYSTEMS
thermal
1
.
the worst ambient temperature conditions that
Standards organizations have also established a
equipment insulated with
105°C. Note that
life
life at
in this class if
expectancy of 20 000 h
a class
A
this classification
insulation system
assumes
by experience or accepted
tests
temperatures above 240°C.
to
40 000
h at the stated temperature. This
would probably
that the insulation
system
last for
is
2 to 5 years
if
oper-
not in contact with corro-
humid, or dusty atmospheres.
For a complete explanation of insulation classes, insulation systems, and temperature indices, see
the
127
companion IEEE Standards Publications Nos. 96. 97, 98, 99, and 101. See also IEEE Std
Underwriters Laboratories publication on insulation systems
UL
1446, 1978.
1
IEEE
Std 1-1969 and
17-1974 and
ELECTRICAL MACHINES AND TRANSFORMERS
128
The temperature of a machine
point, but there are places
warmer
motor, relay, and so forth, he intends to put on the
varies from point to
where the temperature
market. Thus, for class
is
than anywhere else. This hottest-spot tem-
perature must not exceed the
maximum
To show how
allowable
and
Figure 6.7 shows the hot-spot temperature limits
F.
and
H
insulation (curve
1).
built a 10
test the
They
are the temperature limits previously mentioned
maximum
insulation, the
is
(130
-
40)
kW motor using class B
motor he places
it
in
temperature of 40°C and loads
in
10
The maximum ambient temperature of
40°C is also shown (curve 3). The temperature difference between curve
and curve 3 gives the maxSection 6.7.
kW
insulation.
it
up
until
This limiting temperature
delivers
detectors, located at strategic points inside the ma-
chine, record the temperature of the windings. After
the temperatures have stabilized (which
rise
to establish the physical size
may
take
is
called the hot-spot temperature. If the hot-
Class
say
is,
H
180°C
Class F
155°C
165°C
r
/
Class
/
©
©
105°C
//
//
B
130°C
A
/
J /
/
j
145°C
/
hot-spot
120°C
temperature rise
by
//
J/
J
100°C
embedded
average
temperature
by the
thermocouple
rise
resistance method
©
40°C
40°C
limiting
ambient temperature
Figure 6.7
Typical limits of
Shows
Shows
Shows
the
the
noted, and
of the
spot temperature so recorded
Class
is
enables the
this
manufacturer
(3)
it
permissible temperature rise for each insula-
tion class.
(2)
To
of mechanical power. Special temperature
several hours), the hottest temperature
(1)
90°C.
a constant ambient
1
imum
=
the temperature rise affects the size
of a machine, suppose a manufacturer has designed
temperature of the particular class of insulation used.
for class A, B,
B
allowable temperature rise
some dc and ac industrial machines, according to the insulation class:
maximum permissible temperature of the insulation to obtain a reasonable
maximum permissible temperature using the resistance method
the limiting ambient temperature
service
life
147°C, the
EFFICIENCY AND HEATING OF ELECTRICAL MACHINES
manufacturer would not be permitted to
The reason
product.
-
(147°
=
40°)
On
B
the hottest-spot tempera-
if
is
(
100°
The manufacturer immediately
6()°C.
ceives that he can
still
per-
make
rise limits.
For instance, he can reduce the conducthe hot-spot temperature rise
90°C. Obviously,
close to
this
very
is
reduces the weight
and cost of the windings. But the manufacturer also
realizes that the
him
ables
turn,
reduces the amount of iron.
ing the
with a
now
reduced conductor size
reduce the size of the
to
By
en-
thus redesign-
motor, the manufacturer ultimately ends up
machine
The hot-spot temperature
measure because
that operates within the permissible
it
rise is rather difficult to
has to be taken
at
the very inside
of a winding. This can be done by embedding a
small temperature detector, such as a thermocouple
or thermistor. However, this direct method of measuring hot-spot temperature
is
costly,
and
only
is
justified for larger machines.
To simplify
This, in
slots.
Temperature rise
by the resistance method
6,9
per-
remain within the permissible temperature
tor size until
the size
product.
—
more economical design
a
The manufacturer could reduce
standards.
of the motor and thereby market a more competitive
insulation.
only 100°C, the temperature rise
=
and
for class
the other hand,
ture is
40°)
maximum
1()7°C exceeds the
90°C
missible rise of
his
sell
that the temperature rise
is
129
matters, accepted standards permit a
second method of determining temperature
rise.
It
is
based upon the average winding temperature, measured by resistance, rather than the hot-spot temper-
The maximum allowable average winding
ature.
temperature rise limits and has the smallest possible
temperatures for the various insulation classes are
physical size, as well as lowest cost.
In practice,
it
is
shown
in
not convenient to carry out per-
formance
of 4()°C.
tests in a
is
usually loaded to
rated ca-
its
much lower (and more comfortable) am-
pacity in
bient temperatures.
by
tablished
testing
controlled ambient temperature
The motor
Toward
this end,
it
bodies
for
that,
purposes, the ambient temperature
ture rise
is
recorded as before.
under these conditions
90°C
than
is
B
(for class
allowed to
sell his
temperature
to
standards-setting
may
is
If the
tempera-
insulation), the manufacturer
product.
is
assumed
in the
to
correspond to a hot-spot
-
rise
of ( 120°
-
40°)
- 80°C
40°)
=
is
rise
assumed
of ( 30°
1
90°C.
The average temperature of
by the resistance method.
It
a
winding
is
A
75
kW
known winding temagain when the machine is hot. For example, if the winding is made
of copper, we can use the following equation (dethe
winding resistance
at a
perature, and measuring
it
F,
operates
If
U =
(125°
(6.5)
The motor
F
234 =
= 93°C
permissible hot-spot tem-
insulation
average temperature of the winding
when
rise is
32°)
to Fig. 6.7, the
perature rise for class
15°C.
- 234
where
Solution
1
/,)
125°C, does the motor meet the
The hot-spot temperature
=
U = ^(234 +
the hot-
mpe rat Lire stand ard s ?
According
av-
at full-
R,
ambient temperature of 32°C.
spot temperature is
te
its
erage temperature:
motor, insulated class
load in an
found
consists of measuring
rived from Eqs. 6.1 and 6.2) to determine
Example 6-3
case
average winding tempera-
correspond to a hot-spot temperature
lie
equal to or less
120°C
For example,
2, Fig. 6.7.
insulation, an
temperature of 130°C. Consequently, an average
has been es-
anywhere between 10°C and 40°C. The hottest-spot
temperature
ture of
curve
B
of class
is
(155°
-
40°)
easily meets the temperature
R2 =
R =
]
t\
—
hot f°C]
a constant equal to 1/a
=
1/0.004 27
hot resistance of the winding
cold resistance of the winding
temperature of the winding
[°ci
when cold
1
ELECTRICAL MACHINES AND TRANSFORMERS
30
Knowing
winding temperature by the
the hot
we can immediately
sistance method,
Alternatively,
re-
insulation.
calculate the
temperature rise
If this
within the permissible limit (80°C for class
lation), the
product
is
B
falls
insu-
acceptable from a standards
when performance
point of view. Note that
tests are
carried out using the resistance method, the ambient
temperature must again
If the
lie
winding happens
wire, Eq. 6.3 can
must be changed
between 10°C and 40°C,
to be
be used,
still
made of aluminum
but the number 234
to 228.
A dc
motor
be rewound using class F
a very last resort,
A
word of
final
may
caution: temperature rise stan-
also on the type of apparatus (motor, transformer,
relay, etc.), the type of construction (drip-proof, totally
enclosed,
etc.),
and the
of application of
field
the apparatus (commercial, industrial, naval,
be consulted before conducting a heat-run
machine or device
test
been
idle for several
is
days
found
to
in
an
il. The motor then opwhen temperatures have staresistance is found to be 30 il. The
maximum
Although
bilized, the field
basic physical size depends
corresponding ambient temperature
built with class
B
is
24°C.
If
the
allowable temperature rise
power
shown
in Fig. 6.8.
Suppose we have
to build
another generator having the same power and
the winding, at full-
To generate
dards
we
either
have
the
to
same voltage
at
half the speed,
double the number of conductors
on the armature or double the flux from the
Consequently,
we must
practice,
The average temperature of the shunt
full-load
t2
increase both.
We
conclude that for a
(R 2 /R
i
)
true for both ac
- 234
+ 19) - 234
+
(234
(30/22) (234
/,)
The average temperature
its
rise at full-load is
c.
- 24° = 87°C.
The maximum allowable
temperature
resistance for class
B
insulation
is
rise
(120°
-
kW, 2000 r/min motor
by
40°)
80°C. Consequently, the motor does not
meet the standards. Either
its
rating will
have
be reduced, or the cooling system improved,
can be put on the market.
is
machine depends uniquely
torque. Thus, a 100
111°
it
always
and dc machines.
Basically, the size of a
upon
b.
before
is
bigger than a high-speed machine (Fig. 6.9). This
is
=
=
we
given power output, a low-speed machine
field at
= 111°C
=
poles.
either increase the size of
the armature, or increase the size of the poles. In
Solution
a.
volt-
age, but running at half the speed.
load
The full-load temperature rise by the resistance
method
Whether the motor meets the temperature stan-
its
upon the power and
Consider the 100 kW, 250 V, 2000 r/min generator
The average temperature of
es-
rating of a machine,
speed of rotation.
insulation, calculate the
following:
c.
a
and size of a machine
have a
tablishes the nominal
b.
on
(Fig. 6.10).
shunt-field resistance of 22
a.
etc.).
Consequently, the pertinent standards must always
erates at full-load and,
is
size
6.10 Relationship between the speed
that has
ambient temperature of 19°C,
motor
its
dards depend not only on the class of insulation, but
specific
Example 6-4
may
have to be increased.
corresponding temperature rise by subtracting the
ambient temperature.
As
it
to
Figure 6.8
100 kW, 2000 r/min motor; mass: 300
kg.
EFFICIENCY AND HEATING OF ELECTRICAL MACHINES
// 1 te
nn eel la te
A
6-9
1
3
level
V
dc motor connected to a 240
line pro-
duces a mechanical output of 160 hp.
Knowing
6-10
A
1
V
5
1
kW,
that the losses are 12
late the input
power and
calcu-
the line current.
dc generator delivers
20
1
A
to a
load. If the generator has an efficiency of
power
81 percent, calculate the mechanical
needed
6-
Figure 6.9
1
1
it
[hp].
Calculate the full-load current of a 250 hp,
230
100 kW, 1000 r/min motor; mass: 500 kg.
to drive
V
dc motor having an efficiency of 92
percent.
6-12
same physical size as a 10 kW motor
200 r/min because they develop the same
A machine
having class B insulation attains
has about the
a temperature of
running at
torrid
torque.
a.
Low-speed motors are therefore much more costly
than high-speed
for
b.
it
is
often cheaper to use a small
6-
1
3
high-speed motor with a gear box than to use a large
low-speed motor directly coupled to
its
load.
Practical level
6-15
What causes
the losses in a dc motor.
iron losses
Explain
and how can they
why
the temperature of a
What determines
If
the
power
electric
tor in
ma-
the vents in a motor,
its
by
always low
of
6-16
it
delivers an output of
motor driving
m
a skip hoist with-
of minerals from a
1.5 metric tons
deep every 30 seconds.
Thermocouples
power must be reduced. Explain.
per-
the
mo-
in kilowatts.
are used to
measure the
winding temperature of
insulated class
runs at full-load, what
If the
94
power output of
horsepower and
kW ac motor,
out-
nomi-
its
hp.
An
cent, calculate the
machine
rating of a
so,
if
Calculate the efficiency of the motor in
nal hot-spot
we cover up
put
is
hoist has an overall efficiency of
chine?
6-5
the temperature rise?
efficiency of a motor
trench 20
increases as the load increases.
6-4
machine running too hot and,
draws
be reduced?
6-3
Is the
The
1
6-2
8()°C.
is
Example 6-2 when
Questions and Problems
Name
1
What
when it operates at 10 percent
nal power rating. Explain.
6-14
6-1
resistance) in a
how much?
motors of equal power. Consequently,
low-speed drives,
208°C (by
ambient temperature of
is
the
F. If
a
inter-
1200
the motor
maximum
tem-
perature these detectors should indicate in an
6-6
If a
motor operates
may we
load
it
in
above
a cold environment,
its
rated
ambient temperature of 40°C? 3()°C? 14°C?
power?
6-17
Why?
6-7
Name some
A
60 hp ac motor with class F insulation
has a cold winding resistance of 12
of the factors that contribute to
23°C.
the deterioration of organic insulators.
When
it
runs
at rated
load
in
11 at
an am-
bient temperature of 3 °C, the hot winding
1
6-8
A motor is built with class H insulation.
What maximum hot-spot temperature can
withstand?
resistance
it
is
found
to be 17.4 il.
a.
Calculate the hot winding temperature.
b.
Calculate the temperature
rise
of the motor.
1
ELECTRICAL MACHINES AND TRANSFORMERS
132
Could the manufacturer increase
c.
plate rating of the
6-18
An
motor has a normal
electric
when
eight years
is
30°C.
If
it
is
6-24
life
of
is
service
A No.
1
6-25
of the
life
Knowing
The current density [A/mm
b.
The
specific copper losses
is
the conductor
No. 4
A/mm
at
Express the current density
is
its
in the
rise
its
is
losses.
6-27
On
the
reasonably
is
is
tion
1
is
What
is its
rewound using
F
ft
cable
dc circuit to carry
48 A. Assuming a
The power
maximum
loss, in watts, in the
The approximate voltage
6-28
In
Problem 6-27,
if
2-conductor
at the
243
V when
rying a current of 60 A, what
insu-
load end
is
if
V.
the voltage drop in the
cable must not exceed 10
motor having class B insula-
Assume
would normally have a service life of
h, provided the winding tempera-
of 70°C.
by resistance does not exceed 120°C.
By how many hours
duced
RW 75. A 420
V
a max-
No. 6 gauge cop-
the voltage at the service panel
expected
class
Code allows
in a
cable
situ-
20 000
ture
in
it is
car-
minimum
conductor size would you recommend?
kW ac
1
that
the following:
a.
lation?
An
found
operating temperature of 70°C, calculate
deliver at a temperature rise
of two years.
A
being used on a 240
b.
it
By
25°C.
at
it is
AWG wire size, and
Electrical
current of 65
a current of
6.5).
ated in a particularly hot location has a ser-
if
ohms
of 56
per conductor, type
nominal rating
its
electromagnet (insulated class A)
span
27 meters long.
is
of a 4-pole dc motor has a
Determine the
The National
imum
Based on
these facts, if a 20 kW motor has a fullload temperature rise of 80°C, what
life
field
calculate the weight of the wire per pole,
range between 50 percent
example, Fig.
it
AWG that
The shunt
inches.
in circular mils
of a motor
efficiency
and 150 percent of
power can
of 105°C?
6.2, calculate the resistance un-
the bare copper wire has a diameter of 0.04
120°C, calcu-
fW/kg].
roughly proportional to
(see, for
be as high as 70°C.
kilograms.
The temperature
other hand,
AWG conductor in
scraping off the insulation,
b.
constant
may
total resistance
a cur-
2
conductor temperature
it is
der these conditions of a 2-conductor cable
]
per ampere.
6-23
the prop-
lists
of commercially available copper
an area where the operating temperature of
105°C, cal-
conductor operates
If the
life
pounds.
in
AX3
Appendix
proposed to use a No. 4
[W/kgl
a.
vice
a temper-
at
that the
6-26
An aluminum
An
ohms
35
long
level
late the specific losses
6-22
in
Using Eq.
a.
rent density of 2
1
1
conductors. In an electrical installation,
temperature of the conductor
6-2
The Table
erties
culate the following:
6-20
0.
weight of the conductor
60°C,
210m
0 round copper wire
carries a current of 12 A.
Advanced
of No. 2/0 single copper conductor
reel
ature of 25°C. Calculate the approximate
motor?
6-19
A
has a resistance of
installed in a location
new probable
the
is
Industrial application
the ambient temperature
where the ambient temperature
what
name-
the
motor? Explain.
if
the
is
motor runs
ature (by resistance) of
the service life re-
6-29
A dc
a
maximum
operating temperature
busbar 4 inches wide, 1/4 inch thick,
and 30
feet long carries a current of
2500 A. Calculate
the voltage drop
for 3 h at a temper-
temperature of the busbar
200°C?
the
power
loss per
meter?
is
1()5°C.
if
the
What
is
EFFICIENCY AND HEATING OF ELECTRICAL MACHINES
6-30
Equation 6.3 gives the resistance/temperature relationship of
brush pressure:
copper conductors,
1
33
1.5 lbf
brush contact drop:
1
.2
V
namely,
coefficient of friction: 0.2
t2
= R 2 IR
(234
X
+
t
x
)
- 234
Calculate the following:
Using the information given
AX2, deduce
minum
6-3
1
in
Appendix
a.
a similar equation for alu-
b.
conductors.
c.
The commutator of a .5 hp, 2-pole, 3000
r/min dc motor has a diameter of 63 mm.
the contact voltage drop
1
d.
Calculate the peripheral speed in feet per
minute and
6-32
in
miles per hour.
e.
The following information is given on the
brushes used in the motor of Problem 6-3
newtons)
number of brushes: 2
1
5
3/4 in long.
resistivity
The
frictional
brushes
A
(The 5/16
in
when
energy expended by the two
the
commutator makes one revo-
lution (in joules)
brush dimensions: 5/8 in wide, 5/16 in thick,
contact with the
The total electrical power loss (in watts) due to
the two brushes
The frictional force of one brush rubbing
against the commutator surface (in lbf and in
1
f.
current per brush:
The resistance of the brush body in ohms
The voltage drop in the brush body
The total voltage drop in one brush, including
X
5/8 in area
commutator)
of brush: 0.0016 Il.in
g.
is in
h.
The power loss due
of 3000 r/min
The total brush loss
motor rating
to friction,
given the speed
as a percent of the
1
.5
hp
Chapter 7
Active, Reactive,
and Apparent Power
7.0 Introduction
The
7.1
The instantaneous power supplied
concept of active, reactive, and apparent
power plays
a major role in electric
power
Instantaneous power
nology. In effect, the transmission of electrical energy
across
and the behavior of ac machines are often easier
that
to
encouraged
to
The terms
The reader
pay particular attention
active, reactive,
is
which
the
positive value
always expressed
is
means
may be
that
in
The
A
positive or negative.
power flows
into the de-
vice. Conversely, a negative value indicates that
in
power
is
flowing out of the device.
to describe transient-state be-
we apply them to dc circuits.
Our study begins with an analysis of the instantaneous power in an ac circuit. We then go on to define the meaning of active and reactive power and
how to identify sources and loads. This is followed
by a definition of apparent power, power factor, and
the power triangle. We then show how ac circuits
can be solved using these power concepts. In conhavior, nor can
clusion, vector notation
active and reactive
it.
instantaneous power
to this chapter.
and apparent power
voltages and currents are sinusoidal.
They cannot be used
is
terminals times the instantaneous current
watts, irrespective of the type of circuit used.
therefore
apply to steady-state alternating current circuits
its
flows through
Instantaneous power
understand by working with power, rather than dealing with voltages and currents.
to a device
simply the product of the instantaneous voltage
tech-
power
is
in
Example
A
sinusoidal voltage having a peak value of 162
and a frequency of 60 Hz
is
V
applied to the terminals
of an ac motor. The resulting current has a peak
value of 7.5
a.
A and
b.
134
in
terms of the
<J>.
Calculate the value of the instantaneous voltage
and current
circuit.
lags 50° behind the voltage.
Express the voltage and current
electrical angle
used to determine the
an ac
7-1
at
an angle of 120°.
ACTIVE, REACTIVE,
c.
Calculate the value of the instantaneous
p =
power
ei
=
AND APPARENT POWER
X
140.3
= + 989
7.05
1
35
W
at 120°.
d.
Plot the
Because the power
curve of the instantaneous power deliv-
positive,
is
flows
it
at this
instant into the motor,
ered to the motor.
d.
a.
In order to plot the
power,
Solution
Let us
assume
that the voltage starts at zero
We
increases positively with time.
and
can therefore
we
for angles ranging
Table
7A
curve of instantaneous
repeat procedures (b) and (c) above
lists
from
(J)
=
0
to
<J)
=
360°.
part of the data used.
write
e
= Em
sin
§ =
1
62
sin
TABLE 7A
(f>
=
/
b.
At
50°, consequently,
= /m
(J)
e
/
c.
=
sin
(<(>
-
G)
=
162 sin 120°
=
140.3
=
7.5 sin (120°
7.5
=
7.05
Voltage
7.5 sin
(c|>
162 sin
-
degrees
50°)
=
X
162
-
50°)
=
0.866
7.5 sin 70°
0.94
A
The instantaneous power
at
120°
p
USED TO PLOT
in
an ac
circuit.
(]j
Power
Current
7.5 sin
<t|j
-
volts
amperes
50°)
P
watts
0
-5.75
0
-3.17
-218
50
124.1
0
75
156.5
3.17
497
0
115
146.8
6.8
1000
155
68.5
7.25
497
180
0
5.75
0
3.17
-218
205
-68.5
230
-124.1
is
power
AND
68.5
Figure 7.1
Instantaneous voltage, current and
i,
25
0
V
X
Angle
can write
we have
120°
=
=
we
e,
FIG. 7.1
The current lags behind the voltage by an angle
0
VALUES OF
(See Example
7-1
.)
0
0
ELECTRICAL MACHINES AND TRANSFORMERS
136
The
of
power
voltage, current, and instantaneous
The power
plotted in Fig. 7.1.
attains a positive
+ 000 W and a negative peak of - 2
1
ative
power means
that
power is
1
are
peak
W. The neg-
8
actually flowing
0-50°,
180° -230°,
360° -410°.
and
Although a power flow from a device considered
to
be impossible,
it
happens often
in
We
is
given
The simple
to be
20
power cycle
may seem
The
the sections that follow.
quency of
s.
120 Hz, which
the voltage
and current
is
to an ac generator.
and as we would expect
E and
/
The
line
E and
a reading
To
/
are effective values.
the instantaneous values of voltage
did
power
*
con-
the sinusoidal curves of
The peak values are
and V2/ amperes because, as
(Fig. 7.2d).
^2E volts
ously, E and
we
we
will give
watts (Fig. 7.2c).
we have drawn
circuit,
/
it
get a better picture of what goes on in such a
produce the
and
respec-
/,
are in phase (Fig. 7.2b). If
P = El
re-
effective
in a resistive circuit,
nect a wattmeter (Fig. 7.3) into the line,
twice the fre-
that
quite normal: the
voltage and current are designated
phasors
This means that the frequency of
is
is
always twice the
ac circuit of Fig. 7.2a consists of a
connected
ac circuits.
also note that the positive peaks occur at in-
tervals of 1/1
the
in
is
power*
7.2 Active
tively,
reason
phenomenon
frequency.
sistor
a load to a device considered to be a source
this
from
the load (motor) to the source. This occurs during the
intervals
power. Again,
frequency of ac power flow
in
Section 7.1,
we
E
respectively
stated previ-
By multiplying
and current
as
obtain the instantaneous
in watts.
Many persons refer to active power as real power or true
power considering it to be more descriptive. In this book
we use the term active power, because conforms to the
IEEE designation.
it
Figure 7.2
a.
An ac voltage E produces an ac current
/
in this re-
Figure 7.3
Example of a high-precision wattmeter rated 50
V, 200 V; 1 A, 5 A. The scale ranges from 0-50
sistive circuit.
c.
Phasors E and / are in phase.
A wattmeter indicates El watts.
d.
The
b.
tive
active
power
power pulses.
is
composed
V,
100
W to
of
a series of posi-
1000W.
(Courtesy of Weston Instruments)
0-
ACTIVE, REACTIVE,
The power wave consists of a
from zero
pulses that vary
to a
= 2EI = 2P
value of
The
fact that
X
(V2/)
power
is
always positive reveals
watts.
that
always
it
flows from the generator to the resistor. This
pow
properties of what
although
er:
imum,
it
137
series of positive
maximum
(i2E)
of the basic
AND APPARENT POWER
is
is
one
called active
pulsates between zero and
max-
never changes direction. The direction of
it
power flow
is
shown by an arrow P (Fig. 7.2c).
is clearly midway between
The average power
IP and zero, and so
That
watts.
is
=
EI
power indicated by
the
value
its
precisely the
P =
is
2E//2
wattmeter.
The two conductors leading
to the resistor in Fig.
2EJ
power. However, unlike cur-
7.2a carry the active
rent flow,
power does
not flow
down one conductor
Power flows over both conconsequently, as far as power is con-
and return by the other.
ductors and,
cerned,
line,
In
as
we can replace the conductors by
shown in Fig. 7.2c.
general, the line represents any transmission
number of conductors
The generator
an active load.
and the unit
megawatt
is
is
it
may
/
Reactive power consists of a series of positive and
lags 90° behind
E.
are frequently used multiples of
back and forth in this manner is
power (symbol Q), to distinguish it
from the unidirectional active power mentioned before. The reactive power in Fig. 7.4 is also given by
the product EL However, to distinguish this power
from active power, another unit is used the van Its
identical to the resistive
is
7.2a) except that the resistor
by a reactor X
E
.
x
As
is
a result, current
/
now
re-
therefore again
is
twice the line frequency.
Power
that surges
drawn the waveforms for
in
such a
E and
/
power
circuit,
we
and, by again
we
obtain
(Fig. 7.4c).
This
power p consists of a series of identical positive and
negative pulses.
instantaneous
the reactor
The
positive
waves correspond
power delivered by
the generator to
and the negative waves represent instan-
power delivered from
generator.
to
the reactor to the
The duration of each wave corresponds
The
one quarter of a cycle of the line frequency.
megavar (Mvar).
Special instruments, called vanneters, are available to measure the reactive
multiplying their instantaneous values,
curve of instantaneous
—
multiples are the kilovar (kvar) and
lags 90°
(Fig. 7.4b).
To see what really goes on
to
Phasor
frequency of the power wave
behind the voltage
taneous
b.
c.
called reactive
circuit (Fig.
the
this in-
The symbol for active power is P
The kilowatt (kW) and
The circuit of Fig. 7.4a
have
/ in
negative power pulses.
Reactive power
placed
current
circuit.
have.
the watt.
7.3
An ac voltage E produces an ac
ductive
an active source and the resistor
the watt (W).
(MW)
Figure 7.4
a.
connecting two devices, irrespective of the
line
is
a single
(C)
7.5).
A
line voltage
sin 8
if
E
(where B
reading
power
in
a circuit (Fig.
varmeter registers the product of the effective
is
times the effective line current
is
the phase angle
only obtained
between
when E and
/
/
times
E and
/).
A
are out of phase;
they are exactly in phase (or exactly 180° out of
phase), the varmeter reads zero.
Returning
pulse
is
to Fig. 7.4, the dotted area
under each
the energy, in joules, transported in
direction or the other. Clearly, the energy
is
one
deliv-
ered in a continuous series of pulses of very short
duration, every positive pulse being followed by a
ELECTRICAL MACHINES AND TRANSFORMERS
138
By
definition*, a reactor
is
considered to be
a re-
active load that absorbs reactive power.
Example
A
7-2
reactor having an inductive reactance of 4
connected
of a 120
to the terminals
V
0
is
ac generator
(Fig. 7.6a).
a.
Calculate the value of the current in the reactor
b.
Calculate the power associated with the reactor
c.
Calculate the power associated with the ac
generator
d.
Figure 7.5
Varmeter with a zero-center scale.
or negative reactive power flow up
It
to
Draw
the phasor
diagram for the
circuit
indicates positive
100 Mvars.
,
4
Q
30 A|
negative pulse. The energy flows back and forth
between the generator and the inductor without
ever being used up.
What
tive
is
A
the reason for these positive
and nega-
,
energy surges? The energy flows back and forth
because magnetic energy
+
rr^>
120V
alternately being stored
is
3.6 kvar
positive, the magnetic field
is
inside the coil.
A moment
is
when
later
L
l
4J
30 A
up and released by the reactor. Thus, when the
power
/
building up
the
power
negative, the energy in the magnetic field
is
is
E
de-
120
V
creasing and flowing back to the source.
We now
tive
have an explanation for the brief nega-
power pulses
in Fig. 7.1. In effect,
they repre-
30 A
(c)
sent magnetic energy, previously stored up in the
motor windings,
that
is
being returned to the source.
Figure 7.6
See Example
7.4 Definition of a reactive load
7.2.
and
reactive source
Solution
a.
Current
in
the circuit:
Reactive power involves real power that oscillates
back and forth between two devices over a transmission
line.
Consequently,
is
it
whether the power originates
at
/,=
impossible to say
one end of the
line
assume
that
b.
or the other. Nevertheless,
some devices generate
absorb
it.
In other
it is
useful to
reactive
power while others
E
=
XL
120
4
Power associated with
Q=
EI
=
120
V
= 30 A
f1
the reactor:
X 30 - 3600
var
=
3.6 kvar
words, some devices behave like
reactive sources and others like reactive loads.
This definition
is in
agreement with IEEE and 1EC conventions.
ACTIVE, REACTIVE,
power
This reactive
is
AND APPARENT POWER
equal to the current
is
Because the reactor absorbs 3.6 kvar of reactance
times the voltage across
ries
power, the ac generator must be supplying
Consequently, the generator
power:
delivers 3.6 kvar.
it
The
flows therefore in the direction
reactive
shown
power
Q
The
=
EIC
120
reactive
pressed
car-
it
namely
V X
A=
30
power delivered by
in vars
=
3600 var
3.6 kvar
the capacitor
power
or kilovars. Reactive
is
ex-
Q now
flows from the capacitor to the reactor.
We
This phasor
diagram applies
(the reactor)
and the reactive source (the ac genera-
tor) as
Q =
(Fig. 7.6b).
The phasor diagram is shown in Fig. 7.6c.
Current // lags 90° behind voltage E.
d.
terminals,
its
it.
a source of reactive
is
39
source of reactive power. The reactive power deliv-
absorbed by the reactor.
ered by the capacitor
c.
1
to the reactive load
have arrived
very important conclusion:
at a
a source of reactive power.
a reactive power source whenever it is
a capacitor
is
It
acts as
part of a
sine-wave-based, steady-state circuit.
well as the line connecting them.
Let us take another step and remove the reactor
7.5
The capacitor and reactive power
from the
circuit in Fig. 7.7a, yielding the circuit of
Fig. 7.8a.
now
Suppose
that
we add
a capacitor having a reac-
tance of 4 11 to the circuit of Fig. 7.6. This yields the
circuit
of Fig. 7.7a.
pacitor
is /c
expect,
it
=
The
current
120 V/4 il
=
30
/L
.
drawn by
A and,
as
the ca-
we would
leads the voltage by 90° (Fig. 7.7b).
rent of
capacitor
is
now
alone, connected to
It still
30 A, leading the voltage
7.8b). Consequently, the capacitor
E
carries a cur-
by 90°
still
this
power go? The answer
power
delivers reactive
is
to the
(Fig.
acts as a
source of reactive power, delivering 3.6 kvar.
does
The vector sum of I L and / c is zero and so the ac
is no longer supplying any power at all to
The
the terminals of the ac generator.
Where
that the capacitor
very generator to
generator
the circuit.
not
30
However, the current
changed; consequently,
AX
1
Where
can only
V =
20
is
it
in the reactor
has
continues to absorb
3.6 kvar of reactive power.
power coming from?
capacitor, which acts as
this reactive
come from
the
c
It
3.6 kvar
a
-4/
30 A
(a)
'c
30 A
(b)
120 V
varmeter
30 A
(c)
LTi
capacitor
120 V
(b)
Q
= FJ C
Figure 7.8
30 A
Figure 7.7
See Example 7.3.
a.
Capacitor connected to an ac source.
b.
Phasor
c.
Reactive power flows from the capacitor to the
l
c
generator.
leads
Eby
90°.
1
ELECTRICA L MA CHINES A ND TRA NS FORMERS
40
which
it
is
connected! For most people, this takes a
time to accept.
little
How, we may
ask, can a passive
device like a capacitor possibly produce any power?
The answer
energy
is
that reactive
power
really represents
a pendulum, swings back and forth
that, like
without ever doing any useful work. The capacitor
acts as a
temporary energy-storing device repeatedly
accepting energy for brief periods and releasing
as a reactor does, a capacitor stores electrostatic en-
we connect
7.8c),
var,
it
that reactive
from the capacitor
power
is
to the generator.
El
= -3600
The generator
X
= 400
3.5
1
prefer to call
a receiver of reactive power, which, of
same
thing. In
cerned the generator acts as a load.
The
active
power absorbed by
active
power =
1
circuit
W)
power (1304
power (812 var).
tive
7.9).
The
power associated with
L,
C
respective elements
is
a basic difference
remember
is
between active and
most important
re-
thing
one cannot be converted
and reactive powers function
independently of each other and, consequently,
in electric
a
burden on the transmission
line that
whereas active power eventually
3./
produces a tangible result (heat, mechanical power,
light, etc.), reactive
14
A
}=
A
that oscillates
4
-3.5./
power only represents power
back and
forth.
All ac inductive devices such as magnets, trans-
Q
formers, ballasts, and induction motors, absorb
20 A
re-
power because one component of the current
they draw lags 90° behind the voltage. The reactive
power plays a very important role because it proactive
Figure 7.9
duces the ac magnetic field
See Example
a source of ac-
they can be treated as separate quantities
carries them, but,
-+*
16.12
that the
into the other. Active
Both place
-
is
and a receiver of reactive
circuits.
the generator.
2Q
1=3
a source of
between active
and reactive power
There
carry the currents shown. Calculate the active and
reactive
is
7.6 Distinction
to
connected to a group of R,
elements (Fig.
it
304 W.
active power, and perhaps the
is
the resistors must
be supplied by the generator; hence
summary, a ca-
pacitive reactance always generates reactive power.
Example 7-3
An ac generator G
var
is
we sometimes
like reactive load, but
course, amounts to the
2
indeed flowing
now behaving
it
20
In conclusion, the ac generator
a varmeter into the circuit (Fig.
will give a negative reading of
showing
XC =
1
ergy (see Section 2.14).
If
3.5 il capacitor generates reactive power:
2
1
The R, L, C circuit generates a net reactive power
of 400 - 588 = 8 12" var
This reactive power must be absorbed by the
generator; hence, as far as reactive power is con-
it
However, instead of storing magnetic energy
again.
The
Qc =
7.3.
A building,
in
these devices.
shopping center, or
city
may
be con-
sidered to be a huge active/reactive load connected
an electric
Solution
to
The two
tain
resistors absorb active
P =
2
1
2
power given by
+ (16.12 2 X
784 + 520 = 304 W
R =
(14
X
4)
1
The
3 II reactor absorbs reactive
Q L = 1% =
l4
2
X
3
=
588 var
power:
2)
=
utility
system. Such load centers con-
thousands of induction motors and other
elec-
tromagnetic devices that draw both reactive power
(to sustain their
(to
magnetic fields) and active power
do the useful work).
This leads us to the study of loads that absorb
both active and reactive power.
AND APPARENT POWER
ACTIVE, REACTIVE,
7.7
Combined
and reactive
loads: apparent power
The
active
flow
in the
the arrows in Fig.
P
Loads that absorb both active power
power
tance
ple,
Q may
be considered to be
and reactive
made up of a
the circuit
actor are
of Fig.
7.
1
Oa
in
which a
resistor
and
p
,
while the reactor draws a current
/
q
According to our definitions, the resistor
active load
while the reactor
Consequently,
behind.
I
p
is in
is
phase with
The phasor diagram
an
a reactive load.
E while
(Fig. 7.
the resultant line current / lags
.
is
1
/t] lags
90°
Ob) shows that
behind
Furthermore, the magnitude of
E
/ is
and a varmeter
by an an-
7.
into the circuit, the readings will
both be positive, indicating
£/q
re-
connected to a source G. The resistor draws
a current /
gle 6.
resis-
and an inductive reactance. Consider, for exam-
power components P
same direction, as shown by
10c. If we connect a wattmeter
active and reactive
Q both
and
141
P = £V p
Q=
watts and
vans, respectively.
Furthermore,
if
we connect
an ammeter into the
will indicate a current of /
line,
it
sult,
we
amperes. As a
plied to the load
is
equal to EI watts. But
viously incorrect because the
an active component (watts)
is
nent (vars). For this reason the product EI
Apparent power
is
ob-
this is
composed of
and a reactive compopower
called
is
apparent power. The symbol for apparent power
given by
re-
power sup-
are inclined to believe that the
expressed neither
in
is
5*.
watts nor
but in voltamperes. Multiples are the kilo-
in vars,
voltampere (kVA) and megavoltampere (MVA).
7.8 Relationship
between P Q,
3
and S
Consider the single-phase circuit of Fig.
7.
1
1
a
com-
source]
posed of a source, a load, and appropriate meters.
Let us assume that
(b)
E
•
the voltmeter indicates
•
the
•
the wattmeter indicates
•
the varmeter indicates
ammeter
Knowing
indicates /
that
volts
amperes
+P watts
+Q vars
P and Q are
positive,
we know
that
the load absorbs both active and reactive power.
Consequently, the line current
an angle
Current
nents
I
p
/ lags
behind
E. lb
by
8.
/
and
rature, with
values of
/
can be decomposed into two compo/C]
,
respectively in phase, and
phasor
and
/ L|
E
(Fig. 7.1 lb).
in
quad-
The numerical
can be found directly from the
in-
strument readings
/
l
Figure 7.10
a.
Circuit consisting of
a source feeding an active and
reactive load.
Phasor diagram of the voltage and currents.
c.
Active
and reactive power flow from source
to load.
q
= PIE
= QIE
(7.1)
(7.2)
Furthermore, the apparent power 5 transmitted over
the line
b.
P
is
given by S
El,
I
=
from which
S/E
(7.3)
1
ELECTRICAL MACHINES AND TRANS EORMERS
42
Figure 7.11
Instruments used to measure £,
P, and Q in a circuit.
The phasor diagram can be deduced from the instrument readings.
a.
/,
b.
Referring to the phasor diagram (Fig. 7.1 lb),
it
is
Example
7-5
A wattmeter and varmeter are connected into a
obvious that
r-
2
=
/
p
single-phase line that feeds an ac motor.
2
+
/
q
spectively indicate
1
W and 960
800
20 V
1
They
re-
var.
Consequently,
Calculate
->
P
E
s
E
That
2
Q
+
The in-phase and quadrature components
a.
E
and
c.
in
0
2
(7.4)
d.
line current
which
=
P =
Q=
S
We
Solution
apparent power [VA]
Referring to Fig.
power [Wl
reactive power [var]
active
tor,
can also calculate the value of the angle 0 be-
cause the tangent of 6
Thus,
is
obviously equal to
/ //
p
/
a.
/
.
b.
p
q
=
arctan / //
q p
=
arctan
QiP
1
1
,
where the load
is
now
= PIE =
1
= QIE =
960/120
800/120
=
-
15
8
A
A
From
the phasor
= V7 2 +
Example 7-4
alternating-current motor absorbs
40
diagram we have
(7.5)
/
tive
7.
/
2
= Vl5 2 +
8
2
= 17A
kW of ac-
power and 30 kvar of reactive power. Calculate
power supplied to the motor.
c.
The apparent power
is
the apparent
S =
EI=
120
X
17
= 2040 VA
Solution
S
= VP^Vq 2
~
V40 r + 30 2
= 50 kVA
d.
a
mo-
we have
we have
0
An
/
q
The line current /
The apparent power supplied by the source
The phase angle between the line voltage and
b.
is,
2
2
S = P +
/
The phase angle 6 between E and
/ is
(7.4)
6
=
arctan
=
28.1°
QIP =
arctan 960/1800
(7.1)
(7.2)
ACTIVE, REACTIVE,
The power
Example 7-6
A voltmeter and ammeter connected
tive circuit
into the induc-
of Fig. 7.4a give readings of 140
V
and
AND APPARENT POWER
factor of a resistor
cause the apparent power
tive
power.
On
it
equal to the ac-
is
the other hand, the
an ideal coil having no resistance
20 A, respectively.
100 percent be-
is
draws
143
power
factor of
zero, because
is
it
does not consume any active power.
Calculate
To sum
The apparent power of the load
The reactive power of the load
a.
b.
is
parent
The active power of the load
c.
up, the
power factor of a circuit or device
simply a way of stating what fraction of
power
is real,
In a single-phase circuit the
The apparent power
S
and current. Thus, referring
is
Q =
factor
also
is
to Fig. 7.11,
PIS
=
El p IEI
is
=
v
=
=
cos B
El
= 2800
would give a reading of 2800
factor
X 20
140
var
power
=
2.8 kvar
Consequently,
vanneter were connected into the circuit,
If a
power
=
= EI= 140 X 20
= 2800 VA = 2.8 kVA
The reactive power
b.
ap-
a measure of the phase angle 6 between the voltage
Solution
a.
its
or active, power.
power
it
factor
=
cos 6
=
P/S
(7.7)
var.
where
The active power
c.
is
zero.
power
If
a
wattmeter were connected into the
circuit,
factor
0
but
because the current
voltage,
7.9
it
is
is
also equal to
is
2800 VA,
2800
voltage and current
is
the ratio of the active
power
S. It is
power P
to the ap-
said to be lagging
factor
=
leading
(7.6)
power
factor
is
said to be
the current leads the voltage.
in
Example
7-5 and the phase angle between the line voltage
power delivered or absorbed by
circuit or device [W]
active
apparent
Power factor
a
if
Calculate the power factor of the motor
and
=
the current lags behind the
Example 7-7
PIS
where
S
if
voltage. Conversely, the
given by the equation
power
P =
phase angle between the
If we know the power factor, we automatically
know the cosine of the angle between E and / and,
hence, we can calculate the angle. The power factor
var.
The power factor of an alternating-current device or
parent
=
90° out of phase with the
Power factor
circuit is
factor of a single-phase
circuit or device
would read zero.
To recapitulate, the apparent power
= power
it
is
power of
the circuit or device
line current.
the
Solution
[VA]
power
factor
expressed as a simple number, or as
=
PIS
=
1800/2040
-
0.882 or 88.2%
percentage.
Because the active power
apparent
power
S,
it
P can
(lagging)
never exceed the
follows that the power factor
can never be greater than unity (or
100 percent).
cos 0
=
0.882
therefore, 0
=
28.1°
ELECTRICAL MACHINES AND TRANSFORMERS
144
Example 7-8
A single-phase
20 V, 60 Hz
1
motor draws a current of 5 A from a
The power factor of the motor is
line.
65 percent.
Calculate
a.
b.
The
The
power absorbed by the motor
power supplied by the line
active
reactive
Solution
a.
The apparent power drawn by
=
5m
EI
=
120
X
5
the
motor
is
- 600 VA
^(390 W)
The
power absorbed by
active
the motor
is
P m = S m cosQ
= 600 X
b.
The
0.65
- 390
power absorbed by
reactive
(7.7)
Figure 7.12
Power triangle
is
4.
the
motor
VSI^
the
that
power from
amount of nonpro-
Power
Active power
is
considered
P absorbed by
to
Active power
device
is
P
a circuit or device
be positive and
that
is
drawn
hori-
Reactive power
vice
is
delivered by a circuit or
considered to be negative and
drawn horizontally
3.
is
Q
upwards
is
to the left
when
is
drawn
circuit
is
Example 7-8
and
Q
shown in
The power
is
rules.
look like phasors, but they
The concept of
the
power
as convetriangle
is
solving ac circuits that comprise sev-
and reactive power components.
Further aspects of sources
and loads
1
Let us consider Fig. 7.13a in which a resistor and
capacitor are connected to a source.
The
circuit
similar to Fig. 7. 10 except that the capacitor
As
power flows from
is
a
is
re-
a result, reactive
power flows
G
while active
to the source
the source
active and reactive
G
to the resistor.
power components
The
therefore
flow
in
opposite directions over the transmission
line.
A
wattmeter connected into the circuit
give a positive reading
ter will
absorbed by a circuit or de-
considered to be positive and
vertically
S, P,
from the capacitor
zontally to the right
2.
triangle for
However, we can think of them
nient vectors.
useful
by a
downwards
accordance with these
active source.
convention, the following rules apply:
.
are not.
7.1
triangle
2
2
2
The S = P + Q relationship expressed by Eq. 7.4,
brings to mind a right-angle triangle. Thus, we can
show the relationship between S, P, and Q graphically by means of a power triangle. According to
1
in
eral active
ductive power.
7.10
2
1
that is delivered
considered to be negative and
vertically
components
the line than active power. This burdens
the line with a relatively large
is
The power
Fig. 7.
motor draws even more reactive
Q
Reactive power
drawn
(7.4)
= V600 2 - 390 2
= 456 var
Note
a motor. See Example 7-8.
W
or device
Qm =
of
G
P = EI p
power Q. Thus,
Q—
ElLy The source
power P but receives reactive
give a negative reading
delivers active
will
watts, but a varme-
G
is
simultaneously an active
source and a reactive load.
ACTIVE, REACTIVE,
AND APPARENT POWER
145
device (or devices) connected to the receptacle.
If
the device absorbs active power, the receptacle
will provide
G
(a)
E
(source)
if
it;
the device delivers active power,
the receptacle will receive
7
p
simple receptacle outlet
7
g
Y
liver
tive
—
or accept
power
nected to
Q
—
In other words, a
it.
is at all
either active
times ready to de-
power P or
reac-
accordance with the devices con-
in
it.
The same remarks apply
to
any 3-phase 480
V
service entrance to a factory or the terminals of a
(b)
kV
high-power 345
Example
A 50
jjiF
transmission
line.
7-9
paper capacitor
terminals in
Example
is
placed across the motor
7-8.
Calculate
a.
(c)
b.
c.
d.
The reactive power generated by the capacitor
The active power absorbed by the motor
The reactive power absorbed from the line
The new line current
Solution
a.
Figure 7.13
a.
The impedance of
Source feeding an active and reactive (capacitive)
Xc =
load.
b.
Phasor diagram
c.
The
active
of the circuit.
and reactive powers flow
in
opposite
di-
the capacitor
is
1/(2 77/C)
=
1/(217
=
53
(2.11)
x 60 x 50 x
1()"
6
)
il
rections.
The current
It
in
may seem unusual
to
we must
line,
but
P
not the
is
/
have two powers flowing
opposite directions over the
remember that active power
a reactive power Q and that
again
same
as
The
reactive
each flows independently of the other.
tacle in
a
home, also deserves our
such outlets are ultimately
alternators that
power
Qc
active
also
20
V
attention. All
connected
may seem,
an
re-
source (as
we would
as
it
expect), but
behave as an active or reactive load.
factors
huge
to the
can act not only as an active or
Odd
determine whether
way or the other?
It all
it
will
behave
A
£/
-
q
-
120
X
the capacitor
2.26
271 var
recep-
the electrical transmission
and distribution systems.
electrical outlet
1
2.26
is
120/53
power generated by
Speaking of sources and loads, a deceptively
simple electrical outlet, such as the
capacitor
= EIXC =
=
same transmission
in the
it
may
What
in
one
depends upon the type of
b.
The motor continues to draw the same
power because it is still fully loaded.
active
Consequently,
P m = 390
The motor
also draws the
W
same
reactive
power
as before, because nothing has taken place to
change
its
magnetic
field.
Qm =
Consequently,
456 var
is
ELECTRICAL MACHINES AND TRANSFORMERS
146
The motor draws 456 var from
c.
the line, but the
same
power drawn from the
capacitor furnishes 271 var to the
line.
The
line
net reactive
is,
therefore,
Ql = Q m - Q c
= 456 =
The
active
185 var
power drawn from
P\.
d.
271
the line
= 390 W
=
The apparent power drawn from
- V390 +
- 432 VA
2
The new
1
the line
/W390 W)
is
Figure 7.14
Power triangle of a motor and capacitor connected
an ac line. See Example 7-9.
2
85
432/120
nected
in
7. 15a).
We wish
A to 3.6 A by
This represents a big improvement because the line
smaller and the operation of the motor has
factor of the line
4),
= /ys L =
=
<|>
L
=
390/432
We
concerned) of active and reactive power flow
arcos 0.903
=
delivers reactive
arrow
active and reactive
active
powers
power
P.
We
rather
complex
inductive,
cir-
Consider, for example, a group of loads con-
On
to the system.
it
ar-
the other
represents a capacitor,
power
it
The 16 kvar
in
independent) nature of the
powers enables us
to
add
all the
a circuit to obtain the total active
same way, we can add the reactive
power Q. The reapparent power S is then found by
In the
to obtain the total reactive
S
The concept of active and reactive power enables us
some
C
distinct (and
Systems comprising
simplify the solution of
is
directed therefore toward the source.
is
The
powers
the line.
several loads
to
A
the source to the load.
25.5°
sulting total
cuits.
Thus, because load
hand, because load
observe the effect of the capacitor on the ap-
7.12
to
simply draw a block diagram of the individual
row flows from
0.903 or 90.3%
power supplied by
ab-
absorbs reactive power; consequently, the 5 kvar
The power triangle is shown in Fig. 7,14. The repower Qc generated by the capacitor is
drawn vertically downward. By comparing this
power triangle with that in Fig. 7.12, we can visuparent
power
Using the power approach, we do not have
is
is
active
ally
V source (Fig.
worry about the way the loads are interconnected.
(Fig. 7.15b).
cos
380
to a
loads, indicating the direction (as far as the source
not been changed in the least.
The new power
way
to calculate the apparent
by the source.
placing the capacitor in parallel with the motor.
is
a very unusual
sorbed by the system as well as the current supplied
Thus, the line current drops from 5
current
to
line current is
= SJE =
= 3.6 A
/j
is
recall that
=
\
P2
-\
when adding
Q
1
(7.4)
reactive powers,
we
assign a positive value to those that are absorbed by
the system
and a negative value
to those that are
generated (such as by a capacitor). In the same way,
AND APPARENT POWER
ACTIVE, REACTIVE,
Reactive power absorbed by the system:
2.
380 V>
14kW
D
A
8 kvar
9 kvar
(a)
0
+
(5
+
7
Q 2 = (-9 Q
Net reactive power
4.
- -25
16)
Q = + 20
kvar
kvar
absorbed by the system:
=
25)
(
16 kvar
= +20
8)
Reactive power supplied by the capacitors:
3.
5 kvarf
=
G,
2kW
147
5 kvar
kW
8
Apparent power of the system:
5.
S
=
\
P2
7Q
-
24.5
kVA
V
Because the 380
6.
=
1
V24^+7~ 5?
source furnishes the appar-
ent power, the line current
=
I
7.
The power
cos
=
4> L
power, but
it
tem of the
netic
PIS
=
=
500/380
=
24/24.5
64.5
A
is
0.979 (leading)
kW
source delivers 24
of active
receives 5 kvar of reactive power. This
power flows
reactive
comes
= 24
factor of the system
V
The 380
(b)
S/E
is
into the local distribution sys-
company, where
electrical utility
available to create magnetic fields.
fields
may be
it
be-
The mag-
associated with distribution
transformers, transmission lines, or even the elec-
8 kvar
tromagnetic relays of customers connected to the
Figure 7.15
a.
b.
Example of active and reactive loads connected to
a 380 V source.
All loads are assumed to be directly connected to
the
same distribution system.
The power triangle for the system
7.15c.
It
is
we
Thus, starting with the 5 kvar load,
380 V receptacle.
is
shown
move from one
progressively
device to the next around the system.
While so doing, we draw the magnitude and
we
assign a positive value to active
powers
that are
absorbed and a negative value to those that are generated (such as
by an
Note that usually
powers
we cannot add
power
total
apparent
their
power factors are
Let us
I.
now
Active
the apparent
various parts of a circuit to obtain the
in
S.
We
can only add them
power absorbed by
if
meet.
7.
1
5:
(2
+
8
+
14)
right) of
point,
direc-
each power vector,
tail
accordance with the power of each device
When
we
can
starting point to the
end
the selection
draw a power vector from the
is
complete,
which yields the inclined vector having a value
of 24.5 kVA. The horizontal component of
the right,
24
we know
kW and, because
that
by the system. The
it
represents
vertical
it is
this vec-
directed to
power absorbed
component of
5 kvar
is
the system:
directed
P =
left,
tor has a value of
identical.
solve the circuit of Fig.
to head, in
we
alternator).
down,
tion (up,
in Fig.
the graphical solution to our problem.
= +24
kW
tive
downward: consequently,
power generated by
it
the system.
represents reac-
1
48
ELECTRICAL MACHINES AND TRANSFORMERS
5 kvar
starting
point
Figure 7.15c
Power triangle
of the system.
7.13 Reactive
magnetic
We
with the rapidly operating switch rather than with
power without
the resistor
fields
sometimes encounter situations where loads
absorb reactive power without creating any magnetic field at all. This can happen in electronic
power circuits when the current flow is delayed
by means of a rapid switching device, such as a
itself.
Nevertheless, reactive power
consumed just
as surely as
in the circuit.
This switching circuit will be
cussed
in detail in
if
is
a reactor were present
dis-
Chapter 30.
7.14 Solving AC circuits using
the power triangle method
thyristor.
Consider, for example, the circuit of Fig.
7.
16 in
We have seen that active and reactive powers can be
60 Hz source is connected to a resistive load of 10 (2 by means of a sy nchronous mechanical switch. The switch opens and closes its
rather
contacts so that current only flows during the latter
tation.
which a 100
V,
We can
some
to vector (j) no-
calculate the active and reactive powers
associated with each circuit element and deduce the
corresponding voltages and currents. The following
see,
if
we connected
a
wattmeter and varmeter between the source and the
(sometimes called displacement power factor)
of 84.4 percent. The reactive power
example demonstrates the usefulness of
power
this
triangle approach.
W
would respectively read +500
and
corresponds to a lagging power fac-
+ 318 var. This
tor
draw a phasor diagram or resorting
We
to solve
ac circuits without ever having to
almost by intu-
behind the voltage. Indeed,
switch, they
complex
forced delay causes the current to lag
part of each half cycle.
ition, that this
added algebraically. This enables us
is
associated
Example 7-10
In Fig. 7.17a, the voltage
is
60
V.
between terminals
1
and
3
ACTIVE, REACTIVE,
AND APPARENT POWER
149
0)^
7.07
A
(eff
5ft
(a)
12H R
60 V
T
©6
(a)
141
V
S
14.1
©
—
A
9-
700
=
r
VA
S
--^
*l
=
780
VA
1
r^rs_+_
I
(b)
60 V
Figure 7.16
a.
Active
and
reactive
power flow
in
a switched
resis-
tive load.
b.
Figure 7.17
The delayed current flow is the cause
tive power absorbed by the system.
of the reac-
a.
Solving ac circuits by the power triangle method.
b.
Voltages and currents
in
the
circuit.
See Example
7-10.
from which the active power absorbed
Calculate
a.
The current
b.
The voltage between terminals
c.
The impedance between terminals
in
each circuit element
I
P =
and 2
l
S
Solution
We know the impedances of the elements and
(Fig. 7.
60 V exists between terminals 3 and
We now proceed in logical steps, as follows:
a.
The current
in
the capacitor
Ic
=
60/5
=
The current
12
=
X 60 - -720
60/12
=
7b).
The
current
5
A
/,
=
must, therefore, be
=
S/E 3]
The voltage across
A
in the resistor is
/R
= VP + Q = V300 + (-720) 2
= 780 VA
/1
from which the reactive power generated
Qc =
1
that
E 23 = IXL
780/60
=
A
13
the inductive reactance
=13X8=
104
is
V
is
The
var
terminals 1-3:
2
1
2
is
12
W
The apparent power associated with
and 2
I
X 60 = 300
5
is
reactive
reactance
power absorbed by
the inductive
is
G L = E23 X
= +1352
/,
=
var
104
X
13
1
ELECTRICAL MACHINES AND TRANSFORMERS
50
The
power absorbed by
total reactive
Qi. + Qc =
= +632 var
Q=
the circuit
1.25 Mvar
is
1352
0.2
MW
T
- 720
]—
substation
The
total active
power absorbed by
the circuit
15
is
12.47 kV
3
The apparent power absorbed by
2.8
MW
0.75 Mvar
2 Mvar
2
The voltage of
£2 =
,
c.
is
C
10.03 kV
289 A
MW
= \P + Q 2 = V300 2 + 632 2
= 700 VA
S
b.
the circuit
load
Q
2,4
W
P = 300
L2
the line
=
SII
{
10.03 kV
therefore
is
700/13
=
V
53.9
The impedance between terminals 2-1
Z=
E 2\U\. =
12.47 kV
53.9/13
=
289 A
is
4.15 II
289 A
Figure 7.18
Voltages, currents and power.
See Example
7-1
1
Example 7-11
A
kV
single-phase 12.47
transmission line several
C from
kilometers long feeds a load
(Fig. 7.18).
The
reactance of 15
Instruments
at the
inputs to
6
=
arccos 0.833
power dissipated
b) Active
MW and 2 Mvar, respectively.
P = RI2 =
=
Calculate
the line current and
phase angle with respect
its
X
0.2
=
33.6°
in the line:
2.4
Y
a)
at the
substation:
substation in-
power
dicate that the active and reactive
the line are 3
Phase angle between the voltage and current
has a resistance of 2.4 il and a
line
il.
a substation
10
6
X 289 2
= 0.2 MW
Active power absorbed by the load:
to the line voltage at the substation
b)
the active
—
power absorbed by the load
power absorbed by the load
c) the reactive
rL
^sub
=
2.8
MW - 0.2 MW
3
MW
d) the line voltage at the load
e) the
and
phase angle between the voltage
at the load
c)
Q L = XJ =
Solution
a)
Reactive power absorbed by the
2
that at the substation
15
X
289
2
=
1.25
X
line:
10
6
=
1.25
Reactive power absorbed by the load:
Apparent power delivered
VP
S
to the line:
V3 +
+ Q2 =
2
3.60
2
2
Qc
2
=
Qsub
- Gl =
=
0.75
Mvar
2
Mvar -
1.25
Mvar
MVA
d)
Apparent power
at the load:
Line current:
/
=
S
=
3
600 000
E
Power
Mvar
12
470
VA
Sc
=
V
289,4
Voltage
factor at the substation
FP =
P
S
=
3
3.6
MW
MVA
=
0.833
=
-
+ Ql = \fl&
2.90
at the
Ec=
0.75
2
MVA
load end of the line:
Sc
/
2.90
=
MVA
289
A
10.03
kV
AND APPARENT POWER
ACTIVE, REACTIVE,
Power factor
1
5
load end of the line:
at the
P,
2.8
S,
2.90
FP
MW
MVA
=
0.965 ou 96.5%
p
e
—
s=
E.dh
r
*|b
Phase angle between the voltage and current at the
rest of circuit
j
load:
(a)
=
9C
It
=
arccos 0.965
15.2°
follows that the phase angle between the voltage
the substation
at
15.2°)
=
We
that at the load
is
—
(33.6°
18.4°.
Fig. 7.
sis.
and
summarizes the
8
1
S = + E\I*
results of this analy-
could have found the same values using
However, on account of its simpower method of solving this problem
vector algebra.
the
plicity,
is
(b)
very appealing.
7.15
If
rest of circuit
Power and vector notation
vector notation
is
used to solve an ac
can readily determine the active
with
associated
We
sources.
component,
any
current that
P
is
the active
value for
rest of circuit
or
Q
(c)
The vector product
it.'
power S in terms of P + jQ,
power and Q the reactive
P
E4 I*
E
power absorbed (or delivered) by the component.
A positive
S=-
the
the conjugate (/*) of the
flows through
£/* gives the apparent
where
including
simply multiply the phasor voltage
component by
across the
we
circuit,
and reactive power
means
same sequence ab
com-
that the
Figure 7.19
Method of writing power equations.
Z
associated with
The apparent power S
(not ba).
is
therefore written
ponent absorbs active or reactive power. Negative
values
mean
reactive
component
that the
power.
When
It
calculating the £7* vector product,
procedure
very important to follow a standard
der to obtain the correct result.
plies to circuits that
tion or the
is
culate the active
nal
a
Z We
to
7.
The procedure
ap-
1
9a
in
which
a circuit element
of circuit."
We
want
to cal-
and reactive power associated with
note that current
terminal
Consequently,
the
in or-
would be incorrect
In Fig. 7.
that
b,
when
i.e.
in
/
flows from termithe
sequence ab.
calculating the product
subscripts of voltage
E
must be written
in the
a
current has a value
/^B,
its
conjugate /*
= IL~ B.
S = £ ba /.*
to write
9b, sign notation
current
enters
/
Z
is
by
used, and
the
Consequently, the apparent power
S
element
In the case of Fig. 7. 19c,
we want
to,
)
is
seen
given by
is
a
(
+
)
(
+
the
in Fig. 7.
1
we
Zby
write
the
(
—
sign because
)
terminal of
S = — £4 /* be)
terminal.
we can determine
power associated with
from b
(
it
terminal.
Z.
cause the current enters
If
+
= +£,/*
The £]/* product is preceded by
current / is shown as entering by
Thus,
'If
1
use the double subscript nota-
part of a larger "rest
element
is
sign notation (see Sections 2.4 and 2.5).
Consider Figure
Z
it
= £ah /*
S
delivers active or
the apparent
the "rest of circuit" (roc).
9a, because the current circulates
to a in the roc,
we would
write:
1
ELECTRICAL MACHINES AND TRANSFORMERS
52
we would
Similarly, in Fig. 7.19c,
write
Let us illustrate the procedure by a few examples.
Example 7-12
In the circuit
2
of Fig.
the following values are
7. 19c,
given
Figure 7.20
E4 = 70 Z25°
See Example
=
I
7-13.
4 Z40°
Calculate the active and reactive power associated
with element
Z.
The voltage across
Solution
We
have
/
= 4 Z40°;
therefore I*
Since the current flows into the
power equation must bear
5
a
—
(
—
(
)
=
4
the capacitor
E y2 + /(— IO7) =
Z-40°
-£4/*
~
= -7()Z25° X 4 Z.-40 0
= -280 Z-15°
Current
=
terminal
+7
28()(cos(-15°)
= -270.5 +7
sin
(-15°))
in the
We
active
10./
x
=
24.6
Z(-47° +
=
24.6
Z43°
2.46
Z -47°
90°)
capacitor flows from terminal 2 to
with the capacitor
is
72.5
S =
W and Q
~ -270.5
1
Consequently, the power associated
3.
= P+.iQ
Thus P
0
10 J
sign:
)
given by
=
=
£32
terminal, the
is
= +72.5
var
W
that element Z delivers 270 .4
of
power and absorbs 72,5 var of reactive
conclude
£23 /*
= -24.6 Z43° X
= -60.5 Z90°
2.46
- -60.5 (cos
0 - 60.5 7
+
90°
Z47°
7 sin 90°)
=
power.
= P +JQ
Example 7-13
Given the
circuit
.
of Fig. 7.20
which
in
E l2 =
30 Z78°,
determine the power associated with the capacitor
whose reactance
is
Hence P — 0 and Q = —60.5; Consequently, the acpower associated with the capacitor is zero, and
tive
it
delivers 60.5 var of reactive power.
10 O.
Example 7-14
Solution
Going cw around
the circuit,
we can
write (see
The
circuit in Fig. 7.21
sistor
Sections 2.32 to 2.39)
connected
actance.
E2] ~
7
1(1
E2{
_
- 10/ ~
= -2.46 Z
-
10 7)
12.5
133°
-
30
Z
0
a 45 i!
rere-
The source generates
by the phasor
a voltage described
E ab = 159Z65°.
78°
Z -55°
= +
composed of
with a 28 (1 inductive
is
in series
2.46
Z-47°
Calculate
a.
The magnitude and phase of the current
/
AND APPARENT POWER
ACTIVE, REACTIVE,
4512
a
Voltage across the reactance
Ecb =
=
159
[65_°
£\ib
'
28
j28
c.
is
/
X 3^33.11°
j28
= 84^(33.11° +
= 84^123.11°
n
153
90°)
The conjugate /* of the current
/ is
= 3^-33.11°
/*
The apparent power associated with
the resistor
is
Figure 7.21
S r = E.J*
Solving an ac circuit using vector notation.
= (135^33. 11°) (3^-33.11°)
= 405^0°
b.
and across the reactance
the resistor
c.
The active and reactive power associated with
Applying Kirchhoff 's voltage law (see Section
2.32),
we
because there
sin 0°)
jO)
£ba + E ac + Ecb =
-£ab +
-159^65° +
we
phase angle
/
=
+
-
j28
159
53
L
45/
+
j28/
=
0
/(45
+
j28)
-
0
/
-
S
r
=
159
L
45
45
+
j28
45
=
252/190°
-
252 (cos 90°
= 252
-
+
(0
+
j
sin 90°)
jl)
j252
Thus, the reactance absorbs only reactive power
(252
var).
=
65°
3
/ (65°
31.89°
£ ac = 45
(84 Z_123. 11°) (3^-33.11°)
The apparent power associated with
- 53^31.89°
Z_
the reac-
65°
+ 28 = 53
= arctan 28/45 = 31.89°
\
W)
.
=
=
Voltage across the resistor
=
in
2
2
~
~EJ
A
31.89°)
= -477^(65° is
/
X 3^33.11°
135zl33.ll
E\vJ''~
the source
*
= -(159^65°) (3^-33.11°)
= 3^33.11°
=
power (405
real
0
obtain
amplitude
hence 45
no j component
is
Transforming the denominator into polar coordinates,
is
The apparent power associated with
obtain
tance
b.
+
(1
j
Thus, the resistor absorbs only
Solution
and so
+
(cos 0°
= 405
and the source
the resistor, the reactance,
a.
= 405
- 405
The magnitude and phase of the voltage across
0
33.11°)
= -477^31.89°
= -477
(cos 31.89°
- -477
(0.849
= -405 -
j
+
252
j
+
j
sin 31.89°)
0.528)
is
ELECTRICAL MACHINES AND TRANSFORMERS
154
The active and reactive powers are both negative,
which proves that the source delivers an active
power of 405
and a reactive power of 252 var.
W
on sources and loads
7.16 Rules
(sign notation)
We
are often interested in determining whether a de-
vice
an active/reactive source or an active/reactive
is
load without
such as
sis,
making
a complete mathematical analy-
we performed
Section
in
To enable
7. 15.
Figure 7.22
us to positively identify the nature of the source or
load, consider Fig. 7.22 in
line current
/.
The device
The voltage between
which a device
is
part of a circuit.
the terminals
+
A carries a
(
is
£,
and
that are respectively parallel to,
to E. Let /
E.
p
be the component of
out of phase with E.
behind or 90° ahead of
The
1
.
A
voltage
b.
the line current
+
)
A device
when double
/,
re-
is
is
an active source
subscript notation
used. Consider Figure 7.23 in which a device
ries a current /
flowing
in the direction
b
is
enable us to state
3.
/ is
shown
in
phase and
A device
is
shown. The
£ab The
.
voltage
b.
line current /
shown
is
in
phase and
as entering terminal
Otherwise, the device
is
an active source.
A device
a.
is
current
/
is
an active source.
rule also applies:
a reactive load
when
lags 90° behind voltage E. xh
rule also applies:
a reactive load
/c]
when
lags 90° behind voltage
line current /
is
shown
E and
as entering the
(
+)
terminal.
Otherwise, the device
Based on these
rules,
tionships in Fig. 7.22,
active load because
A is
/
p
is
a reactive source.
and observing the phasor
we deduce
is in
that device
A
relais
an
phase with E. Also, device
a reactive source because /
q
is
90° ahead of E.
Figure 7.23
Same
These
rules are in
fol-
an active load when:
£ab and component / p are
a.
The following
as entering the
agreement with IEEE and IEC conventions.
is
A car-
lowing rule applies:
4.
component
b.
whether a device
or active load
terminal.
The following
a.
We can also tell
rule applies*:
Otherwise, the device
2.
on sources and loads
(double subscript notation)
an active load
a.
(
and
rela-
/.
voltage between terminals a and
when
E and component / are
is
between Eand
7.17 Rules
an active load or an active
is
The following
device
in
E.
E
between
whether a device
source.
angles
diagram, together with the phasor
circuit
lationships
at right
/ that is parallel to
phase with, or 180°
Similarly, / can be either 90°
be either
will therefore
It
active/reactive source or
depending upon the phasor
and one
sign.
)
A may be an
active/reactive load
tionship
The phase angle
between E and / can have any value. As a result, /
can be decomposed into two components, I p and /
of the terminals bears a
Device
circuit
as
in Fig.
subscript notation
is
7.22 except that double-
used.
and
a.
ACTIVE, REACTIVE,
b.
shown
line current / is
nal
as entering by termi-
c.
Otherwise, the device
a reactive source.
is
we deduce
tionships in Fig. 7.23,
active
source because
A
Also, device
90° behind
Eab
is
I
p
is
1
that device
A
an
is
80° out of phase with E, lb
a reactive load because /
q
e.
7-1
1
Using the rules given
7. 17,
.
lags
Sections 7.16 and
in
determine which of the devices
in Fig.
7.24a through 7.24f acts as an active (or
power
active)
.
7-12
Questions and Problems
55
The peak power output of the reactor
The duration of each positive power pulse
d.
Based on these rules, and observing the phasor rela-
1
The reactive power absorbed by the reactor
The apparent power absorbed by the reactor
The peak power input to the reactor
a.
b.
a.
AND APPARENT POWER
re-
source.
A single-phase motor draws a current of
2 A at a power factor of 60 percent.
1
Calculate the in-phase and quadrature comPractical level
7-1
What
ponents of current
power? reactive
the unit of active
is
power? apparent power?
7-
7-2
7-3
A
capacitor of 500 kvar
7-5
placed
1
3
in parallel
A
240
V,
60 Hz
line.
A
1
wattmeter
line gives a reading of
the
Name
motor and
a static device that
can generate reac-
power.
Name
7- 14
a static device that absorbs reactive
If a
the reactive
in parallel
Problem 7-13,
What
a.
is
the approximate
power
factor, in
The current
in
What
is
the
power
large
The apparent power of
The line current
e.
The power
7- 5
1
motor absorbs 600
kW at a power
is
An
a.
b.
is
connected across a
1
20
A
10 12 reactance
60 Hz
line.
is
V,
connected to a 120 V,
Calculate:
induction motor absorbs an apparent
at
a
power
factor of 80
percent. Calculate:
generates.
10 (2 resistor
and without drawing any
power of 400 kVA
60 Hz source. Calculate the reactive power
A
14)
circuits in Fig. 7.25.
connected to a 240 V,
60 Hz source. Calculate:
a. The active power absorbed by the resistor
b. The apparent power absorbed by the resistor
c. The peak power absorbed by the resistor
d. The duration of each positive power pulse
7.
phasor diagrams, find the impedance of the
7-16
machine.
it
factor of the motor/capacitor
Using only power triangle concepts
(Section
90 percent. Calculate the apparent
power and reactive power absorbed by the
p,F capacitor
the ac line
combination
factor of
A 200
calculate:
active
d.
motor?
Intermediate level
A
is
with the motor of
power reading of the wattmeter
The total reactive power absorbed by the capacitor and motor
The
b.
a single-phase motor lags
power factor of the
power it absorbs.
capacitor having a reactance of 30 (2
connected
power.
factor of the
7-10
a
2765 W. Calculate
50° behind the voltage.
7-9
to
connected into the
c.
7-8
with respect
apparent power of the group.
candescent lamp?
1-1
/
single-phase motor draws a current of
A from
percent, of a capacitor? of a coil? of an in-
7-6
and
with an inductor of 400 kvar. Calculate the
tive
7-4
is
/
the line voltage.
c.
7-17
A
power absorbed by the motor
power absorbed by the motor
What purpose does the reactive power serve
The
The
active
reactive
circuit
composed of a
1
2 ft resistor in
series with an inductive reactance of 5 (2
carries an ac current of 10 A. Calculate:
a.
The
active
b.
The
reactive
c.
d.
power absorbed by the resistor
power absorbed by the inductor
The apparent power of the circuit
The power factor of the circuit
(0
(b)
(a)
G
E
F
/
/
(d)
(e)
Figure 7.24
See Problem
7-11.
Hf-
1
40 V[
^
ion
4I2(
]
en
5A
J
X
(c)
(b)
(a)
Figure 7.25
See Problem
7-18
7-15.
A coil
having a resistance of 5 ft and an
ductance of 2
H
7-2
in-
carries a direct current of
20 A. Calculate:
a. The active power absorbed
b. The reactive power absorbed
Advanced
7-19
7-20
A
ply voltage
is
connected
200
is
ft. If
the sup-
V, calculate:
a.
The
reactive
power absorbed by
The
reactive
power generated by
the coil
the capacitor
Problem
a
7- 3, if
1
in parallel
we
d.
7_22
power absorbed by
The
The apparent power of the
c.
The power
The power
V
the system
is
0.6 lagging (Fig. 7.26).
calculate:
sy stem
factor of the system
156
coil
factor at the terminals of a
source
Without using phasor diagrams,
with the motor, calculate:
a.
The active power dissipated by the
The apparent power of the circuit
120
place a capacitor of
b.
total active
c.
a.
The value of E
b.
The impedance of
the load
re-
in parallel with
b.
level
motor having
500 var
having a reactance of 1011 and a
a capacitive reactance of 10
power factor of 0.8 absorbs an active power of 1200 W. Calculate
the reactive power drawn from the line.
In
A coil
sistance of 2 ft
Z
ACTIVE, REACTIVE,
AND APPARENT POWER
157
29.
E=
120
V
»E= 120 V
Figure 7.26
/= 5 A
See Problem 7-22.
7-23
In Figs.
7.27a and 7.27b, indicate the mag-
150°
(b)
(a)
nitude and direction of the active and reactive
power flow.
and
/
and
,
q
treat
(Hint:
Decompose
/ into /
5
them independently.)
Figure 7.27
See Problem
Industrial application
7-24
A single-phase
A
p
7-23.
capacitor has a rating of
30 kvar, 480 V, 60 Hz. Calculate
its
capacib.
tance in microfarads.
The
active and reactive
power consumed by
the line
7-25
In
Problem 7-24
calculate:
c.
a.
The peak voltage across the capacitor when
it is connected to a 460 V source
b.
The
active, reactive
and apparent power ab-
sorbed by the load
d.
resulting energy stored in the capacitor
7-28
at that instant, in
The
joules
The voltage across
A2
hp,
230
V,
1
the load
725 r/min 60 Hz single-
phase washdown duty motor, manufactured
7-26
Safety rules state that one minute after a capacitor
is
disconnected from an ac
voltage across
discharge
that
is
it
must be 50
V
or less.
done by means of a
by Baldor Electric Company, has the
line, the
The
Full load current:
resistor
permanently connected across the
is
fol-
lowing characteristics:
efficiency:
capacitor terminals. Based on the discharge
power
1
1.6
A
75.5%
factor:
74%
curve of a capacitor, calculate the discharge
resistance required, in
ohms,
tor in
Problem 7-24. Knowing the
tance
is
when
weight: 80 lb
for the capaci-
a.
resis-
sorbed by this machine
subjected to the service voltage
the capacitor
is in
operation, calculate
b.
7-27
A
3.2 kV,
1
60 Hz single-phase
line con-
(1.
The metering equipment
at
the substation indicates that the line voltage
is
12.5
kV and
that the line
of active power and 2
is
drawing 3
Mvar of reactive
power. Calculate:
a.
The current flowing
operates
at
If a
40 microfarad capacitor
is
connected
current feeding the motor.
The
c.
MW
Will the presence of the capacitor affect the
temperature of the motor?
has a resistance of 2.4 II and a reac-
tance of 12
it
across the motor terminals, calculate the line
nects a substation to an industrial load.
line
when
full load.
wattage rating.
its
Calculate the active and reactive power ab-
7-29
A single-phase heater absorbs 4 kW on a 240
V line. A capacitor connected in parallel with
the resistor delivers 3 kvar to the line.
a.
Calculate the value of the line current.
b. If the capacitor
in the line
new
is
line current.
removed, calculate the
Chapter 8
Three-Phase
8.0 Introduction
8.1
Electric power
We
is
tributed in the
generated, transmitted, and dis-
form of 3-phase power. Homes
and small establishments are wired
power, but
this
power
ferred over single-phase
is
can gain an immediate preliminary understand-
piston
pre-
is
comparable
to a single-phase
the other hand, a 2-cylinder engine
for several impor-
is
Three-phase motors, generators, and transformers are simpler, cheaper,
and more
down
move
efficient
b.
Three-phase transmission lines can deliver
c.
more power for a given weight and cost
The voltage regulation of 3-phase transmission
inside
identical
They
in unison.
power
to deliver
rather than at the
running engine and a
A
cuits
knowledge of 3-phase power and 3-phase
is,
cir-
the
basic
different times.
circuit
will see that
most 3-phase
circuits
we
identical,
can be reduced to
time.
the reader
is
As
this
the reader
may know
produces a smoother
much smoother
output torque.
a 3-phase electrical system, the
As
a result, the total
one phase may be used
behavior of
power at
power flow is
all
to represent the
three.
Although we must beware of carrying analogies
elementary single-phase diagrams. In this regard,
we assume
not
way as
very smooth. Furthermore, because the phases are
techniques used to solve single-phase circuits can be
directly applied to 3-phase circuits. Furthermore,
do
cylinders, but they
three phases are identical, but they deliver
therefore, essential to an understanding of
power technology. Fortunately,
a
and
to the shaft in successive pulses
same
lines is inherently better
in
In
move up
are staggered in such a
from personal experience,
Similarly,
to
6-cylinder
engine could be called a 6-phase machine.
6-cylinder engine identical pistons
a.
machine. On
comparable
The more common
a 2-phase machine.
tant reasons:
common
A single-cylinder engine having one
gasoline engine.
merely represents a tap-off from the
power
Polyphase systems
ing of polyphase systems by referring to the
for single-phase
basic 3-phase system. Three-phase
Circuits
familiar with the previous
too
chapters dealing with ac circuits and power.
far,
system
158
the
is
above description reveals
basically
that a 3-phase
composed of three
single-phase
THREE-PHA SE CIRCUITS
systems that operate in sequence.
much
fact is realized,
Once
this basic
A
winding
stator
59
1
of the mystery surrounding
3-phase systems disappears.
Single-phase generator
8.2
Consider a permanent magnet
stant
NS
revolving
con-
at
speed inside a stationary iron ring (Fig. 8.1).
The magnet
is
driven by an external mechanical
source, such as a turbine.
The
Figure 8.2
ring (or stator) re-
duces the reluctance of the magnetic circuit; consequently, the flux density in the air
if
having terminals a,
insulated
ductors,
A
were absent.
the ring
from
one
in
it.
1
is
Each
each
Ea1 =
At this instant
gap
is
flux
does not cut
greater than
multiturn rectangular coil
mounted
0 because the
the conductors of winding A.
inside the ring but
voltage E. xi
maximum when
is
position of Fig.
turn corresponds to
two con-
8.
the poles are in the
because the flux density
1
On
greatest at the center of the pole.
slot.
the voltage
when
zero
is
is
the other hand,
the poles are in the position
of Fig. 8.2 because flux does not cut the conductors
winding
stator
A
at this
If
tion,
moment.
we
plot E. xl as a function of the angle of rota-
and provided the N, S poles are properly
shaped,
we
obtain the sinusoidal voltage
Suppose
Fig. 8.3. *
shown
in
the alternating voltage has a peak
value of 20 V. Machines that produce such voltages
are called alternating-current generators or syn-
chronous generators. The particular machine shown
Figure 8.1
in Fig. 8.
embedmaximum ( + ).
A single-phase generator with a multiturn
ded
in
two
slots.
At this instant
Ea1
is
called a single-phase generator
is
1
coil
v
+ 20
As
the
magnet
turns,
it
sweeps across the con-
inducing a voltage
in
them according
/
/
/
—y
\
X
\
\
10
ductors,
/
V
to the
/
/
\
equation:
/
\
£al =
B/v
(2.25)
wherein
-
/
KJO
\—
—
90
0
10
2
(
B = instantaneous flux density cutting across
conductors
I
=
=
the
\
\
\
\
\
\
3(i0 degrees 4S >0
/
\
\
—
\
f
/
/
\
\
-
\
/
\
/
20
/
in the slots [T]
Figure 8.3
length of conductors lying in the magnetic
field
v
V]
\
/
\
instantaneous voltage induced in the coil
'0
\
/
igle
\
=
/
N
Voltage induced
in
winding A.
[m]
peripheral speed of the revolving poles [m/s]
The poles shown
The sum of the voltages induced
in all the
ductors appears across the terminals.
con-
The terminal
voltage
in Fig. 8.
composed of
negative pulses.
1
would generate an
alternating
rather brief t'lat-toppped positive and
ELECTRICAL MACHINES AND TRANSFORMERS
160
Power output of a single-phase
generator
8.3
If
a resistor
rent will
is
connected across terminals
The current
/,,
is in
1
a cur-
phase with the voltage and, con-
sequently, the instantaneous
series of positive pulses, as
average power
power
trical
a,
flow and the resistor will heat up (Fig. 8.4).
is
is
power is composed of a
shown in Fig. 8.5. The
one-half the peak power. This elec-
derived from the mechanical power
provided by the turbine driving the generator. As a
result, the turbine
ergy
in pulses, to
must deliver
mechanical en-
its
match the pulsed
electrical output.
This sets up mechanical vibrations whose frequency
is
twice the electrical frequency. Consequently, the
generator will vibrate and tend to be noisy.
8.4
Two-phase generator
Using the same single-phase generator,
a
second winding (B) on the
let
us
mount
Figure 8.5
Graph of the
generator
stator,
is
voltage, current,
and power when the
under load.
identical to
voltage E. tl becomes zero and voltage
maximum
therefore out
by curves
load on phase
Note
A
in Fig.
£ aj
that
positive value
8.6b and by phasors
Eh2 because
before E h2 does.
stator
Figure 8.4
resistor.
its
is
it
in Fig, 8.6c.
reaches
its
peak
called a two-phase generator,
windings are respectively called
phase A and phase
Single-phase generator delivering power to a
attains
value.
leads
This machine
and the
E b2
The two voltages are
of phase by 90°. They are represented
positive
B.
Example 8-1
The generator shown
in Fig.
8.6a rotates
at
6000
r/min and generates an effective sinusoidal voltage
winding A, but displaced from
it
by a mechanical
of 170
V
per winding.
angle of 90° (Fig. 8.6a).
As
the
magnet
rotates, sinusoidal voltages are in-
Calculate
each winding. They obviously have the
a.
same magnitude and frequency but do not reach
maximum value at the same time. In effect, at
b.
duced
in
their
moment when the magnet occupies the position
shown in Fig. 8.6a. voltage
passes through its
c.
positive value, whereas voltage
zero. This
is
conductors
Eh2
is
an-
gle of 90°
the
maximum
The peak voltage across each phase
The output frequency
The time interval corresponding to a phase
Solution
a.
The peak voltage per phase
is
because the flux only cuts across the
in slots 1
after the rotor has
and a
at this instant.
made one
However,
quarter-turn (or 90°),
£ ni = \2E =
= 240 V
1.414
X
170
(2.6}
THREE-PHASE CIRCUITS
b.
One cycle
makes one
is
completed every time the magnet
The period of one cycle
turn.
T=
=
=
s
=
Power output
now
Let us
0.01
will flow in
phase with
/=
c.
A phase
terval
is
\/T=
means
Hz
100
1/0.01
angle of 90° corresponds to a time in-
Consequently, phasor
behind phasor
A and B
each
E. {i
(Fig. 8.7a). Currents
and
/.,
/h
They are respectively in
The currents are, therefore,
resistor.
and
Eh2
-
90° out of phase with each other (Fig. 8.7b). This
of one quarter-revolution, or 10 ms/4
2.5 ins.
connect two identical resistive loads
across phases
s
10 ins
The frequency
of a 2-phase
generator
1/6000 min
60/6000
8.5
is
161
£a]
E h2
lags 2.5
=
that
/.,
reaches
period before
now produces
/b
its
maximum value one quarter-
does. Furthermore, the generator
a 2-phase
power
output.
The instantaneous power supplied
ms
to
each
resis-
tor is equal to the instantaneous voltage times the
.
instantaneous current. This yields the two power
waves shown
of phase
in Fig. 8.8.
A is maximum,
Note
that
that
when the power
B is zero, and
of phase
we add the instantaneous powers of
we discover that the resultant power is
and equal to the peak power P m of one
vice versa. If
both phases,
constant
(a)
1
load on
(a)
phase
\
-h
(b)
0-
90.
if
i
360
21
—t
an
jle
of
450
_
r ot ation
6
i
load on phase B
(c)
n
(b)
Figure 8.6
a.
Schematic diagram of a 2-phase generator.
b.
Voltages induced
c.
Phasor diagram of the induced voltages.
in
a 2-phase generator.
Figure 8.7
Two-phase generator under load.
b. Phasor diagram of the voltages and currents.
a.
A
ELECTRICAL MACHINES AND TRA NSEORMERS
62
1
phase.* In other words, the total
2-phase generator
also constant.
is
benefit,
it
peak power =
it
is
to drive the
less noisy.
'I
1
instantaneous power of phase
in size,
i
except for the addition
of an extra winding.
Three-phase generator
A 3-phase generator is similar to a 2-phase generator,
except that the stator has three identical windings
stead of two.
placed
at
The
three windings a-1, b-2,
120° to each other, as
When
magnet
the
position
in-
maximum
Voltage
in Fig. 8.9a,
E b2
Consequently,
only voltage
will reach
its
Ea
is in
,
is
The
The
i
i
i
i
;
180
270
360
the
*-
angle of rotation
Total instantaneous
power output
it
is
used
to
current
is
one-third of a turn). Similarly, voltage
designate different things.
has to be read
in
context to be understood.
ways
the
in
Figure 8.8
Power produced by a 2-phase generator.
which
out of phase with the voltage (relets to pha-
tain its positive
peak
Ec3
will at-
after the rotor has turned
through 240° (or two-thirds of a turn) from
Consequently, the three stator voltages
three phases of a transmission line (the three conduc-
tors of the line)
£b2
and
,
The phase-to-phase voltage
4.
The phase sequence
(the line voltage)
(the order in
Ec3
They
120°.
3.
its ini-
position.
tial
sor diagram)
2.
90
0
at its
positive peak after
The following examples show some of
the word phase is used.
.
r
i
positive value.
The icrm phase
1
i
windings have the
the rotor has turned through an angle of 120° (or
:;:
i
in Fig. 8.9a.
At the moment when the magnet
shown
i
Instantaneous power of phase B
and c-3 are
effective values, but the peaks occur at differ-
ent times.
Ib =
rotated at constant speed,
is
the voltages induced in the three
same
Eb
. pov rer =
.peak
/MA,
i
shown
A
|
j
8.6
Pm
=
fa
A 2-phase generator does
i
without any increase
Ea
every instant. As a
As an important
produces twice the power output
not vibrate and so
added
power output of the
at
mechanical power needed
result, the
generator
same
the
is
—
are
and as phasors
—
al
,
are respectively out of phase by
shown
as sine
waves
in Fig. 8.9b,
in Fig. 8.9c.
which the phasors follow
each other)
5.
The burned-out phase
(the burned-out
winding of
a
3-phase
8.7
machine)
6.
The 3-phase voltage
(the line voltage of a 3-phase system)
7.
The 3-phase currents
are unbalanced (the currents in a 3-phase
8.
line
or machine are unequal and not displaced
The
pliase-shifi transformer (a device that
at
120°)
(a short-circuit
between two
line
and
E52
conductors)
Phase-to-ground fault
six wires to deliver
phase loads (Fig.
The phase-to-phase fault
(a short-circuit
between
a line or
/c
,
8.
1
power to the individual singleThe resulting currents /.„ / b
0a).
,
are respectively in phase with voltages
and
EcV
Because the
currents have the
same
1.
The phases are unbalanced
(the line voltages, or the line
£al
,
resistors are identical, the
effective values, but they are
winding and ground)
1
to
can change the
voltage)
10.
Let us connect the three windings of the generator
three identical resistors. This arrangement requires
phase angle of the output voltage with respect to the input
9.
Power output of a 3-phase
generator
mutually out of phase by
1
20° (Fig.
8.
1
that they are out of
other)
reach their positive peaks
The
0b).
phase simply means
currents, are unequal or not displaced at 120° to each
fact
that they
at different times.
THREE-PHASE CIRCUITS
The instantaneous power supplied to each resisis again composed of a power wave that surges
tor
between zero and a
(a)
163
maximum
power peaks in the
the same time, due to
P nv However,
value
do not occur
the
three resistors
at
the phase angle
between the
we add the instantaneous powers of all
resistors, we discover that the resulting power
voltages. If
three
is
T
1
Eai
constant, as in the case of a 2-phase generator.
However, the
1
^c3
SN
\
put
\
(b)
0
740
3iiO
50
\/
J
output of a 3-phase generator has
P m Because
is
1
the electrical out-
.
constant, the mechanical
power required
generator does not vibrate. Furthermore, the power
erator to the load,
is
line,
connecting the gen-
constant.
Example 8-2
The 3-phase generator shown
in Fig.
nected to three 20 12 load resistors.
voltage induced in each phase
120°
(c)
I
120°
a.
c.
a.
d.
Three-phase generator.
in a 3-phase generator.
b.
Voltages induced
c.
Phasor diagram
of the
a.
Each
resistor
to
behaves as a single-phase load
in
each resistor
(b)
Three-phase, 6-wire system.
Corresponding phasor diagram.
120 V. calculate
an effective voltage of 120 V. The
power dissipated
b.
con-
The power dissipated in each resistor
The power dissipated in the 3-phase load
The peak power P lu dissipated in each resistor
The total 3-phase power compared to P lu
connected
Figure 8.10
is
the effective
Solution
induced voltages.
a.
is
8.10a
If
the following:
b.
Figure 8.9
to
also constant, and so a 3-phase
flow over the transmission
-N
X
is
drive the rotor
t
J
total
a magnitude of 1.5
is,
therefore,
1
64
ELECTRICAL MACHINES AND TRANSFORMERS
I
i
u
*
U
'b
AI
360
240
0
480
"\
t
x
I
(a)
Figure 8.11
a.
Three-phase, 4-wire system.
b.
Line currents
in
a 3-phase, 4-wire system.
P = E 2 IR =
= 720
b.
The
(all
total
l20
The peak power
/2()
W
power dissipated
three resistors)
2
in the
3-phase load
d.
3
The
ratio of
X 720
is
Pm
to
8.485
is
2 60/1440
1
=
1.5
Thus, whereas the power
The peak voltage across one
resistor
sates
is
tal
E m = <2E = V2 X
- 169.7 V
The peak current
/m
X
69.7
*
is
absolutely constant from instant
to instant.
c.
PT
PT /P m =
= 2160
This power
each resistor
= £ m /m =
= 1440W
is
P v = 3P =
in
in
120
each resistor
= EJR =
= 8.485 A
is
169.7/20
between 0 and a
power
for
all
in
maximum
each resistor
of 1440
three resistors
is
W,
pul-
the
to-
unvarying and
equal to 2 160 W.
Wye
8-8
The
connection
three single-phase circuits of Fig. 8.10 are
electrically
we
independent. Consequently,
can
connect the three return conductors together
form
a single return
conductor (Fig.
to
8.1 la). This
reduces the number of transmission line conduc-
The
tors
from 6
tral
conductor (or simply neutral), carries the sum
to 4.
of the three currents
(/ a
+
/b
+
/c ).
At
first
it
seems
that the cross section of this
conductor should be
three times that of lines
and
diagram of Fig.
'c
return conductor, called neu-
a, b,
8,1 lb clearly
c. However, the
shows that the sum
of the three return currents is zero at every instant. For example, at the instant corresponding to
= / max and I h = / a = -0.5 / max making
+ 4 = 0- We arrive at the same result (and
Figure 8.12
240°, I c
Three-phase, 3-wire system showing source and load.
4+
A>
,
THREE-PHASE CIRCUITS
I
much more simply) by taking the sum of the phaOb. The sum is clearly
(/., + / b + I c ) in Fig. 8.
sors
1
zero.
We
can, therefore,
together without in
remove
the neutral wire al-
any way affecting the voltages
one stroke
or currents in the circuit (Fig. 8.12). In
we accomplish a great saving because the number
of
conductors
line
However, the loads
in
order to
remove
not identical, the
from
drops
six
three!
to
8.11a must be
in Fig.
identical
the neutral wire. If the loads are
absence of the neutral conductor
produces unequal voltages across the three loads.
The
circuit of Fig.
—composed
and load —
8.12
generator, transmission line,
n
3-phase, 3 -wire system.
The generator,
connected
load, are said to be
as the
of the
is
called
as well
in wye, be-
cause the three branches resemble the letter Y.
For equally obvious reasons,
to
some people
prefer
use the term connected In star.
The
circuit of Fig. 8.1 la
4-wire system.
system
is
The
usually the
than the line
is
called a 3-phase,
neutral conductor in such a
same
size or slightly smaller
conductors. Three-phase, 4-wire sys-
tems are widely used to supply electric
power
to
commercial and industrial users. The line conductors are
often called phases,
which
is
the
same
term applied to the generator windings.
8.9
Voltage relationships
Consider the wye-connected armature windings of a
3-phase generator (Fig. 8. 13a).
in
The induced voltage
each winding has an effective value
sented by the length of each phasor
Fig. 8.
are
1
3b.
Knowing
represented
tion
is,
in the
repre-
diagram
in
that the line-to-neutral voltages
by phasors
E
ail ,
Eblv and Ecn the quesE. lb Ehc and
what are the line-to-line voltages
£ca ? Referring
E lN
to Fig. 8.13a,
lowing equations, based on
we
,
,
can write the
fol-
Kirchhoff s voltage law:
Eab = Ean + E nh
= ^an ~ £bn
£ bc = £ bn + E nc
= E hn - Ecn
Eca = Ecn + E na
= Ecn - Ean
(8.1)
(8-0
Figure 8.13
Wye-connected
a.
(8.2)
Line-to-neutral voltages of the generator.
c.
Method
d.
Line voltages
to
determine
line
voltage
£ab £ bC) and £ca
placed at 120°.
(8.3)
3-phase gen-
b.
(8.2)
(8.3)
stator windings of a
erator.
,
£ab
.
are equal and dis-
ELECTRICAL MACHINES AND TRANSFORMERS
166
Referring
we draw phasor
Eq. 8.1,
first to
ex-
E.db
a
actly as the equation indicates:
The
diagram shows that line
by 30° (Fig. 8.13c). Using
resulting phasor
voltage
E
Ean
leads
ilb
simple trigonometry, and based upon the fact that
the length of the line-to-neutral phasors
is
£j N
we
,
have the following:
E
length
x
£ah =
=
of phasor
2
X
2
X £ LN V3/2
£, N cos 30°
Figure 8.14
Voltages induced
in
a wye-connected generator.
- V3 £ LN
Calculate
The
line-to-line
therefore
V3
voltage (called line voltage)
is
times the line-to-neutral voltage:
- V3£ LN
a.
b.
(8.4)
c.
The line-to-neutral voltage
The voltage induced in the individual windings
The time interval between the positive peak
A and
voltage of phase
where
phase
E =
x
£ LN —
V3 =
Due
d.
effective value of the line-to-neutral
Solution
voltage [V|
a.
=
a constant [approximate value
to the
equal to
V3 E
t
N
.
The
of
truth
this
=
can
be seen by referring to Fig. 8.13d, which shows
three phasors: £. l]v
drawn according
The
line
£"
hc
,
and
phasors are
and
8.3, respec-
.
is
The windings
1
3
V
800
are connected in wye; conse-
quently, the voltage induced in each winding
all
£ca The
to Eqs. 8.1, 8.2,
line voltage
line-to-neutral voltage
symmetry of a 3-phase system, we con-
is
peak of
£ln = ^i/^ 3 = 23 900/V3
b.
tively.
The
1.73]
clude that the line voltage across any two generator
terminals
The peak value of the
effective value of the line voltage [V]
the positive
B
13
c.
800
is
V.
One complete
cycle (360°) corresponds to 1/60
s.
Consequently, a phase angle of 120° corresponds
voltages are equal in magnitude and
to an interval of
mutually displaced by 120°.
To
further clarify these results. Fig.
the voltages
8.
between the terminals of
generator whose line-to-neutral voltage
The
line
voltages are
all
equal to
14
is
3-phase system, but the voltage between
b,
c\
b and n, etc.)
is
Example 8-3
3-phase 60
Hz
generator, connected in wye, gen-
erates a line (line-to-line) voltage of 23
900
V.
1
X
1/180
s
60
5.55
ms
positive voltage peaks are, therefore, sepa-
rated by intervals of 5.55 ms.
lines a, b, c, consti-
nevertheless an ordinary single-phase voltage.
A
The
100 V3, or
tute a
b and
=
100 V.
The voltages between
120
360
a 3-phase
173 V.
any two lines (a and
T=
shows
d.
The peak
line voltage is
E m = V2E L
= 1.414 X
= 33 800
The same voltage
23 900
(2.6)
relationships exist in a wye-
connected load, such as
that
shown
in Figs. 8.1
THREE- PHASE CIRCUITS
and
8.
1
2. In
other words, the line voltage
is
1
67
V3
times the line-to-neutral voltage.
Example 8-4
The generator
in Fig.
8.12 generates a line voltage
of 865 V, and each load resistor has an impedance
of 50
fl.
Calculate
The voltage across each resistor
The current in each resistor
The total power output of the generator
a.
b.
c.
Solution
The voltage across each
a.
resistor
is
£ L n = £| A/3 = 865/V3
= 500 V
The current
b.
in
/
each resistor
= ELN /R =
(8.4)
is
500/50
10A
All the line currents are, therefore, equal to 10
Power absorbed by each
c.
resistor
P = ElN I = 500 X
5000
A.
(b)
is
10
W
The power delivered by the generator
to all
three resistors is
P =
3
X 5000 =
15
kW
Figure 8.15
Impedances connected in delta.
b. Phasor relationships with a resistive
a.
8.10 Delta
connection
A
is
3-phase load
said to be
voltages are equal
and the
This corresponds
to
three
balanced when the
line
line currents are equal.
identical
impedances
connected across the 3-phase line, a condition that
is
usually
encountered
in
The three impedances
(as
we
The
3-phase
may
the line; consequently, resistor currents
£bc
,
and
ECiV
h
/2 ,
and
/3
£ab
law, the line currents are given
by
be connected
8.
-h
wye
in
1
(8.5)
5a).
(8.6)
h -
(8.7)
12
shown).
Let us determine the voltage
tionships in
and current
rela-
such a delta connection,* assuming a
resistive load.
The
resistors are
,
Furthermore, according to Kirchhoff's
voltages are produced by an external gen-
erator (not
/
are in phase with the respective line voltages
circuits.
already have seen) or in delta (Fig.
line
load.
connected across
The connection
letter A.
is
so
named because
it
resembles the Greek
ELECTRICAL MACHINES AND TRANSFORMERS
68
1
Let the current
each branch of the delta-connected
in
which corresponds
load have an effective value
the length of phasors
/2 , /3
.
Furthermore,
let
to
the
have an effective value / L which corresponds to the length of phasors /.,, / b /c Referring
line currents
,
,
equation
shows
.
we draw phasor
exactly as the
indicates. The resulting phasor diagram
Eq. 8.5,
first to
that
/
a
/.,
leads
ple trigonometry,
/,
by 30° (Fig.
we can now
8.
1
5b).
Using sim-
write
A
—
10
lx
=
2
X
I,
cos 30°
=
2
X
I,
V 3/2
= V3/
c
o—
Figure 8.15c
See Example
y
.
The
line current
is
therefore
V3
8.5.
times greater than
the current in each branch of a delta-connected load:
/,.
= V3/Z
(8.8)
= l(W3 =
Iy
b.
5.77
A
The voltage across each impedance
is
550
V.
Consequently,
where
=
—
/,
/,
Z=
-
effective value of the line current [AJ
effective value of the current in one
EIIy
=
550/5.77
95 ft
branch of a delta-connected load [A]
—
"V3
a constant [approximate value
The reader can
=
1.73]
position of phasors /b and /c and thereby observe that
20°.
,
the three line currents are equal and displaced by
Table
8A summarizes
1
the basic relationships bein
by a 3-phase
magnitude and
readily determine the
tween the voltages and currents
wye-connected
and delta-connected loads. The relationships are
motor winding, generator wind-
ing, etc.) as long as the
elements
in the three
line
The apparent power supplied by
is
a single-phase
equal to the product of the line voltage
line current
/.
The question now
arises:
line
E times the
What
is
the
apparent power supplied by a 3-phase line having
line voltage
valid for any type of circuit element (resistor, capacitor, inductor,
Power transmitted
8.11
If
we
8. 16a,
E and
a
a line current I?
refer to the
the apparent
wye-connected load of
power supplied
to
Fig.
each branch
is
phases
are identical. In other words, the relationships in
Table
8A
apply to any balanced 3-phase load.
The apparent power supplied
is
Example 8-5
Three identical impedances are connected
across a 3-phase
a.
b.
is
550
V
apparent power
in delta
line (Fig. 8.15c). If the line
S
=
E
,
X
/
X
3
=
Solution
The current
:!:
In
3-phase balanced circuits,
factors. If the
in
each impedance
is
\ 3
El
\3
10 A. calculate the following:
The current in each impedance
The value of each impedance [ft]
is
we can add
ers of the three phases because they
a.
the
.
total
current
to all three branches
obviously three times as great. * Consequently,
power
the apparent pow-
have identical power
factors are not identical, the apparent
powers cannot be added.
THREE-PHASE CIRCUITS
VOLTAGE AND CURRENT RELATIONSHIPS
TABLE 8A
Wye
IN
169
3-PHASE CIRCUITS
Delta connection
connection
7/1.73
Figure 8.16a
Impedances connected
•
The current
current
•
•
in
in
Figure 8.16b
Impedances connected
wye.
each element
is
equal to the line
•
The voltage across each element
line voltage E divided by V 3.
The voltages across
is
equal to the
•
the elements are 120° out of
in the
elements are 120° out of
•
•
case of a delta-connected load (Fig.
the apparent
power supplied
S
= E X
each branch
to
is
equal to the line
The voltage across each element
The
The
8.
1
6b),
is
equal to the
voltages across the elements are 120° out of
currents in the elements are 120° out of
power
/
is
the
same
Consequently, the
as for a
total
in
The relationship between active power P, reactive
power Q, and apparent power S is the same for bal-
wye-connected
apparent power
and apparent
3-phase circuits
8.12 Active, reactive,
is
V3
is
load.
also the
anced 3-phase
We
circuits as for single-phase circuits.
therefore have
2
S = \'P +
same.
We
each element
phase.
phase.
which
in
divided by V3.
phase.
The currents
In the
/
delta
line voltage E.
phase.
•
The current
current
/.
in
therefore have
Q
2
(8. JO)
and
S =
V 3 El
cos 6
(8.9)
=
PIS
(8.
where
where
S
=
total
apparent power delivered by a
3-phase line
E—
/ =
V3 =
[VA]
effective line voltage [V]
effective line current [A)
a constant [approximate value
=
1.73]
= total 3-phase apparent power VA|
=
P
total 3-phase active power [W|
Q = total 3-phase reactive power |var|
cos 0 = power factor of the 3-phase load
0 = phase angle between the line current
S
|
and the line-to-neutral voltage
|°|
1
1
ELECTRICAL MACHINES AND TRANSFORMERS
70
Example 8-6
A
3-phase motor, connected
power
a line current of 5 A. If the
motor
a.
is
The
440V
to a
line,
factor of the
The
total active
The
total reactive
60 Hz
power
power absorbed by
Solution
The
= V3
= V3 X 440 X
EI
=
3811
VA
=
3.81
kVA
total active
See Example
is
O
1
R
power
=
3.05
total reactive
is
X
3.81
The
The
=
\ S
current in each line
is
8
1
V =
also
3.
resistance of each element
is
is
A
3. 15
1
5 A.
0.80
R =
- P2 =
2
each resistor
1
kW
power
in
= P/E= 000 W/3
/
EII
=
-
318/3.15
101 12
Example 8-8
is
In the circuit
Q =
8-7.
The current
P = ScosO =
The
JT
5
b.
c.
O
Figure 8.17
apparent power
total
S
b.
w
line
machine
the
3000
=
9"
V
550
3-phase
apparent power
total
c.
The
A
o
80 percent, calculate the following:
b.
a.
P
draws
\ 3.81
2
-
3.05
2
a.
b.
2.28 kvar
of Fig.
8.
1
8,
calculate the following:
The current in each line
The voltage across the inductor terminals
Solution
8.13 Solving 3-phase circuits
A
a.
balanced 3-phase load may be considered
composed of
three
identical
Consequently, the easiest
cuit
is
to consider
examples
way
to
be
single-phase loads.
method
4
{
to
is
composed of an
inductive reactance
fl in series with a resistance
R=
3
II.
Consequently, the impedance of each branch
Z
to solve such a cir-
only one phase. The following
illustrate the
Each branch
X =
~~~
V4-
3^
=
5
The voltage across each branch
a
(2.i:
is
be employed.
ELN = ELN3 = 440 V/V3 - 254 V
Example 8-7
The current
Three identical
3000
550
V
W
resistors dissipating a total
are connected in
wye
in
/
across a 3-phase
(50.8
A
is
- E LN /Z -
The current
The value of each
in
each
line
resistor
440
Solution
a.
The power dissipated by each
resistor
V
3-phase line
is
P = 3000 W/3 = 1000W
The voltage across
the terminals of each resistor
is
Figure 8.18
E — 550 VA/3 -318 V
See Example
254/5
=
50.8
also the line current.)
Calculate
b.
is
power of
line (Fig. 8. 17).
a.
each circuit element
8-8.
is
A
THREE-PHASE CIRCUITS
The voltage across each inductor
b.
E = IX = 50.8 X
= 203.2 V
V
In a
is
wye connection
1
impedance per phase
the
7
is
understood to be the line-to-neutral impedance. The
4
voltage per phase
is
simply the
line voltage divided
by V3. Finally, the current per phase
equal to the
is
line current.
Example 8-9
A
identical capacitors
If
Hz
3-phase 550 V, 60
the line current
is
line
is
connected
connected to three
in delta (Fig. 8.19).
22 A, calculate the capacitance
of each capacitor.
The assumption of a wye connection can be
made not only for individual loads, but for entire
load centers such as a factory containing motors,
lamps, heaters, furnaces, and so forth.
We
assume
in
that the load center
is
connected
simply
wye and
proceed with the usual calculations.
Solution
The current
/
each capacitor
in
=
/ L /V3
=
22 A/V3
Voltage across each capacitor
Capacitive reactance
X
Xc = EJI =
is
c
-
Example 8-10
A
A
= 550 V
of each capacitor
550/12.7
-
The capacitance of each capacitor
C=
12.7
1/(2tt
=
61.3
V
(line-to-line)
8.20a). If the plant
is
power
factor
total
of 415
kVA
3-phase line (Fig.
is
87.5 percent lag-
ging, calculate the following:
43.3 il
b.
The impedance of the plant, per phase
The phase angle between the line-to-neutral
c.
The complete phasor diagram
a.
is
voltage and the line current
1/2tt/Xc
=
manufacturing plant draws a
from a 2400
X 60 X
43.3)
(2.11)
for the plant
Solution
|jlF
a.
We
assume a wye connection composed of
impedances Z(Fig. 8.20b).
three identical
The voltage per branch
E=
=
The
Figure 8.19
See Example
8-9.
8.14 Industrial loads
In
is
connected
=
in delta or in
capacitors, and so on, often have only three external
and there
is
no way
to tell
how
the inter-
connections are made. Under these circum-
we simply assume
(A wye connection is
connection
stances,
that the
wye.
slightly easier to handle
than a delta connection.)
(8.9)
000/(2400 V 3
=
A
100
=
b.
E/I
=
)
is
1386/100
13.9 12
The phase angle 0 between the line-to-neutral
voltage
386 V) and the corresponding line
current ( 100 A) is given by
(
nal
is
S/(EV3)
Z=
wye. For ex-
V
= 415
ample, 3-phase motors, generators, transformers,
terminals,
1386
The impedance per branch
most cases, we do not know whether a particular
3-phase load
2400/V3
current per branch
/
is
1
is in
cos 6
0
= power factor =
= 29°
0.875
(8.11)
ELECTRICAL MACHINES AND TRANSFORMERS
172
4000
V
3 phase
2400V
y
1
?\
3- phase line
—1((a)
1800 kvar
'Tan
/
a = 100
1386
A
V
Figure 8.21
Industrial motor and
capacitor.
A delta-connected capacitor
bo-
is
See Example
bank rated
also connected to the line. If the
at
8-1
1
1800 kvar
motor produces an
output of 3594 hp at an efficiency of 93 percent and
a
(b)
power
factor of
90 percent
(lagging), calculate the
following:
c.
The
The
The
d.
The apparent power supplied by
a.
b.
power absorbed by the motor
power absorbed by the motor
reactive power supplied by the transmisactive
reactive
sion line
the transmis-
sion line
e.
f.
g.
The transmission line current
The motor line current
Draw
the complete phasor diagram for one
phase
Solution
a.
Figure 8.20
a. Power input
to
a
See Example
wye connection
b.
Equivalent
c.
Phasor diagram
The
factory.
current
in
P 2 = 3594 X
8-10.
and
is
0.746
equivalent to
= 268 kW
1
Active power input to motor:
of the factory load.
of the voltages
Power output of 3594 hp
currents.
= p 2^ = 2681/0.93
= 2883 kW
each phase lags 29° behind the
b.
(3.6)
Apparent power absorbed by the motor:
line-to-neutral voltage.
Sm
c.
The complete phasor diagram is shown in Fig.
8.20c. In practice, we would show only one
phase; for example,
Ean
,
/.„
Q m = \Si=
A
1
P- n
= V3203 T^r 2883 2
395 kvar
.
5000 hp wye-connected motor
4000
2883/0.90
Reactive power absorbed by the motor:
and the phase angle
between them.
Example 8-11
= PJcos 0 =
= 3203 kVA
V (line-to-line),
3-phase, 60
is
Hz
connected to a
line (Fig. 8.2
1
).
c.
Reactive power supplied by the capacitor bank
(see section 7.5):
THREE-PHASE CIRCUITS
Qc = -1800kvar
Total reactive
cause the capacitors are connected
power absorbed by
assumed a wye connection
the load:
in delta,
1
and
1800
+
1395
we
try to
This
is
is
pacitor bank.
an unusual situation because reactive
being returned to the
the capacitor
most cases
line. In
bank furnishes no more than
Qm
The
Active power supplied by the line
P L = P m = 2883
5L
e.
= VPl + Qi = V2883 2 +
= 2911 kVA
Transmission line current
h.
f.
Motor
/m
g.
The
to recognize that
is
in
wye
same reactive power), the line current
would lead £ LN by 90°. Consequently,
A
90° ahead of £, N That is the correct
position for phasor /c no matter how the capacitor
/c
.
is
bank
is
connected
internally.
Phase angle 0 L between the transmission
current and
line is
(
-
405)
£ LN
cos6,,
2
line-to-neutral voltage
= PJS L
= 0.99
eL
=
2883/291
8°
(8.9)
4000)
4000)
462
A
is
£LN = 4000/V3 = 2309 V
420
A
462
A
420
A
462
A
420
A
462
A
Phase angle 9 between the motor current and
the line-to-neutral voltage
4000
is:
V
3-phase
6
line
is:
is
= 5 m /(£L V3)
= 3 203 000/(V3 X
= 462 A
cos 0
/'/
(while gen-
is
= SJ(E^3)
= 2911 000/(V3 X
= 420 A
line current
of 260
kW
Apparent power supplied by the
solution
were connected
erating the
we draw
kilovars of reactive power,
d.
if
follow the actual currents inside the ca-
the capacitors
power
we
for the motor. This can
create unnecessary phase-angle complications
QL = Qc + Q m =
= -405 kvar
73
= power factor =
= 25.8°
0.9
260
260
260
A
A
A
(The motor current lags 25.8° behind the voltage, as
shown
in Fig. 8.22a.)
Line current drawn by the capacitor bank
/c
= Gc/(£lV3)
= 800 000/(V3 X
- 260 A
1
(b)
is
—IH
1800 kvar
4000)
Figure 8.22
Phasor relationships
a.
for
one phase. See Example
8-11.
the
Where should phasor current /c be located on
phasor diagram? The question is important be-
b.
Line currents. Note that the motor currents exceed
the currents of the source.
"1
1
ELECTRICAL MACHINES AND TRANSFORMERS
74
The
line current
(420 A) leads
Eln by
8° be-
cause the kvars supplied by the capacitor bank ex-
ceed the kvars absorbed by the motor.
The phasor diagram
one phase
tor
is
shown
in
Fig. 8.22a.
The circuit diagram and
shown in Fig. 8.22b.
current flows are
We want to emphasize the
ing a
tual
wye
importance of assum-
connection, irrespective of what the ac-
connection
nection for
all
may
By assuming a wye conwe simplify the
be.
circuit elements,
Figure 8.23
The
sequence
are observed
in
the
Figure 8.24
The letters are observed
in
the sequence a-c-b.
in
the
letters
a-b-c.
calculations and eliminate confusion.
As
a final remark, the reader has no doubt no-
ticed that the solution of a 3-phase
problem
in-
volves active, reactive, and apparent power. The
impedance value of devices such
as resistors,
mo-
tors,
and capacitors seldom appears on a nameplate.
This
is
to
be expected because most industrial loads
involve electric motors, furnaces, lights, and so on,
which are seldom described
in
terms of resistance
and reactance. They are usually represented as de-
draw a given amount of power
vices that
power
The
situation
is
somewhat
3-phase transmission
sistances
fixed.
at a
given
factor.
lines.
different in the case of
Here we can define
re-
and reactances because the parameters are
The same remarks apply
to equivalent circuits
describing the behavior of individual machines such
as induction motors
and synchronous machines.
In conclusion, the solution of
may
L,
3-phase circuits
involve either active and reactive power or R,
C elements — and
8.15
sometimes both.
Phase sequence
Figure 8.25
The
In addition to line voltage
letters
are observed
sequence
a-c-b.
and frequency, a 3-phase
system has an important property called phase sequence. Phase sequence
is
important because
it
de-
termines the direction of rotation of 3-phase motors
and whether one 3-phase system can be connected
in parallel
with another. Consequently, in 3-phase
systems, phase sequence
quency and voltage
is
as important as the fre-
understanding of phase
sequence by considering the following analogy.
Suppose the letters a, b, c are printed at 20°
1
tervals
on a slowly revolving disc
in-
(Fig. 8.23). If the
disc turns counterclockwise, the letters appear
in
the sequence a-b-c-a-b-c. Let us call this the posi-
are.
Phase sequence means the order
three line voltages
We can get a quick intuitive
in
which
become successively
the
positive.
tive
sequence.
It
can be described
three ways: abc, bca, or cab.
in
any one of
THREE-PHASE CIRCUITS
If the same disc turns clockwise, the sequence
(Fig. 8.24). We call this
becomes a-c-b-a-c-b
.
.
.
and
the negative sequence,
it
can be described by
rule:
When
175
using the double-subscript notation, the
sequence of the
first
subscripts corresponds to the
phase sequence of the source.
any one of three forms: acb ? cba, or bac. Clearly,
there
a difference between a positive sequence
is
and a negative sequence.
In Fig. 8.17, the
Suppose we interchange any two
in Fig. 8.23,
the result
is
letters
on the disc
while retaining the same counterclock-
wise rotation.
the letters
If
as
shown
becomes c-b-a-c-b-a
a and c
.
.
,
The sequence now
which is the same as the
negative sequence generated by the disc in Fig. 8.24.
We
to
phase sequence of the source
be A-C-B.
Draw
sequence can be converted into a
negative sequence by simply interchanging two
the line voltages.
Solution
The voltages follow the sequence A-C-B, which is
the same as the sequence AC-CB-BA-AC ....
ECB -E BA and
shown
in Fig. 8.27.
is
We
can reverse the phase
line
by interchanging any two
se-
let-
quence of a 3-phase
conductors. Although this
verted into a positive sequence by interchanging
change,
it
may appear to
be a
trivial
can become a major problem when large
letters.
Let us
now
busbars or high-voltage transmission lines have to
consider a 3-phase source having
ter-
minals a, b ? c (Fig. 8.26a). Suppose the line voltages
£ AC -
is
the corresponding phasor diagram
Similarly, a negative sequence can be con-
any two
is
the phasor diagram of
Consequently, the line voltage sequence
conclude that for a given direction of rota-
tion a positive
ters.
known
are interchanged,
in Fig. 8.25.
.
Example 8-12
that
£ab E hc E
,
,
Cil
revolving phasors
sweep past
be interchanged. In practice, measures are taken so
are correctly represented
shown
in
by the
As they
Fig. 8.26b.
be
such drastic mechanical changes do not have to
made
at
the last minute.
major distribution systems
The phase sequence of all
is
known
in
advance, and
the horizontal axis in the conventional
any future connections are planned accordingly.
counterclockwise direction, they follow the se-
quence
If
Eab -E bc -Eca -Eab -E bc ....
we
direct our attention to the first letter in each
subscript,
we find that the sequence
The source shown
sequence a-b-c.
in Fig.
8.26a
is
We can, therefore,
is
a-b-c-a-b-c
.
sequence
.
said to possess the
state the
8.16 Determining the phase
following
Special instruments are available to indicate the
phase sequence, but
ing
we can
also determine
two incandescent lamps and a
it
by us-
capacitor.
three devices are connected in wye. If
The
we connect
the circuit to a 3-phase line (without connecting the
neutral),
3-phase
one lamp
source
der:
(a)
—
—
^/^^^^
Determining the phase sequence of a 3-phase
£"cb
source.
b.
always burn brighter than
(b)
Figure 8.26
a.
will
The phase sequence is in the following
bright lamp
dim lamp capacitor
the other.
Phase sequence depends upon the order
in
the line voltages reach their positive peaks.
which
Figure 8.27
See Example
8-12.
£ ac
or-
ELECTRICAL MACHINES AND TRANSFORMERS
176
3-phase
(a)
line
lamp
Figure 8.29
Method of connecting a single-phase wattmeter.
moves upwhen the connections between source and load
made as shown in Fig. 8.29. Note that the ± cur-
In single-phase circuits the pointer
(b)
scale
are
rent terminal
nal.
Figure 8.28
a.
When
is
connected to the
the wattmeter
upscale reading
Determining phase sequence using two lamps and
means
supply terminals
1,
is
that
±
potential termi-
connected
power
is
way. an
this
flowing from
2 to load terminals
3, 4.
a capacitor.
b.
Resulting phasor diagram.
8,18
Suppose, for example,
cuit
is
connected to a 3-phase
8.28a. If the
brightly, the
lamp connected
phase sequence
ages follow each other
which
is
to say in the
line, as
to
is
phase
shown
C
sequence
is
CB
line volt-
CB-BA-AC,
E BA EAC The
£"
,
.
,
shown
in
in Fig.
burns more
C-B-A. The
sequence
in the
corresponding phasor diagram
8.28b.
in Fig.
In a 3-phase,
3-wire system, the active power sup-
may be measured by two
shown in Fig.
equal to the sum of the two
single-phase wattmeters connected as
8.30.
The
total
power
is
wattmeter readings. For balanced loads,
is
less than
1
if
the
power
00 percent, the instruments
give different readings. Indeed,
Power measurement
in
3-phase, 3-wire circuits
plied to a 3-phase load
factor
8.17
Power measurement
that a capacitor/lamp cir-
if
the
will
power factor is
ac circuits
Wattmeters are used
measure active power
to
in
single-phase and 3-phase circuits.
Owing
is built,
to its external
a wattmeter
meter and ammeter combined
Consequently,
rent terminals.
it
way
connections and the
may be considered
in
it
to be a volt-
same box.
the
has 2 potential terminals and 2 cur-
One of
the potential terminals
one of the current terminals bears a
signs are polarity
marks
that
±
sign.
and
The
±
determine the positive
when
same time
or negative reading of the wattmeter. Thus,
the
±
voltage terminal
as current
is
is
entering the
positive at the
±
current terminal, then
the wattmeter will give a positive (upscale) reading.
The maximum voltage and current the instrument can tolerate are shown on the nameplate.
Figure 8.30
Measuring power in a 3-phase, 3-wire
two-wattmeter method.
circuit
using the
THREE-PHASE CIRCUITS
less
than
50 percent, one of the wattmeters will give
We must then reverse the con-
177
Power measurement
8.19
a negative reading.
in
3-phase, 4-wire circuits
nections of the potential coif so as to obtain a reading of this negative quantity. In this case, the
of the 3-phase circuit
is
power
equal to the difference be-
The two-wattmeter method gives
the
active
balanced or un-
is
single-phase
three
that the
±
±
as
current terminal
potential terminal.
shown
When
is
total
in Fig. 8.3
power.
1
.
Note
again connected to the
the wattmeters are con-
nected this way, an upscale reading means that active
balanced.
Example 8-13
A full-load test on
a 10 hp, 3-phase
following results: P,
the current in
line
circuits,
wattmeters are needed to measure the
The connections are made
tween the two wattmeter readings.
power absorbed whether the load
4-wire
3-phase,
In
voltage
motor yields the
= +5950 W; P 2 = + 2380 W;
each of the three lines
600
is
V. Calculate the
10 A; and the
is
power
factor of
power is flowing from source A, B, C, N to the load.
The total power supplied to the load is equal to
the sum of the three wattmeter readings. The threewattmeter method gives the active power for both
balanced and unbalanced loads.
Some
wattmeters, such as those used on switch-
boards, are specially designed to give a direct read-
the motor.
out of the 3-phase power. Figure 8.32 shows a
Solution
megawatt-range wattmeter
Apparent power supplied to the motor
^3EI = V3 X 600 X
S =
=
Active
10 390
power
is
circuit that
a generating station.
The
measures the
current trans-
formers (CT) and potential transformers (PT) step
10
down
VA
power supplied
in
the line currents and voltages to values
com-
patible with the instrument rating.
to the
motor
is
SSI
P = 5950 + 2380
= 8330 W
cosO = P/S= 8330/10 390
=
0.80, or 80 percent
p.
Example 8-14
>
When
the
the line
motor
Example 8-13 runs
in
current drops to 3.6
readings are
P,
A
at
p
and the wattmeter
- +1295 W; P 2 =
Calculate the no-load losses
no-load,
and power
LOAD
.
..
•
:
p>
845 W.
factor.
Solution
Apparent power supplied to motor
)
S = V3 EI = V3 X 600 X
= 3741 VA
3.6
Figure 8.31
Measuring power
in
a 3-phase, 4-wire
circuit.
No-load losses are
P =
P,
+ P2 =
= 450
1295
-
8,20 Varmeter
845
A varmeter indicates
W
It is
Power factor
=
P/S
=
450/3741
=
0.12
= \2%
built the
the reactive
same way
power
as a wattmeter
ternal circuit shifts the line voltage
in a circuit.
is,
but an in-
by 90° before
it
ELECTRICAL MACHINES AND TRANSFORMERS
178
Figure 8.32
Measuring active power
is
in
applied to the potential
employed
tions
in the control
a high-power
coil.
circuit.
Varmeters are mainly
rooms of generating
and the substations of
electrical utilities
load resistance (Fig. 8.33). Furthermore, given
and
the
In 3-phase, 3-wire
balanced
We
circuits,
the
we
can cal-
two wattmeter
simply multiply the
differ-
ence of the two readings by V 3. For example,
two wattmeters
indicate
=
6 76 vars. Note that
1
surement
is
if
is
this
(5950
—
2380)
,
phase sequence of the
1
it
is
line voltages
£ l2
connected as indicated.
capacitive and
If the
in-
ductive reactances are interchanged, the 3-phase
system becomes completely unbalanced.
X
method of var mea-
only valid for balanced 3-phase circuits.
Example 8-15
A
800
phases
kW
1
= 440 Z
single-phase load
connected between
is
and 2 of a 440 V, 3-phase
0,
E23 = 440 Z -
line,
E ]2
wherein
E3l = 440 Z
120,
12(X
Calculate the load currents and line currents
8.21
It
A
remarkable single-phase
to 3-phase transformation
sometimes happens
unity
power
3-phase
line.
that a large single-phase
a.
This can create a badly unbalanced
it
three phases perfectly
is
possible to balance the
by connecting
the single-phase load
b.
When
is
connected alone
line
balancing reactances are added across
the remaining lines, as
shown
in Fig.
8.34
Solution
a.
The
resistance of the single-phase load
a capacitive
reactance and an inductive reactance across the
other two lines.
When
on the 3-phase
factor load has to be connected to a
system. However,
,
essential that the three impedances be
the
+5950 W and +2380 W repower
spectively, the reactive
V3
-2-3-
,
power from
readings (Fig. 8.30).
l
£23 E3l
large industrial consumers.
culate the reactive
impedances V 3 times greater than the value of the
sta-
The reactances must each have
R
E2
P
=
800 000
0.242
n
is
THREE-PHASE CIRCUITS
The current
in the
1
79
load and in two of the three
lines is
The
current in the third line
3-phase system
b.
By
is
and so the
zero,
is
badly unbalanced,
introducing capacitive and inductive reac-
tances having an impedance of 0.242
we
0.4 19 fl,
V3 =
obtain a balanced 3-phase
line, as
demonstrated below. Taking successive loops
around the respective
Section 2.32),
E ]2 -
elements
in Fig.
2.38
£ 3] -
j
obtain the following results:
= 0 /. /, = 4A3E l2 =
X 440Z0 = 1817Z0
4.13
£23 +
we
0.242/,
£12
2.38
circuit
and using Kirchhoff's voltage law (see
8.34,
= 0
j 0.419 U
X 440Z(-120 +
0.419
/3
-
0
=
j 2.38 £ 23
= 1047^-30
U =
.'.
90)
/3
X 440Z(120 + 90 -
=
£ 31 =
= 1047Z30
-j 2.38
180)
£23
Applying Kirchhoff's current law
and
Figure 8.33
A single-phase resistive load can be transformed
a balanced 3-phase load.
3,
we
to
nodes
1
,
2,
obtain
into
U
=
/,
~
/3
= 1817Z0 - 1047Z.30
= 1817 - 907 - j 523
= 1047/1-30
/b
Ic
=
=
=
=
=
h ~ h
1047Z-30 - 1817Z0
907 - j 523 - 1817
-907 — j 523
1047Z210
= h ~ h
= 1047Z30 - 1047Z.-30
= 907 + j 523 - 907 + j 523
=1047j
= 1047Z90
Figure 8.34
Thus, / A / B /c make up a balanced 3-phase system because they are equal and displaced at 120°
See Example 8-14.
to
,
,
each other (Fig. 8.35).
180
ELECTRICAL MACHINES AND TRANSFORMERS
Tm
e.
8-6
The ohmic value of each
a.
What
b.
Could we reverse
is
resistor
the phase sequence in Fig. 8.10?
it
by changing the direction
of rotation of the magnet?
8-7
A 3-phase
draws a
motor connected
line current
to a
600
V
line
of 25 A. Calculate the
apparent power supplied to the motor.
8-8
Three incandescent lamps rated 60 W, 120
are connected in delta.
needed so
£ 23
8-9
8-14.
b.
fl resistors are
new power
Questions and Problems
8-10
If
one
cut,
is
Practical level
8-
1
wye-connected generator
V
in
each of
its
windings.
8-11
voltage of 100
terminals
is
in Fig. 8.9
V
What
is
1,
a
to a
8-3
8-12
at
240°
each of these
b.
If
c.
R =
is
£al Could we
ahead of£ al ?
.
Eb2
is
120°
lines a-b-c of Fig.
15 (2,
what
resistors are
line voltage is 13.2
is
a.
b.
c.
d.
the line current if the resistors are
in delta?
known
to
be connected
in
wish to apply full-load
100 kVA,
to a
each resistance
the elements are connected
wye
a.
In
b.
In delta
The windings of a 3-phase motor
nected
in delta. If the resistance
two terminals
is
0.6
II,
what
is
are con-
between
the resis-
the voltage across each resistor?
is
the current in
tance of each winding?
each line?
Calculate the power supplied to the 3-phase
Three
the resistors are
also say that
8-14
Three 24 fl resistors are connected
across a
load.
8-5
is
if
wye?
the resistors are
If
We
if
1
What
the line current
in
kW when
V, 3-phase line.
load. Calculate the value of
The voltage between
8.12 is 620 V.
a.
What
208
4 kV, 3-phase generator using a resistive
8- 3
8-4
is
wye, calculate the resistance of each.
instants?
Referring to Fig. 8.9c, phasor
is
c.
the instantaneous value of the volt-
behind phasor
Eh2
240°, and 330°.
each of these instants?
is
connected
the polarity of terminal a with re1 at
What
connected
b.
at 0°, 90°, 120°,
age across terminals 2, b
same
conductor of a 3-phase line
the load then supplied by a single-
heater dissipates 15
generates a peak
per phase.
spect to terminal
c.
3-phase load?
supplied to the load.
A 3-phase
Calculate the instantaneous voltage between
What
to the
in delta
burns out, calculate the
line
connected
a.
The generator
b.
connected
in-
Calculate the line voltage.
a.
line
V
is
phase voltage or a 2-phase voltage?
A 3-phase
duces 2400
8-2
one
the fuse in
If
line voltage
lamps burn normally?
on a 208 V, 3-phase line.
a. What is the power supplied
Figure 8.35
See Example
Three 10
that the
What
connected
kV and
in delta
V, 3-phase line. Calculate the
resistance of three elements connected in
in delta. If the
wye
the line current
1202 A, calculate the following:
The current in each resistor
The voltage across each resistor
The power supplied to each resistor
The power supplied to the 3-phase load
600
8-15
that
A 60 hp
would
dissipate the
3-phase motor absorbs 50
from a 600 V, 3-phase
b.
kW
line. If the line cur-
is 60 A, calculate the following:
The efficiency of the motor
The apparent power absorbed by the motor
rent
a.
same power.
THREE-PHASE CIRCUITS
The reactive power absorbed by
The power factor of the motor
c.
d.
8-16
Three 15
voltage
If the line
530
is
motor
Q reac-
in Fig. 8.18.
The
8-24
V, calculate the
An
industrial plant
2.4
kV
8-
1
7
The voltage across each
W lamps and a
Two 60
connected
in
wye. The
to the terminals
outlet.
nal Y,
What
a.
1
resistor
b.
0
|jlF
circuit
capacitor are
8-
1
8
connected
is
V
X-Y-Z of a 3-phase 20
1
8-25
wye
1
Two
kW,
the phasor diagram for the line voltages.
b.
The
0 jxF capacitors are connected
An
In
Problem
to terminal
8-20
Hz
line.
Three
1
5
If the
power generated
1
is
X, which lamp will be brighter?
connected
resistors
to a
in
in 3
8-27
absorb 60
3-phase
line. If
a.
b.
H resistors (R) and three 8 fl reline.
Without draw-
8-28
R and X in series, connected in wye
R and X in parallel, connected in delta
R connected in delta and X connected in wye
8-23
In
is
in
wye and
that the
motor
R
with an inductive reactance X.
Calculate the values of
R and
X.
power
output.
energy does the motor consume
h?
The
The
load
power
factor;
line current if the line voltage
630
is
V.
A 20 H resistor is connected
A and B of a 3-phase, 480 V
between
line.
lines
Calculate
Two 30
il resistors are
AB
and
BC
connected between
of a 3-phase 480
V
line.
Calculate the currents flowing in lines A,
B, and C, respectively.
is
8-30
A
1
50 kW, 460
V, 3-phase heater
stalled in a hot water boiler.
that each branch can
be represented by a resistance
a.
motor has an efficiency of 85 percent,
balanced, calculate the following:
phases
50 Hz instead of 60 Hz.
Problem 8-15, assume
connected
V
from a 600
respectively.
8-29
In Fig. 8.19, calculate the line current if the
frequency
A
line.
the currents that flow in lines A, B, and C,
rent for each of the following connections:
8-22
the load
Industrial application
ways
ing a phasor diagram, calculate the line cur-
c.
.5
1
16 A,
is
The wattmeters in Fig. 8.30 register +35
kW and —20 kW, respectively. If the load
is
wye, calculate the
absorbed.
across a 530 V, 3-phase
b.
kW and
motor having a cos 0 of 82 per-
How much
connected
actors (X) are connected in different
a.
line indicate 3.5
calculate the mechanical
8- 7, if the capacitor
new power
1
V
Calculate the active power supplied to the
a.
Calculate
b.
Three delta-connected
kW when
deter-
8,
motor.
they are reconnected
8-2
electric
3-phase
in
c.
8-19
1
respectively. If the line current
The apparent power
The power factor of
b.
line current
reactive
8.
the resistance and reactance.
cent draws a current of 25
across a 2300 V, 60
The
can be represented by
calculate the following:
a.
the following:
a.
that the plant
wattmeters connected into a 3-phase,
3-wire 220
level
Three
a
imped-
the equivalent line-to-neutral
mine the values of
8-26
Advanced
kVA from
factor of 80 percent
an equivalent circuit similar to Fig.
the phase sequence?
is
is
Assuming
b.
The capacitor is connected to termiand the lamp that burns brighter is
What
Draw
draws 600
power
ance of the plant?
connected to terminal X.
a.
line at a
lagging.
and apparent power sup-
active, reactive,
plied to the 3-phase load
b.
the phase angle between the line cur-
is
and the corresponding line-to-neutral
voltage?
following:
a.
What
b.
rent
and three 8
ft resistors
connected as shown
tors are
the
181
does
in series
8-3
1
it
produce
Three 5
if
the line voltage
il resistors are
across a 3-phase 480
V
connected
line.
is
in-
What power
is
in
470 V?
wye
Calculate the
1
82
ELECTRICA L MACHINES AND TRANSFORMERS
has a full-load efficiency of 93.6% and a
current flowing in each. If one of the resistors
is
that flows in the
8-32
One
power
disconnected, calculate the current
remaining two.
of the three fuses protecting a 3-phase
200 kW, 600 V is removed so as to reduce the heat produced by
the boiler. What power does the heater de-
factor of
83%. Calculate
the following:
b.
The active power drawn by the motor;
The apparent power drawn by the motor;
c.
The
a.
full-load line current;
electric heater rated at
A 450 kW,
duces
1
300
a.
575
lb
V,
3-phase steam boiler pro-
8-34
is
612
produced
if
A 40 hp, 460 V,
TEFC, premium
1 1
Company
V, 3-phase,
Assuming
the
is
32
in,
450 kg Square
D
used to drive a 1600 hp,
60 Hz squirrel-cage motor.
motor has
a
minimum
96%
effi-
and 90%,
What
is
the reactive
power drawn from
the
line at full-load?
80 r/min, 3-phase, 60 Hz
manufactured by Baldor Electric
X
in
delivered by the controller.
b.
efficiency induction motor
X 24
respectively, calculate the full-load current
the line
V.
in
ciency and power factor of
of steam per hour. Estimate
the quantity of steam
voltage
A 92
motor controller
2400
velop under these conditions?
8-33
8-35
c.
What
is
the phase angle
neutral voltage
and the
between the
line current?
line-to-
Chapter 9
The Ideal Transformer
9.0 Introduction
frequency/
The
transformer
ful electrical
is
probably one of the most use-
devices ever invented.
It
can raise
or lower the voltage or current in an ac circuit,
isolate circuits
from each
other,
and
it
it
can
can increase
or decrease the apparent value of a capacitor, an inductor, or a resistor Furthermore, the transformer
enables us to transmit electrical energy over great
distances and to distribute
it
safely in factories
and
homes.
We
will study
transformers
some of
the basic properties of
in this chapter. It will
help us under-
stand not only the commercial transformers covered
in later
chapters but also the basic operating princi-
ple of induction motors, alternators,
and synchro-
nous motors. All these devices are based upon the
Figure
laws of electromagnetic induction. Consequently,
a.
we encourage
9.1
b.
covered here.
Voltage induced
in
a coil
A
itive
9.1
voltage
able
the reader to pay particular attention
to the subject matter
A
is
induced
in
a
coil
when
coil of Fig. 9. la,
links) a variable flux
soidally at a
The
which surrounds
(or
links a vari-
sinusoidal flux induces a sinusoidal voltage.
and negative peaks
$ max
.
The
induces a sinusoidal ac voltage
Consider the
it
flux.
effective value
is
alternating flux
in the coil,
whose
given by
flux alternates sinu-
E=
frequency/ periodically reaching pos-
183
4.44/7V<D max
(9. 1)
1
ELECTRICAL MACHINES AND TRANSFORMERS
84
where
9.2 Applied voltage
E =
effective voltage induced [V]
/=
frequency of the flux [Hz]
= number of turns
/V
^max = peak value of
=
4.44
It
on the
and induced voltage
and draws a current / m
[Wb]
the flux
a constant [exact value
=
is
It
even by an ac current that flows
— N
is
As
in
any inductive
is in
<l>
The
Thus,
derived from Faraday's law equation
which AcjVAr
A(|)/Ar in
of flux and c
b,
1
when
and so the voltage
AcJ)/Ar
is
is
change
the instantaneous induced voltage.
is
in Fig. 9.
the flux
the rate of
is
the flux
time, the rate of change Ac|)/Ar
is
is
is
increasing with
greater than zero
positive. Conversely,
when
decreasing with time, the rate of change
less than zero; consequently, the voltage
negative. Finally,
when
the flux
is
neither in-
creasing nor decreasing (even for one microsecond), the rate of change Act>/A/
voltage
given by
is
circuit, / m lags
90° behind
phase with the current (Fig. 9.2b).
detailed behavior of the circuit can be ex-
c|>
density
sinusoidal current
,
which
Consequently,
I
is
peak flux
Bm
.
]X
is
which,
use the peak
rms value? The reason
m
induces an effective voltage
of the
coil,
whose value
E
E must be
is
cIJ
milx
given by Eq.
is
The
.
E a and
flux
9.
1
.
On
the
the induced
identical because they appear be-
tween the same pair of conductors. Because
we can
<J>.
across the terminals
other hand, the applied voltage
voltage
a sinusoidal
sinusoidal flux
called the magnetizing cur-
£u =
= 4.44/MP max
from which we obtain
is
(9.2)
4.44 JN
determines the
level of saturation.
Example 9-1
The coil in Fig.
.
9.1
possesses 4000 turns and links
an ac flux having a peak value of 2
quency
is
mWb.
If
the fre-
60 Hz, calculate the effective value and
(a)
frequency of the induced voltage E.
Sol til ion
4.44/W* inax
(9.1)
4.44
X 60 X 4000 X
2131
V
0.002
(b)
The induced voltage has an
of 2131
age
is
V
effective or
RMS
Figure 9.2
value
a.
and a frequency of 60 Hz. The peak volt-
2131 V2
= 3014
V.
£,
write
proportional to the peak flux
in iron cores,
m produces
rent.The peak value of this ac flux
zero, and so the
why do we
also arises:
/
in turn creates a
zero.
max instead of the
that the
is
The
mmf /V7 m
E,
is
The question
flux
Xm
of the coil
plained as follows:
Eq. 9.1
is
If the resistance
.
negligible, the current
and
in the coil
itself.
e
to a si-
2 tt/V2)
be created by a moving magnet, a nearby ac
coil, or
connected
coil of /V turns
coil
does not matter where the ac flux comes from:
may
shows a
Fig. 9.2a
nusoidal ac source £„. The coil has a reactance
The voltage E induced
plied voltage
b.
Eg
in
a
coil is
equal
to the
ap-
.
Phasor relationships between
£fg! E,
/
m and
,
<l>.
THE IDEAL TRANSFORMER
This equation shows that for a given frequency and
a given
to the
number of
turns,
(I>
lllax
varies in proportion
applied voltage E„. This means that
if
£g
is
kept constant, the peak flux must remain constant.
For example, suppose we gradually
a.
b.
c.
d.
The
The
The
The
185
peak value of flux
mmf
peak value of the
inductive reactance of the coil
inductance of the coil
insert an
Solution
iron core into the coil while
keeping E„ fixed (Fig.
The peak value of
the ac flux will remain ab-
solutely constant during this operation, retaining
original value
$ max
even when the core
is
its
com-
pletely inside the coil. In effect, if the flux increased
(as
we would
also increase.
at
But
this is
we
said,
E„
is
is
therefore the same.
netizing current
core
is
flux, a
The peak current
is
lm
is
much
Wik) = ^2/ =
=
kept fixed.
For a given supply voltage E„, the ac flux
and 9.3
b.
E would
impossible because E — E„
(9.2)
90)
expect), the induced voltage
every instant and, as
9.2
= EJ(4MJN)
= 120/(4.44 X 60 X
= 0.005 - 5 mWb
<D max
a.
9.3).
in Figs.
The peak
However, the mag-
smaller
when
the iron
U=
same
-
inside the coil. In effect, to produce the
smaller magnetomotive force
is
in Fig. 9.3 is
much
X 4
A
5.66
is
NI m = 90 X 5.66
509.
1
needed with
The
an iron core than with an air core. Consequently, the
magnetizing current
mmf U
>/2
smaller than
c.
in Fig. 9.2.
flux
is
equal to 5
mmf is
the coil
509.
1
mWb at the
The inductive reactance
d.
The inductance
when
is
X m = EJI m =
= 30
instant
ampere-turns.
120/4
SI
is
L = XJ2irf
=
30/(2tt
(2.10)
X
60)
- 0.0796
=
9,3
79.6
mH
Elementary transformer
In Fig. 9.4, a coil
having an
air
core
an ac source E„. The resulting current
(b)
a total flux
around the
<1>,
which
we
coil. If
Figure 9.3
The
a.
is
flux in the coil
remains constant so long as
Eg
first,
flux.
constant.
An
second
Phasor relationships.
b.
the
it
A
coil
Hz
having 90 turns
is
in
produces
the space
bring a second coil close to
ac voltage
and
its
E2
is
therefore induced
in
the
value can be measured with a
The combination of the two coils is
The coil connected to the
called a transformer.
is
connected
to a
120 V, 60
source. If the effective value of the magnetizing
current
dispersed
excited by
/m
will surround a portion 4> ml of the total
coil
voltmeter.
Example 9-2
is
is
4 A, calculate the following:
source
is
called the primary winding (or primary)
and the other one
(or secondary).
is
called the secondary winding
1
ELECTRICAL MACHINES AND TRANSF ORMERS
86
Figure 9.5
Terminals having the
Figure 9.4
marked
Voltage induced
<I>
m1
leakage
;
in
a secondary winding. Mutual
flux is
<I>
f1
flux is
.
value
voltage exists only between primary terminals
is
and secondary terminals 3-4, respectively.
voltage exists between primary terminal
ondary terminal
trically isolated
The
flux
up into two
<I>
3.
The secondary
is
No
and sec-
1
therefore elec-
from the primary.
parts: a
mutual flux
4> ml
,
which
links the
same
does. Suppose, during
instant as
apart, the
mutual flux
total flux
<t>,
two
we
coils
is
is
very small compared to the
then say that the coupling between
weak.
We
two
coils closer together.
two
coils touch, the
)
by bring-
However, even
mutual flux will
small compared to the total flux
pling
is
weak, voltage
E2
is
<J>.
When
still
be
the cou-
relatively small and,
when
still,
load
connected across the secondary terminals. In
most
it
collapses almost completely
industrial transformers, the
and
is
a
primary and sec-
ondary windings are wound on top of each other
positive with respect to sec-
(Fig. 9.5).
Terminals
1
and
3 are
terminal
1
and another large dot beside secondary
The dots are called polarity marks.
The polarity marks in Fig. 9.5 could equally
terminal
3.
well
be placed beside terminals 2 and 4 because, as the
voltage alternates, they too,
become simultaneously
positive, every half-cycle. Consequently, the polar-
marks may be shown beside terminals
beside terminals 2 and
1
and 3 or
4.
9,5 Properties of polarity
A
transformer
is
marks
usually installed in a metal enclo-
sure and so only the primary and secondary terminals
are accessible, together with their polarity marks. But
although the transformer
may
lowing rules always apply
to
not be visible, the fol-
to polarity
marks;
improve the coupling between them.
A current entering a polarity- marked terminal
produces a mmf that acts in a "positive'' direc-
9.4 Polarity of a transformer
tive" direction* (Fig. 9.6). Conversely, a current
1
.
tion.
In Fig. 9.4 fluxes
On
and
3> ml are both
stant.
It
They
also pass through zero at the
follows that voltage
E2
will reach
same
its
in-
peak
As
a result,
it
produces a flux
in the "posi-
flowing out of a polarity-marked terminal pro-
produced by
magnetizing current /nr Consequently, the fluxes are
in phase, both reaching their peak values at the same
instant.
1
that
if
worse
is
primary terminal
then said to possess the same polarity. This sameness
ity
E2
bring the secondary right up to the primary so
that the
secondary terminal 3
ondary terminal 4
can obtain a better cou-
pling (and a higher secondary voltage
ing the
that
positive with respect to primary terminal 2
which
links only the turns of the primary. If the coils are far
we
at the
can be shown by placing a large dot beside primary
created by the primary can be broken
turns of both coils; and a leakage flux
the
instantaneous polarity are
one of these peak moments,
A
1-2
same
with a dot.
*
"Positive" and "negative" are
cause
we can
shown
in
quotation marks be-
rarely look inside a transformer to see in
which direction
the flux
is
actually circulating.
THE IDEAL TRANSFORMER
187
9.6 Ideal transformer at no-load;
voltage ratio
Before undertaking the study of practical, commer-
we
cial transformers,
shall
examine
the properties
of the so-called ideal transformer. By definition, an
ideal transformer has
no losses and
nitely permeable. Furthermore,
by the primary
ondary,
is
its
core
is infi-
any flux produced
completely linked by the sec-
and vice versa. Consequently, an
ideal
transformer enclosure
transformer has no leakage flux of any kind.
current entering a polarity-marked terminal pro-
duces a
have properties which ap-
Practical transformers
Figure 9.6
A
flux in
a "positive" direction.
proach those of an ideal transformer. Consequently,
our study of the ideal transformer will help us understand the properties of transformers
Figure 9.8a shows an ideal transformer
duces a mint and flux
in the
"negative" direction.
Thus, currents that respectively flow into and out
2.
turns.
The primary
of polarity-marked terminals of two coils pro-
E„ and
The
If
one polarity-marked terminal
is
momentarily
positive, then the other polarity-marked terminal
to
momentarily positive (each with respect
is
its
in
primary and secondary respectively possess
duce magnetomotive forces that buck each
other.
in general.
is
4
the
and
N2
connected to a sinusoidal source
the magnetizing current / U1 creates a tlux
flux
is
<t>
m
it
is
a mutual flux.
flux varies sinusoidally, and reaches a peak value
)
mrtx
.
According
to Eq. 9.
1,
we
can therefore write:
other terminal). This rule enables us to re-
late the
*m-.^.
phasor voltage on the secondary side
ti
with the phasor voltage on the primary side.
For example,
with phasor
in Fig. 9.7,
£ ab
phasor
£dc
is in
phase
.
(a)
(a)
t:
g
,
A',
(b)
(b)
4>rr
Figure 9.7
a.
Instantaneous polarities
b.
Phasor
rent
is
increasing.
relationship.
when
the magnetizing cur-
Figure 9.8
The ideal transformer at no-load. Primary and secondary are linked by a mutual flux.
b. Phasor relationships at no-load.
a.
.
completely linked by the primary and sec-
ondary windings and, consequently,
The
which
/V,
1
ELECTRICAL MACHINES AND TRANSFORMERS
88
=
£,
4.44/JV,<l> lllilx
(9.3)
Calculate:
a.
The
effective voltage across the secondary
and
terminals
E2 =
From
4.44./yV2
we deduce
these equations,
the voltage ratio
O max
(9.4)
b.
The peak voltage across
the expression for
c.
The instantaneous voltage across the secondary
when the instantaneous voltage across the primary is 37 V
and turns ratio a of an ideal trans-
former:
the secondary terminals
Solution:
a.
The
turns ratio
N2 /N
where
£|
The secondary voltage
N2 =
numbers of turns on
voltage induced
secondary [V]
in the
on the primary
turns
25 times more
£ 2 = 25 X
E
Furthermore, because the primary
<F m they are necessarily in phase.
The phasor diagram at no load is given
is in
If
the transformer has fewer turns
b.
The voltage
on the sec-
As
is
£,
shorter
any inductor, current / m lags
90° behind applied voltage Zs The phasor reprethan phasor
£",.
obviously
phase with magne-
senting flux <P m
is
which produces
However, because
magnetic circuit
this is
is
.
in
varies sinusoidally; consequently,
P cak)=
V 2£ = V 2 X 3000
=4242 V
c.
The secondary voltage
at
e 2 = 25
permeable and so
diagram of such a transformer
9.8b except that phasor / m
is
is
identical to Fig.
infinitesimally small.
primary and 2250 turns on the secondary
source.
tween the primary and secondary
is
4 A.
x
37
E
x
V
Pursuing our analysis,
I 2 will
I2
con-
Does
The coupling be-
swer
is
is
perfect, but the
let
us connect a load
Z across
A
the secondary of the ideal transformer (Fig. 9.9).
transformer having 90 turns on the
magnetizing current
=
e
current ratio
secondary current
Hz
when
X 37 = 925 V
Example 9-3
nected to a 120 V, 60
25 times greater than
9.7 Ideal transformer under load;
required to produce the
is
is
every instant. Consequently,
it.
Thus, under no-load conditions, the phasor
A not quite ideal
is:
an ideal transformer,
infinitely
no magnetizing current
$m
V
E 2 = 3000 V
.
tizing current I m
flux
90/2250
in
tT
the
apply Eq. 9.5:
l
=
the peak secondary voltage
2(
E2
we can
/E 2 =N /N 2
\20/E2
180° out of phase) as indicated by the polarity
marks.
120
in Fig.
phase with phasor £, (and not
ondary than on the primary, phasor
]
which again yields
,
E2
X
25
Instead of reasoning as above,
equal to the ratio of the
is
and secondary voltages are induced by the same
9.8b. Phasor
=
Ei
-3000 V
and secondary voltages
turns.
Consequently:
turns.
turns ratio
number of
therefore 25 times greater
is
than the primary voltage because the secondary has
the secondary
This equation shows that the ratio of the primary
mutual
(9.5)
voltage induced in the primary [V]
numbers of
=
-2250/90
x
= 25
=
E2 =
N\ =
a
is:
E2
immediately flow, given by:
= E2 IZ
we connect the load? To anwe must recall two facts. First, in
change when
this question,
an ideal transformer the primary and secondary windings are linked by a mutual flux
<I>
m and by no
,
other
THE IDEA L TRANSFORMER
same
the
<*W:
when
time. Thus,
maximum +
is
be
E
To
N2
N,
{
mark on
when
effect,
on the primary
/,,
)
/,
lags behind
behind
Ee
,
we can now draw
facts,
polarity
a resistive-inductive load, current
E2
by an angle
Flux
6.
but no magnetizing current
produce
the phasor
under load (Fig.
ideal transformer
Assuming
9.9b).
must flow out of the
the secondary side (see Fig. 9.9a).
Using these
to
(
flows into a polarity mark
/[
side, I 2
diagram of an
I2
maximum +
phase. Furthermore, in order to produce the
in
bucking
(a)
I 2 is
89
In other words, the currents must
).
(
goes through zero,
/2
goes through zero, and when
1
this flux
because
this is
<I>
/
lags 90°
m
m
is
needed
an idea trans-
former. Finally, the primary and secondary currents
(b)
According
are in phase.
to Eq. 9.6, they are related
by the equation:
Figure 9.9
a.
b.
AT,
under load. The mutual
mains unchanged.
Phasor relationships under load.
Ideal transformer
(9.7)
flux re-
where
f = primary
flux. In
other words, an ideal transformer, by defini-
tion,
has no leakage flux. Consequently, the voltage
ratio
under load
is
the
same
/Vj
as at no-load, namely:
the
if
E„
the supply voltage
is
E
induced voltage
primary
E2
E2
also remains fixed.
remains fixed whether a load
Let us
remains fixed.
{
now examine
the
is
It
mmf would produce
in current
not.
the mutual flux
change under
<L>
m But we just saw
.
load.
only remain fixed
We
if
conclude
If
it
acted
that
<E>
in
m does not
<I>
m can
the primary develops a
mmf
which exactly counterbalances
Thus, a primary current
/j
that flux
N2 I 2
at
every
must flow so
AVi = N 2 I 2
To obtain
.
profound change
instant.
f and
I2
what we gain
and vice versa. This
is
see that the
in voltage,
the primary
E2 l 2
we
lose
consistent with the
Ef
to
x
must equal the apparent power output
power inputs and outwould mean that the
transformer itself absorbs power. By definition, this
of the secondary.
If the
puts were not identical,
is
impossible
in
it
an ideal transformer.
Example 9-4
ideal transformer
having 90 turns on the primary
and 2250 turns on the secondary
is
connected
to a
(9.6)
200
the required instant-to-instant bucking ef-
fect, currents
ratio. In effect,
we
the inverse of the volt-
is
requirement that the apparent power input
An
that:
9.5 and Eq. 9.7,
transformer current ratio
age
mmf N2 I2
a
turns ratio
Comparing Eq.
We conclude that
magnetomotive forces
current I 2 produces a secondary
—
connected or
created by the primary and secondary windings. First,
alone, this
turns on the secondary
kept fixed, then
Consequently, mutual flux <P m also remains fixed.
follows that
number of
on the primary
x
a
Second,
secondary current [A]
= number of turns
N2 =
- N /N2
£,/£ 2
=
I2
current |A)
must increase and decrease
at
V,
50 Hz source. The load across the secondary
draws a current of 2
cent lagging (Fig.
A
at
9. 10a).
a
power
factor of 80 per-
90
1
ELECTRICAL MACHINES AND TRANSFORMERS
Figure 9.10
See Example 9-4.
b. Phasor relationships.
a.
current in the secondary. Therefore
Calculate:
a.
The
b.
The instantaneous
effective value of the primary current
current
in the
100
mA,
primary when
/,
1/
100
is
mA
c.
The peak
d.
Draw
flux linked
by the secondary winding
the phasor diagram
c.
a.
The
is
The peak
turns ratio
= ?s
/
— '2
instantaneous
= 25 X 0.
= 2.5 A
the
same
as that linking the primary.
flux in the secondary
=
<P max
1/25
current ratio
is
= E^/(4MfN )
= 200/(4.44 X 50 X
= 0.01
]
therefore 25 and because the
10
primary has fewer turns, the primary current
25 times greater than the secondary current.
25
X
2
d.
by means of Eq.
N,/,
90/,
/,
b.
To draw
= 50 A
Instead of reasoning as above,
the current
mWb
the phasor diagram,
lows: Secondary voltage
=
we can
The instantaneous current
E2
is
in
Phase angle between
in the
power
primary
is
al-
ways 25 times greater than the instantaneous
as fol-
25
X 200
phase with £, indicated by the polarity
marks. For the same reason,
2
we reason
is
E2 = 25 X Ei =
= 5000 V
calculate
9.6.
= N 2 l2
- 2250 X
= 50 A
90)
is
Consequently:
/,
is
is:
^=yv,/yv 2 -90/2250
The
=
In an ideal transformer, the flux linking the sec-
ondary
Solution:
l2
is:
instantaneous
1
the instantaneous current in the secondary
when
/j
is
in
E 2 and
l 2 is
factor
=
cos 6
0.8
=
cos e
9
=
36.9°
phase with
I2
.
THE IDEA L TRA NS FORMER
Figure 9.11
Symbol for an
b. Symbol for an
a.
ideal transformer
ideal transformer
and phasor diagram using sign notation.
and phasor diagram using double-subscript
The phase angle between £, and /, is also 36.9°.
The mutual tlux lags 90° behind E„ (Fig. 9.10b).
In
and
for
an
1
To highlight the bare
former,
it is
9.1 lb),
ideal transformer
best to
/,
la,
draw
in
it
symbolic form. Thus,
in-
and the mutual tlux
ing primary
4> m
,
we
(Fig.
9.
1
1).
to indicate the
direction of current flow as well as the polarities of
E
voltages
into
,
and
£2 F° r example,
-
a current
one polarity-marked terminal
panied by a current
is
ity-marked terminal. Consequently,
ways
in
{
flowing
always accom-
/,
and
U
are al-
phase.
/N 2 =
a,
we
and
E2
are always in phase,
and so are
double-subscript notation
£ab
/2
and
Ecd
are
always
(which
in
used (Fig.
is
phase and so are
.
may sometimes
the nature of the load
be a source) connected to
the secondary side.
9.9
Impedance
ratio
Although a transformer
is
generally used to trans-
form a voltage or current,
it
also has the impor-
tant ability to transform an
impedance. Consider,
for example, Fig. 9. 12a in
which an
former
T
load Z.
The
is
ideal trans-
connected between a source E„ and a
ratio of transformation
is
a,
and so we
can write
Furthermore,
N
/[
flowing out of the other polar-
I2
x
simply show a box hav-
and secondary terminals
marks are added, enabling us
Polarity
and
E
The angle a depends upon
essentials of an ideal trans-
stead of drawing the primary and secondary windings
notation.
/ 2 .*
If the
symbol
9.8 Circuit
9
an ideal transformer, and specifically referring to
Fig. 9.
/,
1
if
we
let
the ratio of transformation
obtain
£,
*
= a£ 2
Some
texts
show
the respective voltages
and currents as being
180° out of phase. This situation can arise depending upon
how
the behavior of the transformer
is
described, or
how
the
By
us-
voltage polarities and current directions are assigned.
and
ing the methodology
/,
=
/ 2 /a
never any doubt as to
we have adopted
how the phasors
in this
book, there
should be drawn.
is
92
1
ELECTRICAL MACHINES AND TRANSFORMERS
impedance across
the actual
tical to
the secondary
terminals multiplied by the square of the turns
ratio.
The impedance transformation is real, and not
illusory like the image produced by a magnifying
An
glass.
(a)
ideal transformer can
of any component, be
ductor. For example,
modify the value
a resistor, capacitor, or in-
it
a 1000 12 resistor
if
placed
is
across the secondary of a transformer having a pri-
mary
secondary turns ratio of
to
across the primary as
X
(1/5)"
= 40
12.
if
it
Similarly,
a reactance of 1000 fl
ondary,
it
1
:5,
will appear
it
had a resistance of 1000
is
if
a capacitor having
connected to the sec-
appears as a 40 il capacitor across the
primary. However, because the reactance of a capacitor
—
(X c
Figure 9.12
a. Impedance transformation using a transformer.
b. The impedance seen by the source differs from
is
inversely proportional to
l/2ir/'C), the
the primary terminals
actual value.
Z
We
its
capacitance
apparent capacitance between
is
25 times greater than
its
can therefore artificially increase
(or decrease) the microfarad value of a capacitor
by means of a transformer.
EJE,
impedances from
secondary to primary
and vice versa
9.10 Shifting
and
LIU =
As
far as the
ance
source
Z x between
is
l/a
concerned,
it
sees an imped-
the primary terminals given by:
Zx = EJ1
As
a further illustration of the impedance-changing
properties of an ideal transformer, consider the circuit
of Fig.
9.
1
3a.
It is
composed of
a source E„, a
!
transformer T, and four impedances Z, to
On the other hand,
Z given by
the secondary sees an
impedance
transformer has a turns ratio
We can progressively
to 9.l3e.
Zx
can be expressed
^
=
a£\
in
another way:
/ 2 /a
2
Z
As
the
impedances are shifted
x
1
3b
way,
the
impedances are transferred
a~.
to the pri-
A,
mary
side, the ideal
this position the
a~Z
(9.8)
This means that the impedance seen by the source
times the real impedance (Fig.
ideal transformer has the
9.
in this
transformer ends up
treme right-hand side of the circuit (Fig.
Z =
2
secondary imped-
impedance values are multiplied by
If all
Consequently,
a
shift the
the circuit configuration remains the same, but the
shifted
a~£\
a
/,
.
ances to the primary side, as shown in Figs.
Z = E 2 /I 2
However,
Z4 The
a.
amazing
9.
1
is
2b). Thus, an
ability to increase
is
iden-
1
3d). In
secondary of the transformer
is
on
open-circuit. Consequently, both the primary and
secondary currents are zero.
move
In
der
We
can therefore
re-
the ideal transformer altogether, yielding the
equivalent circuit
or decrease the value of an impedance. In effect, the
impedance seen across the primary terminals
at the ex-
9.
comparing
how
a circuit
shown
Figs. 9.
1
in Fig. 9.l3e.
3a and
9.
which contains a
1
3e,
real
we may wonT
transformer
THE IDEAL TRANSFORMER
Z,
a
2
193
Z3
(e)
Figure 9.13
a.
The
and
actual circuit showing the actual voltages
currents.
Impedance Z 2
b.
is
c.
Impedance Z3
Impedance Z4
is
rents
All
in
/2
.
change
E3 and
in
/3
Note
.
Note
.The cur-
shifted to the primary side.
the corresponding
e.
E 2 and
in
shifted to the primary side.
is
the corresponding
d.
Note
shifted to the primary side.
the corresponding changes
T are now
change
E4 and
in
/4
zero.
the impedances are
now
transferred to the
mary side and the transformer
pri-
no longer needed.
is
can be reduced to a circuit which has no transformer
any meaningful relationship
at all. In effect, is there
between the two circuits? The answer
is
yes
—
there
is
a useful relationship between the real circuit of Fig.
9.
13a and the equivalent circuit of Fig. 9.l3e. The
reason
that the voltage
is
£ across each element in the
secondary side becomes
a£ when
the element
shifted to the primary side. Similarly, the current
each element
the element
On
in the
is
secondary side becomes
account of this relationship,
simply reduce
in Fig. 9.
rents.
by
I
1
it
Z4
easy to solve
is
it
shown
in Fig. 9.
to the equivalent
3e and solve for
3a.
the voltages and cur-
all
/a
and by
a,
which yields the actual voltages
illustrate,
in Fig. 9.
14
in the
suppose that the
is
£4
volts
and
amperes. Then,
secondary
that the real current
it
cuit, the
voltage across the a"Z4 impedance
£4 X
is /4
a volts.
On
through the impedance
side.
real voltage across
through
to
1
form shown
These values are then respectively multiplied
and currents of each element
To
when
//a
shifted to the primary side.
a real circuit such as the one
We
is
/ in
in
the equivalent ciris
equal
the other hand, the current
is
equal to
/4
^
a amperes
ELECTRICAL MACHINES AND TRANSFORMERS
194
Figure 9.14
Actual voltage and current
in
Z4
impedance
a
2
.
Z4
a£ 4
Figure 9.15
Equivalent voltage and current
(c)
.
z,
1?
words, whenever an impedance
(Fig. 9. 15). In other
is
Z4
in
transferred to the primary side, the real voltage
across the impedance increases by a factor a, while
the real current decreases by the factor a.
general,
In
ferred
whenever an impedance
from one side of a transformer
the real voltage across
the turns ratio. If the
side
it
changes
impedance
is
is
trans-
to the other,
transferred to the
where the transformer voltage
is
higher, the
voltage across the transferred impedance will also
be higher. Conversely,
ferred to the side
if
the
J
proportion to
in
impedance
is
Figure 9.16
a.
where the transformer voltage
is
b.
lower than the
the ratio of the
In
real voltage
—
again, of course, in
c.
some cases
it
is
useful to shift
is
Impedance Z
1
is
The source
is
impedances
secondary side (Fig. 9.16a). The procedure
in
to the
is
the
d.
and
in £-,
A,.
transferred to the secondary side.
in
£g Note
.
also
T are zero.
All the impedances and even the source are now
on the secondary side. The transformer is no
longer needed because its currents are zero.
that the currents in
from the primary side
and
transferred to the secondary side.
Note the corresponding change
number of turns.
the opposite way, that
actual circuit, showing the real voltages
Note the corresponding change
lower, the voltage across the transferred impedance
is
The
currents on the primary side.
trans-
THE IDEA L TRA NS FORMER
same, but
all
vided by a
2
impedances so transferred are now
(Fig. 9.16b).
We
can even
source E„ to the secondary side, where
source having a voltage
is
now
Eg /a. The
it
Z-
becomes a
ideal transformer
\
R 2 + (X L - Xc
= V4 2 +
primary of
circuit (Fig. 9.16c). In this position the
circuit in Fig. 9. 18
95
is
shift the
located at the extreme left-hand side of the
the transformer
+
\ 16
-
(5
-
2
(2. 17)
)
2
2)
9
on open-circuit. Consequently,
is
both the primary and secondary currents are zero.
before,
The impedance of the
di-
1
we can remove
=
As
the transformer completely,
The
s
n
current in the circuit
is
leaving us with the equivalent circuit of Fig. 9.16d.
= E/Z =
/
Example 9-5
The voltage across
Calculate voltage
Fig. 9.17,
E
knowing
and current
/ in the circuit
that ideal transformer
primary to secondary turns ratio of
1
:
T
£7 100
has a
100.
The
the
impedances
to solve this
to the
problem
is
,
= 8V
therefore,
100
- 800 V
Questions and Problems
impedance values are
or 10 000. Voltage
E becomes
£7100, but current / remains unchanged because
9-
1
The
already on the primary side (Fig. 9.
1
500 turns and
coil in Fig. 9.2a has
a re-
actance of 60 il but negligible resistance.
it
it
is
X
is,
is
primary side of the trans-
turns than the secondary, the
2
E
A
to shift all
former. Because the primary has 100 times fewer
divided by 100
8
2
2X4
=
//?
actual voltage
E=
way
easiest
=
=
the resistor
of
Solution
The
10/5
8).
is
connected to a 120 V, 60 Hz source
If
£u
,
calculate the following:
a.
The
effective value of the magnetizing cur-
rent / m
9-2
b.
The peak value of / ni
c.
d.
The peak and mm!' produced by
The peak flux 4> m:ix
In
Problem
to
40
9-3
,
if
the voltage E„
What
coil
is
new
reduced
is
mmf developed
and the peak flux
(I)
I1UIX
.
meant by mutual flux? by leakage
flux?
Figure 9.17
See Example
1
V, calculate the
by the
1:100
9-
the coil
9-5.
9-4
The
ideal transformer in Fig. 9.9 has
turns on the primary and
secondary.
of 12
Figure 9.18
Equivalent circuit of Fig. 9.17.
a voltage
a resistance
Calculate the following:
The voltage E 2
b.
The
c.
e.
The current /,
The power delivered to
The power output from
the
secondary
In
Problem
the
impedance seen
d.
9-5
11.
Z is
500
turns on the
The source produces
E„ of 600 V, and the load
a.
300
current I 2
9-4,
by the source
what
is
the primary
[
WJ
W]
|
196
9-6
ELECTRICAL MACHINES AND TRANSFORMERS
In Fig. 9. 17, calculate the voltage across
the capacitor
through
9-9
and the current flowing
Two
coils are set
up as shown
in Fig. 9.4.
Their respective resistances are small and
may be
it.
1,
neglected.
The
having terminals
coil
2 has 320 turns while the coil having
ter-
Industrial application
minals
9-7
The nameplate on a 50 kVA transformer
shows a primary voltage of 480 V and a sec-
when
1
is
mine the approximate number of turns on the
primary and secondary windings. Toward
this
wound around
the
end, three turns of wire are
and a voltmeter
external winding,
across this 3-turn coil.
then applied to the
1
connected
is
V winding,
voltage across the 3-turn winding
found
to
How many turns are there on the
480 V and 120 V windings (approximately)?
A coil with an air core has a resistance of
is connected to a 42 V, 60
14.7 fl. When
it
Hz
ac source,
it
draws a current of
1
.24 A.
Calculate the following:
a.
b.
c.
The impedance of the coil
The reactance of the coil, and its inductance
The phase angle between the applied voltage
(42 V) and the current
(
1
.24 A).
Hz
It is
voltage
is
found
that
applied to
ter-
voltage across terminals 3-4
V. Calculate the
peak values of <)>,
<(>, ,,
.
|jlF,
600
V
paper capacitor
we need one having
jjlF. It is
is
available,
a rating of about
proposed to use a transformer
modify the 40
fxF so that
it
appears as
fxF. The following transformer ratios
are available: 120 V/330 V; 60 V/450 V;
480 V/1 50 V. Which transformer is the most
300
be 0.93 V.
9-8
1-2, the
4> ml
A 40
but
to
and the
is
^
300
A voltage of 76 V is
20
22
and
4 has 160 turns.
a 56 V, 60
minals
We wish to deter-
ondary voltage of 20 V.
3,
appropriate and what
is
the reflected value
of the 40 fxF capacitance? To which side of
the transformer should the
be connected?
40 p,F capacitor
Chapter
1
Practical Transformers
10.0 Introduction
In
we
Chapter 9
discovered
real
its
with an imperfect core
studied the ideal transformer and
basic properties.
However,
in the
The
world transformers are not ideal and so our
happens
account. Thus, the windings of practical transform-
have resistance and the cores are not
iron
infinitely
is
not completely captured by the sec-
And
losses,
a practical
we
primary
which con-
The
transformer can be described by an
resistance
current
is in
is
phase with
E
current
its
£p
that
1
0.
1
a).
The
produces a
x
(Fig. 10.1b).
Xm
a
is
measure of
the permeability of the transformer core. Thus,
age regulation and the behavior of transformers that
also used to illustrate
with the primary
To furnish these
drawn from the line. This
The magnetizing reactance
cir-
The per-unit method
mode of application.
We
represents the iron losses and
/,
the permeability
in parallel.
Rm
losses a small current
developed from fundamental concepts. This
connected
in parallel
excited by a source
is
enables us to calculate such characteristics as volt-
are
low?
the resulting heat they produce.
equivalent circuit comprising an ideal transformer
cuit is
rather
is
voltage £],
discover that the properties of
and resistances and reactances. The equivalent
and eddy-current
hysteresis
whose permeability
R m and X m
elements
tribute to the temperature rise of the transformer.
In this chapter
What
replaced by an
terminals of the ideal transformer (Fig.
finally, the iron cores pro-
duce eddy-current and hysteresis
having
core
is
can represent these imperfections by two circuit
ondary. Consequently, the leakage flux must be
taken into account.
the previous
in
such a perfect core
if
losses and
permeable. Furthermore, the flux produced by the
primary
transformer studied
ideal
chapter had an infinitely permeable core.
simple analysis must be modified to take this into
ers
Ideal transformer
10.1
/m
is
flowing
X m is relatively
through X m represents
low,
if
The
mag-
low.
the
netizing current needed to create the flux 3> m in the
is
core. This current lags 90° behind
197
E
.
}
1
ELECTRICAL MACHINES AND TRANSFORMERS
98
= 0
/1
Jo
/o=0
A
5
120 V
60 Hz
A
An
ideal
X
T
Figure 10.1a
An
imperfect core represented by a reactance
a resistance
Rm
Xm and
Figure 10.2a
See Example
10-1
.
Xm
The values of the impedances R m and
can be
found experimentally by connecting the trans-
transformer
is
shown
former to an ac source under no-load conditions and
<t>
measuring the active power and reactive power
absorbs.
m =
is
The peak value
again given by Eq. 9.2:
E f(AMfN
x
x
(9.2)
)
it
The following equations then apply:
R m = E* 2 /P m
Xm = Er/Q m
in Fig. 10.1b.
of the mutual flux <£ m
Example 10-1
(10-1)
A
(10.2)
exciting current 70 of 5
large transformer operating at no-load
A when
draws an
the primary
is
where
a wattmeter test
R [u =
Xm =
resistance representing the iron losses [Cl]
equal to 180
magnetizing reactance of the primary
Calculate
winding
Ei
=
Pm —
Q in =
The
[flj
primary voltage [V]
c.
iron losses
reactive
power needed
IvarJ
total
/ in
.
[W]
flux
<t>
m
to set
up the mutual
It is
ally a small
is
equal to the phasor
called the exciting current 7 0
.
<t>
m
at
no-load for
The reactive power absorbed by
The value of R m and X m
The value of 7f 7 m and 7
sum
of 7f
It is
usu-
,
,
the core
t)
The apparent power supplied
Sm
in
percentage of the full-load current. The
phasor diagram
W.*
Solution
a.
current needed to produce the flux
an imperfect core
and
a.
b.
it is
=
E,7C
=
120
to the core
X
is
5
= 600 VA
The
iron losses are
P m = 80 W
this less-than-ideal
1
The
reactive
Qm =
power absorbed by
^^~Pi
= 572
Figure 10.1b
Phasor diagram
of
a practical transformer
at no-load.
con-
Hz source (Fig. 10.2a). From
known that the iron losses are
nected to a 120 V, 60
var
the core
= V600 r^780I
is
PRACTICAL TRANSFORMERS
The impedance corresponding
b.
99
1
to the iron losses
is
2
Rm = E
=
{
120 /180
The magnetizing reactance
c.
2
}
= 0
is
2
/Q m = 120 /572
25.2 ft
The current needed
If
1
a
so
Xm = E
/
2
=
IP m
supply the iron losses
to
=
E\/R m
=
1.5
=
is
120/80
Figure 10.3
A
Transformer with
The magnetizing current
/m
= E IX m =
= 4.8 A
across the primary
120/25.2
The exciting current
-
V/?
5
is
x
$m
its
h =
infinitely
+
/t)
i
peak value
Vl.5
2
+
4.8
2
cause
it
A
E2
to the current
is
given
in Fig. 10.2b.
Ep
and
it
sets
up
a
mutual flux
given by ^> nikl
is
Ep and
= £ p /(4.44 /TV,).
infinitely
is
permeable and be-
/, = 0.
E2 = (N2 /N E p Owing
zero, no mmf is available to
has no losses, the no-load current
The voltage
The phasor diagram
at no-load.
This flux lags 90° behind
a in the core.
Because the core
is
=
II
permeable core
is
is
given by
being
drive flux through the
air;
{
)
.
consequently, there
is
no
leakage flux linking with the primary.
If
= 1.5
A
120
V
Let us
now connect
a load
Z
across the sec-
ondary, keeping the source voltage
A'L^Si
= 4.8
/m
10.4).
/«
~ 5
A
Ep
fixed (Fig.
This simple operation sets off a train of
events which
we
list
as follows:
Figure 10.2b
Phasor diagram.
10.2 Ideal transformer
with loose coupling
We have just seen how an ideal transformer behaves
when it has an imperfect core. We now assume a
transformer having a perfect core but rather loose
coupling between
ings.
We
also
its
primary and secondary wind-
assume
that the
primary and sec-
ondary windings have negligible resistance and the
turns
areN ,N2
Consider the transformer
to a
source E„ and operating
Figure 10.4
Mutual fluxes and leakage fluxes produced by a trans-
.
l
in Fig. 10.3
at
no-load.
connected
The voltage
former under load. The leakage fluxes are due to the
imperfect coupling between the
coils.
ELECTRICAL MACHINES AND TRANSFORMERS
200
1
Currents
.
and
/,
2.
I2
flow
to
= N2 /N
hence
:
{
produces an
mmf /V,/,.
equal and
/V,/|
mmf N2 I 2
= N2 I2
while
-
/|
produces an
These magnetomotive forces are
in direct
when
opposition because
flows into the polarity-marked terminal
flows out of polarity-marked terminal
The mmfiV 2 / 2 produces
3.
portion of
C I>
(
2
(
<t>,
.
mmf /V]/,
A portion
Flux
<I>,
of
(<I> t 2
-
produces a
(<I> ml
is
,
)
called the
The magnetomotive
the magnetic field
)
does
we
can
Figure 10.5
not.
A
$
ni
links with the sec-
P n ) does
analyze
this
and
/,
In general,
1
of two parts; a
same
the
new mutual
flux
(The mutual flux
<!>,-,.
<I> inl
total flux
posed of a mutual flux
<I>
A voltage E n
.
composed
and
we combine
is
£n
A voltage E
2.
produced by
mutual flux
<I>
m
ni
,
and
(Fig. 10.5). This
/2
com-
is
is
we
flux
<I>
n
is
mutual flux
is
NJ
phase with
2
1
.
<I>
n and
induced by mutual flux
x
(10.5)
4> in
and
=
4.44./N
<I>
l
Ep =
six basic facts,
(10.6)
m
applied voltage E^.
we now proceed
to
note that the primary leakage flux
/ lr
and leakage flux
O
l2
10-3 Primary
E
We can better identify the four induced voltages
E2 £ n and £ t2 by rearranging the transformer cir-
induced
composed of two
A voltage E [2
,
,
s
in the
secondary
is
cuit as
parts:
induced by leakage flux
winding
<I> I2
and secondary
leakage reactance
is in
.
Fifth, the voltage
actually
Using these
.
I
primary
in the
=4.44.^,*,-,
Sixth, induced voltage
cre-
h while the secondary leakage
created by /V 2 / 2 Consequently, leakage flux
is in
induced
parts:
develop the equivalent circuit of the transformer.
created by
phase with
are not in phase.
Ep
3> m2 into a single
ated by the joint action of the primary and sec-
Fourth,
(10.4)
given by
.
ondary mmfs.
(J>,-,
E2
#m
given by
not
m2 and a leakage flux <£ (2
<t>
4.44/7V2
induced by leakage flux
£,
Third,
m and
and a leakage
(p mJ in Fig. 10.4
as ( F mla in Fig. 10.3.)
Second, the
E [2
composed of two
reason as follows:
is
tp
induced by mutual flux
Similarly, the voltage
is
/]
E2
E2 =
how
is,
situation?
we
voltage
upset
/2
that existed in the core be-
produced by
flux.
given by
(
new
First, the total flux
A
2.
c
(
primary leakage flux.
forces due to
|.
Referring to Fig. 10.4,
flux
transformer possesses two leakage fluxes and a
ac flux
total
was connected. The question
fore the load
A
.
^ m 2) links with the primary
ondary winding, while another portion
not.
3.
mutual
Similarly, the
4.
t
called the secondary leakage flux.
2 is
<I>,
/
1, 1 2
a total ac flux 4> 2
winding while another portion
Flux
in
by the ideal-transformer equation
related
/,// 2
immediately begin
I2
and secondary windings. They are
the primary
and
10.6.
to
.
4.44/N 2 <I\2
(10.3)
by two fluxes,
clearly
<J>
r2
and
This rearrangement does not change the value
of the induced voltages, but
Er
Thus, the secondary
show even more
that the jV 2 turns are linked
$m
given by
shown in Fig.
is drawn twice
age stand out by
itself.
it
Thus,
does make each
it
becomes
volt-
clear that
PRACTICAL TRANSFORMERS
201
Figure 10.6
Separating the various induced voltages due to the mutual flux and the leakage fluxes.
Figure 10.7
Resistance and leakage reactance
E l2
is
of the
primary and secondary windings.
drop across a reactance. This
really a voltage
X
secondary leakage reactance
-
{
2 is
given by
Example 10-2
The secondary winding of
180 turns.
X = Ef2 /I2
(1
l2
0.7)
When
a transformer possesses
the transformer
is
under load, the
secondary current has an effective value of 8 A, 60
1
The primary winding
also
is
shown
twice, to
separate £, from E n Again, it is clear that E n is
simply a voltage drop across a reactance. This pri.
mary leakage reactance
Xn
Xn
is
=£
fl
shown
in
Figure 10.7.
We
(10.8)
//
l
tive
which, of course, act
windings.
mWb. The
m has a peak value of
secondary leakage flux
<I>,
2
has a
mWb.
Calculate
a.
in series
and
b.
with the respec-
c.
/?,
The voltage induced
by
have also added the
primary and secondary winding resistances
Ri,
20
peak value of 3
<I>
given by
The primary and secondary leakage reactances
are
Hz. Furthermore, the mutual
its
in the
secondary winding
leakage flux
The value of the secondary leakage reactance
The value of E 2 induced by the mutual
ELECTRICAL MACHINES AND TRANSFORMERS
202
—Oi
Eo
1
'lm
Figure 10.8
Complete equivalent
/f
X
a practical transformer. The shaded box T
circuit of
If
Solution
a.
The
effective voltage induced by the secondary
leakage flux
an
is
we add
ideal transformer.
elements
circuit
sent a practical core,
we
Xm
and
Rm
to repre-
obtain the complete equiv-
alent circuit of a practical transformer (Fig. 10.8).
is
In this circuit
E l2 =
=
4.44/yV 2
f2
X 60 X
4.44
=
O
143.9
X
180
T
is
an ideal transformer, but only the
(10.3)
primary and secondary terminals
0.003
cessible;
I
-2
and 3-4 are ac-
other components are "buried" inside
transformer
the
V
all
However, by appropriate
itself.
we can find the values of all the circuit elements that make up a practical transformer.
Table 10A shows typical values of /?,, R 2 X n
X r2 X m and R m for transformers ranging from kVA
tests
b.
The secondary leakage reactance
is
X = EnJI 2
(10.7)
{2
=
143.9/18
,
,
I
,
400 MVA. The nominal primary and secondary
£np and £ ns range from 460 V to 424 000 V.
The corresponding primary and secondary currents
/ np and / ns range from 0.4 7 A to 29 000 A.
The exciting current /0 for the various transto
= 8ft
c.
The voltage induced by
voltages
the mutual flux
is
1
E2 = 4.44^2*.
=
(10.4)
X 60 X
4.44
1
X
80
0.02
formers
also shown.
is
is
It
always much smaller
than the rated primary current
= 959 V
Note
that in
where S n
is
the rated
/,
EnpInp = En J ns —
each case
5n
power of the transformer.
10.4 Equivalent circuit
TABLE 10A
of a practical transformer
The
circuit of Fig.
1
0.7
and inductive elements
is
composed of
(/?,,
R 2 X n X i2
,
,
resistive
,
Z) cou-
m which links the
primary and secondary windings. The leakagepled together by a mutual flux
free
<t>
in
same properties and obeys
For example, we can
primary
(Ay/V 2 )
side
2
,
as
we
by
shift
multiplying
did before.
I
10
2400
2400
1
400000
100
1000
2470
69000
13800
V
A
A
460
347
600
6900
424000
0.4 17
4.17
8.02
14.5
29000
2.17
28.8
167
145
943
a
58.0
5.16
11.6
27.2
0.0003
pos-
R,
fi
1.9
0.095
0.024
0.25
0.354
X~n
a
32
4.3
39
151
0.028
Chapter
Xa
xm
Q.
1.16
0.09
0.09
1.5
27
150000
460
Km
A»
52.9
impedances
their
'ns
kVA
V
same
It
the
rules as the ideal transformer discussed in
9.
/„p
the dotted
actually an ideal transformer.
is
sesses the
Ens
,
magnetic coupling enclosed
square
£n P
ACTUAL TRANSFORMER VALUES
to the
values
by
a
200000 29000
400000 51000
220000
505000
432000
A
0.0134 0.0952
0.101
0.210
317
,
PRACTICAL TRANSFORMERS
10.5 Construction of a
power
Winding
resistances
2
to reduce the J
transformer
R
loss
R and R 2
:
1
0.9a
Power transformers
are usually designed so that
a
characteristics
approach those of an ideal
ondary are wound on one
their
transformer. Thus, to attain high permeability, the
made of
The resulting
magnetizing current / m is at least 5000 times
smaller than it would be if an air core were used.
Furthermore, to keep the iron losses down, the core
core
is
is
iron (Fig.
1
0.9a).
laminated, and high resistivity, high-grade silicon
steel is used.
Consequently, the current
supply the iron losses
than
/m
is
/r
needed
to
usually 2 to 4 times smaller
power transformer
and secondary
leg. In practice, the
amount of copper. For
built
up so as
Figure
1
to
is
0.9b shows
how
2. 10a).
small transformer are stacked to build up the core.
Figure
1
0.9c shows the primary winding of a
voltages.
them as closely together as insulation considera-
The
coils are carefully insulated
from each other and from the core. Such
coils
is
means
that the
tight
when
It
also guarantees
a load
is
connected
secondary terminals.
cou-
secondary
almost exactly equal to
times the primary voltage.
voltage regulation
1
the laminations of a
ondary coils on top of each other, and by spacing
voltage at no-load
the
not square (as shown) but
be nearly round. (See Fig.
ondary windings depends upon
between the
sec-
primary
reason, in larger transformers the cross section
Leakage reactances X n and X r2 are made as
small as possible by winding the primary and sec-
pling
ensure
much
bigger transformer.
.
tions will permit.
to
a simplified version of
coils are distributed over both core
of the laminated iron core
is
is
which the primary and
in
legs in order to reduce the
same
are kept low, both
and resulting heat and
high efficiency. Figure
203
N2 /N
i
good
to the
The number of
turns on the primary and sectheir
than a low-voltage winding.
current in a
HV
winding
is
On
the other hand, the
much
smaller, enabling
us to use a smaller size conductor.
amount of copper
windings
respective
A high-voltage winding has far more turns
is
in
As
a result, the
the primary and secondary
about the same. In practice, the outer
coil (coil 2, in Fig. I0.9a)
length per turn
is
greater.
weighs more because the
Aluminum
ductors are used.
laminated iron core
Figure 10.9a
Figure 10.9b
Construction of a simple transformer.
Stacking laminations inside a
coil.
or copper con-
ELECTRICAL MACHINES AND TRANSFORMERS
204
subtractive polarity
additive polarity
x2 «
Figure 10.10
Additive and subtractive polarity
H
cation of the
1
-X
additive polarity
1
depend upon the
lo-
terminals.
when
terminal H,
is
diagonally
opposite terminal X,. Similarly, a transformer has
subtractive polarity
when
terminal Hj
terminal X, (Fig. 10.10). If
we know
is
adjacent to
that a
power
transformer has additive (or subtractive) polarity,
we do
not have to identify the terminals by symbols.
Subtractive polarity
is
standard for
all
single-
phase transformers above 200 kVA, provided the
high-voltage winding
is
rated
above 8660
V. All
other transformers have additive polarity.
Figure 10.9c
Primary winding
290 A.
of
a large transformer;
rating
128
kV,
10.7 Polarity tests
To determine whether
(Courtesy ABB)
a transformer possesses ad-
ditive or subtractive polarity,
A
ther
transformer
is
reversible in the sense that ei-
winding can be used as the primary winding,
1
where primary means the winding that
is
we proceed
as follows
(Fig. I0.ll):
.
connected
Connect
the high-voltage
winding
to a
low-
voltage (say 120V) ac source E„.
to the source.
2.
10.6 Standard terminal
saw
in
markings
Section 9.4 that the polarity of a trans-
a
jumper
J
between any two adjacent
HV and LV terminals.
3.
We
Connect
Connect a voltmeter
adjacent
HV
and
LV
Ex
between the other two
terminals.
former can be shown by means of dots on the primary
and secondary terminals. This type of marking
on instrument transformers.
On power
is
used
j
transformers,
however, the terminals are designated by the symbols
H, and H 2 for the high-voltage (HV) winding and by
X and X 2 for the low-voltage (LV) winding. By con(
vention, H, and Xi have the
Although the
H2
bols H,,
,
polarity
X,, and
power transformers
the four
dard
X2
it is
is
same
polarity.
known when
the
sym-
are given, in the case of
common
practice to
mount
terminals on the transformer tank in a stan-
way
so that the transformer has either additive
or subtractive polarity.
A transformer is said to have
Figure 10.11
Determining the polarity of a transformer using an ac
source.
PRA CTICA L TRA NS FORMERS
Connect another voltmeter
4.
winding.
polarity
Ex
If
Ep across
additive. This tells us that H,
is
HV
the
gives a higher reading than
Ep
ing.
,
the
and X,
to the
On the other hand, if £x
than E the polarity is subp
are diagonally opposite.
gives a lower reading
jumper J
In this polarity text,
E
secondary voltage
the
Ep
voltage
tracts
s
E
E
s
with the primary
terminal of the voltmeter
is
Ex = Ep + £ or Ex =
the polarity. We can now see
is
marked H,
.
10,8 Transformer taps
Due
to
age
in a particular
may
consistently be lower than normal. Thus, a dis-
voltage drops in transmission lines, the volt-
s
region of a distribution system
tribution transformer having a ratio of
may be connected
voltage
the polarity test, an ordinary 120 V,
60 Hz source can be connected
2400 V/120
to a transmission line
V
where the
never higher than 2000 V. Under these con-
is
its
to the
HV
winding,
nominal voltage may be several
During a polarity
test
on a 500 kVA, 69 kV/600
motors
may
1
1
V
stall
10. 13).
this
problem taps are provided on the
Taps enable us
8 V,
Ex =
We
percent.
119 V. Determine
markings of the terminals.
is
additive because
HV
may
be
4'/>, 9,
Ex
LV
is
greater than
voltage
is
only 2076
would use terminal
(or
H 2 and
and
Some
X,).
Figure 10.12 shows another circuit that
may
used to determine the polarity of a transformer.
source, in series with an open switch,
LV winding
a voltage
A dc
connected
of the transformer. The trans-
former terminal connected
is
is
to the positive side
marked X,. A dc voltmeter
is
of the
Determining the polarity
/,, 9,
ratio so
or
13'/>
is
1
V
1
0.
(instead of
and tap
1
3, if the line
2400 V), we
5 to obtain 120
V
whenever
above or below a preset
the secondary voltage
level.
Such tap-chang-
ing transformers help maintain the secondary volt-
age within
±2
percent of
its
rated value throughout
the day.
connected
2
O-
Figure 10.13
of
a transformer using a dc
on
transformers are designed to change the
taps automatically
HV terminals. When the switch is closed,
is momentarily induced in the HV wind-
Figure 10.12
source.
be
1
or 13'/2 percent below normal. Thus,
the secondary side.
source
by 4
secondary voltage, even though the primary voltage
terminals con-
Consequently, the
across the
change the turns
can therefore maintain a satisfactory
nected by the jumper must respectively be labelled
to the
to
referring to the transformer of Fig.
Solution
X2
electric
under moderate overloads.
as to raise the secondary voltage
the following readings
10.11),
Ep =
were obtained:
H, and
cook food, and
electric stoves take longer to
To correct
transformer (Fig.
polarity
consider-
primary windings of distribution transformers (Fig.
Example 10-3
the polarity
is
ably less than 120 V. Incandescent lamps are dim,
hundred kilovolts.
.
H2
marked
ditions the voltage across the secondary
making
even though
Ep
)
the terms additive and subtractive originated.
In
The
+
(
and the other
either adds to or sub-
In other words,
,
the pointer of the voltmeter
upscale, the transformer terminal connected
effectively connects
in series
Consequently,
.
from
Ep — Es depending on
how
,
and terminals H, and X, are adjacent.
tractive,
moment,
this
If, at
moves
205
Distribution transformer with taps at
2184
V,
and 2076
V.
2400
V,
2292
V,
ELECTRICAL MACHINES AND TRANSFORMERS
206
Losses and transformer
10.9
rating
Like any electrical machine, a transformer has
They
losses.
are
composed of
The nameplate of a distribution transformer indicates
250 kVA, 60 Hz, primary 4160 V, secondary 480 V.
the following:
Calculate the nominal primary and secondary
a.
2
1.
2.
3.
I
R
__
Example 10-4
losses in the windings
currents.
we apply 2000 V to the 4 60 V primary, can
we still draw 250 kVA from the transformer?
If
b.
Hysteresis and eddy-current losses in the core
1
Stray losses due to currents induced in the tank
and metal supports by the primary and sec-
Solution
ondary leakage fluxes
a.
The
duce
l)
form of heat and pro-
losses appear in the
an increase
efficiency.
temperature and 2) a drop
in
Under normal operating conditions,
efficiency of transformers
is
very high;
it
may
Nominal current of
L
np
nominal S
=
99.5 percent for large power transformers.
—
ns
The heat produced by the iron losses depends
upon the peak value of the mutual flux <E> m which
in turn depends upon the applied voltage. On the
,
E np
4160
b.
we
...
apply 2000
V
to the primary, the flux
rent should not
we must
set limits to
at
an accept-
two
limits
determine the nominal voltage
/ np
E np
The power rating of a transformer
is
product of the nominal voltage times the nominal
current
of the
However, the
primary or secondary
result
is
winding.
not expressed in watts, be-
cause the phase angle between the voltage and current
may have any
voltage
and
equal to the
value
at all,
depending on the na-
ture of the load. Consequently, the
capacity of a transformer
is
power-handling
expressed
in
voltam-
S = 2000
V X
is
directly related to the
through
it.
apparent power
This means that a 500
will get just as hot feeding a
as a
500
kW
kVA
that
flows
transformer
500 kvar inductive load
resistive load.
The rated kVA, frequency, and voltage are always shown on the nameplate. In large transformers the
corresponding rated currents are also shown.
A=
120
kVA
Ep
on the
pri-
mary of a transformer, with the secondary open-circuited.
creases
As
the voltage rises, the mutual flux
in direct
Om
in-
proportion, in accordance with Eq.
9.2. Exciting current / u will therefore increase but,
when
the iron begins to saturate, the magnetizing
current
/m
has to increase very steeply to produce
/Q ,
of a transformer
60
Let us gradually increase the voltage
voltamperes (MVA), depending on the size of the
rise
using this far lower
10.10 No-load saturation curve
the required flux. If
The temperature
cur-
nominal value, other-
is
peres (VA), in kilovoltamperes (kVA) or in mega-
transformer.
However, the load
its
maximum power output
the
of the transformer winding
(primary or secondary).
exceed
and
and
wise the windings will overheat. Consequently,
both the applied
voltage and the current drawn by the load. These
nominal current
A
521
.
the iron losses will be lower than normal
Consequently,
keep the transformer temperature
is
n
the core will be cooler.
to
= 60 A
s
other hand, the heat dissipated in the windings de-
able level,
winding
is
S — 250 X 1000
— —
—
£
480
£ns
pends upon the current they
carry.
V
480
n..
nominal
If
the
nominal S
—
/
winding
250 X 1000
Nominal current of
the
reach
V
4160
Sn
~
£p
nominal
in
the
we
pass
we draw
a graph of
Ep
versus
see the dramatic increase in current as
the
normal
operating
point
Transformers are usually designed
peak flux density of about
1
.5 T,
(Fig.
we
10.14).
to operate at a
which corresponds
roughly to the knee of the saturation curve. Thus,
when nominal
voltage
is
applied to a transformer,
the corresponding flux density
is
about
1.5 T.
We
can exceed the nominal voltage by perhaps 10 percent, but if
we were
to
apply twice the nominal
PRA CTICA L TRA NS FORMERS
207
kV
20
18
16
-
no 'mat oper atincjpoi it
14
En 12
10
—
nc>min al cu rrem
8
6
4
2
0
0
0,5
6
5
4
1
A
exciting current /Q
Figure 10.14
No-load saturation curve of a 167 kVA, 14.4 kV/480
V,
60 Hz transformer.
Figure 10.15
Single-phase dry-type transformer, type AA, rated
voltage, the exciting current could
become even
greater than the nominal full-load current.
The nonlinear
shows
Xm
relationship between
that the exciting branch
in Fig. 10. la) is
effect,
although
Rm
not as constant as
is
Ep
and
/0
(composed of R m and
it
Xm
de-
creases rapidly with increasing saturation. However,
and so
R m and X m
terials inside
close to rated voltage,
remain essentially constant.
10.11 Cooling
To prevent rapid
at
methods
deterioration of the insulating
ma-
kVA
air.
The
metallic housing
is
fitted
the
can be
di-
with ventilating
may
flow over
windings and around the core (Fig.
it
away
in
a
to the tank,
dissipated by radiation and convection to
Oil
1
consequently,
is;
0.
1
6).
is
a
much
it
is
better in-
invariably
used on high-voltage transformers.
As
the
power
1
0. 15).
filled tank (Fig.
rating increases, external radiators
10.17). Oil circulates
around the
transformer windings and moves through the radiators,
air.
where
For
the heat
still
is
again released to surrounding
higher ratings, cooling fans blow
over the radiators (Fig.
For transformers
may be
Such dry-type transformers are used inside build-
oil
hostile atmospheres.
is
are
and enclosed
oil
sulator than air
Larger transformers can be built the same way, but
away from
mineral
in
the outside air (Fig.
forced circulation of clean air must be provided.
ings,
for
are added to increase the cooling surface of the oil-
by the natural flow of the surrounding
louvres so that convection currents
immersed
steel tank. Oil carries the heat
a transformer, adequate cooling of the
Indoor transformers below 200
200 kVA
Distribution transformers below
usually
where
windings and core must be provided.
rectly cooled
V,
appears. In
reasonably constant,
most transformers operate
at
60 Hz, insulation class 150°C
indoor use. Height: 600 mm; width: 434 mm; depth:
230 mm; weight: 79.5 kg.
(Courtesy of Hammond)
15 kVA, 600 V/240
10.
in the
1
air
8).
megawatt range, cooling
effected by an oil-water heat exchanger. Hot
drawn from
the transformer tank
heat exchanger where
it
is
pumped
to a
flows through pipes that are
ELECTRICA L MA CHINES AND TRANSFORMERS
208
Figure 10.16
Two single-phase
transformers, type OA, rated 75
60 Hz, 55°C temperature rise,
pedance 4.2%. The small radiators at the side inkVA, 14.4 kV/240
V,
im-
Figure 10.17
crease the effective cooling area.
Three-phase, type
OA
grounding transformer, rated
1900 kVA, 26.4 kV, 60 Hz. The power of this transformer is 25 times greater than that of the transformers
in
contact with cool water. Such a heat exchanger
is
very effective, but also very costly, because the water
itself
used. Thus, a transformer
18 000/24 000/32
.
is still
self-cooled. Note,
the transformer
much room as
itself.
method of cooling
may have
a triple rating of
000 kVA depending on whether
by the natural circulation of
it
The type of transformer cooling
is
designated by
the following symbols:
air
(AO)
(
1
8
AA-dry-type, self-cooled
000
AFA-dry-type, forced-air cooled
or
OA-oil-immersed, self-cooled
by forced-air cooling with fans (FA) (24 000
kVA)
3.
it
cooled
kVA)
2.
10.16, but
big transformers are designed to have a
multiple rating, depending on the
1
in Fig.
has to be continuously cooled and recirculated.
Some
is
shown
however, that the radiators occupy as
OA/FA-oi I- immersed,
or
by the forced circulation of
forced-air cooling
(FOA)
oil
(32
air
accompanied by
000 kVA).
self-cooled/forced-
cooled
A O/FA/FOA-oil -immersed,
self-cooled/forced-
air cooled/forced-air, forced-oil
These elaborate cooling systems are nevertheless
cooled
economical because they enable a much bigger output from a transformer of a given size and weight
The temperature
(Fig. 10.19).
transformers
is
rise
by resistance of oil-immersed
either
55°C or 65° C. The tempera-
PRA CTICA L TRA NS FORMERS
209
Figure 10.19
Three-phase, type OA/FA/FOA transformer rated
36/48/60 MVA, 225 kV/26.4 kV, 60 Hz, impedance
Figure 10.18
7.4%. The circular tank enables the
Three-phase, type FOA, transformer rated 1300 MVA,
the temperature rises
24.5 kV/345 kV, 60 Hz 65°C temperature
oil in
ance:
1 1
imped-
contact with
air.
Other
nuclear power generating station,
The
is
one
forced-oil circulating
below the cooling
(Courtesy of Westinghouse)
just
of the largest
weight of core and
turc
must be kept low
By
fans.
weight of
to preserve the quality
coils:
37.7
t
weight of tank and accessories: 28.6
pumps can
coil
(44.8
Total weight: 104.5
oil.
details:
.5%. This step-up transformer, installed at a
units ever built.
be seen
rise,
oil to expand as
and reduces the surface of the
m3
):
38.2
t
t
t
of the
contrast, the temperature rise of a dry-type
transformer
may be
as high as
1
80°C, depending on
the type of insulation used.
10.12 Simplifying the equivalent
circuit
N,
The complete equivalent circuit of the transformer
as shown in Fig.
0.8 gives far more detail than is
needed in most practical problems. Consequently,
1
let
us try to simplify the circuit
former operates
I
.
l
)
at
At no-load (Fig. 10.20)
because
T
is
when
no-load and 2)
1 2 is
/n
flows
a transformer
at no-load.
the trans-
drop across them
zero and so
These impedances are so small
circuit of
at full-load.
current in
is /,
an ideal transformer. Consequently,
only the exciting current
Figure 10.20
Complete equivalent
N2
in
R and
}
Xn
.
that the voltage
is
negligible. Furthermore, the
R 2 and X [2
is
zero.
We
can, therefore,
neglect these four impedances, giving us the
much simpler
tio,
a
= N /N2
]
circuit of Fig. 10.2
,
is
1
obviously equal
.
The
turns ra-
to the ratio
of
ELECTRICAL MACHINES AND TRANSFORMERS
210
primary
the
to
secondary voltages
Ep/E meas
sured across the terminals.
2.
hi full-load
I
p
Consequently,
least
is at
we can
20 times larger than
sponding magnetizing branch. The resulting
shown
cuit
is
cuit
may
/()
a£ s
in Fig. 10.22.
This simplified
when
be used even
the load
is
a
cir-
2
Z
J
cir-
only 10
percent of the rated capacity of the transformer.
We
1
.
neglect /0 and the corre-
Figure 10.23
Equivalent circuit with impedances shifted to the
can further simplify the circuit by
shift-
mary
pri-
side.
ing everything to the primary side, thus elimi-
nating transformer
T (Fig.
nique was explained
in
10.23). This tech-
Section 9.10. Then, by
summing the respective resistances and reacwe obtain the circuit of Fig. 10.24. In
tances,
this circuit
Rp =
fl
t
+ crR 2
Xn = X n +
a
2
(10.9)
Xr
(10.10)
.
where
Rp =
total
transformer resistance referred to the
total
of
a large transformer
is
mainly reactive.
primary side
Xp =
Figure 10.24
internal impedance
The
transformer leakage reactance referred
to the
primary side
The combination of R p and X constitutes the totransformer impedance Z referred to the prip
mary side. From Eq. 2. 2 we have
tal
1
\Rcn
VR- + X;
L,
Impedance Z p
a:1
N,
N2
when
one of the important parameters of
is
the transformer.
It
produces an internal voltage drop
the transformer
is
loaded. Consequently,
Zp
affects the voltage regulation of the transformer.
Figure 10.21
Transformers above 500
Simplified circuit at no-load.
Xp
reactance
Rp
*f2
*2
as
.
In such transformers
voltages
tance
Xp
10.25).
a transformer
at full-load.
we can
neglect
Rp
,
is
as far
and currents are concerned.* The
is
thus reduced to a simple reac(Fig.
quite remarkable that the relatively
circuit of Fig.
simple reactance
Figure 10.22
possess a leakage
between the source and the load
It
complex
kVA
that is at least five times greater than
equivalent circuit
Simplified equivalent circuit of
(I0.ll)
10.8 can be reduced to a
in series
with the load.
PRACTICAL TRANSFORMERS
Solution
a.
Rated primary current
/np
= 5 n /Enp =
3
000 000/69 000 = 43.5
A
Rated secondary current
Ais
Figure 10.25
The internal impedance
b.
of
a large transformer
= SJEns =
3
000 000/4160 = 721
A
Because the transformer exceeds 500 kVA, the
windings have negligible resistance compared to
is
their leakage reactance;
mainly reactive.
we can
therefore write
Zp = X p = 127(1
Referring to Fig. 10.26a, the approximate imped-
ance of the 2000
10.13 Voltage regulation
An
important attribute of a transformer
is its
Z = E~IP = 4160
= 8.65 il
volt-
age regulation. With the primary impressed voltage
held constant
at its rated value, the
tion, in percent, is
voltage regula-
Load impedance
defined by the equation:
a
voltage regulation
=
—
E
l
— E
- X
100
2
Z=
/
p
2
secondary voltage
X
(69/4. 16)
/2
secondary side
000 000
8.65
= 2380
12
we have
28.95
A
no-load [V]
at
a£ at full-load
s
[V]
2
(a Z) /
p
= 2380 X
28.95
= 68 902 V
The voltage
regulation depends upon the
power
power
factor
factor of the load. Consequently, the
must be specified.
load voltage
If the
load
may exceed
is
capacitive, the no-
the full-load voltage, in
which case the voltage regulation
E =
s
negative.
at
3000 kVA, 69
X
Because the primary voltage
it
=
(4.16/69)
is
4154
held constant
follows that the secondary voltage
4160
is
68 902
at
at
single-phase transformer rated
kV/4. 6 kV, 60
1
of 127
(1,
Hz
has a total internal impedance
V.
x9
= 127
n
Ij_
Zp
referred to the primary side.
\
kV
Calculate
a.
b.
I
The rated primary and secondary currents
The voltage regulation from no-load to fullload for a 2000 kW resistive load, knowing
that the
primary supply voltage
is
fixed
.
a = t-t^ =
at
4.16
69 kV
c.
The primary and secondary
ondary
is
currents
if
accidentally short-circuited.
the sec-
Figure 10.26a
See Example
10-7.
16.58
V
69 kV,
no-load
Example 10-5
A
is
= 69 000/V127 2 + 2380 2
=
secondary voltage
2
referred to primary side:
Referring to Fig. 10.26b
(10.12)
where
£ NL =
£ hl =
kW load on the
1
z
T
is
ELECTRICAL MACHINES AND TRANSFORMERS
2 2
1
Voltage regulation
is;
X
voltage regulation
4160
^
100
41 54
X
(10.12)
100
4154
0.14%
The voltage regulation
c.
is
excellent,
Referring again to Fig. 10.26b,
is
accidentally short-circuited,
/
p
= E p/Xp =
= 543 A
if
the secondary
aEs =
0 and so
Figure 10.27
Open-circuit test and determination of
69 000/127
The corresponding current
/s
turns
Rm Xm
,
,
and
ratio.
on the secondary
and 10.24 by means of an open-circuit and a
side:
short-
circuit test.
=
/s
a/
p
=
(69/4.16)
X 543
During the open-circuit
= 9006 A
test,
rated voltage
and current
plied to the primary winding
/0
,
is
ap-
voltage
E p and active power P m are measured (Fig. 10.27).
The secondary open-circuit voltage E s is also mea,
These
sured.
test results
give us the following
in-
formation:
active
power absorbed by core =
apparent power absorbed by core
power absorbed by core
reactive
where
Figure 10.26b
See Example
short-circuit currents in both the primary
secondary windings are
rated values.
1
2
The
I
R
1
Rm
R m = Ep
losses are, therefore, 12.5
2
or
imme-
diately to prevent overheating. Very powerful elec-
Turns
a
ratio
(10.1)
X m = Ep 2 /Q m
(10.2)
Xm
is
a
= NJNo
56 times greater than normal and, unless the wind-
may be
During the short-circuit
is
or torn apart.
is
For a given transformer,
values of
Xm
,
/?
,
less than 5 percent
much lower
/ sc
should be less than
its
than
of rated voltage)
The primary
nominal value
to
prevent overheating and, particularly, to prevent a
we can determine
m R p and
the secondary winding
applied to the primary (Fig. 10.28).
current
impedances
test,
Ep/E,
short-circuited and a voltage E„
normal (usually
10.14 Measuring transformer
is
is
tromagnetic forces are also set up. They, too, are
ings are firmly braced and supported, they
2
/P m
Magnetizing reactance
56 times greater than normal. The circuit-breaker
damaged
P~m
corresponding to the core loss
and
2.5 times greater than the
or fuse protecting the transformer must open
1
-
= S m = Ep Ia
= Qm
10-7.
Resistance
The
Qm =
f* in
X p shown
the actual
in Figs.
10.21
rapid change in winding resistance while the test
being made.
is
PRACTICAL TRANSFORMERS
The voltage E sc current
/ sc
,
measured on the primary side
,
and power
(Fig. 10.28)
P sc
are
Transformer impedance referred to the primary
and the
Zp = £ sc //sc =
- 650 (2
following calculations made:
Total transformer
mary side
impedance referred
to the pri-
is
RJtJ =
Rp
1
is
2600/4
Resistance referred to the primary
(10.13)
is
2400/16
= ison
Total transformer resistance referred to the primary
side
2
is
Leakage reactance referred
to the
primary
is
(10.14)
primary side
V650 2 - 150 2
Xp
Total transformer leakage reactance referred to the
- 632
a
is
VZ;
Example 10-6
(10.11)
__
During a short-circuit
test
on a transformer rated
500 kVA, 69 kV/4. 6 kV, 60 Hz, the following
1
age, current, and
Terminals X,,
2600 V
£\
(
volt-
power measurements were made.
were in short-circuit (see Fig,
X2
Figure 10.29
See Example 10-6.
10.28):
2600
V
4A
2400
Example 10-7
W
An
Calculate the value of the reactance and resistance
of the transformer, referred to the
HV
side.
Solution
shop, a 69
Referring to the equivalent circuit of the trans-
former under short-circuit conditions (Fig. 10.29),
we
was conducted on the transExample 10-6. The following results were obtained when the low-voltage winding
was excited. (In some cases, such as in a repair
open-circuit test
former given
in
kV
voltage
may
not be available and the
open-circuit test has to be done by exciting the
winding
LV
at its rated voltage.)
find the following values:
E = 4160 V
/n
K
Using
this
acteristics
a.
= 2A
5000
W
information and the transformer char-
found
in
Example
the values of X m and
Rm
10-6, calculate:
on the primary side
(Fig. 10.21)
b. the
efficiency of the transformer
plies a load of
80
%
when
250 kVA, whose power
it
(lagging).
Solution
a.
Applying Eq.
10.1 to the
secondary side:
Figure 10.28
Short-circuit test to
winding resistance.
determine leakage reactance and
= 4160 2 /5000 =
3461 (1
sup-
factor
is
ELECTRICAL MACHINES AND TRANSFORMERS
214
The apparent power S m
= E
5m
s
= 4160 X
I {)
2
and R p Let us assume
= 8320 VA
load
is
4 60
6650
The load current
11
/?
Xm
m and
000/4 160)
2
11
referred the primary side
=
275 times
greater.
The
Rm =
X 2602
275
275 X
=
II
=
3461 fl
715
X
10
952 X
10
3
turns ratio
=
715
kH
(1
- 952
k(l
the
time. Thus,
when we
=
with cos 6
is
about 250
about
that
loads and voltages fluctuate
b. Industrial
0.8,
it
all
state that a load is
is
kVA and
power
The
total
swer, even
in
if
iron loss
is
=
3.62
X
is
150
1966
same
the
rated voltage on the
is
LV
as that
measured
at
side of the transformer.
Total losses are
certain
it.
= 6966
The
circuit of the transformer
and
5000
bosses
Knowing this,
assumptions that make it much
able to give
active
+
1966
W = 7 kW
power delivered by
the transformer
its
add the
The values of R p
known, and so we only have to
magnetizing branch. To simplify the calcu-
lations,
we
is
Rp
2
is
calculating efficiency, there
we were
The equivalent
Xp
Il
=
The
is
easier to arrive at a solution.
load
A
3.62
arriving at a precise mathematical an-
we can make
and
-
P iron = 5000W
Consequently,
in
60/16.59
2
=
about 69 kV.
no point
=
is:
copper loss (primary and secondary)
250 kVA,
factor
the primary side
/2 /a
Copper
the
Furthermore, the primary voltage
0.8.
=
16.59
if
understood that the load
that the
is
The current on
/,
would have been found
primary had been excited at 69 kV.
These are the values
000/4160
= 69W/4I60 V -
a
(1
3
is
The
values on the primary side are therefore:
Xm =
calculate the efficiency of
= SIE, = 250
= 60 A
12
= 4160 2 /6650 = 2602
will be (69
Xp
the transformer.
= V8320 2 - 5000 2
The values of
We now
V.
1
greater than
that the voltage across the
.
^s^pI
Qm =
much
cause these impedances are
is:
to the load
is
represented by Fig. 10.30.
PG = S cosG = 250 X
are already
shifted
input terminals
1
1
1,
50f2
X m and R m from points 3, 4 to the
2.
632
This change
a
is
/,
justified be-
=
200
The
active
power received by
60 A
Note
efficiency
that in
is
7
therefore
= PJP - 200/207
= 0.966 or 96.6%
X
making
the calculations,
sider the active power.
transformer and
10-7.
is
3
ti
Figure 10.30
the transformer
=
+ bosses = 200 +
= 207 kW
The
See Example
0.8
kW
its
ciency calculations.
The
reactive
we
only con-
power of the
load does not enter into
effi-
PRACTICAL TRANSFORMERS
The
10.15 Introducing the per-unit
best approach
calculate
is
often encountered
when
dealing
with transformers and other electrical machines.
reason
is
relative
rents
employ
to
The
that per-unit values give us a feel for the
its
ohmic value and use
For example,
former
listed in
_ E ns _
" L ~
and kilowatts, we simply work
volts
we don't have to carry
when per-unit values are used.
The per-unit method as applied to transformers
V
8 A
28
'
12.012
Using
value of the secondary resistance
who are
easy to understand. However, readers
useful to read Sections 1.9 to 1.11 in Chapter
1
tual values
R
of
{
,
at
Table
be-
1
OA which
,
,
,
transformers ranging from
scanning through the table,
1
kVA
we
is
400
to
ances vary from 505 000 17 to 0.0003
MVA.
In
12, a
range
over
all
that the various voltages,
currents and impedances are expressed in actual
values using volts, amperes and ohms.
ACTUAL TRANSFORMER VALUES
/„s
*i
Ri
*n
Q
Q
n
n
4.17
/ np
Using
this load
V
A
2400
,
impedance
576
as a reference, the rela-
value of the primary resistance R\
/?,(pu)
is
5.16(2
=
0.0090
576
The
l
1000
100
10
12
and R 2 (pu) are pure
relative values R\(p\i)
numbers because they are the ratio of two quantities
that bear the same unit.
Circuit elements on the primary side are always
compared with
the nominal load
impedance Z np on
400000
2400
2400
12470
69000
13800
460
347
600
6900
424000
0.417
4.17
8.02
14.5
29000
167
145
943
.6
27.2
0.0003
17
28.8
58.0
5.16
1.9
0.095
0.024
0.25
0.354
32
4.3
39
151
0.028
0.09
0.09
1.5
27
2.
1.
16
1
1
a
Q
200000
29000
150000
505000
460
Rm
400000
51000
220000
432000
317
A,
A
0.0134
0.0952
0.101
0.210
52.9
R\,R 2 X n X l2 X tn and R m
in ohms, we could express them relative to another
ohmic value. The question is: what value should we
,
,
on
the
secondary side are compared with the nominal load
impedance Z ns on the secondary
Proceeding
of the 10
kVA
values X, |(pu),
The
relative
in this
side.
for the other impedances
we
transformer,
obtain the relative
m (pu), etc. displayed in Table 10B.
impedances of the other transformers
/?
are calculated the
tive
way
same way.
In
nominal load impedances
as the reference impedances.
each case, the respec-
Z np and Z ns are chosen
Using the rated voltage
and power of the transformer, they are given
E
_np Instead of expressing
pri-
12
the primary side. Similarly, circuit elements
kVA
/np
E
no
recognizable pattern to the values; they are
impedance on the
is:
in
is
TABLE 10A
12.017
~ np
tive
is
0.0079
(pu)
of five
excess of a billion to one. Furthermore, there
is
side
R2
0.095 12
2
re-
see that the imped-
as a reference, the relative
Similarly, the nominal load
displays the ac-
It
R 2 Xn X (2 X m and R m
map. The reason
R
mary
Let us begin by looking
ohmic value
it
fore proceeding further.
produced here for convenience.
this
not
yet familiar with per-unit calculations will find
V
V
A
A
trans-
347
with numbers. Consequently,
Ens
can
is
along units
£„p
kVA
case of the 10
in the
We
as a reference.
Table 10A, the nominal load im-
pedance on the secondary side
and powers. Thus, instead of dealing with
ohms, amperes,
the
it
magnitudes of impedances, voltages, curZ,ls
is
the nominal load
(voltage and current) of the transformer.
method
Per-unit notation
is
215
*«P.
- E
by:
1
(HU5a)
sa
,
En,
(10.15/;)
choose as a basis of comparison?
s
tt
ELECTRICAL MA CHINES AND TRANSFORMERS
2 6
1
symbols #|(pu),
7\?
2 (pu),
R2 X
values of/?,,
In practice, the relative
<
tl
,
etc.,
and are designated by the
are called per-unit values
X n (pu),
The
etc.
quantities
Example 10.8^
A
transformer rated 250 kVA, 4 60 V/480 V, 60 Hz
1
has an impedance of 5.1 %. Calculate:
used as references are called base quantities. Thus,
Z np Z ns S n E
,
a.
£ ns
.
,
,
/
/ ns listed in
,
Table 10B are
impedance on
the base
base quantities.
all
examining Table OB. the reader
I
will note that
former referred
for a given transformer, the values of
X, ,(pu)
and
X r2 (pu)
kVA
V
V
A
2400
100
1000
400000
2400
12470
69000
13800
460
347
600
6900
424000
0.417
4.17
8.02
14.5
29000
17
28.8
167
145
943
5760
576
1555.
4761
0.4761
211.6
12.0
3.60
47.61
449.4
R\ (pu)
0.0101
0.0090
0.0075
0.0057
0.00071
/? 2
(PU)
0.0090
0.0079
0.0067
0.0053
0.00079
X
X rj
X ni
(pu)
0.0056
0.0075
0.0251
0.0317
0.0588
(pu)
0.0055
0.0075
0.0250
0.0315
0.0601
(pu)
34.7
50.3
96.5
106
966
/? ni
Znp = E p
- 69
2
/S n
= 4160
2.
69.4
(pu)
0.032
A,<P»)
88.5
141.5
0.023
0.013
90.7
0.015
666
0.0018
2
is
/250 000
tt
Base impedance on the secondary side
2
n
Q
zns
Base impedance on the primary side
Table 10A.
in
10
I
A
A,s
z„ P
ri
of the trans-
Solution
a.
PER UNIT TRANSFORMER VALUES
TABLE 10B
Aip
Zp
primary side
to the
AVpu) and
are nearly the same. This pat-
does not show up
tern of similarity
£„»
and sec-
are nearly the same. Similarly, the values of
/? 2 (pu)
t„p
impedance
the total internal
b.
In
the primary
ondary side
Zns = E /S n = 480
= 0.92 n
2
s
b.
The
actual value of
Zp =
Zp
% X Z np =
5.1
is
/250 000
on the primary side
0.051
X 69
(1
=
is:
3.52 11
10,17 Typical per-unit impedances
We
have seen
ative
that
we can
get a better idea of the
rel-
magnitude of the winding resistance, leakage
actance,
etc.,
re-
of a transformer by comparing these im-
pedances with the base impedance of the transformer.
There
is
even a similarity between the per-unit
whose
values of transformers
ferent.
ratings are quite dif-
For example, the R\(pu) of the
former (0.0101
)
is
as the /?,(pu) of the
I
kVA trans-
of the same order of magnitude
1000
kVA transformer
despite the fact that the latter
is
(0.0057)
1000 times more
powerful and the voltages are vastly different.
Clearly, the per-unit
method
offers insights that
In
making
the comparison, circuit elements located
on the primary side are compared with the primary
base impedance. Similarly, circuit elements on the
secondary side are compared with the secondary base
impedance. The comparison can be made either on a
percentage or on a per-unit basis;
ter.
kVA
transformers ranging from 3
example, the table shows
would otherwise not be evident.
we
shall use the lat-
Typical per-unit values are listed in Table IOC for
to 100
MVA.
For
that the per-unit resistance
of the primary winding of a transformer ranges from
10,16 Impedance of a transformer
The total internal impedance Zp of a transformer
was defined in Section 10. 12 and highlighted in
Fig. 10.24. In power and distribution transformers
its
value
However,
is
it
always indicated on the nameplate.
is
expressed as a percent of the nominal
load impedance. Thus,
3.6
%. the per
if
the nameplate
unit value of
Zp
is
0.036.
is
marked
0.009
to
and 100
0.002 for
all
MVA. Over
the per-unit resistance
power
this
7?,
ratings
between 3
kVA
tremendous power range,
of the primary or secondary
windings varies only from 0.009 to 0.002 of the base
impedance of
the transformer.
Knowing
the base im-
pedance of either the primary or the secondary winding,
we can
readily estimate the order of magnitude of
the real values of the transformer impedances. Table
IOC
is,
therefore, a useful source of information.
PRACTICAL TRANSFORMERS
TABLE 10C
217
TYPICAL PER-UNIT VALUES OF TRANSFORMERS
Figure 10.31
Equivalent circuit of a transformer.
Typical per-unit values
Circuit element (see Fig.
/?,
0.3
1
3
1
orR 2
.
X n orX n
kVA
to
250
kVA
l
MVAio I00MVA
0.009-0.005
0.005-0.002
0.008-0.025
0.03-0.06
20-30
50-200
20-50
100-500
0.05-0.03
0.02-0.005
Example 10-9
Using the information given
the
approximate
0.35
SI
1.7 SI
Table 10C, calculate
in
23
ma
4.6 mSl
impedances of a
real values of the
250 kVA, 4160 V/480 V, 60 Hz distribution
trans-
former.
Solution
We
first
determine the base impedances on the
mary and secondary
Example
10-8,
we
From
side.
the
results
pri-
of
Figure 10.32
have
See Example
Z np =
Z
lls
=
69
0.92
n
This example shows the usefulness of the per-
We now calculate the real impedances by multiplying Znp and Z ns by the per-unit values given in
Table IOC. This yields the following results:
=
0.005
X 69
R2 =
Xn =
X r_ =
Xm =
Rm =
0.005
X
0.025
X 69
0.025
X
/?,
30
50
(2
0.92
tt
10-9.
tt
=
0.35 12
H=
=
0.92 (2
1
=
4.6
mil
alent circuit of the
range of transformers.
.7 12
23 m()
X 69 fl = 2070 12 =
X 69 n - 3450 (2 -
method of estimating impedances. The equiv250 kVA transformer is shown
in Fig. 10.32. The true values may be 20 to 50 percent higher or lower than those shown in the figure. The reason is that the per-unit values given in
Table 10C are broad estimates covering a wide
unit
2 kil
3.5 kI2
Example 10.10
The 500 kVA, 69 kV/4160 V, 60 Hz transformer
shown in Fig. 10.30 has a resistance R of 150 (2
ELECTRICAL MACHINES AND TRANSFORMERS
218
and a leakage reactance
unit
a.
X p of 632 f2.
Using the per-
G(pu)
method, calculate:
= V0.5 2 when
the voltage regulation
kVA
tween zero and 250
= VS 2 (pu) - P 2 (pu)
the load varies be-
=
The
80%
b.
the actual voltage across the
c.
the actual line current I
250
kVA
R
The
Solution
clear that the pres-
is
it
R p and Xp
voltage drop across
magnetizing branch does not affect the voltage
We now draw
regulation.
in
To determine
fer all voltages,
HV
(69
kV)
terminals
1
is
We
assume the voltage between
69 kV, and that it remains fixed.
The base power P H
The base voltage E
is
is
ti
kVA
69 kV
500
Consequently, the base current
'b
= Pb/Ek = 300
= 7.25 A
and the base impedance
Z B = E B /I H = 69
The
will re-
impedances, and currents to the
side.
2
,
we
the voltage regulation,
per-unit value of
Rp
Figure
Note
terminals
per-unit value of
Xp (pu) The
Xp
it
4 of the
circuit
250
=
4
The
Z ]2 (pu) =
primary
Figure 10.30.
The
per-unit im-
is
2/136.87°
1.2
0.0158
+
1.6
+
j(1.2
+
1,
0.0664
0.0664)
is
= 69 000/69 kV =
A-p(pu)
flp(pu)
1.0
j
0.0664
power absorbed
= 250 kVA/500 kVA =
A'.(pu)
0.5
j
per-unit value of the active
the load
3.33
power absorbed by
is
P(pu)
=
5(pu) cos 9
=
0.5
X
0.8
-
0.4
Figure 10.33
The
per-unit value of the reactive
by the load
is
is
5(pu)
The
2
= 1.616 + j 1.266
= 2.053Z 38.07°
is
E i2
in
impedance between terminals
per-unit
is
=
not
X j 3.33
+ j 3.33
= 1.6+j
per-unit value of the apparent
by the load
3,
2.50
Z34 (pu)
0.0158
The
shown
equivalent circuit diagram.)
is
632/9517
is
does not enter into the calcula-
is
ft
shown
(These terminals are not accessible; they exist only
in the
= 9517
3.333
The magnetizing branch
pedance between terminals
000/7.25
=
Q is
0.3
that the load appears across the
3,
000/69 000
per-unit value of voltage
E, 2 (pu)
Uf_
the equivalent per-unit circuit
R p (pu) = 150/9517 = 0.0158
The
corresponding to
v
"
0(pu)
10.33.
shown because
tions.
0.4
£2 (pu) _
X L (pu)
Consequently, the
,
X
2
2.50
~
P(pu)
P is
corresponding to
1.0
)
(pu)
v
per-unit load reactance
ence of the magnetizing branch does not affect the
/?,
£ 2 (pu _
.
Fig. 10.30,
0.3
per-unit load resistance
load
{
examining
2
lagging power factor
at a
of
In
0.4
is
power absorbed
Per-unit equivalent circuit of a
feeding a 250
kVA
load.
500 kVA transformer
PRACTICAL TRANSFORMERS
The per-unit current
I
219
ensure proper load-sharing between the two trans-
is
{
formers, they must possess the following:
/i(pu)
.0
-
Z 12 (pu)
L
2.053
a.
38.07°
b.
The same primary and secondary voltages
The same per-unit impedance
= 0.4872^-38.07°
Particular attention must be paid to the polarity
The
per-unit voltage
£34 (pu) =
/j(pu)
£34
across the load
of each transformer, so that only terminals having
is
10.34).
- (0.4872/1-38.07°) (2^36.87°)
=
same
the
X Z34 (pu)
circuit as
0.9744/1 -1.20°
polarity
An
soon as the transformers are excited.
In order to calculate the currents flowing in each
transformer
The
per-unit voltage regulation
E 34 (pu)
at
£34 (pu)
is
must
— £34 (pu)
no-load
at
Consider
0.0263
is
former referred
therefore 2.63
is
first
mary voltage
0.9744
The voltage regulation
when
we
they are connected in parallel,
determine the equivalent circuit of the
the equivalent circuit
when
a single
transformer feeds a load Z, (Fig. 10.35a). The
1^ -j0.9744_
a.
first
system.
full-load
full-load
at
connected together (Fig.
are
error in polarity produces a severe short-
£p
pri-
and the impedance of the trans-
to the
primary side
is
Zpl
.
If the ratio
%.
2
We
can
now
calculate the actual values of the volt-
O-
age and current as follows:
Voltage across terminals
4
3,
A
is
O-
£ 34 = EM (pu) X £ B
= 0.9744 X 69 000
=
b.
=
£34
X
67.23
is
B
(4160/69 000)
X
10
3
X
>H 2
Actual line current
/,
Figure 10.34
Connecting transformers
a
..
in parallel to
share a load.
is
=
/,(pu)
=
0.4872
=
3.53
X
X
/B
7.246
A
10.18 Transformers
When
X 1#
0.0603
= 4054 V
c.
* 2# .
2
kV
67.23
Actual voltage across the load
£ S6 =
*H
1
in parallel
growing load eventually exceeds the power
rating of an installed transformer,
connect a second transformer
we sometimes
in parallel
with
it.
To
Figure 10.35a
Equivalent circuit of a transformer feeding a load
ZL
.
ELECTRICAL MACHINES AND TRANSFORMERS
220
of transformation
that
shown
in Fig.
is
1
a, the circuit
can be simplified to
0.35b, a procedure
we are
already
portional to the respective
we want
kVA ratings. Consequently,
to fulfill the following condition:
familiar with.
If
is
a second transformer having an impedance
connected
circuit
the
in parallel
becomes
with the
shown
that
first,
spectively
/,
and
/2
.
(I0.18)
2
in Fig. 10.35c. In effect,
impedances of the transformers are
The primary currents
Z
the equivalent
in the
From
Eqs.
and 10.18
10. 17
can readily be
it
in parallel.
proved
transformers are re-
Because the voltage drop £, 3
is the same, we can write
across the impedances
that the desired condition
is
met
if
the trans-
formers have the same per-unit impedances. The
following example shows what happens
when
the
per-unit impedances are different.
(10.16)
Example 10-11
that
is.
A
1
00 kVA transformer
an existing 250
(10.17)
The
of the primary currents
is
connected
in parallel
with
supply a load of
330 kVA. The transformers are rated 7200 V/240
but the
ratio
is
kVA transformer to
1
V,
kVA unit has an impedance of 4 percent
250 kVA transformer has an impedance of
00
therefore de-
while the
termined by the magnitude of the respective primary
impedances
—and
not by the ratings of the
two
6 percent (Fig. 10.36a).
trans-
formers. But in order that the temperature rise be the
Calculate
same
a.
for both transformers, the currents
must be pro-
The nominal primary
current of each trans-
former
c.
The impedance of the load referred to the primary side
The impedance of each transformer referred to
d.
The
b.
the primary side
actual primary current in each transformer
Solution
a.
Nominal primary current of
former
the
250 kVA
trans-
is
Figure 10.35b
Equivalent circuit with
all
impedances
referred to the
=
250 000/7200
34.7
A
primary side.
X
t H,
-
1:250
kVA
-d H 7
1
>
240
j:
V
z
a Z
y
,
<\
H,
X,
100 kVA:
Z p2
d :H 2
=
4%
X2
Figure 10.35c
Equivalent circuit of two transformers
a load
Zv All
impedances
in parallel
feeding
referred to the primary side.
load
;=
X 2 b-
:
7200
6%
::|Zpr
Figure 10.36a
Actual transformer connections.
b
V
330
kVA
PRACTICAL TRANSFORMERS
Nominal primary current of the 100 kVA
former
=
and
A = 46 -
side, is
im-
0.36b,
we
justified be-
46
find that the
Load impedance referred
primary side
to the
is
=
A
28.8
17.2
which
is
A
seriously overloaded
is
25 percent above
unit
is
its
A
28.8
20.7)
carries a primary current of
it
The 250 kVA
cause the transformers are fairly big.
+
20.7/(12.4
The 100 kVA transformer
because
are considered to be enis
= 46 X
/,
that transformer
This assumption
tirely reactive.
A
primary
to the
Note
Zp2
13.9
of the two transformers
circuit
in Fig. 10.35c.
Zpl
=
100 000/7200
and the load, referred
pedances
1
1
load current divides in the following way:
The equivalent
given
Referring to Fig.
d.
is
/ n2
b.
trans-
22
17.2 A,
rated value of 13.9 A.
not overloaded because
carries a current of 28.8
A
versus
its
it
only
rated value of
34.7 A. Clearly, the two transformers are not carry2
2 = E p /S| olld = 7200
= 157 n
The approximate
2
/330 000
ing their proportionate share of the load.
The 100 kVA transformer
of
load current
its
c.
=
JE p
S hr
= 330 000/7200 = 46 A
The base impedance of
the
kVA
207
unit
its
Transformer impedance referred
tends to carry
proportionate share of the load.
If the
is
percent impedances were equal, the load would be
shared between the transformers
il
power
their respective
to the
in
proportion to
ratings.
primary
is
Zpl =
X 207 =
0.06
Base impedance of the 100
Questions and Problems
12.4(1
kVA
Practical level
unit
is
1
Znp2 = 7200 2 /10() 000 = 518
Transformer impedance referred
side
to the
transformer (6 percent).
A low-impedance transformer always
more than
250 kVA
Z npt = 7200 2 /250 000 -
side
overloaded because
is
impedance of the 250
A.
is
low impedance (4 percent), compared
0-
1
12
to the
Name
the principal parts of a
transformer.
primary
10-2
Explain
how
a voltage
is
induced
in the
secondary winding of a transformer.
is
Z p2 =
0.04
X 518 =
10-3
20.7 il
The secondary winding of a transformer
has twice as
many
turns as the primary.
Is
the secondary voltage higher or lower
than the primary voltage?
/„,
/,
= 34.7 A
= 28.8
10-4
A
Which winding
is
connected
to the load:
the primary or secondary?
ma a
z pi
=
Z pl
= 20.7
46
A
10-5
State the voltage and current relationships
between the primary and secondary wind-
7200
V
n
157
I 2 = 17.2
/n ,
A
7
= 13.9 A
Figure 10.36b
Equivalent
circuit.
transformer
is
Calculations
show
seriously overloaded.
that the
ings of a transformer under load. The primary and secondary windings have /V|
a
100 kVA
and
N2
turns, respectively.
10-6
Name
10-7
What purpose does
the no-load current of
a transformer serve
?
the losses
produced
in a
transformer.
222
10-8
ELECTRICAL MACHINES AND TRANSFORMERS
Name
in
three conditions that must be
met
order to connect two transformers
10-15
What
the purpose of taps
is
on a
trans-
Name
10-16
three
methods used
10-17
The primary of a transformer is connected
to a 600 V, 60 Hz source. If the primary
has 1200 turns and the secondary has 240,
the
peak flux
a 60
in
Hz
ac supply voltage
10-18
The transformer
by a
20
1
V,
fixed.
is
10.37
in Fig.
is
excited
60 Hz source and draws
a no-
load current / 0 of 3 A. The primary and
secondary windings respectively possess
low-voltage winding
200 and 600
turns. If
mary
linked by the secondary, cal-
V
is
excited by a
source, calculate the voltage
across the
A 6.9 kV
HV
winding.
transmission line
a transformer having
is
a.
connected to
b.
1500 turns on the
c.
If
d.
the load across the secondary has an im-
pedance of 5 H, calculate the following:
a. The secondary voltage
b. The primary and secondary currents
The primary of a transformer has twice as
many turns as the secondary. The primary
voltage is 220 V and a 5 17 load is con-
as the primary
Figure 10.37
See Problem
10-18.
is
40 percent of
the pri-
the transformer, as well
and secondary currents.
The voltage indicated by the voltmeter
The peak value of flux 4>
The peak value of <t> m
Draw
the phasor
/ tr 4> m
10-19
In Fig.
1
,
and
0.38,
diagram showing
when 600 V
is
applied to
V
is
measured
H2
across terminals X,,
a.
What
and
is
E h £2
the voltage
,
80
X2
.
between terminals H,
X2 ?
b. If terminals H,,
X, are connected together,
calculate the voltage across terminals
c.
,
<£,
terminals H] and
nected across the secondary. Calculate the
power delivered by
tlux
culate the following:
primary and 24 turns on the secondary.
10-14
why
Explain
in the core.
The windings of a transformer respectively have 300 and 7500 turns. If the
2400
10-13
Problem 10-11, calculate the peak
transformer remains fixed as long as the
calculate the secondary voltage.
10-12
In
value of the tlux
to cool trans-
formers.
10-11
nominal
Intermediate level
former?
10-10
a ratio of
to 2.4 kV. Calculate the
current of each winding.
parallel.
10-9
A 3000 kVA transformer has
60 kV
in
Does
H2 X2
,
the transformer have additive or sub-
tractive polarity?
.
PR A CTICA L TRA NS FORMERS
^ = 300
/,
TV,
223
= 1200
(96 A)
Figure 10.38
See Problem
10-20
10-19.
what would happen
a. Referring to Fig. 10.34,
if
we
reversed terminals H, and
H2
of trans-
Would
X2
The primary
if
H h H2
terminals
of transformer
B were
and
Advanced
10-27
level
Referring to Fig. 10.39, calculate the peak
value of flux
Explain
leg
reversed?
Explain.
10-21
wound on one
is
other.
the operation of the transformer
bank be affected
X,,
10-33.
and the secondary on the
former B?
b.
Figure 10.39
See Problem
why
the secondary voltage of a
in the
supplied by a 50
is
core
Hz
if
the transformer
source.
practical transformer decreases with in1
creasing resistive load.
The impedance of a transformer
0-28
as the coupling
10-22
10-23
What
is
meant by
the following terms:
a.
Transformer impedance
b.
Percent impedance of a transformer
The transformer
in
0-24
A 2300 V
the im-
is
R\
=
18 12
R 2 = 0.005
xn =
connected
to terminals
40
10-25
A
66.7
kVA,
calculate the following:
a.
The percent impedance of the transformer
The impedance [12] referred to the sec-
d.
The percent impedance
in the
f.
transformer un-
Calculate the losses and efficiency
the transformer delivers 66.7
when
MVA to a
load having a power factor of 80 percent.
10-26
If the
transformer shown
were placed
ture rise
Explain.
in a
tank of
would have
to
The total copper losses at full load
The percent resistance and percent
reac-
tance of the transformer
der these conditions.
b.
referred to the sec-
ondary side
e.
Calculate the losses
referred to
ondary side
percent.
a.
[12]
c.
an efficiency
when it delivers full power
load having a power factor of 100
of 99.3 percent
to a
The transformer impedance
the primary side
connected across the secondary
MVA transformer has
a
s
1
b.
is
12
E p = 14.4 kV (nominal)
£ = 240 V (nominal)
x n_ = o.oi n
the transformer has a nominal rating of 75
and 4 in Fig. 10.13. Calculate the following:
a. The voltage between terminals X, and X 2
b. The current in each winding, if a 12 kVA
load
given for the
is
transformer circuit of Fig. 10.22.
If
line
The following information
10-29
pedance 1(2] referred to:
a. The 60 kV primary
b. The 2.4 kV secondary
1
increases
reduced between the
primary and secondary windings. Explain.
Problem 10-15 has an
impedance of 6 percent. Calculate
is
in Fig.
oil,
10.15
the tempera-
be reduced to 65°.
10-30
During a short-circuit
test
on
a 10
66 kV/7.2 kV transformer (see
the following results
were obtained:
Eg = 2640 V
/ sc - 72 A
R = 9.85 kW
.
(
MVA,
Fig. 10.28),
ELECTRICAL MACHINES AND TRANSFORMERS
Calculate the following:
a.
The
Industrial application
and the
total resistance
actance referred to the 66
b.
total
leakage
kV primary
The nominal impedance of
re-
10-34
side
A
transformer has a rating 200 kVA,
400 V/277
14
the transformer
The high-voltage windWhat is the
V.
ing has a resistance of 62 H.
referred to the primary side
c.
The percent impedance of
V
approximate resistance of the 277
the transformer
winding?
10-31
In
Problem 10-30,
if
rated voltage are 35
the iron losses at
kW,
calculate the full-
load efficiency of the transformer
power
factor of the load
is
if
10-35
The primary winding of the transformer
Problem 10-34
the
is
wound
with No.
1
in
gauge
1
AWG wire. Calculate the approximate
85 percent.
cross section (in square millimeters) of the
10-32
a.
The windings of a transformer operate
current density of 3.5
made of copper and
ture of
A/mm
2
.
If
at a
conductors
1
operate
0-36
tempera-
at a
If
75°C, calculate the copper loss per
aluminum windings were
0-33
If
cording to Fig. 10.39,
it
would have very
poor voltage regulation. Explain
propose a method of improving
why and
it.
kVA
weighs
1
transformer rated
whereas a 100
18 kg,
of the same kind weighs
in
watts per kilogram in each case.
10-37
a transformer were actually built ac-
10
445 kg. Calculate the power output
same condi-
tions.
1
secondary winding.
oil-filled distribution
kVA transformer
used, calculate
the loss per kilogram under the
An
at
kilogram.
b.
in the
they are
The transformer shown in Fig. 10. 13 has
rating of 40 kVA. If 80 V is applied between terminals X, and
will
X
2,
a
what voltage
appear between terminals 3 and 4?
If
a single load
is
applied between terminals
what
is
the
3 and 4
current that can be
maximum
drawn?
allowable
Chapter
1
Special Transformers
windings, each rated
11.0 Introduction
connected
Many transformers are designed to meet specific
we
industrial applications. In this chapter
at
1
20
V.
The windings
and so the
in series,
total
tween the lines is 240 V while that between the
lines and the center tap is 20 V (Fig.
). The
center tap, called neutral, is always connected to
study
1
some of the
special transformers that are used in dis-
tribution systems,
neon
signs, laboratories, induction
furnaces, and high-frequency applications.
they are special, they
1
.
2.
The ampere-turns of
I
.
I
on the high-voltage winding
to the neutral terminal
is
usu-
of the secondary
windings are connected
that both
to
ground.
the transformers are under load:
The voltage induced in a winding is
proportional to the number of turns,
quency, and the flux in the core.
bonded
winding so
a result, the following approximations can be
made when
H2
Terminal
ally
of the standard transformers discussed in Chapter 10.
As
I
ground.
Although
possess the basic properties
still
are
voltage be-
The nominal
rating of these distribution trans3
mounted on poles of
the fre-
pany
(Fig.
1
1
.2) to
kVA
500 kVA. They are
comsupply power to as many as 20
formers ranges from
directly
to
the electrical utility
customers.
the primary are equal and
The load on
opposite to the ampere-turns of the secondary.
distribution
transformers varies
greatly throughout the day, depending on customer
3.
4.
The apparent power input to the transformer
equal to the apparent power output.
The exciting current
may be
in the
is
demand.
In residential districts a
peak occurs
morning and another peak occurs
primary winding
noon. The power peaks never
neglected.
in the
in the late after-
last for
more than one
or two hours, with the result that during most of the
24-hour day the transformers operate far below
11.1
their
Dual-voltage distribution
transformer
normal
rating.
tem, every effort
Transformers that supply electric power to
dential
areas
generally
have
Because thousands of such
transformers are connected to the public
small. This
resi-
two secondary
is
is
made
utility sys-
keep the no-load losses
achieved by using special low-loss
silicon-steel in the core.
225
to
ELECTRICAL MACHINES AND TRANSFORMERS
226
—
I
A
Figure 11.3
Autotransformer having
N2 turns
A/-,
turns on the primary
and
on the secondary.
H2
11.2 Autotransformer
Figure 11.1
a.
Distribution transformer with
ondary.
b.
Same
The
1
central conductor
20 V/240 V secis
distribution transformers
the neutral.
turns,
mounted on an
ing
connected
reconnected to give
only 120 V.
N
Consider a single transformer winding having
is
iron core (Fig.
1 1
to a fixed-voltage ac
.3).
{
The windand
source
the resulting exciting current /(> creates an ac flux <& m
in the core.
the flux
is
As
in
Suppose a tap
there are
any transformer, the peak value of
fixed so long as E\
N2
C
is
is
fixed (Section 9.2).
taken off the winding, so that
turns between terminals
A
and C.
Because the induced voltage between these terminals
is
proportional to the
number of
turns,
E2
is
given by
E 2 = (N2 /N0 X
£,
(ll.l)
Clearly, this simple coil resembles a transformer
having a primary voltage £, and a secondary voltage
E 2 However,
.
A and the
A are no longer isolated from
the primary terminals B,
secondary terminals C,
common
each other, because of the
If
we connect
terminal A.
a load to secondary terminals
CA,
the resulting current I2 immediately causes a pri-
mary current /] to flow (Fig. 1 .4).
The BC portion of the winding obviously carries current /,. Therefore, according to Kirchhoff s
1
CA portion carries a current (I 2 —
/]). Furthermore, the mmf due to /, must be equal
and opposite to the mmf produced by (7 2 — /]). As
current law, the
Figure 11.2
a result,
we have
Single-phase pole-mounted distribution transformer
rated:
100 kVA, 14.4 kV/240 V/120
V,
60 Hz.
(N
I
]
l
- N2 =
)
(I 2
- I,)N2
SPECIAL TRANSFORMERS
221
B
O—
:
(N,-N 2
c
)
(h
-fx)
E2
A
-o-
load
Figure 11.4
Autotransformer under load. The currents flow
which reduces
opposite directions
in
3.6
to
I
]
assuming
N = I2 N 2
that both the transformer losses
b.
and exciting current are negligible, the apparent
c.
power drawn by the load must equal the apparent
power supplied by the source. Consequently,
1
1
.
1
,
1
1
1
1
.3
CA
and
Solution
= E 2 I2
and
.2,
connected across the secondary,
is
The secondary voltage and current
The currents that How in the w inding
The relative size of the conductors on windings
BC
(U.3)
a.
Equations
load
(ll. 2)
]
£,/,
kW
the upper and lower windings.
calculate:
a.
Finally,
in
The secondary voltage
are identical to those
of a standard transformer having a turns ratio
N /N2
is
E2 = 80% X 300 =
240
V
.
i
However,
in
winding
actually part of the primary winding. In
effect,
is
this
autotransformer the secondary
separate secondary winding.
As
difference in size
cal
a result, autotrans-
becomes
power
and
2.
On
isolation
windings
is
a
E
= PIE 2 = 3600/240 =
The current supplied by
]
/E 2
The
/,
important
lies
=
=
PIE,
the current in
between
between the primary and secondary
serious drawback
in
Autotransformers are used to
some
start
secondary
applications.
induction
mo-
voltage of transmission lines,
and, in general, to transform voltages
to
is
15
A
the source
(Fig.
1
1.5).
is
the current in
=
A
winding BC = 12 A
winding CA = 15 —
3600/300
12
12
=
3
A
the other hand, the absence of electri-
tors, to regulate the
mary
output.
particularly
the ratio of transformation
b.
current
and cheaper than
lighter,
standard transformers of equal
0.5
I2
an autotransformer eliminates the need for a
formers are always smaller,
when
The secondary
ratio
Example 11-1
The autotransformer
is
close to
in Fig.
when
the pri-
The conductors
in the
secondary winding
can be one-quarter the size of those
ing
BC
because the current
(see Fig.
winding
1
1.5).
BC
is
However,
is
in
CA
wind-
4 times smaller
the voltage across
equal to the difference be-
tween the primary and secondary voltages,
I
11.4 has an 80 per-
cent tap and the supply voltage £,
c.
is
300
V. If a
namely (300 - 240) = 60 V. Consequently,
winding CA has four times as many turns as
winding BC. Thus, the two windings require
essentially the same amount of copper.
ELECTRICAL MACHINES AND TRANSFORMERS
228
-12A
/,
B
-O
2 =
/
G
:
15A
300 V
1
3A
A
240
<\
V
Figure 11.5
Autotransformer of Example 11-1.
11.3 Conventional transformer
6.
The voltages add when terminals of opposite
X2
polarity (H, and
connected as an
autotransformer
conventional
changed
into
may add to,
is
in series.
Depending
made, the secondary
volt-
or subtract from, the primary voltage.
basic operation and behavior of a transformer
unaffected by a mere change
in
is
external connections.
Consequently, the following rules apply whenever a
conventional transformer
is
connected as an auto-
transformer:
1
.
2.
3.
in
If
nominal voltage
1
1
.6
We
Hz.
wish to reconnect
in three different
age
a.
shown in
kVA, 600 V/l 20 V, 60
ways
it
as an autotransformer
to obtain three different volt-
ratios:
600
600
20
1
to
720
V
V
secondary
primary
V primary
to
480
V
secondary
V
V
primary to 480
maximum
secondary
load the transformer can
carry in each case.
rating.
rated current flows in one winding, rated cur-
ing (reason:
4.
If
The ampere-turns of the windings
always equal).
rated voltage exists across
one winding, rated
voltage automatically exists across the other
winding (reason: The same mutual flux links
2
both windings).
5.
If the
i
current in a winding flows from Hj to
the current in the other
X 2 to X,
and
5
1
rent will automatically flow in the other wind-
are
H2
single-phase transformer
has a rating of a
Calculate the
rating.
The voltage across any winding should not exits
Fig.
c.
any winding should not exceed
nominal current
ceed
(or
are connected together.
)
Example 11-2
The standard
b.
The current
its
and Xj) are con-
an autotransformer by connecting the
upon how the connection
age
X2
two-winding transformer can be
primary and secondary windings
The
H2
or
when H, and X,
voltages subtract
A
,
nected together by means of a jumper. The
H2
winding must flow from
and vice versa.
— 120 v-^o
o
,
Figure 11.6
Standard 15 kVA, 600 V/120 V transformer.
SPECIAL TRANSFORMERS
229
Solution
Nominal current of the 600 V winding
= SIE =
/,
=
000/600
15
{
is
Nominal current of the 120 V winding
I2
a.
=
=
S/E 2
To obtain 480
15
000/120
=
A
25
is
125
A
secondary voltage (120 V)
V, the
X2
between terminals X,,
must subtract from
we
the primary voltage (600 V). Consequently,
connect terminals having the same polarity
shown
gether, as
in Fig.
ing schematic diagram
Note
.7.
1
The correspond-
given
that the current in the 120
same
the
1
is
as that in the load.
to-
in Fig.
V
1
1
winding
Because
this
Figure 11.7
Transformer reconnected as an autotransformer to
.8.
give a ratio of
600 V/480
V.
is
wind-
ing has a nominal current rating of 125 A, the
maximum
load can draw a
S.,
The
.
A X
25
480
V=
60
kVA
currents flowing in the circuit at full-load
shown
are
1
1
power.
If
in Fig.
we assume
from X,
25
A
to
1
1
Note the following:
.8.
that the current
X2
in the
of
1
25
A flows
winding, a current of
H2
must flow from
to
H,
in the
other
winding. All other currents are then found
by applying Kirchhoff 's current law.
2.
The apparent power supplied by
is
S
b.
=
100
To obtain a
= 720
V.
site polarity
gether, as
same
600 V/720
=
V, the
of Fig.
1 1
.7
showing voltages and
secondary
primary voltage: 600
to the
Figure 11.8
Schematic diagram
current flows.
V = 60kVA
600
+
Consequently, terminals of oppo-
(H, and
X2
in the
must be connected
)
in Fig.
1
1
secondary winding
load
125
is
is
again
and therefore the
is
again 125 A.
720
V = 90kVA
load current
to-
.9.
as that in the load,
maximum
maximum
Sb
add
shown
The current
the
A X
ratio of
voltage must
120
the source
equal to that absorbed by the load:
The
now
A X
The previous examples show
ventional transformer
is
that
when
a con-
connected as an auto-
Figure 11.9
Transformer reconnected to give a
ratio of
600 V/720
V.
ELECTRICAL MACHINES AND TRANSFORMERS
230
69 kV
line
primary
\C
c/__
25
A
distributed
\^
capacitance
voltmeter
Oto 150 V
secondary
grounded
Figure 11.10
Transformer reconnected
to give
a
ratio of
120 V/480
V.
Figure 11.11
on a 69 kV line. Note
between the windings.
Potential transformer installed
transformer,
it
can supply a load far greater
As
than the rated capacity of the transformer.
mentioned
earlier, this
is
the distributed capacitance
one of the advantages
of using an autotransformer instead of a con-
most exactly
ventional transformer. However, this
is
The nominal secondary voltage
ways
example
the case, as the next part of our
not al-
be.
To obtain
we
the desired ratio of 120
to
480
X,X 2
is
now connected
same
transmission
600
V
winding
maximum
ilar to that
load current cannot exceed 25 A.
AX
and
Sc
=
25
This load
is
less than the
480
the
isolate
V = 12kVA
full line
nominal rating (15
winding
voltage on the
In this regard,
is
HV
to withstand the
one terminal of the secondary
always connected
to
ground
ing these three autotransformer connections.
to be isolated
The temperature rise of the transformer is the
same in each case, even though the loads are
respectively 60 kVA, 90 kVA, and 12 kVA. The
pacitance between the two windings
that the currents in the
windings and
to eliminate
when touching one of the
secondary leads. Although the secondary appears
from the primary, the distributed ca-
visible connection
between
voltage
ground.
the
secondary
By grounding one of
nals, the highest voltage
and so the losses are the same.
lines
and ground
The nominal
usually less than
insulation
is
is
makes an
in-
which can produce a very high
the flux in the core are identical in each case
11.4 Voltage transformers
sim-
side.
the danger of a fatal shock
is
is
between the primary and secondary
windings must be particularly great
kVA) of the standard transformer.
We want to make one final remark concern-
reason
metering
of conventional transformers. However,
the insulation
therefore,
is,
to
lines (Fig. 11.11).
The construction of voltage transformers
is
as that in the load; consequently, the
The corresponding maximum load
to
measure or monitor the voltage on
lines
equipment from these
in the
ir-
may
This permits standard instruments and relays
ers are used to
a),
to terminals
(Fig. 11. 10).
This time, the current
usually 115 V,
be used on the secondary side. Voltage transform-
V,
again connect H, and X, (as in solution
but the source
the
V
is
respective of what the rated primary voltage
shows.
c.
phase with the primary voltage.
in
winding
and
the secondary termi-
between
the secondary
limited to 115 V.
rating of voltage transformers
500 VA. As
a result, the
often far greater than the
volume of
Voltage transformers (also called potential trans-
formers) are high-precision transformers
the ratio of primary voltage to
a
known
load."'
constant,
in
which
secondary voltage
which changes very
little
Furthermore, the secondary voltage
steel.
is
with
is
copper or
al-
In the
case of voltage transformers and current transformers,
the load
is
called burden.
is
volume of
SPECIAL TRANSFORMERS
Voltage transformers installed on
ways measure
the
line-to-neutral
eliminates the need for two
one side of the primary
is
late the
kV
lines al-
to
ground. For
transformer
shown
11.12 has one large porcelain bushing to iso-
HV line from
the grounded case.
The
latter
houses the actual transformer.
The
basic impulse insulation (BIL) of
Current transformers
1 1 .5
This
bushings because
connected
example, the 7000 VA, 80.5
in Fig.
HV
HV
voltage.
23
650 kV
expresses the transformer's ability to withstand
Current transformers are high-precision transform-
which the
ers in
rent
a
is
of primary to secondary cur-
ratio
known
constant that changes very
little
with the burden. The phase angle between the
pri-
mary and secondary current is very small, usually
much less than one degree. The highly accurate current ratio
and small phase angle are achieved by
keeping the exciting current small.
Current transformers are used to measure or
lightning and switching surges.
monitor the current
ondary
in a line
and
to isolate the
equipment connected
tering and relay
me-
to the sec-
The primary is connected in series with
shown in Fig. 1. 3. The nominal sec-
side.
the line, as
1
ondary current
mary current
1
usually 5 A, irrespective of the pri-
is
rating.
Because current transformers (CTs) are only
used for measurement and system protection, their
power
rating
small
is
and 200 VA. As
—generally
in the
formers, the current ratio
to the
number of
ondary windings.
tio
of 150 A/5
A
between 15
VA
case of conventional transis
inversely proportional
turns on the primary and sec-
A current transformer having a rahas therefore 30 times more turns
on the secondary than on the primary.
For safety reasons current transformers must
ways be used when measuring currents
mission lines. The insulation between
in
HV
the primary
and secondary windings must be great enough
withstand the
line surges.
stand
is
full line-to-neutral
The maximum voltage
the
CT can
line
load'
primary
C
:-j-:c
/ distributed
capacitance
Figure 11.12
7000 VA, 80.5 kV, 50/60 Hz potential transformer having an accuracy of 0.3% and a BIL of 650 kV. The primary terminal at the top of the bushing is connected
to the HV line while the other is connected to ground.
The secondary is composed of two 115 V windings
each tapped at 66.4 V Other details: total height: 2565
mm; height of porcelain bushing: 1880 mm; oil: 250 L;
weight: 740 kg.
Figure 11.13
(Courtesy of Ferranti- Packard)
Current transformer installed on a 69 kV
secondary
secondary
grounded
to
voltage, including
always shown on the nameplatc.
69 kV
al-
trans-
-
line.
with-
ELECTRICAL MACHINES AND TRANSFORMERS
232
the
As in the case of voltage transformers (and for
same reasons) one of the secondary terminals is
always connected
to
ground.
ing serves to isolate the
HV
1
kV
line
line.
from
The
A
current
large bush-
the ground.
The
terminals that are connected in series with the
The
line current
Hows
HV
current
c.
CT
a typical installation
is
shown
By way of comparison,
former shown
because
it
is
in Fig.
shown
is
1
1
.
17
in Fig. 11.15
is
is
50
in-
a.
The current
in Fig.
trans-
The
turns ratio
=
400/5
N^/Nn
11,17 has a rating
I2
=
80
is
smaller, mainly
It is
imped-
ratio is
and
The secondary current
of 50 VA, 400 A/5 A, 36 kV. 60 Hz.
total
the transmission-line
Solution
insulated for only 36 kV.
Example 1 1-3
The current transformer
If
il.
The secondary current
The voltage across the secondary terminals
The voltage drop across the primary
/,// 2
VA current
much
2
280 A, calculate
in Fig. 11.16.
the
1.
the
up the bushing and out by the other terminal. The
of a
a.
line.
bushing, through the primary of the transformer, then
ternal construction
3.
wires on the secondary side possess a
b.
one terminal, down
into
1
ance (burden) of
CT is housed in the grounded steel case at the lower end
of the bushing. The upper end of the bushing has two
manner similar to that shown in
The ammeters, relays, and connecting
4.4 kV, in a
Fig. II.
Figure 11.14 shows a 500 VA, 100 A/5
transformer designed for a 230
into an ac line, having a line-to-neutral voltage of
=
=
1/80
is
280/80
=
3.5
A
connected
!
Figure 11.14
500 VA, 100 A/5 A, 60 Hz current transformer, insulated
for a 230 kV line and having an accuracy of 0.6%.
(Courtesy of Westinghouse)
N
Figure 11.15
Current transformer
in
the
final
(Courtesy of Ferranti-Packaro)
process
of construction.
SPECIAL TRANSFORMERS
M
"
233
Mm
'
Figure 11.17
Epoxy-encapsulated current transformer rated 50 VA,
A, 60 Hz and insulated for 36 kV.
(Courtesy of Montel, Sprecher & Schuh)
400 A/5
/,
Current transformer
220
kV,
3-phase
in
one phase
series with
line inside
The voltage across
c.
IR
=
X
3.5
The secondary voltage
The primary voltage is
E,
=
4.2/80
=
is
1
.2
=
4.2
V
than normal.
is
=
52.5
a miniscule voltage drop,
the 14.4
kV
cle.
mV
The
the
is
compared
to
line-to-neutral voltage.
Opening the secondary
of a CT can be dangerous
and
falls
<t»
much
s
first
falls,
for
but
it
most of
remains
at a
the time.
The
thing happens during the second half-cycle.
During these saturated
intervals, the induced volt-
age across the secondary winding
cause the flux changes very
little.
is
negligible be-
However, during
tremely high
primary
flowing
is
accidentally opened, the primary current
in the
circuit. If the sec-
to-
primary current
the unsaturated intervals, the flux changes
is
is
half cycle, flux <P
secondary circuit of a current transformer while
ondary
higher
of every half cy-
II. 18, as the
core also rises and
in the
1
so large that the core
during the
fixed, saturation level
secondary
may be 00
Every precaution must be taken to never open the
current
to that
normal exciting cur-
core reaches peaks
flux
Referring to Fig.
rises
same
11-6
to the
tally saturated for the greater part
/,
This
200 times greater than
rent, the flux in the
therefore 4.2 V.
0.0525
no further bucking effect due
ampere-turns. Because the line current
is
to
E2 =
compared
the exciting current of the transformer because there
a substation.
the burden
negligible
is
of the electrical load. The line current thus becomes
of a
is
b.
continues to flow unchanged because the imped-
ance of the primary
Figure 11.16
rate,
at
an ex-
inducing voltage peaks of several
hundred volts across the open-circuited secondary.
This
is
a
dangerous situation because an unsuspect-
ELECTRICAL MACHINES AND TRANSFORMERS
234
ing operator could easily receive a bad shock.
voltage
is
The
particularly high in current transformers
having ratings above 50 VA.
view of
In
ondary
must
the above,
first
if
CT
circuit of a
a meter or relay in the sec-
we
has to be disconnected,
secondary winding and
short-circuit the
then remove the component. Short-circuiting a current transformer
does no harm because the primary
current remains unchanged and the secondary current
can be no greater than that determined by the turns
ra-
winding may be
re-
tio.
The
moved
short-circuit across the
after the secondary circuit
is
again closed.
11-7 Toroidal current transformers
Figure 11.18
When
the line current exceeds 100 A,
we can some-
times use a toroidal current transformer.
It
consists
Primary current,
CT
is
flux,
and secondary voltage when a
open-circuited.
of a laminated ring-shaped core that carries the sec-
ondary winding. The primary
is
composed of a
sin-
gle conductor that simply passes through the center
of the ring (Fig.
conductor
is
1
1
.
1
9).
The
unimportant as long as
it
is
secondary possesses
less centered. If the
the ratio of transformation
is
having a ratio of 1000 A/5
200 turns
position of the primary
more
N
N. Thus, a toroidal
A
or
turns,
CT
has 200 turns on the
secondary winding.
Toroidal
widely used
(MV)
CTs are
indoor installations. They are also incorporated
in circuit-breaker
(Fig.
simple and inexpensive and are
low-voltage (LV) and medium-voltage
in
1
bushings to monitor the line current
1.20). If the current
limit, the
CT causes the
Figure 11.19
Toroidal transformer having a ratio of
nected to measure the current
in
a
1000 A/5
A, con-
line.
exceeds a predetermined
circuit-breaker to
trip.
Example 11-4
A potential
transformer rated 14 400 V/l 15
current transformer rated 75/5
V
and a
A are used to measure
the voltage and current in a transmission line. If the
voltmeter indicates
1
1
V
1
and the ammeter reads
3 A, calculate the voltage and current in the line.
Solution
cm
The voltage on
E=
The current
111
the line
x
is
(14 400/115)
=
13
900V
Figure 11.20
in the line is
Toroidal transformer surrounding a conductor inside a
/=
3
X
(75/5)
=
45
A
bushing.
SPECIAL TRANSFORMERS
11.8 Variable autotransformer
1
1.22).
a fixed
A
variable autotransformer
is
often used
when we
wish to obtain a variable ac voltage from a fixedvoltage ac source.
a
single-layer
The transformer
toroidal iron core.
A
composed of
movable carbon brush
in slid-
winding serves as a variable
ing contact with the
tap.
is
winding wound uniformly on a
The brush can be
set in
235
The input voltage £, is usually connected to
90 percent tap on the winding. This enables
E 2 to vary from 0 to
1
10 percent of the input voltage.
Variable autotransformers are efficient and provide
good voltage regulation under variable
ondary
line
loads.
The
sec-
should always be protected by a fuse or
circuit-breaker so that the output current
/2
never ex-
ceeds the current rating of the autotransformer.
any position between 0
and 330°. Manual or motorized positioning
may
be
used (Figs. 11.21 and 11.23).
As
the brush slides over the bared portion of the
winding, the secondary voltage
portion to the
Figure 11.21
Cutaway view
number of
of a
E2
turns
increases in pro-
swept out
(Fig.
manually operated 0-140 V, 15 A
showing (1) the laminated
variable autotransformer
toroidal core; (2) the single-layer winding; (3) the
mov-
able brush.
{Courtesy of American Superior Electric)
Figure 11.23
200 A, 0-240 V,
50 A, 120 V units,
connected in series-parallel. This motorized unit can
vary the output voltage from zero to 240 V in 5 s.
Dimensions: 400 mm x 1500 mm.
{Courtesy of American Superior Electric)
Variable autotransformer rated at
50/60 Hz.
Figure 11.22
Schematic diagram
ing a fixed
90%
tap.
of
a variable autotransformer hav-
It
is
composed
of eight
236
ELECTRICAL MACHINES AND TRANSFORMERS
11.9
High-impedance transformers
The primary winding P is connected to a
V ac source, and the two secondary windings
240
The transformers we have studied so far are alt designed to have a relatively low leakage reactance,
ranging perhaps from 0.03 to
0.
S are connected
tube.
per unit (Section
1
<T>
10.
1
However, some
3).
applications require
and commercial
industrial
much higher reactances, some-
across the long neon
the secondary voltage
of the transformer (Fig.
circuit voltage (20
following
in the
in series
to the large leakage fluxes
E2
tp.,
falls rapidly
as
typical applications:
kV)
1
soon as the neon tube
current
is
1
.24c).
The high open-
initiates the discharge, but
lights up, the
automatically limited to 15
secondary
mA. The
electric toys
arc welders
corresponding voltage across the neon tube
fluorescent lamps
electric arc furnaces
to
neon signs
reactive
oil
and
with
increasing current, as seen in the regulation curve
times reaching values as high as 0.9 pu. Such high-
impedance transformers are used
b
,
Owing
falls
The power of these transformers ranges
from 50 VA to 1500 VA. The secondary voltages
power regulators
1
5 kV.
burners
Let us briefly examine these special applications.
1
.
A
toy transformer
is
often accidentally short-
by children
circuited, but being used
ther practical nor safe to protect
Consequently, the transformer
that
its
leakage reactance
is
is
it
it
nei-
is
with a fuse.
designed so
so high that even a
permanent short-circuit across the low- voltage
secondary will not cause overheating.
The same remarks apply
to
some
bell trans-
formers that provide low-voltage signalling
power throughout a home.
If
a short-circuit oc-
curs on the secondary side, the current
is
auto-
matically limited by the high reactance so as
not to burn out the transformer or
damage
the
fragile annunciator wiring.
2.
Electric arc furnaces
and discharges
in
gases
possess a negative E/I characteristic, meaning
that
once the arc
struck, the current increases
is
as the voltage falls.
To maintain
a uniform discharge,
ance
in series
ance
may
a steady arc, or
we must add
an imped-
with the load. The series imped-
be either a resistor or reactor, but
prefer the latter because
it
consumes very
we
little
active power.
However,
the load,
it
is
if
a transformer
usually
is
used
to
more economical
porate the reactance in the transformer
designing
it
itself,
have a high leakage reactance.
to
typical
example
shown
in Fig.
1
is
1
the neon-sign transformer
.24.
0
supply
to incor-
by
A
Figure 11.24
a. Schematic diagram
—
15
of a
30
mA
neon-sign transformer.
b.
Construction of the transformer.
c.
Typical E-I characteristic of the transformer.
SPECIAL TRANSFORMERS
range from 2
kV
to
20 kV, depending mainly
ings are very loosely coupled.
mary windings
upon the length of the tube.
Returning to Fig.
1
1
.24a,
we
note that the
center of the secondary winding
(typically
is
only one-half the voltage across the
neon tube. As a
result, less insulation is
have properties similar
to
neon-sign
HV
1
.25).
The
controller permits
more
or less sec-
causing the leakage flux
to flow,
A change
in the
flux produces a corresponding
active
power absorbed by
leakage
change
in
transformer, incorporated in a static var
improve the power factor of the
pensator,
is
the re-
The
com-
the transformer.
transformers. Capacitors are usually added to
total circuit.
line
765 kV) while
windings (typically 6 kV)
to vary accordingly.
Fluorescent lamp transformers (called ballasts)
1
ondary current
needed
for the high-voltage winding.
kV and
between 230
are connected to an electronic controller (Fig.
This ensures that the secondary line-to-ground
voltage
three pri-
are connected to the
the three secondary
grounded.
is
The
237
further discussed in Section 25.27.
Oil-burner transformers possess essentially
the
same
about 10
A
kV
the oil jet.
3.
Some
oil
Q
secondary open-circuit voltage of
The
M
'4
arc continually ignites the va-
while the burner
is
in
operation.
between two carbon
taining an intense arc
A relatively
low secondary voltage
leakage flux
elecis
used
tertiary
and the large secondary current
is
^
primary
electric furnaces generate heat by main-
trodes.
kV
Q
M
two closely
immediately above
creates an arc between
spaced electrodes situated
porized
3-phase primary input 230
characteristics as neon-sign trans-
formers do.
winding
limited by the
leakage reactance of the transformer. Such trans-
formers have ratings between 100
MVA.
In
kVA and 500
secondary
leakage flux
very big furnaces, the leakage reactance
of the secondary, together with the reactance of
the conductors,
usually sufficient to provide
is
the necessary limiting impedance.
4.
Arc- welding transformers are also designed to
have a high leakage reactance so as to stabilize
the arc during the welding process.
The open-
about 70 V, which
facilitates
circuit voltage
striking the arc
is
when
the electrode touches the
work. However, as soon as the arc
lished, the
secondary voltage
15 V, a value that
the arc
5.
As
and the
is
falls to
about
we mention
the
units that absorb reactive
transmission
line.
enormous 3-phase
power from a 3-phase
These transformers are
tionally designed to
11.10 Induction heating
transformers
welding current.
a final example of high-impedance trans-
formers,
leak-
estab-
depends upon the length of
intensity of the
Figure 11.25
Three-phase static var compensator having high
age reactance.
inten-
produce leakage flux and,
consequently, the primary and secondary wind-
High-power induction furnaces also use the
former principle to produce high-quality
other alloys.
The induction
stood by referring to Fig.
frequency 500
that
Hz
1
trans-
steel
and
principle can be under1.26.
ac source
is
A
relatively high-
connected to a
coil
surrounds a larse crucible containing molten
ELECTRICAL MACHINES AND TRANSFORMERS
238
primary
molten iron
Figure 11.26
Coreless induction furnace. The
currents
in
the molten metal.
the reactive
The
iron.
coil is the primary,
itself.
O
produces eddy
capacitor furnishes
power absorbed by the
coil.
and the molten iron
secondary turn, short-circuited
acts like a single
upon
flux
The
Consequently,
carries a very large
it
secondary current. This current provides the energy
that
keeps the iron
scrap metal as
15
it is
melting other
in a liquid state,
added
to the pool.
Such induction furnaces have ratings between
kVA and 40 000 kVA. The operating frequency
becomes progressively lower
power
as the
Figure 11.27
Channel induction furnace and
is
required to drive the flux through
the molten iron and through the
must remember
far
air.
In this regard,
that the temperature of
above the Curie
point,
far as permeability
is
and so
it
we
molten iron
The magnetizing
is
like air as
why
these
Capacitors are installed close to the coil to sup-
In
power
it
known
as a
channel fur-
nace, a transformer having a laminated iron core
made
to link with a channel of
in Fig.
1
1
ted to the
.27.
The channel
Hows
molten
iron, as
Hz
is fit-
coil is
channel and through the molten
iron in the crucible. In effect, the channel
lent to a single turn short-circuited
coil.
is
On
is
the
large because the sec-
obviously not tightly coupled to the
Nevertheless, the power factor
II. 26,
and 80 percent. As a
is
higher
being typically between 60
result, a
smaller capacitor bank
required to furnish the reactive power.
Owing
to the
very high ambient temperature, the
primary windings of induction furnace transformers are
always made of hollow, water-cooled copper
on
is
itself.
aluminum, copper, and other metals,
1
1.28
shows
as well as iron.
a very special application of
is
source, and the secondary current
in the liquid
primary
is
than that in Fig.
Figure
shown
a ceramic pipe that
bottom of the crucible. The primary
excited by a 60
12
is
other hand, the leakage flux
ondary turn
low because the flux
is
permeable iron core.
conductors. Induction furnaces are used for melting
absorbs.
another type of furnace,
current
to a highly
is
furnaces are often called coreless induction furnaces.
ply the reactive
confined
is
behaves
concerned. That
water-cooled
rating
60 Hz is used when
the power exceeds about 3000 kVA.
The power factor of coreless induction furnaces is
very low (typically 20 percent) because a large magincreases. Thus, a frequency of
netizing current
its
transformer.
equiva-
the induction heating principle.
11.11 High-frequency transformers
In electronic
power supplies there is often a need to
from the input and to reduce the
isolate the output
weight and cost of the
such as
in
aircraft,
unit. In
there
is
other applications,
a strong incentive to
SPECIAL TRANSFORMERS
239
order to illustrate the reason for this phenomenon,
limit
we
how
avoid a cumbersome theoretical analysis,
take a practical transformer and observe
when
haves
the frequency
Consider Fig.
tional
20 V/24
1
ing of 36
in the
II. 29,
be-
1
.5
T.
a rat-
0.5 kg
The
flux
core attains a peak of 750 fjiWb. The lami-
nated core
is
about
made of
1
ordinary silicon steel having
mm
(12 mils) and the
W. The
and
the primary
1
.5
A
current rating
silicon
A
>~
1
t
core
mA for
12 mil
1.5
600
total
300
cm
mA
120 V
60 Hz
is
for the secondary.
core: 6 x 5 x 2.5
300
it
which shows a conven-
peak flux density of
at a
to
will
raised.
60 Hz transformer having
V,
a thickness of 0.3
is
is
VA. This small transformer weighs
and operates
loss
we
our discussion to transformers. Furthermore,
120
24 V
t
T
.
J
P = 36 VA
B=1.5T
0max = 75OAWb
core loss =
1
W
Figure 11.29
Figure 11.28
Special application of the transformer effect. This picture
of
shows one stage
in
a steam-turbine generator.
the diameter of a 5
t
It
consists of expanding
coil-retaining ring.
A
coil of
as-
its
temperature up to 280°C
expansion enables the
ends, where
it
ring to
in
h.
The
main
resulting
be slipped over the
rise of the large
this
it
in-
mass.
(Courtesy of ABB)
means
that the
can be increased
E=
=
=
which
4.44,/TV]
400 Hz, while
in
electronic
to
to,
typically
power supplies
quency may range from 5 kHz
An
is
the
at a fre-
same peak
^ max
will re-
to Eq. 9.3,
corresponding primary voltage
to
the fre-
50 kHz.
increase in frequency reduces the size of such
devices as transformers, inductors, and capacitors. In
<I>
(9.3)
max
X 6000 X 600 X 750 X
000 V
12
is
6
l()
100 times greater than before! The sec-
becoming 2400
frequency
Assuming
4.44
by using a relatively high frequency compared
in aircraft the
it
00 times higher than
750 |xWb. However, according
at
ondary voltage
60 Hz. Thus,
1
follows that the flux
minimize weight. These objectives are best achieved
say,
for.
is
coil-
clean and produces a very uniform
is
was designed
it
flux density,
the ring, bringing
about 3
cools and contracts. This method of
duction heating
temperature
in
to the transformer,
us consider the effect of operating
what
is
which induces large eddy currents
let
quency of 6000 Hz, which
wound around the ring and
connected to a 35 kW, 2000 Hz source (left foreground). The coil creates a 2000 Hz magnetic field,
bestos-insulated wire
Without making any changes
the construction of the rotor
will likewise be 100 times greater,
The operating conditions are
The primary and secondary
currents remain unchanged and so the power of the
transformer is now 3600 VA, 00 times greater than
shown
in Fig.
V.
11.30.
1
in Fig.
1
1
.29. Clearly, raising the
a very beneficial effect.
frequency has had
ELECTRICAL MACHINES AND TRANSFORMERS
240
However, the advantage
because
at
6000 Hz
700 W), due
to the increase in
is
eddy current and hys-
Thus, the transformer
teresis losses.
not feasible because
To
it seems
enormous (about
it
in Fig.
we
of 12 mil silicon
cording
to
1
.5
T to 0.04 T. As
be
a result, ac-
Eq. 9.3, the primary and secondary volt-
ages will have to be reduced to 320
spectively.
as
reduction in
steel, this requires a
from
the flux density
The new power of
P = 320 X
0.3
=
96
V
and 64 V,
re-
the transformer will
VA (Fig.
11.31
).
This
is al-
most 3 times the original power of 36 VA, while
same temperature rise.
By using thinner laminations made of
re-
it
is
special
possible to raise the flux density
above 0.04 T while maintaining the same core
losses.
Thus,
if
we
replace the original core with
this special material, the flux density
to 0.2 T.
=
cl>
max of
750 |xWb X
(0.2 T/1.5 T)
means
primary voltage can be raised to
that the
100
£ = 4.4477V, * nlllx
= 4.44 X 6000 X 600 X
= 1600 V
core: 6 x 5 x 2.5
12 mi
silicon
300
120/
silicon
12
600
Figure 11.31
interested, of course, in maintaining the
1
V to 24 V. This
20
read-
is
number of turns on the primary will be reduced
to 600 t X (120 V/1600 V) = 45 turns,
while the secondary will have only 9 turns. Such a
drastic reduction in the
wire
the
Bearing
former
can
size
mind
in
is still
number of turns means
be
increased
that the capacity of the trans-
480 VA,
it
follows that the rated pri-
mary current can be raised to 4 A while that
secondary becomes 20 A. This rewound
same
size
in
the
trans-
11.33) has the
special core (Fig.
its
that
significantly.
and weight as the one
in
Fig.
11.29.
Furthermore, because the iron and copper losses
same
are the
both cases, the efficiency of the
in
high frequency transformer
It is
now obvious
is better.
that the increase in
has permitted a very large increase
pacity of the transformer.
It
in the
frequency
power
ca-
follows that for a given
A
cm
special core
J
'
= 3600
VA
300
mA
1
ff=1.5T
2400 V
^max^SOjUWb
1600^
core loss = 700
6 kHz ^
W
600
r
120
5
p = 480 VA
B = 0.2T
A
t+
t
«W= lOO^Wb
320 V
r
core: 6 x 5 x 2.5
cm
core loss =
1
W
t
120/
cm
special core
1.5
.
is
.32).
from 600
mil
mA
6 kHz
1
Figure 11.32
core: 6 x 5 x 2.5
32*0V^
1
power output a high frequency transformer is much
smaller, cheaper, more efficient, and lighter than a
60 Hz transformer.
Figure 11.30
300
VA (Fig.
480
core: 6 x 5 x 2.5
—
/
I(T
A=
the
cm
15A
600
X
100
.5
achieved by rewinding the transformer. Thus,
I
mA
12 kV
6 kHz
^Wb, which
V,
ily
can be raised
This corresponds to a peak flux
1
former with
taining the
nickel-steel,
V X
We are
320
original voltage ratio of
same
320
is
and so the enhanced capacity of the transformer
is
Based upon the properties
1.29.
1
.30
can reduce the
flux density so that the core losses are the
in Fig.
1
1
will quickly overheat.
get around this problem,
they were
The corresponding secondary voltage
not as great as
is
the core loss
64
A
vh
Y
P = 96 VA
# = 0.04T
0max = 2OAiWb
core loss =
1
W
4
20 A
A
B=
120~\T
6 kHz
1
45/
Figure 11.33
9/
24
V
T
0.2
T
<^max = lOO^Wb
core loss = 1
W
SPECIA L TRA NS FORMERS
Questions and Problems
Calculate the voltage across the secondary
a.
winding
Practical level
11-1
What
is
What
the difference between an auto-
is
former?
1
1
-3
Why
1
-4
11-5
Calculate the voltage drop the transformer
c.
If the
produces on the
Of a
is
looped four
times through the toroidal opening, calcu-
current transformer?
new
late the
current ratio.
Industrial application
why the secondary winding
PT must be grounded.
Explain
CT or
A toroidal
of a
11-12
The nameplate of a small transformer
dicates 50 VA, 120 V, 12.8 V When
1
current transformer has a ratio of
How many
turns does
have?
it
current transformer has a rating of 10
VA, 50 A/5 A, 60 Hz,
1
V
8.8
would
2.4 kV. Calculate
V
13.74
If
V were
120
at
no-load
available,
what
Why
the secondary voltage be?
this voltage
the nominal voltage across the primary
in-
applied to the primary, the
is
voltage across the secondary
is
A
line conductor.
primary conductor
must we never open the secondary of
1500 A/5 A.
11-6
ammeter has an impedance
the
b.
the purpose of a voltage trans-
a current transformer?
1
if
of 0.1 5 n.
transformer and a conventional transformer?
11-2
24
is
higher than the indicated
nameplate voltage?
winding.
11-13
Intermediate level
11-7
A
single-phase transformer has a rating of
100 kVA, 7200 V/600 V, 60 Hz.
If
re-
it is
connected as an autotransformer having a
ratio
it
1
1-8
In
can carry.
Problem
11-14
1
1-7,
how
should the trans-
H2
,
X,,
X2
)
be con-
in
Advanced
11-10
A
of 6.6 k V/600
how
is
Problem
1
1-7
is
V
What
load can
it
120
V
winding wound
Many
airports use series lighting systems in
rent
winding, or vice versa?
sion line
it
where
is
installed
numin
60 Hz
kept constant
at
20 A. The secondary
A
to a
incandescent lamp.
a.
Calculate the voltage across each lamp.
b.
The
0.07
resistance of the secondary winding
0
while that of the primary
Knowing
It
that the
is
is
0.008 U.
magnetizing current and
the leakage reactance are both negligible,
has a primary to secondary capacitance
If
is
100 W, 6.6
should the connections be made?
of 250 pF.
large
windings are individually connected
carry
100 VA, 2000 A/5 A, 60 Hz, 138 kV.
calculate the voltage across the primary
on a transmis-
winding of each transformer.
the line-to-neutral voltc.
If
140 lamps, spaced
138 kV, calculate the capacitive
vals, are
leakage current that flows to ground (see
11-11
V
the 12.8
current transformer has a rating of
is
il. Is the
upon
source. In one installation, the primary cur-
recon-
level
age
0.306
the pri-
5.2 11 and that of the secondary
1
ber of current transformers are connected
nected again as an autotransformer having a
and
mary
series across a constant current,
The transformer
ratio
However, the resistance of
which the primary windings of a
nected?
1-9
epoxy and cannot be
seen.
is
of 7800 V/7200 V, calculate the load
former terminals (H,,
1
Referring to Problem 11-12, the windings
are encapsulated in
at
in series
power
The
a temperature of 105°C.
toroidal current transformer of Fig.
1
000 A/5 A. The
conductor carries a current of 600 A.
line
in inter-
using
wire, calculate the
Fig. M.13).
11.19 has a ratio of
every 50
No 14
minimum voltage of the
Assume the wire operates at
connected
11-15
A
source.
no-load
test
on a
1
5
kVA. 480 V/l 20
60 Hz transformer yields
the following
V,
242
ELECTRICAL MACHINES AND TRANSFORMERS
saturation curve data
winding
is
The primary
a.
Draw
is
If the
the
1
20
V
c.
known
to
have 260
Draw
the saturation curve at
flux in
60 Hz (peak
mWb versus current in
mA). At
what point on the saturation curve does
turns.
uration
the saturation curve (voltage versus
become important?
sat-
Is the flux dis-
torted under these conditions?
mA).
current in
b.
when
excited by a sinusoidal source.
experiment were repeated using a
50 Hz source, redraw the resulting saturation curve.
E
14.8
31
A)
59
99
49.3
144
66.7
210
90.5
430
110
120
130
136
142
V
700
1060
1740
2300
3200
mA
Chapter 12
Three-Phase Transformers
connected
12.0 Introduction
is distributed throughout North America by
means of 3-phase transmission lines. In order to
transmit this power efficiently and economically, the
Power
vice versa.
kV
3.8
power
to
that has to be transmitted
and the distance
is
also
A
it
uniform, ranging from
systems to 600
V
1
20/240
shift
from one
this re-
ratio
of the
the primaries and secon-
enables us to change the number of
if
there
were a
12-phase system.
practical application for
it,
we
could even convert a 3-phase system into a 5-phase
can be achieved by using three
system by an appropriate choice of single-phase
transformers and interconnections.
form a 3-phase transformer bank.
In
making
tant to
When
how
into a 2-phase, a 6-phase, or a
Indeed,
single-phase transformers connected together to
12.1
The amount of phase
depends again upon the turns
phases. Thus, a 3-phase system can be converted
windings mounted on a 3-legged core. However,
result
between the 3-phase input voltage and
shift feature
level to another.
The transformers may be inherently 3-phase,
same
shift
daries are interconnected. Furthermore, the phase-
having three primary windings and three secondary
the
are connected.
transformers, and on
quires the use of 3-phase transformers to transform
the voltages
may
wye, or
3-phase transformer bank can also produce a
the 3-phase output voltage.
V single-phase
3-phase systems. Clearly,
in
a result, the ratio of the 3-phase input
upon how they
phase
the appropriate
voltage levels used in factories and homes. These are
fairly
As
and the secondaries
only upon the turns ratio of the transformers, but
765 kV) depend upon the amount of
has to be carried. Another aspect
in delta
voltage to the 3-phase output voltage depends not
voltages must be at appropriate levels. These levels
(1
ways. Thus, the primaries
in several
be connected
the various connections,
observe transformer
polarity
may produce
polarities.
it
is
An
imporerror in
a short-circuit or unbalance
Basic properties of 3-phase
transformer banks
the line voltages
three single-phase transformers are used to
former banks can be understood by making the
The
transform a 3-phase voltage, the windings can be
and currents.
basic behavior of balanced 3-phase trans-
lowing simplifying assumptions:
243
fol-
ELECTRICAL MACHINES AND TRANSFORMERS
244
1
.
2.
3.
The exciting currents
nected
are negligible.
The transformer impedances, due
to the resis-
is
Terminal H, of each trans-
H2
connected to terminal
of the next
X2
tance and leakage reactance of the windings,
transformer. Similarly, terminals X, and
are negligible.
cessive transformers are connected together.
The
total
tual physical layout of the transformers
apparent input power to the trans-
former bank
Fig. I2.l.
is
put power.
Furthermore,
when
single-phase transformers
are connected into a 3-phase system, they retain
their basic single-phase properties,
voltage ratio, and flux
ratio,
polarity
marks X,,
X2
in
and Hj,
between primary and secondary
that
Ex
x
is in
Fig.
phase with
EH
the core.
H2
is
,
Given the
C
AO
the phase shift
Ht
X
Ho
X;
-O-i
1
2
balanced
BOH
.
three-phase
CO
R
of
to a level appropriate for
the outgoing transmission line
is
The schematic diagram is drawn in such a way to
show not only the connections, but also the phasor
H2
X;
Hi
X,
Ho
X2
load
I
transform the voltage of the incoming
transmission line A, B,
line
l,
2, 3.
The incoming
connected to the source, and the outgoing
connected
to the load.
The transformers
\ 3/„
line
are con-
Figure 12.1
Delta-delta connection of three single-phase trans-
formers.
The incoming
lines (source) are A, B,
the outgoing lines (load) are
V3/
1
,
2, 3.
s
load
Figure 12.2
Schematic diagram
is
in Fig. 12.2.
zero, in the sense
three single-phase transformers P, Q, and
12.1
The corresponding schematic diagram
all
such as current
12.2 Delta-delta connection
The
is
of suc-
The acshown in
equal to the total apparent out-
given
is
in delta-delta.
former
of
a delta-delta connection and associated phasor diagram.
C and
THREE-PHASE TRANSFORMERS
relationship between the primary and secondary
c.
drawn
d.
corresponding primary winding to
e.
voltages. Thus, each secondary
parallel to the
which
is
it
winding
coupled. Furthermore,
duces voltages
£AB £ BC ECA
,
,
if
is
G
source
in-
current in the
HV
current in the
LV
lines
lines
currents in the primary and secondary
windings of each transformer
pro-
according to the
The
The
The
245
f.
The load
carried by each transformer
dicated phasor diagram, the primary windings are
same way, phase by phase. For example, the primary of transformer P between lines A
and B is oriented horizontally, in the same direction
Solution
oriented the
£AB
as phasor
a.
The apparent power drawn by
S
.
Because the primary and secondary voltages
EH
and
Hn
phase,
Ex
En
(secondary voltage of
b.
E AB (primary
£2 3 s m phase
transformer P) must be in phase with
of the same transformer). Similarly,
with
£ BC
,
and
E3i
ECA
with
i-
tween the respective incoming and outgoing
mission lines are
If
in
small.
is
c.
connected to lines
1
The
-2-3, the
incoming
=
(24.4
=
102
in the
/
V3 times greater than the respective
and
/s
ondary windings
the transformer
flowing
(Fig.
bank
in the
The
rating of
I2
single transformer.
Note
tutes
that
arrangement, each transformer,
considered alone, acts as
phase
H2
circuit.
in the
rent / s
if
it
were placed
Thus, a current
primary winding
flowing from
X2
to
is
X|
/
p
MVA.
is
(8.9)
X
10 )/(V3
X
138 000)
A
LV
lines is
=
5/(V3 E)
=
(24 A
6
X
10 )/(V3
X 4160)
= 3386 A
although the transformer bank consti-
a 3-phase
also 24.4
line
6
current in the
three times the rating of a
is
is
HV
primary and sec-
The power
12.2).
HV line
current in each
/,=5/(V3£)
d.
currents
MVA
follows that the apparent power fur-
in any delta connection, the line
produces balanced line currents
A-B-C. As
It
nished by the
trans-
phase.
a balanced load
lines
24.4
(7.7)
The transformer bank itself absorbs a negligible
amount of active and reactive power because
the 1~R losses and the reactive power associated
resulting line currents are equal in magnitude. This
currents are
21/0.86
=
is
with the mutual flux and the leakage fluxes are
,
such a delta-delta connection, the voltages be-
In
P/cos 0
of a given transformer must be in
Xi
follows that
it
=
=
the plant
e.
Referring to Fig.
mary winding
in a single-
flowing from H] to
/
p
1
2.2, the current in
each
pri-
is
=
1
02/V 3
=
58.9
A
associated with a curin the
The
secondary.
current in each secondary winding
/s
= 3386/V3 =
1955
is
A
Example 12-1
Three single-phase transformers are connected
delta-delta to step
down
a line voltage of
4160 V to supply power
The plant draws 21
MW
1
38
kV
in
to
to a
manufacturing plant.
at a
lagging power factor
f.
Because the plant load
is
balanced, each trans-
former carries one-third of the
24.4/3
The
-
8.13
total load, or
MVA.
individual transformer load can also be
obtained by multiplying the primary voltage
of 86 percent.
times the primary current:
Calculate
a.
b.
The apparent power drawn by the plant
The apparent power furnished by the HV
S =
line
=
E p Ip =
8.13
138 000
MVA
X
58.9
ELECTRICA L MACHINES AND TRANSFORMERS
246
Note
that
we can
do not know how the
fect, the plant load
3-phase load
(shown
composed of hundreds of
which
are
connected
as a
is
connected. In ef-
box
in Fig.
individual loads,
in
delta,
we
others
12.2)
is
some of
wye.
in
Furthermore, some are single-phase loads operating
at
much lower
voltages than
4160
V,
powered by
smaller transformers located inside the plant.
sum total of these
The
calculate the line currents and the
currents in the transformer windings even though
The
relative values of the currents in the trans-
former windings and transmission
Fig.
C
1
are
The
2.4.
Thus,
the line currents in
V 3 times the currents
line currents in
in the
phases
lines are given in
phases A, B, and
primary windings.
2, 3 are the
l,
same
A
shift
delta-wye connection produces a 30° phase
between the
of the incoming and
line voltages
outgoing transmission
lines.
Thus, outgoing
line
loads usually results in a reasonably
well-balanced 3-phase load, represented by the box.
AO-
12.3 Delta-wye connection
When
the transformers are connected in delta-wye,
the three primary windings are connected the
way as
in Fig.
1
2.
1
.
together, creating a
BO
same
H,
X
H2
M
X;2
X
f
1
—
C—-—
balanced
2
:
I
H,
Xt
H2
X;
H,
X
H2
x 2 |—
-O(
•
three-phase
load
However, the secondary windings
are connected so that
all
the
common
X2
terminals are joined
neutral
N
CO
(Fig. 12.3). In
3
-O--I
such a delta- wye connection, the voltage across each
primary winding
age.
is
equal to the incoming line volt-
However, the outgoing
line voltage is a/3
the secondary voltage across each transformer.
times
as
the currents in the secondary windings.
Figure 12.3
Delta-wye connection of three single-phase
transformers.
load
Figure 12.4
Schematic diagram of a delta-wye connection and associated phasor diagram. (The phasor diagrams on the
mary and secondary sides are not drawn to the same scale.)
pri-
THREE-PHA SE TRANSFORMERS
375 A
3932 A
247
1
80 kV
90
MVA
13.2kV«
375 A
3932 A
Figure 12.5
See Example
E i2
voltage
£ AB
,
12-2.
is
30° ahead of incoming
as can be seen
The voltage across
line voltage
from the phasor diagram.
The voltage between
outgoing line feeds an isolated group of loads, the
phase
shift creates
no problem. But,
if
and 3
the outgoing
be connected in parallel with a line comfrom another source, the 30° shift may make
such a parallel connection impossible, even
line voltages are
One
that
is
it
s
b.
1,
2,
The load
C
is
kV
139
carried by each transformer
S
=
90/3
is
= 30 MVA
wye conThe
reduces the amount of insulation
needed inside the transformer. The
to
the
therefore,
the outgoing lines
E = 80 V3 =
otherwise identical.
of the important advantages of the
nection
if
is,
is
line has to
ing
the secondary
80 kV.
If the
current in the primary winding
HV winding has
/
be insulated for only 1/V3, or 58 percent of the
The
p
is
= 30 MVA/ 13.2 kV = 2273 A
current in the secondary winding
is
line voltage.
/s
Example 12-2
c.
Three single-phase step-up transformers rated
MVA,
a 13.2
a 90
1
3.2
kV
kV/80 kV
at
The current
If
in
each incoming
/=
they feed
MVA load, calculate the following:
The
2273 V 3
b.
c.
The secondary line voltage
The currents in the transformer windings
The incoming and outgoing transmission
line
A, B,
= 3937 A
current in each outgoing line
/
a.
A
375
40
are connected in delta-wye on
transmission line (Fig. 12.5).
= 30 MVA/80 kV =
=
375
1
,
2, 3 is
A
12.4 Wye-delta connection
line
The currents and voltages
currents
in a wye-delta connection
are identical to those in the delta-wye connection of
Solution
The
easiest
way
Section
to solve this
the windings of only
former
problem
is
one transformer,
say, trans-
P.
The voltage across
ously 13.2 kV.
2.3.
The primary and secondary connec-
tions are simply interchanged. In other words, the
H 2 terminals are connected together to create a neuand the X X 2 terminals are connected in delta.
tral,
a.
1
to consider
the primary winding
is
obvi-
[
,
Again, there results
a
30° phase
shift
between the
voltages of the incoming and outgoing lines.
ELECTRICAL MACHINES AND TRANSFORMERS
248
ac source
|
Figure 12.6
Wye-wye connection
ac source
with neutral of the primary connected to the neutral of the source.
'
Figure 12.7
Wye-wye connection using a
12.5
tertiary winding.
Wye-wye connection
to a delta-delta connection, except that
former
When
cial
transformers are connected
in
wye-wye, spe-
precautions have to be taken to prevent severe
One way
distortion of the line-to-neutral voltages.
to prevent the distortion is to
connect the neutral of
the primary to the neutral of the source, usually
way
of the ground (Fig. 12.6). Another
way
is
by
to
provide each transformer with a third winding,
called tertiary winding.
three
1
2.7).
The
tertiary
transformers are connected
They
windings of the
in
delta
(Fig.
often provide the substation service
is
delta connection
is
is
ample,
if
two 50 kVA transformers are connected
bank
as
it
is
obviously 2
may seem,
it
X 50 =
100 kVA. But, strange
can only deliver 86.6
The open-delta connection
emergency
situations.
Thus,
if
mainly used
is
b
n^-Mj
Q
I
possible to transform the voltage of a 3-phase
The open-delta arrangement
is
in
identical
Figure 12.8a
Open-delta connection.
o
[h
x~|
in
three transformers
A
O-
12.6 Open-delta connection
open-delta.
kVA before
the transformers begin to overheat.
0—4
It is
in
open-delta, the installed capacity of the transformer
transformer.
system by using only 2 transformers, connected
only 86.6 per-
cent of the installed transformer capacity. For ex-
Note that there is no phase shift between the incoming and outgoing transmission line voltages of
wye-wye connected
trans-
seldom used because the load
capacity of the transformer bank
voltage where the transformers are installed.
a
one
absent (Fig. 12.8). However, the open-
THREE-PHASE TRANSFORMERS
are connected in delta-delta and
comes defective and has
ble to feed the load
to
one of them be-
be removed,
it
is
249
Thus, the ratio
possi-
maximum
on a temporary basis with the
kVA
300 kVA
260
load
installed transformer rating
two remaining transformers.
=
Example 12-3
Two
single-phase 150
formers are connected
maximum
V
kVA, 7200 V/600
in
0.867, or
86.7%
12.7 Three-phase transformers
trans-
open-delta. Calculate the
3-phase load they can carry.
A transformer bank
transformers
former
Solution
may
(Fig.
1
composed of three single-phase
be replaced by one 3-phase trans-
2.9).
The magnetic core of such
a
transformer has three legs that carry the primary and
Although each transformer has a rating of 50 k VA,
1
two together cannot carry a load of 300 kVA.
The following calculations show why:
secondary windings of each phase. The windings are
the
The nominal secondary current of each
former
connected
internally, either in
wye
the result that only six terminals
or in delta, with
have to be brought
trans-
outside the tank. For a given total capacity, a 3-phase
is
transformer
=
/s
1
50 kVA/600
V=
250
A
is
single-phase
always smaller and cheaper than three
transformers.
Nevertheless,
single-
phase transformers are sometimes preferred, partic-
The current
ceed 250 A
mum
/s in lines 1, 2, 3
(Fig.
cannot, therefore, ex-
12.8b). Consequently, the maxi-
load that the transformers can carry
when
a replacement unit
one 3-phase 5000
(8.9)
259 800
Figure 12.8b
Associated schematic and phasor diagram.
VA
is
essential.
For ex-
kVA. To guarantee continued service we can
is
S = ^3EI
= V3 X 600 X 250 =
= 260 kVA
ularly
ample, suppose a manufacturing plant absorbs 5000
ond one as
kVA
a spare. Alternatively,
we can
single-phase transformers each rated
plus one spare.
install
transformer and keep a sec-
at
install three
1667 kVA,
The 3-phase transformer option
is
250
ELECTRICA L MACHINES AND TRANSFORMERS
more expensive (total capacity: 2 X 5000 = 10 000
kVA) than the single-phase option (total capacity:
4 X 1667 = 6667 kVA).
Fig. 12.10 shows successive stages of construction of a
3-phase
1
10
MVA,
222.5 kV/34.5
changing transformer/ Note that
1
three
A
main
legs, the
'
in
kV
tap-
addition to the
tional lateral legs.
They enable
the designer to re-
duce the overall height of the transformer, which
simplifies the
problem of shipping.
In effect,
when-
ever large equipment has to be shipped, the designer
is
faced with the problem of overhead clear-
ances on highways and
rail lines.
magnetic core has two addi-
1 £
lap-changing transformer regulates the secondary voltage
by automatically switching from one tap to another on the pri-
mary winding. The tap-changer
is
a motorized device under the
control of a sensor that continually monitors the voltage that
hns to be held constant.
3 I
1—t w V
Figure 12.10a
Core of a 1 1 0 MVA, 222.5 kV/34.5 kV, 60 Hz, 3-phase
transformer. By staggering laminations of different
widths, the core legs can be made almost circular. This
reduces the coil diameter to a minimum, resulting in
2
less copper and lower l R losses. The legs are tightly
bound to reduce vibration. Mass of core: 53 560 kg.
Figure 12.9
Three-phase transformer for an electric arc furnace,
rated 36 MVA, 1 3.8 kV/1 60 V to 320 V, 60 Hz. The
secondary voltage is adjustable from 1 60 V to 320 V
by means of 32 taps on the primary winding (not
shown). The three large busbars in the foreground deliver a current of 65 000 A. Other characteristics: impedance: 3.14%; diameter of each leg of the core: 71
mm; overall height of core: 3500 mm; center line distance between adjacent core legs: 1220 mm.
(Courtesy of Ferranti- Packard)
Figure 12.10b
Same
transformer with coils
windings are connected
in
in
place.
wye and
The primary
the secondaries
in
Each primary has 8 taps to change the voltage
in steps of ±2.5%. The motorized tap-changer can be
seen in the right upper corner of the transformer.
Mass of copper: 1 5 230 kg.
delta.
THREE-PHASE TRANSFORMERS
The 34.5 kV windings (connected in delta) are
to the core. The 222.5 kV windings
(connected in wye) are mounted on top of the 34.5 kV
mounted next
windings.
A space of several centimeters separates the
two windings
cool
251
to
ensure good isolation and to allow
flow freely between them. The
oil to
ings that protrude
nected to a 220
bushings are
kV
much
HV
bush-
from
the oil-filled tank are con-
line.
The medium voltage (MV)
smaller and cannot be seen
in the
photograph (Fig. 12.10c).
12.8 Step-up
and step-down
autotransformer
When the
voltage of a 3-phase line has to be stepped
up or stepped down by a moderate amount,
nomically advantageous
to
it
is
eco-
use three single-phase
transformers to create a wye-connected autotrans-
The
former.
in Fig.
1
agram
is
2.
1
actual physical connections are
1
a,
given
shown
and the corresponding schematic
in Fig.
12. lib.
The respective
di-
line-
to-neutral voltages of the primary and secondary
are obviously in phase. Consequently, the incoming
Figure 12.10c
Same
has been subjected to a 1050 kV impulse test on the HV side and a
similar 250 kV test on the LV side. Other details:
power rating: 110 MVA/146.7 MVA (OA/FA); total
mass
transformer ready
including
oil:
158.7
for shipping.
t;
It
overall height: 9 m; width:
and outgoing transmission
phase.
tral,
The
neutral
is
line
connected
voltages are
to the
otherwise a tertiary winding must be added
to
prevent the line-to-neutral voltage distortion mentioned previously (Section 12.5).
8.2 m, length: 9.2 m.
{Courtesy of ABB)
A
B
O2
-o
H
Figure 12.11a
Wye-connected autotransformer.
in
system neu-
TN
Figure 12.11b
Associated schematic diagram.
load
252
ELECTRICAL MACHINES AND TRANSFORMERS
Figure 12.11c
Single-phase autotransformer (one of a group of three) connecting a 700 kV, 3-phase, 60 Hz transmission line to
an existing 300 kV system. The transformer ratio is 404 kV/173 kV, to give an output of 200/267/333 MVA per
transformer, at a temperature rise of 55°C. Cooling is OA/FA/FOA. A tertiary winding rated 35 MVA, 1 1.9 kV
maintains balanced and distortion-free line-to-neutral voltages, while providing power for the substation. Other
and windings: 132 t; tank and accessories: 46 t; oil: 87 t; total
265 t. BIL rating is 1950 kV and 1050 kV on the HV and LV side, respectively. Note the individual 700 kV
(right) and 300 kV (left) bushings protruding from the tank. The basic impulse insulation (BIL) of 1950 kV and
1050 kV expresses the transformer's ability to withstand lightning and switching surges.
(Courtesy of Hydro-Quebec)
properties of this transformer: weight of core
weight:
THREE-PHASE TRANSFORMERS
For a given power output, an autotransformer
former (see Section
1
1
.2).
This
is
the ratio of the
incoming
line voltage lies
between 0.5 and
particularly true
winding rated
1
1
.9
kV.
kV
an existing 300
kV
The
cerned)
=
3
66.3
with a tertiary
MVA.
basic transformer rating (as far as size
is
considerably less than
MVA.
This
lies
between 0.5 and
kV
to
kV
transmission
230 kV
line has to be
supply a load of 200
MVA.
totransformers are to be used. Calculate the basic
power and voltage rating of each transformer, assuming they are connected as shown in Fig.
Solution
199
stations
ing
is
and
in
-
199
H2
between H, and
230/V3
to
I,
SN3E
=
(200
X
=
335
lines
1
is
and
lx
power flow over
trans-
the phase shifting principle, con-
between phases B and
12. 12).
As we
slide contact
A
o
£
kV
AP
P
A is
= 66 kV
—
—
o
to
B
:
rheostat
line
C
pri-
66 kV.
line
is
(8.9)
6
1()
)/(V3
X 345
000)
output
A
voltage
The power associated with winding X,X 2
S
converter
special electric controls. Phase shift-
QMne a
voltage
each phase of the outgoing
=
Such multi-
is
kV
133
secondary voltage rating of 133
in
line.
in large electronic
also used to control
output
=
133
H2
This means that each transformer has an effective
The current
phase an-
shifting en-
mission lines that form part of a power grid.
line
mary
.5)
kV
The voltage of winding X,X 2 between
=
1
us consider only
between X, and
=
345/V3
line-to-neutral voltage
£, A
=
say).
line-to-neutral voltage
£AN =
let
shift the
Such phase
systems from an ordinary 3-phase
phase systems are used
of a 3-phase line (Fig.
To simplify the calculations,
The
(345/230
ables us to create 2-phase, 6-phase, and 12-phase
sider a rheostat connected
Em =
keeping with the
2.0.
3-phase system enables us to
To understand
12.11b.
The
con-
12.9 Phase-shift principle
A
Three single-phase transformers connected as au-
one phase (phase A,
is
load-carrying
system.
a 3-phase,
stepped up to 345
in
fact that the ratio of transformation
gle of a voltage very simply.
Example 12-4
The voltage of
is
its
part of a 3-phase trans-
It is
former bank used to connect a 700
line to
basic rating of the 3-phase transformer bank
X
22.1
capacity of 200
2.
Figure 12.11c shows a large single-phase auto-
transformer rated 404 kV/173
is
if
outgoing
line voltage to
The
is
smaller and cheaper than a conventional trans-
253
= 66 000 X
335
-
22.
1
Winding H,H 2 has the same power
is
MVA
rating.
sic rating
of each single-phase transformer
fore 22.1
MVA.
The
is
ba-
there-
Figure 12.12
EAP can be phase -shifted
by means of a potentiometer.
Voltage
with respect to
EAC
C
P
ELECTRICA L MA CHINES AND TRANSFORMERS
254
from phase B toward phase C, voltage
both
amplitude and phase.
in
moving from one end of
shift in
to the other.
changes
a 60° phase
the potentiometer
we move from B
Thus, as
E AP
We obtain
C, voltage
to
EAP gradually advances in phase with respect to
£AB At the same time, the magnitude of E AP varies
slightly, from E (voltage between the lines) to 0.866
.
E when
the contact
Such
circuits
draws
is
in the
middle of the rheostat.
a simple phase-shifter can only be used in
where the load between terminals
few milliamperes.
a
If a
IR drop
plied, the resulting
pletely changes the voltage
A
and P
heavier load
ap-
is
in the rheostat
com-
and phase angle from
what they were on open-circuit.
we connect a mulautotransformer between phases B and C
12.13). By moving contact P, we obtain the
To get around
titap
(Fig.
this
nals
shifts as be-
but this time they remain essentially un-
changed when a load
A and
P.
Why
is
is
connected between termi-
this
so? The reason
flux in the autotransformer
fixed.
mains
As
is
is
that the
E BC
fixed because
(both
in
magnitude
and
phase)
whether the autotransformer delivers a current
3 tapped autotransformers con-
nected between lines A, B, and C. Contacts P,, P 2
P 3 move
in
obtain a
maximum
tandem as we switch from one
set
We now
to source
of 60° as
ABC. We
we move
the autotransformers to the
some
practical applications
of the phase-shift principle.
12.10 Three-phase to 2-phase
transformation
2-phase system are equal but dis-
in a
placed from each other by 90°. There are several
ways
to create a
source.
One
2-phase system from a 3-phase
of the simplest and cheapest
single-phase autotransformer having taps
cent and 86.6 percent.
We
two phases of a 3-phase
line,
If the
A
discuss
shift
,
of
taps to the next. This arrangement enables us to cre-
line
phase
from one extremity of
other.
The voltages
shows
,
to
the load or not.
Fig. 12. 14
P h P 2 P 3 whose phase angle
ate a 3-phase source
changes stepwise with respect
is
a result, the voltage across each turn re-
fixed
shifter.
problem,
same open-circuit voltages and phase
fore,
Figure 12.14
Three-phase phase
connect
as
voltage between lines A,
is
to use a
at
50 per-
between any
it
shown
B,C
is
in Fig.
100
1
2.
1
5.
V, voltages
EAT and £ NC are both equal to 86.6 V. Furthermore,
they are displaced from each other by 90°. This relationship can be seen by referring to the phasor dia-
gram
1
.
(Fig.
1
Phasors
2.
1
5c) and reasoning as follows:
EAB £ BC
,
,
and
ECA
are fixed by the
source.
2.
U^A-^^^A/->^^^A/'•^l
O
line
B
1
1
3
4
b
b
Figure 12.13
Autotransformer used as a phase-shifter.
7
n
line
Phasor
E AN
cause the
is in
phase with phasor
same ac
£ AB
be-
flux links the turns of the au-
<
totransformer.
3.
Phasor
same
E AT
is
reason.
in
phase with phasor
£ AB
for the
THREE-PHASE TRANSFORMERS
and the other an 86.6 percent tap on
tap
the
255
primary
The transformers are connected as shown in
12.16. The 3-phase source is connected to termi-
winding.
Fig.
nals A, B,
C
secondary
and the 2-phase load
windings.
The
ratio
connected to the
is
of transformation
(3-phase line voltage to 2-phase line voltage)
by
EAB /E l2
.
The
is
given
Scott connection has the advantage of
isolating the 3-phase
and 2-phase systems and provid-
ing any desired voltage ratio between them.
load
Except for servomotor applications, 2-phase
1
systems are seldom encountered today.
Example 12-5
A 2-phase,
kW (10 hp), 240 V, 60 Hz motor has
7.5
an efficiency of 0.83 and a power factor of 0.80.
is
to
It
be fed from a 600 V, 3-phase line using a Scott-
connected transformer bank
(Fig. 12.16c).
<b)
Calculate
a.
b.
c.
The apparent power drawn by the motor
The current in each 2-phase line
The current in each 3-phase line
Solution
a.
-NC
The
power drawn by
active
the
The apparent power drawn by
Figure 12.15
a. Simple method to obtain a 2-phase system from a
3-phase line, using a single transformer winding.
b. Schematic diagram of the connections.
c. Phasor diagram of the voltages.
From KirchhofPs
ECA =
Loads
I
E AN + £ NC +
phasor £ NC must have
and direction shown
=
S
=
The
ratio
b.
The current
in
1
00/86.6
-
the
fixed and given by
EAH /EAV =
to
is
9036/0.8
VA
295/2
is
= 5648 VA
each 2-phase
line is
c.
The transformer bank
tle
itself
consumes very
lit-
active and reactive power; consequently, the
3-phase line supplies only the active and reac-
1.15.
Another way
295
=
motor
f=S/E = 5648/240
= 23.5 A
other,
of transformation (3-phase voltage to
is
1
<|>
the
in the figure.
and 2 must be isolated from each
2-phase voltage)
1
5=11
such as the two windings of a 2-phase induction motor.
P/cos
The apparent power per phase
voltage law,
Consequently,
0.
the value
is
P = PJt) = 7500/0.83
= 9036 W
(c)
4.
motor
produce a 2-phase system
Scott connection.
It
consists of
two
is lo
use
identical
single-phase transformers, the one having a 50 percent
power absorbed by the motor. The
power furnished by die 3-phase
therefore,
295 VA.
tive
total ap-
parent
line
1
1
is,
ELECTRICAL MACHINES AND TRANSFORMERS
256
86.6%
phase
AO
1
OB
Figure 12.16c
See Example
12-5.
The 3-phase
=
/
line current
5/(V3 E)
=
l
1
is
295/(V3
X
600)
= 10.9A
Figure
1
2.
line voltages
12.1
1
1
6c shows the power circuit and the
and currents.
Phase-shift transformer
A phase-shift transformer is a special type of 3-phase
autotransformer that shifts the phase angle between
incoming and outgoing
the
lines without
changing
the voltage ratio.
Consider a 3-phase transmission
to the terminals
(b)
b.
Scott connection.
Phasor diagram
(Fig.
incoming
Figure 12.16
a.
former
of the Scott connection.
1
A, B,
2.
1
7).
C
The transformer
line voltages
however, changing their
1
,
2, 3 are shifted
connected
twists
a
magnitude. The
through an angle
that all the voltages of the
line
line
of such a phase-shift transall
the
without,
result
is
outgoing transmission
with respect to the voltages of
the incoming line A, B, C.
The angle may be
lead-
THREE-PHASE TRANSFORMERS
257
Example 12-6
phase-shift
A phase-shift transformer is designed
transformer
MVA on a 230 kV,
3-phase
±
variable between zero and
a.
to control
50
1
The phase angle
line.
is
15°.
Calculate the approximate basic power rating
of the transformer.
Calculate the line currents
b.
outgoing transmission
Tap changer
incoming and
in the
lines.
Solution
a.
Figure 12.17a
The basic power
Phase-shift transformer.
ST
Note
0.025 S L a max
=
56
The
X
0.025
(12.1)
X
150
15
MVA
power
rating
is
much
that the transformer carries.
feature of
b.
is
=
=
that the
power
rating
less than the
This
a
is
autotransformers.
all
line currents are the
same
both lines, be-
in
cause the voltages are the same. The line current
is
I
Figure 12.17b
Phasor diagram showing the range over which the
phase angle of the outgoing line can be varied.
SL
=
(150
=
377
Fig. 12,18a
ing or lagging,
and
is
usually variable between zero
and ±20°
The phase angle is sometimes varied in discrete
means of a motorized tap-changer.
The basic power rating of the transformer
(which determines
power
its
size)
depends upon the ap-
carried by the transmission line, and
upon the phase
shift.
For angles
less than 20°,
it
is
given by the approximate formula
ST
=
0.025 S L
a max
leg.
PN
=
basic
power
rating of the 3-phase
000)
A
an example of a 3-phase transformer
with a tap brought out
shift of, say,
a, 3.
S,
0.025
=
The incoming
line
and the outgoing
result
Similarly,
is
£2 n
ECN
l
is
connected
to terminals
line to terminals
1,
A, B,
a£ s
C
£, N lags 20° behind £ AN
20° behind E HN and £ 1N lags
that
.
,
(Fig. 12.18c).
principle of obtaining a phase shift
connect two voltages
in
E ]b
apparent power carried by the trans-
mission line [VA]
generated by phase A. The values of E,> N and
B
is
is
to
series that are generated
ated by phase
an approximate coefficient
A and
2, 3.
by two different phases. Thus, voltage
shift |°]
has one
The wind-
transformer bank VA]
a max = maximum transformer phase
20
ings of the three phases are interconnected as shown.
[
=
A
terminal
at
a second winding having terminals
The basic
Sv
X 230
The transformer has two windings on each
20° behind
where
(8.9)
h
10 )/(V3
Thus, the leg associated with phase
winding
The
(I2.l)
is
X
could be used to obtain a phase
degrees.
steps by
parent
that
N3E
=
connected
in series
selected so that the output voltage
is
gener-
E PN
E ]b are
with
equal to the in-
put voltage while obtaining the desired phase angle
_\-SS
ELECTRICAL MACHINES AND TRANSFORMERS
coming terminals are
A, B, C; the outgoing terminals
Figure 12.18c
Phasor diagram
are 1,2,3.
a transformer that gives a phase-
of
shift of 20°.
- E
EPN
=
\.\4E
= 0.40E
In practice, the internal circuit
phase-shift transformer
However,
it
rests
upon the
The purpose of such transformers
just discussed.
will
of a tap-changing,
much more complex.
basic principles we have
is
be covered
in
Chapter 25.
12.12 Calculations involving
3-phase transformers
The behavior of a 3-phase transformer bank is calculated the same way as for a single-phase transformer. In making the calculations, we proceed as
Figure 12.18b
Schematic diagram
follows:
of the transformer in Fig. 12.18a.
1.
We
assume
that the
primary and secondary
windings are both connected
between them.
In
our particular example,
line-to-neutral voltage of the
incoming
if
E is
the
line, the re-
spective voltages across the windings of phase
A are
in
wye, even
if
they are not (see Section 8.14). This eliminates
the
problem of having
to deal with delta-wye
and delta-delta voltages and currents.
THREE-PHASE TRANSFORMERS
2.
We
consider only one transformer (single phase)
of this assumed
3.
The primary voltage of
former
is
coming
4.
wye-wye transformer
259
Z T (pu)
MP")
= tO.115
bank.
this hypothetical trans-
the line-to-neutral voltage of the in-
line.
The secondary voltage of
this
transformer
is
the
line-to-neutral voltage of the outgoing line.
5.
The nominal power
rating of this transformer
is
one-third the rating of the 3-phase transformer
bank.
6.
Figure 12.19
The load on
transformer
this
is
See Example
one-third the
12-7.
load on the transformer bank.
This
Example 12-7
The 3-phase step-up transformer shown
10.18 (Chapter 10)
is
rated 1300
impedance
kV, 60 Hz,
MVA,
a.
kV
11.5 percent.
Fig.
in
It
steps
The
power
a
S,
power
810
therefore,
j
0.115
The voltage
=
81 0/3
= 270
1
2.
9.
1
MVA
£, across the load
is
ter-
HV side of the transformer
MVA at 370 kV with a lagging
minals when the
delivers
is,
The equivalent circuit is shown in Fig.
b. The power of the load per phase is
circuit of this trans-
Calculate the voltage across the generator
almost entirely reactive.
is
impedance
ZT (pu) =
former, per phase.
b.
per-unit
up
line.
Determine the equivalent
a very large transformer; consequently, the
24.5 kV/345
the voltage of a generating station to
345
is
transformer impedance
£,
The
= 370 kV/V3 = 213.6W
per-unit
power of
the load
is
factor of 0.90.
MVA =
5 L (pu) = 270 MVA/433.3
0.6231
Solution
a.
First,
we
By
note that the primary and second-
ary winding connections are not specified.
We
don't need this information. However,
we assume
in
selecting £, as the reference phasor, the
per-unit voltage across the load
that both
£ L (pu) =
windings are connected
We
this
shall use the per-unit
We
problem.
the secondary
select the
method
EB =
The
to solve
£B
/.(pu)
L F
=
199.2
kV
The power
/u
Ratio of transformation
a
=
per-unit current in the load
=
is
345/24.5
=
5,(pu)
11
£ L (pu)
is
345/V3
=
is
0.6231
£L
=
0.581
1300/3
=
I
1.0723
is
0.9.
Consequently.
by an angle of arcos 0.90
=
25.84°.
Consequently, the amplitude and phase of the
14.08
is
given by
rating of the transformer
will be used as the base
SB
=
factor of the load
lags behind
per-unit load current
The nominal power
kV
0
1.0723Z.0
nominal voltage of
winding as our base voltage,
The base voltage
213.6 kV/1 99.2
=
wye.
is
power 5 B Thus,
/,
.
433.3
MVA
The
(pu)
-
per-unit voltage
0.581 1Z. -25.84°
£
s
(Fig. 12.19)
is
ELECTRICAL MACHINES AND TRANSFORMERS
260
E
s
(pu)
= EJpu) +
/ L (pu)
X ZT (pu)
Hi
= 1.0723^0° + (0.5811/1 -25.84°)
X (0.1 15/190°)
=
1.0723
+
0.0668Z.64.16
=
1.0723
+
0.0668(cos 64.16°
j
X1
0
+
sin 64.16°)
=
1.1014
=
1.103/13. 12°
+
j
0.0601
Therefore,
E =
s
The
1.103
X 345 kV =
per-unit voltage on the primary side
Ep =
The
381 kVZ.3.12°
1.
is
also
103/13. 12°
effective voltage across the terminals of the
generator
is,
therefore,
£g =
=
=
£p(pu)
X £B
1.103
X
27.02
kV
(
primary)
24.5
kV
Figure 12.20
Polarity
12.13 Polarity marking
marking
of
3-phase transformers.
of 3-phase transformers
The HV terminals of a 3-phase transformer are
marked H h H 2 H 3 and the LV terminals are marked
X h X 2 X 3 The following rules have been stan-
ondary
dardized:
are
,
.
,
1
.
If the
wye-wye
terminals on the
or delta-delta, the
voltages between similarly-marked terminals
are in phase. Thus,
is
in
phase with
E xx
£ HiH|
is
in
phase with
Exx
EH
is in
phase with
Ex
£H
H
H
x
If the
,
HV
side
LV
side.
Thus,
E HH
leads
£ X|Xi
by 30°
EH
H|
leads
£ XiX|
by 30°
£H
Hi
leads
Ex
30°
x by
Fig. 12.20
shows two ways of representing
the
delta-wye terminal markings.
primary and secondary windings are con-
nected
internal connections
and so on.
and so on.
2.
The
so that the voltages on the
always lead the voltages of similarly-marked
primary windings and secondary wind-
ings are connected
line voltages.
made
in
wye-delta or delta-wye, there results
a 30° phase shift
between the primary and sec-
3.
These rules are not affected by the phase
sequence of the line voltage applied to the
primary
side.
THREE-PHASE TRANSFORMERS
Questions and Problems
1
2-7
In order to
261
meet an emergency, three
single-phase transformers rated
Practical level
1
2-
1
that the transformer terminals
H2
have polarity marks H,,
X,,
,
X2
,
make
wye-delta on a 3-phase 18
a.
What
Delta-wye
b.
Open-delta
1
2-2
250 kVA, 7200 V/600
in
V,
at
load
450 kVA,
is
In the
a.
12-9
incoming and outgoing transmis-
In the
of 36
MVA,
in Fig. 12.9
13.8
nominal currents
ondary
12-4
kV/320
in the
has a rating
V Calculate the
lines.
in Fig.
mode
ates in the forced-air
225
current
during the
600
to
is
1,
V.
kV
6.9
2,
and 3
Then,
in
a
P
are
by mistake con-
Determine the voltages measured between
lines 1-2, 2-3,
Draw
the
and
3-1.
new phasor diagram.
Three
1
50 kVA, 480 V/4000
V,
60 Hz
sin-
is
if
the primary line volt-
kV and
12-11
when
The
exciting
the transformers are
The core
loss in a
300
kVA
distribution transformer
the primary line
is
3-phase
estimated to
be 0.003 pu. The copper losses are
150 A.
in
line.
operating at no-load.
0.0015 pu.
the transformer overloaded?
If
the time, and the cost of electricity
4.5 cents per
kWh,
is
calculate the cost of
the no-load operation in the course of
connected?
Calculate the line currents for a 600
load.
the transformer operates
effectively at no-load 50 percent of
Problem 12-2 are
V line to 7.2 kV.
How must they be
kVA
load the bank
current has a value of 0.02 pu. Calculate
600
c.
balanced and equal
on a 4000 V, 3-phase
10.19 oper-
used to raise the voltage of a 3-phase
b.
maximum
lines
the line current
The transformers
a.
the
Industrial application
Calculate the currents in the sec-
ondary lines
12-6
is
A-B-C
b.
morning peaks.
Is
What
gle-phase transformers are to be installed
The transformer shown
b.
400 kVA.
and the voltage between
a.
Intermediate level
is
250 kVA,
the transformers overloaded?
ings of transformer
12-10
age
at
are connected in open-
Referring to Figs. 12.3 and 12.4, the line
Calculate the nominal currents in the pri-
a.
V
nected in reverse.
mary and secondary windings of the transformer shown in Fig. 10.18, knowing that
12-5
transformer bank?
to the
similar installation the secondary wind-
primary and sec-
the windings are connected in delta-wye.
line.
the outgoing line voltage?
transformers rated
Are
kV
load that can be
voltage between phases
primary and secondary windings
The transformer
maximum
can carry on a continuous basis?
is
12-3
is
kV/600
b.
sion lines
b.
Two
calculate the
following currents:
the
delta to supply a load of
60 Hz, are con-
wye-delta on a 12 470 V, 3-phase
line. If the
a.
2-8
2.4
Three single-phase transformers rated
nected
What
b.
a.
is
connected
nections:
are connected
in
schematic drawings of the following con-
1
kV
100 kVA, 13.2 kV/2.4
Assuming
12-12
one
year.
The
bulletin of a transformer manufacturer
kVA, 230 V/208
Calculate the corresponding primary
indicates that a 150
and secondary currents.
60 Hz, 3-phase autotransformer weighs
V,
262
ELECTRICAL MACHINES AND TRANSFORMERS
310
lb,
whereas a standard 3-phase trans-
12-14
former having the same rating weighs
1220
12-13
1b.
Why
1
kVA, 480 V/l 20
5
in delta to
formers on a 600
H2 X h X2
,
V,
source.
60 Hz are con-
12-15
Then
can be drawn from the 600
calculate the
maximum
that the autotransformer can carry.
You wish
to operate a
40
hp,
460
V,
3-phase motor from a 600 V, 3-phase sup-
V
ply. The full-load current of the motor
42 A. Three 5 kVA, 20 V/480 V,
3-phase
line.
The H,,
marks appear on the
is
1
single-phase transformers are available.
the transformers should be
How would
you connect them? Are they
connected.
able to furnish the load current
b.
Determine the 3-phase voltage output
the
c.
Determine the phase
of the transformer.
shift
between the
3-phase voltage output and the 600 V,
3-phase input.
V
load
function as autotrans-
polarity
Show how
(kVA)
at
metal housing.
a.
Problem 12-13 calculate the maximum
line current that
this difference 9
Three single-phase transformers rated
nected
In
motor without overheating?
drawn by
Chapter 13
Three-Phase Induction Motors
that ranges
13.0 Introduction
Three-phase
induction
motors are the motors
most frequently encountered
and easy
are simple, rugged, low-priced,
They run
tain.
to
constant speed from
The speed
frequency-dependent
consequently,
these
is
motors
are
not
supports
main-
at essentially
zero to full-load.
and,
They
in industry.
to control the
the 3-phase induction
its
We
the basic principles of
its
behavior.
general construction and the
We
way
use two types of rotor windings: (1) conven-
3-phase windings made of insulated wire and
then
cage induction motors (also called cage motors)
the
and wound-rotor induction motors.
A squirrel-cage
is
composed of bare cop-
into the slots. The opposite ends are welded
two copper end-rings, so that all the bars are
short-circuited together. The entire construction
pushed
thousand horsepower permit the reader to see that
same
rotor
per bars, slightly longer than the rotor, which are
motors ranging from a few horsepower to several
operate on the
The type of winding
gives rise to two main classes of motors: squirrel-
Squirrel-cage, wound-rotor, and linear induction
all
circumference of the lam-
(2) squirrel-cage windings.
windings are made.
they
A number of evenly spaced slots,
internal
of rotor slots to provide space for the rotor winding.
motor and develop the funda-
mental equations describing
discuss
that
tions.
speed of commercial induction
we cover
frame
made up of
The rotor is also composed of punched laminaThese are carefully stacked to create a series
fre-
tional
chapter
steel
core
inations, provide the space for the stator winding.
motors.
In this
hollow, cylindrical
stacked laminations.
quency electronic drives are being used more and
more
a
punched out of the
easily
adapted to speed control. However, variable
mm to 4 mm, depending on the
from 0.4
power of the motor.
The stator (Fig. 13.2) consists of a
to
basic principles.
(bars and end-rings) resembles a squirrel cage,
from which the name
Principal
13.1
A 3-phase
components
induction motor (Fig.
parts: a stationary stator
rotor
is
and
1
3.
1
)
small and
in-
aluminum, molded
The
show progressive
13.3a). Figs.
to
form an
13.3b and 13.3c
stages in the manufacture of a
squirrel-cage motor.
263
In
are
die-cast
tegral block (Fig.
a revolving rotor.
derived.
made of
has two main
separated from the stator by a small air gap
is
medium-size motors, the bars and end-rings
ELECTRICAL MACHINES AND TRANSFORMERS
264
A wound
rotor has a 3-phase winding, similar
the one on the stator.
The winding
tributed in the slots and
is
is
uniformly
usually connected in 3-
wire wye. The terminals are connected to three
rings,
which
revolving
turn with the rotor (Fig.
slip-rings
and
13.4).
associated
slip-
The
stationary
brushes enable us to connect external resistors
ries
to
dis-
in se-
with the rotor winding. The external resistors are
mainly used during the start-up period; under normal
running conditions, the three brushes are short-circuited.
13.2 Principle of operation
Figure 13.1
Super-E, premium efficiency induction motor rated
10 hp, 1760 r/min, 460
V,
3-phase, 60 Hz. This
to-
tally-enclosed fan-cooled motor has a full-load current of
tor of
1
2.7 A, efficiency of 91 .7%, and
81%. Other
power
fac-
characteristics: no-load current:
5 A; lockedrotor current: 85 A; locked rotor torque:
breakdown torque: 3.3 pu; service factor
90 kg; over-all length including
shaft: 491 mm; overall height: 279 mm.
(Courtesy of Baldor Electric Company)
2.2 pu;
1
.15; total weight:
The operation of a 3-phase induction motor is based
upon the application of Faraday's Law and the
Lorentz force on a conductor (Sections 2.20, 2.2
1,
and 2.22). The behavior can readily be understood
by means of
the following example.
Consider a series of conductors of length
/,
whose extremities are short-circuited by two bars A
and B (Fig. 3.5a). A permanent magnet placed
above this conducting ladder, moves rapidly to the
1
right at a
speed
r,
so that
across the conductors.
its
magnetic
field
B sweeps
The following sequence
of
events then takes place:
Figure 13.2
Exploded view of the cage motor of Fig. 13.1, showing the stator, rotor, end-bells, cooling
and terminal box. The fan blows air over the stator frame, which is ribbed to improve heat
(Courtesy of Baldor Electric Company)
fan, ball bearings,
transfer.
THREE-PHASE INDUCTION MOTORS
moving magnet
field
is
replaced by a rotating
windings, as
in the stator
13.3
is
produced by the 3-phase currents
The
we
now
will
265
field.
The
flow
that
explain.
rotating field
Consider a simple stator having 6 salient poles, each
of which carries a coil having 5 turns (Fig.
1
3.6).
Coils that are diametrically opposite are connected
in series
by means of three jumpers
connect terminals
a-a, b-b,
and
that respectively
c-c.
This creates
AN, BN, CN,
that
are mechanically spaced at 120° to each other.
The
three identical sets of windings
Figure 13.3a
Die-cast aluminum squirrel-cage rotor with integral
cooling fan.
(Courtesy of Lab-Volt)
1
A
.
E—
voltage
while
it
Blv
is
induced
in
each conductor
being cut by the flux (Faraday's law).
is
The induced voltage immediately produces a
/, which flows down the conductor un-
2.
current
derneath the pole-face, through the end-bars,
and back through the other conductors.
Because the current-carrying conductor
3.
the magnetic field of the
lies in
permanent magnet,
it
experiences a mechanical force (Lorentz force).
4.
The force always
acts in a direction to drag the
conductor along with the magnetic
field
(Section 2.23).
If the
celerate
move,
conducting ladder
is
toward the
However, as
right.
free to
it
it
will ac-
picks up
speed, the conductors will be cut less rapidly by the
moving magnet, with
age
E and
the current
the force acting
If
the ladder
magnetic
the result that the induced volt/
will diminish. Consequently,
on the conductors
were
field, the
to
move
at the
will also decrease.
same speed
and the force dragging the ladder along would
come
as the
induced voltage E, the current
all
/,
be-
zero.
In an
itself to
Figure 13.3b
(1),
induction motor the ladder
is
form
I
a squirrel-cage (Fig.
closed upon
3.5b) and the
in the manufacture of stator and
Sheet steel is sheared to size
punched (3), blanked (4), and
Progressive steps
rotor laminations.
blanked
punched
(2),
(5).
(Courtesy of Lab-Volt)
ELECTRICAL MACHINES AND TRANSFORMERS
266
upper
Figure 13.3c
Progressive steps
a.
in
Molten aluminum
the injection molding of a squirrel-cage
is
poured
into
a cylindrical
cavity.
rotor.
The laminated
rotor stacking
is
firmly held
between
two molds.
b.
Compressed
rams the mold assembly into the cavity. Molten aluminum is forced upward through the
and into the upper mold.
Compressed air withdraws the mold assembly, now completely filled with hot (but hardened) aluminum.
The upper and lower molds are pulled away, revealing the die-cast rotor. The cross section view shows
that the upper and lower end-rings are joined by the rotor bars. (Lab-Volt)
air
rotor bar holes
c.
d.
two
coils in
each winding produce magnetomotive
forces that act in the
The
same
three sets of windings are connected in
thus forming a
perfectly
common
symmetrical
always flow
neutral N.
Owing
arrangement,
the
wye,
to the
line-to-
bers,
suppose that the peak current per phase
Thus,
when
/a
= +7
pere-turns
balanced 3-phase system.
Because the current
ings.
The
displaced
/.„
/h ,
and
/c will
currents will have the
in
flow
In
sume
B, C,
flux.
It is
this flux
we
same value but
in turn,
are interested
will
be
create a
we
corresponding
a
as-
by the arrows)
A will
is
1
0 turns
= 70 amof
value
positive, the flux
is
flux.
directed
upward, according to the right-hand
As time goes
by,
we
rule.
can determine the instanta-
neous value and direction of the current
each
in
winding and thereby establish the successive flux
patterns. Thus, referring to Fig.
rent /a has a value of
in.
order to follow the sequence of events,
that positive currents (indicated
vertically
the wind-
time by an angle of 120°. These currents
produce magnetomotive forces which,
magnetic
in
and
10 A.
is
A, the two coils of phase
mmf of 7 A X
together produce an
we connect a 3-phase source to terminals A,
to
Furthermore, to enable us to work with num-
neutral impedances are identical. In other words, as
If
line to neutral.
line.
regards terminals A, B, C, the windings constitute a
alternating currents
windings from
in the
Conversely, negative currents flow from neutral
direction.
A X
1
0 turns
1
3.7 at instant
10 A, whereas
fb
and
1
/c
,
cur-
both
—5 A. The mmf of phase A is
= 00 ampere-turns, while the mmf
have a value of
10
+
1
Figure 13.4b
of the slip-ring end of the rotor.
(Courtesy of Brook Crompton Parkinson Ltd)
Close-up
267
ELECTRICAL MACHINES AND TRANSFORMERS
268
length
/
Figure 13.5a
Moving magnet cutting across a conducting ladder.
ength
Figure 13.5b
Ladder bent upon
of phases
B and C
rection of the
B
/
form a squirrel-cage.
itself to
are each 50 ampere-turns.
mmf depends
upon
current flows and, using the right-hand rule,
that the direction of the resulting
shown
in Fig. I3.8a.
Note
The
di-
the instantaneous
magnetic
we
find
is
as
that as far as the rotor
is
field
concerned, the six salient poles together produce a
magnetic
having essentially one broad north
field
pole and one broad south pole. This
means
that the
6-pole stator actually produces a 2-pole field.
combined magnetic
At instant
tains a
2,
peak of
field points
one-sixth cycle
—
10 A, while
/a
The
upward.
later,
and
current
/b
/ L at,
both have a
value of
+5 A
new
has the same shape as before, except that
it
field
has
(Fig.
1
3.8b).
moved clockwise by an
We
discover that the
angle of 60°. In other
words, the flux makes 1/6 of a turn between instants
l
and
cycle,
we find that the magnetic
plete turn during
field
one cycle (see Figs.
makes one com1
3.8a to I3.8f).
The rotational speed of the field depends, therefore, upon the duration of one cycle, which in turn
depends on the frequency of the source.
2.
Proceeding
Figure 13.6
Elementary stator having terminals A, B, C connected
to a 3-phase source (not shown). Currents flowing
from line to neutral are considered to be positive.
in this
instants 3, 4, 5, 6,
and
way
7,
for each of the successive
quency
is
60 Hz, the resulting
separated by intervals of 1/6
in 1/60
s,
that
is,
field
If the fre-
makes one
turn
3600 revolutions per minute. On
Figure 13.7
Instantaneous values of currents and position of the flux
269
in Fig.
13.6.
ELECTRICAL MACHINES AND TRANSFORMERS
270
A
A
g
O
I'-
N
N
Figure 13.8d
Figure 13.8c
Flux pattern at instant
Flux pattern at instant 3.
A
4.
A
g
N
Figure 13.8e
Figure 13.8f
Flux pattern at instant 5.
Flux pattern at instant
the other hand, if the frequency were 5 Hz, the field
would make one turn in 1/5 s, giving a speed of only
300 r/min. Because the speed of the rotating field is
change any two of the
the
new phase sequence
necessarily synchronized with the frequency of the
the
same
we
find that the field
source,
it
is
called synchronous speed.
produces a
speed
field that rotates clockwise. If
line
positive crests of the currents in Fig.
in the
we
connected to the
will be
interstator,
A-C-B. By following
of reasoning developed in Section
in the opposite,
therefore, reverse
The
lines
now
revolves
at
1
3.7 follow
order A-B-C. This phase sequence
its
3.3,
or counterclockwise direction.
will,
direction of rotation.
Although early machines were
poles, the stators of
1
synchronous
Interchanging any two lines of a 3-phase motor
13.4 Direction of rotation
each other
6.
built with salient
modern motors have
internal di-
THREE-PHASE INDUCTION MOTORS
ameters that are smooth. Thus, the salient-pole stator
of Fig. 13.6
as
shown
is
1
3.6, the
are replaced
Note
that
two
each
coils of phase
A (Aa and An)
slots
coil
in Fig.
coil pitch is
covers
more
duces more flux per turn,
terminal
shown
A to
the neutral
coils of phases
1
80° of the circum-
A
current
only 60°.
because
efficient
/a
pro-
it
flowing from
N yields the flux distribution
are identical to those
Fig.
120° to each other.
The
netic field
at
due
1
3.9b, they are
resulting
mag-
to all three phases again consists of
In practice, instead
shown
in Fig.
two, three or
more
of using a single coil per pole
13.9a, the
coils
coil
lodged
in
is
subdivided into
adjacent slots.
The
staggered coils are connected in series and constitute
what
known as a phase group. Spreading the coil
way over two or more slots tends to create a
is
in this
sinusoidal flux distribution per pole,
the
in 5 suc-
in Fig. 13.20.
of poles
Soon after the invention of the induction motor, it
was found that the speed of the revolving flux could
be reduced by increasing the number of poles.
To construct a 4-pole stator, the coils are distributed as shown in Fig. I3.l0a. The four identical
A now span only 90° of the stator cir-
cumference. The groups are connected
in
such a
way
that adjacent
tomotive forces acting
when
other words,
winding of phase
two poles.
as
be placed
in series to
synchronous speed
groups of phase
B and C
of phase A and, as can be seen in
displaced
Number
13.5
13.9a.
in the figure.
The
shown
is
on the inner surface of the
ference whereas the coils in Fig. 13.6 cover
The 180°
gered coils connected
cessive slots
and 13.24a.
two
by the two coils shown
are lodged in
stator.
replaced by a smooth stator such
in Figs. 13.2
In Fig.
They
now
271
which improves
performance of the motor and makes
it
less noisy.
A phase group (or simply group) composed of 5
stag-
A
in
in series
and
groups produce magneopposite directions. In
a current
flows
/,,
(Fig. 13. 10a),
it
in the stator
creates four al-
N-S poles.
The windings of the other two phases
ternate
cal but are displaced
are identi-
from each other (and from
When
phase A) by a mechanical angle of 60°.
wye-connected windings are connected
source, a revolving field having four poles
(Fig.
13. 10b).
This field rotates
speed of the 2-pole
will shortly explain
field
why
shown
at
the
to a 3-phase
is
created
only half the
in Fig.
13.9b.
We
this is so.
group
phase
Figure 13.9a
Phase group 1 is composed of a single coil lodged
two slots. Phase group 2 is identical to Phase
group 1 The two coils are connected in series. In
practice, a phase group usually consists of two or
more staggered coils.
in
.
(/ c
1
C
= - 5 A)
Figure 13.9b
Two-pole,
magnetic
A and
/
b
full-pitch,
=
/c
lap-wound stator and resulting
when the
= -5 A.
field
current
in
phase A = +10
ELECTRICAL MACHINES AND TRANSFORMERS
272
phase group
group
1
group
1
phase B
1
Figure 13.10a
The
phase groups
four
pole magnetic
group
of
phase A produce a
4-
field.
1
group
1
-5
A.
rent flow in the three phases, let us restrict our at-
tention to phase A. In Fig. 13.11 each phase group
covers
a
mechanical angle of 360/8
=
45°.
Suppose the current in phase A is at its maximum
positive value. The magnetic flux is then centered
on phase A, and the N-S poles are located as
shown
in Fig.
13.12a. One-half cycle later, the
current in phase
tive value.
Figure 13.10b
fore,
Four-pole, full-pitch, lap-wound stator
magnetic
field
when
/a
= +10 A and
and
/b
=
The
A
will reach
its
maximum
except that
all
N
the
poles will
resulting
lc
= -5
A.
S poles and vice versa (Fig. 13.12b). In comparing the
two
figures,
it is
netic field has shifted
clear that the entire
by an angle of 45°
this gives us the clue to finding the
We
we
can increase the number of poles as
much
as
please provided there are enough slots. Thus,
Fig.
13. II
shows a 3-phase, 8-pole
stator.
phase consists of 8 groups, and the groups of
phases together produce an 8-pole rotating
When
connected to a 60
like the
Hz
spokes of a wheel,
Each
all
the
field.
source, the poles turn,
at
a synchronous speed
of 900r/min.
How
will
can
we
tell
what the synchronous speed
be? Without going into
all
nega-
same as behave become
flux pattern will be the
the details of cur-
tion.
cles
mag-
— and
speed of rota-
The flux moves 45° and so it takes 8 half-cy(= 4 cycles) to make a complete turn. On a
60 Hz system the time
fore 4
X
1/60
=
1/15
to
s.
make one
turn
is
there-
Consequently, the flux
r/s or 900 r/min.
The speed of a rotating field depends therefore
upon the frequency of the source and the number of
poles on the stator. Using the same reasoning as
turns at the rate of 15
above,
we
can prove that the synchronous speed
always given by the expression
is
THREE-PHASE INDUCTION MOTORS
phase group
phase group
1
maximum
1
Figure 13.12b
Figure 13.12a
Flux pattern
273
when
the current
in
phase A
is
at
when
Flux pattern
its
maximum
positive value.
in Fig.
the current
negative value.
13.12a but
it
The
phase A
in
pattern
is
is
the
at
its
same as
has advanced by one pole
pitch.
ing field created by the stator cuts across the rotor
bars and induces a voltage in
P
This
where
is
cut, in rapid succession,
= synchronous
ns
/=
speed fr/min]
the
creases with frequency and decreases with the
N
is
pole followed by a
number of
N
and S poles
that
sweep across
always equal
is at rest,
it
a
is
frequency of the source.
to the
in-
Because the rotor bars are short-circuited by the
num-
end-rings, the induced voltage causes a large cur-
ber of poles.
rent to flow
— usually
bar in machines of
Example 13-1
duction motor having 20 poles
when
it is
connected
several hundred amperes per
medium power.
The current-carrying conductors
Calculate the synchronous speed of a 3-phase in-
Hz
by a
conductor per second; when the rotor
This equation shows that the synchronous speed
a 50
of them.
S pole. The frequency of the voltage depends upon
frequency of the source [Hz]
p = number of poles
to
all
an ac voltage because each conductor
the flux created
by the
stator,
are in the path of
consequently, they
all
experience a strong mechanical force. These forces
source.
tend to drag the rotor along with the revolving
Solution
In
ns
= 120///7 = 120 X
= 300 r/min
50/20
1
.
field.
summary:
A
revolving magnetic field
3-phase voltage
is
is
set
up when
a
applied to the stator of an
induction motor.
13.6 Starting characteristics
of a squirrel-cage
2.
motor
Let us connect the stator of an induction
motor
The revolving
field
induces a voltage
in the ro-
tor bars.
to a
3-phase source, with the rotor locked. The revolv-
3.
The induced voltage
rents
which flow
creates large circulating cur-
in the rotor bars
and
end-rinizs.
ELECTRICAL MACHINES AND TRANSFORMERS
274
The current-carrying
4.
magnetic
the
5.
rotor bars are
by the
field created
immersed
in
they are
stator;
equal to the load torque.
therefore subjected to a strong mechanical force.
constant
The sum of the mechanical forces on
motor only turns
which tends
tor bars produces a torque
the rotor along in the
volving
same
the ro-
all
to drag
direction as the re-
at
—slip
load.
The moment this state of equilibrium
as the rotor
released,
is
it
rapidly acceler-
ates in the direction of the rotating field.
up speed,
As
picks
it
the relative velocity of the field with re-
spect to the rotor diminishes progressively. This
causes both the value and the frequency of the
duced voltage
to
decrease because the rotor bars are
more slowly. The
cut
first,
in-
will continue to increase, but
never catch up with the revolving
it
rotor bars
field. In effect, if
same speed as the field (synflux would no longer cut the
and the induced voltage and current
fall to
to
produce a current
overcome
tor bars sufficiently large to
usually less than
machines
That
(
motors
field (called slip),
0.1% of synchronous
in
speed
is
small:
motor
will begin to slow
is initially
down and
at
greater motor torque.
running
at no-load. If
the revolving field will
rate.
The
in-
at synchronous speed, they
sometimes called asynchronous machines.
never actually turn
and
13.9 Slip
The
state
are
speed
slip
slip s
of an induction motor
is
the difference be-
tween the synchronous speed and the rotor speed,
expressed as a percent (or per-unit) of synchronous
speed.
The
per-unit slip
=
=
—
s
ns
n
The
given by the equation
is
motor comes
slip
synchronous speed [r/minl
rotor speed fr/minj
zero
slip is practically
equal to
l
100%) when
(or
no-load and
at
the rotor
is
is
locked.
.
motor
hp, 6-pole induction
phase, 60
Hz
source.
excited by a 3-
is
the full-load speed
If
is
1
140
r/min, calculate the slip.
Solution
The synchronous speed of
ns
producing a greater and
The question
is,
for
how
long
can this go on? Will the speed continue to drop un-
the
seldom exceeds 5%.
the
motor
is
resulting current in the bars
will increase progressively,
No;
it
constant speed machines. However, because they
A 0.5
a higher and higher
duced voltage and the
the
and more) rarely
speed.
apply a mechanical load to the shaft, the motor
cut the rotor bars
til
).
induction motors are considered to be
Example 13-2
the
kW
000
(1
kW and less),
10
why
is
the ro-
in
13.8 Motor under load
we
1
motors run very
loads, induction
exceeds 0.5% of synchronous speed, and for small
the braking
At no-load the percent difference
between the rotor and
Suppose
1
zero.
chronous speed so as
torque.
upset, the
is
3.
close to synchronous speed. Thus, at full-load, the
Under these conditions the force
acting on the rotor bars would also become zero and
the friction and windage would immediately cause
the rotor to slow down.
The rotor speed is always slightly less than synwould
is
will
the rotor did turn at the
chronous speed), the
torque
its
rotor current, very large at
decreases rapidly as the motor picks up speed.
The speed
change (Section
will start to
slip for large
As soon
when
constant speed
exactly equal to the torque exerted by the mechanical
Under normal
13.7 Acceleration of the rotor
a
at
very important to understand that a
rate. It is
motor speed
field.
When this state is reached, the
speed will cease to drop and the motor will turn
The
=
=
\20flp
=
120
X
60/6
(13.1)
1200 r/min
difference between the synchronous speed of
to a halt?
motor and the mechanical load
the revolving flux and rotor speed
is
the slip speed:
will reach a
of equilibrium when the motor torque
is
exactly
ns
-
n
=
1200
-
1140
=
60 r/min
THREE-PHASE INDUCTION MOTORS
The
slip is
c.
s
=
(« s
=
0.05 or
-
500 r/min
at
60/1200
(13.2)
d.
5%
Motor turning
2000 r/min
at
From Example
The voltage and frequency induced in the rotor both
slip. They are given by the follow-
depend upon the
a.
At
K-
=
s
(13.3)
sf
(approx.)
=
motor speed n
0.
slip is
=
n)/n s
-
(1200
0)/1200
the induced current)
h
in
b.
When
=
=
4'
X 60 = 60 Hz
1
motor turns
the
same
in the
motor speed n
field, the
frequency of the source connected to
s
the stator [Hz]
=
E2 =
Eoc =
slip
at rest
would be induced
bars were disconnected from
the
£oc
is
c.
When
= tf=
the
=
direction as the
positive.
-
(1200
the end-rings. In
n
in the
motor speed
= —500. The
-
(-500)1/1200
should be noted that Eq. 13.3 always holds
(1200
+
500)/ 1200
but Eq. 13.4
=
1.417
valid only
if
the revolving flux
(expressed in webers) remains absolutely constant.
A slip
However, between zero and full-load the actual
value of
E2
is
Hz
negative; thus,
is
[1200
—
35
slip is
=
is
1
opposite direction
=
/V 3 times the voltage between the open-
circuit slip-rings.
It
500)/ 200
X 60 =
0.583
(n s
1
slip is
is
—
is
The
0.583
motor turns
to the field, the
s
voltage
true,
=
n)/n s
in the rotor bars
wound-rotor motor the open-circuit
the case of a
700/1200
fi
the voltage that
if
=
[V]
cage motor, the open-circuit voltage
In a
(« s
the rotor current)
open-circuit voltage induced in the ro-
when
-
=
is
The frequency of the induced voltage (and of
voltage induced in the rotor at slip s
tor
1
is
the rotor [Hz]
s
=
The frequency of the induced voltage (and of
(13.4)
frequency of the voltage and current
synchronous speed of the
13-2, the
standstill the
where
/=
direc-
1200 r/min.
is
Consequently, the
ing equations:
f2 =
same
Solution
motor
E 2 = sEoc
in the
tion as the revolving field
13.10 Voltage and frequency
induced in the rotor
f2 =
opposite
in the
direction to the revolving field
=
n)/n,
Motor turning
215
/?)//? s
greater than
=
1700/1200
implies that the motor
1
is
operating as a brake.
only slightly less than the value given
The frequency of the induced voltage and
by the equation.
current
f2 =
Example 13-3
rotor
is
f=
s
X 60 =
1.417
85 Hz
The 6-pole wound-rotor induction motor of Example
13-2
the
is
excited by a 3-phase 60
Hz
d.
source. Calculate
frequency of the rotor current under the follow-
ing conditions:
The motor speed
turns in the same
n = +2000. The
s
a.
At
b.
Motor turning
500 r/min
tion as the revolving field
in the
same
direc-
positive because the rotor
direction as the field:
slip is
=
(n s
=
(1200
standstill
at
is
-
n)ln s
-
2000)/ 1200
= -800/1200 = -0.667
ELECTRICAL MACHINES AND TRANSFORMERS
276
A negative
slip implies that the
motor
actually
is
1.
operating as a generator.
The frequency of
current
the induced voltage
means
negative frequency
reversed. Thus,
is
in the rotor
wind-
stator
quency
is
mutual flux
is
can say that the frequency
is
a transformer (Fig.
1
lists
I
kW
simply 40 Hz.
no-load
therefore low;
is
machines
efficiency
that the current
rents are
is
the full-load current
is
compared
to
it.
is
to
it.
and
all
Finally, the base
it
to 0.05 for large
the
is
is
zero.
under load,
mmf which
tends
m This sets up an opposthe stator. The opposing mmfs of
change the mutual flux
4>
.
in a transformer.
are created, in
The total
m
power needed to produce these three fluxes is
slightly greater than when the motor is operating at
no-load. However, the active power (kW) absorbed
addition to the mutual flux
<f>
(Fig. 13.14).
reactive
the syn-
is
to create
within ac-
machines. The
motor
mmfs of the secondary and primary
As a result, leakage fluxes <J> n and
other torques are
speed
it
the rotor and stator are very similar to the opposing
other cur-
all
3).
needed
zero because the output power
ing current flow in
and
Similarly, the base torque
the full-load torque and
compared
to
The base
torque are expressed in per-unit values.
current
1
links both the
similar to the
is
Motor under load. When
2.
power range between
in the
and 20 000 kW. Note
m
chanical tolerances will permit.
the typical properties of squirrel-
cage induction motors
3.
is
the current in the rotor produces a
Table I3A
<E>
it is
made as short as meThe power factor at
ranges from 0.2 (or 20%)
ceptable limits, the air gap
we
Characteristics of squirrelcage induction motors
1
in
flux
the revolving field and, in order to keep
for small
13.1
The
consequently
rotor;
Considerable reactive power
concerned, a negative frequency gives the same
reading as a positive frequency. Consequently,
and the
that sup-
friction losses in the rotor plus
the iron losses in the stator.
the phase sequence
itive,
windage and
plies the
A-B-C when the frequency is posis A-C-B when the frenegative. As far as a frequency meter is
rotor voltages
component
ing flux 3> m and a small active
the phase sequence of the
if
(of
of a magnetizing component that creates the revolv-
phase se-
that the
the motor runs at no
between 0.5 and 0.3 pu
full-load current).
is
quence of the voltages induced
ings
When
at no-load.
The no-load current is similar to the
exciting current in a transformer. Thus, it is composed
and rotor
f2 = sf= -0.667 X 60 = -40 Hz
A
Motor
load, the stator current lies
chronous speed of the motor. The following explanations will clarify the meaning of the values given
by the motor increases
in
in the table.
the mechanical load.
follows that the power factor
TABLE 13A
size
—
No-load
*
Small
Current
Torque
Slip
(per-unit)
(per-unit)
(per-unit)
means under
factor
Big*
Small
Big
Small
Big
Small
Big
Small
Big
l
1
1
1
0.03
0.004
0.7
0.96
0.8
0.87
to
to
to
to
0.9
0.98
0.85
0.9
-0
0
0
0.2
0.05
1
0
0
0.4
0.1
0.5
rotor
Power
Efficiency
Small*
Full-load
Locked
almost direct proportion
TYPICAL CHARACTERISTICS OF SQUIRREL-CAGE INDUCTION MOTORS
Loading
Motor
It
0
0.3
0
5
4
to
to
to
to
6
6
3
1
11
kW
(15 hp); big
1
.5
0.5
means over
1
1
120
kW
(1500 hp) and up
to
25 000
hp.
to
THREE-PHASE INDUCTION MOTORS
Figure 13.13
Figure 13.14
At no-load the flux
tual flux
tive
(t>
power
the motor
in
m To create
.
is
this flux,
is
mainly the mu-
At full-load the mutual flux decreases, but stator
considerable reac-
needed.
of the motor improves dramatically as the mechanical
load increases. At full-load
it
98%
3.
and rotor leakage fluxes are created. The reactive
power needed is slightly greater than in Fig. 13.13.
where
ranges from 0.80 for
small machines to 0.90 for large machines.
ciency
211
at full-load is particularly
high;
it
The
/ =
Ph =
E=
600 =
effi-
can attain
for very large machines.
full-load current [A]
output power [horsepower)
rated line voltage (V)
empirical constant
Locked-rotor characteristics. The locked-rotor
current
the I
The
2
is
R
5 to 6 times the full-load current,
losses 25 to
rotor
36 times higher than normal.
must therefore never remain locked for
more than a few seconds.
Although the mechanical power
is
needed
Recalling that the starting current
that
the
no-load current
we can
0.3 pu,
lies
is
5 to 6 pu
and
between 0.5 and
readily estimate the value of these
currents for any induction motor.
at standstill is
motor develops a strong torque. The power
zero, the
factor
making
low because considerable reactive power
to
produce the leakage flux
stator windings.
and
Example 13-4
a.
3-phase induction motor having a rating of
because the stator and
500 hp, 2300
windings are not as tightly coupled (see
Section 10.2).
Calculate the approximate full-load current,
locked-rotor current, and no-load current of a
These leakage fluxes are much
larger than in a transformer
the rotor
in the rotor
is
V.
b.
Estimate the apparent power drawn under
c.
State the nominal rating of this motor,
locked-rotor conditions.
expressed
13.12 Estimating the currents
in an induction motor
Solution
a.
The
full-load current of a 3-phase induction
may be
motor
means of
the following ap-
= 600 P h /E
(13.5)
calculated by
proximate equation:
I
in kilowatts.
The
full-load current
/
is
= 600 P h /E
- 600 X 500/2300
= 130 A (approx.)
(13.5)
ELECTRICAL MACHINES AND TRANSFORMERS
278
The no-load
current
=
/0
=
0.3/
X
0.3
1
energy
30
39
A
starting current
=
/ LR
(approx.)
6/
-
X
6
The remaining
the
watts,
it
and not
of
this
fol-
power
stator.
f
js
is
owing
active
power P v
gap and transferred
Due
dissi-
windings. Another portion
dissipated as heat in the stator core,
P
Due
00
kVA
power of a motor
always
(8.9)
is
relates to the
motor expressed
in
expressed
car-
is
to the rotor
jr
to the
PR
losses in the rotor, a third portion
dissipated as heat, and the remainder
in kilo-
The nominal
SI units
is
is,
rating
therefore,
windage and bearing-friction
tain
PL
,
the mechanical
is
finally
power P m By
.
f v representing
losses, we finally ob-
subtracting a small fourth portion
mechanical output
to the electrical input.
P
available in the form of mechanical
(approx.)
,
power available
at the shaft
to drive the load.
The power flow diagram of Fig. 13.15 enables
P = 500/1.34
= 373 kW (see Power
Appendix AX0)
to identify
conversion chart in
ties
Voltages, currents, and phasor diagrams enable us
understand the detailed behavior of an induction
Figure 13.15
Active
power flow
in
a 3-phase induction motor.
and
Efficiency.
motor
is
us
to calculate three important proper-
of the induction motor: (1)
power, and (3)
L
13.13 Active power flow
to
is
{
to the
by electromagnetic induction.
EI
= V3 X 2300 X 780
When
electrical
flows through the ma-
in the
iron losses.
1
it
pated as heat
ried across the air
3
how
copper losses, a portion
130
is
=
easier to see
to the stator
tions
= V3
is
flows from the line into the 3-phase
The apparent power under locked-rotor condi-
S
c.
it
converted into mechanical energy by
chine. Thus, referring to Fig. 13.15, active
Pc
is
= 780 A (approx.)
b.
is
lowing the active power as
=
The
motor. However,
is
its
its
efficiency, (2)
its
torque.
By
definition, the efficiency of a
the ratio of the output
power
to the input
power:
efficiency Cn)
- PJP e
(13.6)
THREE-PHASE INDUCTION MOTORS
R losses in the rotor. It can
rotor rR losses P-. are related
2.
/
power P T by
Motor
be shown * that the
4.
to the rotor input
motor
„
the equation
The torque Tm developed by
torque.
any speed
at
9.55
279
is
the
given by
Pm
(3.5)
Pjv = sP
n
(13.7)
T
9.55
where
rotor
]V
—
s
PR
losses
[W]
P
r
9.55
transmitted to the rotor
T
Equation 13.7 shows that as the
PR
A
nous speed
=
(s
slip increases, the
0.5) dissipates in the
locked
is
ted to the rotor
is
(s
=
1), all
/n s
it
the
(13.9)
s
P
power transmitted
r
"s
9.55
When
=
=
power transmit-
The
Mechanical power. The mechanical power P m
is equal to the power transmitted to the rotor minus its PR losses. Thus,
developed by the motor
any
[W]
to the rotor
synchronous speed [r/minj
multiplier to take care of units [exact
value: 60/2tt|
dissipated as heat.
3.
at
speed [N-ml
form of heat
receives.
PJn
torque developed by the motor
rotor turning at half synchro-
50 percent of the active power
r
where
consume a larger and larger propower P r transmitted across the air
to the rotor.
the rotor
[
W]
losses
portion of the
P
therefore.
slip
P = power
gap
-v)
s)
9.55
P =
rotor
(1
« s (l
actual torque 7,
slightly less than
overcome
the
available
Tm due
,
windage and
at
the shaft
friction losses.
most calculations we can neglect
in
is
to the torque required to
However,
this
small
difference.
Pm = P
1
1
= P
r
Equation 13.9 shows that the torque
J'*
sP r
r
(13.7)
proportional to the active
tor.
whence
The
actual mechanical
the load
is
needed
to
losses.
In
(13.8)
power available
P m due
to drive
power
overcome the windage and friction
most calculations we can neglect this
slightly less than
,
must absorb a large amount of active power. The
form of
latter is dissipated in the
to the
small loss.
mechanical
electromagnetic
power
heat, consequently,
the temperature of the rotor rises very rapidly.
Example 13-5
A
3-phase induction motor having a synchronous
speed of 1200 r/min draws 80
power output
directly
to the ro-
Thus, to develop a high locked-rotor torque, the
rotor
s)P T
is
power transmitted
kW
from a 3-phase
electrical
X
speed of flux
losses
transferred
electromagnetic torque
p,
of rotor
to rotor
Pm - P
1
9.55
in rotor
i
9.55
but from Eq. 3.5
_
(iii)
p,
(i)
1
rotor speed
Pm ~
Tm must
but the mechanical torque
X mechanical
torque
the electromagnetic torque
9.55
Timc
equal
.
Thus
Hence,
rm = Tmia
_ nT^
Also from Eq. 3.5 we can write
tn)
Substituting
Py
=
(ii), (iii),
SP,
(iv)
and
(iv) in
(i).
we
find
ELECTRICAL MACHINES AND TRANSFORMERS
280
feeder.
tor
The copper losses and iron losses in the stato 5 kW. If the motor runs at
52 r/min,
amount
1
e.
The efficiency
is
= PJPC =
1
T)
calculate the following:
70/80
0.875 or 87.5%
a.
b.
c.
d.
The active power transmitted to the rotor
The rotor ER losses
The mechanical power developed
The mechanical power delivered to the load,
knowing that the windage and friction losses
are equal to 2
e.
kW
The efficiency of
Example 13-6
A
3-phase, 8-pole squirrel-cage induction motor,
connected
line,
possesses a synchronous
and iron losses
the copper
motor
the
60 Hz
to a
speed of 900 r/min. The motor absorbs 40 kW, and
5
kW
and
1
kW,
in the stator
amount
to
respectively. Calculate the torque
developed by the motor.
Solution
a.
Active power to the rotor
Solution
is
The power transmitted across the
=
b.
The
-
80
=
5
75
kW
Pr
slip is
=
s
=
(n s
-
Tm =
n)/n s
-
(1200
Rotor
I
R
Note
torque)
]v
c.
v
X
75
=
3
The mechanical power developed
P^n
d.
0.04
=
P\
=
75
~
-
2
I
3
kW
windage
rotor
P
9.55
X
=
361
N-m
T
P>„
-
P,
P mt due
=
is
72
13-7.
this
problem
(the
at
a standstill or running at
power P transmitted
v
full
to the
equal to 34 kW, the motor develops a torque
Example 13-7
to the load
to the friction
2
to
independent of the speed of rotation. The
and
A 3-phase induction motor having a nominal rating of
100 hp (-75
= 70 kW
is
kW)
and a synchronous speed of 1800
connected to a 600
V
source (Fig. 13.16a).
The two-wattmeter method shows a
HP(^75kW)
1783 r/min
Figure 13.16a
(13.9)
34 000/900
solution
the
that
is
100
See Example
kW
fn s
losses.
=
is
of 361 N-m.
r/min
Pi
1
9.55
motor could be
rotor
P losses in
- 72 kW
x
slightly less than
5
speed, but as long as the
is
The mechanical power P delivered
is
to the rotor
0.04
losses are
P = sP =
~ Pi
- = 34
gap
Pjs
=
52)/ 1200
11
= 48/1200 2
= Pc ~
= 40 -
air
total
power con-
THREE-PHASE INDUCTION MOTORS
sumption of 70 kW, and an ammeter indicates a
measurements give
current of 78 A, Precise
line
b.
The
slip
is
a rotor
s
known about
stator iron losses
windage and
P =
between two
resistance
kW
2
r
= (1800 = 0.0205
the motor:
friction losses
Px =
.2
1
kW
stator terminals
Rotor
=
0.34 (2
2
J
R
b.
Rotor I~R losses
c.
Mechanical power supplied
Mechanical power developed
=
to the load, in
Py
~
P)v
=
e.
Torque developed
Solution
Power supplied
to the stator
P c = 70
is
d.
kW
R
0.34/2
wye con-
e.
Torque
63.5
kW
=
62.3
kW =
=
83.5 hp
-
/\
63.5
62.3
-
X
1.2
1.34 (hp)
is
62.3/70
-
0.89 or
89%
1763 r/min:
at
=
T0.17 ft
9.55
R =
=
3.1
kW
P =
2
]s
Iron losses
{
Power supplied
=
Pc
=
(70
-
X
3
2
(78)
X
0.17
y
in s
=
9.55
X 64 900/1800
The torque developed by
to the rotor:
^js
3.1
-
summarized
a
Pi'
-
in
motor depends upon
speed, but the relationship between the
2)
Figure 13.16b
Example
calculations are
Fig. 13.16b.
13.14 Torque versus speed curve
kW
13-7.
its
two cannot
be expressed by a simple equation. Consequently,
=
64.9
3.1
in
P
= 344 N m
The above
2
3 I
Power flow
-
= PJP e -
=
1.33
losses are
P =
Pr
is
is
R =
2
kW
to the load:
Efficiency of the motor
T]
Stator resistance per phase (assume a
PL
63.5
=
1763 r/min
at
-
"4.9
Mechanical power
P,
Efficiency
Stator f
1.33
to the rotor
d.
nection)
=
64.9
r
horsepower
a.
763)/ 1800
P = sP = 0.0205 X
c.
Power supplied
1
losses:
]r
Calculate
a.
n)fn s
(n s
speed of 1763 r/min. In addition, the following characteristics are
281
kW
kW
we
2kW
prefer to
show
the relationship in the
form of a
ELECTRICAL MACHINES AND TRANSFORMERS
282
curve. Fig. 13.17
shows
inal
full-load torque
T and
1.5
torque)
the
is
is
following example illustrates the changes that occur.
starting torque
is
the
it
breakdown torque.
motor runs at a speed
is
rings.
is
Figure
accelerat-
ing from rest to the
10
n. If
motor torque
torque.
As soon
motor
will
as the
two torques
breakdown
if
torque), the
motor
Small motors (15 hp and
breakdown torque
at
20
breakdown torque
(
at
develop
of about
/z d
1
98%
that
it
from no-load
to full-load,
N-m (-73.7
fHbf).
The
full-load current
is
is
100 A. The ro-
This can be achieved by using a material of
in
Figure
1
3.
1
8b.
It
can be seen that the
is
start-
ing torque doubles and the locked-rotor current de-
00 A to 90 A. The motor develops its
breakdown torque at a speed /Vd of 500 r/min, compared to the original breakdown speed of 800 r/min.
creases from
If
we
becomes
imum
es-
is
motor having a syn-
and the locked-rotor current
shown
of syn-
rotor resistance of a squirrel-cage rotor
V
and end-rings. The new torque-speed curve
of
13.15 Effect of rotor resistance
sentially constant
380
higher resistivity, such as bronze, for the rotor bars
chronous speed.
The
A
2.5.
500 hp and more)
about
8a shows the torque-speed curve of a
Let us increase the rotor resistance by a factor of
T
their
80%
1
tor has an arbitrary resistance R,
lower
will quickly stop.
less)
a speed
synchronous speed. Big motors
attain their
of 100
are in balance, the
the load torque exceeds 2.5
3.
1
again equal to the load
turn at a constant but slightly
speed. However,
(the
is
1
kW (13.4 hp), 50 Hz,
chronous speed of 000 r/min and a full-load torque
me-
the
chanical load increases slightly, the speed will drop
until the
and end-
minimum
The
torque developed by the motor while
At full-load the
in the rotor bars
torque (called breakdown
Pull-up torque
T.
nom-
The torque-speed curve is greatly affected by
such a change in resistance. The only characteristic
that remains unchanged is the breakdown torque. The
T.
maximum
2.5
aluminum, or other metals
the torque-speed curve of a
conventional 3-phase induction motor whose
of 70
except
A
increases with temperature. Thus, the resis-
1
again double the rotor resistance so that
5 R, the locked-rotor torque attains a
value of 250
A (Fig.
N-m
it
max-
for a corresponding current
13.18c).
further increase in rotor resistance decreases
tance increases with increasing load because the
both the locked-rotor torque and locked-rotor cur-
temperature
rent.
rises.
For example,
if
the rotor resistance
is
in-
In designing a squirrel-cage motor, the rotor resis-
creased 25 times (25 R), the locked-rotor current
tance can be set over a wide range by using copper,
drops to 20 A, but the motor develops the same
breakdown torque
2.5
T
2
T
,
|
locked-rotor torque
1.5 Ti
o
^pull-up
0.5
full
torq ue
nominal
T
T
20
60
40
—
p~
rotational
speed
Figure 13.17
Typical torque-speed curve of a
3-phase squirrel-cage induction motor.
80
100
%
load
Nm
N-m
Figure 13.18
Rotor resistance affects the motor characteristics.
283
ELECTRICAL MACHINES AND TRANSFORMERS
284
Nm),
torque (100
starting
as
locked-rotor current was 100
In
summary,
because
it
relatively
A
it
when
did
the
1
.
a high rotor resistance
is
The locked-rotor current can be
rotor torque will
low
squirrel-cage motor.
Unfortunately,
it
current
(Fig.
resistors in
series with the rotor. Nevertheless, the locked-
desirable
produces a high starting torque and a
starting
drastically re-
duced by inserting three external
(Fig. 13.1 8d).
13.18c).
also produces a rapid fall-off in
2.
still
be as high as that of a
The speed can be varied by varying
the exter-
speed with increasing load. Furthermore, because
the slip at rated torque
The
losses are high.
and the motor tends
is
high, the
efficiency
is
R
therefore low
3.
it is
preferable to have
up
and the
slip
We
is
is
small. Consequently, the effi-
high and the motor tends to run cool.
which require
is
a diagram of the circuit used to start
nected to three wye-connected external resistors by
set
of slip-rings and brushes. Under
locked-rotor (LR) conditions, the variable resistors
and a low running resistance by designing the rotor
bars in a special
However,
are set to their highest value.
way
(see Fig. 14.5,
Chapter
14).
up, the resistance
the rotor resistance has to be varied
if
over a wide range,
it
may
a long time to bring
wound-rotor motor. The rotor windings are con-
means of a
can obtain both a high starting resistance
ideally suited to accelerate high-
to speed.
Fig. 13.19
at rated torque
is
The speed de-
less with increasing load,
a
ciency
The motor
inertia loads,
a low rotor resistance (Fig. 13.18a).
much
I
nal rotor resistors.
to overheat.
Under running conditions
creases
motor
2
load speed
be necessary to use a
is
is
As
the
motor speeds
gradually reduced until
full-
reached, whereupon the brushes are
By properly selecting the resistance
we can produce a high accelerating torque
short-circuited.
wound-rotor induction motor. Such a motor envalues,
ables us to vary the rotor resistance at will by
means of an external
with a stator current that never exceeds twice
full-
rheostat.
load current.
To
13.16 Wound-rotor motor
We
explained the basic difference between a
squirrel-cage motor and a wound-rotor motor in
Section
1
3.
more than
1
.
Although a wound-rotor motor costs
a squirrel-cage motor,
it
start
large
motors,
we
offers the fol-
lowing advantages:
a large thermal capacity.
A
liquid
electrolyte.
To vary
its
resistance,
trodes.
The
is
com-
a suitable
we simply
vary
large thermal capacity of the electrolyte
Figure 13.19
to the three slip-rings of
in
the level of the electrolyte surrounding the elec-
speed controller
connected
use
liquid rheostat
posed of three electrodes immersed
starting rheostat
External resistors
often
rheostats because they are easy to control and have
a wound-rotor induction motor.
and
THREE-PHASE INDUCTION MOTORS
limits the
temperature
rise.
For example,
in
conjunction
bring a large
synchronous machine up
We
used
one ap-
kW wound-rotor motor to
plication a liquid rheostat
with a 1260
in
is
to speed.
can also regulate the speed of a wound-rotor
motor by varying the resistance of the rheostat. As
we
increase the resistance, the speed will drop. This
method of speed control has the disadvantage
of heat
lot
ciency
is
that a
dissipated in the resistors; the effi-
is
therefore low. Furthermore, for a given
rheostat setting, the speed varies considerably
if
the
rating of a self-cooled wound-rotor
motor depends upon the speed
at
Thus, for the same temperature
rise,
40
which
it
operates.
a motor that can
lOOkWat 800 r/min can deliver only
develop
kW at 900 r/min.
it
60
lodged
coils,
However,
the
if
can deliver 50
motor
is
in
60
slots.
total
The
X
of (4
and are staggered
intervals (Fig.
The
13.20).
X
coils in each
are connected in series
may
3
at
5)
group
one-slot
coils are identical
possess one or more turns.
=
The width
and
of each
coil is called the coil pitch.
Such
a distributed
winding
is
obviously more
costly to build than a concentrated winding having
only one coil per group. However,
starting torque
it
improves the
and reduces the noise under running
conditions.
from a
the stator windings are excited
3-phase source, a multipolar revolving
field
duced. The distance between adjacent poles
the pole pitch.
It
is
is
is
pro-
called
equal to the internal circumfer-
about
ence of the stator divided by the number of poles. For
cooled
example, a 12-pole stator having a circumference of
1
by a separate fan,
group must have a
When
mechanical load varies.
The power
coils per
285
kW at 900 r/min.
600
mm has a pole-pitch of 600/12 or 50 mm.
In practice, the coil pitch
is
between
80% and
100% of the pole pitch. The coil pitch is usually
made less than the pole pitch in order to save copper
13,17 Three-phase windings
named
and
to
Nikola Tesla invented the 3-phase induction motor.
The
shorter coil width reduces the cost and weight
In
883 a 27-year-old Yugoslav
1
His
first
model had a
similar to the one
scientist
salient-pole stator winding
shown
in Fig.
1
3.6.
Since then the
design of induction motors has evolved consider-
modern machines
ably;
are built with lap windings
distributed in slots around the stator.
A
lap
winding consists of a
set
tribution
improves the torque during
much
To
Fig. 13.21a.
=
poles
X
coils
is
it
follows that the
equal to the
number of groups.
A
stator
must have
at least
2, 3.
or
more
coils per
The number of
tion.
coils
12 slots.
group
and
in
coils are held upright, with
one
at least
of
12 coils.
many
5 coils per group
has coils. Consequently, a 4-pole, 3-phase
designers have discovered that
us
4-pole,
Furthermore, in a lap winding the stator has as
it
let
shown
at least
minimum number
3-phase stator must therefore have
slots as
laid out flat as
4X3=12
phase groups. Because a group must have
coil,
The 24
is
phases
Thus, a 4-pole, 3-phase stator must have
one
and
easier to insert in the slots.
get an overall picture of a lap winding,
The number of groups
groups
start-up,
2-pole machines, the shorter pitch also makes the
suppose a 24-slot stator
given by the equation
the air gap.
often results in a quieter machine. In the case of
evenly distributed around the stator circumference.
is
in
of the windings, while the more sinusoidal flux dis-
coils
of phase groups
improve the flux distribution
it
is
However, motor
preferable to use
rather than only one.
slots increases in propor-
For example, a 4-pole. 3-phase stator having 5
Figure 13.20
The five coils are connected
phase group.
in
series to create
one
ELECTRICAL MACHINES AND TRANSFORMERS
286
coil side set in
each
the windings are
slot. If
now
laid
c.
down so that all the other coil sides fall into the slots,
we obtain the classical appearance of a 3-phase lap
winding having two
The
coils are
coil sides per slot (Fig.
1
3.2 b).
1
Number of groups
poles
d.
The pole
one for each phase. Each wind-
ing consists of a
One
ber of poles.
(say) to slot 13.
around the circumference of the
rically distributed
The following examples show how
stator.
e.
this is
done.
of
= 40
4-
=
10
4.
pitch corresponds to
pole pitch
number of groups equal to the numThe groups of each phase are symmet-
= number
per phase
10
Coils per group
connected together to create three
identical windings,
—
=
-
=
slots/poles
120/10
12 slots
pole pitch extends therefore from slot
1
The coil pitch covers 10 slots (slot to slot
The percent coil pitch = 10/12 = 83.3%.
1
The next example shows
in
greater detail
1
1
).
how
the coils are interconnected in a typical 3-phase sta-
Example 13-8
The stator of a
possesses 120
tor winding.
3-phase, 10-pole induction motor
slots. If a lap
winding
is
used, calcu-
a.
b.
c.
d.
e.
The total number of coils
The number of coils per phase
The number of coils per group
The pole pitch
The coil pitch (expressed as a percentage of the
pole pitch),
1
to slot
if
Example 13-9
A
late the following:
the coil width extends from slot
stator
having 24 slots has to be
1
.
2.
The connections between
The connections between
are standing upright, with
We
bution for phase
Solution
a.
A
b.
Coils per phase
120
-
3
-
the coils
the phases
coils.
one
Assume
coil side in
that they
each
slot
will first determine the coil distri-
A and
then proceed with the con-
nections for that phase. Similar connections will
120-slot stator requires 120 coils.
=
with a
Solution
The 3-phase winding has 24
(Fig. 13.22).
1
wound
3-phase, 4-pole winding. Determine the following:
then be
40.
made
for phases
B and C. Here
is
the line of
reasoning:
a.
The revolving
p
field creates
4 poles; the motor
4 groups per phase, or 4
therefore has
phase groups in
all.
Each rectangle
X
3
in Fig.
=
1
13.22a
I
represents one group. Because the stator contains
lln iVr
1/2
20 21 22 23 24
3
4
5
6
24
7
/
slot
number
coils,
each group consists of 24/12
2
b.
The groups
(poles) of each phase
24
must be uni-
The group
distribution for phase A is shown in Fig.
13.22b. Each shaded rectangle represents two
Figure 13.21a
in
2 con-
secutive coils.
formly spaced around the
Coils held upright
=
stator slots.
stator.
upright coils connected in series, producing
the
two terminals shown. Note
that the
me-
chanical distance between two successive
20 21 22 23 24
1
2
3
4
5
6
groups always corresponds to an electrical
7
phase angle of 180°.
Figure 13.21b
Coils laid
down
c.
to
make a
typical lap winding.
Successive groups of phase
site
magnetic
polarities.
A must have oppo-
Consequently, the four
^each
each group
one group of one phase
coils are
ot two
composed of
coils in series
AAA
Figure 13.22a
The 24
is
grouped two-by-two
to
make 12
groups.
BtJtfMtzftlMt
(—180°
Figure 13.22b
The four groups
of
(electrical)—
phase A are selected so as
start of
phase
to
be evenly spaced from each
other.
A
Figure 13.22c
The groups
of
phase A are connected
start of
in
series to create alternate
phase
,
start of
N-S
poles.
phase C
1
1
cil
:b;
120°
240°—
Figure 13.22d
The
start of
phases B and
A
C
C
begins 120° and 240°, respectively, after the start of phase A.
'4
Figure 13.22e
When
all
phase groups are connected, only
six leads remain.
287
ELECTRICAL MACHINES AND TRANSFORMERS
288
Figure 13.22f
The phase may be connected
A are
groups of phase
wye
in
connected
duce successive N-S-N-S poles
Phase
A now
1
A2
B and C
same way around
the stator.
and three leads are brought out
delta,
pro-
f.
are spaced
(Fig.
The groups
series in the
in
phases B and
same way
C
A are
A,A 2
,
.
sulting 3 wires corresponding to the 3 phases
are brought out to the terminal
box of
the
ma-
chine (Fig. 13.22f). In practice, the connections
made, not while the coils are upright
shown) but only
in
the slots.
according to Fig.
450
after they
first coil
of
1
and connections
3.22e.
and 13.24b show the
kW (600 hp)
and
coil
induction motor.
procedure used
kW (50 hp)
in
stator.
are connected in
as those of phase
B|B 2 and C)C 2 They may be connected either
in wye or in delta inside the machine. The re,
Thus, the
shortened
the first and sixth slots
winding a smaller 37.5
(Fig. 13.22e). This yields six terminals:
are
in
Fig. 13.25 illustrates the
(Fig. 13.22d).
e.
to slot 6).
1
lodged
Figs. 13.24a
A
of phase
(
suit
stator of a
located at 120° and 240° (electrical) with re-
A
is
span of
to a
may be
3.23). All the other coils
1
follow
However, the
the terminal box.
6 slots, the coil pitch
A
phase
.
starting terminals B, and C, are respectively
spect to the starting terminal
=
to 5 slots (slot
ter-
to
Because the pole pitch corresponds
24/4
3.22c).
has two terminals, a starting
The phase groups of phases
the
in
in series to
(Fig.
minal A, and a finishing terminal
d.
or
have been
(as
laid
down
13.18 Sector motor
Consider a standard 3-phase, 4-pole, wye-connected
motor having a synchronous speed of 1800 r/min.
Let us cut the stator
is
left
in
in half,
so that half the winding
removed and only two complete
(per phase). Next,
let
N
and S poles are
us connect the three phases
wye, without making any other changes to the ex-
isting coil connections. Finally,
nal rotor
gap
above
we mount
this sector staton leaving a
(Fig. 13.26).
the origi-
small
air
THREE-PHASE INDUCTION MOTORS
should be reduced to half
the stator winding
nal
now
289
original value because
its
has only one-half the origi-
number of turns. Under
these conditions, this re-
markable truncated sector motor still develops about
20 percent of
The
moves
original rated power.
its
sector motor produces a revolving field that
the
at
same peripheral speed as
However,
the original 3-phase motor.
making
the flux in
instead of
a complete turn, the field simply travels
continuously from one end of the stator to the other.
13.19 Linear induction motor
It is
flat,
obvious that the sector stator could be
laid out
without affecting the shape or speed of the
field. Such a flat stator produces a field
moves at constant speed, in a straight line.
Using the same reasoning as in Section 3.5, we can
magnetic
Figure 13.24a
that
Stator of a 3-phase,
Hz
induction motor.
108 preformed
One
coil
side
450 kW, 1 1 80 r/min, 575 V, 60
The lap winding is composed of
coils
having a pitch from slots
falls into
the bottom of a slot
other at the top. Rotor diameter: 500
460 mm. (Courtesy
mecaniques Roberge)
length:
mm;
1
to 15.
and the
1
prove that the flux travels
at a linear
synchronous
speed given by
axial
vs
= 2wf
(13.10)
of Services Eiectro-
where
\\
=
linear
vr
—
width of one pole-pitch
/=
Note
synchronous speed |m/s]
frequency [Hz]
that the linear
speed does not depend upon the
number of poles but only on
is
[in]
the pole-pitch. Thus,
moving
at
the
stator (say),
If
a
flat
flat stator,
along with
same speed
as that of a 6-pole linear
provided they have the same pole-pitch.
squirrel-cage winding
is
brought near the
the travelling field drags the squirrel cage
it
use a simple
(Section 13.2). In practice,
aluminum
we
generally
or copper plate as a rotor (Fig.
13.27). Furthermore, to increase the
power and
duce the reluctance of the magnetic path, two
tors are usually
sides of the
Figure 13.24b
Close-up view of the preformed
coil in Fig.
13.24a.
it
possible for a 2-pole linear stator to create a field
to re-
flat sta-
mounted, face-lo-face, on opposite
aluminum
plate.
a linear induction motor.
The combination
The
is
direction of the
called
motor
can be reversed by interchanging any two stator leads.
If
we connect
the stator terminals to a 3-phase,
60 Hz source, the rotor will again turn
1800
r/min.
To prevent
saturation,
at
close to
the
voltage
In
many
practical applications, the rotor
is
sta-
tionary while the stator moves. For example, in
some high-speed
trains, the rotor is
composed of
a
Figure 13.25
50
Stator winding of a 3-phase,
carrying 48 coils connected
a.
Each
composed
coil is
in
hp,
575
V,
60 Hz, 1764
r/min induction motor.
of 5 turns of five No. 15
copper wires connected
with a high-temperature polyimide insulation. Five No. 15 wires
b.
c.
One
coil
fore,
from
Each
coil
side
threaded
stator
possesses 48
slots
into slot
1
(say)
in parallel.
in parallel is
and the other side goes
The wires are covered
equivalent to one No. 8 wire.
into slot 12.
The
coil pitch
is,
there-
to 12.
1
side
side placed
is
The
wye.
in
and 4 empty
fills
the
does not touch the second coil
shows 3 empty and uninsulated slots
a composition paper liner. The remaining 10 slots each carry one coil
a
slot
and
same
slot.
Starting from the top, the photograph
half
slots insulated with
is
covered with a paper spacer so that
it
side.
d.
A
varnished cambric cloth, cut
in
the shape of a triangle, provides extra insulation between adjacent phase
groups.
(Courtesy of Services Efectromecaniques Roberge)
290
THREE-PHASE INDUCTION MOTORS
13.20 Traveling
We
sometimes
are
291
waves
with the impression that
left
when
the flux reaches the
there
must be a delay before
more
at
end of
the beginning. This
a linear stator,
returns to restart once
it
is
not the case.
ear motor produces a traveling
wave of
The
lin-
which
flux
moves continuously and smoothly from one end of
shows how the
flux moves from left to right in a 2-pole linear motor. The flux cuts off sharply at extremities A, B of
the stator to the other. Figure 13.28
Figure 13.26
Two-pole sector induction motor.
However,
the stator.
pears
at
the right,
it
N
as fast as a
or S pole disap-
builds up again at the
left.
linear rotor
Properties of a linear
induction motor
13.21
(aluminum, copper or
iron plate)
The
properties of a linear induction motor are al-
most
identical to those of a standard rotating
ma-
chine. Consequently, the equations for slip, thrust,
power,
L
etc.,
are also similar.
Slip
Slip,
is
defined by
=
s
(v s
-
(I3.ll)
v)/v s
where
s
\\
Figure 13.27
Components
of
a 3-phase linear induction motor.
aluminum
thick
tending over the
stator
is
plate fixed to the
full
v
ground and ex-
length of the track.
The
linear
bolted to the undercarriage of the train and
straddles the plate. Train speed
is
varied by chang-
ing the frequency applied to the stator (Fig.
1
3.3
2.
1
slip
synchronous linear speed [m/s]
speed of rotor (or stator) [m/s]
Active power flow. With reference to Fig.
active
way
it
3.7,
power flows through
a linear
motor
and 13.8 apply
tween consecutive phase groups of phase
mm,
A
is
300
calculate the linear speed of the magnetic field.
(13.6)
(\
-s)P
is
given by:
F= P
pitch
vs
is
300 mm. Consequently,
=
2vv/
=
2
=
45 m/s or
X
X
1
75
62 km/h
T
(13.12)
/v s
where
(13.10)
0.3
(13.8)
r
thrust or force developed by a lin-
ear induction motor
Solution
The pole
The
13.6,
(13 .7)
T
Pm =
Thrust.
that the
to both types of machines:
= PJP c
P}r = sP
3.
same
Consequently, Eqs.
flat.
t\
Example 13-10
The stator of a linear induction motor is excited
from a 75 Hz electronic source. If the distance be-
3. 15,
1
in the
does through a rotating motor, except
and rotor are
stator
1
).
—
=
=
F = thrust [Nl
P = power transmitted to the rotor W]
v = linear synchronous speed |m/s|
v
s
(
292
ELECTRICAL MACHINES AND TRANSFORMERS
Example 13-11
An
overhead crane
in a factory is driven horizon-
by means of two linear induction motors
tally
whose
two
rotors are the
I-beams upon which
steel
The 3-phase, 4-pole
the crane rolls.
linear stators
(mounted on opposite sides of the crane and facing
the respective
pitch of 8
webs of
the I-beams) have a pole
cm and are driven by a variable frequency
electronic source. During a test
tors, the
on one of the mo-
following results were obtained:
Hz
stator frequency: 15
power
kW
to stator: 5
copper loss
+
iron loss in stator:
1
kW
crane speed: 1.8 m/s
Calculate
a.
Synchronous speed and
b.
Power
c.
I
d.
Mechanical power and thrust
2
R
slip
to the rotor
loss in rotor
Solution
a.
Linear synchronous speed
= 2wf
- 2 X 0.08 X
= 2.4 m/s
vs
The
Power
=
(v s
=
0.25
- v)/v
= (2.4 - 1.8)/2.4
to the rotor
/>
js
r
=
c.
!
R
s
(13.11)
is
P = Pc - 5 -
2
15
slip is
j
b.
(13.10)
- P
f
(see Fig. 13.15)
1
4kW
loss in the rotor
is
Figure 13.28
Shape of the magnetic field created by a 2-pole,
3-phase linear stator, over one complete cycle. The
successive frames are separated by an interval of
time equal to 1/6 cycle or
60%.
Pir
= *P
= 0.25 X
= kW
(13.7)
r
1
4
THREE-PHASE INDUCTION MOTORS
d.
Mechanical power
cause the flux density
is
The
(Fig- 13.15)
}r
sulting
value
induced current reaches
nnn
and
1
time. This current, re-
and
creates magnetic
3,
shown
as
sss
the re-
maximum
its
Fig.
in
13.29.
is
According
F = PJv s
= 4000/2.4
=
same
at virtually the
turning by conductors
poles
thrust
greatest at the center of
magnet moves very slowly,
the pole. If the
Pr-P
4-1
= 3kW
Pm =
=
is
293
1667
(13.12)
to the
laws of attraction and repulsion,
while the rear half
N=
1.67
magnet
the front half of the
attracted
is
repelled upward
is
downward. Because
the distribution of the nnn and sss poles
kN (-375
is
sym-
lb)
metrical with respect to the center of the magnet,
the vertical forces of attraction and repulsion are
13.22 Magnetic levitation
In
we saw
Section 13.2
moving permanent
that a
magnet, sweeping across a conducting ladder, tends
to
drag the ladder along with the magnet.
now show
We
that this horizontal tractive force
accompanied by a
is
also
to
Referring to Fig. 13.29, suppose that conduc-
1,2,3 are three conductors of
ladder. The center of the N pole of
the stationary
the
sweeping across the top of conductor
age induced
in
this
conductor
is
2.
magnet is
The volt-
maximum
be-
But suppose now
rapidly.
magnet
(stationary)
\
s
s
s
s
/
/
that the
nil.
magnet moves very
its
maximum
con-
value a fraction of a
after the voltage has attained
Consequently, by the time the current
its
in
maximum.
conductor 2
maximum, the center of the magnet is already
some distance ahead of the conductor (Fig. 13.30).
The current returning by conductors and 3 again
is
1
creates nnn and sss poles; however, the
is
now
directly
N
pole of the
above an nnn pole, with the
low
result that a strong vertical force tends to
magnet upward.* This
effect
push the
called the principle
is
of magnetic levitation.
777TTT\
n n
ss
is
to its inductance, the current in
speed
N
conducting ladder
Owing
ductor 2 reaches
magnet
::
force
vertical
only a horizontal tractive
is
force.
second
push the magnet away from the ladder.
tors
Consequently, there
will
which tends
vertical force,
and the resulting
equal,
\
1
Magnetic
n n n
0'
levitation
is
used
in
some
ultra-high-
speed trains that glide on a magnetic cushion rather
than on wheels.
A powerful electromagnet fixed un-
derneath the train moves above a conducting
ducing currents
Figure 13.29
ladder.
Currents and magnetic poles at low speed.
in the rail in the
The force of
levitation
same way
is
rail in-
as in our
always accompa-
nied by a small horizontal braking force which
must, of course, be overcome by the linear motor
magnet
high
that propels the train.
See Figs. 13.31 and 13.32.
speed
N
The
current
is
always delayed (even
terval of time A/,
m
of the
rotor.
This delay
the
is
low speeds) hy an
the
in-
IJR time constant
so brief that, at low speeds, the
maximum at virtually the same lime and
voltage does. On the other hand, at high speeds,
current reaches
place as the
at
which depends upon
its
same delay At produces
a significant shift in
space be-
Figure 13.30
tween the points where the voltage and current reach
Currents and magnetic poles at high speed.
respective
maximum
values.
their
Figure 13.31
This 17
t
track.
is driven by a linear motor. The motor consists of a stationary rotor and a flat stator
undercarriage of the train. The rotor is the vertical aluminum plate mounted in the center of the
electric train
fixed to the
The 3-tonne
stator
is
varied from zero to 105 Hz.
energized by a 4.7 MVA electronic dc to ac inverter whose frequency can be
The motor develops a maximum thrust of 35 kN (7800 lb) and the top speed
200 km/h. Direct-current power at 4 kV is fed into the inverter by means
with 6 stationary dc busbars mounted on the left-hand side of the track.
means
of a
brush assembly
in
is
contact
a superconducting electromagnet. The magnet
is 1300
magnet are maintained at a
2
temperature of 4 K by the forced circulation of liquid helium. The current density is 80 A/mm and the resulting flux density is 3 T. The vertical force of repulsion attains a maximum of 60 kN and the vertical gap between the magnet and the reacting metallic track varies from 100 mm to 300 mm depending on the current.
Electromagnetic levitation
nun long, 600
mm wide,
is
obtained by
and 400
mm deep,
of
and weighs 500
kg.
The
coils of the
,
(Courtesy of Siemens)
294
THREE-PHASE INDUCTION MOTORS
295
superconducting electromagnet
conducting track
guide and support
wheels
linear
motor
(stator)
conducting plate
(rotor)
brush assembly for power
input,
from 4 kVdc source
Figure 13.32
Cross-section view of the main components of the high-speed train
(Courtesy of Siemens)
Questions and Problems
shown
in Fig.
1
3.31
Give two advantages of a wound-rotor
13 -8
motor over a squirrel-cage motor.
Practical level
13-1
Name
the principal
components of an
Both the voltage and frequency induced
13-9
in
in-
the rotor of an induction
duction motor.
motor decrease
as the rotor speeds up. Explain.
13-2
Explain
how
a revolving field
is
set
up
in
13-10
a 3-phase induction motor.
1
3-3
If
we double
the
number of poles on
stator of an induction motor, will
its
A 3-phase,
connected
the
syn-
chronous speed also double?
20-pole induction motor
to a
a.
What
is
b.
If the
voltage
600
V.
is
60 Hz source.
the synchronous speed?
is
reduced to 300 V, will the
synchronous speed change?
13-4
The
rotor of an induction should never be
c.
locked while
to the stator.
1
3-5
Why
full
voltage
is
13-
Explain.
13-7
1
1
groups are there, per phase?
Describe the principle of operation of a
linear induction motor.
does the rotor of an induction motor
turn slower than the revolving field?
13-6
How many
being applied
13- 12
Calculate the approximate values of the
What happens to the rotor speed and rotor
current when the mechanical load on an
starting current, full-load current, and
induction motor increases?
575
Would you recommend using
a 50 hp
induction motor to drive a 10 hp load?
Explain.
no-load current of a 150 horsepower,
13- 3
1
V,
Make
3-phase induction motor.
a drawing of the magnetic field cre-
ated by a 3-phase, 12-poIe induction motor.
296
ELECTRICAL MACHINES AND TRANSFORMERS
13-14
How
wc change
can
the direction of rota-
the
motor?
tion of a 3-phase induction
mmf developed by the windings,
the resulting mmf point in a direction
Does
c.
intermediate between the
Intermediate level
13-15
sponding
Calculate Ihe synchronous speed of a
a.
1
3-phase, 12-pole induction motor that
b.
cited
by
What
is
load
a
3-2
1
ex-
is
if
produces
the slip at full
synchronous speed
a
when connected to
number of
source. Calculate the
6 percent?
is
slots
of 900 r/min
nominal speed
mmf s corre-
and 4?
3-phase lap- wound stator possessing
72
60 Hz source,
the
A
to instants 3
60 Hz
a
coils
per phase group as well as the probable
1
3-
1
6
A
3-phase 6-pole induction motor
Hz
nected to a 60
induced
supply.
in the rotor bars
is
con-
The voltage
is 4 V when the
is
locked. If the motor turns in the
13-17
its
13-22
The 3-phase, 4-pole
stacking (axial length) of 200
At 1000 r/min
maximum
At 1500 r/min
calculate the following:
a.
Calculate the approximate values of
b.
a.
full-
The
kW. 4000
V, 3-phase.
1
9
A
900
b.
c.
3-phasc, 75 hp, 440
V
2 percent.
1
3-23
when
the stator
induction motor
A large
4000
3-phase,
A
and a
385
and a power factor of 83 percent.
when
Calculate the nominal current per phase.
sponding speed
open-circuit voltage of
is
240 V appears
total active
found
to
is
wye and
tween two
stator terminals
total iron losses are
windage and
motor turns
same
direction as the
r/min. in the
same
direction as the
rotating field
b.
At 900
rotating field
c.
a.
of
b.
field
Referring to Fig. 13.7, calculate the instan-
taneous values of
I5()
/.,,
/h ,
and
/c
for an angle
u
.
Determine the actual direction of current flow
in the
three phases at this instant
10
friction losses are 12
il.
The
kW.
c.
d.
The load mechanical power [kW|, torque
[kN-m], and efficiency
1
At 3600 r/min. opposite to the rotating
is 0.
kW and the
23.4
The power factor at full-load
The active power supplied to the
2
The total I R losses in the rotor
a.
b.
in the
At 600 r/min,
is
the resistance be-
Calculate the following:
open-circuit voltage and frequency across
a.
kW
corre-
be 709.2 r/min. The stator
duction motor when the rotor is locked.
The stator has 6 poles and is excited by
60 Hz source. If the rotor is driven by a
a
The
accurately measured and
in
the dc
squirrel-
power of 2344
operating at full-load.
connected
if
bars
in the rotor
60 Hz
V,
across the slip-rings of a wound-rotor in-
the slip-rings
0.7 T,
connected to a
is
The peak voltage induced
The pole-pitch
has a full-load efficiency of 91 percent
An
is
a
the
cage induction motor draws a current of
variable-speed dc motor, calculate the
13-20
flux density per pole
If
60 Hz source
Calculate the nominal full-load speed and
is
mm and
mm.
peripheral speed |m/s] of the revolving
field
induction motor,
60Hz
torque knowing that the slip
3-
stator of Fig. 13.25
has an internal diameter of 250
frequency:
c.
r/min,
1
Fig. 13.22.
b.
current of a 75
con-
coil
ro-
load current, starting current, and no-load
13-18
complete
same
direction as the flux, calculate the approxi-
mate voltage induced and
a. At 300 r/min
the
nection diagram, following steps (a) to
(f) in
tor
Draw
coil pitch.
and calculate
3-24
If
we
rotor
(
\
7c\
slightly increase the rotor resistance
of an induction motor, what effect does
this
a.
have (increase or decrease) upon
Starting torque
b.
Starting current
c.
Full- load speed
d. Efficiency
THREE-PHASE INDUCTION MOTORS
e.
Power
f.
Temperature
13-30
factor
of the motor
rise
rated
at its
power output
1
3-25
13-26
voltage
Explain the principle of magnetic levitation.
Advanced
In Fig. 13.5a the
The
permanent magnet has
and moves at 30 m/s.
mm
flux density in the air
gap
is
0.5
T
1
3-3
1
and the effective resistance per rotor bar
is
mil. Calculate the current
1
3-27
If the
conducting ladder
in Fig.
3.5a
1
is
pulled along with a force of 20 N, what
1
3-28
A
5000
3-phase,
hp,
6000
V,
60 Hz
1
What
r/min.
I~R losses
1
3-29
at
The motor
at
0.0073
3-28 has the
1
=
13-33
17°C
=
V
1600
= 6000 V
= 100 A
line-to-line stator voltage
6.
active
7.
windage and
8.
iron losses in the stator
9.
locked-rotor current
2207
power supplied
to
motor
at
450
3.
1
9).
r/min, calculate the
Voltage between the slip rings
Rotor resistance (per phase) and the
tolal
Approximate
rotor current, per phase
The train shown in Fig. 13.31 moves at
200 km/h when the stator frequency is
By supposing
a negligible slip,
motor [mm].
A 3-phase,
300 kW, 2300
V,
at
=
The motor has
load efficiency and
power
1
V =
2760
terminal voltage rises to
kW
effect (increase or decrease)
1800
A
to stator with rotor locked
=
kW
a full-
factor of 92 per-
cent and 86 percent, respectively.
kW
39
6000
5
used to
drive a compressor.
motor operates
=
friction losses
60 Hz,
is
no-
kW
power
a speed of
1
develop a torque of 20
1780 r/min induction motor
no-load stator current, per phase
10. active
at
the linear
5.
91
kN-m
to
calculate the length of the pole-pitch of
=
17°C
4.
=
motor has
105 Hz.
rings with rotor locked
load
inserting resis-
fol-
open-circuit voltage induced between slip-
3.
Problem 13-29 by
in
If the
c.
Problem
11 at
wish to control the speed of the motor
at
13-32
dc resistance between rotor slip-rings
2.
We
are the approximate rotor
dc resistance between stator terminals
0.112H
Torque developed by the rotor
power dissipated
lowing characteristics:
1.
Mechanical power output
e.
a.
2-
rated load?
in
the stator
in
d.
b.
pole wound-rotor induction motor turns
594
Active power supplied to the rotor
following:
is
on the magnet?
the braking force exerted
I~R losses
c.
tors in series with the rotor (see Fig.
tractive force.
1
(locked-rotor) conditions:
b.
given
and the
/
LR
in
full-
Reactive power absorbed by the motor
a.
level
a width of 100
Referring to the motor described
Problem 13-29, calculate under
297
at full-load,
V
If
the
while the
determine the
upon
a.
Mechanical power delivered by the motor
b.
Motor torque
c.
Rotational speed
d. Full-load current
e.
Power
f.
Starting torque
factor and efficiency
Calculate
a.
Rotor and stator resistance per phase
75°C (assume
b.
it
turns at
at
g.
wye connection)
Voltage and frequency induced
when
c.
a
200 r/min and
Breakdown torque
i.
Motor temperature
in the rotor
at
Starting current
h.
rise
594 r/min
j.
Reactive power absorbed by the motor to
k.
Flux per pole
Exciting current
create the revolving field, at no-load
2
d.
I
at
e.
1.
R
losses in the stator
when
the
Iron losses
motor runs
no-load (winding temperature 75°C)
Active power supplied to the rotor
at
no-load
13-34
A
3-phase, 60
Hz
linear induction
motor
has to reach a top no-load speed of 12 m/s
298
ELECTRICAL MACHINES AND TRANSFORMERS
and
must develop a
it
standstill thrust
of
Calculate
10 kN. Calculate the required pole-pitch
and the
minimum
2
I
R
a.
b.
loss in the rotor, at
c.
standstill.
d.
Industrial application
1
3-35
A
1
e.
0 hp, 575 V,
1
1
60 r/min, 3-phase, 60 Hz
induction motor has a rotor
made of alushown in Fig.
minum, similar to the rotor
3.3a. The end-rings are trimmed
1
f.
is
1
3-37
What
making them
effect will this
a.
The
The starting torque
The temperature rise
c.
full
hp,
standstill
less thick.
1
1
83 r/min, 575 V, 3-phase, 60
320
V
is
3-36
The
stator of a
600
that the
RMS
about 0.6 V. Estimate
the no-load speed of the motor.
1
3-38
The
rotor of a
60 hp,
1
760 r/min, 60 Hz
at full-load
induction motor has
1
known
It is
brush voltage drop
load speed of the motor
Hz
at
between the open-circuit
lines of the rotor.
have on the following:
b.
A 25
density
T
0.54
wound-rotor induction motor produces
in a lathe,
cutting off the cooling fins and also a portion of the rings,
The number of coils on the stator
The number of coils per phase
The number of coils per group
The coil pitch (in millimeters)
The area of one pole
The flux per pole if the average flux
hp,
1
1
60 r/min, 575
V,
3-phase, 60 Hz induction motor has
90 slots, an internal diameter of 20 inches,
and an axial length of 16 inches.
ter
of
1
1
1
17 bars
and a diame-
inches. Calculate the average
force on each bar (in newtons)
motor
is
running
at full-load.
when
the
Chapter 14
Selection
and Application
of
Three-Phase Induction Motors
be replaced by that of any other manufacturer,
14.0 Introduction
When purchasing a 3-phase induction motor
we
a particular application,
several types can
have to
make
fill
The
for
shaft height, or the type of coupling.
often discover that
selection
is
establishes limiting values for electrical,
generally
simplified because the manufacturer of the lathe, fan,
pump, and so
forth indicates the type of
motor
best suited to drive the load. Nevertheless,
ful to
know something about
it
that
is
is
use-
must
satisfy
also cover
some
to
motors
Motors are grouped
Standardization and
classification of
1.
*
Manufacturers
Motors and Generators. Standards
publication
in
Canada
Drip-proof motors. The frame
motors
of one manufacturer can
(NEMA)
which they have
We limit our discussion to five important
and solid particles which
in a
MG-I
fall at
drip-proof
liquid drops
any angle between
vertical. The
means of a fan directly couCool air, drawn into the motor
are cooled by
pled to the rotor.
Standards in the United Slates are governed by National
Electrical
in
0 and 15 degrees downward from the
motors under 500 hp
have standardized dimensions. Thus, a 25 hp,
Hz motor
categories, de-
into several
motor protects the windings against
induction motors*
725 r/min, 60
environment and
pending upon the environment
in general.
to operate.
1
rise.
cooling methods
classes.
industrial
to starting
14.2 Classification according
special appli-
interesting devices will enable the reader to gain a
better understanding of induction
all
requirements as
in-
nous generators and frequency converters. These
The frames of
minimum
locked-rotor current, overload capacity,
and temperature
cations of induction machines, such as asynchro-
14.1
mechan-
and thermal characteristics. Thus, motors
the basic construction
and characteristics of the various types of 3-phase
we
stan-
ical,
torque,
duction motors that are available on the market.
In this chapter
The
dardization covers not only frame sizes, but also
we
the need. Consequently,
a choice.
without having to change the mounting holes, the
governed by Canadian Standards Association (CSA) publication
titled
are similarly
299
C
154.
The two standards
arc essentially identical.
ELECTRICAL MACHINES AND TRANSFORMERS
300
through vents
in the
frame,
is
and then expelled. The
ings
temperature
rise
resistance)
ing
blown over
the wind-
4.
maximum
allowable
and high-power motors
(measured by the change
may
be 6()°C
in
wind-
80°C 105°C
125°C, depending on the type of insulation used
the windings. Drip-proof motors (Fig. 14.1
used
2.
in
)
ribbed motor frame (Fig.
in
a splash-
proof motor protects the windings against liquid
fall at
any angle be-
tween 0 and 100° downward from the
temperature
rise
These motors are mainly used
3.
vertical,
similar to that in drip-proof motors and
maximum
the
to the shaft,
in
is
An
external
air
over the
1
4.3).
A
concentric outer
shield prevents physical contact with the fan and
serves to channel the airstream.
drops and solid particles that
is
coupled
in
can be
air.
blows
usually cooled by an external blast of
fan, directly
Medium-
enclosed are
that are totally
or
most locations.
Splash-proof motor. The frame
Cooling
Totally enclosed, fan-cooled motors.
perature rise
5.
is
the
same
The permissible tem-
as for drip-proof motors.
Explosion-proof motors. Explosion-proof mo-
tors are
used
highly inflammable or explosive
in
surroundings, such as coal mines,
grain elevators.
They
oil refineries,
and
are totally enclosed (but not
also the same.
wet locations.
Totally enclosed, nonventilated motors.
These
motors have closed frames that prevent the free
exchange of
air
between the inside and the outside
They are designed for very wet and
dusty locations. Most are rated below 10 kW beof the case.
cause
it
is
difficult to get rid
of the heat of larger
machines. The motor losses are dissipated by natural
convection and radiation from the frame. The
permissible
1
temperature
rise
is
65°C,
85°C,
10°C, or I30°C, depending on the class of insu-
lation (see Fig. 14.2).
Figure 14.2
Two
enclosed nonventilated (TENV) 2 hp, 1725
cage motors are shown in foreground and two
30 hp, 1780 r/min totally enclosed blower-cooled motors (TEBC) in background. These 3-phase, 460 V
motors are intended to operate at variable speeds
totally
r/min
ranging from a few revolutions per minute to about 3
times rated speed.
The 2 hp motors have a full-load current of 2.9 A,
and power factor of 76 per-
efficiency of 84 percent
cent. Other characteristics: no-load current:
locked-rotor current:
pu;
tal
26
1
.7 A;
A; locked-rotor torque: 4.2
breakdown torque: 5.0
pu; service factor: 1.0; to-
weight: 39 kg; overall length including shaft:
mm;
overall height:
The 30 hp motors have a
efficiency of
377
235 mm.
34 A,
84 per-
full-load current of
93 percent, and power factor
of
cent. Other characteristics: no-load current: 12 A;
tion
Energy efficient drip-proof, 3-phase squirrel-cage inducmotor rated 230 V/460 V, 3 hp, 1750 r/min, 60 Hz.
locked-rotor current: 214 A; locked rotor torque: 1 .6 pu;
breakdown torque: 2.84 pu; service factor: 1 .0; total
weight: 200 kg; overall length including shaft: 834 mm;
overall height: 365 mm.
{Courtesy of Gould)
(Courtesy of Baldor Electric
Figure 14.1
Company
)
SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS
301
14.3 Classification according to
electrical
and mechanical
properties
enclosures just mentioned,
In addition to the various
3-phase squirrel-cage motors can have special electrical
and mechanical characteristics, as
listed
below.
Motors with standard locked-rotor torque
(NEMA Design B). Most induction motors belong
to this group. The per-unit locked-rotor torque decreases as the size of the motor increases. Thus, it
ranges from .3 to 0.7, as the power increases from
20 hp to 200 hp (15 kW to 150 kW). The corre/.
l
Figure 14.3
Totally
enclosed fan-cooled induction motor rated 350
440
(Courtesy of Gould
hp,
1760
r/min,
V,
sponding locked-rotor current should not exceed
6.4 times the rated full-load current. These general-
3-phase, 60 Hz.
purpose motors are used to drive fans, centrifugal
pumps, machine
2.
and so
tools,
forth.
High starting-torque motors
(NEMA Design C).
These motors are employed when
Pumps and
are difficult.
that
have
to start
starting conditions
piston-type compressors
under load are two typical applica-
tions. In the
range from 20 hp to 200 hp, the locked-
rotor torque
is
200%
of full-load torque, which cor-
responds to a per-unit torque of
The locked-rotor
2.
current should not exceed 6.4 times the rated full-
load current.
equipped with
In general, these motors are
double-cage
rotor.
double-cage rotor
lowing
Figure 14.4
a.
Totally enclosed, fan-cooled, explosion-proof motor.
Note the particularly rugged construction
of this type
b.
facts:
The frequency of the rotor
as the motor speeds up
A conductor that
of motor.
(Courtesy of Brook Crompton-Parkinson Ltd
(cage
)
airtight)
the
and the frames are designed
enormous pressure
that
may
to an internal explosion.
the flanges
on the end-bells are made extra long
Furthermore,
in
order to cool any escaping gases generated by such
may
permissible temperature rise
tally
the
is
enclosed motors (see Fig.
1
be initiated by
same
4.4).
lies
close to the rotor surface
has a lower inductive reactance than
The
as for to-
The conductors of cage
1
are
When the motor is connected
build up inside the
motor due
an explosion. Such explosions
)
much
2)
smaller
than those of cage 2
to withstand
the spark or a short-circuit within the windings.
1
current diminishes
one buried deep inside the iron core (cage
c.
a
The excellent performance of a
(Fig. 14.5) is based upon the fol-
to the line with the
rotor at standstill, the frequency of the rotor current
is
equal to line frequency.
Owing
ductive reactance of squirrel-cage
to the high in2, the rotor cur-
rent flows mainly in the small bars of cage
effective
motor resistance
is
essentially equal to that of cage
high starting torque
is
1.
The
therefore high, being
1.
developed.
Consequently,
a
ELECTRICAL MACHINES AND TRANSFORMERS
302
Figure 14.5
Typical torque-speed curves of
NEMA
design B, C, and
D
minimum NEMA
Hz squirrel-
motors. Each curve corresponds to the
values of locked-rotor torque, pull-up torque, and breakdown torque of a 3-phase 1800 r/min, 10 hp, 60
cage induction motor. The cross-section
As
falls,
the
of the respective rotors indicates the type of rotor bars
motor speeds up, the rotor frequency
with the result that the inductive reactance of
tors are usually
windings
that the reactance of both
The
rotor current
tance of cage
1
so low (typically
is
is
Hz)
I
negligible.
then limited only by the resis-
is
and cage 2 operating
in parallel.
The
rated speed
is
1
,
the effective rotor resistance at
much lower than
at standstill.
For
this
reason the double-cage rotor develops both a high
starting torque
and a low
slip at full-load.
Despite their high torque. Design
recommended
not
reason
is
that
punch holes
for starting high-inertia loads.
3.
in
cage
1
.
Owing to its small
may melt.
The
start-
size,
tends to overheat and the bars
High-slip
(NEMA
motors
rated speed of high-slip, Design
lies
between
85%
and
95%
These motors are used
in sheet metal.
ates the operation, a clutch
When
the
Punching a hole requires a tremendous amount
son
is
that the
punching energy
fraction of a second.
punch does
The energy
delivered in a
is
is
furnished by the
in a
itself.
As
the
work, the speed of the flywheel
lot
of kinetic
very short time. The speed of the motor
also drops considerably, along with that of the fly-
wheel. However, the Class
D
will not
The high-remade of brass, and the mo-
its
drops immediately, thus releasing a
energy
the
drawn from
exceed
As soon
of synchronous speed.
to accelerate high -inertia
initi-
engages the flywheel,
of power, sometimes exceeding 1000 hp. The rea-
the current
motors usually
worker
is
that
causing the punch to descend and pierce the sheet.
Design D). The
loads (such as centrifugal dryers), which take a rel-
motor
its
design ensures that
rated value.
as the hole
is
D
the line at the lower speed
is
pierced, the only load on
the flywheel,
which
is
now
gradually
brought back up to speed. During the acceleration
atively long time to reach full speed.
period, the
sistance squirrel cage
thus restoring the energy
is
machine tools
flywheel rather than by the motor
motors are
most of the rotor tR losses during
up are concentrated
it
C
to
large drop in speed with increasing load
also ideal to drive impact-type
Because the conductors of cage 2 are much larger
than those of cage
designed for intermittent duty
prevent overheating.
both squirrel-cage windings diminishes. At rated
speed the rotor frequency
used.
motor delivers energy
it
lost
to the flywheel,
during the impact.
A
SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS
powerful motor will quickly accelerate the
motor, and
fly-
wheel, permitting rapid, repetitive operation of the
punch
is
press.
low, a
On
much
the other hand,
if
2.
it
and
and
D
it
the characteristics of
motors.
The
(as a percentage of full-load
Fig. 14.5 enable us to
NEMA
rotor construction
of equal power.
Design B, C,
is
if
increased (by using brass instead
of copper or aluminum), the locked-rotor torque
creases, but the speed at rated torque
is
By way
also shown,
obtained by changing the rotor design. For example,
is
always greater
torque) than that of a similar low-speed motor
can be seen that the distinguishing properties are
the rotor resistance
in-
two 10
motor speed
The choice of motor speed
synchronous speeds. The difference
would
is
rather limited because
by quantum jumps, depending upon the frequency
and the number of poles. For example,
sible to build a
and running
ing an acceptable efficiency
say,
it
impos-
is
conventional induction motor hav-
of 2000 r/min on a 60
Hz
supply.
at a
speed,
totally en-
in price
justify the use of a high-speed
alone
motor and a
900
to drive a load operating at, say,
speeds (100 r/min or
r/min.
has to operate at very low
a gearbox
less),
mandatory.
is
The gears are often an integral part of the motor,
making for a very compact unit (Fig. 14.6).
synchronous speed of induction motors changes
the
60 Hz,
hp, 3-phase,
closed, fan-cooled induction motors having different
gearbox
lower.
of example. Table 14A compares the
properties of
When equipment
14.4 Choice of
factor are
The locked-rotor torque of a high-speed motor
is
The torque-speed curves of
power
will
only take longer to bring the flywheel up to speed.
compare
efficiency and
higher.
the repetition rate
smaller motor will suffice;
its
303
A
gearbox
is
when equipment
also mandatory
has to run above 3600 r/min. For example, in one
industrial application a large gear unit
is
used to
5000 r/min centrifugal compressor
a 3560 r/min induction motor.
drive a 1200 hp,
coupled to
Such a motor
would require 2 poles and a corresponding synchronous speed of 3600 r/min. The
2000)/3600
=
slip
of (3600
-
0.444 means that 44.4% of the power
14,5
The
Two-speed motors
stator of a squirrel-cage induction
supplied to the rotor would be dissipated as heat.
designed so that the motor can operate
(See Section
ent speeds.
1
3. 13.)
The speed of a motor
the speed of the
for
machine
low-speed machines,
a high-speed
rectly
is
it
obviously determined by
it
has to drive. However,
is
motor and a gearbox instead of
.
di-
coupling a low-speed motor to the load.
There are several advantages
1
often preferable to use
to using a
is
to
wind the
pumps. One way
stator with
that only
one winding
only half the copper
gearbox:
is
in
Power
being utilized.
However, special windings have been invented
whereby the speed
TABLE 14A
is
changed by simply changing
The synchronous
COMPARISON BETWEEN TWO MOTORS OF DIFFERENT SPEEDS
Synchronous
Power
speed
factor
drill
obtain two
operation at a time and so
the external stator connections.
of a low-speed
to
two separate windThe problem is
in the slots is
high-speed motor
less than that
differ-
ings having, say, 4 poles and 6 poles.
For a given output power, the size and cost of a
is
two
Such motors are often used on
presses, blowers, and
speeds
motor can be
at
Efficiency
Locked-rotor
Mass
torque
Price
(2002)
hp
kW
r/min
%
9c
%
kg
U.S. $
l()
7.5
3600
89
90
150
50
650
10
7.5
900
75
85
1
70
2000
25
1
ELECTRICAL MACHINES AND TRANSFORMERS
304
ac source
Figure 14.6
Gear motor rated at 2.25 kW, 1740 r/min, 60 Hz. The
output torque and speed are respectively 172 N-m
and 125 r/min.
(Courtesy of Reliance
speeds
obtained
Electric)
usually
are
speed
the
in
(3600/1800 r/min, 1200/600 r/min,
etc.).
ratio
2:1
The lower
produced by the creation of consequent
is
poles.
Consider, for example, one phase of a two-pole,
3-phase motor (Fig, 14.7a).
connected
in series to a
flows into terminal
of terminal
As
2.
When
and current
1
two poles
the
are
60 Hz ac source, current
a result, one
N
I2
(=
l\)
I
{
tlows out
pole and one S pole
Figure 14.7
Two
a.
are created
and the flux has the pattern shown. The
synchronous speed
=
ns
is
120f/p
that
stator
This
120
X
60/2
percent of the pole-pitch.
shown
rent
/,
minal
now
in Fig.
As a
windings.
When
the coils are connected
motor
is
produced. Two
Because every
in parallel, as
14.7b. In this case, at the instant cur-
result,
two
S pole,
the
connect the two poles
flows into terminal
2.
series produce a
of the
in parallel,
a 4-pole
poles are conse-
each pole covers only one-quarter of the
achieved by using a coil pitch equal to 50
Let us
in
r/min
circumference instead of the usual one-half.
is
connected
quent poles.
= 3600
Note
b.
-
short-pitch coils
two-pole motor.
1
N
,
current
I2
flows into
ter-
poles are created by the
it
two
nious
N
pole must be accompanied by a
follows that two S poles will appear between
N poles. The
way
south poles created in this inge-
are called consequent poles.
nection produces 4 poles in
speed
is
all,
The new con-
and the synchronous
1800 r/min. Thus, we can double the number
of poles by simply changing the stator connections.
is
upon
this principle that
2-speed motors are
built.
It
SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS
now
again produces 4 poles per phase, but
possess the same polarity (Fig.
Two-speed motors have a
1
they
all
4.8b).
relatively lower effi-
ciency and power factor than single-speed motors
They can be designed
do.
develop
to
both
(at
speeds) either constant power, constant torque, or
variable torque.
that has to
The choice depends upon
The 2-speed motors described so
tios
of
2:
l
.
the load
be driven.
If
the
motor drives a
The reason
big a change in speed.
far
have pole
fan. this
is
may
that the
ra-
be too
power of
a fan varies as the cube of the speed. Consequently,
if
the speed
is
reduced by
one-eighth, which
To overcome
power drops
half, the
to
often too low to be of interest.
is
this
problem, some 3-phase wind-
ings are designed to obtain lower pole ratios such as
8/10,
1
4/ 6, 26/28,
10/ 4,
1
1
larly useful in driving
and 38/46. These pole
PAM, motors
amplitude modulation, or
2-speed fans
horsepower range and more.
moderate reduction
in fan
PAM
are particuthe
in
hundred
motors enable a
power by simply recon-
necting the windings to give the lower speed.
motor
characteristics under
14.6 Induction
various load conditions
The complete torque-speed curves displayed
1
Figure 14.8
a. High-speed connection of a 3-phase stator, yielding 4 poles.
b. Low-speed connection of same motor yielding
4.5 are important, but
most of the time a motor runs
Figure
1
4.8
shows
the stator connections for a
2-speed, 4-pole/8-pole,
ing.
to
I
to 6, are
3-phase motor. Six leads,
brought out from the stator wind-
For the high-speed connection, power
is
applied
terminals 1-2-3, and terminals 4-5-6 are open.
resulting delta connection produces
having two
Tn
two S poles
connected
(Fig.
.
It
so happens that between
circuiting terminals 1-2-3
The
1
4.8a).
Note
is
made by
and applying power
resulting
4.9).
is
essentially a
The slope of
pends mainly upon the rotor resistance
the line de-
—
the lower
the resistance, the steeper the slope.
In effect,
the slip
tance
R
s,
it
can be shown that
torque
T,
at rated
line voltage E,
frequency,
and rotor
resis-
are related by the expression
4 poles per phase
= kTR/E 2
(14.
1)
that
where k
in series.
The low-speed connection
minals 4-5-6.
1
s
N and
the four poles are
The
that
close to synchro-
nous speed, supplying a torque that varies from zero
to full-load torque
straight line (Fig.
numbered
at
these limits the torque-speed curve
8 poles.
in Fig.
must be recognized
it
shortto ter-
double-wye connection
is
a constant that
depends upon the con-
struction of the motor.
This expression enables us
formula showing
how
to establish a
simple
the line voltage and rotor
ELECTRICA L MA CHINES A ND TRA NS FORMERS
306
Figure 14.9
The torque-speed curve
is
resistance affect the behavior of the
load. In effect,
motor
once
between the no-load and rated torque operating
essentially a straight line
we know
motor under
the characteristics of a
for a given load condition,
we can
predict
its
speed, torque, power, and so on, for any other load
condition. These quantities are related by the formula
an accuracy of better than 5 percent which
cient for
most
suffi-
Example 14-1
A
V
3-phase, 208
induction motor having a syn-
chronous speed of 1200 r/min runs
to a
V
215
line
stant torque load. Calculate the
A-
is
practical problems.
when connected
Ax
points.
increases to
240
at
1
140 r/min
and driving a con-
speed
if
the voltage
V.
where
n
=
Solution
subscript referring to the
initial,
conditions (the given conditions
spond
to the
=
subscript referring to the
s
=
slip
E=
may
The
slip at
215 Vis
corres
nominal rating of the motor)
x
T=
R =
or given load
new
load conditions
=
(n s
=
(1200
=
0.05
n)ln s
-
11
40)/ 1200
torque [N-m]
When the voltage
rotor resistance
rises to
stator voltage [V]
applying the formula, the only restriction
.
V, the load torque
we can
write
is
new torque Tx must not be greater than
Tn (EJEn ) 2 Under these conditions Eq. 14.2 yields
that the
240
and
rotor resistance remain the same. Consequently, in
applying Eq. 14.2,
In
—
.v
x
=
sn
=
0.04
(E n /Ex )
2
=
0.05 (215/240)
2
-
SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS
The
slip
speed
Example 14-3
therefore,
is,
X 1200 = 48
0.04
A 3-phase, 4-pole
r/min
rating of
The new speed
nx
=
240
at
1200
V
- 48 =
1152 r/min
Under
is
Hz
line.
The
it
wye
across the
Nm
speed of 1000 r/min.
at a
N
Calculate the speed for a torque of 400
b.
Calculate the value of the external resistors so
kW
at
200
1000)/ 1800
=
0.444
motor develops 10
that the
m.
r/min.
Solution
to
a.
The given conditions
are
machine has run for several
after the
Tn = 300 N
hours.
.v
Calculate
The hot
in
is
initial
23°C. The speed drops
connected
these conditions the motor de-
a.
com-
8-pole induction motor driving a
connected to a fixed 460 V, 60
rotor resistance in terms of the cold re-
n
torque of 400
The approximate hot temperature of the
bars, knowing they are made of copper
m
- (1800 -
All other conditions being fixed,
sistance
b.
wound-rotor induction motor has a
kWJ 760 r/min, 2.3 kV, 60 Hz. Three ex-
velops a torque of 300
cold rotor temperature
a.
0
1
rotor slip-rings.
pressor runs at 873 r/min immediately after
864 r/min
1
ternal resistors of 2 f i are
is
Example 14-2
A 3-phase,
307
we have
for a
N m the following:
rotor
-v
x
=
-
sn
(TJTn ) = 0-444 (400/300)
0.592
Solution
a.
The synchronous speed
ns
The
=
120
initial
=
///;
and
120
X
= 900
slip
speed
(900
-
873)/900
=
0.03
sx
=
(900
-
864)/900
=
0.04
speed change
is
in rotor resistance.
x
=
0.04
=
.s
due
entirely
.v
n
to the
change
b.
is
The hot
R,
734 r/min
1066
kW at 200
is
9.55 Pin
9.55
X
(3.5)
10 000/200
= 478 N-m
)
The
greater than the
rated torque
7Vaicd
=
is
9.55 Pin
-
R
(234
9.55 X 110 000/1760
= 597 N m
is
+ T
)
- 234
(6.5)
Because Tx
is
less than r, atcd
,
we can
apply
x
Eq. 14.2.
\
=
1066 r/min.
is
=
rotor temperature
t,=
=
to 10
r/min
cold rotor resistance.
b.
1800
load.
=
33%
-
=
The torque corresponding
Tx =
1.33 /?„
hot rotor resistance
-
1800
734 r/min with increasing
therefore write
(RJR n
X
speed drops from 1000 r/min to
that the
(R x /R n )
0.03
Rx =
Note
are fixed; consequently,
We can
0.592
r/min
n
=
=
Consequently, the motor speed
final slips are
The voltage and torque
The
60/8
n
.v
the
The
is:
1.33 (234
= 108°C
+
23)
- 234
The
slip is
sx
= (1800 -
200)/ 1800
=
0.89
ELECTRICAL MACHINES AND TRANSFORMERS
308
we
All other conditions being fixed,
have, from
Rule
1
The
-
heat dissipated in the rotor during
the starting period (from zero speed to final rated
Eq. 14.2:
sx
=
*n
speed)
(TJTn ) (RJR n )
all
0.89
=
0.44 (478/300) (RJ2)
is
equal to the final kinetic energy stored
This rule holds
true, irrespective
of the stator volt-
age or the torque-speed curve of the motor. Thus,
and so
in
the revolving parts.
if
a motor brings a massive flywheel up to speed, and
rx =
Three 2.5
ft
n
wye-connected
tor circuit will
10
2.5
if
the energy stored in the flywheel
joules, the rotor will have dissipated
resistors in the ro-
enable the motor to develop
kW at 200r/min.
the
form of heat. Depending upon the
tor
and
its
is
then 5000
5000
joules in
size of the ro-
cooling system, this energy could easily
produce overheating.
14.7 Starting an induction
motor
14.8 Plugging an induction motor
High-inertia loads put a strain on induction motors
some
because they prolong the starting period. The starting
In
current in both the stator and rotor
and
is
high during
this
that the revolving field
lem. For motors of several thousand horsepower, a
prolonged starting period
may even
is
installed.
many
loads.
To
The
line voltage
may
fall
overload the
motor
riod, the
It
below normal
chanical
relieve the problem, induction motors are
rotor.
often started on reduced voltage. This limits the
During
its
speed to
power P m
is
fall.
The associated me-
entirely dissipated as heat in the
Unfortunately, the rotor also continues to re-
P
r
the line voltage drop as well as the heating rate of the
which
windings. Reduced voltage lengthens the start-up
Consequently, plugging produces
time, but this
is
whether the start-up time
remembering
is
long or short,
the following rule for a
it
is
motor
also
from the
heat
/"/?
(Fig.
is
plugged, the rotor
2
l
stator,
14. 10).
losses in the
may
is
locked.
melt the rotor bars or
overheat the stator winding. In this regard
Figure 14.10
a 3-phase induction motor
as
even exceed those when the rotor
high rotor temperatures
is
not loaded mechanically:
When
dissipated
Motors should not be plugged too frequently because
worth
that
is
rotor that
usually not important. However,
oppo-
in the
plugging pe-
acts as a brake.
ceive electromagnetic power
the motor, and consequently reduces
this
absorbs kinetic energy from the still-revolving
load, causing
seconds, thus affecting other connected
power drawn by
suddenly turns
site direction to the rotor.
transmission line feeding the plant where the motor
for
motor
industrial applications, the induction
load have to be brought to a quick stop. This
can be done by interchanging two stator leads, so
becomes a major prob-
interval so that overheating
its
R losses
are very high.
it is
worth
SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS
remembering the following
tions for a
Rule 2
motor
that
The
-
is
heat dissipated in the rotor during
is
three times the original kinetic energy of
the revolving parts.
all
ergy dissipated
original
it
produces
In effect, the en-
only equal to the
in the rotor is
energy stored
kinetic
that
is
does plugging.
far less heat than
the plugging period (initial rated speed to zero
speed)
The advantage of dc braking
rule for plugging opera-
not loaded mechanically:
309
in
the revolving
masses, and not three times that energy. The energy dissipated
in the rotor is
independent of the
magnitude of the dc current. However, a smaller
Example 14-4
A 100 kW, 60 Hz,
dc current increases the braking time. The dc cur175 r/tnin motor
1
is
coupled
to a
flywheel by means of a gearbox. The kinetic energy
of
is 300 kJ when the motor
The motor is plugged to a stop
the revolving parts
all
runs at rated speed.
and allowed
to run
up
to
1
175 r/min in the reverse
two or three times the rated current of
Even larger values can be used, prothe stator does not become too hot. The
rent can be
the motor.
vided that
braking torque
proportional to the square of the
is
dc braking current.
direction. Calculate the energy dissipated in the rotor if the
flywheel
is
the only load.
Example 14-5
A
Solution
During the plugging period, the motor speed drops
from
175 r/min to zero. The heat dissipated
1
rotor
is
3
erates to
X 300
kJ
= 900
nominal speed
kJ.
The motor then
kJ.
this
+ 300 =
1200
is
The
period
By reversing the speed this way, the total
dissipated in the rotor from start to finish
900
accel-
in the reverse direction.
energy dissipated in the rotor during
300
in the
25 kg
62 A.
24
heat
therefore
a.
14.9 Braking with direct current
high-inertia load can
also be brought to a quick stop by circulating dc
current in the stator winding.
Any two
having a
is
The dc
0.32
We
moment
of inertia of
between two
stator ter-
and the rated motor current
12,
want
total
resistance
to stop the
is
motor by connecting a
battery across the terminals.
Calculate
b.
its
V
.
is
kJ.
induction motor and
m2
minals
c.
An
50 hp, 1760 r/min, 440 V, 3-phase induction mo-
tor drives a load
The dc current in the stator
The energy dissipated in the rotor
The average braking torque if the stopping time
is 4 min
Solution
a.
The dc current
stator termi/
is
— EIR —
24/0.32
= 75 A
nals can be connected to the dc source.
The
in
direct current produces stationary N, S poles
This current
The number of poles created is equal
number of poles which the motor develops
to the
how
the
motor
ter-
When
the rotor
an ac voltage
is
sweeps past the stationary
induced
in the rotor bars.
The
field,
volt-
age produces an ac current and the resulting rotor
2
I
R
losses are dissipated at the expense of the
kinetic energy stored in the revolving parts.
motor
finally
comes
to rest
when
all
b.
The
kinetic energy in the rotor
and load
at
1760
is
5.48
X 10" 3
5.48
X
10" 3
- 424
kJ
Ek =
=
Jn
2
X 25 X 1760
(3.8)
2
The
the kinetic en-
ergy has been dissipated as heat in the rotor.
is
short.
r/min
minals are connected to the dc source.
slightly higher than the rated
not overheat, because the braking time
normally. Thus, a 3-phase, 4-pole induction motor
produces 4 dc poles, no matter
is
current of the motor. However, the stator will
the stator.
Consequently, the rotor absorbs 424 kJ during
the braking period.
ELECTRICAL MACHINES AND TRANSFORMERS
310
c.
The average braking torque
7"
can be calculated
An =
1760
-
factor 1.15. The allowable temperature
10°C higher than
from the equation
tors operating at
9.55 TAt/J
9.55
T=
19.2
T X
(3.14)
X
(4
vided. This
is
because even
Abnormal conditions
14.10
in the stator,
to internal
overheating of
the bearings, etc.) or to external conditions. External
problems may be caused by any of the following:
1
.
not
much
as 125 percent, as
if
recommended
the external frame
perature of the windings
Abnormal motor operation may be due
problems (short-circuit
overloads as
to carry
long as supplementary external ventilation
N-m
may
pro-
is
for long periods
is
cool, the tem-
be excessive.
14.12 Line voltage changes
The most important consequence of a line voltage
change is its effect upon the torque-speed curve of
the motor. In effect, the torque at any speed
Mechanical overload
mo-
normal load.
During emergencies a drip-proof motor can be
made
60)/25
then
rise is
that permitted for drip-proof
pro-
is
portional to the square of the applied voltage. Thus,
2.
Supply voltage changes
3.
Single phasing
4.
Frequency changes
the stator voltage decreases by
if
voltage drop
to the
We
will
examine
the nature of these
problems
in
the sections that follow.
According
motor
shall
±10%
of the nominal voltage, and for any frequency within
of the nominal frequency.
If
the voltage and
frequency both vary, the sum of the two percentage
changes must not exceed 10 percent. Finally,
designed
tors are
up
to
1000
to the
mo-
much
On
when
sea level. At higher altitudes the
the permissible limits
poor cooling afforded by the thinner
due
air.
Mechanical overload
as twice their rated
power
for short periods,
As
may
the line.
less than its rated value.
the motor
is
is
too high
running, the flux per pole will be
at full-load, this
current, with the result that the temperature increases
slightly
and the power factor
is
slightly reduced.
the 3-phase voltages are unbalanced, they can
produce a serious unbalance of the three
rents.
line cur-
This condition increases the stator and rotor
losses, yielding a higher temperature.
Although standard induction motors can develop as
much
drawn from
the other hand, if the line voltage
A voltage
un-
3.5% can cause the temperature to increase by 15°C. The utility company
should be notified whenever the phase-to-phase
line voltages differ by more than 2 percent.
balance of as
14.11
often produced during start-up, due
increases both the iron losses and the magnetizing
If
may exceed
is
starting current
above normal. For a motor running
to operate satisfactorily at altitudes
m above
temperature
all
heavy
at
A line
a result of the lower voltage, the starting torque
be
to national standards, a
operate satisfactorily on any voltage within
±5%
10%, the torque
every speed will drop by approximately 20%.
little
as
they should not be allowed to run continuously be-
yond
their rated capacity.
heating,
Overloads cause over-
which deteriorates the insulation and
duces the service
life
of the motor. In practice, the
overload causes the thermal overload relays
starter
fore
its
box
re-
to trip, bringing the
motor
in the
to a stop be-
temperature gets too high.
Some
drip-proof motors are designed to carry a
continuous overload of 15 percent. This overload
capacity
is
shown on
the nameplate by the service
14.13 Single-phasing
If
one
or
if
line
of a 3-phase line
a fuse
ning, the
is
accidentally opened,
blows while the 3-phase motor
machine
is
run-
will continue to run as a single-
phase motor. The current drawn from the remaining
two
lines will
almost double, and the motor will be-
gin to overheat.
motor
will
The thermal
eventually
trip
relays protecting the
the
circuit-breaker,
thereby disconnecting the motor from the
line.
SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS
%
311
breakdown torque
250
'X
200
3-phase
P"
^
150
i
^pull-up torq ue
nominal torque T
100
full
—
load
i\
i
50
sirigle-
phase
—
i
i
i
\
\
\
\
\
!
\
i
40
20
80
60
100
%
speed
\
Figure 14.11
Typical torque-speed curves
when
a 3-phase squirrel-cage motor operates normally and
when
it
operates on
single-phase.
The torque-speed curve
a 3-phase
down
seriously affected
torque decreases to about
value,
all.
is
when
motor operates on single phase. The break-
40%
and the motor develops no
of
its
original
starting torque at
Consequently, a fully loaded 3-phase motor
simply stop
if
one of its
lines
is
resulting locked-rotor current
normal 3-phase
enough
LR
current.
may
suddenly opened. The
is
about
It
is
to trip the circuit breaker or to
90%
of the
therefore large
blow the
1
4. II
each other closely
single-phase
until the
breakdown
is
50 Hz may cause problems when they
nected to a 60
torque approaches the
in
some
to gear
1
large distribution system, except during a
is
major dis-
However, the frequency may vary
electrical
signif-
energy
generated by diesel engines or gas turbines. The
in a hospital, the electrical
system on a ship, and the generators
camp, are examples of
this
is
in
a
lumber
type of supply.
The most important consequence of
change
may
20%
not be acceptable
we
either have
motor speed or supply an expen-
50 Hz source.
well on a 60
Hz
line,
but
terminal voltage should be raised to 6/5 (or
its
of the nameplate rating. The
20%)
torque
is
new breakdown
then equal to the original breakdown
torque and the starting torque
is
only slightly re-
remain satisfactory.
A 60 Hz
Important frequency changes never take place on a
emergency power supply
this
duced. Power factor, efficiency, and temperature
torque.
on isolated systems where
and
A 50 Hz motor operates
rise
icantly
the
are con-
system. Everything runs
applications. In such cases
down
sive auxiliary
14.14 Frequency variation
turbance.
Hz
faster than normal,
fuses.
shows the typical torque-speed curves
of a 3-phase motor when it runs normally and when
Note that the curves follow
it is single-phasing.
Fig.
Machine tools and other motor-driven equipment imported from countries where the frequency
a frequency
the resulting change in motor speed:
if
the
frequency drops by 5%, the motor speed drops by 5%.
but
its
motor can also operate on
83%) of its nameplate
and
fore,
a
50 Hz
line,
terminal voltage should be reduced to 5/6 (or
value.
The breakdown torque
same as be-
starting torque are then about the
and the power
ture rise
factor, efficiency,
and tempera-
remain satisfactory.
14.15 Induction motor operating
as a generator
Consider an
electric train
cage induction motor that
wheels.
As
powered by a
is
directly
the train climbs up a
hill,
squirrel-
coupled
the
to the
motor
will
ELECTRICAL MACHINES AND TRANSFORMERS
312
Figure 14.12
The
makes
electric train
the round
trip
be-
tween Zermatt (1604 m) and Gornergrat
(3089 m) in Switzerland. The drive is provided by four 3-phase wound-rotor induction motors, rated 78 kW, 1470 r/m, 700 V,
50 Hz. Two aerial conductors constitute
phases A and B, and the rails provide
phase C. A toothed gear-wheel 573 mm in
diameter engages a stationary rack on the
roadbed to drive the train up and down the
steep slopes. The speed can be varied from
zero to 14.4 km/h by means of variable resistors
in
the rotor
circuit.
The
rated thrust
is
78 kN.
(Courtesy of ABB)
run
than synchronous speed, devel-
at slightly less
overcome both
oping a torque sufficient
to
and the force of
At the top of the
gravity.
level ground, the force of gravity
friction
hill,
to overcome the
The motor runs at
and the motor has only
into play
friction of the rails
light load
and the
and very close
What happens when
to
air.
move
begins
it
to
rotate
is
to a
it
gasoline engine (Fig.
as the engine speed exceeds the
synchronous speed, the motor becomes a generator,
delivering active
it
is
power P
to the electrical
system
connected. However, to create
its
to
mag-
in-
speed. Thus, a higher engine speed produces a
mo-
develops a counter torque that opposes the
same
As soon
effect as a
coupled to the
speed. However, as soon as this takes place, the
tor
to
above synchronous
crease in speed. This torque has the
and coupling
line
14.13).
motor has to absorb reactive power
power can only come from the ac line, with
the result that the reactive power Q flows in the opposite direction to the active power P (Fig. 14. 3).
The active power delivered to the line is directly proportional to the slip above synchronous
the train begins to
accelerate and because the motor
phase
which
synchronous speed.
downhill? The force of gravity causes the train
wheels,
on
no longer comes
We can make an asynchronous generator by connecting an ordinary squirrel-cage motor to a 3-
netic field, the
Q. This
1
brake. However, instead of being dissipated as heat,
mechanical braking power
the
3-phase line
in the
is
3-phase system
returned to the
form of electrical energy.
An
in-
P
duction motor that turns faster than synchronous
speed
acts, therefore, as a generator,
it
^
gasoline
converts the
engine
mechanical energy
and
this
chine
is
energy
receives into electrical energy,
returned to the
induction
line are rarely
motors
Such
a
ma-
a
motor
In cranes, for
to run
running
off
it
to the line.
the
G
squirrel-cage
induction motor
applications that
above synchronous speed.
motor receives power from
f
a
example, during the lowering cycle,
"load" and returns
ooo
QHill
used to drive trains (Fig.
14. 12), there are several industrial
may cause
the
line.
called an asynchronous generator.
Although
3-phase
is
it
mechanical
Figure 14.13
Gasoline engine driving an asynchronous generator
connected
to a
3-phase
opposite directions.
line.
Note that
Pand Qflow
in
SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS
313
Solution
a.
The apparent power drawn by
when it operates as a motor is
=
=
=
S
1.73
X 440 X
31.2
kVA
=
for
put
reached
is
very small
at
3%.
The reactive power may be supplied by a group
of capacitors connected to the terminals of the moWith
this
is
slightly
When
least
r/min
/=
produces
pni\2{)
=
4
a
at
frequency
a speed of
slightly
X 2400/120 = 80
The terminal voltage of
with the capacitance, but
its
less
the
less
connected
reactive
when
power
as
2400
than
Consequently, the capaci-
/c
phase
is
= QIE = 5700/440
=
13
A
magnitude
is
limited by
is
The capacitive reactance per phase
insuffi-
Xc ~ Ell =
= 34 il
at least as
the machine normally ab-
is
440/13
The capacitance per phase must be
at least
operating as a motor.
l/2ir/X c
=
1/(2tt
We
=
78 [xF
40 hp, 1760 r/min, 440 V, 3-phase
squirrel-cage induction motor as an asynchronous
to use a
The
motor
84%.
rated current of the
and the full-load power factor
is
is
41 A,
Figure 14.15 shows
connected. Note that
tive
a.
The voltage
in delta.
Example 14-6
generator.
2
because the capacitors are
C=
wish
26.2
the generator increases
capacitor bank must be able to supply
much
2
5.7 kvar per phase.
440
tive current per
generator voltage will not build up. The
sorbs
is
V
is
Hz.
saturation in the iron. If the capacitance
cient, the
power absorbed
machine operates as an asynchronous
17^-3 =
per phase
than that corresponding to the speed of rotation.
Thus, a 4-pole motor driven
reactive
generator, the capacitor bank must supply at
arrangement we can supply a 3-phase
The frequency generated
kW
1
load without using an external 3-phase source (Fig.
14. 14).
is
0.84
= V31.2 2 = 7 kvar
slips, typically less
than
tor.
26.2
Q = \ S2 - P
However, the rated out-
41
power absorbed
active
P = S cos B
= 31.2 X
The corresponding
greater electrical output.
machine
V3 EI
The corresponding
Figure 14.14
Capacitors can provide the reactive power for any
asynchronous generator. This eliminates the need
a 3-phase external source.
the
Calculate the capacitance required per phase
X 60 X
how
if
34)
the generating system
is
the load also absorbs reac-
power, the capacitor bank must be increased to
if
provide
it.
the capacitors are connected in delta.
b.
At what speed should the driving engine run
generate a frequency of 60
Hz?
to
b.
The driving engine must
turn at slightly
more
than synchronous speed. Typically, the slip
ELECTRICAL MACHINES AND TRANSFORMERS
314
14.16 Complete torque-speed
characteristic of an
induction machine
30
= =
Q
7kVar
kW
t
^
OOP
4D=
1840 r/min
I
^>25 kW
On
j^r
B
We
have seen
that a
3-phase squirrel-cage induction
motor can also function as a generator or
as a brake.
— motor,
generator,
These three modes of operation
and brake
— merge
into each other, as
can be seen
from the torque-speed curve of Fig.
curve, together with the adjoining
Figure 14.15
See Example 14-6.
grams,
14. 16.
power flow
illustrates the overall properties
This
dia-
of a 3-phase
squirrel-cage induction machine.
should be equal to the full-load
machine operates as
slip
=
when
the
a motor. Consequently,
1800
= 40
The engine should
slip
-
1760
r/min
We see,
for example, that when the shaft turns in
same direction as the revolving field, the induction machine operates in either the motor or the
generator mode. But to operate in the generator
the
mode, the
therefore run at an approxi-
must turn
mate speed of
n
shaft
must turn
faster than
at less
than synchronous speed.
Finally, in order to operate as a brake, the shaft
- 800 + 40
= 1840 r/min
1
must
turn in the opposite direction to the revolving
flux.
V-
u
BRAKE
synchronous
speed. Similarly, to operate as a motor, the shaft
n
m
r
|
speed
stator
rotor
+ 2 h
i
rotor
stator
T = torque developed
stator
n = speed of rotation
by the machine
Figure 14.16
Complete torque-speed curve
of
a 3-phase induction machine.
n s = synchronous speed
of the revolving field
SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS
14.17 Features of a wound-rotor
induction motor
sult,
it
imum
possible for the motor to develop
is
far,
we have directed our attention
cage induction motor and
generator and brake.
motor
is
the
its
to the squirrel-
that this type
is
of
in industry.
However, the wound-rotor induction motor has cer-
trial
1
applications.
make
attractive in special indus-
it
These may be
listed as follows:
final
We
have already seen
that for a given load, an in-
crease in rotor resistance will cause the speed of an
induction motor to
we
Thus, by varying the exter-
fall.
wound-rotor motor we can obtain
want, so long as
nous speed. The problem
Frequency converter
We now examine
in the shortest
14.19 Variable-speed drives
any speed
Variable-speed drives
3.
speed can be reached
possible time.
nal resistors of a
Start-up of very high-inertia loads
.
2.
max-
related properties as a
The reason
one most frequently used
tain features that
its
torque during the entire acceleration period.
Thus, the
So
315
as heat in the resistors
it
is
that the
is
makes
for a very inefficient
system, which becomes too costly
these applications.
below synchro-
power dissipated
when
the motors
have ratings of several thousand horsepower.
14.18 Start-up of high-inertia loads
around
this
problem by connecting the
We get
slip-rings to
an electronic converter. The converter changes the
We
recall that
whenever a load
is
brought up to
speed by means of an induction motor, the energy
dissipated in the rotor
is
equal to the kinetic energy
power
at
system (Fig.
this
14. 17).
imparted to the load. This means that a high-inertia
speed control system
load will produce very high losses in the rotor, caus-
that
become excessively
ing
it
the
wound-rotor motor
in the
to
is
hot.
itself
is
dissipated
is
is
a result, such a variable-
very efficient,
is lost in
the
form of
in the
sense
heat.
that the external resistors
can be varied as the motor picks up speed.
14.20 Frequency converter
A conventional
remains cool.
Another advantage
power
As
The advantage of
that the heat
external resistors connected to the slip-rings.
Thus, the rotor
little
into power at line frepower back into the 3-phase
low rotor frequency
quency and feeds
As
a re-
wound-rotor motor may be used as
a frequency converter to generate a frequency different
from
that
of the
utility
company. The
stator of
Figure 14.17
The water supply
in
the City of Stuttgart, Germany,
provided by a pipeline that
11 0
km
long.
Constance
is
in
is
1
.6
m
in
is
diameter and
The water is pumped from Lake
the Alps. The pump in the background
driven by a wound-rotor induction motor rated at
3300 kW, 425 to 595 r/min, 5 kV, 50 Hz. The variable
speed enables the water supply to be varied according to the needs of the city.
The enclosed motor housing seen in the foreground contains an air/water heat exchanger that
uses the 5°C water for cooling purposes. During
connected to the slipup to speed the sliprings are connected to an electronic converter which
feeds the rotor power back into the line.
(Courtesy of Siemens)
start-up, liquid rheostats are
rings, but
when
the motor
is
ELECTRICAL MACHINES AND TRANSFORMERS
316
R
n
R
R
wound-rotor
induction
"motor"
Figure 14.18
Wound-rotor motor used as a frequency converter.
Figure 14.19
Power flow in a frequency converter when the output
frequency
machine
the wound-rotor
line,
and the rotor
by a motor
is
M (Fig.
is
driven
14.
1
connected
at
rotor supplies
power
E 2 and
to the 3 -phase load at a voltage
both of which depend upon the
slip.
frequency / 2
Thus, accord-
Calculate
a.
,
b.
we have
ing to Eqs. 13.3 and 13.4,
greater than the line frequency.
an appropriate speed
The
8).
is
to the utility
The
The
is
turns ratio of the stator to rotor windings
when the rotor
same direction as
rotor voltage and frequency
driven
720 r/min
at
in the
the revolving field
h
=
(13.3)
sf
c.
(13.4)
In general, the desired
times that of the
13.3, in
utility
frequency
is
company. According
be positive and greater than
1
.
As
is
to Eq.
rotor voltage and frequency
driven
a.
The
turns ratio
a result, the shaft
a
ing flux.
the frequency converter
is
when
to the
the rotor
revolving
Solution
must be rotated against the direction of the revolv-
The operation of
720 r/min opposite
at
field
two or three
order to attain this frequency the slip must
The
b.
The
slip at
is
= E]/Eoc = 2300/500
= 4.6
720 r/min
is
-
(1
then identical to that of an induction motor operating as a brake.
However,
the
dissipated as heat in the rotor,
power P ]n usually
is
now
.v
=
available to
supply power to the load. The converter acts as a
generator, and the active
in
Fig.
14.19.
power flow
Note how similar
is
as
shown
this
is
to the
power flow when an induction motor runs
=
The
(;? s
n)ln,
=
0.6
rotor voltage at
720 r/min
E 2 = sEoc =
= 300 V
as a
800 - 720)/l80()
0.6
is
X 500
brake (Fig. 14.16).
The
Example 14-7
A3-phase wound-rotor induction motor has
of 150 hp
(—110 kW), 1760
r/min, 2.3 kV,
60 Hz.
ro-
The
ro-
tor voltage
tor
is
between the slip-rings
is
500
V.
driven by a variable-speed dc motor.
is
f2 = sf= 0.6 X 60
= 36 Hz
a rating
Under locked-rotor conditions, the open-circuit
rotor frequency
c.
The motor speed is considered to be negative
( — ) when it turns opposite to the revolving
field. The slip at —720 r/min is
SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS
=
(n - n)ln
= (1800 - (-720))/1800
= (1800 + 720)/ 1800
s
s
The converter must
s
at a
speed of 1800 r/min. The negative sign indicates that the rotor must run opposite to the re-
volving
1.4
The
therefore be driven
317
field.
The induction motor driving
the
converter must, therefore, have a synchronous
rotor voltage and frequency at
speed of 1800 r/min.
720 r/min
b.
are
E2 =
=
sE„ c
1
X 500
.4
The
rotor delivers an output of
corresponds to Ppated in the rotor,
= 700 V
f2 = sf= 1.4 X 60
= 84 Hz
P
useful power delivered
The power P r transferred
is
]V
to a load (Fig. 14.20).
from the
stator to the rotor
P,
60 kW. This
but instead of being dissi-
= P vh =
20
is
60/3
(13.7)
kW
Example 14-8
We
wish
rotor
60
to
use a 30
kW, 880
motor as a frequency converter
kW at an
The power
60 Hz wound-
r/min,
converter
(F) to generate
c.
frequency
is
60 Hz,
calculate the following:
a.
The speed of the induction motor (M)
b.
The
Fig. 14. 19
active
power delivered
to the stator of the
c.
d.
Will the frequency converter overheat under
Solution
To generate
1
80 Hz the
slip
the small cop-
Fig. 14.20,
we can
see
how
must be
rotor receives 20 kW of
power from the stator and 40 kW of
mechanical power from the driving motor M.
The rotor converts this power into 60 kW of
electrical power at a frequency of 180 Hz.
Induction motor M must therefore have a rating of 40 kW, 60 Hz, 800 r/min.
summary, the
1
f2 =
(13.3)
sf
180
X 60
s
load
from which
f
s
The
stator
is still
=
= 60 Hz
3
fed from the 60
Hz
line,
con-
sequently, the synchronous speed of the converter
ven
is
at a
900
r/min.
The converter must be
dri-
speed n given by
s
=
(/? s
3
=
(900
-
h)//? s
-
(13.2)
frequency
converter
driving
«)/900
motor
from which
Figure 14.20
n
= -1800
r/min
the
into (and out of) the con-
electrical
these conditions
a.
kW plus
the stator.
verter.
In
The power of the induction motor (M)
and
power flows
active
frequency converter
in
The remaining power input to the rotor amounting to (60 - 20) = 40 kW, is derived from the
mechanical input to the shaft. By referring to
that dri-
ves the frequency converter
equal to 20
per and iron losses
approximate frequency of 180 Hz (see
Fig. 14.18). If the supply-line
input to the stator of the frequency
is
See Example
14-8.
3
1
d.
ELECTRICA L MA CHINES A ND TRA NSEORMERS
8
The
because the 20
its
14-10
stator of the converter will not overheat
kW
it
absorbs
much
is
nominal rating of 30 kW. The rotor
overheat either, even though
The increased power
arises
from the
the voltage induced in the rotor
higher than
at standstill.
The
is
and the rotor
stator
how much
60 kW.
fact that
14-11
more
at
A
is
By
breakdown torque and
are the
nected to a 520
180
V
line.
Explain
is
con-
how
the
following parameters are affected:
twice
effective,
a.
Locked-rotor current
b.
Locked-rotor torque
c.
No-load current
d.
No-load speed
e.
Full-load current
f.
Full-load
iron losses in the stator are normal.
Questions and Problems
g.
14-12
Practical level
50 r/min
1
line.
3-phase squirrel-cage induction motor
60 Hz, consequently, the
is
1
having a rated voltage of 575 V,
iron losses in the
is
V,
locked-rotor torque reduced?
probably not overheat. The
will
frequency
is
60 Hz,
will not
rotor will be high because the frequency
normal speed, the cooling
50 hp, 440
connected to a 208 V, 3-phase
three times
Hz; however, because the rotor turns
standard squirrel-cage induction motor
less than
delivers
it
A
rated at
a.
power
factor
Full-load efficiency
Referring to Fig. 14.6,
if
we
eliminated the
gearbox and used another motor directly
14-1
What
is
between a drip-proof
the difference
coupled
motor and an explosion-proof motor?
14-2
What
is
the approximate
life
expectancy
of a motor?
14-3
Explain
a
NEMA Design D
unsatisfactory for driving a
14-4
Identify the
b.
14-13
why
to the load,
motor
How many
Draw
poles would the motor have?
the
pump.
motor components shown
14.5).
in
Give the values of the LR, pull-up,
and breakdown torques and the correflow of active power
3 -phase induction
a.
b.
14-6
As
As
a
motor when
it
operates
14-14
motor
A 300
b.
14-8
14-9
to start
If the line
on such a line?
b.
c.
14-15
relates to induction motors.
can bring an induction motor to a
quick stop either by plugging
from
method produces the
citing the stator
in the
60 Hz
it
motor? Explain.
Which
amount of
2
I
R
the
losses.
voltage then drops to 1944 V,
The new speed, knowing
We
The new power output
2
The new I R losses in the
heat
that the load
rotor
wish to make an asynchronous generator
using a standard squirrel-cage induction
tor rated at
40
(Fig. 14. 14).
or by ex-
a dc source.
least
590 r/min. Calculate
torque remains the same
for the following applications:
Give some of the advantages of standard-
We
V, 3-phase,
calculate the following:
A saw in a lumber mill
A variable speed pump
it
2300
if
type of ac motor would you recom-
ization as
and r/minl.
approximate value of the rotor
a.
a.
hp,
full-load speed of
Will a 3 -phase motor continue to rotate
What
mend
lbf
squirrel-cage induction motor turns at a
a brake
motor be able
[ft
in a
one of the lines becomes open? Will the
14-7
squirrel-cage induction
motor, rated at 30 hp, 900 r/min (see Fig.
sponding speeds
Show
power
its
the typical torque-speed curve of a
NEMA Design C
is
Fig. 14.3.
14-5
what would
output have to be [hp]?
hp,
208
V,
mo-
870 r/min, 60 Hz
The generator
is
driven
at
2100
r/min by a gasoline engine, and the load
consists of three 5 12 resistors connected in
wye. The generator voltage builds up when
three 100 fxF capacitors are connected in
SELECTION AND APPLICATION OF THREE PHASE INDUCTION MOTORS
wye
is
a.
across the terminals.
520
c.
d.
The
tor
e.
voltage
14-19
is
A 30 000
hp—
Advanced
14-20
best suited to drive the
60 Hz
hp, 13.2 kV, 3-phase,
A
plant.
The motor runs
at
a.
an
it
has an efficiency of 98.
LR
are respectively 0.7
1
comThe motor
power
drives the
speed of 4930 r/min.
% and a
c.
d.
e.
14-21
in
Which
A
The
is
of the two rotors will be the hottest,
60 Hz, drives
rotor
is
in
speed?
a belt conveyor.
connected
the slip rings
in
wye and
a.
b.
is
530
V. Calculate the
The
The
rotor
winding resistance per phase
resistance that must be placed in series
with the rotor (per phase) so that the motor
knowing
14-22
A
150 hp,
40 hp
at a
speed of 600 r/min,
that the line voltage
water tempera-
is
load, close to
its
is
running
The motor and compressor
in
Problem
at 1.3
14-16 are started on reduced voltage,
the acceleration period
compressor has
a
The squirrel-cage
2
000 lb -ft
a J of 18
How
What
long will
is
clocked
(
1
.2 pu),
is
equal to
calculate the
b.
The magnitude of the plugging torque
The moment of inertia of the rotor
In
Problem 14-22 calculate the energy
a.
motor
rotor alone has
is
that the torque exerted
14-23
.
it
dissipated in the rotor during the plugging
take to bring the motor
and compressor up
b.
no-
following:
of inertia of
lb ft~ referred to the
Assuming
the starting torque
0.25 pu. The
is
moment
s.
during the plugging interval
and the average starting torque during
a.
at
synchronous speed of
reversed, and the stopping time
shaft.
kV
1200 r/min. The stator leads are suddenly
exchanger.
130 000
2.4
165 r/min, 440 V, 60 Hz,
1
3-phase induction motor
ture as the water flows through the heat
14-18
the
following;
cooled by
350 gallons (U.S.) of water
through the heat exchanger per minute.
has the shortest acceleration
nominal open-circuit voltage between
circulating
Calculate the increase
motor and,
3-phase, wound-rotor induction motor
2.3 kV,
pu and 4.7 pu.
Problem 14-16
D
class
.4
this
having a rating of 150 hp, 1760 r/min,
torque and current
The full-load current
The total losses at full load
The exact rotor TR losses if the windage
and friction losses amount to 62 kW
The LR current and torque
The torque developed at the compressor shaft
The motor
1
1800 r/min. Could
to
Which motor
will deliver
14-17
inertia load of
after reaching the no-load
Calculate the following:
a.
B induction
hp. squirrel-cage. Design
1
time from zero to 1200 r/min?
b.
by means of a gearbox,
The
motor deliver
if so,
exact full-load speed of 1792.8 r/min and
factor of 0.90.
[hp] can the
motor be replaced by a
a turbo compressor in a large oxygen-
at a
should be used, and what
level
kgnr, from 0
air-to-water cooled induction motor drives
pressor
line.
motor accelerates an
manufacturing
kW,
without overheating?
following gasoline engines are
which one
10
at
to be con-
approximate speed of the motor?
What power
b.
Hz
line voltage
will be the
generator?
b.
What
a.
stator current
If the
induction motor rated
nected to a 60
bank
available— 30 hp, 100 hp, and 150
14-16
A 3-phase
1450 r/min, 380 V, 50 Hz has
The approximate frequency generated
active power supplied to the load
reactive power supplied by the capaci-
The
The
b.
If the line
V, calculate the following:
to speed, at
interval.
no-load?
the energy dissipated in the rotor
during the starting period [Btu]?
14-24
A 3-phase,
rating of
8-poIe induction motor has a
40
hp, 575 V.
60 Hz.
It
drives a
320
ELECTRICAL MACHINES AND TRANSFORMERS
steel
flywheel having a diameter of 3
inches and a thickness of 7.875
1
.5
The
in.
Industrial application
14-27
torque-speed curve corresponds to that of
D
a design
a.
in Fig. 14.5.
of inerlia
d.
Calculate the rated speed of the motor and
Motor B: 75
Draw
[ft-lbf].
hp,
900
day.
How
erates about 6 hours per day.
In
b.
Using Eq.
time.
r/min; lubricate
Motor A runs continually, 24 hours per
Motor B drives a compressor and op-
at 0, 180,
Problem 14-24 calculate the average
a.
lubricate
every 10 000 hours of running time.
360.540. 720, and810r/min.
14-25
to be
40 hp
the torque-speed curve of the
motor and give the torques [N-m]
motor have
Motor A: 75 hp, 3550 r/min;
every 2200 hours of running
|lb-ft~|.
Calculate the loeked-rotor torque
a
its
the corresponding torque [ft-lbf].
c.
in
following schedule applies to two motors:
Calculate the mass of the flywheel and
moment
b.
motor given
The bearings
greased regularly, but not too often. The
often
should the bearings of each motor be
torque between 0 and 180 r/min.
3.
r/min,
c.
greased per year?
14 calculate the time required
to accelerate the
flywheel from 0 to 180
14-28
mium
the flywheel at 180 r/min.
540
that this
N-m
in
14-29
A standard 40
hp motor, similar
mass of
line current
The energy required
to
to
is
14-30
the av-
if
A constant
has windings similar to those shown
60 kg
minals
to
make
tion
the
min]
when
is
The
resistance
V
in Fig.
is
and 2
1
measured between
12
il.
in the
ter-
What
high-speed connec-
resistance
would you
expect to measure between terminals 4 and
Assuming
energy
horsepower, 2-speed induction
motor rated 2 hp, 1760/870 r/min, 460
Gornergrat |MJ]
|
mo-
climb from Zermatt
The minimum time required
trip
g.
one
the energy sav-
ings that accrue to the high-efficiency
14.8.
f.
to the
Problem 14-28, costs $1723
82%. Calculate
factor of
at full-load
the loaded train
erage weight of a passenger
e.
in
tor during the 3-year period.
The approximate transmission
when the motors are operating
total
is
and has an efficiency of 90.2% and power
The speed of rotation of the gear wheel
when the train moves at 9 miles per hour
The speed ratio between the motor and the
The
at full-
$0.06/kWh.
cost of energy
described
gear wheel
d.
$2243, runs
during a 3-year period, knowing that the
Calculate the following:
c.
at
Calculate the cost of driving the motor
addition to
The train in Fig. 14. 12 has a mass of
78 500 lb and can carry 240 passengers.
b.
and an efficiency of 93.6%.
load 12 hours per day, 5 days a week.
the flywheel load.
a.
pre-
time the load exerts a fixed
counter-torque of 300
14-26
86%
The motor, priced
knowing
r/min.
V, 3-phase,
energy induction motor has a power
factor of
d. Calculate the time required to accelerate
the flywheel from 0 to
1780 r/min, 460
60 Hz, drip-proof Baldor Super E
assuming no other load on the motor.
Using Eq. 3.8 calculate the kinetic energy
in
A 40 hp,
that
80 percent of the
the train
is
going uphill and that 80
percent of the mechanical energy
verted to electrical energy
6
electrical
converted into mechanical energy
is
14-31
A
in the
low-speed connection?
150 hp
7
1
175 r/min, 460 V, 3-phase,
60 Hz induction motor has the following
recon-
properties:
when going
downhill, calculate the total electrical en-
ergy consumed during a round
trip
[kW-h].
no-load current: 71
A
full-load current: 183
A
SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS
locked-rotor current:
1
550
A full-load torque:
886
breakdown
2552
A
a.
ft-lbf
b.
torque:
ft-lbf
A
power
factor:
from the main panelboard
motor, 850
elboard
is
ft
away. The voltage
480
V
ture of the cable
to the
at the
mo-
the
motor
is
impedance
is
purely re-
approximate current
started up across the line.
Estimate the resulting starting torque.
d.
Compare
it
with the rated starting torque,
percentagewise.
pan-
and the average temperais
the cable
c.
3-conductor 250 kcmil copper cable
stretches
Assuming
when
32%
circuit of the
under locked-rotor conditions.
sistive, calculate the
locked-rotor torque: 1205 ft-lbf
locked-rotor
Determine the equivalent
tor
321
estimated to be 25°C.
i
-32
In
Problem 14-3
1
express the currents and
torques in per-unit values.
Chapter 15
Equivalent Circuit of the
Induction Motor
15.0 Introduction
secondary windings
cal
—one
The preceding three chapters have shown that we
single primary winding
can describe the important properties of squir-
we want
if
pensible. In this chapter'"
circuit
and observe
tor
Finally,
We
indis-
We
low-power and high-power mo-
we develop
its
proper-
is
the
same
in
equivalent
cir-
as that of a transformer,
Chapter
10, Fig. 10.20.
assume a wye connection
E„
=
r\
—
stator
a-,
=
stator leakage reactance
x2
=
rotor leakage reactance
r2
—
rotor winding resistance
is
very simi-
for the stator
The
and
circuit para-
identical
who
winding resistance
externa] resistance, effectively connected
slip-ring
and the neutral of the
rotor
primary windings and 3 identi-
This chapter can be skipped by the reader
source voltage, line to neutral
between one
construction to a 3-phase transformer. Thus, the
to
)
meters, per phase, are identified as follows:
the equivalent circuit of an
3-phase wound-rotor induction motor
have lime
1
acts exactly like
its
their basic differences.
The wound-rotor induction
motor
motor has 3
5.
the rotor, and a turns ratio of 1:1.
Rx =
lar in
1
previously developed
under load.
15.1
A
cuit (Fig.
then analyze the
asynchronous generator and determine
ties
it
a conventional transformer, and so
we develop the equivalent
from basic principles.
characteristics of a
is
On
consider a
and a single secondary wind-
When the motor is at standstill,
to
gain even a better understanding of the properties of
the motor, an equivalent circuit diagram
we can
ing in analyzing the behavior of the motor.
rel-cage and wound-rotor induction motors without
using a circuit diagram. However,
each phase.
set for
account of the perfect symmetry,
Xm =
magnetizing reactance
Rm =
resistance
does not
iron losses
whose
losses correspond to the
and windage and
friction losses
study the more theoretical aspects of induction
T=
motor behavior.
322
ideal transformer
having a turns ratio of
1
:
EQUIVALENT CIRCUIT OF THE INDUCTION MOTOR
323
Figure 15.1
Equivalent circuit of a wound-rotor induction motor at standstill.
Figure 15.2
Approximation of the equivalent
for
3-phase trans-
the case of a conventional
In
acceptable
circuit is
load current
true: / 0
may
/
.
p
compared
negligible
is
However,
in a
motor this
be as high as 40 percent of
the air gap. Consequently,
we cannot
to the
no longer
is
shown
we
in Fig.
can
1
E2 U
,
on the primary and secondary side of
in the
secondary winding will become
these
new
to Ei (the turns ratio
This greatly simplifies the equa-
induced
is
Fig.
1
5.2
motor when the rotor
How
when
affected
the
Suppose the motor runs
the rotor speed
is
ns
(
I
—
motor
starts
a slip
at
s),
s,
where
meaning
/7
S
is
The frequency
locked.
is
turning?
that
the syn-
chronous speed. This will modify the values of
be used.
hp the exact
circuit of Fig.
1
5.
1
if
the
slip
is
v,
l
;
is
sfand
I
be equal
motor were
sta-
the actual voltage
sEi
this
changes
the
imped-
ance of the secondary leakage reactance from
j.\ 2
to
jsx 2 Because resistors are not frequency-sensitive,
.
the values of r2 and
the
two together
tance
For motors under
is
shows
5.3
E2 would
)
E2 =
a true rep-
is
resentation of the
is it
is
But because the
compromising accuracy.*
/
Directing our attention to the secondary side, the
amplitude of the induced voltage
tions that describe the behavior of the motor, with-
out
1
where
operating conditions.
tionary.
it
sfi
the frequency of the source E„. Fig.
to the input terminals, as
shift
5.2.
and
the ideal transformer T. Furthermore, the frequency
because of
p
eliminate the
magnetizing branch. However, for motors exceeding 2 hp,
I
x
we would be justified in removing the magnetizing branch composed of jX m and R m because
former,
the exciting current /0
motors above 2 hp.
R2
,
R K remain
to
the same. Let us
form a single secondary
lump
resis-
given by
should
R,
=
(I5.l)
ELECTRICAL MACHINES AND TRANSFORMERS
324
external
resistance
frequency
frequency
f
sf
Figure 15.3
Equivalent circuit of a wound-rotor motor
the stator
in
The
is
details of the
Fig. I5.4a,
it
is
running at a
secondary circuit are shown
and the resulting current
^~ =
when
But the frequency of the voltages and currents
f.
sE, Z_
a-£,
„
A-
=
•
R2
VRj +
+js.x 2
slip s.
in
The frequency
the rotor
of the voltages
in
I 2 is
^
- B
/U'2
(15.2)
2
Cvx,)
where
total
(3
=
resistance
arctan sx 2 IR 2
(
The corresponding phasor diagram
Fig.
5.4b.
1
and currents
is sf.
diagram
It is
1
5.3)
circuit
shown
is
of rotor
in
important to realize that this phasor
relates to the
frequency sf Consequently,
frequency = sf
it
cannot be integrated into the phasor diagram on the
primary
there
is
side,
fect, the
.
/,
and
/,
(frequency /)
absolute value of
that of U.
/2
is
a direct relationship between
in the rotor
£, and
where the frequency
Nevertheless,
I2
(frequency
Furthermore, the phase angle
is
exactly the
same
(a)
sf)
in the stator. In ef-
exactly the
is
/,
/
as that
(3
same
as
between
between
E2
and
This enables us to draw the phasor diagram for £,
and
/,
as
shown
in Fig. 15.5.
To summarize:
1
.
The
effective value of
tive value of
2,
is
/,
equal to the effec-
even though
their frequencies
(b)
are different.
2.
The
effective value of
tive value of
3.
E2
E
equal to the effec-
is
{
divided by the slip
The phase angle between
as that between E 2 and I 2
E
.
and
]
/]
Figure 15.4
a.
,s.
is
the
same
b.
Equivalent circuit of the rotor; E2 and / 2 have a frequency sf
Phasor diagram showing the current lagging behind the voltage by angle
(3.
EQUIVALENT CIRCUIT OF THE INDUCTION MOTOR
we can
Thus, on the primary side
= U =
/,
R2 +
325
write
(15.4)
jsx 2
Therefore,
(15.5)
R2
Zn
jx2
S
The impedance Z2 seen between
nals
1,
2 of the ideal transformer
the primary termiis,
therefore,
Figure 15.6
Equivalent
circuit of
a wound-rotor motor referred to
the primary (stator) side.
Z2 =
1
=
+
jx 2
(15.6)
/,
As
a result,
we can
This equivalent circuit of a wound-rotor induc-
simplify the circuit of Fig.
motor
The leakage reactances jx Jx 2 can now be lumped together to create
tion
a single total leakage reactance jx.
motor
15.3 to that
shown
in Fig. 15.6.
it
is
is
so similar to that of a transformer that
not surprising that the wound-rotor induction
{
total
It is
equal to the
leakage reactance of the motor referred to the
stator side.
is sometimes called a rotary transformer.
The equivalent circuit of a squirrel-cage induction
motor is the same, except that R 2 is then equal to the
equivalent resistance r2 of the rotor alone referred to
the stator, there being
15.2
no external
resistor.
Power relationships
some
power relationships for the
3-phase induction motor. The following equations
The equivalent
circuit enables us to arrive at
basic electromechanical
can be deduced by visual inspection of the equiva-
V
P =
lent circuit of the
wound-rotor motor
(Fig. 15.7):
(sx 2 h
arctan sx 2 IR 2
Figure 15.5
voltage and current in the stator are separated by
the same phase angle p, even though the frequency
The
is
different.
The
final
equivalent circuit of the wound-rotor
induction motor
is
shown
in Fig. 15.7. In this di-
agram, the circuit elements are fixed, except for
the resistance
R 2 /s.
Its
value depends upon the slip
and hence upon the speed of the motor. Thus, the
value of
R 2 ls
will vary
from
motor goes from start-up
speed
(s
=
0).
(.v
R 2 to infinity as the
= 1) to synchronous
Figure 15.7
The primary and secondary leakage reactances x-,
and x2 are combined to form an equivalent total leakage reactance x.
ELECTRICA L MACHINES AND TRANSFORMERS
326
1
.
Active power absorbed by the motor
Efficiency of the motor
10.
is
PJP
n =
2.
Reactive power absorbed by the motor
Note: The preceding quantities are "per phase";
is
some must be
Q = E^IX m + Irx
3.
motor
Apparent power absorbed by the motor
=
S
2
\
P + Q
is:
multiplied by 3 to obtain the total
quantities.
is
15,3 Phasor diagram
2
of the induction
4.
Power
factor of the
motor
If
cos 6
=
PIS
we use current /, in Fig. 5.7 as the reference phawe obtain the complete phasor diagram of the
1
sor,
wound-rotor motor shown in Fig. 5.8. In this diagram (and also in future calculations) it is useful to
define an impedance Z, and angle a as follows:
5.
Line current
6.
Active power supplied to the rotor
1
is
is
r
Power
cuit
dissipated as
I
R
losses in the rotor cir-
V>~ +jc 2
a —
arctan xlr
In these equations r,
is
{
]V
2
I
R 2 -sP
Because
stator.
v
and
Mechanical power developed by the motor
is
a are fixed,
r,
and jx are
than r and so the angle
x
=
9.
P,
(
1
-
lation of
Torque developed by the motor
j
9.55
-
Pm
111
9.55
P
-
I
ns
n
_
9.55 Pl\
^
is
(\
s)'
s)
(1
5.7b)
motor referred
fixed,
is
to the
follows that
it
1
000
hp, jx
is
a approaches
In the equivalent circuit
.V)
5.7a)
Z
of Fig.
1
much
larger
90°.
5.7, the calcu-
mechanical power, torque, and speed de-
and R 2 is. The magnetizing branch
pends upon
r jx,
R m and jX m
does not come into play. Consequently,
{
for these calculations the equi valent circuit and cor-
responding phasor diagram can be simplified to that
shown
in Figs. 15.9
and
15. 10.
v
Figure 15.9
As far as mechanical power,
Figure 15.8
Phasor diagram of the voltages and currents in
Fig. 1 5.7. The power factor of the motor is cos tt.
{
irrespective of the speed of the motor.
motors above
In large
(1
the stator resistance and jx
the total leakage reactance of the
is
P =
8.
2
=
Z,
P = ICR 2 /s
7.
motor
is
torque,
and speed are
concerned, we can neglect the magnetizing branch
Xm
Rm This yields a simpler
motor behavior.
and
sis of
.
circuit for
the analy-
EQUIVALENT CIRCUIT OF THE INDUCTION MOTOR
1
Phasors
.
AB
angle
2.
BC
and
have the same length and
between them
the angle
CAB =
angle
327
is
(
80
1
ACB =
—
a)°.
a/2
Consequently,
/t
hR^s
/,r,
Figure 15.10
Phasor diagram of the circuit of Fig. 1 5.9. Note that
phasor / Z is always a degrees ahead of phasor
1
15.4
Breakdown torque and speed
We have seen that the torque developed by the motor
is given by T = 9.55 P /n where P is the power der
livered to the resistance
to a basic
imum
power
s
v
R 2 is
transfer theorem, the
(and therefore the torque
the value of R 2 /s
is
is
breakdown torque
slip at
= R 2 IZ
sh
The current
is
(15.9)
X
breakdown torque
at the
is
According
power
is
max-
maximum) when
=
/, b
£,/(2Z, cos a/2)
The breakdown torque
(I5.10)
is
equal to the absolute value of im-
pedance Z|. Thus, for
maximum
R 2 /s = Z
Under these circumstances,
voltage drop across Z,
We
(Fig. 15.9).
The
= EJ2
cos ^
/,Z,
/-,.
1
is
torque
-
(15.8)
l
the
magnitude of the
R 2 /s.
We
note that the magnitudes of both the breakdown
Tb and
torque
circuit resistance
However, the
R2
pends upon
The phasor diagram corresponding to
condition is shown in Fig. 5.
It is a
1
1
.
breakdown current
the
/ ib
are fixed,
sense that they are independent of the rotor
in the
can therefore write
(I5.ll)
%
(4Z, cos- a/2)
S
/?
equal to that across
1
9.55
Th =
.
R2
.
breakdown torque de-
slip at the
Indeed,
if
R2 =
Z,, the
breakdown
torque coincides with the starting torque because s b
this special
special case
of the phasor diagram of Fig. 15.10. Simple
geom-
etry yields the following results:
then equal to
l
.
These conclusions are
all
is
borne out by
the torque-speed curves in Fig. 13. 18 (Chapter 13).
In the case of squirrel-cage motors, the resistance
R 2 becomes equal
to r 2
,
which
is
the resistance of the
rotor alone reflected into the stator. In practice, the
angle
a
lies
correspond
tors. In
between 80° and 89°. The larger angles
to
medium- and high-power cage mo-
R 2 /Z
such machines the ratio
as 0.02. Consequently, the
at slips as
small as 2 percent.
15.5 Equivalent circuit of
practical
Figure 15.11
Phasor diagram when the motor develops its maximum torque. Under these conditions R2 ls = Z,.
can be as low
}
breakdown torque occurs
two
motors
The impedances and resulting equivalent circuits of
two squirrel-cage motors, rated 5 hp and 5000 hp
are given in Figs.
1
5.
1
2 and
1
5.
1
3,
together with the
ELECTRICAL MACHINES AND TRANSFORMERS
328
motor
The motors
ratings.
wye and
are both connected in
3.
a/2
=
4.
The
slip at
impedances are given per phase.
the
15.6 Calculation of the
We
now
will
calculate the
2
l
+
.5
6
2
=
6.
=
76°
1
8
nb
for
arclan
.v/r,
arctan 6/1.5
l. 2/6.
The
-
n s (\
=
1450r/min
.v
440V/V 3
0.194)
is
=
a/2
6900 V/V'3
Motor
rating:
60 Hz,
1
800 r/min, 440
full-loiid cuitciu: 7
V,
5000
3-phase
A
lneked-rotor current: 39
A
hp,
60 Hz. 600
r/min,
358
r2
rotor resistance
j.x
-
total
1
1
.5 12
.2 12
reactance
1
6 6
1
A
r2
rotor resistance 0.080 (2
j.x
=
total
=
magnetizing reactance 46
leakage reactance 2.6 12
12
no-load losses resistance 600 (2
(The no-load losses include the iron losses plus
The no-load
windage and
15
friction losses.)
1
stator resistance 0.083 fi
Rm =
no-load losses resistance 900 12
3-phase
=
/X m
10 (1
V,
=
/*,
leakage reactance 6 12
6900
A
locked-rotor current:
stator resistance
jXm = magnetizing
rating:
full-load current:
=
Rm =
b)
-
1800(1
2 Z, cos
5 hp,
0.194
is
=
=
breakdown
current at
/,K
Motor
=
8
1
O
6.
a
is
breakdown
hp motor.
V
I.
at
breakdown torque Th
/, b
0.788
]
The speed n b
5.
and the corresponding speed n h and current
the 5
breakdown
= R 2 /Z =
sb
breakdown torque
-
38°, cos a/2
losses of 26.4
kW for windage
and
kW
(per phase) consist of
friction
and
1
1
.4
kW
for the
iron losses.
Figure 15.13
Figure 15.12
Equivalent circuit of a 5 hp squirrel-cage induction
Because there
R2 ^ r2
motor.
rotor,
.
is
no external resistor
in
the
Equivalent circuit of a 5000 hp squirrel-cage induction motor.
Although
this
powerful than the motor
gram remains the same.
motor
in Fig.
is
1000 times more
15.12, the circuit dia-
EQUIVALENT CIRCUIT OF THE INDUCTION MOTOR
-
7.
A3
—
=
2X6.18X0.788
The power
to the rotor
2
P =
=
I
r
{
R 2 ls =
2
26.
X
26.1
A
2
I\
Z
6.18
9.55
P
9.55
X 4210
W
is
r
200
0
800
600
400
The
total
is
torque
r/min
SPEED
the torque developed
is,
therefore, 3
X
22.3
per phase.
= 67 N-m.
Figure 15.14
Torque-speed curve
The same
15.7 Torque-speed curve
We can determine the complete torque-speed curve of
hp motor by selecting various values of slip and
solving the circuit of Fig. 15. 12.
Table 15 A, and the curve
in
is
The
given
5 hp,
440
V,
r/min,
5000 hp
1
5. 15
to the
breakdown torque. These
characteristics are
typical for large squirrel-cage induction motors.
in Fig. 15.14.
TORQUE-SPEED AND LOAD
CHARACTERISTIC
5000
60 Hz squirrel-cage induction
hp,
6900
V,
600
r/min,
60 Hz squirrel-cage induction
motor
/,
IWj
[A]
s
T
n
[N-ml
[r/min J
Torque
IkN-ml
0.0125
2.60
649
3.44
1777
5.09
1243
6.60
1755
0.026
5.29
1291
6.85
1753
0.6
0.05
9.70
2256
12.0
1710
0.4
2
1
1.49
/
p
4.9
1617
0
1616
8.2
23.4
1614
1610
1120
360
10.6
40.8
2921
480
17.7
64.7
1593
26.8
6095
540
30.8
80.4
1535
42.1
10114
570
51.7
89.5
1
0.03077
47.0
11520
581.5
68.2
93.1
1133
0.02
43.1
10679
588
79.8
95.1
878
0.0067
19.9
5000
596
90.
96.6
358
0.0033
10.2
2577
598
85.1
95.4
198
7.39
0.2
26.4
4196
22.3
1440
0.1
0.4
33.9
3441
18.3
1080
0.05
0.6
36.6
2674
14.2
720
0.8
37.9
2150
11.4
360
38.6
1788
0.026.
-377 -600
6.3
14.4
=
[A|
[9*1
0
0.2
0
|
240
1620
9.49
[r/min
[hpl
0
18.8
at s
ElTcy
500
3547
developed
cos a
2.98
17.2
s
Speed
4.95
0.1
The rated power of 5 hp
Total
power
0.025
1
for the
results are listed
motor
s
made
the results and Fig.
the near-synchronous speed from no-load right up
TORQUE-SPEED CHARACTERISTIC
1800
lists
shows the torque-speed curve. Note the relatively
low starting torque for this large motor, as well as
TABLE 15B
TABLE 15A
a 5 hp motor.
of
calculations are
motor. Table 15B
and other characteristics
the 5
1000 1200 1400 1600 1800
N-m
1800
that this
N-m
X
= 4210
22.3
Note
22.3
(per
is
The breakdown torque Tb
8.
breakdown torque
phase )
440
329
363
ELECTRICA L MA CHINES A ND TRA NSEORMERS
330
breakdown torque = 47 kN
m
(per phase)
254 V
j110
Figure 15.16
Equivalent circuit of a 5 hp motor operating as an
asynchronous generator. Note that a negative resis-
tance
SPEED
Figure 15.15
Torque-speed curve
1
.
is
reflected into the primary circuit.
Net resistance of branch
R n = -48 +
of
a 5000 hp motor.
2.
Z = \Rl +
asynchronous generator
have already learned
tion
motor can
act as a generator if
lent circuit for the 5
hp motor,
it
-
it
is
driven above
3.
Now
power
= V
that a squirrel-cage induc-
synchronous speed.
that
Current
its
= -46.5
(
-
46.88
/,
other prop-
is
H
is
.r
46.5)
2
+
6
2
0
branch 1-2-3-4
in
we have the equivawe can calculate the
can generate, together with
-2-3-4
Impedance of branch 1-2-3-4
15-8 Properties of an
We
I.5
l
is
- EIZ = 254/46.88
- 5.42 A
erties as a generator.
Let us connect the motor to a 440 V, 3-phase
line
and drive
it
at a
speed of
1
845 r/min, which
4.
Active power delivered to the rotor
P = IcR 2 /s =
is
is
5.42- (-48)
{
45 r/min above synchronous speed. The
s
=
(/? s
=
(1800
-
slip is
= -1410
n)ln s
-
This negative power means that 1410
1845)/ 1800
flowing from the rotor
- -0.025
5.
The value of /? 2 Av
in the
equivalent circuit
W
is,
The
2
l
R
there-
P - Ifr 2 =
]r
=
6.
The negative resistance indicates that power is
flowing from the rotor to the stator rather than from
the stator to the rotor. Referring to Fig. 15.16
make
the following calculations:
we
35.2
5.42"
P
v
X
1.2
W
The mechanical power
equal to
is
losses in the rotor are
fore,
R 2 ls = 1.2/(- 0.025)
= -48 n
W
to the stator.
input to the shaft
is
plus the losses P-]r in the rotor:
p
= p + p.
= 1410 +
= 1445 W
35.2
EQUIVALENT CIRCUIT OF THE INDUCTION MOTOR
7.
The
2
I
R
2
F =
l,
js
=
=
r,
44.
5.42
2
X
1.5
W
1
= PJS = 1294/1502
0,861 = 86.1%
cos 0
losses in the stator are
16.
The
efficiency of the asynchronous generator
useful electric
8.
The
windage and
iron plus
mechanical input
9.
The
to the line feeding
rotor to stator
1
.
3
=
phases
3
9.55
=
X
1
X
3
1445
2
=
/,
=
176 var
a-
=
5.42
2
X
No-load
2
1
5,6, this
/X m
= 254
at
/110
power absorbed by the motor
X nv R m
The power
p
that the value of
/, is
negligible
Xm R m
,
at
at
no-
R 2 /s
is
compared
very togh
to / 0 Thus,
.
.
Their values can be de-
no-load, as follows:
Measure the
stator resistance
r,
R u between any
is
generator terminals A,
762
2
VA
'"I
Run
the
motor
is
£ NL
load current
/ NI
=
RlA.ll
no-load using rated line-to(Fig. 15.
and the
1
7).
total
Measure
the no-
3-phase active
.
The following calculations of total apparent power
total reactive power Q N] are then made:
S NL and
1502/254
factor at the generator terminals
at
line voltage,
power P Nl
p
an induction motor runs
exceedingly small. Referring to Fig.
means
value of
b.
= S/E=
= 5.91 A
in
the
no-load the circuit consists essentially of the
power
762 vars
2
line current /
and x
two terminals. Assuming a wye connection, the
VP 2 + Q 2 = V1294 +
1502
,
by means of
is
a.
at the
,
termined by measuring the voltage, current, and
Qi
/
When
is
magnetizing branch
var
is
The
test
load, the slip
2
r2
tests.
and so current
Apparent power
=
N-m
the equivalent circuit
following
is
Total reactive
S
22.3
6
Reactive power absorbed by the magnetizing
B
-
is:
854
the equivalent circuit can be found
+ Qi
= 176 + 586 =
15.
hp
The approximate values of
Q =
14.
X P
9.55
is
1445/746
Torque exerted by the driving motor
X 1294 - 3882 W)
is
= 586
3.
5.81
X
3
Reactive power absorbed by the leakage reac-
Q2 = E
1
=
=
tance
reactance
12.
3
to drive the generator
15,9 Tests to determine
2,
1
=
n
W
(P c for the
10.
The horsepower needed
T=
v
1294
8.
1
= P - P]S - P - P v
= ]410 - 44.1 - 71.7
=
7.
1
minus
losses
v
= 89.5%
0.895
is
Pc = power delivered from
ir
1445
power delivered
active
motor
the
1294
W
71.7
is
Pc
P
power
friction losses are
P f + P v = E2 /R m = 254 2 /900
=
331
is
ML
ELECTRICAL MACHINES AND TRANSFORMERS
332
Figure 15.17
A
Figure 15.18
no-load test permits the calculation of
Xm
and
tfm of
the magnetizing branch.
A
locked-rotor test permits the calculation of the total
leakage reactance
From these
P + Pv
windage,
t
The
Rm
resistance
representing
circuit of
and iron losses
friction,
P +
r
/\
results
xand the total resistance
+ r2
we can determine the equivalent
the induction motor.
Hence,
is
/WO 'ud ~
The magnetizing reactance
).
More
is:
r,
elaborate tests are conducted on large ma-
chines, but the above-mentioned procedure gives
adequate
results that are
Locked-rotor
Under
test
motor
the rotor of an induction
current
I
almost
is
p
Furthermore, the slip
that r2 /s
is
s is
locked, the stator
is
times
six
its
rated
value.
equal to one. This means
equal to r2 where r 2
,
the rotor reflected into the stator.
is
the resistance of
Because
greater than the exciting current 70
,
/ is
p
we can
1
5.9,
composed of the leakage reactance
and the reflected rotor
x, the stator resistance r h
sistance
R 2 /s =
r 2 /\
mined by measuring
—
re-
r2 Their values can be deter.
the voltage, current, and
most cases.
Example 15-1
A no-load test conducted on a 30 hp, 835 r/min, 440 V,
3-phase, 60
Hz
squirrel-cage induction motor yielded
the following results:
much
neglect
the magnetizing branch. This leaves us with the circuit of Fig.
in
when
rated line voltage,
No-load voltage
(line-to-line):
No-load current: 14
440
V
A
No-load power: 1470
W
Resistance measured between two terminals:
0.5 (2
power
The locked-rotor
under locked-rotor conditions, as follows:
test,
conducted
at
reduced
volt-
age, gave the following results:
a.
Apply reduced 3-phase voltage
that the stator current
is
to the stator so
about equal to
its
rated
Take readings of
total
E
l
R (line-to-line),
3-phase power P K (Fig.
]
/,
R and the
,
Locked-rotor current:
JS' LR
are then
= F LR I
60 A
15. 18).
Determine the equivalent
The following calculations
V
W
Locked-rotor power: 7200
value.
b.
Locked-rotor voltage (line-to-line): 163
circuit of the motor.
made:
Solution
R
[
\v J
3
Assuming
the stator windings are connected
wye, the resistance per phase
= Cir/3/ lr
3/T R (r, + r 2 = P LR
r,
*
)
From
-
0.5
the no-load test
a/2
we
=
find
is
0.25 II
in
EQUIVALENT CIRCUIT OE THE INDUCTION MOTOR
Rx =
=
'
10 669
P m = 1470
2
5-
1
e Il /p nl = 440
= i46a
2
/(1470
^lr'lr V3
/10 568
-3 X
14
2
X
0.25)
A
we
/,
R
X 60V
163
=
60
A
2
- 7200
_
G,.r
15
c.
-
0.67 (2
=
0.25
H
=
0.67
-
The value of Z, and the angle a
The speed when the breakdown torque
is
The current
I
\
at the
breakdown torque
(see
The value of
a.
Problem 15-2, draw the equivalent
circuit if the motor runs at 950 r/min in
is
/
2
breakdown torque [N-m]
ator? Calculate the torque of the
machine.
is
„= 7200/(3 X 60 2
,
the
In
the same direction as the revolving flux.
Does the machine operate as a gener-
b.
r/3
V, calculate the
d.
1.42(2
2
Total resistance referred to stator
+ ^ =
346
reached
2
333
X 60
3
3/iLR
is
5 12 and the line-
is
Fig. 15.9)
Total leakage reactance referred to stator
—
voltage
to- neutral
15-3
x
leakage reactance
following:
Pl R = V16 939
333 var
the
12. If
find
b.
15
and
wye-connected squirrel-cage motor
equivalent rotor resistance of 0.5
a.
=
to the text, explain the
the impedances, currents,
having a synchronous speed of 900 r/min
VA
VS
V
°LR
in Fig. 15.19.
voltages in Fig. 15.2.
W
2
shown
is
Without referring
total
939
because
has a stator resistance of 0.7 il and an
the locked-rotor test
16
circuit
meaning of
15-2
7200
r2
15. 1.)
Questions and Problems
10 568 var
x m = ^\[/Q\l = 440
= 18.3 n
From
seeEq.
The equivalent
W
1
R,
0;
VA
V10 669 2 - 1470 2
-
R2 =
(In a squirrel-cage motor,
440 X 14V 3
333
Draw
runs
)
flux.
the equivalent circuit
if
the
motor
950 r/min opposite to the revolving
Does the machine operate as a genera-
at
tor? Calculate the torque.
r
{
1
0.25
0.25
=
0.42
5-4
n
Q
A 550
780 r/min, 3-phase, 60 Hz
squirrel-cage induction motor running
no-load draws a current of 12 A and a
total power of 1500 W. Calculate the
V,
value of
1
Xm
and
Rm
at
per phase (see
Fig. 15.2).
15-5
440
18.3
#n
146
f3
Q
The motor
rent of
X
30
in
Problem 15-4 draws
A and
when connected
a
power of 2.43
to a
90
a cur-
kW
V, 3-phase line
under locked-rotor conditions. The
resis-
tance between two stator terminals
is
Figure 15.19
0.8 fl. Calculate the values of r h r 2 and
Determining the equivalent
x and
induction motor (see
,
circuit of
Example
15-1).
a squirrel-cage
the locked-rotor torque [N
rated voltage.
m|
at
334
1
5-6
ELECTRICAL MACHINES AND TRANSFORMERS
If
motor in
6200 V, calculate
the line voltage for the
Fig. 15.15
the
dropped
new breakdown
to
/ nd us trio I
15-9
torque and starting
5-7
Consider the 5 hp motor whose equivalent
circuit
torque.
1
a.
A 440 V,
3-phase,
motor has the following
characteristics:
If the
=
1
x
=
6
.2 il
c.
n
15-8
In
compare
magnetizing branch can be neglected,
connected
in series
if
a 4.5 fl resis-
with each
Problem 15-7 calculate the
is
connected
in series
line.
starting
torque and the breakdown torque
reactor
Calculate the 50
to obtain the
calculate the value of the starting torque
is
Determine the values of the leakage reacat
a
frequency of 50 Hz.
and the breakdown torque
tor
in Fig. 15.12.
tance and the magnetizing reactance
1.5 fi
x
r2
shown
ing reactances.
b.
=
r
is
Calculate the values of the inductances (in
millihenries) of the leakage and magnetiz-
800 r/min squirrel-cage
1
application
if
a 4.5 ft
with each
15-10
The
5
V
line-to-neutral voltage
with the voltage
at
60 Hz.
hp motor represented by the equiva-
lent circuit
503
it
Hz
same magnetizing current and
of Fig.
1
5.
1
2
is
connected
(line-to-line), 3-phase,
to a
80 Hz
source. The stator and rotor resistances are
assumed to remain the same.
a. Determine the equivalent circuit when the
line.
motor runs
b.
at
2340 r/min.
Calculate the value of the torque [N-m] and
power
[hp]
developed by the motor.
Chapter
1
Synchronous Generators
some
16.0 Introduction
other source of motive power.
3-phase voltage
Three-phase synchronous
primary source of
all
generators
consume. These machines are the
verters in the world.
are
the
the electrical energy
largest
energy con-
into electrical energy, in
powers ranging up
will study the construction
In this chapter
we
and characteristics of these
are based
upon
large,
modern
in
view
this materia]
upon the dc exciting
The frequency of the
speed and the number of
used when the power output
However,
generators.
may wish
rotates, a
the speed of rotation and
voltage depends upon the
for greater outputs,
and more practical
the elementary principles cov-
Section 8.6, and the reader
ered
it
poles on the field. Stationary-field generators are
1500
to
As
induced, whose value depends
current in the stationary poles.
They convert mechanical energy
MW.
They
upon
we
is
A
to re-
to
is
less than 5
employ a revolving dc
tor
field.
revolving-field synchronous generator has a
stationary armature called a stator.
before proceeding further.
kVA.
cheaper, safer,
is
it
winding
is
The 3-phase
sta-
directly connected to the load, with-
out going through large, unreliable slip-rings and
16.1
A
Commercial synchronous
brushes.
generators
insulate the windings because they are not sub-
stationary stator also
makes
jected to centrifugal forces. Fig. 16.
Commercial synchronous generators
are built with
The
The field is excited by a dc generator,
mounted on the same shaft. Note that the
brushes on the commutator have to be connected to
cut by a re-
another set of brushes riding on slip-rings to feed
as a dc generator.
which
is
usually
The armature possesses a
3-phase winding whose terminals are connected to
three slip-rings mounted on the shaft. A set of
volving
armature.
the dc current I x into the revolving field.
16.2
brushes, sliding on the slip-rings, enables the armature to
is
Number
of poles
an external 3-phase load.
The number of poles on a synchronous generator de-
driven by a gasoline engine, or
pends upon the speed of rotation and the frequency
be connected
The armature
a schematic
alternator.
A stationary -field synchronous generator has the
salient poles create the dc field,
is
easier to
diagram of such a generator, sometimes called an
either a stationary or a rotating dc magnetic field.
same outward appearance
1
it
to
335
ELECTRICA L MA CHINES A ND TRA NSEORMERS
3 36
ils
pilot exciter
25
kW
3-phase alternator
500 MW, 12 kV, 60 Hz
Figure 16.1
Schematic diagram and cross-section view of a typical 500
synchronous generator and its 2400 kW dc exciter.
The dc exciting current lx (6000 A) flows through the commutator and two slip-rings. The dc control current lc from
the pilot exciter permits variable field control of the main exciter, which, in turn, controls /x
MW
.
we wish
to produce. Consider, for
conductor
that
is
p =
example, a stator
successively swept by the
N
and S
120.////
= 20 X
60/200
1
poles of the rotor.
when an N
ilar
If
a positive voltage
is
induced
pole sweeps across the conductor, a sim-
negative voltage
is
crosses the conductor, the induced voltage goes
through a complete cycle. The same
other conductor on the stator;
that the alternator
is
true for every
we can
therefore de-
frequency
is
given by
From an
(,6I)
120
is
frequency of the induced voltage [Hz]
that carry a
is
is
always connected
connected
to
.
The voltage per phase
is
between the
the voltage
200 r/min
nected to a synchronous generator.
voltage has a frequency of 60 Hz,
If
is
con-
the induced
how many
poles
in
16.3).
wye and
the
A wye connection is
only 1/V3 or
lines.
between a
line voltage.
amount of
We can
58%
of
This means that
stator
is
only
conductor
58%
of the
therefore reduce the
insulation in the slots which, in turn,
enables us to increase the cross section of the
conductors.
A
larger conductor permits us to in-
crease the current and, hence, the power output
Solution
Eq. 16.
ground.
and the grounded stator core
at
composed of
3-phase lap winding (Figs. 16.2,
the highest voltage
hydraulic turbine turning
It is
preferred to a delta connection because
1
Example 16-1
From
S poles
identical to that of a 3-phase
induction motor (Section 13. 17).
The winding
p = number of poles on the rotor
n = speed of the rotor [r/min]
does the rotor have?
N and
electrical standpoint, the stator of a syn-
chronous generator
neutral
A
8 pairs of
16.3 Main features of the stator
where
/=
1
a cylindrical laminated core containing a set of slots
pn
f=
36 poles, or
induced when the S pole
speeds by. Thus, every lime a complete pair of poles
duce
=
1
,
we have
of the machine.
SYNCHRONOUS GENERATORS
337
Figure 16.2a
Stator of a 3-phase,
500 MVA, 0.95 power factor, 15
2350 mm; 378
effective axial length of iron stacking:
kV,
60 Hz, 200 r/min generator.
Internal diameter:
9250 mm;
slots.
(Courtesy of Marine Industrie)
2.
When
a synchronous generator
voltage induced
in
and the waveform
distortion
is
is
under load, the
each phase becomes distorted,
is
no longer sinusoidal. The
mainly due
to
an undesired third
harmonic voltage whose frequency
that
is
three times
of the fundamental frequency. With a
wye
connection, the distorting line-to-neutral harmonics
do not appear between the
lines
because they
effectively cancel each other. Consequently, the
line voltages
remain sinusoidal under
conditions. Unfortunate! v.
when
all
load
a delta connec-
tion
is
used, the harmonic voltages
but add up. Because the delta
is
do not cancel,
closed on
itself,
they produce a third-harmonic circulating current,
which increases the
The nominal
erator depends
greater the
However,
exceeds 25
its
losses.
kVA
synchronous gen-
rating. In general, the
rating, the higher the voltage.
nominal line-to-line voltage seldom
kV because
takes up valuable space
conductors.
R
line voltage of a
upon
power
the
I
the increased slot insulation
at
the
expense of the copper
Figure 16.2b
The copper bars connecting successive
19 250 A per phase.
stator poles are
designed
to carry a current of
3200
A.
The
total
output
is
(Courtesy of Marine Industrie)
Figure 16.2c
The stator is built up from toothed segments
silicon-iron steel laminations (0.5
an insulating varnish. The
mm
slots are 22.3
mm deep. The salient poles of the
much
thicker (2
mm)
thick),
of high-quality
covered with
mm wide
and 169
composed of
These laminations
rotor are
iron laminations.
are not insulated because the dc flux they carry does not
vary.
the
The width of the poles from tip to tip is 600 mm and
gap length is 33 mm. The 8 round holes in the face
air
of the salient pole carry the bars of
338
a squirrel-cage winding.
Figure 16,3
MVA, 3600 r/min, 19 kV, 60 Hz steam-turbine generator during the construction phase.
The windings are water-cooled. The stator will eventually be completely enclosed in a metal housing (see background). The housing contains hydrogen under pressure to further improve the cooling.
{Courtesy of ABB)
Stator of a 3-phase, 722
339
ELECTRICA L MA CHINES A ND ERA NS FORMERS
340
16.4 Main features of the rotor
made
ing, the field coils are
of bare copper bars,
with the turns insulated from each other by strips of
Synchronous generators are
rotors: salient-pole rotors
rotors.
built with
two types of
mica
and smooth, cylindrical
Salient-pole rotors are usually driven by
have
to turn at
Most hydraulic
rotors.
(Fig.
turbines
dc field winding,
in series,
embedded
1
6.6).
Under normal
we
in the
often add
pole-faces
conditions, this winding
does not carry any current because the rotor turns
at
generator changes suddenly, the rotor speed begins
is
directly coupled to
required, a large
number of poles
to fluctuate,
50 Hz
producing momentary speed variations
above and below synchronous speed. This induces a
are re-
voltage
in
the squirrel-cage winding, causing a large
quired on the rotor. Low-speed rotors always pos-
current to flow therein.
sess a large diameter to provide the necessary space
magnetic
for the poles.
connected
maximum power from
the waterwheel, and because a frequency of
is
coils are
synchronous speed. However, when the load on the
a waterfall. Because the rotor
Hz
The
low speeds (between 50 and 300
r/min) in order to extract the
or 60
6.5).
a squirrel-cage winding,
are driven by high-speed steam turbines.
Salient-pole
1
In addition to the
low-speed hydraulic turbines, and cylindrical rotors
/.
(Fig.
with adjacent poles having opposite polarities.
The
large circular steel
salient poles are
frame which
ing vertical shaft (Fig.
1
6.4).
is
mounted on a
dampen
fixed to a revolv-
The
field of the stator,
the oscillation of the rotor. For this reason,
the squirrel -cage winding
To ensure good cool-
current reacts with the
producing forces which
is
sometimes called a
damper winding.
Figure 16.4
This 36-pole rotor
is
being lowered into the stator shown
a 330 V, electronic rectifier. Other details are: mass: 600
(Courtesy of Marine Industrie)
in Fig.
t;
16.2.
moment
The 2400 A dc
of inertia: 41
40
exciting current
tm 2
;
air
gap: 33
is
supplied by
mm.
Figure 16.5
tor is
made
width of 89
Figure 16.6
a 250 MVA salient-pole generaof 18 turns of bare copper bars having a
This rotor winding
mm
for
and a thickness
of 9
Salient-pole of a
250
MVA
generator showing 12 slots
t0 carry the squirre |. cage winding,
mm.
Figure 16.7a
Rotor of a 3-phase steam-turbine generator rated 1530 MVA, 1500 r/min, 27 kV, 50 Hz. The 40 longitudinal slots
are being milled out of the solid steel mass. They will carry the dc winding. Effective axial magnetic length: 7490
mm; diameter: 1800 mm.
(Courtesy of Allis-Chalmers Power Systems Inc., West Mis, Wisconsin)
34
ELECTRICAL MACHINES AND TRANSFORMERS
342
Figure 16.7b
Rotor with
rent of
1 1
its
.2
4-pole dc winding. Total mass: 204
kA
is
supplied by a 600
V dc
Inc.,
lines,
even
anced load conditions.
2.
Cylindrical rotors.
;
It
is
well
known
than low-speed turbines.
The same
is
that high-
true of high-
speed synchronous generators. However, to generfrequency
we cannot
air
gap: 120
of the
N
main
mm. The dc
exciting cur-
shaft.
use less than
to
and S poles.
The high speed of rotation produces strong cenwhich impose an upper
trifugal forces,
speed steam turbines are smaller and more efficient
ate the required
2
and retained by high-strength end-rings, * serve
create the
due to unbal-
the line currents are unequal
85 t-m
the end
of inertia:
West Allis, Wisconsin)
also tends to maintain bal-
anced 3-phase voltages between the
when
moment
brushless exciter bolted to
(Courtesy of Allis-Chalmers Power Systems
The damper winding
t;
diameter of the
at
rotor. In the
3600 r/min, the
limit
on the
case of a rotor turning
elastic limit
of the
steel requires
the manufacturer to limit the diameter to a maxi-
mum of
erful
m.
.2
l
1000
On
the other hand, to build the
MVA to
1500
MVA
pow-
generators the vol-
2 poles, and this fixes the highest possible speed.
ume
On
high-power, high-speed rotors have to be very long.
a
60 Hz system
speed
is
it
is
3600
r/min.
The next lower
It
follows that
1800 r/min, corresponding to a 4-pole ma-
chine. Consequently, these steam-turbine generators possess either 2 or
The
of the rotors has to be large.
4 poles.
rotor of a turbine-generator
The dc
is
cylinder which contains a series of longitudinal
slots
milled out of the cylindrical mass (Fig. I6.7).
wedged
field excitation
and exciters
of a large synchronous
a long, solid
steel
Concentric field coils, firmly
16.5 Field excitation
into the slots
generator
is
*
1.
See Fig.
1
an important part of
28 (Chapter
II).
its
overall design.
SYNCHRONOUS GENERATORS
stationary field
—o A
rrrr
air
3-phase
alternator
-OB
-oC
gap
343
terminals
pole
bridge
rectifier
/
pilot exciter
^.exciting coil
3\
3-phase rotor
3-phase stator winding
r—T
main exciter
y
alternator
Figure 16.8
Typical brushless exciter system.
The reason
is
must ensure not only
that the field
stable ac terminal voltage, but
to
sudden load changes
tem
order to maintain sys-
in
Quickness of response
stability.
a
must also respond
one of the
is
voltage
may have
as
as
little
300
to
to rise to twice
normal value
its
400 milliseconds. This
in
represents a
very quick response, considering that the power of the
exciter
may be
several thousand kilowatts.
important features of the field excitation. In order
to attain
it,
two dc generators
and a pilot
citer
no rotating parts
The main
slip-rings.
voltage
lies
main ex-
at all are also
involve
employed.
exciter feeds the exciting current to the
of the synchronous generator by
field
and
are used: a
exciter. Static exciters that
way
Under normal conditions
between 125
V
and 600
V.
of brushes
/c ,
produced by the
1000
kVA
kW
alternator
exciter
needed
(2.5% of
its
1
6.
1
).
whereas
a
2500 kW exciter suffices for an alternator of
(only 0.5% of its rating).
500
Under normal conditions the excitation is varied
MW
automatically.
It
responds to the load changes so as to
maintain a constant ac line voltage or to control the
active
power delivered
re-
to the electric utility system.
serious disturbance on the system
may produce
a
A
sud-
den voltage drop across the terminals of the alternator.
The
exciter must then react very quickly to keep
the ac voltage
from
falling.
wear and carbon
dust,
we
constantly
and replace brushes, slip-rings,
and commutators on conventional dc excitation
sys-
systems have been developed. Such a system con-
to excite a
rating)
to clean, repair,
regulated
It is
pilot exciter (Fig.
is
to brush
have
tems. To eliminate the problem, brushless excitation
The power rating of the main exciter depends
upon the capacity of the synchronous generator.
Typically, a 25
Due
the exciter
manually or automatically by control signals that vary
the current
16.6 Brushless excitation
For example, the exciter
sists
of a 3-phase stationary-field generator whose
ac output
is
rectified
by a group of
output from the rectifiers
rectifiers.
The dc
fed directly into the field
is
of the synchronous generator (Fig.
1
6.8).
The armature of the ac exciter and the
are mounted on the main shaft and turn
rectifiers
together
with the synchronous generator. In comparing the
excitation system of Fig.
we
can
see
they are
1
6.8 with that of Fig. 16.
3-phase rectifier replaces the commutator,
rings,
and brushes.
(which
is
really a
by an electronic
In other
slip-
words, the commutator
mechanical
rectifier.
1,
except that the
identical,
rectifier) is replaced
The
result
is
that
the
brushes and slip-rings are no longer needed.
The dc
control current
/c
from the
regulates the main exciter output
/v .
pilot exciter
as in the case of
ELECTRICA /. MA CHINES A ND ERA NS FORMERS
344
Figure 16.9
This brushless exciter provides the dc current for
the rotor
of
4
1
is*
shown in Fig. 16.7. The exciter consists
kVA generator and two sets of diodes.
a 7000
Each set, corresponding respectively to the posiand negative terminals, is housed in the circular rings mounted on the shaft, as seen in the
center of the photograph. The ac exciter is seen
to the right. The two round conductors protruding
tive
from the center
of the shaft (foreground) lead the
exciting current to the
1530
MVA
generator.
(Courtesy of Allis-Chalmers Power Systems
Inc.,
West All is, Wisconsin)
The frequency of
a conventional dc exciter.
main exciter
is
the
generally two to three limes the syn-
chronous generator frequency (60 Hz). The
crease
in
frequency
is
in-
obtained by using more poles
on the exciter than on the synchronous generator.
Fig.
1
shows
6.9
the rotating portion of a typical
brushless exciter. Sialic exciters that involve no rotating parts at
all
employed.
are also
16.7 Factors affecting the size
of
synchronous generators
The prodigious amount of energy generated by elecutility companies has made them very con-
trical
scious about the efficiency of their generators. For
example,
station
if
the efficiency of a
improves by only
1
%,
1
it
000
MW generating
represents extra rev-
enues of several thousand dollars per day.
gard, the size of the generator
tant
because
as the
power
its
is
In this re-
particularly impor-
efficiency automatically improves
increases. For example,
if a
small
l
kilowatt synchronous generator has an efficiency of
50%,
l()
a larger, but similar
model having a capacity of
MW inevitably has an
This improvement
in
efficiency of about
efficiency with size
why synchronous
is
MW and
000
up possess efficiencies of the order of 99%.
Another advantage of large machines is
son
generators of
1
power output per kilogram increases
increases. For example,
if
a
l
20 kg (yielding 1000W/20 kg
90%.
the rea-
Figure 16.10
Partial
view
of
87 MVA, 428
a 3-phase, salient-pole generator rated
r/min,
are water-cooled.
that the
as the
power
kW generator weighs
= 50 W/kg). a lOMW
and the use
50 Hz. Both the
The high
rotor
resistivity of
of insulating plastic tubing
water to be brought
into direct
parts of the machine.
{Courtesy of ABB)
and
stator
pure water
enables the
contact with the
live
S YNCHRONO US
GENERA TORS
345
generator of similar construction will weigh only
ticated cooling techniques (Figs. 16.10
20 000 kg, thus yielding 500 W/kg. From a power
Other technological breakthroughs, such as better
standpoint, large machines
materials,
weigh
relatively less than
small machines; consequently, they are cheaper.
Section 16.24
why
at
the end of this chapter explains
Everything, therefore, favors the large machines.
in size,
we
run into seri-
ous cooling problems. In effect, large machines
herently produce high
power
in-
losses per unit surface
area (W/m~); consequently, they tend to overheat.
To prevent an unacceptable temperature
must design
efficient cooling
we
become
rise,
systems that
ever more elaborate as the power increases. For ex-
ample, a circulating cold-air system
cool synchronous generators
but between 50
the 1000
hollow,
point
MW
is
MW
range have to be equipped with
water-cooled
conductors.
Ultimately,
exceeds the savings made elsewhere, and
upper limit to
the
a
reached where the increased cost of cooling
is
To sum
1).
in
modifying the design of early ma-
chines (Fig. 16. 12).
As
regards speed, low-speed generators are
al-
power. Slow-speed bigness simplifies the cooling
problem; a good air-cooling system, completed with
a heat exchanger, usually suffices. For example, the
large,
slow-speed 500
generators installed
are air-cooled
500
MVA, 200 r/min
in a typical
1
to
plant
whereas the much smaller high-speed
MVA, 800 r/min units installed
have
synchronous
hydropower
in a
steam plant
be hydrogen-cooled.
adequate to
whose rating is below
and 300 MW, we have
hydrogen cooling. Very big generators
to resort to
in
1
ways bigger than high-speed machines of equal
However, as they increase
MW,
16.
and novel windings have also played a
major part
the efficiency and output per kilogram increase
with size.
50
and
this fixes
size.
16.8 No-load saturation curve
Fig. 16.
3a shows a 2-pole synchronous generator
It
is
driven
at
constant speed
by a turbine (not shown). The leads from the
3-phase, wye-connected stator are brought out to
terminals A, B. C, N, and a variable exciting current
/x
up, the evolution of big alternators has
mainly been determined by the evolution of sophis-
1
operating at no-load.
produces the flux
in the air
gap.
Let us gradually increase the exciting current
while observing the ac voltage
£ 0 between terminal
Figure 16.11
The
electrical-
200/1 15
the
is
V,
energy needed on board the Concord
12 000 r/min, 400 Hz. Each generator
enormous power developed by
used
to cool the
generator and
(Courtesy of Air France)
is
is
aircraft is
supplied by four 3-phase generators rated 60 kVA,
driven by a hydraulic motor, which absorbs a small portion of
the turboreactor engines.
The
then recycled. The generator
hydraulic fluid streaming from the hydraulic motor
itself
weighs only 54.5
kg.
346
ELECTRICAL MACHINES AND TRANSFORMERS
Figure 16.12
was
This rotating-field generator
system. The alternator
was
first
in 1888. It was used in a 1000-lamp street lighting
steam engine and had a rated output of 2000 V, 30 A at a frewhich represents 26 W/kg. A modern generator of equal speed and power
installed
driven by an 11
00
in
North America
r/min
quency of 1 10 Hz. It weighed 2320 kg,
produces about 140 W/kg and occupies only one-third the
A, say, and the neutral N. For small values of / x the
,
floor
space.
Fig. 16. 13c
ator
ing current. However, as the iron begins to saturate,
phases on the
the voltage rises
/x
.
If
we
much
less for the
plot the curve of
£u
same increase
versus
the no-load saturation curve of the
generator.
It
/x ,
we
a schematic diagram of the gener-
the
revolving rotor and the three
stator.
in
obtain
16.9
synchronous
Synchronous reactance
equivalent circuit of an ac
generator
similar to that of a dc generator
is
is
showing
voltage increases in direct proportion to the excit-
(Section 4. 13).
Fig. 16.13b
curve of a 36
shows
the actual no-load saturation
Consider a 3-phase synchronous generator having
MW,
3-phase generator having a
terminals A, B,
nominal voltage of 12
kV
(line to neutral).
about 9 kV, the voltage increases
in
Up
put of 12 kV, but
if
1
age rises only to 15 kV.
is
driven by a turbine (not
proportion to
shown), and
is
chine and
load are both connected in wye, yield-
00 A produces an out-
the current
feeding a balanced 3-phase load
(Fig. 16. 14).
the current, but then the iron begins to saturate.
Thus, an exciting current of
C
The generator
to
doubled, the volt-
its
is
excited by a dc current
Ix
.
The ma-
ing the circuit of Fig. 16.15. Although neutrals N,
and
N2
tential
are not connected, they are at the
because the load
is
same po-
balanced. Consequently,
SYNCHRONOUS GENERATORS
\
I;
B
:
•
I
i
347
|
C
i
i
!
i
Figure 16.13c
Electric circuit representing the generator of
Fig. 16.13a.
is
an alternating-current machine, the inductance
X
manifests itself as a reactance
X =
s
,
given by
277/L
s
where
X =
s
=
E —
/'
synchronous reactance, per phase [H]
generator frequency [Hz]
apparent inductance of the stator winding, per
phase HJ
[
The synchronous reactance of
ternal
Figure 16.13
Generator operating
a.
b.
impedance, just
like
its
a generator
is
an
in-
internal resistance R.
The impedance is there, but it can neither be seen
nor touched. The value of X s is typically 10 to 100
times greater than R; consequently,
at no-load.
No-load saturation curve
3-phase generator.
of
a 36 MVA, 21
kV,
neglect the resistance, unless
we
we can always
are interested in
efficiency or heating effects.
We
can simplify the schematic diagram of
16.16 by showing only one phase of the
we could connect them
together (as indicated by the
fect, the
two other phases
Fig.
stator. In ef-
are identical, except that
short dash line) without affecting the behavior of the
their respective voltages (and currents) are out of
voltages or currents in the circuit.
phase by
The
field carries
an exciting current which pro-
duces a flux O. As the
duces
in the stator three
field revolves, the flux in-
equal voltages
Eu
that are
120° out of phase (Fig. 16.16).
Each phase of the stator winding possesses a resistance R and a certain inductance E. Because this
1
20°. Furthermore,
tance of the windings,
cuit
of Fig.
1
6.
1
7.
we
if
we
neglect the resis-
obtain the very simple
fore be represented
by an equivalent
of an induced voltage
E0 in
circuit
O
which induces
composed
series with a reactance
In this circuit the exciting current / x
flux
cir-
A synchronous generator can thereX
s
.
produces the
the internal voltage
E<r
For a
348
ELECTRICAL MACHINES AND TRANSFORMERS
/
A
\
B
load
C
load
alternator
Figure 16.14
Generator connected
to
a load.
Figure 16.17
Equivalent
only
circuit of
a 3-phase generator, showing
one phase.
E
given synchronous reactance, the voltage
E0
terminals of the generator depends upon
E0
load Z. Note that
ages and
/ is
E
and
are line-to-neutral volt-
16,10 Determining the value of
X
We can determine the unsaturated value of X
by the
During the open-circuit
at rated
test the
sponding exciting current
En
are recorded.
The
excitation
age
is
/ xn
is
is
driven
raised until
The
attained.
is
s
s
test.
generator
speed and the exciting current
the rated line-to-line voltage
Electric circuit representing the installation of Fig. 16.14.
the
the line current.
following open-circuit and short-circuit
Figure 16.15
at
and the
corre-
and line-to-neutral
volt-
then reduced to zero and the
three stator terminals are short-circuited together.
With the generator again running
exciting current
value
/ xn
The
tor
is
at rated
gradually raised to
speed, the
its
original
.
resulting short-circuit current / sc in the sta-
windings
is
measured and
X
s
is
calculated by us-
ing the expression
X = £ n //
s
(16.2)
Sc
where
X =
synchronous reactance, per phase
En —
rated open-circuit line-to-neutral voltage
s
[11]*
LV|
Figure 16.16
Voltages and impedances
its connected load.
:;:
in
a 3-phase generator and
This value of
reactance.
behavior.
It
X
is
h
corresponds
widely used
to the direct-axis
to describe
synchronous
synchronous machine
SYNCHRONOUS GENERATORS
/ sc
=
When
short-circuit current, per phase, using
the
same
exciting current
En
required to produce
The synchronous reactance
/ Xll that
the terminals are short-circuited, the only
impedance limiting
was
[A]
to the
the current flow
X - EJ1 =
- 5n
is
its
heavily saturated, the value of X s
may
the iron
be only half
unsaturated value. Despite this broad range
usually take the unsaturated value for
due
that
4000/800
s
When
is
synchronous reactance. Consequently,
not constant, but
is
varies with the degree of saturation.
349
X
s
The synchronous reactance per phase
we
because
fore 5
it
yields sufficient accuracy in most cases of interest.
The equivalent
b.
is
there-
il.
phase
circuit per
shown
is
in
Fig. 16.18a.
The impedance of the
Example 16-2
A
3-phase
synchronous generator produces an
open-circuit line voltage of
citing current
is
6928
50 A. The ac terminals
=
are then
short-circuited, and the three line currents are
be 800 A.
a.
Calculate the synchronous reactance per phase.
b.
Calculate the terminal voltage
The
resistors are
connected
if
wye
in
current
/
three 12 il
across the
= EJZ=
minals.
E=
1
X;
\
2
+
5
(2.12)
2
n
3
is
The voltage across
ter-
2
\ 12
=
found
to
\R
Z
V when the dc ex-
circuit is
1R
=
4000/13
=
A
308
the load resistor
X
308
1
2
is
- 3696 V
Solution
a.
The
line-to-neutral induced voltage
The
line voltage
under load
E,
E0 = E L /V3
= 6928/V3
(8.4)
= 4000 V
- V3 E
= V3 X 3696
- 6402 V
The schematic diagram of
x
s
= 5
is
is
sualize
a
what
16.11
We
is
happening
Fig. 16.18b helps us viin the actual circuit.
Base impedance,
recall that
first select
when using
a base voltage
per-unit
we
EH
power per phase as the base power.*
lows that the base impedance Z B is given by
the rated
1
line voltage =
=
we
In the
use the rated
line-to-neutral voltage as the base voltage
t
s
the per-unit system
and a base power.
case of a synchronous generator,
Z,
X
^
It
and
fol-
(16.3)
6394 V
alternator
:;:
Figure 16.18
a. See Example 16-2.
b.
Actual line voltages and currents.
In
many power studies the base power is selected to be equal
power of the generator and the base voltage is
to the rated
the linc-to-linc voltage. This yields the
base impedance.
same value Z n
for the
ELECTRICAL MACHINES AND TRANSFORMERS
350
where
ZB =
base impedance (line-to-neutral) of
the generator [ft]
EB =
Note
base voltage (line-to-neutral) fV]
[
The
d.
for other
may
X
s
(pu)
lies
between
Note
P =
nous reactance of
60 Hz ac generator has a synchro1
.2
from
are
2
/
line to
(pu) R(pu)
pu and a resistance of 0.02
2
l
X
=
0.02
0.02
that at full-load the per-unit value
equal to
Example 16-3
5 kV,
=
be expressed as
=
of
I is
1
The copper
1
7.5
per-unit copper losses at full-load are
=
depending upon the design of the machine.
A 30 MVA,
X
0.02
that the generator possesses.
.
2,
ZB =
impedance values
that all
P(pu)
impedances
a per-unit value of ZB In general,
and
0.15 ft
used as a basis of comparison
is
Thus, the synchronous reactance
0.8
0.02
=
neutral.
S B — base power per phase VA]
The base impedance
=
losses for
0.02 S B
600
=
all
0.02
3 phases are
X
30
=
0.6
MW
kW
pu.
16.12 Short-circuit ratio
Calculate
a.
b.
c.
d.
The base
voltage, base
power and base imped-
Instead of expressing the synchronous reactance as
ance of the generator
a per-unit value of
ZB
The
The
The
sometimes used.
the ratio of the field current / x!
actual value of the synchronous reactance
actual
needed
winding resistance, per phase
age
copper losses
total full-load
EH
a.
is
SH
Eq.
000/V3
15
= 30MVA/3 =
= 10 7 VA
10
is
1
6.2.
Thus,
if
X
as defined in
s
X
s
is
l
.2,
the
1/1.2 or 0.833.
is
many
=
(16.3)
types of loads, but they can
all
be reduced to
two basic categories:
7.5 ft
The synchronous reactance
is
X = X (pu) X ZB
= 1.2ZB = 1.2 X
= 9(2
s
The resistance per phase
7.5
is
Figure 16.19
R =
The
short-circuit.
exactly equal to the
the per-unit value of
is
volt-
produce
The behavior of a synchronous generator depends
upon the type of load it has to supply. There are
ti
s
|// X 2)
to
is
16.13 Synchronous generator
under load
MVA
Z = E H 2 /S H
- 8660 2 /10 7
c.
on a sustained
short-circuit ratio
is
The base impedance
b.
,
armature
needed
reciprocal of the per-unit value of
E B = £,/V3 =
= 8660 V
The base power
the short-circuit ratio
to the field current I x2
short-circuit ratio (7 X
The base voltage
,
to generate rated open-circuit
rated current / B
Solution
It is
R(pu)
X ZB
Equivalent
circuit of
a generator under load.
SYNCHRONOUS GENERATORS
1
.
2.
The
Isolated loads, supplied by a single generator
The
infinite
bus
diagram
resulting phasor
Note
16.20.
E
that
t)
given
is
E
leads
by
5
begin our study with isolated loads, leaving
greater than the terminal voltage, as
the discussion of the infinite bus to Section 16. 16.
In
Consider a 3-phase generator that supplies power
to a load
having a lagging power
some cases
the load
is
somewhat
that current / leads the terminal voltage
factor. Fig. 16. 19
represents the equivalent circuit for one phase. In or-
What
effect
The answer
does
.
E0
in Fig.
capacitive, so
by an angle
The
16.21.
across the synchronous reactance
the following facts:
list
Current
/
lags behind terminal voltage
E by
0.
voltage
EK
to the pha-
sum of £ and Ex However, the terminal voltage is
now greater than the induced voltage £0 which is a
sor
angle
0.
90° ahead of
is still
£0 is again equal
the current. Furthermore,
an
is
have on the phasor diagram?
this
found
is
Fig.
we would expect.
der to construct the phasor diagram for this circuit,
1
in
degrees.
Furthermore, the internally-generated voltage
We
we
35
.
,
2.
3.
Cosine 6
Voltage
= power
Ex
4.
EK
Voltage
the phasor
5.
Both
the
£0
by 90°.
/
- jIX s
E0
very surprising result. In effect, the inductive reac-
across the synchronous reactance
leads current
sion
factor of the load.
It is
given by the expres-
sum of £
Ex
plus
£x
O
is
equal to
<I)
rent / x
is
that
s
enters into partial resonance with the capacit
may appear we
something for nothing, the higher terminal
more power.
voltage does not yield any
load
is
entirely capacitive, a very high ter-
.
minal voltage can be produced with a small excitare voltages that exist inside
synchronous generator windings and cannot
Flux
X
reactance of the load. Although
If the
ing current.
However,
in later
that such under-excitation
be measured directly.
6.
itive
are getting
.
generated by the flux
and
tance
is
chapters,
we
will see
undesirable.
produced by the dc exciting cur-
Example 16-4
.
A
36
MVA,
20.8 kV, 3-phase alternator has a syn-
chronous reactance of 9
1
Cl
and a nominal current of
kA. The no-load saturation curve giving the
En
tionship between
If the excitation is
age remains fixed
and
/ x is
given
rela-
in Fig. 16.13b.
adjusted so that the terminal voltat
current required and
21 kV. calculate the exciting
draw
the phasor
diagram for
the following conditions:
Figure 16.20
Phasor diagram
for
a lagging power factor load.
a.
No-load
b.
Resistive load of 36
c.
Capacitive load of 12
MW
Mvar
Solution
We
/
shall
immediately simplify the circuit
to
show
only one phase. The line-to-neutral terminal voltage
for
all
cases
is
fixed at
E=
E
a.
20.8/V3
At no-load there
is
=
12
kV
no voltage drop
chronous reactance; consequently,
Figure 16.21
Phasor diagram
for
a leading power factor load.
E = £ =
tl
12
kV
in the
syn-
ELECTRICAL MACHINES AND TRANSFORMERS
352
Q=
E,E 0
1
2/3
-
4 Mvar
X
10 712
^12kV
The
Figure 16.22a
Phasor diagram
line current
= Q/E =
- 333 A
/
at no-load.
is
The voltage across
The exciting current
/x
b.
=
The
/
36/3
100
-
(see Fig. I6.l3b)
£ x =JIX = 1000 X
,/
This voltage
The voltage £ generated by
()
phasor sum of
sor diagram,
£o =
\
E and £ x
its
value
.
1000
A
is
equal to the
/ x is
kV
Figure 16.22c
Phasor diagram with a capacitive
E0 generated
£
The voltage
£<,
1
6.22c).
/x
kV
by
E
»
Note
that
£0
is
=
kV
12
load.
/ x is
equal to the
.
12
+ (-3)
70
A
is
(see Fig. 16.13b)
again less than the terminal volt-
age £.
The required exciting current
is
The phasor diagram
= 200 A
/x
c.
kV/!90 o
The corresponding exciting current
given by
15
x
= £ + £x =
= 9kV
Referring to the pha-
T~£; = Vl2 2 + 9 2 =
3
»
kV^90°
9
-
E0
9 kV
phasor sum of £ and
=
9
i
90° ahead of/.
is
is
Ex
the terminal voltage.
9
00Q
(333 A
is
The current is in phase with
The voltage across Xs is
S
s
(
leads / by 90° (Fig.
6.22a.
3
X K/712 000 =
12
1
As before £x
MW
12
full-load line current
= PIE =
X
£x = JIX,=j333 X
is
The phasor diagram is given in Fig.
With a resistive load of 36 MW:
The power per phase is
P =
4
(see Fig. 16.13b)
The phasor diagram is given in Fig. 16.22b.
With a capacitive load of 12 Mvar:
The reactive power per phase is
given
for this capacitive load
is
in Fig. 16.22c.
16.14 Regulation curves
When a single
load,
we
voltage
The
synchronous generator feeds a variable
are interested in
£ changes
relationship
knowing how
the terminal
as a function of the load current
between
£ and / is called the
/.
regula-
tion curve. Regulation curves are plotted with the field
excitation fixed and for a given load
Fig.
MVA,
E
1
1
Figure 16.22b
Phasor diagram with a
kA
unity
12
power
kV
factor load.
1
power
factor.
6.23 shows the regulation curves for the 36
21
kV,
3-phase
generator
discussed
in
Example 16-4. They are given for loads having
unity power factor, 0.9 power factor lagging* and
0.9 power factor leading, respectively. These curves
were derived using the method of Example 16-4,
SYNCHRONOUS GENERATORS
povver factor
"-^
The percent regulation
x
0.9 lacigincJ
%
.
regulation
353
is
=
X
100
rated load
10 Lr\/ mnn a
/
y lea dine
(\5-
=
12)
X
100
- 25%
12
I
We
note that the percent regulation of a synchro-
nous generator
erator.
much
is
The reason
is
greater than that of a dc gen-
the high impedance of the syn-
chronous reactance.
750
500
250
Load current
1250
1000
16.15 Synchronization
of a generator
/
Figure 16.23
We often
Regulation curves of a synchronous generator at
three different load
power
E 0 was
kept fixed instead of E. In each
power requirements of
connected
starting point for all the curves
Later,
terminal voltage
000 A).
The change
rent
load
(
(
was set so that the
was the rated line2 kV) at rated line cur-
1
is
voltage between no-load and
full-
expressed as a percent of the rated terminal
voltage.
load.
The percent regulation
is
given by the
a large utility system
to the
when
system to provide the extra power.
power demand
falls,
selected gen-
from the sys-
until
the
power again builds up
the following day.
Synchronous generators are therefore regularly being connected and disconnected from a large
grid in response to
En
"
regulation
many
X
100
power
customer demand. Such a grid
said to be an infinite bus because
equation
%
in
For example, as
erators are temporarily disconnected
tem
1
in
common
build up during the day, generators are successively
of the three cases, the value of E0
to- neutral
connect two or more generators
to
factors.
the
except that
have
parallel to supply a
it
is
contains so
generators essentially connected
in parallel
that neither the voltage nor the frequency of the grid
can be altered.
where
Before connecting a generator to an
=
EH =
(or in parallel with another generator),
no-load voltage VJ
[
synchronized.
rated voltage [VI
nized
1.
Example 16-5
power
factor curve in Fig. 16.23.
it
A
generator
meets
all
is
it
bus
must be
said to be synchro-
the following conditions:
The generator frequency
is
equal to the system
frequency.
Calculate the percent regulation corresponding to
the unity
when
infinite
2.
The generator voltage
is
equal to the system
voltage.
Solution
The
rated line-to-neutral voltage at full-load
is
3.
The generator voltage
is in
phase with the sys-
tem voltage.
EB =
12
kV
4.
The no-load terminal voltage
15
is
kV
The phase sequence of the generator
same as that of the system.
To synchronize an
alternator,
is
we proceed as
the
follows:
ELECTRICAL MACHINES AND TRANSFORMERS
1
.
Adjust the speed regulator of the turbine so that
the generator frequency
is
3.
ment has
frequency.
2.
En
is
E0 and E
(Fig. 16.24).
by
This
instru-
a pointer that continually indicates the
phase angle between the two voltages, covering
Adjust the excitation so that the generator volt-
age
Observe the phase angle between
means of a synchroscope
close to the system
the entire range
equal to the system voltage E.
from zero
to
360 degrees.
Although the degrees are not shown, the
dial has a
when the voltages are in
when we synchronize an alter-
zero marker to indicate
phase. In practice,
nator, the pointer rotates slowly as
it
tracks the
phase angle between the alternator and system
voltages. If the generator frequency
is
slightly
higher than the system frequency, the pointer rotates clockwise, indicating that the generator has a
tendency to lead the system frequency.
Conversely,
if
the generator frequency
is
slightly
low, the pointer rotates counterclockwise.
bine speed regulator
is
The
A fi-
that the pointer barely creeps across the dial.
nal
check
is still
4.
The
made
to see that the alternator voltage
equal to the system voltage. Then,
moment
Figure 16.24
Synchroscope.
is
the pointer crosses the zero
line circuit
tur-
fine-tuned accordingly, so
breaker
is
at the
marker
.
.
closed, connecting
the generator to the system.
{Courtesy of Lab- Volt)
Figure 16.25
This floating
r/min,
oil
derrick provides
60 Hz supply
all
its
own energy needs. Four
board are thyristor-controlled dc motors.
(Courtesy of Siemens)
diesel-driven generators rated
the electrical energy. Although ac power
is
1200 kVA, 440 V, 900
all the motors on
generated and distributed,
~
SYNCHRONOUS GENERATORS
In
is
modern generating
stations, synchronization
usually done automatically.
If
will
We
seldom have
except
parallel
in isolated locations (Fig,
As mentioned previously,
to
it
is
in
16.25).
An
bus
own
its
many
alternators
is
system so powerful
a
that
its
terminals.
Once con-
nous generator becomes part of a network comprising hundreds of other generators that deliver
thousands of loads.
It is
power
impossible, therefore, to
specify the nature of the load (large or small, resistive or capacitive)
connected
to the terminals of this
What,
then, determines the
particular generator.
tion,
=
(E0
-
E)/X s
Because the synchronous reactance
£x
the current lags 90° behind
(Fig.
is
1
therefore 90° behind E, which
is
inductive,
6.26b). The
machine delivers? To answer
we must remember
means
that
that both the value
we
were an induc-
it
we
Consequently, when
tive reactance.
a synchronous generator,
to the infinite bus.
The
it
over-excite
power
power increases as
supplies reactive
reactive
we raise the dc exciting current. Contrary
we might expect, is impossible to make a
it
tor deliver active
now
Let us
power by
raising
its
ques-
and the
can vary only two
genera-
excitation.
decrease the exciting current so that
Ea becomes smaller than E. As a result, phasor E =
£0 — £" becomes negative and therefore points to the
left (Fig.
6.26c). As always, current / = EJX lags
90° behind E x However, this puts / 90° ahead of E,
x
S
.
which means
it
that the alternator sees the
system as
if
were a capacitor. Consequently, when we under-
excite an alternator,
machine parameters:
what
to
1
this
frequency of the terminal voltage across the generator are fixed. Consequently,
if
it
nected to a large system (infinite bus), a synchro-
the
therefore circulate in the circuit
the generator sees the system as
voltage and frequency upon any
apparatus connected to
power
E
£o
it.
infinite
imposes
to
/
current
to
/ will
s
much more common
bus) that already has
(infinite
current
s
X
given by
given by
connect a generator to a large power system
connected
E
experience a difference of potential
A
connect only two generators
to
increase the exciting current, the volt-
increase and the synchronous reactance
=
Synchronous generator
on an infinite bus
16.16
we now
Ea will
age
355
it
draws reactive power from the
system. This reactive power produces part of the
1
.
2.
The
exciting current
magnetic
/x
The mechanical torque exerted by
Let us see
fects the
how
der
the turbine
field required
by the machine; the remain-
supplied by exciting current
is
—
16.18 Infinite bus effect of varying
the mechanical torque
—
16.17 Infinite bus effect of varying
the exciting current
Let us return to the situation with the synchronous
generator floating on the
connect
is
equal
it
after
we synchronize
to an infinite bus, the
to,
and
E of the system
in
a generator
and
induced voltage
£
()
phase with, the terminal voltage
(Fig.
1
6.26a). There
is
no difference
of potential across the synchronous reactance and,
consequently, the load current
the generator
no power;
it
is
is
.
a change in these parameters af-
performance of the machine.
Immediately
/x
/ is
zero.
connected to the system,
said to float
on the
line.
Although
it
delivers
and
in
phase. If
we open
line,
the
E0
and
E being
equal
steam valve of the
tur-
bine driving the generator, the immediate result
an increase
in
mechanical torque (Fig.
1
is
The
6.27a).
E0 will atmaximum value a little sooner than before.
£0 will slip ahead of phasor E, leading by
rotor will accelerate and, consequently,
tain
its
Phasor
a phase angle 8.
same
it
Although both voltages have
the
value, the phase angle produces a difference
Figure 16.26a
Generator
floating
on an
infinite
bus.
Figure 16.26b
Over-excited generator on an
infinite
bus.
Figure 16.26c
Under-excited generator on an
infinite
bus.
turbine
Figure 16.27
a.
Turbine driving the generator.
b.
Phasor diagram showing the torque angle
5.
356
SYNCHRONOUS GENERATORS
E = £ — E across the
of potential
x
()
synchronous
re-
actance (Fig. 16.27b).
A
current / will flow (again lagging 90° behind
£J, but
this
time
it is
almost
in
phase with E.
It
fol-
power delivered by
the generator also increases.
To
understand the physical meaning of the diagram,
let
us
tor will
continue to accelerate, the angle 8 will con-
tinue to diverge,
system
the
and the
power delivered to
up. However, as soon
electrical
will gradually build
power delivered to the system is
power supplied by the turwill cease to accelerate. The generator
examine the
and position of the
currents, fluxes,
poles inside the machine.
Whenever 3-phase
lows that the generator feeds active power into the
system. Under the driving force of the turbine, the ro-
currents flow in the stator of a
generator, they produce a rotating magnetic field
identical to that in an induction motor. In a synchrothis field rotates at the same speed
same direction as the rotor. Furthermore,
same number of poles. The respective
nous generator
and
in the
has the
as the electrical
it
equal to the mechanical
fields
bine, the rotor
stationary with respect to each other.
again run
will
angle 8 between
It
is
synchronous speed, and the Torque
at
Ea
and
£ will
remain constant.
important to understand that a difference of
potential
is
created
when two equal
voltages are out
of phase. Thus, in Fig. 16.27, a potential difference
of 4
kV
exists
between
Ea
and
E,
although both
produced by the rotor and
When
tween them.
the stator current /
veloped.
The only
16,19 Physical interpretation
of alternator behavior
ator (by admitting
it
the
phase angle between
value of
E
x
E0
and
E
increases, the
increases and, hence, the value of / in-
creases. But a larger current
means
that the active
may
be set up be-
the generator floats on the line,
zero and so no forces are de-
is
flux
that created
is
induces the voltage
If a
6.27b shows that when
Depending on
hand and the rotor poles on the other hand, powerful
forces of attraction and repulsion
and
1
stator are, therefore,
the relative position of the stator poles on the one
voltages have a value of 12 kV.
The phasor diagram of Fig.
357
£
mechanical torque
()
is
(Fig.
1
by the
rotor,
6.28a).
applied to the gener-
more steam
to the turbine), the
rotor accelerates and gradually advances by a
mechanical angle a, compared
sition (Fig.
begin
1
to its original po-
6.28b). Stator currents immediately
to flow,
owing
to the electrical
5 between induced voltage
E 0 and
phase angle
terminal volt-
age E. The stator currents create a revolving field
I:
I
/
Figure 16.28a
The N poles of the
of the stator.
rotor are lined
up with the S poles
Figure 16.28b
The N poles of the
the stator.
rotor are
ahead
of the
S poles
of
ELECTRICAL MACHINES AND TRANSFORMERS
358
and a corresponding
set
of
N
and S poles. Forces
P =
of attraction and repulsion are developed be-
-
sin
8
(16.5)
tween the stator poles and rotor poles, and these
magnetic forces produce a torque that opposes
where
P —
En =
E=
X =
8 =
mechanical torque exerted by the turbine.
the
When
the electromagnetic torque
is
equal to the
mechanical torque, the mechanical angle will no
longer increase but will remain
at
a constant
active power, per phase
There
is
a direct relationship
a and
chanical angle
between the me-
the torque angle 8, given
b
=
pa/2
by
(16.4)
[
V
]
terminal voltage, per phase [V]
synchronous reactance per phase
s
value a.
[W]
induced voltage, per phase
torque angle between
En
This equation can be used under
tions, including the case
when
and
all
E
[il]
[°]
load condi-
the generator
is
con-
nected to an infinite bus.
where
To understand
8
=
torque angle between the terminal
voltage
E0
E and
the excitation voltage
meaning, suppose a generator
infinite
generator
is
bus having a voltage E.
kept constant so that
The term E0 E/X S is then
power which the alternator
the generator
mechanical angle between the centers
vary directly with sin
of the stator and rotor poles [mechanical
its
connected to an
Furthermore, suppose that the dc excitation of the
[electrical degrees)
p — number of poles on
a ~
is
degrees]
we admit more
Thus, as
and
so, too, will the active
is
constant.
delivers to the bus will
8, the sine
gle.
£0
fixed, and the active
of the torque an-
steam, 8 will increase
power output. The relatwo is shown graphically in
Note that between zero and 30° the
tionship between the
Example 16-6
The rotor poles
shift
by
1
.
Fig.
of an 8-pole synchronous generator
0 mechanical degrees from no-load
to full-
load.
gle of 30°.
Calculate the torque angle between
a.
16.29.
power increases almost linearly with the torque
angle. Rated power is typically attained at an an-
E0
and the
E at full-load.
E or £u is leading?
terminal voltage
Which
b.
voltage,
,
Solution
a.
The torque angle
8
b.
When
is:
= pa/2 =
= 40°
8
X
10/2
a generator delivers active power,
ways leads
E0
al-
E.
16.20 Active power delivered
by the generator
We
can prove (see Section 16.23) that the active
power delivered by a synchronous generator
given by the equation
is
Figure 16.29
Graph showing the relationship between the active
power delivered by a synchronous generator and the
torque angle.
SYNCHRONOUS GENERATORS
=
However, there is an upper limit to the active
power the generator can deliver. This limit is
reached when 8 is 90°. The peak power output is
then
P m AX — E„E/X
.
.
S
If
we
the infinite bus.
The
tion
is
trip as
flow
in
try to
exceed
and
X
(3
13.3)
= 4()MW
the stator. In practice, this condi-
soon as synchronism
16.21 Control of active
large, pulsating cur-
never reached because the circuit breakers
power
is,
this limit
rotor will turn faster than the
is lost.
We
resynchronize the generator before
liver
the alternator
therefore,
and lose synchronism with
rotating field of the stator,
rents will
MW
The peak power output of
(such as by admitting more steam to the turbine),
the rotor will accelerate
13.3
359
it
then have to
can again de-
When
synchronous generator
a
system,
speed
its
is
connected
to a
kept constant by an extremely
is
sensitive governor. This device can detect speed
changes as small as 0.01%.
system sensitive
to the grid.
power
An
automatic control
such small speed changes im-
to
mediately modifies the valve (or gate) opening of
Example 16-7
the turbine so as to maintain a constant speed and
A
constant
36
MVA,
21 kV, 1800 r/min, 3-phase generator
On
connected to a power grid has a synchronous reactance of 9 (1 per phase. If the exciting voltage
is
kV (line-to-neutral), and the system voltage
17.3 kV (line-to-line), calculate the following:
12
is
in
b.
before
We
network
have
is
is
done
as
more elaborate systems
under the control of a computer.
individual overspeed detectors are
always ready
to
particularly
a generator, for
if
station so that
and transmission of energy
efficiently as possible. In
the entire
stations.
communicate with each other
modify the power delivered by each
In addition,
Solution
a.
station operators
the generation
of step (loses synchronism)
falls out
it
power delivered by
advance between the various generating
The
The active power which the machine delivers
when the torque angle 8 is 30° (electrical)
The peak power that the generator can deliver
output.
each generator depends upon a program established
to
a.
power
a big utility network, the
respond to a large speed change,
one reason or another,
should suddenly become disconnected from the
Ea =
E=
8 =
kV
12
17.3
system. Because the steam valves are
kV/V3 =
10
kV
attain a
30°
onds.
The
active
power delivered
P = (E„EIXS)
=
The
total
6.67
X
10/9)
X
grid
is
X
centrifugal forces at synchronous speed
any excess speed can quickly create a very
situation.
Consequently, steam valves
gencies. At the
- 20
all
three phases
is
burners must be shut
MW
sin
90
10/9)
X
same
is
attained
time, the pressure build-up
off.
16.22 Transient reactance
synchronous generator connected
to a
system
subject to unpredictable load changes that
1
in
the steam boilers must be relieved and the fuel
A
X
may
to 5 sec-
must immediately be closed off during such emer-
(E0 EIX S )
(12
4
0.5
The maximum power, per phase,
when 8 = 90°.
P =
=
The
in
wide
are already close to the limit the materials can with-
dangerous
MW
6.67)
speed 50 percent above normal
stand, so
8
power delivered by
(3
b.
(12
sin
power
to the
still
open, the generator will rapidly accelerate and
times occur very quickly.
In
is
some-
such cases the simple
ELECTRICAL MACHINES AND TRANSFORMERS
360
X'
short-
direct bearing
circuit
at the
on the capacity of the
circuit breakers
generator output. In effect, because they must
short-circuit
a
interrupt
cycles,
it
three
in
to
six
follows that they have to interrupt a very
normal
high current.
load
On
the other hand, the low transient reactance
simplifies the voltage regulation
problem when the
load on the generator increases rapidly. First the
drop due
ternal voltage
-short-circuit-
load
would be
ing.
if
to X' ci
the synchronous reactance
X
Second,
f
in-
smaller than
is
X
s
were
below
stays at a value far
X
it
act-
for a
s
sufficiently long time to quickly raise the exciting
current
/x
.
which helps
•V'd
-
/
Raising the excitation
increases
E0
.
to stabilize the terminal voltage.
•
Example 16-8
A 250 MVA,
time
25 kV, 3-phase steam-turbine gener-
ator has a synchronous reactance of 1.6 pu and a
Figure 16.30
transient reactance X' d of 0.23 pu.
Variation of generator reactance following a short-
rated output at a
circuit.
circuit
power
factor of
suddenly occurs on the
It
delivers
100%.
line,
A
its
short-
close to the
generating station.
equivalent circuit
flect the
shown
in Fig.
does not
16. 17
re-
behavior of the machine. This circuit
only valid under steady-state conditions or
is
when
the load changes gradually.
tance X'
X
s
X
f
varies
when
a generator
is
suddenly short-circuited. Prior to the short-circuit,
the synchronous reactance
is
X
simply
However,
s
much lower value X' d
creases gradually until
upon
100
T.
it
is
kVA
it
only
in
lasts a fraction
the 1000
It
.
again equal to
The duration of the
the size of the generator. For
machines
the circuit breakers should fail to
a.
The base impedance of
then in-
X
interval
s
after a
depends
it
The
sy
2
=
25 000 /(250
=
2.5 fl
nchronous reactance
X
10
6
)
is
may
last as
X =X
s
=
long
s
(pu)
1.6
X
ZB
2.5
= 4il
The reactance X' d
of the alternator.
It
is
may
called the transient reactance
be as low as
synchronous reactance.
short-circuit current
to the
is
machines below
of a second, but for
MVA range
the generator
at
as 10 seconds.
sponding
open
Solution
immedi-
the instant of short-circuit, the reactance
time interval
The induced voltage E0 prior to the short-circuit
The initial value of the short-circuit current
The final value of the short-circuit current if
varies as a function of time.
shows how
ately falls to a
c.
must be replaced by another reac-
whose value
Fig. 16.30
a.
b.
For sudden load current changes, the synchronous reactance
Calculate
is
1
5 percent of the
Consequently,
much higher
the
The
initial
rated line-to- neutral voltage per phase
E = 25/\3 =
14.4
kV
than that corre-
synchronous reactance
JVS
.
This has a
The
rated load current per phase
is
is
SYNCHRONOUS GENERATORS
36
rated
—-load —I—
short-circuit
kA
,47.3
-i
Change
Figure 16.31
See Example 16-8.
current
in
5/V3
E
6
= 250 X
X
10 /(1.73
which
5
6
s
short-circuit occurs across
See Example
drop
Ex
shows
and during the
is
23.1
T of
terval
= 5774 X
IX S
= EJX, =
= 6.8 kA
16-8.
27.2/4
only 1.2 times rated current.
is
Fig. 16.32
internal voltage
£x =
=
1
1
1
4
25 000)
= 5774 A
The
when a
the terminals of a generator.
/
=
1
2
3
* time
1
Figure 16.32
5780 A
/
1
1
0
4
We
assume
to,
a time in-
5 seconds. Note that in practice the cir-
would
cuit breakers
kV
the generator current prior
short-circuit.
certainly trip within 0.
1
s
after
the short-circuit occurs. Consequently, they have to
The current is in phase with E because
power factor of the load is unity. Thus,
interrupt a current of about
the
ring to the phasor diagram (Fig. 16.31),
47 kA.
refer-
E0
is
16.23 Power transfer between
E =
i}
\
E2 +
E\
2
V14.4 +
The
2
The
kV
27.2
b.
two sources
23.1
transient reactance
circuit of Fig.
because
it is
0.23
6.33a
is
particularly important
in the
study of generators,
synchronous motors, and transmission
is
we
such circuits
=
1
encountered
X
In
lines.
are often interested in the active
power transmitted from a source A to a source B
vice versa. The magnitude of voltages £, and E 2
2.5
<
=
The
0.575
initial short-circuit
n
or
as
well as the phase angle between them, are quite ar-
current
bitrary.
Applying Kirchhoff s voltage law
to this
is
= EJX
= 27.2/0.575
circuit,
we
obtain the equation
l1
=
47.3
If
kA
we assume
angle 6 and
which
c.
If
is
8.2 times rated current.
the short-circuit
tion
is
is
sustained and the excita-
unchanged, the current
E
that / lags
leads
]
E2
behind £\ by an arbitrary
by an angle
phasor diagram shown (Fig.
leads
/
1
<X
we
obtain the
6.33b). Phasor IX
by 90°. The active power absorbed by B
is
will eventually
level off at a steady-state value:
P = E2 I cos
(16. 6)
ELECTRICAL MACHINES AND TRANSFORMERS
362
The active power alway s flows from
the lagging voltage. In Fig.
leads
E2
;
1
6.33,
it
hence power flows from
is
the leading to
obvious
that
E
{
left to right.
Example 16-9
Referring to Fig. 16.33a, source
age
age
A generates
a volt-
E = 20 kV Z 5° and source B generates a
E 2 — 15 kV Z 42°. The transmission line
volt-
{
con-
necting them has an inductive reactance of 14
11.
Calculate the active power that flows over the line
and specify which source
is
actually a load.
Solution
The phase angle between the two sources is 42° —
5° = 37°. The voltage of source B leads that of
source A because its phase angle is more positive.
Consequently, power Hows from B to A and so A is
actually a load. The active power is given by:
E,E2
X
Figure 16.33
Power flow between two voltage sources.
.
- sin
o
20 kV X
(16.8)
kV
15
sin
37°
14
From
the sine law for triangles,
we have
20 000
/X/sin 8
E,/sin
=
£,/sin(90
= EJcos
Consequently,
Substituting
(
/
cos 0
=
P =
i\f
+
9)
12.9
Note
we
E =
E2 =
]
8
=
X=
active
active
(16.7)
E2
-
sin 8
(16.8)
power transmitted [W]
1
The physical
[VJ
phase angle between
E
and
x
E2
f°l
reactance connecting the sources
may seem, power flows
(
15
kV)
to
cost,
[11
The magnitude off
E
x
and
does not have
is
is
E2
machines
machine has a pro-
power output, relaand temperature rise. The following
its
why
efficiency,
these characteristics are
]
Let us consider a small ac generator having the
equal to that
:
determined by
the angle 6 be-
to be specified.
inti-
mately related.
following characteristics:
power P received by B
between
it
size of an electrical
found effect upon
analysis reveals
/
6
16.24 Efficiency, power, and size of
tive
and
10
the one having the higher voltage (20 kV).
find
delivered by A, because the reactance consumes no
tween
that, strange as
from the source having the lower voltage
voltage of source 2 [V]
the phase angle
X
12.9
1
X
voltage of source
active power.
=
MW
electrical
P =
000
B
where
The
15
14
£, sin h/X
16.7) in Eq. 16.6,
X
0.602
=
kW
power output
I
rated voltage
120 V, 3 phase
rated current
4.8
rated speed
1
A
800 r/min
SYNCHRONOUS GENERATORS
efficiency
73%
input torque
7.27
morhent of
0.0075 kg*m
inertia
Under these conditions, we can
N-m
of the generator as
erties
2
its
size
For example, suppose that
m
m
80
predict the prop-
is
increased.
dimen-
the linear
all
The volume
363
external diameter
0.
external length
0.15
mass
20 kg
mass
power output/mass
50 W/kg
Using
we can
The mass of the bigger machine will
therefore be 27 X 20 kg = 540 kg. The losses will
rise to 27 X 0.37 kW = lOkW.
The slots are 3 times wider and 3 times deeper.
As a result, the cross section of the conductors is 9
times greater which means they can carry 9 times
more current. The larger machine can therefore de-
this information,
1
sions are tripled.
crease by a factor of 3
P"
=
-
X
calculate the losses
100
kW
x
73
eq. 3.6
100
liver a current
As
power
input
=
P,
-
losses
1.37
kW
is
kW -
1.37
1
.0
losses comprise the
kW
that
kW
0.37
PR losses
in the
and eddy-current losses
the hysteresis
and the windage and
windings,
in the iron
way
that
actly the
same proportion, while keeping
materials throughout. Thus,
iron lamination
type
is
used
was used
in the larger
of insulation
is
if
such
in
dimensions are raised
linear
its
the
the
same
keep the same
current densities (A/itT) as in the original machine.
las)
in the
flux densities (tes-
various parts of the magnetic circuit
(core, air gap, stator teeth, etc.).
As
a result, the
losses per
cm
3
R
will be
original machine.
and
2
I
It
losses per
cm 3
B
is
flux density in the larger
windage and
assume
unchanged.
However,
before.
it
recall
machine
length
the
/
is
the
has
it.
same
tripled.
As
has the same
number of conductors
and be-
as before
cause they are connected the same way, the genera-
X
produce a voltage of 9
tor will
120
V =
1080
V.
Thus, by tripling the linear dimensions, the voltage and current both increase by a factor of
9.
This
power output increases 9X9 = 81
times. The power output of the new generator is
therefore 81 X
kW = 81 kW.
The power input needed to drive the ac generathat the
1
tor is P,
:
=
81
kW +
losses
kW. The new efficiency
=
is
81
kW +
10
kW =
therefore:
in the
71
=
°
vol-
81
friction
91
its
X
100
eq. 3.6
P;
=
kW
x
kW
0.89
100
- 89%
that the
(
left
the speed at
We
Blv.
the length of the conduc-
/ is
The
is
E =
which the flux cuts across
v
follows that the copper losses
same way.
number of slots, conductors and interconnections remain the same as
before and that the speed of rotation
800 r/min) is
further
the flux density,
and
losses also increase the
We
43.2 A.
and the iron
everywhere the same as
that the
A=
times because the diameter of the rotor has tripled.
91
iron losses will increase in proportion to
ume. Let's assume
4.8
regards the generated voltage per conductor,
means
same
X
creases by a factor of 9. Because the larger generator
nuts and bolts.
will
of 9
same
magnifying everything, including the bearings,
We
as
also used, thereby duplicating and
will also maintain the
so, too, will
a result, the voltage generated per conductor also in-
machine. The same type
we
27 and
ex-
in
a particular type of
in the stator,
In this larger generator
27. Consequently, the
Furthermore, the peripheral speed v has increased 3
friction losses.
Let us increase the size of the machine
a
=
a factor of
determined by equation (2.25)
tor
The
by
the losses.
of the machine:
r\
will increase
3
will therefore in-
1
The
which
the
efficiency has increased from
is
a dramatic improvement.
power output has increased
73%
to
The reason
89%
is that
81 times, while the
ELECTRICA L MA CHINES AND TRANSFORMERS
364
losses increased only
27 times. Consequently, the
bound to
ciency of the machine was
effi-
increase with size.
about 20()°C. Consequently, the cooling of large
machines
a very important matter.
is
The original machine produced an output of 50
W/kg. The larger machine has a mass of 540 kg and
produces 81 kW. Consequently, it produces 81
regarding physical size, power output, efficiency,
kW/540 kg = 150 W/kg which
including ac and dc motors and transformers.
3 times greater
is
In
conclusion, the general principles covered here
temperature
rise
and so
forth,
apply to
machines,
all
than before.
The
proof,
generator
larger
if
eighty-one
produce 81 kW,
X 20
therefore
is
relatively
Questions and Problems
and cheaper than the smaller machine. As
lighter
=
kg
kW
1
their
1620 kg.
Practical level
generators were used to
combined mass would be 81
This generating center would
1
6-
1
Why
obviously be more costly and take up more floor
space than the single 81
As another matter of
moment of
mass and
inertia
,/
kW
of a rotor
the square of
we
is
proportional to
Hence, when linear dimensions are
= nir —
243.
The moment of
therefore 243
The
X
inertia
tripled,
X
27
2
3
J
=
3
5
=
1.8
kgm 2
16-3
is
the original
1
kW
1080 V,
rated voltage
43.2
rated current
3
A
1800r/min
89%
moment
of inertia
Nm
1.8
kg-nr
external length
0.45
1
m
m
linear
dimensions are
temperature
rise.
the losses increase 27 times. Hence, the
power
dis-
sipated per square meter increases by a factor of
better
prevent
damage
is
directly-coupled genera-
If the
must generate a frequency of 60 Hz,
The number of poles on
The exaci turbine speed
An
the rotor
isolated 3-phase generator produces a
a load
If
is
cooling
hound
maintain the same line voltage?
What
conditions must be met before a gener-
Calculate the
number of poles on
system?
the genera-
means
to the insulating materials, the
3.
are
Calculate the
number of poles on
shown in Fig. 16.
6-8
A 3-phase
generator turning
Hz.
How
fected
to
its
if
200 r/min
will the terminal voltage be af-
terminals?
Resistive load
tem-
b.
Inductive load
c.
Capacitive load
of
1
1
the following loads are connected
a.
maximum
at
the air1
generates a no-load voltage of 9 kV, 60
To
to be hotter.
perature rise has to be limited to a
6-7
craft generator
1
unless
6-6
When
tripled, the heat-dissipating
used, the larger machine
found
close to
tor in Fig. 16.12 using the information given.
150W/ks
is
is
it
350 r/min.
at
ator can be connected to a 3-phase
surface area of the machine increases 9 times but
Consequently,
6-5
540 kg
big problem
site,
should turn
to
1
The one
analyzing a hydropower
connected to the machine, must the exci-
1
mass
power output/mass
the larger?
tation be increased or decreased in order
483
0.54
is
having a lagging power factor of 0.8
rated speed
external diameter
For a given power output, which of
no-load line voltage of 13.2 kV.
phase
efficiency
input torque
In
a.
16-4
kW
main differences between steam-
that the turbines
b.
81
wye?
calculate the following:
are in striking contrast to
machine.
power output
State the
tors
.
characteristics of the larger generator are
summarized below. They
synchronous generators?
in large
the stator always connected in
these machines
=
of the larger machine
0.0075 kg-nr
is
ators.
its
will
are the advantages of using a stationary
turbine generators and salient-pole gener-
recall that the
radius (see Table 3A).
its
increase by a factor of J
16-2
generator.
interest,
What
armature
SYNCHRONOUS GENERATORS
16-9
In
Problem
16-8,
if
the field current
Calculate
is
kept constant, calculate the no-load volt-
a.
age and frequency
b.
the speed
if
is
The synchronous impedance Z per phase
The total resistance of the circuit, per
s
lOOOr/min
a.
phase
5 r/min.
b.
365
The
c.
total reactance
of the circuit, per
phase
Intermediate level
16-10
What
meant by the synchronous
is
tance of a 3-phase generator?
16-11
reac-
Draw
nator
meaning of
all
the parameters.
State the advantages of brushless excita-
Using a schematic
how
circuit diagram,
the rotor in Fig. 16.7
16-17
show
needed
a
a.
24.2
12.1
A 3-phase
£0
16-18
Ev
kV
16
generator possesses a synchro-
is
100
(1,
calculate the value of
per phase.
,
The generator
in Fig.
is
3
kV
connected to an
Q and the excitation
per phase
to
load having a lagging
1
6.2 has a synchro-
nous reactance of 0.4 H, per phase.
nous reactance of 6
voltage
KVA,
factor of 0.8. If the synchronous re-
actance
kV
kV
generator rated 3000
2400 KVA,
power
to generate a no-
load line voltage of
b.
A 3-phase
20 kV, 900 r/min, 60 Hz delivers power
excited.
is
the volt-
age across the load
Referring to Fig. 16.13, calculate the exciting current
16-13
The phase angle between E0 and
f.
the
tion systems over conventional systems.
16-12
h.
e.
equivalent circuit of a generator and explain the
g.
The line current
The line-to-neutral voltage across the load
The line voltage across the load
The power of the turbine driving the alter-
d.
It is
bus having a
infinite
line
voltage of 14 kV, and the excitation volt-
(ref. Fig.
age
is
adjusted to
1
.
14 pu.
16.19). Calculate the line-to-neutral volt-
age
E for
a resistive load of 8 (2 and draw
Calculate
the phasor diagram.
a.
16-14
a.
In
Problem 16-13, draw
the curve of
E
for the following resistive loads: infin-
sus
/
ity,
24. 12, 6,
3,0 ohms.
power P per phase
b. Calculate the active
b.
c.
in
inside stator circumference) corresponding
each case.
c.
Draw
to this
the curve of
E
versus
value of load resistance
a
is
P.
the
For what
power output
16-19
Referring to Fig.
1
6.2, calculate the length
of one pole-pitch measured along the
16-16
on the 500
[in].
M VA alternator of
Fig. 16.2 yields the following results:
maximum?
ternal
displacement angle
A test taken
1.
16-15
The torque angle 8 when the generator delivers 420
The mechanical displacement angle a
The linear pole shift (measured along the
MW
ver-
circumference of the
in-
2.
stator.
The 3-phase generator shown
Open-circuit line voltage
is
15
kV
for a dc
exciting current of 1400 A.
Using the same dc current, with the armature short-circuited the resulting ac line current is 21
in Fig.
000 A.
16.16 has the following characteristics:
Calculate
E0 = 2440 V
X =
s
R =
load impedance
Z=
a.
144(1
17 12
175 (2 (resistive)
The base impedance of
the generator, per
phase
c.
The value of the synchronous reactance
The per-unit value of X s
d.
The
b.
short-circuit ratio
366
ELECTRICAL MACHINES AND TRANSFORMERS
Advanced
16-20
16-25
level
b.
d.
tween
going
incoming
the cool
air, if
the air flow
air
is
of 500 A.
1
.3
b.
be-
280 nrVs
16-26
gap
(See Section
16-22
2.
1
the line circuit breakers suddenly trip, cal-
second
1
later,
assum-
By how many mechanical degrees do
1
how many
degrees?
A 400 Hz
electrical
By
alternator has a 2-hour rating of
nal diameter of
mmf re-
length of 9.5
80
1
V,
7).
and has an
slots
22 inches and an
in.
The
rotor
for a field current of 3
Referring to Fig. 16.17, the following
the
-second interval?
position) during the
stator possesses
quired for the iron portion of the magnetic
circuit.
.
80
percent power factor (Fig. 16.34a). The
gap length
no-load. Neglect the
at
lb
kg nr.
75 kVA, 1200 r/mim 3-phase, 450
inches, calculate the flux density in
the air
6
poles advance (with respect to their normal
and warm out-
that the air
10
10
culate the speed of the generating unit (tur-
and carries a dc current
Knowing
X
6
driven
is
whose moment of
2
ft
The rotor has a J
ing that the wicket gates remain wide open.
Referring to Fig. 16.4, each coil on the rotor has 21 .5 turns,
If
X
54
bine and alternator)
The torque developed by the turbine
The average difference in temperature
c.
is
of 4.14
a.
Problem 16-20
in
a hydraulic turbine
inertia is
The total losses in the machine
The copper losses in the rotor
a.
16-21
The generator
by
The synchronous generator in Fig. 16.2
has an efficiency of 98.4% when it delivers an output of 500 MW. Knowing that
the dc exciting current is 2400 A at a dc
voltage of 300 V, calculate the following:
1
is
A at
inter-
axial
designated
1
1
5 V.
in-
Calculate
formation
is
given about a generator:
c.
The number of poles on the rotor
The number of coils on the stator
The number of coils per phase group on
d.
The
a.
Ea =
E=
12
kV
14
kV
x =
2
b.
the stator
s
Ea
leads
a
E
e.
a.
length of one pole pitch, measured
along the circumference of the stator
by 30°
Calculate the total active power output of
The
tor
resistance of the dc winding on the ro-
and the power needed to excite
it
the generator.
1
6-23
b.
Draw
c.
Calculate the power factor of the load.
the phasor
diagram for one phase.
The steam-turbine generator shown
16.3 has a
The
excitation voltage
pu and the machine
is
E0
is
16-27
in Fig.
synchronous reactance of
1
.3
pu.
adjusted to 1.2
connected
to
an
hid List rial a pplica tion
A 33.8
kVA, 480
operate
at
a
bus of 19 kV.
If
the torque angle 5
16-24
active
b.
The
line current
c.
Draw
is
given:
83.4%
Weight: 730
Wk
2
lb
(moment of
Insulation: class
the phasor diagram, for
80 percent. The
is
power output
The
factor of
diesel-
designed to
infi-
20°, calculate the following:
a.
power
60 Hz
is
following additional information
Efficiency:
nite
V, 3-phase,
driven emergency alternator
one phase
In Problem 16-23. calculate the active
power output of the generator if the steam
valves are closed. Does the alternator receive or deliver reactive power and how
much?
inertia)
:
15.7
2
lb. ft
B
Calculate
a.
The minimum horsepower
rating of the
diesel engine to drive the generator
b.
The maximum allowable temperature of
windings, using the resistance method
the
SYNCHRONOUS GENERATORS
16-28
A 220 MVA, 500
0.9
power
r/min, 13.8 kV,
factor, water-turbine
50 Hz,
Efficiency
nous generator, manufactured by Siemens,
Insulation class:
mass of
Total
(t
=
stator:
Unsaturated synchronous reactance: 1.27 pu
Runaway speed
158
2
mass of
rotor:
Static excitation
t
current
270
in
generator mode:
890 r/min
metric ton)
Total
factor:
Transient reactance: 0.37 pu
F
of inertia: 525 t*m
power
98.95%
synchro-
has the following properties:
Moment
at full-load, unity
367
is
is
used and the excitation
2980 A under an
excitation volt-
age of 258 V.
t
Figure 16.34a
Rotor and stator of a 75 kVA, 1200 r/min, 3-phase, 450
driven by a 100 hp, 1200 r/min synchronous motor.
V,
400 Hz
alternator for shipboard use.
The
alternator
is
Figure 16.34b
Stator and rotor of the 100 hp, 1200 r/min, 60
serves as a base
for
the alternator.
The
Hz synchronous motor. The
rotor is
duction motor. (Courtesy of Electro-Mecanik)
equipped with a
squirrel
stator is mounted on a bedplate that also
cage winding to permit starting as an in-
368
ELECTRICAL MACHINES AND TRANSFORMERS
The generator
also designed to oper-
is
16-29
In industry application
Problem 16-28,
ate as a motor, driving the turbine as a
calculate the following:
pump. Under these conditions,
a.
The horsepower
b.
when it runs as a pump motor
The kinetic energy of the rotor when
develops an output of 145
the
motor
MW.
Both the stator and rotor are water-
at rated
cooled by passing the water through the
c.
hollow current-carrying conductors. The
The
is
treated so that
less than 5 (JiS/cm.
through the stator
its
conductivity
a rate of 8.9
second and through the rotor
at
liters
runs
when
kinetic energy of the rotor
its
maximum
it
allowable runaway
is
speed
The pure water Hows
at
it
speed
reaches
water
rating of the generator
d.
per
The time
to reach the
runaway speed
event that a short-circuit occurs
5.9 liters
erator
is
delivering
its
when
rated load,
in the
the gen-
and assum-
per second. Given the above information,
ing that the water continues to flow unchecked
make
through the turbine (gates wide open)
a.
the following calculations:
The
rated active
unity
power
power
output, in
MW
factor and at 0.9 lagging
at
power
factor
b.
c.
d.
The rated reactive power output, in Mvar
The short circuit ratio
The value of the line-to-neutral synchronous reactance, per phase
e.
16-30
In
Problem 16-28 calculate the power
dis-
sipated in the rotor windings and the
power
loss per pole.
Knowing
water flow and that the
is
inlet
the rate of
temperature
26°C, calculate the temperature of the
water flowing out of the rotor windings.
minimum
the total losses of the generator at full load
What
and unity power factor
the circulating water?
is
the
resistivity
(H-m) of
Chapter 17
Synchronous Motors
17.0 Introduction
motor speed stays constant, irrespective
fixed, the
of the load or voltage of the 3-phase
The synchronous generators described in the previous chapter can operate either as generators
or as motors.
When
they run
operating as motors (by con-
field.
The speed of
Most synchronous motors
synchronism with the revolving
rotation
is
150
therefore tied to the
frequency of the source. Because the frequency
constant speed but because they possess
We
will study
these features in this chapter.
synchronous motors. As the name implies, synchroin
at
other unique electrical properties.
necting them to a 3-phase source), they are called
nous motors run
However,
line.
synchronous motors are used not so much because
kW (200 hp) and
speeds ranging from
is
1
15
are
rated
between
MW (20 000 hp) and turn
at
50 to 800 r/min. Consequently,
1
these machines are mainly used in heavy industry
Figure 17.1
Three-phase, unity power factor synchro-
nous motor rated 3000 hp (2200 kW), 327
r/min, 4000 V, 60 Hz driving a compressor
used in a pumping station on the Trans-
Canada
pipeline.
Brushless excitation
provided by a 21 kW, 250
fier,
which
is
V
mounted on the
shaft
the bearing pedestal and the main
{Courtesy of General
369
is
alternator/recti-
Electric)
between
rotor.
ELECTRICAL MACHINES AND TRANSFORMERS
370
(Fig.
we
in
1
7.
1
).
At the other end of the power spectrum,
find tiny single-phase synchronous motors used
They
control devices and electric clocks.
cussed
17.1
in
Chapter
are dis-
rent
is
fed into the winding from an external exciter.
Slots are also
punched out along
ing similar to that in a 3-phase induction motor. This
18.
damper winding serves to start the motor.
Modern synchronous motors often employ
Construction
less excitation, similar to that
Synchronous motors are
identical in construction to
salient-pole ac generators.
The
stator
is
composed
of a slotted magnetic core, which carries a 3-phase
lap winding.
Consequently, the winding
is
also
identical to that of a 3-phase induction motor.
The rotor has
by a dc current
connected
a set of salient poles that are excited
(Fig.
in series to
the circumference
of the salient poles. They carry a squirrel-cage wind-
17.2).
two
The exciting
slip-rings,
coils are
and the dc cur-
used
in
brush-
synchronous
generators. Referring to Fig. 17.3, a relatively small
3-phase generator, called
fier are
mounted
at
dc current / x from the
salient-pole
exciter,
and a 3-phase
one end of the motor
rectifier
windings,
is
recti-
shaft.
The
fed directly into the
without
going
through
brushes and slip-rings. The current can be varied by
controlling the small exciting current Ic that flows in
the stationary field winding of the exciter. Fig. 17.4
Figure 17.2
Rotor of a 50 Hz to 1 6 2/3 Hz frequency
converter used to power an electric railway.
The 4-pole rotor at the left is associated
with a single-phase alternator rated 7000
kVA, 1 6 2/3 Hz, PF 85%. The rotor on the
right is for a 6900 kVA, 50 Hz, 90% PF
synchronous motor which drives the single-phase alternator. Both rotors are
equipped with squirrel-cage windings.
(Courtesy of ABB)
Figure 17.3
Diagram showing the main components
chronous generator.
of
a brushless exciter
for
1
-
dc control source
2
-
stationary exciter poles
3
-
alternator (3-phase exciter)
4
-
3-phase connection
5
-
bridge rectifier
6
-
dc
7
-
rotor of synchronous
8
-
stator of
9
-
3-phase input to stator
line
a synchronous motor.
motor
synchronous motor
It
is
similar to that of a syn-
SYNCHRONOUS MOTORS
shows how
mounted
The
the exciter, rectifier,
in a
and
kW synchronous
salient poles are
motor.
rotor and stator always have the
ber of poles.
the
3000
As
number of
in the
same num-
poles determines the synchronous
Example 17-1
Calculate the number of salient poles on the rotor of
the synchronous motor shown in Fig. 17.4a.
Solution
120
/
P
=
120,///?
frequency of the source [HzJ
poles
s
200 =
p =
speed fr/min]
p — number of
at
200 r/min;
(17.1)
/z
/=
60 Hz and runs
consequently,
where
= motor
at
The motor operates
-
s
J
case of an induction motor,
speed of the motor:
//
37
The
rotor possesses
1
(120
X
60)//?
36 poles
8 north poles and
1
8 south poles.
Figure 17.4a
Synchronous motor rated 4000 hp (3000
kW), 200 r/min, 6.9 kV, 60 Hz, 80%
power factor designed to drive an ore
crusher.
The brushless exciter (alternamounted on the overhung
tor/rectifier) is
shaft
and
is
rated
50 kW, 250
(Courtesy of General
V.
Electric)
Figure 17.4b
Close-up
of the
50
kW
exciter,
showing
the armature winding and 5 of the 6
diodes used
to rectify the
(Courtesy of General
ac current.
Electric)
ELECTRICAL MACHINES AND TRANSFORMERS
372
synchronous motor
17.2 Starting a
A
synchronous motor cannot
quently, the rotor
is
When
slightly
is
can
it
the stator
start
of stator
conse-
up as an induction
connected
is
motor accelerates
line, the
itself;
axis of S pole
usually equipped with a squirrel-
cage winding so that
motor.
by
start
axis of N pole
of rotor
until
to the
3-phase
reaches a speed
it
below synchronous speed. The dc excitation
suppressed during
While
this starting period.
the rotor accelerates, the rotating flux cre-
ated by the stator sweeps across the slower
salient poles.
moving
Because the coils on the rotor possess
number of turns, a high voltage is
rotor winding when it turns at low
a relatively large
induced
in the
speeds. This voltage appears between the slip-rings
Figure 17.5
The poles
of the rotor are attracted to the opposite
poles on the stator. At no-load the axes of the poles
coincide.
and
decreases as the rotor accelerates, eventually
it
becoming negligible when
synchronous speed. To
we
prove the starting torque,
slip-rings or connect
the rotor approaches
and
limit the voltage,
them
either short-circuit the
to
an auxiliary resistor
power capacity of
we sometimes have
ited,
to the stator.
As
in the
the supply line
is
lim-
reduced voltage
to apply
case of induction motors,
we
use either autotransformers or series reactors to
Chapter 20). Very
limit the starting current (see
synchronous motors (20
large
sometimes brought up
called a
tor,
lations the
more) are
speed by an auxiliary mo-
to
pony motor.
MW and
some
Finally, in
motor may be brought up
to
big instal-
pull-in torque of a
right
moment. For example,
motor
will
immediately slow
breakers will
chronous speed, the rotor
trip. In
This produces alternate
N
moment happen
at this
tion
is set
on the
to
at close to
syn-
excited with dc current.
and S poles around the
circumference of the rotor (Fig.
site polarity
1
strong magnetic attrac-
up between them. The mutual
attraction
locks the rotor and stator poles together, and the rotor
is literally
field.
yanked
the circuit
ment when excitation should be applied. The motor
with the revolving
Once
is
carries
the
induced
no
field.
motor turns
at
synchronous speed, no
in the squirrel-cage
winding and so
current. Consequently, the behavior of a
is
entirely different
from
that of
into step with the revolving
The torque developed
because of the magnetic attraction between the
poles of the rotor and the opposite poles of the
To reverse
the direction of rotation,
we
interchange any two lines connected to the
stator.
simply
stator.
7.5). If the poles
be facing poles of oppo-
stator, a
down and
practice, starters for synchro-
an induction motor. Basically, a synchronous motor
running
is
re-
then pulls automatically and smoothly into step
rotates
is
magnetic
nous motors are designed to detect the precise mo-
voltage
17.3 Pull-in torque
motor
should happen that
stator, the resulting
synchronous motor
as the
it
pulsion produces a violent mechanical shock. The
it
As soon
if
is
at the
emerging N, S poles of the rotor are opposite the
N, S poles of the
speed by a
variable-frequency electronic source.
synchronous motor
powerful, but the dc current must be applied
the
during the starting period.
If the
The
im-
to
at this
propriately called the pull-in torque.
moment
is
ap-
17.4 Motor under load
general description
When
a synchronous motor runs at no-load, the ro-
tor poles are directly opposite the stator poles and
their axes coincide (Fig.
1
7.5).
However,
ply a mechanical load, the rotor poles
if
we
ap-
fall slightly
SYNCHRONOUS MOTORS
373
a bigger mechanical load, and the increased power
can only
come from
the 3-phase ac source.
17.5 Motor under load
simple calculations
We
can get a better understanding of the operation
of a synchronous motor by referring to the equiva-
shown
lent circuit
in Fig.
1
7.7a.
It
phase of a wye-connected motor.
represents one
It
identical to
is
the equivalent circuit of an ac generator, because
both machines are built the same way. Thus, the
Figure 17.6
The rotor poles are displaced with respect to the axes
of the stator poles when the motor delivers mechani-
flux
<£>
rent
cal power.
/x
As
.
E0
Consequently,
a between
the poles increases progressively as
increase the load (Fig.
more powerful torque
is
to the
and the motor develops an ever
field,
But there
we
Nevertheless, the
17.6).
magnetic attraction keeps the rotor locked
revolving
to
The mechanical angle
E0 is in
motor comes
step creates a
away from
to a halt.
the stator poles
circuit breakers
and
A motor that pulls out of
major disturbance on the
immediately
trip.
line,
and the
This protects the
motor because both the squirrel-cage and
stator
windings overheat rapidly when the machine ceases
to run at
The
Under
7.7b).
E0 =
If,
we
in addition,
adjust the
motor "floats" on the
E, the
line current /
practically zero. In ef-
is
is
to
supply the small
friction losses in the motor,
and so
it is
negligible.
the shaft?
if
we
The motor
apply a mechanical load to
will begin to
causing the rotor poles to
poles by an angle a.
E0
reaches
its
Due
fall
to this
maximum
electrical degrees
mechanical
value a
behind
E.
slow down,
behind the stator
E
is
i}
E0
tial
Ex
than
now
The mechanical
placement a produces an electrical phase
between
shift,
little later
before. Thus, referring to Fig. I7.7c,
8
dis-
shift 5
and E.
The phase
upon the magneto-
these conditions, in-
phase with the line-to-neutral
only current needed
windage and
synchronous speed.
pull-out torque depends
1
What happens
ceeds the pull-out torque of the motor, the rotor
the
and the
fect, the
as the angle increases.
a limit. If the mechanical load ex-
poles suddenly pull
voltage £(Fig.
excitation so that
line
varies with the excitation.
already mentioned, the rotor and stator poles
duced voltage
turn at synchronous speed.
in
depends on the dc exciting cur-
are lined up at no-load.
behind the stator poles, but the rotor continues
£0
created by the rotor induces a voltage
the stator. This flux
shift
produces a difference of poten-
across the synchronous reactance
X
s
given by
motive force developed by the rotor and the stator
poles.
The
mmf of the
dc excitation
/x ,
rotor poles
while that of the stator depends
upon the ac current flowing
pull-out torque
is
in
E x — E — E0
depends upon the
The
nom-
the windings.
usually 1.5 to 2.5 times the
Consequently, a current
j'X,
inal full-load torque.
The mechanical angle a between
stator poles has a direct bearing
As
to
the rotor and
/
must flow
given by
=
E*
from which
on the stator current.
the angle increases, the current increases. This
is
be expected because a larger angle corresponds to
/
= -jEJX
= -j(E -
s
E„)/X x
in the circuit,
374
ELECTRICAL MACHINES AND TRANSFORMERS
Example 17-2a
A 500 hp, 720 r/min synchronous motor connected to
a
3980
age
V, 3-phase line generates an excitation volt-
£n of
1790
ing current
II
is
V
(line-to-neutral)
when
the dc excit-
25 A. The synchronous reactance
and the torque angle between
Ea
and
E
is
22
30°.
is
Calculate
a.
b.
Figure 17.7a
c.
Equivalent circuit of a synchronous motor, showing
d.
one phase.
The value of £ x
The ac line current
The power factor of the motor
The approximate horsepower developed by
the
motor
e.
E Eo
The approximate torque developed
at the shaft
Solution
This problem can best be solved by using vector no-
Figure 17.7b
Motor
tation.
at no-load, with
EQ
adjusted to equal E.
a.
The voltage E
(line-to-neutral) applied to the
motor has a value
Ex
—
E - E0
E = EjV3 = 3980/V3
4
= 2300 V
Let us select
E
E
as the reference phasor,
whose
angle with respect to the horizontal axis
sumed
is
as-
to be zero. Thus,
E = 2300Z0 0
It
follows that
E0
is
given by the phasor
Ea = 1790/1-30°
Figure 17.7c
Motor under load
17.7b, but
it
The equivalent
EQ
has the same value as
lags behind E.
circuit per
phase
Moving clockwise around
The current lags 90° behind E x because Xs is inductive. The phasor diagram under load is shown in
Because
/ is
nearly in phase with E, the
motor absorbs active power. This power
is
entirely
transformed into mechanical power, except for the
relatively small
copper and iron losses
In practice, the excitation voltage
to
in the stator.
E0
is
adjusted
be greater or less than the supply voltage E.
value depends upon the
and the desired power
power output of
factor.
the
given
the circuit
plying KirchhofTs voltage law
Fig. I7.7c.
is
in Fig.
17.8a.
in Fig.
-E + E + E0 =
x
EX =
E-
write
0
E0
= 2300^0° - 1790Z-30 0
= 2300 (cos0° + ;sin0°) 1790 (cos -30°
= 2300 -
1550
= 750 + j
895
Its
motor
and ap-
we can
- 1168Z50 0
+j
+7
sin
895
-30°)
SYNCHRONOUS MOTORS
375
Ex
1168 V
Figure 17.8a
Equivalent circuit of a synchronous motor connected
to a
source
E.
Thus, phasor
leads phasor
b.
The
Ex has a value
E by 50°.
of
1
1
68
V
and
it
1790
Figure 17.8b
fl2I = Ex
_
~
1
See Example
A 50 °
A 90°
22
Approximate torque:
e.
9.55
T=
A and
Thus, phasor / has a value of 53
40° behind phasor E.
The power
motor
factor of the
it
=
is
current
E across
/.
Hence,
power
X
9.55
3715
X
1
10
3
720
N-m
Example 17-2b
The motor in Example 17-2a
has a stator resistance
of 0.64 Q. per phase and possesses the following
factor
=
cos 6
=
=
0.766, or
losses:
cos 40°
2
I
76.6%
R
losses in the rotor:
3,2
Stator core loss:
The power
280.
given by the
motor terminals and the
the
X P
n
lags
cosine of the angle between the line-to-neutral
voltage
17-2.
168
= 53^-40°
c.
V
given by
line current / is
factor
is
lagging because the current
Windage and
3.3
friction loss:
1
.5
kW
kW
kW
lags behind the voltage.
The complete phasor diagram
is
shown
in Fig.
17.8b.
d.
Calculate
a.
Total active
power input
to the stator:
b.
c.
Pi
=
3X
-
3
L
Solution
cos 40°
a.
W=
142
280.
Power
input to the stator
lkW
Stator/
Neglecting the
shaft
E, N / cos 6
X 2300 X 53 X
= 280
The actual horsepower developed
The actual torque developed at the
The efficiency of the motor
2
I
R
2
/?
losses
-
3
X
is
53
2
280.
X
Total stator losses
=
power transmitted across
is 280.
kW.
Power transmitted
to the rotor
stator, the electrical
the airgap to the rotor
1
=
271.4
+
kW
Q = 5.4 kW
= 8.7 kW
= 280. - 8.7
0.64
losses and iron losses in the
5.4
1
3.3
1
kW
Approximate horsepower developed:
P - 280.1 X
3
10 /746
=
375 hp
The power at the shaft
minus the windage and
is
the
power
to the rotor
friction losses.
The
rotor
ELECTRICAL MACHINES AND TRANSFORMERS
376
IR
losses are supplied by an external dc source
and so they do not affect the mechanical power.
chanical power.
by
Power
available at the shaft:
P0 =
271.4
-
269.9
X
l(r
-
Example
The corresponding torque is:
value calculated in
b.
T -
X P
9.55
P=
361.8 hp
H
-
therefore expressed by
"
sin 8
(17.2)
of the motor, per
phase [Wl
X
269.9
10*
720
//
E E
P = mechanical power
17-2a.
X
9.55
is
where
very close to the approximate
is
synchronous motor
kW
269.9
746
This power
a
the equation
=
1.5
is available in the form of meThe mechanical power developed
across the air gap
En =
E—
line-to-neutral voltage of the source [V]
X =
synchronous reactance per phase
s
= 3580 N-m
=
8
line-to-neutral voltage induced by
torque angle between
£
(>
and
[V]
/x
E
[electrical degrees]
c.
=
Total losses
Total
Total
4-
3.3
4
3.2
4
1.5
=
13.4
power input = 280.1 4- 3.2 = 283.3
power output = 269.9 kW
Efficiency
Note
5.4
=
=
269.9/283.3
=
0.9527
that the stator resistance of 0.64
compared
small
to
95.3
Q
Consequently, the true phasor diagram
close to the phasor diagram of Fig.
kW
is
%
very
of 22
reactance
the
kW
1
is
Q.
This equation shows that the mechanical
power increases with the torque angle, and its
maximum value is reached when 8 is 90°. The
poles of the rotor are then midway between the N
and S poles of the stator. The peak power P max
(per-phase)
very
far as torque
tional to the
Power and torque
speed
When
a synchronous motor operates under load,
draws active power from
given by the same equation
The power
the line.
we
t>
As
in the
s )
is
sin 8
voltage E, the excitation voltage
PR
power
is
This
is
If
we
and iron losses
to
the
is
directly propor-
fixed.
The torque
is
derived from Eq. 3.5:
where
T=
P =
(16.5)
power
£
()
,
and the phase
neglect the relatively
in the
stator,
all
the
transmitted across the air gap to the rotor.
analogous
it
is
torque, per phase [N-m]
mechanical power, per phase [W]
ns
=
synchronous speed [r/min]
9.55
=
a constant [exact value
absorbed by the motor depends upon the supply
small
concerned,
previously used for
case of a generator, the active
angle 8 between them.
is
mechanical power because the rotor
it
the synchronous generator in Chapter 16:
P = (£ £/X
given by
7.8b.
As
17.6
is
The maximum torque
=
60/2 tt|
motor can develop
is
called the pull-out torque, mentioned previously.
It
occurs
when 8 = 90°
the
(Fig. 17.9).*
power P r transmitted
across the air gap of an induction motor (Section
13.13).
tor
PR
However,
losses
in
are
a synchronous motor, the roentirely
source. Consequently,
all
supplied by the dc
the
power transmitted
The remarks
rotors.
in ihis
section apply to motors having
Most synchronous motors have
case the pull-out torque occurs
at
smooth
salient poles: in this
an angle of about 70°.
S YNCHRONO US
MOTORS
P
T
8
90
800
100
120
86.6
693
150
50
400
180
0
0
These values are plotted
377
in Fig. 17.9.
The torque curve can be found by applying Eq.
b.
17.4:
T=
Figure 17.9
Power and torque per phase as a function of the torque
angle 8. Synchronous motor rated 150 kW (200 hp),
1200 r/min, 3-phase, 60 Hz. See Example 17-3.
1
1
If
9.55 /VI 200
=
/VI 25
the
output:
7mux = 800 N-m
r/min.
460
synchronous
V, 3-phase
motor has a synchronous reactance of 0.8
phase.
=
The pull-out torque TnvAX coincides with
c.
maximum power
Example 17-3
A 50 kW, 200
9.55 Pin s
the excitation voltage
En
is
fixed
at
12,
per
300
V,
The
the
per phase, determine the following:
actual pull-out torque
N-m) because
this is a
3 times as great
is
(2400
3-phase machine. Similarly,
power and torque values given
in Fig.
1
7.9 must
also be multiplied by 3. Consequently, this 150
a.
b.
c.
The power versus 8 curve
The torque versus 8 curve
The pull out torque of the motor
maximum
motor can develop a
or about
400
kW
output of 300 kW,
hp.
Solution
a.
The
line-to-neutral voltage
17.7 Mechanical
is
As
corresponding values of
is
a
and the number of poles
p.
It
given by
(17.2)
selecting different values for 8,
late the
case of synchronous generators, there
a, the torque angle 8
is
sin 5
= (266 X 300/0.8)
= 99 750 sin 8 [W]
= 100sin8[kWl
By
in the
precise relationship between the mechanical angle
is
P = (E0 E/XJ
electrical
angles
E = EjV3 = 460/ V3
= 266 V
The mechanical power per phase
and
8
=
/x*/2
(1
7.5)
sin 8
Example 17-4
A 3-phase, 6000
we
P and
can calcuT,
per phase.
tor
P
T
PI
[kW]
[N*m
0
0
0
30
50
400
60
86.6
693
?
4 kV, 180 r/min, 60 Hz mo1.2
°
1
from
At
their no-load position. If
the line-to-neutral excitation
late the
£0 =
2.4 kV, calcu-
mechanical power developed.
Solution
The number of poles
(continued)
(1.
full-load the rotor poles are displaced by a me-
chanical angle of
5
kW
has a synchronous reactance of
\20fln s
is
=
120
X 60/180 = 40
ELECTRICAL MACHINES AND TRANSFORMERS
378
The
axis of
electrical torque angle is
N
pote
of rotor
8
Assuming
a
motor
to the
=
pa/2
wye
=
X
(40
)/2
1
=
axis of S pole
!
of stator
;
20°
connection, the voltage
E
applied
is
E = E,/V3 =4kV/V3
= 2.3 kV
- 2309 V
and the excitation voltage
is
Ea = 2400 V
Figure 17.10a
The mechanical power developed per phase
P = (E E/X
=
power =
Total
(y
s)
(2400
X
=
1
=
1573
X
3
5
sin
The flux produced by the stator
gap through the salient poles.
is
flows across the air
(17.2)
2309/1.2) sin 20°
axis of N pole
of rotor
573 300
J
axis of S pole
!
of stator
\
kW
1573
= 4719 kW (-6300
hp)
17.8 Reluctance torque
If
we
gradually reduce the excitation of a synchro-
nous motor when
that the
it
is
running
motor continues
to
no-load,
at
speed even when the exciting current
reason
is
that the flux
to cross the short
produced by the
gap between the
much
the stator rather than the
tween the poles.
In other
Fig.
1
7. 10a.
On
is
find
is
zero.
The
stator prefers
is
less in the axis
this
of the
phenomenon,
in
the
tract the salient poles in
1
7. 10c).
tains
a mechanical load
is
applied to the shaft, the
have the shape shown
maximum
Fig. 17,11
rotor poles will fall behind the stator poles, and the
stator flux will
its
in Fig.
1
7.
1
0b.
positive value at 8
maximum
is
namely
at
8
=
90°.
the reluctance torque as a func-
The torque reaches a maximum
=
45°. For larger angles
negative value
at
8
=
it
attains
135°. Obviously,
Thus, a considerable reluctance torque can be de-
to run as a reluctance-torque motor, the angle
veloped without any dc excitation
lie
at all.
The reluctance torque becomes zero when the
midway between the stator poles.
The reason is that the N and S poles on the stator atrotor poles are
is
zero
where the regular torque T at-
value,
shows
tion of the angle 8.
a
opposite directions (Fig.
Consequently, the reluctance torque
precisely at that angle
motor develops a reluctance torque.
If
salient poles are attracted to the stator poles,
thus producing a reluctance torque.
and
salient poles
concentrated as shown
account of
Figure 17.10b
The
longer air gap be-
words, because the reluc-
tance of the magnetic circuit
salient poles, the flux
we
run at synchronous
must
between zero and 45°. Although a positive torque
still developed between 45° and 90°, this is an un-
stable region of operation.
The reason
angle increases the power decreases.
is
that as the
S YNCHRONO US
MOTORS
379
Figure 17.12
Figure 17.10c
The reluctance torque is zero when the
are midway between the stator poles.
salient poles
In
a synchronous motor the reluctance torque
plus
(1)
the smooth-rotor torque (2) produce the resultant torque
(3).
Torque
can be seen
is
due
(2) is
to the
dc excitation
in Fig. 17.12.
of the rotor.
However, the difference
not very great, and for this reason
we
shall con-
tinue to use Eqs. 17.2 and 17.5 to describe synchro-
nous motor behavior.
Losses and efficiency
synchronous motor
17.9
Figure 17.11
of a
Reluctance torque versus the torque angle.
In order to give the reader a sense of the order of
As
magnitude of the pull-out torque, resistance, reaccase of a conventional synchronous
in the
tance,
motor, the mechanical power curve has exactly the
and losses of a synchronous motor, we have
drawn up Table 7A. It shows the characteristics of
a 2000 hp and a 200 hp synchronous motor, respectively labeled Motor A and Motor B.
1
same shape
sence
as the torque curve. Thus, in the ab-
of dc
Does
8
at
=
mechanical
the
excitation,
reaches a peak
power
45°.
power
The answer is
the saliency of the poles modify the
and torque curves shown
in Fig. 17.9?
The following points should be noted:
1.
yes. In effect, the curves
shown
in Fig.
1
7.9 are those
27° and 37°.
of a smooth-rotor synchronous motor. The torque of a
salient-pole
rotor
motor
is
1
7.
1
1
.
It
at full-load
corresponds
ranges between
to the electrical an-
gle 8 mentioned previously.
equal to the
component and
nent of Fig.
sum of the smoothreluctance-torque compo-
The torque angle
the
2.
(4.2
Thus, the true torque curve of a syn-
chronous motor has the shape
(3)
The peak reluctance torque
given
is
of the peak smooth-rotor torque.
the
in Fig. 17. 12.
As
is
about 8 per-
3.
The
is
to excite the
2000 hp motor
only about twice that needed for
200 hp motor
per-unit
a result, the
cent greater than that of a smooth-rotor motor, as
kW)
larger the
about 25 percent
peak torque of a salient-pole motor
The power needed
(2.
1
kW).
In general, the
synchronous motor the smaller
power needed
total losses
to excite
of Motor
four times those of Motor
A (38 kW)
B
(9.5
is
the
it.
are only
kW)
despite the
ELECTRICAL MACHINES AND TRANSFORMERS
380
CHARACTERISTICS OF TWO
TABLE 17A
SYNCHRONOUS MOTORS
MOTOR A
NAMEPLATE RATING
power
[hp]
power
[
k
W
2000 hp
492 k
W
1
]
4000 V
220 A
line voltage
line current
speed
MOTOR
B
440 V
A
208
90()r/min
60 Hz
3
3
.0
l
pull-out torque (pu)
dc exciter power
17.10 Excitation and reactive
power
Consider a wye-connected synchronous motor connected to a 3-phase source whose line voltage
fixed (Fig. 17.13).
voltage
wye
wye
kW
gap
V
25
10
mm
V
25
1
mm
6
stray losses
stator/
rotor I
2
2
kW
kW
kW
3.5 kW
2 kW
9.5 kW
97.5%
94.0%
1
/?
R
total losses
efficiency
IMPEDANCES AND VOLTAGES
stator
X
s
XJR
0.0638
l
2
phase voltage
fact that
E
E0
is
therefore created by the
U
v
E.
<I>
induces a line-to-
in the stator. If
in the stator,
However, because
E
com-
.
is
it
fixed,
we
neglect the
follows that E.d
<I> is
=
also fixed, as
case of a transformer (see Section 9.2).
mmf
needed to create the constant flux
may be produced
by both.
either
by the
<4>
stator or the rotor or
If the rotor exciting current / x is zero, all the
Ex.
0.0262
254
285
V
V
ten times as powerful. This
another property of large motors: the more
horsepower they develop, the smaller the
tive losses are.
As
in
power. Compare the
effi+
two motors: 97.5% versus 94.0%.
The synchronous reactance
much
rela-
a result, the efficiencies im-
prove with increase
ciencies of the
4.
produce
On the other
23
V
2873 V
is
line currents /
O
H
0.62
il
2309
Motor A
The
l
122
S
phase voltage
is
Rs
is
x
(line-to-neutral values)
7.77(2
stator resistance
ratio
1
total flux
very small IR drop
in the
kW
kW
4 kW
10.3 kW
4.2 kW
38 kW
8.5
friction
also fixed.
neutral voltage
The
stator core loss
is
E
follows that the line-to-neutral
Pursuing our reasoning, flux
kW
2.1
The
r.
LOSSES
windage and
E
It
bined action of U.d and
2.2
27°
1
U
.0
36.7°
4.2
dc exciter voltage
l
.4
1
at full-load
connection
air
motor
hand, the rotor produces a dc magnetomotive force
factor
torque angle
as far as
s
a magnetomotive force U.d in the stator.
LOAD CHARACTERISTICS
power
R
concerned.
49 k
60 Hz
phases
the effect of
is
200 hp
1
I800r/min
frequency
ways neglect
performance
X
s
per phase
is
122 times larger than
R As
s
.
a result,
X
6-
Figure 17.13
larger than the resistance of the stator
winding. Note that for the 2000 hp motor
6
s
we can
total flux 4> is due to the mmf produced by the ro(U ) plus the mmf produced by the stator (Ua ). For
The
is
tor
al-
a given
r
EL
,
the flux
<E>
is
essentially fixed.
S YNCHRONO US
produced by the
flux has to be
The stator must
power from the
stator.
then absorb considerable reactive
we
3-pfiase line (see Section 7.9). But
if
rotor with a dc current / x
mmf helps
the rotor
,
duce part of the flux O. Consequently,
power
drawn from
is
the ac line. If
The
ical
active power absorbed
power of the motor.
is
MOTORS
38
equal to the mechan-
excite the
pro'p
less reactive
we
gradually
raise the excitation, the rotor will eventually pro-
duce
the required flux
all
by
The
itself.
stator then
draws no more reactive power, with the
power
the
What happens
tive
motor becomes unity
factor of the
critical level?
we
if
The
of reactive power, just as
if
it
motor
this
we can cause
power
to correct the
same time
the
at
agram
the
current
q
,
and
motor designed
p
/
line
36.87°. This cur/
p
and
power P
q
is
=
0.8/ s
(17.7)
=
0.6/ s
(17.8)
given by
P = £ab / p =
The
=
broken up into two components
/
active
The
in Fig. 17.14.
clear that
power facme-
power
for a
power equal
factor of 0.8 (leading).
power
reactive
0.8
£ab /
power delivered by
Q = £ab /q =
0.6
the
£ab /
(17.9)
s
motor
is
(17.10)
s
to
can supply
A
factor of 0.8 can deliver
75 percent of
kW
rated
its
me-
motor shown
75% X 3000 = 2250
in
kvar to
same time as it develops its rated mekW. Motors designed to opcrate at leading power factors are bigger and more
costly than unity power factor motors are. The reathe line at the
chanical output of 3000
is
is
rating
chanical load. Thus, the 3000
son
it
motor
factor motor
develops the same me-
leads E. lb by arcos 0.8
/s
rent can be
/
as the
It
power, they are usually designed to op-
erate at a full-load
Fig. 17.4
power
chanical
as they furnish
Most synchronous motors are designed to operate at
unity power factor. However, if they also have to de-
reactive
shows an 80% power
15
to the load they are driving.
Power factor
liver reactive
7.
1
also operating at full-load.
The
17.11
di-
at full-load.
Fig.
to the
source
important property, synchronous
motors are sometimes used
chanical
like a
absorb or deliver reactive power.
to either
of a plant
this
were a capacitor.
Thus, by varying the dc excitation
Because of
.0).
Figure 17.14
Unity power factor synchronous motor and phasor
of absorbing reac-
The motor then behaves
line.
1
motor above
excite the
stator, instead
(
power, actually delivers reactive power
3-phase
tor
result that
that for a
given horsepower rating, both the dc
Figure 17.15
80 percent power factor synchronous motor and phasor diagram at full-load.
It
follows from Eqs. 17.9 and 17. 10 that
exciting current and the stator current are higher.
Q=
This can be explained as follows.
Fig. 17. 14
is
the schematic
power factor motor operating
to-neutral voltage
The
active
is E,db
and the
is,
of rated mechanical output
line-
was
as
line current
power absorbed per phase
P = Eay Jp
The
P
— 15%
diagram of a unity
at full-load.
0.75
is /
p
stated previously.
.
If
we compare
therefore,
I
p
with
/s
,
we
find that / s
=
i
.25 /
p
.
Thus, for the same mechanical power output, a mo(17.6)
tor
designed for a leading power factor of
80%
has
ELECTRICAL MACHINES AND TRANSFORMERS
382
25%
to carry a line current that is
power
that operates at unity
greater than one
,'s2
factor.
's2
\36.9°
17.12 V-curves
200
P = 800
Suppose a synchronous motor
rated mechanical load.
We
havior as the excitation
is
in
wish
is
operating
to
examine
power remains
power
1
6.
will
We
assume
we reduce
If
S
=
/x
power.
1000 kVA. As a
is
(b)
shown
Figure 17.18
in
200 A but
Field excitation raised to
a.
with
same me-
chanical load. Motor delivers reactive power to the
unity,
line.
Fig.
b.
= 00 A and P = 800 kW.
Phasor diagram shows current leading the voltage.
1
the excitation to
draw reactive power from
to the active
mechan-
factor
thus yielding the phasor diagram
7.
S = 1000 kVA
(a)
fixed. Let us begin by adjusting
the excitation I x so that the
1
its
be-
its
Because a change
varied.
excitation does not affect the speed, the
ical
at
A
kW
We
assume
70 A,
the
motor
the line in addition
that
S increases
to
result, the line current will in-
crease from
ponent of
/
sl
p
to / sl (Fig.
1
phase with
in
because the motor
is still
7.
1
7).
£. lb
is
Note
the
that the
same
com-
as before
developing the same me-
chanical power.
Current
E
b
o
—v
I
M
y
P = 800 kW
S = 800 kVA
A
4' 100
A
/s)
lags behind £"ab
motor
the rotor
smaller than before, but the apparent
is
is
power S absorbed by
If
(b)
(a)
Figure 17.16
Synchronous motor operating at unity power factor
with a mechanical ioad of 800 kW. Field excitation
is 100 A.
b. Phasor diagram shows current in phase with the
voltage.
we
lagging.
the stator
is
field current I x in
greater.
increase the excitation to
motor delivers reactive power
is
a.
and so the power
,
The
factor of the
/x
= 200 A,
to the line to
the
which
connected (Fig. 17.18). The apparent power
again greater than
We
assume S
comes /s2 and
=
in the unity
power
1000 kVA. The
it
is
factor case.
line current be-
However, the in-phase
it leads ELXh
component of /s2 is still equal to I p because the mechanical power is the same.
.
By varying the excitation this way, we can
power of the synchronous motor
the apparent
plot
as a
function of the dc exciting current. This yields a
V-shaped curve
Ix
=
70
A
36.9°
P = 800 kW
S = 1000 kVA
(a)
tive
(b)
the
b.
reduced to 70 A but with same meMotor absorbs reactive power from
Field excitation
chanical load.
line.
Phasor diagram shows current lagging behind the
voltage.
The V-curve
the V-curve corresponds to full-load.
V-curve
Figure 17.17
a.
(Fig. 17.19).
is
always
displayed for a fixed mechanical load. In our case,
is
power
The no-load
also shown, to illustrate the large reacthat
can be absorbed or delivered by
simply changing the excitation.
Example
A
1 7-5
4000 hp (3000 kW), 6600
synchronous motor operates
power
1
1
(2,
V,
60 Hz, 200 r/min
at full-load at a
leading
factor of 0.8. If the synchronous reactance
calculate the following:
is
SYNCHRONOUS MOTORS
383
kVA
^
1000
full-load
x
O+
'
source E
yrio-load
400
O-
200
Figure 17.20
synchronous motor connected to a source
E. Note the arbitrary (+) polarity marks and arbitrary
direction of current flow. See Example 17-5.
Circuit of a
200
120
80
40
-'x
Figure 17.19
No-load and full-load V-curves
nous motor.
E = 3815^0°
of
a 1000 hp synchroIt
c.
The apparent power of the motor, per phase
The ac line current
The value and phase of E0
d.
Draw
e.
Determine the torque angle 8
a.
b.
the phasor
=
given by
328/136.9°
Writing the equation for the circuit
-E+jlX + E0 =
diagram
s
shall
find
0
thus
immediately change the given values
correspond
The
to
E0 = E-jIX
= 3811/10° - j (328/136.9°) 11
= 3811^0° - 3608/1(36.9° + 90°)
= 3811 (cos 0° + y sin0°) -
active
to
one phase of a wye-connected motor.
power per phase
P=
3000/3
is
= lOOOkW
The apparent power per phase
3608 (cos 126.9°
line-to-neutral voltage
(8.11)
d.
Consequently,
is
EQ
lags 26° behind E,
complete phasor diagram
E=
The
line current
I
/
EjV3 =
leads
the value
,
\
1
1
V
e.
The torque angle
8
is
is
shown
and the
in Fig. 17.21.
26°.
17,13 Stopping synchronous
motors
=
Owing
36.9°.
to the inertia
of the rotor and
its
load, a large
synchronous motor may take several hours
and phase of the excita-
£ we draw
(>
38
1250 X 1000/3811
an angle of arcos 0.8
To determine
tion voltage
=
C
is
= S/E=
= 328 A
E by
6600/ V3
sin 126.9
3811 + 2166 -; 2885
= 5977 -y2885
= 6637/1-26°
is
= 1250kVA
The
+;
=
S = P/cosd = 1000/0.8
c.
we
s
We
b.
/ is
/
Solution
a.
follows that
after
the equivalent circuit
time,
being disconnected from the
we
line.
to stop
To reduce
the
use the following braking methods:
of one phase (Fig. 17.20). This will enable us to
write the circuit equations. Furthermore,
lect
E as
the reference phasor
and so
we
se-
I.
Maintain
full
short-circuit.
dc excitation with the armature
in
ELECTRICAL MACHINES AND TRANSFORMERS
384
e.
The time required for the speed
600 r/min to 150 r/min
to fall
from
Solution
a.
In Fig. 17.22a the
motor has just been discon-
nected from the line and
generator
is
now
operating as a
The speed
frequency is 60 Hz.
in short-circuit.
r/min, and the
is still
Consequently, the impedance per phase
Z=VR
2
+ Xi
\ 0.2
6637 V
=
2
600
is
(2.X 12)
+
16
2
16 ft
Figure 17.21
See Example
The current per phase
17-5.
/
is
= EJZ=
2400/16
= 150A
2.
Maintain
full
dc excitation with the armature
connected to three external
3.
Apply mechanical braking.
In
methods
it
I
and
2, the
The power
resistors.
r/min
P=
motor slows down because
functions as a generator, dissipating
the resistive elements of the circuit.
its
energy
is
Mechanical
usually applied only after the motor has
reached half-speed or
less.
A
lower speed prevents
undue wear of the brake shoes.
Example 17-6
A 1500 kW, 4600
2
3I
R =
13.5
in
b.
braking
dissipated in the 3 phases at
3
X
150
E0
2
X
is
fixed, the in-
0.2
kW
Because the exciting current
duced voltage
is
proportional to the speed.
Consequently, when the speed has dropped to
150 r/min,
Ea = 2400 X
600 r/min, 60 Hz synchronous motor possesses a synchronous reactance of
The frequency
16 ft and a stator resistance of 0.2 ft, per phase.
and so
= 600 V
(150/600)
V,
is
also proportional to the speed,
The excitation voltage E 0 is 2400 V, and the moment of inertia of the motor and its load is 275
kg-irr. We wish to stop the motor by short-cir-
The synchronous reactance
cuiting the armature while keeping the dc rotor
the frequency; consequently,
/=
60
X
(15/60)
=
is
15
Hz
proportional to
current fixed.
16U
Calculate
a.
The power dissipated
in the
armature
at
600
in the
armature
at
150
r/min
b.
The power dissipated
r/min
c.
d.
The
The
600
is
kinetic energy at
600 r/min
kinetic energy at 150 r/min
Figure 17.22a
Motor turning at 600 r/min (Example
17-6).
S YNCHRONO US
4
P =
LI
13.5
150 A
600 V
whence
0.2 Q.
150 r/min
Note
that the
t
MOTORS
3 85
Wit
(3.4)
=
508.6//
=
37.7
s
motor would stop much sooner
if
external resistors were connected across the stator terminals.
Figure 17.22b
Motor turning at 150 r/min (Example
17-6).
The synchronous motor
17.14
versus the induction motor
X,
=
X
16
=
(15/60)
4
0
We
Referring to Fig. 17.22b the
phase
at
150 r/min
new impedance
per
But
is
have already seen
that induction
motors have
600
excellent properties for speeds above
at
r/min.
lower speeds they become heavy, costly, and
have relatively low power factors and efficiencies.
Z = V0.2 2 +
4
2
=
a
4
Synchronous motors
are particularly attractive
for low-speed drives because the
The current phase
power
factor can
is
always be adjusted
/= £0 /Z =
600/4
=
A
150
to
1
.0
Although more complex
and the efficiency
is
high.
weight and
to build, their
cost are often less than those of induction motors of
Thus, the short-circuit current remains un-
changed
r/min to 150 r/min. The power dissipated
3 phases
therefore the
is
equal power and speed. This
motor decelerates from 600
as the
P =
same
13.5
is
particularly true for
speeds below 300 r/min.
in the
as before:
A synchronous motor can
tor
kW
of
a
plant
Furthermore,
its
while
improve the power
carrying
starting torque
its
rated
fac-
load.
can be made consid-
The
erably greater than that of an induction motor.
c.
The
kinetic energy at
5.48
d.
The
A
10
-3
=
5.48
=
542.5 kJ
X
10
"
is
reason
Jrr
(3.8)
X 275 X
of the squirrel -cage wind-
at
5.48
X
10
3
150
2
motors
in
decelerating from
-
=
542.5
=
508.6 kJ
The time
in
and
kilns,
A synchronous capacitor is essentially a synchronous
motor running
armature
for the speed to drop
is
ultra-low speeds. Thus, huge motors
33.9
lost as heat in the
r/min to 150 r/min
at
MW range drive crushers, rotary
17.15 Synchronous capacitor
kl
is
the 10
variable-speed ball mills.
W = E -Ek2
This energy
rating.
very low frequencies enable us to run synchronous
33.9 kJ
The loss in kinetic energy
600 r/min to 150 r/min is
motor and
same nominal
The biggest difference is in the starting torque.
High-power electronic converters generating
is
X 275 X
effi-
synchronous speed. Figure 17.23 compares
a synchronous motor having the
kinetic energy at 150 r/min
tance.
that the resistance
ing can be high without affecting the speed or
ciency
600-
is
the properties of a squirrel-cage induction
E k2 =
=
e.
X
600 r/min
given by
resis-
from 600
at
no-load.
Its
only purpose
is
to ab-
sorb or deliver reactive power on a 3-phase system,
in
order to stabilize the voltage (see Chapter 25).
machine
acts as an
The
enormous 3-phase capacitor
(or
ELECTRICAL MACHINES AND TRANSFORMERS
386
I
i
i
s ynchrcDnous
motor
97
induction
>
motor
/J
96
rf
//
//
//
o
c
a)
it
95
!y
if
if
if
if
(a)
if
94
93
92
91,
25
50
*-
75
100
125
%
mechanical power
Figure 17.24a
%
Three-phase, 16
250
rated
-200 Mvar
kV,
900
r/min
synchronous capacitor
(supplying reactive power) to
+300
Mvar (absorbing reactive power). It is used to regulate
the voltage of a 735 kV transmission line. Other characteristics: mass of rotor: 143 t; rotor diameter: 2670
mm; axial length of stator iron: 3200 mm; air gap
synchronous motor
200
= 150
length: 39.7
mm.
(b)
100
induction motor
^
50
60
40
20
80
100
%
speed
Figure 17.23
Comparison between the
torque
(b) of
efficiency (a)
synchronous motor, both rated
r/min, 6.9 kV, 60 Hz.
inductor)
and
starting
a squirrel-cage induction motor and a
whose
reactive
at
4000
hp,
1800
power can be varied by
changing the dc excitation.
Most synchronous capacitors have ratings that
range from 20 Mvar to 200 Mvar and many are
hydrogen-cooled (Fig. 17.24). They are started up
like synchronous motors. However, if the system
cannot furnish the required starting power, a pony
motor is used to bring them up to synchronous
speed. For example, in one installation, a 160
Mvar
mm
Figure 17.24b
Synchronous capacitor enclosed in its steel housing
containing hydrogen under pressure (300 kPa, or"
2
about 44 lbf/in ).
{Courtesy of Hydro-Quebec)
SYNCHRONOUS MOTORS
synchronous capacitor
is
and brought up
started
speed by means of a 1270
387
to
kW wound-rotor motor.
Example 17-7
A
synchronous capacitor
16 kV,
actance of 0.8 pu and
is
Calculate the value of
EQ
a.
Absorbs 160 Mvar
b.
Delivers 120
It
160 Mvar,
rated at
is
1200 r/min, 60 Hz.
has a synchronous
connected
to a 16
kV
re-
line.
so that the machine
Q=
120 Mvar<feMjl
Mvar
/
Solution
a.
i 4335
The nominal impedance of the machine
Zn = En 2 /S n
= 16 000 2 /(160 X
=
(16.3)
10
6
5550 V
9250
V
14 800
V
)
a
1.6
A
is
Figure 17.25b
Over-excited synchronous capacitor delivers reactive
The synchronous reactance per phase
X = X s (pu)Zn =
s
=
The
i.28
0.8
X
is
power (Example
1.6
From
Fig.
17-7).
7.25a
1
n
we can
-£+ 77X
line current for a reactive load
of 160
Mvar
s
write
+ £o =
0
hence
is
E„
/n
= 5 n /(V3£n
-
160
X
c
X
10 7( .73
1
16 000)
= 5780 A
The drop across the synchronous reactance
is
b.
Ex =
IX S
= 5780 X
=
E- jIX
1.28
Note
that the excitation voltage
less than the line voltage
The load current when
ing 120
IU
line-to-neutral voltage
E = EjV3 =
= 9250 V
Selecting
E as
E=
The current
machine
is
/
This time
we have
0°
lags 90° behind
the
absorbing reactive power; conse-
quently,
/
= 5780^-90°
850 V)
10 7(1.73
machine
X
/
E by
leads
is
deliver-
16 000)
90° and so
= 4335Z90 0
Fig. 17.25b
we can
write
En = E-jIX
= 9250^0° - 4335 X 1.28/180°
= (9250 + 5550)Z()°
s
=
14 800/10°
is
(9250 V).
)
(
X
/
From
E because
1
90°)
is
= Q/(V3EU
120
the
(
-
= 4335 A
16 000/1.73
9250,/
Mvar
-
is
the reference phasor,
1.28^(90°
much
= 7400 V
The
s
= 9250^0° - 5780 X
= 1850^0°
)
ELECTRICAL MACHINES AND TRANSFORMERS
388
5780
motor [hp] knowing
A
it
has an efficiency
of 95%.
17-8
A synchronous
operates
at
a
motor driving a pump
power factor of 100%.
What happens
if
the dc excitation
is
in-
creased?
17-9
mm^ Q= 160 Mvar
7400
A 3-phase,
to a
4 kV, 60 Hz
current of
320
A and
line
draws a
absorbs 2000 kW.
E
V
y
>
I
9250
Calculate
V
a.
1850
V
b.
c.
5780
225 r/min synchronous motor
connected
A
d.
Figure 17.25a
Under-excited synchronous capacitor absorbs reactive power (Example 17-7).
17-10
The apparent power supplied to the motor
The power factor
The reactive power absorbed
The number of poles on the rotor
A synchronous
3-phase
motor draws
line. If the
1
50
A from
exciting current
raised, the current drops to 140 A.
a
is
Was
the
motor over- or under-excited before the
was changed?
excitation
The
excitation voltage (14
800 V)
is
now
con-
siderably greater than the line voltage (9250 V).
Intermediate level
17-11
a. Calculate the
approximate full-load current
of the 3000 hp motor
Questions and Problems
in Fig. 17.1, if
it
has
an efficiency of 97%.
Practical level
1
1
7-
1
7-2
Compare
7-3
the construction of a synchro-
17-12
7-4
Referring to Fig. 17.2,
how
Explain
a synchronous motor starts up.
17-13
should the dc excitation be applied?
Why does the
Name some
What
A 3-phase
a.
What
b.
If
is
factor.
speed of a synchronous motor
1
.8
is
we
used for?
meant by an under-excited syn-
17-14
does
1
7-7
over-excite a synchronous motor,
its
power
line voltage
at
motor draws 2000
factor of
the approximate
90%
leading. Calculate
the
power
kV, but the exciting current remains
how
the following
a.
Motor speed and mechanical power output
b.
Torque angle 8
c.
Position of the rotor poles
d.
Power
e.
Stator current
factor
A synchronous
motor has the following
parameters, per phase (Fig. 17.7a):
kVA at
power developed by
unity
suddenly drops to
mechanical power output increase?
A synchronous
a
60 Hz operates
quantities are affected:
to a squirrel-
chronous motor?
The
unchanged. Explain
of the advantages of a syn-
it
synchronous motor rated 800
hp, 2.4 kV,
meant by a synchronous capacitor
is
and what
7-6
what speed must
quencies?
cage induction motor.
1
at
a squirrel -cage induction motor.
chronous motor compared
17-5
the value of the field resistance?
is
the rotor turn to generate the indicated fre-
remain constant even under variable load?
1
What
nous generator, a synchronous motor, and
When
1
b.
E=
x =
s
/
2.4 kV;
2
a
= 900 A.
Ea =
3
kV
SYNCHRONOUS MOTORS
Draw
the phasor diagram and determine:
17-19
nous motor rated 400 hp, 2300
b.
Active power, per phase
r/min, 80 A,
c.
Power
The
d.
Reactive power absorbed (or delivered),
factor of the
a. In
motor
is
new
the mechani-
citation
its
power
excitation
is
is
adjusted
17-20
0.6
increased without making any
is
is
the effect
The synchronous capacitor
(I,
phase
upon the
per phase.
0.007
is
to a stop,
it
by the motor
dc excitation
initial line
motor
the
tor
line current
reactive
power absorbed
level
10 H, per phase.
The
stator is
a.
at full-load
b.
(4000 hp) with a leading power factor of
0.89. If the efficiency
is
97%,
calculate
c.
the following:
c.
The apparent power
The line current
The value of E0 per phase
The mechanical displacement of the poles
The
their no-load position
total reactive
electrical
f.
power supplied
power
b.
A so
that the
its
is
rated value, or 1600 V, at
d.
The total braking power and braking torque
at 900 r/min
The braking power and braking torque at
450 r/min
The average braking torque between 900
r/min and 450 r/min
The time for the speed to fall from 900 r/min
to 450 r/min, knowing that the moment of
inertia
of the rotor
is
1
.7
6
X
1()
lbfr.
17-21
A 500 hp,
3-phase, 2200 V, unity power
factor synchronous motor has a rated current of 103 A.
It
can deliver
its
rated out-
put so long as the air inlet temperature
1
7- 7
1
we wish
40°C or
less.
The manufacturer
is
states that
to adjust the
the output of the
motor must be decreased
factor to unity.
by
1
percent for each degree Celsius
above 40°C.
Calculate
a.
250
voltage across the resistors
system
The approximate maximum power the
motor can develop, without pulling out of
Problem
fixed at
H
wye. The
Industrial application
to the
step [hp]
In
is
in
r/min.
,
from
e.
connected to three large 0.6
Calculate
connected
wye, and the motor operates
machine coasts
in Fig.
17.4 possesses a synchronous reactance of
d.
is
900
kV motor shown
hp, 6.9
resistance per
the
braking resistors connected
(or delivered)
one-tenth of
The 4000
The
11. If
der to shorten the stopping time, the sta-
The torque angle
power absorbed by
17.24
will run for about 3 h. In or-
d.
active
in Fig.
possesses a synchronous reactance of
unity. If the ex-
c.
a.
to
synchronism
The
The
The
b.
adjusted
,
c.
following:
in
is
iy
The value of Xs and of £ 0 per phase
The pull-out torque [ft-lbfj
The line current when the motor is about
a.
b.
power absorbed
reactive
factor
other change, what
Advanced
E
hp synchronous motor drives a
so that the
a.
synchronous reactance of
Calculate
pull out of
compressor and
b.
450
the line current
if
(or delivered) by the motor, per phase.
17-18
stator has a
0.88 pu, and the excitation
suddenly removed,
b. Calculate the
A 500
V,
60 Hz, drives a compressor.
to 1.2 pu.
Problem 17-14 calculate
cal load
17-17
factor synchro-
Torque angle 8
and the new torque angle 5
17-16
power
unity
a.
per phase
17-15
A 3-phase,
389
The exciting voltage £0
The new torque angle
required, per phase
If the air inlet
46°C, calculate the
motor
current.
temperature
maximum
allowable
is
390
1
7-22
ELECTRICAL MACHINES AND TRANSFORMERS
An 8800 kW.
6.0 kV,
1
500 r/min, 3-phase,
50 Hz, 0.9 power factor synchronous motor
manufactured by Siemens has the
fol-
sing the above information, calculate the fol-
wing:
a.
1.
962
Rated current:
A
Rated torque:
Pull-out torque:
4.
Locked-rotor current:
5.
Excitation voltage:
56.0
b.
kNm
2.
3.
c.
1.45 pu
160
V
387
A
Full-load efficiency, excluding excitation
losses:
9.
Temperature
11.
f.
520 kg-m~
Maximum
g.
12.
13.
14.
its
in
gallons (U.S.)
total
moment
which the motor can
of inertia
(in
pull into syn-
The
The
total losses
of the motor
total efficiency
at full-load
of the motor
at full-
moment
2
The
reactive
power delivered by
the
motor
full-load
If the iron losses are
equal to the stator cop-
per losses, calculate the approximate resis-
465 L/min
permissible external
1370 kg-m
The maximum
at
25°C
of cooling water:
32°C
Flow of cooling water:
inertia:
motor including
load
of inertia of rotor:
rise
e.
97.8%
to
10.
the
chronism
d.
Excitation current:
Moment
mass of
The flow of cooling water
lb-ft'),
4.9 pu
7.
8.
total
per minute
6.
system
The
enclosure, in metric tons
lowing properties:
of
tance between two terminals of the
stator.
h. Calculate the resistance of the field circuit.
Mass of rotor: 6. 10 t (t = metric
Mass of stator: 7.50
Mass of enclosure: 3.97
t
t
ton)
Chapter
1
Single-Phase Motors
18.0 Introduction
Fig. 18.2
shows
the progressive steps in wind-
ing a 4-pole, 36-slot stator. Starting with the lami-
Single-phase motors
electric
are the
most familiar of all
motors because they are used
in
nated iron stator, paper insulators
home
ers
—
are
first
inserted
in
—called
the
slots.
r/min,
60 Hz
slot lin-
The main
appliances and portable machine tools. In general,
employed when 3-phase power
they are
is
not
available.
There are many kinds of single-phase motors on
the market, each designed to
cation.
However, we
meet a specific appli-
will limit our study to a
few
basic types, with particular emphasis on the widely
used split-phase induction motor.
18.1
Construction of a singlephase induction motor
Single-phase induction motors are very similar to
3-phase induction motors. They are composed of a
squirrel-cage rotor (identical to that in a 3-phase
motor) and a stator (Fig.
1
8. 1).
main winding, which creates a
The
set
stator carries a
of N, S poles.
It
also carries a smaller auxiliary winding that only
when
the motor
auxiliary winding has the
same num-
operates during the brief period
starts up.
The
Figure 18.1
Cutaway view of a 5 hp, 1725
phase capacitor-start motor.
ber of poles as the main winding has.
(Courtesy of Gould)
391
single-
ELECTRICAL MACHINES AND TRANSFORMERS
392
Figure 18.2a
Bare, laminated stator of a 1/4 hp
squirrel-cage rotor
is
(1
87 W), single-phase motor. The 36
3-phase motor.
slots are insulated with a
paper
liner.
The
identical to that of a
Figure 18.2c
Figure 18.2b
Four poles of the main winding are inserted
in
the slots.
Four poles
of the auxiliary
winding straddle the main
winding.
(Courtesy of Lab-Volt)
winding
1
is
then
laid
the
in
8.2b). Next, the auxiliary
that
(Fig.
its
1
slots
winding
is
(Figs.
I8.2a,
embedded
so
poles straddle those of the main winding
8.2c).
The reason
be explained shortly.
for this
arrangement will
Each pole of the main winding consists of
group of four concentric
(Fig.
1
8.3a).
coils,
connected
Adjacent poles are connected so as
produce alternate N, S
polarities.
the center of each pole
The empty
(shown as
a
in series
to
slot in
a vertical dash
SINGLE-PHASE MOTORS
393
one pole pitch
—(180°)
—H
10 20
25 30 turns No. 16 wire
Figure 18.4
Main and auxiliary windings
motor.
iary
The
winding opens
mounted on the
nous speed.
line)
in
stationary contact
and the
when
shaft,
a 2-pole single-phase
series with the auxil-
in
the centrifugal switch,
reaches 75 percent of synchro-
partially filled slots
on either side of
are used to lodge the auxiliary winding.
The
has only two concentric coils per pole (Fig.
Fig.
1
shows a 2-pole
8.4
8.3c).
1
main
stator; the large
winding and the smaller auxiliary winding are
at right
18.2
Synchronous speed
in the
speed of
dis-
angles to each other.
placed
As
it
latter
case of 3 -phase motors, the synchronous
all
single-phase induction motors
is
given
by the equation
ns
=
120/
1
(17.1)
P
where
ns
90°-*!
center of
main
winding
=
/=
center of
auxiliary
p
winding
The
synchronous speed [r/min]
frequency of the source [Hz]
= number of poles
rotor turns at slightly less than synchronous
speed, and the full-load slip
5 percent for fractional
Figure 18.3
Main winding of a 4-pole, 36-slot motor
showing the number of turns per coil.
b. Mmfs produced by the main winding.
a.
c.
laid
out
typically 3 percent to
flat,
Position of the auxiliary winding with respect to the
main winding.
is
horsepower motors.
Example 18-1
Calculate the speed of the 4-pole single-phase
shown in
percent. The
tor
Fig. 18.1 if the slip at full-load
line
frequency
is
60 Hz.
is
mo3.4
ELECTRICAL MACHINES AND TRANSFORMERS
394
Solution
produces an ac
to the stator. the resulting current / s
The motor has 4
=
ns
poles, consequently,
120///?
=
(120
X
flux
//
is
s
.
The
flux pulsates back
and forth
but, unlike
the flux in a 3-phase stator, no revolving field
60)/4
duced. The flux induces an ac voltage
= 1800r/min
The speed
<£>
is
pro-
in the station-
ary rotor which, in turn, creates large ac rotor currents.
In effect, the rotor
given by:
behaves
like the short-circuited
secondary of a transformer; consequently, the motor
s
=
(n s
0.034
=
(1800
n
=
1739 r/min
-
n)!n s
-
(17.2)
has no tendency to
h)/1800
However,
the other,
As
spin.
18.3 Torque-speed characteristic
if
we
start
by
itself (see Fig.
1
8.5
a schematic diagram of the rotor and main
is
will continue to rotate in the direction of
it
ates until
tor
reaches a speed slightly below synchro-
it
winding of a 2-pole single-phase induction motor.
turn. Fig.
when
is
locked. If an ac voltage
is
applied
that the
develops a positive torque as soon as
Suppose the rotor
8.6).
a matter of fact, the rotor quickly acceler-
nous speed. The acceleration indicates
Fig.
1
spin the rotor in one direction or
8.6
1
the
shows the
is
mo-
begins to
typical torque-speed curve
main winding
starting torque
it
is
zero, the
excited.
Although the
motor develops a power-
rotor current
ful
torque as
it
approaches synchronous speed.
18.4 Principle of operation
The
principle of operation of a single-phase induc-
tion
motor
is
quite complex, and
by the cross-field theory
As soon
E
120 V, 60 Hz
is
may be
explained
*
as the rotor begins to turn, a speed
induced
in the rotor
emf
conductors as they cut the
ac source
stator flux <D S (Fig. 18.7). This voltage increases as
Figure 18.5
Currents
The
is
in
the rotor bars
when
resulting forces cancel
the rotor
is
locked.
each other and no torque
produced.
*6
ac source
Figure 18.7
Currents induced
They produce a
in
the rotor bars
due
to rotation.
flux <I\ that acts at right
angles to the
stator flux
speed
*
Figure 18.6
Typical torque-speed curve of a single-phase motor.
The double revolving Held theory (diseussed
1
8.
18)
is
in
Section
also used to explain the behavior of the single-
phase motor.
SINGLE-PHASE MOTORS
= r/4
t
/
395
=
3774
Figure 18.8
Instantaneous currents and
flux in a single-phase momain winding excited. The duration of one
T seconds, and conditions are shown at suc-
tor with the
cycle
is
cessive quarter-cycle intervals.
a.
Stator current
b.
Stator current
however,
c.
d.
e.
/s is
is
maximum,
rotor current
zero, rotor current
is
smaller than
<I>
S
is
zero.
is
/r
maximum;
.
maximum, but negative.
Rotor current is maximum, but negative.
After one complete cycle (t = T) the conditions
Stator current
is
re-
peat.
f.
Resulting flux <b in the air gap rotates ccw at synchronous speed. Its amplitude varies from a maximum of <I> S to a minimum <I>
r
the rotor speed increases.
flow
currents produce an ac flux
same time
behind
does not reach
/,
to
r
which
acts at right
its
is
the
maximum value at the
as 3> s does. In effect, <I\ lags almost 90°
due
<t> s ,
to the
The combined
volving magnetic
motor.
<l>
Equally important
angles to the stator flux
<!>,.
causes currents
It
bars facing the stator poles. These
in the rotor
fact that
.
inductance of the
action of
4\ and
4> r
rotor.
produces a
field, similar to that in a
The value of
<I>
r
re-
3-phase
increases with increasing
speed, becoming almost equal to 3> s
why
speed. This explains in part
at
synchronous
the torque
in-
creases as the motor speeds up.
We can
understand
duced by referring
the currents
tor
/s
= + 10
A
r
=
T
and
sume
how
the revolving field
to Fig. 18.8.
It
and fluxes created respectively by
stator, at
that the
successive intervals of time.
motor
is
is
pro-
gives a snapshot of
the ro-
We
as-
running far below synchronous
ELECTRICAL MACHINES AND TRANSFORMERS
396
speed, and so
much
is
<!>,.
smaller than
<J) s
.
is
obvious that the combination of <I> S and
a revolving field. Furthermore, the flux
izontally
and
shown
wise
it
in Fig.
18.8f.
The
strong hor-
Thus, the
elliptic pattern
flux rotates counterclock-
iary
rotor.
As
motor approaches syn-
the
tioned previously)
is
and
standstill
torque
T—
=
=
/s
Fig.
1
8.9.
sina
*/ a / s
(1 8. 1)
locked-rotor torque [N mJ
locked-rotor current
in the
auxiliary
in the
main
]
locked-rotor current
winding [A|
a starting torque in a single-phase motor,
When the main
at
low speeds. The locked-rotor
at
winding [A
create a revolving field. This
(men-
<!>,.
result, <P.d strength-
where
18.5 Locked-rotor torque
we must somehow
weak. As a
T=
produced.
done by adding an auxiliary winding, as shown
during the
a
given by
is
/.,
To produce
is
<I>
the rotor flux
thereby producing a powerful torque both
ens
chronous speed, CI\ becomes almost equal to <P S and
a nearly perfect revolving field
when
acceleration period
direction as the rotor. Furthermore,
speed of the
immediately see that the auxil-
will
winding produces a strong flux
synchronous speed, irrespective of the ac-
rotates at
tual
vertically.
low speed follows the
same
in the
weak
relatively
field strength at
it
produces
is
The reader
By observ-
ing the flux in the successive pictures of Fig. 18.8,
is
a = phase angle between
in
k
=
and auxiliary windings are
and
/s
/.,
[°|
a constant, depending on the design
of the motor
connected to an ac source, the main winding pro-
duces a flux
duces a flux
<P S
,
the
If
two fluxes
so that <1\ either lags or leads
set up.
manner
To
while the auxiliary winding pro-
<E> S ,
The 2-phase revolving
are out of phase,
a rotating field
field
is
obtain the desired phase shift between
(and hence between
pedance
is
created in a
similar to the revolving field of a 3-phase
motor (see Section
/.,
Z in
and
4> a ),
The
rise to various types
many cases
pedance
in the
self, as
is
incorporated
the desired im-
auxiliary winding
it-
explained below.
special switch
is
also connected in series with
the auxiliary winding.
when
and
depending upon the desired starting torque.
of split-phase motors. In
A
/s
an im-
resistive, inductive, or capaci-
The choice of impedance gives
13.3).
we add
series with the auxiliary winding.
impedance may be
tive,
<J\
the
chronous
It
disconnects the winding
motor reaches about 75 percent of synspeed.
A
speed-sensitive
switch mounted on the shaft
purpose (Fig.
1
is
centrifugal
often used for this
8. 10).
18.6 Resistance split-phase motor
i
i
The main winding of a single-phase motor is always
2
made of relatively large wire, to reduce the I R
losses (Fig. 18. la). The winding also has a relatively large number of turns. Consequently, under
1
o-
Figure 18.9
Currents and fluxes
ac source
locked-rotor conditions, the inductive reactance
at standstill
when the main and
An elliptical revolv-
auxiliary windings are energized.
ing field
is
«-o
produced.
high and the resistance
locked-rotor current
applied voltage
/s
E (Fig.
is
low.
As
a result,
is
the
lags considerably behind the
18.11 b).
SINGLE-PHASE MOTORS
397
Figure 18.10
a.
Centrifugal switch
position.
b.
The
in
Centrifugal switch
position.
Due
the closed, or stopped,
stationary contact
in
is
closed.
the open, or running,
to centrifugal force, the rectan-
gular weights have
swung
out against the
straining tension of the springs. This
caused the
plastic collar to
move
re-
has
to the
left
along the shaft, thus opening the stationary
contact
series with the auxiliary winding.
In a resistance split-phase
centrifugal
motor (often simply
called split-phase motor), the auxiliary winding has
switch
)
E
in
)
)
auxiliary winding
70 turns per pole
No. 22 wire
a relatively small
resistance
number of
higher and
is
turns of fine wire.
Its
reactance lower than that
its
of the main winding, with the result that the lockedrotor current
/ a is
more nearly in phase with E. The
a between /a and / s produces
resulting phase angle
the starting torque.
The
main winding
/s
120 turns per pole
and
line current
/.,.
At
/ L is
start-up,
it
equal to the phasor
is
sum
of
usually 6 to 7 times the
nominal current of the motor.
No. 16 wire
Owing
to the
small wire used on the auxiliary
winding, the current density
is
high and the winding
heats up very quickly. If the starting period lasts for
a=
more than
25°
5 seconds, the
and may burn
winding begins
out, unless the
motor
is
to
smoke
protected by
a built-in thermal relay. This type of split-phase
tor is well
mo-
suited for infrequent starting of low-
inertia loads.
Example 18-2
A
resistance split-phase
motor
is
rated at
1/4
hp
(187 W), 1725 r/min, 115 V, 60 Hz. When the rotor
is locked, a test at reduced voltage on the main and
(b)
auxiliary windings yields the following results:
Figure 18.11
a.
b.
Resistance split-phase motor (1/4 hp, 115 V, 1725
r/min, 60 Hz) at standstill.
Corresponding phasor diagram. The current in the
auxiliary winding leads the current
winding by 25°.
in
the main
main
auxiliary
winding
applied voltage
current
active
power
E=
/ =
V
4 A
23
60
W
winding
£=
/, =
P., =
23
1
.5
30
V
A
W
ELECTRICAL MACHINES AND TRANSFORMERS
398
Calculate
v s 2 - Pi
G =
s
a.
b.
The phase angle between / a and /
The locked-rotor current drawn from
s
= V92 2 - 60 2 =
the line at
V
115
= VSJ -
Oa
69.7 var
Pi
Solution
17.0 var
We first calculate the phase
E of the main winding.
a.
The apparent power
=
5S
-
£V S
angle
<$>
s
between
Is
and
The
Q=
is
X
23
4
=
The
The power
cos
factor
<|> s
+ Ga
69.7 + 17.0 =
- PJS =
-
60/92
S
S
0.65
= VP 2 + Q 2
V
=
4> s
We now calculate the phase angle
and E of the auxiliary winding.
=
The power
cos
EI,
=
factor
=
4> a
=
/,
<T>
a
between
r
SIE
-V$6J* = 125 V
V
current at 23
=
125/23
-
=
5.44
X
-
(115/23)
is
5.44
The locked-rotor current drawn
/,
The apparent power
W
The locked-rotor
49.6°
lags 49.6° behind the voltage E.
5a
is
is
thus,
/s
is
86.7 var
apparent power absorbed
total
motor
the
Qs
=
VA
92
power absorbed by
total reactive
at
A
1
V
15
is
A
27.2
is
23
X
1.5
=
34.5
VA
Due
to their
low
cost, resistance split-phase in-
duction motors are the most popular single-phase
motors. They are used where a moderate starting
is
PJS.A
=
30/34.5
=
torque
0.87
is
required and where the starting periods are
infrequent.
They drive
fans,
chines, oil burners, small
thus,
pumps, washing ma-
machine
tools,
and other
devices too numerous to mention. The power rating
=
*a
/.,
29.6°
usually lies between
a =
=
cj)
s
-
<\>.
A
=
/s
and
49.6°
-
W and 250 W (1/12 hp
to
1/3 hp).
lags 29.6° behind the voltage.
The phase angle between
60
Ia is
29.6°
18.7 Capacitor-start motor
The
20.0°
capacitor-start
motor
is
identical to a split-phase
motor, except that the auxiliary winding has about as
b.
To determine
we first caloiP and Q drawn by both
the total line current,
culate the total value
windings and then deduce the
power S.
The total
total
apparent
many
turns as the
series with the auxiliary
The capacitor
active
power absorbed
is
main winding
has. Furthermore, a
capacitor and a centrifugal switch are connected in
80°,
which
is
is
winding
chosen so that
(Fig. 18. 2a).
1
/.,
leads
/s
by about
considerably more than the 25° found
in
a split-phase motor. Consequently, for equal starting
=
The
reactive
powers
60
Q
auxiliary windings are
s
+ 30 = 90 W
and Q.A of the main and
torques, the current in the auxiliary winding
about half that
in a split-phase motor.
It
is
only
follows that
during the starting period the auxiliary winding of a
capacitor motor heats up less quickly. Furthermore,
SINGLE-PHASE MOTORS
voltage, electrolytic capacitors are
centrifugal
much
399
smaller
and cheaper than paper capacitors. However,
elec-
can only be used for short
peri-
trolytic capacitors
ods
ac circuits whereas paper capacitors can op-
in
on ac
erate
indefinitely. Prior to the
development of
electrolytic capacitors, repulsion-induction
had
motors
be used whenever a high starting torque was
to
Repulsion-induction
required.
motors possess a
commutator and brushes that require considerable maintenance. Most motor manufacturers
special
have stopped making them.
when a high
They are built in sizes
kW (~I/6 hp to 10 hp).
Capacitor-start motors are used
starting torque
ranging from
1
is
required.
20
W to 7.5
Typical loads are compressors, large fans, pumps,
and high-inertia loads.
Table
motor
1
8A
gives the properties of a capacitor-start
having
a
rating
W(l/3
250
of
1760 r/min, 115 V, 60 Hz. Fig.
18.
1
hp),
shows
3
the
torque-speed curve for the same machine. Note that
during the acceleration phase (0 to 1370 r/min) the
main and auxiliary windings together produce a very
high starting torque.
When
the rotor reaches 1370
r/min, the centrifugal switch snaps open, causing the
motor
to operate
along the torque-speed curve of the
main winding. The torque suddenly drops from 9.5
(b)
Nm to 2.8 Nm, but the motor continues to accelerate
Figure 18.12
until
a.
Capacitor-start motor.
b.
Corresponding phasor diagram.
reaches 1760 r/min, the rated full-load speed.
it
and power factor of
single-phase induction motors
18,8 Efficiency
the locked-rotor line current /L
cally
is
smaller, being typi-
4 to 5 times the rated full-load current.
Owing to the
high starting torque and the relatively
low value of /a the capacitor- start motor
to applications
is
well suited
involving either frequent or prolonged
starting periods.
Although the starting characteristics
of this motor are better than those of a split-phase
tor,
mo-
both machines possess the same characteristics
under load. The reason
identical
circuit
is
that the
main windings
and the auxiliary winding
when
the
is
motor has come up
no longer
are
in the
is
full-load a 186
a direct
W motor (1/4 hp) has an efficiency
and power factor of about 60 percent. The low power
factor
is
mainly due to the large magnetizing current,
which ranges between 70 percent and 90 percent
full-load
Consequently, even
current.
at
these motors have substantial temperature
The
relatively
these motors
to speed.
The wide use of capacitor-start motors
The efficiency and power factor of fractional horsepower single-phase motors are usually low. Thus, at
horsepower
is
rises.
low efficiency and power
a
consequence of
ratings.
Integral
of
no-load
factor of
their fractional
horsepower
single-
result of the availability
of small, reliable, low-cost
phase motors can have efficiencies and power fac-
electrolytic capacitors.
For given capacitance and
tors
above 80 percent.
ELECTRICAL MACHINES AND TRANSFORMERS
400
1370
r/min
a
auxiliary
S
winding'
in circuit
c
300 r/min
;entrifuga
svvitch oper s
\
centrifuga
:
\
I
i
clos es
S\/vitch
2.8
no minal
ill
Figure 18.13
Torque-speed curves
^*"*
"
400
600
ki.
full
load
/bu r/mi n
\\
'///// V////.
800
1000
> speed n
of a capacitor-start motor, rated 1/3
1200
1400
hp (250 W), 1760
r/min,
1800 r/min
1600
1
15
V,
60 Hz, class A
insulation.
CHARACTERISTICS OF A CAPACITOR -START MOTOR
TABLE 18A
Rating:
N-m
V////
200
0
^
tore
m
1.35
ji
250 W, 1760
r/min,
115
V,
60 Hz, Insulation Class 105°C
No-load
Full -load
115
V
voltage
115
power
250
W
current
4.0
V
A
current
5.3
A
losses
105
W
RF.
64%
voltage
Locked
63.9%
efficiency
speed
1760 r/min
torque
1.35
Nm
115
23
7a
current 7 L
torque
3.4
current
Nm
1600 r/min
speed
%
13
A
V
A
19 A
29 A
voltage
current 7 S
current
Breakdown
rotor
torque
capacitor
6Nm
320
[jlF
*
SINGLE-PHASE MOTORS
18.9 Vibration of single-
phase motors
If
we
we
touch the stator of a single-phase motor,
note that
it
vibrates rapidly, whether
full-load or no-load.
operates
it
These vibrations do not
at
exist in
2-phase or 3-phase motors; consequently, single-
phase motors are more noisy.
What causes
electric
to the fact
the deceleration intervals coincide with the negative
peaks. Consequently, the acceleration/deceleration
power whereas
it
properties given in Table 18A.
is
5.3
draw
A and
the
it
+ 1000
the
W
The
We
line.
it
is
current,
power P supplied
find that
218 W.
When
P
we
to the
we
can
mo-
oscillates
between
power
positive
the
motor receives energy from the
when
full-load current
lags 50° behind the line voltage. If
18. 14).
W and
mechan-
motor having the
waveshapes of voltage and
plot the instantaneous
tor (Fig.
is
It
delivers constant
power. Consider the 250
ical
due
power is positive, negative, or zero, the mechanical
power delivered is a steady 250 W.
The motor will slow down during the brief periods when the electric power it receives is less than
250 W. On the other hand, it will accelerate whenever
the electric power exceeds the mechanical output
plus the losses. The acceleration intervals coincide
with the positive peaks of the power curve. Similarly,
motor always receives pulsating
this vibration?
that a single-phase
40
line.
is
Conversely,
negative the motor returns energy to the
However, whether the instantaneous
electric
occur twice per cycle, or 120 times per sec-
intervals
ond on
60 Hz system. As a
a
and rotor vibrate
The
stator
at
vibration and noise.
is
18.15).
both the stator
are
transmitted to the
in turn,
generates additional
vibrations
mounting base which,
motor
result,
twice the^line frequency.
To eliminate
the problem, the
often cradled in a resilient mounting (Fig.
It
consists of
two
soft rubber rings placed
between the end-bells and a supporting metal
bracket. Because the rotor also vibrates, a tubular
rubber isolator
is
sometimes placed between
Figure 18.14
The instantaneous power absorbed by a single-phase motor varies between +1000
output is constant at 250 W; consequently, vibrations are produced.
W and
-218 W. The power
the
ELECTRICAL MACHINES AND TRANSFORMERS
402
shaft
and the mechanical load, particularly when
the load
a fan.
is
Two-phase and 3-phase motors do not
because the
total
from
phases
all the
is
vibrate
power they receive
instantaneous
constant (see Section 8.7).
18,10 Capacitor-run motor
The capacitor-run motor is essentially a 2-phase
motor that receives its power from a single-phase
source. It has two windings, one of which is directly
connected to the source. The other winding is also
connected to the source, but
capacitor (Fig. 18.
Figure 18.15
Single-phase capacitor-start motor supported
in
a
re-
and noise
silient-mount cradle to reduce the vibration
transmitted to the mounting surface. Motor rated at 1/3
hp,
1725
r/min,
230
V,
60 Hz has a
3.0 A, efficiency of 60 percent,
full-load current of
and power
factor of
60
percent. Other characteristics: no-load current: 2.6 A;
has a large
compared
1
with a paper
in series
The capacitor-fed winding
6).
number of turns of relatively small
wire,
connected winding.
to the directly
This particularly quiet motor
is
used to drive fixed
loads in hospitals, studios, and other places where
silence
is
important.
It
has a high power factor on
account of the capacitor and no centrifugal switch
is
locked-rotor current: 13 A; locked rotor torque: 3.6 pu;
breakdown torque: 3.0
pu; service factor:
required.
1
278 mm;
232 mm.
(Courtesy of Baldor Electric Company)
The motor acts
when it operates at
low.
Capacitor-run motor having a NEMA rating
Corresponding phasor diagram at full load.
of
30 millihorsepower.
motor only
full-load (Fig. 18.16b).
*
Figure 18.16
b.
is
as a true 2-phase
these conditions, fluxes
(a)
a.
starting torque
.35; total
weight: 10 kg; overall length including shaft:
overall height:
However, the
ct>
s
a
and
<t s created
Under
by the
SINGLE-PHASE MOTORS
two windings are equal and out of phase by 90°. The
motor
is
The shaded-pole motor is very popular for ratings
below 0.05 hp (—40 W) because of its extremely
below 500 W.
simple construction (Fig.
Reversing the direction
winding
order to reverse the direction of rotation of the
In
we have
discussed so
far,
we have
to inter-
change the leads of either the auxiliary winding or
main winding.
the
However,
if
a single-phase motor
with a centrifugal switch,
versed while the motor
is
its
is
equipped
same
In the case
running because both windings are
times. In the case of very small motors, the ro-
in Fig.
1
8.
1
7. In
in position
1,
while winding
winding
B
is in
A
is
When
the switch
is
thrown
is
directly across the line,
series with the capacitor.
connection the motor turns clockwise.
switch
is
the circuit
such a motor, the main and
auxiliary windings are identical.
this
in
can be reversed by using a double-throw switch
shown
to position 2, the role
reversed and the motor will
<J>
a that lags
behind
tor is
as
and
<!>!, <t> 2 ,
<J> 3
all in
,
come
<X>
3)
<£>
and
starts the
and
2
<J>
a
3
This current produces a flux
.
x
Consequently,
The combined
.
motor.
The
<I>
+
(<J>i
/ b in
<I>
3)
<1>
2
.
and
As
is
from the
(I>
b
the
produces
rotation
and then
run up to speed in the opposite direction.
copper ring
(auxiliary winding)
/b
main winding
r\
%J
•
q n n o
\J
xJ
ac source
Figure 18.17
Reversible single-phase motor using a 2-pole switch
Figure 18.18a
and capacitor.
Fluxes
in
a shaded-pole motor.
—
sJ
-
right.
and the
a
revolving field that drives the rotor clockwise.
o
+
which
before, the
of the windings
to a halt
(<t> 2
field,
the ring,
With
When
also lags
up by the pole on the
is set
b lags behind
bined action of
a
shaded (ring side) of the pole.
Flux 4> 2 induces a current
sulting flux
<I>
action of
direction of rotation
to the
similar torque
links the
pole, inducing a
produces a weak revolving
unshaded side
A
<J>
phase. Flux
on the left-hand
<&
at all
tation
nents
/.,.
18. 16)
a copper ring surrounding
The main winding is a simple coil connected to
The coil produces a total flux <£> that
may be considered to be made up of three compo-
behind
can be changed while the mo-
basically a
is
the ac source.
rather large current
of a capacitor-run motor (Fig.
It
which the auxiliary
in
a portion of each pole.
the main wind-
If
direction.
the direction of rotation
composed of
rotation cannot be re-
ing leads are interchanged, the motor will continue
to turn in the
is
short-circuited ring
running.
18. 18).
small squirrel-cage motor
of rotation
motors
18.12 Shaded-pole motor
then essentially vibration-free. Capacitor-
run motors are usually rated
18.11
403
re-
comweak
ELECTRICAL MACHINES AND TRANSFORMERS
404
TABLE 18B
Properties of a Shaded-Pole Motor, having 2 poles,
Rated 6 W, 115
V,
60 Hz.
No-load
input
A
0.26
current
power
W
15
3550 r/min
speed
Locked rotor
input
A
0.35
current
power
torque
24
W
10
mN-m
Full-load
input
power
2900
torque
2600 r/min
breakdown speed
power
18.13 Universal motor
starting
torque,
efficiency,
and
factor are very low, the simple construction
and absence of
a centrifugal switch give this
marked advantage
in
8.
1
9.
entire
very similar to
basic construc-
motor
is
is
shown
in Fig.
laminated to
re-
low-power applications. The
it
fixed by the position of the copper rings. Table
18B gives the
The
is
The
magnetic circuit
tion of a small universal
motor
direction of rotation cannot be changed, because
is
universal motor
a dc series motor (Section 5.8).
1
a
mN-m
21
V,
The single-phase
Although the
mN-m
W
6
breakdown torque
5 millihorsepower, 115
r/min
19
mechanical power
at
W
21
speed
Figure 18.18b
Shaded-pole motor rated
60 Hz, 2900 r/min.
(Courtesy of Gould)
A
0.33
current
ac flux
typical properties of a 2-pole shaded-
pole motor having a rated output of 6
W.
Example 18-3
Calculate the full-load efficiency and slip of the
shaded-pole motor whose properties are listed
in
Table USB.
Solution
The
efficiency
is
(PJP,)
X
X
(6/21)
100
(3.6)
100
28.6%
(/7 S
—
(3600
0.194
n)/n s
- 2900)/3600
= 19.4%
Figure 18.19
Alternating-current series motor, also called universal
motor.
SINGLE-PHASE MOTORS
duce eddy-current losses. Such a motor can operate
on either ac or dc, and the resulting torque-speed
about the same
in
each case. That
is
why
it
is
is
called
Series motors are built in
motors formerly used
When
the motor
is
connected
to
an ac source, the
ac current flows through the armature and the series
field.
The
field
in
produces an ac flux
<E>
electric locomotives.
performance curves of a
8000 r/min, universal motor
115 V,
The
hp.
different sizes,
very large traction
to
some
Fig. 18.20 gives the ac
a universal motor.
many
from small toy motors
starting
full-load current
405
175
is
rated at 1/100
mA.
that reacts
with the current flowing in the armature to produce
18.14 Hysteresis motor
a torque. Because the armature current and the flux
reverse simultaneously, the torque always acts in
the
same
in this
direction.
No
revolving field
is
produced
type of machine; the principle of operation
is
same as that of a dc series motor and it possesses
the same basic characteristics.
The main advantage of fractional horsepower
the
universal motors
ing torque.
is
speed and high
their high
They can
start-
therefore be used to drive
high-speed centrifugal blowers
in
vacuum
cleaners.
The high speed and corresponding small size for a
given power output is also an advantage in driving
portable tools, such as electric saws and drills. Noload speeds as high as 5000 to 5 000 r/min are pos-
To understand
sis
motor,
the operating principle of a hystere-
us
let
first
consider Fig.
1
8.2
1
.
It
shows
a
stationary rotor surrounded by a pair of N, S poles
that
can be rotated mechanically
rection.
ial
The
rotor
is
in
composed of
of high coercive force. Thus,
magnet material whose
As
in
it
resistivity
of an insulator. Consequently,
up eddy currents
a
such a
it
is
a clockwise di-
ceramic materis
a permanent
approaches that
impossible to
set
rotor.
the N, S field rotates,
it
magnetizes the rotor;
consequently, poles of opposite polarity are contin-
uously produced under the moving N, S poles. In
effect, the
revolving field
is
continuously reorient-
1
sible but, as in
any series motor, the speed drops
ing the magnetic
individual
rapidly with increasing load.
domains
in the rotor. Clearly, the
domains go through
a
complete cycle (or
hysteresis loop) every time the field
makes one
complete revolution. Hysteresis losses are therefore
produced
in the rotor,
proportional to the area of the
r/mm
hysteresis loop (Section 2.26).
14 000
These losses are
dis-
sipated as heat in the rotor.
Let us assume that the hysteresis loss per revo-
12 000
lution
%
is
Eh
joules and that the field rotates
50-
10 000
efficie
ncy
mA
40-
8000
\
Spe
400
o
stationary rotor
30-|- 300
6000
/
CI rrent
^
5=
LU
4000
2000
—
-
rate
j
20-
200
10-
100
torque
010
30
20
40
mN m
Torque
Figure 18.20
Characteristics of a small
tor
1
15
having a full-load rating of
60 Hz universal mo1/100 hp at 8000 r/min.
V,
Figure 18.21
Permanent magnet
revolving
field.
rotor
and a mechanically-driven
at
ELECTRICAL MACHINES AND TRANSFORMERS
406
n revolutions per minute.
the rotor per minute
The energy
The corresponding power
nE h
(dissipated as heat)
Ph =
Wit
=
the mechanical
the N, S poles. This
power
P =
Because P
[W]
in the rotor
power used
can
to drive
given by
is
/z779.55
= P h we
is
(3.4)
/z£ h /60
However, the power dissipated
come from
hysteresis
motor
W=
only
dissipated in
is
(3.5)
have
,
=
/z779.55
speed
Figure 18.22
Typical torque-speed curves of two capacitor-run
/?£ h /60
motors:
whence
T= £
h /6.28
a.
Hysteresis motor
b.
Induction motor
(18.2)
where
T =
Eh —
torque exerted on the rotor [N-m]
hysteresis energy dissipated in the ro-
per turn
tor,
=
6.28
[J/r]
=
constant [exact value
2tt]
Equation 18.2 brings out the remarkable feature
that the torque
18.21)
is
tion. In
needed
to drive the
magnets
other words, whether the poles just barely
creep around the rotor or whether they
move
speed, the torque exerted on the rotor
is
same.
It
(Fig.
constant, irrespective of the speed of rota-
is
this basic
at
high
always the
property that distinguishes hys-
Figure 18.23
teresis
motors from
all
other motors.
In practice, the revolving field
3-phase
stator, or
auxiliary winding.
When
inside such a stator,
it
is
by a single-phase
it
Single-phase hysteresis clock motor having 32 poles
produced by a
stator
a hysteresis rotor
is
is
essentially constant as
(a) in Fig.
18.22. This
is
until
Thanks
it
the curve
entirely different
squirrel-cage induction motor,
toward zero as
shown by
from a
whose torque
falls
approaches synchronous speed.
to the fixed
frequency of large distribution
systems, the hysteresis motor
is
employed
ferrite rotor.
placed
immediately accelerates
reaches synchronous speed. The accelerating
torque
and a
having an
in electric
clocks, and other precise timing devices (Fig.
1
8.23).
It
is
also used to drive tapedecks, turntables, and
other precision audio equipment. In such devices
the constant speed
looking
for.
is,
However,
of course, the feature
the hysteresis
ticularly well suited to drive
of their high
inertia. Inertia
motor
we
is
are
par-
such devices because
prevents
many
synchro-
nous motors (such as reluctance motors) from coming up to speed because to reach synchronism, they
have
to
suddenly lock
in
with the revolving
field.
SINGLE-PHASE MOTORS
No
such abrupt transition occurs
motor because
to
in the hysteresis
When
c.
develops a constant torque right up
it
at
synchronous speed.
In
some
W=
The power dissipated
function as a vibration-free capacitor-run motor.
is
accelerating,
its full
torque
P =
available to carry the mechanical load and to
overcome
Once
inertia.
the
motor runs
reaches synchronous
it
speed, the rotor poles are
still
There
d.
magnetized and so
is
=
Wit
no energy
motor runs
at
to the rotor.
therefore,
0.8
=
180
J
in the rotor is
180/60
=
3
W
loss in the rotor
when
the
synchronous speed because the
magnetic domains no longer reverse.
an ordinary permanent-magnet
like
is,
X
225
moves
The en-
the rotating field
stalls,
ergy loss per minute
enhanced by designing the motor to
While the motor
motor
225 r/min with respect
turntable audio equipment these fea-
tures are further
is
the
407
sychronous motor. The rotor poles will lag behind
the stator poles
by a certain angle, whose magni-
18.15
tude depends upon the mechanical torque exerted
Synchronous reluctance
motor
by the load.
We
Example 18-4
A small 60 Hz
poles. In
hysteresis clock
motor possesses 32
making one complete
turn with respect to
to 0.8
b.
c.
the
motor
when
when
rotor losses
the motor
is
stalled
motor runs
the
at
syn-
it
pull-in
up as a standard
salient poles lock with the re-
and so the motor runs
at
synchronous
speed. Both the pull-in and pull-out torques are
to those of a hysteresis
motor of
accelerate high-inertia loads to synchronous speed.
The
and pull-out torques are about
T= £
=
poles
s
-
=
120///;
=
225 r/min
The maximum power
P =
/?
779.55
= 3W
=
0.8/6.28
at
stator poles are
X
(18.2)
60/32
is
(225
(or 3/746
=
X
0.127)/9.55
1/250 hp)
1
).
a rate that corresponds to the slip. If the
is
120
18.22.
corre-
slipping past the rotor
Nm
0.127
The synchronous speed
/?
=
h /6.28
18.24
approaches syn-
sponding to full-load torque (operating point
equal in a hysteresis motor:
b.
starts
when
The reason can be seen by referring to Fig.
Suppose the motor has reached a speed //,
Solution
The
field,
weak, compared
rotor losses
the stator. Fig.
equal size. Furthermore, reluctance motors cannot
chronous speed
a.
Such a reluctance motor
volving
number
poles must be equal
rotor milled out to create four salient poles.
chronous speed, the
The pull-in and pull-out torques
The maximum power output before
The
The
number of poles on
shows a
squirrel-cage motor but,
stalls
d.
number of
J.
Calculate
a.
of salient poles. The
to the
the revolving field, the hysteresis loss in the rotor
amounts
can build a synchronous motor by milling out a
standard squirrel-cage rotor so as to create a
Figure 18.24
Rotor of a synchronous reluctance motor.
ELECTRICAL MACHINES AND TRANSFORMERS
408
rotor
so
is
to
in the
lock with the revolving field,
time
it
this interval {At),
it
chronous speed
//
s
will
,
sweep
It is
particularly well adapted to variable-frequency
electronic speed control. Inertia
is
then no problem
because the speed of the revolving field always
never be achieved. The
tracks with the speed of the rotor. Three-phase re-
is
going from speed
that in
is
must do
not achieved during
past a rotor pole. If pull-in
problem
it
takes for one stator pole to
/?,
to syn-
the kinetic energy of the re-
motors of several hundred horsepower
luctance
have been
using this approach.
built,
volving parts must increase by an amount given
by Eq. 3.8:
AE =
k
5.48
X \0~'J(n
18.16 Synchro drive
2
-
s
2
/;,
(18.3)
)
In
where
./ is
moment of
the
inertia.
Furthermore, the time interval
At
=
60/(/z s
-
is
some remote-control systems we may have
move the position of a small
given by
meters away. This problem
(18.4)
/?,)/?
flexible shaft.
But
if
must develop an accelerating power P d of at
tor
P = AEJAt
X
1.8
10
4
n s (n s
- n\Y Jp
power P
will
never pull into
step. In essence, a reluctance
motor can only synchronize when the
small and the
moment
How does
of inertia J
is
slip
speed
is
impractical.
Two
rotors are also connected in parallel
a single-phase source.
about
this
arrangement
is
and energized
The remarkable
that the rotor
other.
Thus,
if
we slowly turn rotor A clockwise
B will move clockwise through
17°.
Obviously, such a system enables us to control
a rheostat
from
a
remote
'location.
receiver
a synchro system.
feature
on one ma-
chine will automatically track with the rotor on the
»/b
of
in
phases of the respective
~U
Figure 18.25
then
rheostat
motors whose 3-phase stators are connected
from
is
cheaper than any other type of synchronous motor.
Components and connections
We
knob and
through 17°, rotor
low.
Despite this drawback, the reluctance motor
away, the
becomes
100
such a shaft work?
parallel (Fig. 18.25).
demanded by the load. If the sum of
P u + P L exceeds the power capacity of the motor, it
by using a
m
is
Consider two conventional wound-rotor induction
x
easily solved
electrical shaft to tie the
(approx)
Furthermore, the motor must continue to supply
the
together.
(18.5)
Ll
=
least
employ an
one or two
the rheostat
flexible-shaft solution
Consequently, to reach synchronous speed, the mo-
is
to
is
rheostat that
SINGLE-PHASE MOTORS
Two
One
409
miniature wound-rotor motors are required.
the stator voltages are again in balance (phase by
coupled to a control knob,
phase), and the torque-producing currents disappear.
(the transmitter)
is
and the other (the receiver)
is
coupled
Synchros are often employed
to the rheo-
to indicate the po-
The 5-conductor cable (conductors a-b-c-1-2)
linking the transmitter and receiver constitutes the
with the result that the torque requirements are
flexible electrical shaft.
small.
stat.
The behavior of
system
this selsyn or synchro control
Assume
explained as follows.
is
that the
sition of an antenna, a valve, a
gun
turret,
Such transmitters and receivers
and so on,
are built with
watch-like precision to ensure that they will track
with as
error as possible.
little
transmitter and receiver are identical and the rotors
are in identical positions.
cited, they
behave
When
the rotors are ex-
like the primaries of
two
EQUIVALENT CIRCUIT OF A
SINGLE-PHASE MOTOR
trans-
formers, inducing voltages in the respective stator
windings. The voltages induced
in the three stator
windings of the transmitter are always unequal
Chapter
In
(Fig.
we developed
15
5.6) for
1
the equivalent circuit
one phase of a 3-phase induction mo-
because the windings are displaced from each other
tor.
by 120°. The same
exception that the magnetizing branch has been
in the stator
is
true for the voltages induced
moved
of the receiver.
Nevertheless, no matter what the respective stator
voltages of the transmitter and receiver
are identical in both
the rotors
may
be, they
synchros (phase by phase) when
occupy the same
position.
The
stator volt-
ages then balance each other and, consequently, no
The
current flows in the lines connecting the stators.
rotors,
however, carry a small exciting current
Now
if
we
This circuit
/0
They
will
reproduced
with the
in Fig. 18.26,
between
to the technically correct position
points
1
and
2.
The reason
for the
change
is
that
most single-phase motors are fractional horsepower
machines for which the exact
needed
this
circuit
diagram
to get reasonably accurate results.
we now develop
model,
is
Using
a similar equivalent
circuit for a single-phase motor.
.
turn the rotor of the transmitter,
three stator voltages will change.
is
its
18.17 Magnetomotive force
no longer
distribution
balance the stator voltages of the receiver; consequently, currents
necting the
/.„ / b , / L
,
will
flow
in the lines
con-
two devices. These currents produce a
torque on both rotors, tending to line them up. Since
the rotor of the receiver
with the transmitter.
is
free to
As soon
move,
it
will line
up
as the rotors are aliened.
In order to optimize the starling torque, efficiency,
power
tor,
tor
and noise
factor,
level of a single-phase
mo-
magnetomotive force produced by each stapole must be distributed sinusoidally across the
the
pole face. That
number of
is
the reason for using the special
turns (10, 20, 25, and 30) on the four
concentric coils
shown
Let us examine the
poles
when
I
8.3(a).
the concentric coils carry a peak current
2 amperes. Table
of, say,
mmf, using
of the
in Fig.
mmf created by one of the four
1
8C shows
the distribution
numbers
as a measure of
the slot
distance along the pole. For example, the 25-turn coil
lodged
in slots
these slots an
2 and 8 (Fig.
mmf of 25 X
8.27). produces
1
2
=
between
50 amperes
(or
am-
pere-turns). Similarly, the 10-turn coil in slots 4 and 6
2
The
Figure 18.26
Equivalent circuit of one phase of a 3-phase cage motor referred to the
produces between these
primary (stator) side.
Fig.
1
8.27.
the pole
slots
distribution of these
is
an
mmf of 20 A.
mmfs
is
illustrated in
The total mmf produced in the middle of
60 + 50 + 40 + 20 = 70 A and it drops
1
ELECTRICAL MACHINES AND TRANSFORMERS
4 0
1
soidal, but the
TABLE 18C
only 0.4
1-9
slot
slot
3-7
slot
4-6
9
30
2-8
slot
Mml'
Turns
Coil pitch
25
2
20
2
X
10
40
-
10
mmf in
85 turns
the center of the pole will be
=
the current reverses and
mmf
X 30 = 60 A
X 25 = 50 A
X 20
A X
20
equal
to, say,
—
1
,2
A, the
mmf will
However, the
still
be distributed sinusoidally but with a peak value
the center of
A
We
A
l
.2
A X
in
102 A.
conclude that the ac current produces a pul-
mmf, which is distributed
each pole and whose amplitude
3 - phase stator. the
mmf
sinusoidally across
varies sinusoidally
mmf
Thus, unlike the
time.
in
= -
85 turns
sating
170 ampere turns
85 turns
will also reverse.
34 A. Subsequently, when
is
does not rotate but remains fixed
18.18 Revolving
produced by
a
of a single-phase stator
mmfs
in place.
a single-
in
phase motor
It
can be proved mathematically that a stationary pul-
mmf
sating
having a peak amplitude
placed by two
volving
in
M can
be
re-
mmfs having a fixed amplitude Mil
re-
opposite directions
synchronous speed.
at
Referring to our previous example, a 4-pole pulsat-
mmf that reaches positive and negative peaks of
A at a frequency of 60 Hz can be replaced by two
4- pole mmfs having a constant amplitude of 85 A roing
123456789
-
-
1
1
^
pole pitch
70
tating in opposite directions at
1800 r/min. The
re-
mmfs are also distributed sinusoidally in
space. As the oppositely moving mmfs take up successive positions, the sum of their magnitudes at any
volving
Figure 18.27
Distribution of the
pole
when
current
magnetomotive force across one
is
2 A.
point in space
is
equal to the pulsating
point. This can be seen
mmf at
that
by referring to Fig. 18.28,
which shows a portion of the forward and backward
off in steps on either side of center. Adjacent poles
have the same
magnetic
mmf
distribution but with opposite
polarities.
We have superposed upon this figure a smooth
mmf having a perfectly sinusoidal distribution. It
reveals that the stepped mmf produced by the four
concentric coils tracks the sine
Indeed,
soidal
we could
mmf without
The
wave very
replace the stepped
closely.
mmf by
a sinu-
introducing a significant error.
current flowing in the four coils alternates
sinusoidally (in time) at the line frequency of 60 Hz.
Consequently, as the current varies, the
in
proportion. For example,
mentarily 0.4 A, the
when
mmf varies
the current
mmf distribution
is
mo-
remains sinu-
revolving fields
(mmfF and mmfB
the stationary but pulsating
sweeping
),
past
mmf.
The revolving mmfs respectively produce the
same effect as the revolving mmf created by a
3-phase stator. Consequently, we would expect the
circuit diagram of a single-phasor motor to resemble that of a 3-phase motor. However, since the
mmfs
slip
.v
the rotor has a
if
with respect to the forward-moving mmf,
will automatically
spect to the
The
ing
on
rotate in opposite directions, their effect
the rotor will be different. Thus,
slip
of (2
s)
with
it
re-
backward-moving mmf.
circuit
mmf
have a
diagram as regards the forward-mov-
having a
slip s
is
shown
in Fig.
18.29a.
SINGLE-PHASE MOTORS
Figure 18.29
50
mmfB ^
r'
a.
/\
v
\^
~
mmf f
Equivalent circuit as regards the forward-moving
mmf.
b.
Equivalent circuit as regards the backward-moving
mmf.
-90°
0
6
.
+90°
Similarly, the circuit diagram for the backward-
~
revolving
Fig.
1
mmf having
8.29b. For the
the physical
to
^0A
-90°
/0\
/t=
+90°
—
a slip (2
moment we
meaning of
r,, r 2 ,
Vj,
s) is
shown
in
will not define
.\
2 , etc.,
except
say that they are related to the stator and rotor
resistances and reactances.
How
should
wc merge
these two diagrams into a single diagram to represent the single-phase motor?
4"
Figure 18.28
pulsating mmf having a peak amplitude of 170 A
can be represented by a forward and backward revolving mmf having a fixed amplitude of 85 A. Shown are
successive positions of mmf F and mmf B and the corresponding amplitude of the stationary, pulsating mmf.
18.19 Deducing the circuit diagram
of a single-phase motor
The
First,
the
/ jF
we know
that the oppositely rotating
same magnitude. Consequently,
and
/
cuits can
,
B are identical, which
be connected
means
in series.
mmfs have
the stator currents
that the
two
cir-
Second, the forward
ELECTRICAL MACHINES AND TRANSFORMERS
412
2
2jx 2
2./.V,
r,
2
Figure 18.31
Equivalent circuit of a single-phase motor at standstill.
In practice
we assume x = x 2
.
x
The above analysis
ances r,,.Vj, etc., shown
Thus,
Equivalent circuit of a single-phase motor.
of r
E¥
is
associated with
EB
backward voltage
Because the
mmfK
associated
is
circuits are in series, the
voltages must be equal to the voltage
stator.
It
,
while the
with
mmfB
sum of
r h r2
,
-V], .v 2 ,
etc.,
which case the
meaning of the
E applied to the
circuit
suppose the motor
slip s
=
1
.
is
Example 18-5
A test
motor behaves
like a
(the
represent the following physical elements:
Draw
stator resistance
2r 2
=
rotor resistance referred to the stator
2/.X-,
=
stator leakage reactance
2jx 2
=
rotor leakage reactance referred to the
2jX m
—
and iron
resistance corresponding to the
and iron losses
magnetizing reactance
4
il
3 11
600
losses:
O
60
il
the equivalent circuit diagram and determine
the
factor of
Solution
circuit
the values of the listed
friction,
single-
results:
power output, efficiency, and power
motor when it turns at 1725 r/min.
The equivalent
stator
windage,
725 r/min
2 11
magnetizing reactance:
the
2R m =
1
n
tion,
reveals that the parameters
etc.,
=
60 Hz,
V,
resistance corresponding to the windage, fric-
sim-
r,
2r,
3
in
short-circuit.
,
20
rotor leakage reactance referred to the stator:
cir-
is in
.V|
1
stator leakage reactance:
the
rotor)
,
on a 1/4 hp,
phase motor reveals the following
stationary, in
which the secondary winding
It
imped-
the equivalent circuit.
stator resistance:
of Fig. 18.30 therefore reduces to that shown
ple transformer in
10 ohms, the value
is
forth, for the other
rotor resistance referred to the stator:
Under these conditions,
Fig. 18.31. In essence, the
ohms, and so
parameters
forward and backward circuits are identical. The
cuit
in
the stator resistance
5
.
these
follows that the equivalent circuit of the sin-
interpret the
if
is
{
ances
gle-phase motor can be represented by Fig. 18.30
To
18.29 to 18.31 are
in Figs.
equal to one-half of the actual physical quantities.
Figure 18.30
stator voltage
indicates that the imped-
The
slip is s
=
(1800
We first determine
circuit
between points
diagram
(Fig. 18.32)
shows
impedances divided by .two.
the
1
725)/ 1800
=
0.0417.
impedance of the forward
1, 3:
SINGLE-PHASE MOTORS
4
1
1.5
J
1
1
+
j
.
=
1
=
14.89
+
+
1.5
j
+
j
300
30
+
13.89
j
1
.5
19.53
j
21.03
The impedance of the backward
points 3, 2
+
48
between
circuit
120 V
is
1
+
1
1.5
j
1
1
1
+
j
=
+
1
=
j
1.93
The current
/
=
1
+
j
+
0.93
+
j
+
1.02
j
1.5
.45
1
2.95
in the stator is
£/(Z F
/
.5
300
30
+ ZB =
=
+j
16.82
12()/(
)
23.98)
Figure 18.32
120/(29.29^54.95)
See Example
= 4.097Z -54.95
The forward voltage between points
1,
3
18-5.
is
E v = IZ ¥ = (4.097Z -54.95) X (14.89 +
= 4.097/1-54.95 X 25.77Z54.7
Backward
rotor current:
21.03)
j
l
In
=
I
I
= 105.6Z-0.25
The backward voltage between
points 3, 2
E B = IZ H = 4.097^-54.95 X (1.93 +
= 4.097^-54.95 X 3.52Z56.8
=
is
j
l
2.95)
14.42/1 1.85
=
3.89
'
1
30
4.097
300
Z
+
48
1
.5
.5
1
(0.93
+
j
1
.45)
Z - 54.9 5 X 1,72 Z 57 .32
1.81 Z 55.78
+
j
Pv =
1.5
54.95
(
1
Z- 54.95
X
3.89
+
23.96
48.02/1 1.79
2.16
to rotor:
1.5
j
1
9.53)
2
lv
9.55 P.
Z 54.58
X
Forward torque
1
2.044/1
j
j
Z- 53.4
Forward power
j
48.02/1 1.79
4.097
+
+
1
1
48
.02
.02
L81Z55.78
Forward rotor current:
=
1
Z - 54.95
4 .097
4.09 7
'f
I
300
30
j
/i
48
7,
=
1.02
2
X 48 =
W
200.5
:
X
9.55
s
X
2.044
1
Backward power
I pj'~
H
=
to rotor
=
200.5
=
1.064
N m
800
3.89
PB
2
X
:
1.02
=
15.4
W
1
ELECTRICAL MACHINES AND TRANSFORMERS
414
Backward torque Tn
Pn
9.55
//
motors?
X
9.55
-
15
such a mounting necessary on 3-phase
:
15.4
-
0.082
N-m
1
800
1
s
What
1
- Tu =
1.064
-
0.082
-
0.982
8-8
N-m
X
1725
0.982
177
~
9.55
b.
W
c.
9.55
d.
Horsepower:
e.
177
f.
0.24 hp
746
Active power input to
=
X
120
is
in this
best suited to drive the follow-
A small portable drill
A 3/4 hp air compressor
A vacuum cleaner
A 1/100 hp blower
A 1/3 hp centrifugal pump
A 1/4 hp fan for use in a hospital
g.
An
h.
A
ward
electric timer
hi-fi turntable
stator:
=
4.097 cos 54.95
282.3
W
Intermediate level
1
Power
main advantage of a capacitor-
ing loads:
a.
nT _
e
the
Which of the motors discussed
chapter
Mechanical power output P:
£/cos
is
run motor?
Net torque:
7V
8-7
8-9
factor:
Referring to Fig.
1
8.
1
1 ,
the effective im-
pedance of the main and auxiliary wind-
cos 54.95
-
0.57
ings under locked-rotor conditions are
= 57%
given as follows:
Efficiency:
177
0.627
= 62.7%
Effective
Effective
resistance
reactance
282
Main winding
4
Auxiliary winding
7.5 il
fl
4
7.5 11
il
Questions and Problems
If the line
Practical level
18-1
A 6-pole
to a
single-phase motor
60 Hz source. What
connected
a.
synchronous
b.
is
is its
speed?
1
8-2
What
is
the purpose of the auxiliary wind-
c.
d.
The power
the rotation of such a
State the
18-10
What
8-5
main difference between a
18-6
The palm of the human hand can just
how
1/4
a shaded-pole motor
the properties
hp motor
is
64°C
and advan-
a.
Can
b.
Is
a person keep his
the
Referring to Fig.
18. 13, if the
stant at
havior of the motor
some single-phase motors
resilient
mounting?
Is
hand on the frame?
motor
connected to a load whose torque
Why
equipped with a
an ambient tem-
motor running too hot?
tages of a universal motor.
are
in
perature of 76°F,
18-11
some of
If
split-
are their relative advantages?
Explain briefly
List
factor under locked-rotor con-
barely tolerate a temperature of 130°F.
operates.
1
/s
the full-load temperature of the frame of a
phase motor and a capacitor-start motor.
18-4
/,,
ditions
motor?
8-3
119 V, calculate the
is
The magnitude of and / s
The phase angle between /u and
The line current I L
ing in a single-phase induction motor?
How can we change
1
voltage
following:
on the
is
is
con-
4 N-m, explain the resulting be-
line.
when
it
is
switched
SINGLE-PHASE MOTORS
18-12
a.
A single-phase
moior vibrates
quency of 100 Hz. What
*
power
the
'
b.
c.
a.
motor does not have
set in a resilient
mounting.
to
be
b.
Why?
at
c.
1600 r/min. Calculate the hys-
Referring to the 6
Table
a.
b.
c.
d.
W shaded-pole motor
18-18
in
and the voltage
at the
The
[N m]
starting torque
A
3 hp,
1
725 r/min 230 V, totally-enclosed,
fan-cooled, capacitor-start, capacitor- run,
single-phase motor manufactured by
The rated power output in millihorsepower
The full-load power factor
The slip at the breakdown torque
The per unit no-load current and locked-ro-
Baldor Electric
Company
has the follow-
ing properties:
A
no-load current: 5
tor current
18-14
resistance of the transmission line
starting current
Industrial appl ication
8B, calculate the following:
1
the Appendix, calcu-
in
following:
when
in-lb
teresis loss per revolution [J],
18-13
The
The
AX3
1
motor terminals
60 Hz single-phase hysteresis
motor develops a torque of 6
running
late the
line?
A capacitor-run
A 4-pole,
Using Table
at a fre-
the frequency of
is
4
locked-rotor current: 90
Referring again to Fig. 18.13, calculate
full -load current:
A
A
15
the following:
a.
The locked-rotor torque
b.
The
per-unit value of the
c.
The
starting torque
winding
d.
The
e.
How
is
locked-rotor torque: 30 lbfft
(t't-Ibf
|
LR
torque
when only
the
main
breakdown torque: 20
full-load
breakdown torque
are the lorque-speed curves affected
the line voltage falls
from
1
V
15
to
100
if
power
18-15
Table
In
service factor:
1.15
V?
a.
mass: 97
1
9 lbf
The voltage across
1b
Using the above information, calculate the
the capacitor under
following:
The corresponding phase angle between
and
18-16
ft
8 A, calculate the following:
locked-rotor conditions
b.
87%
factor:
full-load torque:
level
lbfft
excited
per-unit
Advanced
79%
full-load efficiency:
a.
/s
The
per-unit values of locked-rotor torque,
locked-rotor current, and breakdown torque
/.,
Referring to Fig. 18.16,
if
b.
the capacitor-
The
full-load torque expressed in
run motor operates at full-load, calculate
c.
The capacitor
could be added across
that
the following:
the stator so that the full- load
a.
The
line current
/,
rises
b.
c.
d.
18-17
The power factor of the motor
The active power absorbed by each winding
The efficiency of the motor
The motor described in Table 8A has an
LR power factor of 0.9 lagging. It is installed in a workshop situated 600 ft from
a home, where the main service entrance
is located. The line is composed of a 2conductor cable made of No. 12 gauge
copper. The ambient temperature is 25°C
newton-
meters
18-19
A
from
87%
to
power factor
90%
3/4 hp, 1725 r/min, 230 V, totally-
enclosed, fan-cooled, capacitor-start,
single-phase motor manufactured by
1
and the service entrance voltage
is
122 V.
Baldor Electric
Company
has the fol-
lowing properties:
no-load current: 4.4
A
locked-rotor current:
f u 1 1 - load c u rre nt
:
5 3
.
locked-rotor torque:
30
A
A
9.5 lbf
ft
416
ELECTRICAL MACHINES AND TRANSFORMERS
locked-rotor
full- load
power
factor:
efficiency:
breakdown torque:
58%
long and
fed from the service entrance
is
where the voltage
66%
is
230 V ±5%.
Using the above information, determine
6.
1
lbfft
the following:
full-load
power
service factor:
factor:
1
68%
a.
b.
full-load torque:
mass:
29
The motor
ft
We
wish
motor to
lb
starting torque (newton-meters),
a cable temperature of
to raise the
90%
pacitor across
power
at full-load
its
25
n
C
factor of the
by installing a ca-
terminals. Calculate the
approximate value of the capacitance,
is
copper cable
Code
2.25 lbf
The lowest
assuming
.25
fed by a 2-conductor No. 12
that has a National Electrical
rating of
20 A. The cable
is
240
feet
microfarads.
in
Chapter 19
Stepper Motors
writers, tapedecks, valves,
Stepper
motors are special motors
used
that are
In this chapter
when motion and position have to be precisely
controlled. As their name implies, stepper motors
rotate in discrete steps,
a pulse that
is
each step corresponding
supplied to one of
Depending on
its
its
18°, or
by
as
little
cover the operating princi-
will
and
By varying
the pulse rate, the
Elementary stepper motor
A very
simple stepper motor
is
made of
a 2-pole rotor
Stepper motors can turn clockwise or counter-
in Fig. 19.
1
soft iron.
The windings can
means of three switches A, B, C.
When the switches are open,
clockwise, depending upon the sequence of the
pulses that are applied to the windings.
any position. However,
The behavior of a stepper motor depends greatly
upon the power supply that drives it. The power sup-
sulting magnetic field created
ply generates the pulses,
a
which
microprocessor.
The pulses
while counterclockwise (ccw) pulses are (-).
number of steps is known exactly
follows that the number of revolutions
(
rotate
As
the rotor can take up
switch
will line
A
is
closed, the re-
by pole
up with pole
60°, Next,
if
2. In
we open
we now
so doing,
it
will
B and siwill turn ccw
switch
3.
in
60° steps by closing and opening the switches
in
this
we can make
the rotor advance
the sequence A, B, C, A, B,
417
If
ccw
Clearly,
step.
will attract
time lining up with pole
by an additional 60°,
This permits the motor to be used as a precise posi-
1
up as shown.
multaneously close switch C, the rotor
a
is al-
ccw by
if
and simultaneously close switch B,
the rotor will line
)
times.
accuracy of one
A
+
at all
to an
open switch
it
are
result, the net
ways precisely known
the rotor and so
in turn are usually
counted and stored, clockwise (cw) pulses being
It
shown
consists of a stator having three salient poles and
be successively connected to a dc power supply by
r/min.
by
will also discuss
19.1
It
a time, or to rotate stepwise at speeds as high as
initiated
We
limitations.
the types of drives used to actuate these machines.
as a fraction
advance very slowly, one step
4000
plotters, type-
printers.
stator windings.
motor can be made
to
we
and
more common stepper motors, together with
their properties
to
of a degree per pulse.
at
ple of the
design, a stepper motor can ad-
vance by 90°, 45°,
X-Y
tioning device in machine tools,
19.0 Introduction
C
Furthermore,
ELECTRICAL MACHINES AND TRANSFORMERS
418
opposite (cw) direction. Picking up speed,
again overshoot the center line of pole
upon
will
it
where-
2,
the magnetic field will exert a pull in the
ccw
direction.
The rotor will therefore oscillate like a pendulum around the center line of pole 2. The oscillations will gradually die out because of bearing
friction. Fig.
1
9.2
shows the angular position of
The rotor starts at
the rotor as a function of time.
0° (center of pole
Figure 19.1
rotor
which each step moves the
in
switches
to a halt (at 3 ms).
line at
reverse
by operating the
rotation
the
reverse sequence A, C, B, A, C,
in the
B
the last switch that
was closed
in a
switching se-
quence must remain closed. This holds the rotor
its last
position and prevents
it
in
from moving under
the influence of external torques. In this stationary
state the
motor
will
remain locked provided the
external torque does not exceed the holding torque
moving from one
tion of the rotor will
position to the next, the
mo-
be influenced by the inertia and
the frictional forces that
amine
/
>
come
into play.
We now ex-
amplitude
in
rotor has a
low
ing friction.
inertia
It is
at
facing pole
l.
Let this cor-
respond to the zero degree (0°) angular position. At
the
moment
switch
A opens and switch B closes,
rotor will start accelerating
ccw toward pole
the
2.
It
rapidly picks up speed and soon reaches the center
line
of pole
2,
where
However, the rotor
able speed and
it
does
so, the
it
is
it
should
now moving
come
to
rest.
with consider-
will overshoot the center line.
magnetic
field
of pole 2 will pull
As
it
in
the opposite direction, thereby braking the rotor.
The
rotor will
come
to a halt
comes
to rest
The reader will note that in Fig. 19.2 we have
drawn the instantaneous speed of the rotor as a
function of time. The speed can be given in revolutions per second, but for stepper motors it is more
meaningful to speak of degrees per second. The
also
speed
momentarily zero
is
at
t
=
and becomes permanently zero
is
ter line
greatest
of pole
whenever the
2.
3 ms, 5 ms, 7 ms,
at
/
>
10 ms. The
rotor crosses the cen-
Clearly, the oscillations last a rel-
down.
atively long time before the rotor settles
Without making any other changes, suppose we
when
no-load and that the
and a small amount of bear-
initially
until the rotor
10 ms.
shaft.
We
discover that both the pe-
riod and the amplitude of the oscillations increase
19.2 Effect of inertia
motor operates
rotor
4 ms.
wheel on the
the
The
reverse and again crosses the center
in
increase the inertia of the rotor by mounting a fly-
the nature of these forces.
Suppose
line
overshoots the center line
oscillations continue this way, gradually di-
minishing
at
speed
of the motor.
=
t
The
.... In order to fix the final position of the rotor,
In
by 30° before coming
by 60°.
we can
and reaches 60° (center
It
now moves
Simple stepper motor
l)
of pole 2) after 2 ms.
and
start
moving
in the
the inertia increases. In Fig. 19.3, for
exam-
time to reach the 60° position has
ple, the
creased from 2
ms
to
in-
4 ms. Furthermore, the am-
plitude of the oscillations has increased.
also takes a longer time to settle
down
The
(20
rotor
ms
in-
stead of 10 ms).
The
oscillations can be
For example,
the friction.
damped by
if
raised sufficiently, the oscillations
19. 3
increasing
the bearing friction
shown
is
in Fig.
can be suppressed so as to give only a single
overshoot,
shown
damping
accomplished by using an eddy-current
is
in
Fig.
brake or a viscous damper.
fluid
such as
oil
19.4.
In
practice,
the
A viscous damper uses a
or air to brake the rotor
whenever
STEPPER MOTORS
deg
,
ing effect
60
zero
30
proportional to speed;
is
when
the rotor
10
—
*-
12
14
16
19.3 Effect of a mechanical load
18
overshoot.
Let us return to the condition
shown
where the rotor has low
and a small amount
inertia
of viscous damping due to bearing
rotor
is
coupled
moving, the effect
would expect,
it
deg
w)
to
I
angu lar p os
i
tic
ms
4
)n
in Fig.
shown
is
t
8
10
12
14
16
18
ms
ms
in Fig.
greater.
19.2, except that the iner-
in Fig.
The overshoot
is
greater and the rotor
9.2 with
is
the mechanical load and the
The
oscillations
down.
order to obtain fast stepping response,
in
its
load) should be as
pressed by using a viscous damper.
The time
to
move from one
position to the next
can also be reduced by increasing the current
deg
in the
winding. However, thermal limitations due to l~R
90
-antjular pos
lion
losses dictate the
60
CO
1
small as possible and the oscillations should be sup-
takes longer to settle down.
g
is
damped more
are
oscillations
the inertia of the rotor (and
Figure 19.3
tia is
it
As we
also prolong the time before the rotor settles
Therefore,
conditions as
9.5.
Furthermore, the overshoot
inertia increase the stepping time.
—» time
8.
19.5).
summary, both
In
!
Same
1
quickly.
pee(
0
W
in Fig.
takes longer for the motor to attain
smaller and the
60
19.2,
friction. If the
90
»
in Fig.
mechanical load while
to a
the 60° position (compare 2
»
therefore
is
at rest.
is
time
Figure 19.2
In moving from pole 1 to pole 2, the rotor oscillates
around its 60° position before coming to rest. The
speed is zero whenever the rotor reaches the limit of
?
that the brakit
speed
0
its
moving. Viscous damping means
it is
angular position
90
419
~*
maximum
current that can be
used.
speed
30
Returning to Fig. I9.l,
1
.
-
0
8
i
10
12
—*
time
14
16
18
ms
Same
shows
and the
in-
stantaneous position of the rotor (as well as
its
1
Figure 19.4
9.6
conditions as
in Fig.
19.3, except that viscous
We
assume
and that
90
at the
ngul ar po sitio n
c
V)
to
it
have
/.,,
/h
that the
driving a mechanical load. Note that
is
is
zero
—"
time
14
16
18
ms
a duration of 8 ms. Consequently, the step-
Same
conditions as
coupled
to
is,
0.048
is
1
in Fig.
19.2, except that the rotor
a mechanical load.
s to
000/8
=
125 steps per second.
complete one
therefore,
minute.
Figure 19.5
is
=
beginning and
at the
In this figure the pulses
revolution requires 6 steps, and so
12
/c
makes one-half revolution.
stepper motor has some inertia
end of each pulse.
ping rate
10
,
the motor
the speed of the rotor
deg
a.
the current pulses
when
speed)
damping has been added.
us excite the wind-
let
ings in succession so that the motor rotates. Fig.
60/0.048
However,
start-stop
the
=
turn.
1
takes 6
One
+ 25
1
The average speed
250 revolutions per
stepper motor rotates
jumps and not smoothly
motor would.
it
in
as an ordinary
420
ELECTRICAL MACHINES AND TRANSFORMERS
current
71
a,
10A
S777777777Z7A
-fr77777777777i^.
J777777,
i
10A
deg
240
—
/
© 180
« 120
60
0
settling time
H
6
-
ms
8
0
24
16
-
ms
time
24
16
time
Figure 19.6
Graph of current pulses, angular position, and instantaneous speed
(24 ms) produce one half-revolution.
of rotor during the
When the motor
continue to flow
is
first
four steps.
at rest, a
in the last
Three steps
holding current must
winding
that
was excited
so that the rotor remains locked in place.
19,5 Start-stop stepping rate
When
motor inches along
the stepping
stop fashion
shown
in Fig.
1
9.6, there
is
in the start-
an upper limit
to the permissible stepping rate. If the pulse rate of the
current in the windings
is
too
fast, the rotor is
unable
to accurately follow the pulses, and steps will be
-
lost.
current
This defeats the whole purpose of the motor, which
Figure 19.7
Graph of pull-over torque versus current
motor; diameter: 3.4 inches; length: 3.7
to correlate
of
in;
19.4 Torque versus current
this
As mentioned
rate
motor depends upon the current.
shows the relationship between the two
stepper
Fig. 19.7
for a typical stepper motor.
8 A, the
motor develops
the torque that the
When
a torque
of 3 N-m. This
it
is
pulses. In order to maintain
is
settle
down
before ad-
that the interval
at least
between successive
6 ms, which means
limited to a
maximum
of
1
9.6,
=
167 steps
per second (sps).
Bearing
in
mind what was
clear that the
said in Section 19.2,
maximum number
of steps
it
is
second depends upon the load torque and the
called the
1
steps
that the stepping
000/6
is
motor can exert while moving
from one position to the next, so
pull-over torque.
the current
)
to the next position. Referring to Fig.
means
must be
previously, the torque developed by
—
synchronism, the rotor must
vancing
is
instantaneous position (steps) with the
number of net ( + and
a stepper
weight: 5.2
Ibm.
a
its
tia
is
pet-
iner-
of the system. The higher the load and the greater
the inertia, the lower will be the allowable
of steps per second.
number
STEPPER MOTORS
deg
1
mode
start-stop
300
42
J
240
CD
/
C
slew
CO
mg nnode
180
slew curve
I
120
/
60
/
0
4
2
8
6
10
time
speed
start-stop
k
0
6
4
2
10
8
ms
16
14
12
*-
s,ewspeed
ms
16
14
12
time
800 sps
400 500 600
200
0
Figure 19.9
speed
»-
Angular position versus time curve when the stepper motor operates in the start-stop mode and the
a.
Figure 19.8
and slewing characteristic of a typical stepper motor. Each step corresponds to an advance of 1 .8
slewing mode. Stepping rate
is
same
the
in
both
Start-stop
cases.
Instantaneous speed versus time curve
b.
degrees.
curve
1
curve