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ELECTRICAL ENGINEERING
FIRST COURSE
McGraw-Hill BookCompany
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ELECTRICAL
ENGINEERING
FIRST COURSE
BY
ERNST JULIUS BERG, Sc. D.
PROFESSOR OP ELECTRICAL ENGINEERING
UNION COLLEGE, SCHENECTADY, N. Y.
AND
WALTER LYMAN UPSON, E. E., M.
ASSOCIATE PROFESSOR OF ELECTRICAL ENGINEERING
UNION COLLEGEJ|SCHENECTADY, N. Y.
FIRST EDITION
McGRAW-HILL BOOK COMPANY, INC.
239 WEST 39TH STREET, NEW YORK
6 BOUVERIE STREET, LONDON, E. C.
1916
COPYRIGHT, 1916, BY THE
MCGRAW-HILL BOOK COMPANY, INC.
THK MAPI.* PHKS8 YORK
PREFACE
A text-book in electrical engineering emanating from Union
College may be the occasion of some surprise to those who have
been conversant with the development of the electrical course
in that institution. The authors have, it is true, recorded their
objections to the use of a prescribed text. These objections still
hold good. In brief, they are, first, that a prescribed text tends
to take the life out of the class room, whether the course be con-
ducted by lectures or recitations, and second, that it tends to take
the life out of the study by relieving the student of responsibility
of continued effort.
At Union College the fundamental aim is that the student
shall first comprehend, and then, create. Comprehension comes
through directed effort. This, the student acquires readily in
the laboratory, but in the class room, it is not so easy. The
recitation falls short because it deals with the individual rather
than the class. The lecture fails when the student knows he can
fall back upon the text-book. The fault, however, is not with
the text-book itself, but with the use that is made of it.
Obviously, then, its proper use is as a means of directing the
student's effort toward comprehension. Indeed, it should com-
pel effort, not in order to make up for an author's failure to ex-
press himself clearly, but in order that the ideas shall sink in and
make permanent impressions on the mind. The book should,
therefore, be so constructed and used that it shall be an additional
aid to the student in creating his own expression of the ideas
with which he is brought into contact in the lecture, the recitation
and the laboratory. It is desirable that fundamental ideas shall
become fixed and clear in the student's mind as soon as possible,
thus leaving him in a position to exert his full mental effort on
that which is more advanced.
As he progresses, he should acquire, more and more, the power
of self-direction, that is, the power to create or construct his own
ideals. Creative work finds its primary impulse in imitation.
The student should have before him, at the outset, a model,
which he is faithfully to copy.
vi PREFACE
In attempting to embody these principles in the present
volume, the authors have sought to maintain a harmonious inter-
relationship between the book and the class room. The lectures
which form the basis of the book were first delivered eight
years ago. In that, and subsequent years, students have had to
rely for assistance upon their own notes and on help received
individually from instructors. Many of the problems assigned
are now worked out completely or in part in the text. They
thus cease to be available for assignment, but the ideas contained
in them have been extended to form new problems whose solutions
will be obtained only after study of the problems solved.
These new problems have generally remained unsolved, in the
past, owing to lack of available time. It is believed that they
may now be carried through with fair completeness and, indeed,
that many other suggestions coming from them may be followed.
It is the belief of the authors that no book on Electrical Engi-
neering can now be produced which does not bear testimony to the
pioneer work of such writers as Fleming, Silvanus Thompson,
Bedell and Crehore, Steinmetz and McAllister.
In addition, the authors desire to acknowledge their indebted-
ness to Dr. A. S. McAllister who has critically gone over the
manuscript, to Mr. N. S. Diamant for suggestions relating to
material contained in the earlier portions of the book, and to Mr.
E. S. Lee for assistance in reading the proof.
CONTENTS
PAGE
PREFACE v
CHAPTER I
UNITS - 1
Development of Ohm's Law Two Resistances in Parallel
General Solution of a Network by KIRCHOFF'S Laws Effects of
Current in a Wire Power.
CHAPTER II
FORM OF WORK 9
Object of Problems Summary.
CHAPTER III
*
MAGNETISM 14
Cylindrical Poles Flat Poles Magnets as Commonly Used in
Meters Energy Density in a Field Limits of Pole Intensity
The Magnetic Cycle Permeability.
CHAPTER IV
PRINCIPLE OF THE ELECTRIC MOTOR 20
Determinations of Magnetic Intensity Magnetic Intensity at Any
Point along the Axis of a Coil Magnetic Intensity in the Center
of a Long Coil Application of Magnetic Formulae to Instruments.
CHAPTER V
DESIGN OF A LIFTING MAGNET 26
CHAPTER VI
GENERATION OF ELECTROMOTIVE FORCE IN A DYNAMO 29
E.m.f. Waves in Fields that are not Uniform E.m.f. Wave when
the Coil is Wound on an Iron Core Additional Problems for
the Determination of E.m.f. Waves.
CHAPTER VII
INDUCTANCE 33
The Rate of Energy Supply or Power Equation Starting and
Stopping Current in an Inductive Circuit.
vii
viii CONTENTS
CHAPTER VIII
PAGE
ALTERNATING CURRENTS 38
Average Value of a Sine Wave Effective Value of a Sine Wave.
CHAPTER IX
DIRECT-CURRENT GENERATORS 44
Homopolar Generators Direct-current Machines with Commu-
tators Types of Direct-current Commutator Machines Arma-
ture Reaction Characteristics of Direct-current Generators
Numerical Application.
CHAPTER X
A STUDY OF THE DESIGN OF A DIRECT-CURRENT GENERATOR .... 55
Flux Calculation The Magnetic Circuit Area of Flux Path
through Teeth Area of Flux Path through Gap Areas of
Armature Core, Pole Core and Yoke Materials No-load and
Full-load Saturation Curves Armature Reaction The Shunt
Field Winding The Series Field Winding Armature Resistance
Brush Resistance Series Field Resistance Commutator and
Brushes Flux Distribution around the Armature Losses and
Efficiency Copper Losses Core Loss Summary of Losses,
Output and Efficiency Temperature Rise.
CHAPTER XI
ELECTRICAL CONSTANTS OF A DIRECT-CURRENT GENERATOR HAVING
COMMUTATING POLES AND COMPENSATING WINDING 74
Commutation.
CHAPTER XII
DIRECT-CURRENT GENERATORS IN PARALLEL AND SERIES 88
Direct-current Generators in Series The Three-wire System
Boosters.
CHAPTER XIII
DIRECT-CURRENT MOTORS 97
Types of Direct-current Motors Speed Characteristics of Direct-
current Motors Power and Torque Torque Characteristics.
CHAPTER XIV
THEORY OF THE BALLISTIC GALVANOMETER 102
CHAPTER XV
VECTOR REPRESENTATION OF ALTERNATING-CURRENT WAVES . . . 105
Use of the Symbol j Circuit of Resistance in Series with an
Inductive Impedance Impedances in Parallel.
./ '1
CONTENTS ix
CHAPTER XVI
PAGE
THE SYMBOLIC METHOD IN TRANSMISSION LINE CALCULATION ... 113
Addition Multiplication Power Average Value of Power dur-
ing a Period Power Factor Transmission Line Calculation
Power of Generator.
CHAPTER XVII
CONSTANT POTENTIAL CONSTANT CURRENT TRANSFORMATION . . . 120
CHAPTER XVIII
CAPACITY AND CAPACITY REACTANCE 123
Condenser Expression of Condensive Impedance Circuit Con-
taining Resistance, Inductance and Capacity in Series Resonance.
CHAPTER XIX
PARALLEL CIRCUITS 129
Transmission Line Supplying Power to Parallel Loads Approxi-
mate Transmission Line Calculation.
*
CHAPTER XX
DISTORTED WAVES RESONANCE EFFECTS 133
E.m.f. Which Causes Distorted Current waves.
CHAPTER XXI
CONSTANT POTENTIAL CONSTANT CURRENT TRANSFORMATION (Con-
tinued from Chapter XVII) 140
Power and Wattless Components of Volt-amperes.
CHAPTER XXII
THEORY AND USE OF THE WATTMETER 146
Wattmeter Connections.
CHAPTER XXIII
SIMPLE PROBLEMS IN ELECTRO-STATICS 152
Potential Intensity Capacity of a Sphere Potential Gradient
Capacity of a Spherical Concentric Condenser The Capacity of
a Concentric Cylinder Capacity of Two Parallel Plates so Large
that the Effects of their Edges may be Neglected Capacity of
a Transmission Line Capacity of a Three-phase Cable Induc-
tance of a Concentric Cable Inductance of a Transmission Line.
CHAPTER XXIV
DISTRIBUTED INDUCTANCE AND CAPACITY .... 162
x CONTENTS
CHAPTER XXV
PAGE
NOTES ON THE MATHEMATICS OF COMPLEX QUANTITIES 169
Representation of Complex Quantities Addition of Two Complex
Quantities Multiplication of Two Complex Quantities Division
of Two Complex Quantities Multiplication Involution and Evo-
lution The Roots of a Complex Quantity Exponential Repre-
sentation of Complex Quantities Differentiation of a Complex
Number or Vector Logarithm of a Complex Number or Vector.
CHAPTER XXVI
THE TRANSFORMER ; 174
The Transformer Diagram Equivalent Transformer Circuit
Example of Transformer Calculation Approximate Method of
Determining the Regulation, Efficiency and Power Factor of
Transformers.
CHAPTER XXVII
HYSTERESIS AND EDDY-CURRENT LOSSES 186
Hysteresis Loss Eddy-current Loss.
CHAPTER XXVIII
WAVE DISTORTION IN TRANSFORMERS 189
Effect of Hysteresis Dependence of Core Loss on the Shape of the
E.m.f. Wave.
CHAPTER XXIX
DISTORTED WAVES 196
Application of FOURIER'S Theorem to Wave Analysis.
CHAPTER XXX
MECHANICAL STRESSES IN TRANSFORMERS . 202
Determination of the Leakage Inductance of the Primary and
Secondary Calculation of Stresses.
CHAPTER XXXI
GENERAL PRINCIPLES OF TRANSFORMER DESIGN 209
Type Efficiency Losses B and V Hysteresis and Eddy-
current Loss per Cubic Inch Magnetizing Current Number of
Turns, Total Flux, Area and Length of Magnetic Circuit Resist-
ance, Length of Mean Turn, Total Length and Size of Windings
Per Cent. Magnetizing Current and Core Loss Efficiency
Regulation Heating Weight and Cost of Material.
CONTENTS xi
CHAPTER XXXII
PAGE
COMBINATIONS IN MULTIPHASE TRANSFORMER SYSTEMS ...... 226
The Three-phase System Voltage Waves in Three-phase, Four-
wire System Three-phase, Y-connected Transformers Three-
phase A-connected Transformers Voltage Waves with Y-con-
nected Transformers Three-phase Transformers Shell-type
Three-phase Transformers Open Delta Transformer Connection
T-connection of Transformers Rating of T-connected Trans-
formers Two-phase, Three-phase Transformation rAuto-trans-
formers Compensators for Two-phase, Three-phase Transfor-
mation Dissimilar Transformers in Series Dissimilar Trans-
formers in Parallel Three-phase Connection for Dissimilar
Transformers.
CHAPTER XXXIII
ALTERNATORS 246
One-, Two- and Three-phase Connection Voltage to Neutral
Rating of Alternators.
CHAPTER XXIV
ARMATURE REACTION 252
CHAPTER XXXV
CHARACTERISTICS OF ALTERNATORS WITH DEFINITE POLES .... 258
CHAPTER XXXVI
APPROXIMATE DETERMINATION OF THE SELF-INDUCTION OR LOCAL
MAGNETIC LEAKAGE REACTANCE OF AN ALTERNATOR 270
CHAPTER XXXVII
ARMATURE REACTION IN MULTIPHASE MACHINES 278
Effect of Distributed Winding on the Armature Reaction.
CHAPTER XXXVIII
HUNTING 283
CHAPTER XXXIX
STUDY OF THE DESIGN CONSTANTS OF ALTERNATORS 289
General Constants Slot Dimensions Flux Determination Air
Gap Teeth Flux Density Armature Length Armature Resist-
ance Magnetic Circuit Dimensions The Main Field Magneto-
motive Force Losses and Efficiency Temperature Rise
Regulation.
xii CONTENTS
CHAPTER XL
PAGE
SHORT-CIRCUIT OF ALTERNATORS 301
Stresses on End-connections of the Armature Coils Multiphase
Short-circuits Armature Reaction Electromotive Force and
Current Induced in the Field Windings.
CHAPTER XLI
SYNCHRONOUS MOTORS 324
Synchronous Motor Equations.
CHAPTER XLII
INDUCTION MOTORS . . . ; 342
The Rotary Field Theory of Operation Case 1. Armature at
Standstill Case 2. Armature at about Half-speed Motor As-
sumed Y-connected Motor Assumed A-connected Motor and
Transmission Line Motor with Auto-transformer.
CHAPTER XLIII
STUDY OF THE DESIGN CONSTANTS OF AN INDUCTION MOTOR . . . 365
Air Gap Rotor Diameter Stator Slots per Pole Slot and Tooth
Dimensions Main Flux Stator Length Rotor Slots Slot
and Tooth Dimensions Rotor Secondary Resistance Leakage
Reactance.
CHAPTER XLIV
ROTARY OR SYNCHRONOUS CONVERTERS 383
Voltage and Current Ratios Voltage Control Heating of the
Armature Voltage Control Numerical Application Voltage
Control by Use of the "Split Pole" Transformer Connections for
Rotary Converters Synchronous Condensers.
CHAPTER XLV
SINGLE-PHASE ALTERNATING-CURRENT MOTORS 400
The Series Motor Compensated Series Motor Repulsion Motor.
.411
ELECTRICAL ENGINEERING
CHAPTER I
UNITS
As it is assumed that the student has had an elementary
course in Physics, it seems feasible to omit herein the definition
of the fundamental mechanical and electrical units. However,
before taking up the electrical engineering problems, it is essential
that a review be made of the chapters in physics relating to these
units.
The student should be able to present, not only by means of
equations, but in words for this is far more important the
relations between force, work, energy*, power, torque, etc.
In regard to electrical units it is assumed that he is already
familiar with such terms as " current" and "electromotive force"
and appreciates that . . .
Current is analogous to water flowing. The absolute unit of
current is the abampere. The practical unit is the ampere. One
abampere is 10 amperes.
Quantity, likewise, is analogous to water at rest. The practical
unit of quantity is the coulomb, which is the amount of electricity
involved when 1 amp. flows throughout 1 sec., or, 1 amp. sec.
Difference of potential is analogous to pressure-difference and is
the electromotive force which causes current to flow in a circuit.
The absolute unit of potential-difference is the abvolt. The
practical unit is the volt, which is 10 8 abvolts.
Resistance is that property of the material of a circuit which im-
pedes the flow of electricity. The absolute unit of resistance is
the abohm. The practical unit is the ohm, which is 10 9 abohms.
Resistance depends on material and temperature. With constant
temperature,
-
2 ELECTRICAL ENGINEERING
where
R = resistance of a given conductor,
p = specific resistance or resistivity of the material,
Z = length, and A = area of cross-section, of the conductor.
Specific resistance or resistivity, is the resistance of a unit cube of
"any material taken between opposite faces. !
In practice it is sometimes convenient to use the resistance of a
wire 1 ft. long and 0.001 in. in diameter as the unit of resistivity.
This unit is called the circular-mil-foot.
In problems involving resistance, it is frequently convenient to
use the reciprocal of resistance, known as the conductance.
G = o> where G is the conductance of a circuit of resistance R.
K
Likewise the reciprocal of resistivity, called conductivity, is often
used.
The resistance of a wire at any temperature t, when its resistance
at any other temperature is known can be calculated by the
following equation
R t = Rt.ll + a tl (t - t,)}
When ti = 0C. then R t = R (l + aj)
where R t is the required resistance at any temperature, t, R in
this case is the resistance at 0, and a is a constant, called the
temperature coefficient.
For copper, a = 0.004 (approximately) when t is given in
Centigrade degrees.
At any other temperature the value of a is:
1
234.5 + t
where t is the temperature in degrees C.
Since a depends upon the temperature, in all calculations in-
volving a its value is calculated for that temperature at which the
resistance is known.
Knowing the resistance Ri at a temperature ti the resistance
Rz at temperature tz is thus accurately determined from the
following relation:
234.5 + t, 234.5 +
UNITS 3
TABLE I
Table I gives approximately the temperature coefficients and resistivities
in ohms per centimeter cube of some of the more common electrical con-
ductors at ordinary temperature.
Conductor
Temp, coefficient a
Resistivity
Aluminium
0.0042
2.9 X 10~ 6
Carbon
00052
720 X 10~
Copper
0.004
1.6 X 10~ 6
German silver
0.00027
20.9 X 10~ 6
Iron
0.0046
9 7 X 10~ 6
Nickel
0.0062
12.4 X 10~
Platinum . . .
0.0036
9 X 10-
Silver .
0.004
1.5 X 10~ 6
Tungsten
0.005
5 X 10-
Development of Ohm's Law. According to OHM'S law the
current in a circuit at any instant is equal to the potential differ-
ence divided by the resistance, or,
7 = R'
Obviously, where a number of resistances are in series, the
total resistance is the sum of the individual resistances, or,
Rtotal ~ 2r = TI + TZ + 7*3 -f- . . .
Two Resistances in Parallel. To find the total current 7, and
the currents 7i, Iz in the resistances r\ and r 2 , when a potential
difference E is applied (Fig. 1). /
By OHM'S law,
Tjl Tjl
r Hi j
and
FIG. 1.
E E
To find a single resistance, r , which shall be the equivalent of
ri and r 2 in parallel, evidently
'
4 ELECTRICAL ENGINEERING
Whence,
r a =
r 2
Having two resistances in parallel, in series with a third resist-
ance (Fig. 2), to find the combined resistance. Let the combined
resistance of r*i and r 2 be r . Then r = -
T\ -J- 7*2
The condition is, then, that of two re-
i : Sr i r sistances r and r 3 in series and the total
j>*\ r >i 3
n resistance R = r + r 3 .
+ J
Hence
r
* / - - -
-L """ yv
FIG. 2. 72 r + r 3
To find /i and /2.
It is evident that / = 7 3 .
Knowing 7 3 and r 3 , we may at once determine E$ which is the
potential difference, or drop, across r 3 . Thus, by OHM'S law,
E z = / 3 r 3 .
It is evident that the potential difference E , across r\ and T* is
E E$.
' T _ E . T - ^
- Tt> /2 ~ r 2
General Solution of a Network by Kirchoff's Laws. In cir-
cuits or networks of a more complicated nature in which the
resistances and electromotive forces are known, the currents in
the various branches may be calculated by the application of
KIRCHOFF'S laws which may be stated as follows:
Law I. The algebraic sum of all the currents flowing toward
a branch point is equal to zero. _
Law II. The algebraic sum of all the
e.m.fs. acting around a closed circuit is
equal to the sum of the products, ri,
around the mesh. Or the impressed
e.m.f. is equal to the sum of all e.m.fs.
consumed by the resistances.
For example, let the circuit be as
shown in Fig. 3 where arrows represent arbitrarily chosen direc-
tions of current. For the points A, B, C, D, applying Law I,
equations may be written:
UNITS
A.
B.
C.
D.
- i -
is - i* - t' 4 =
ii + 12 - i s =
i 4 + *B - i = 0.
(1)
(2)
(3)
(4)
Applying Law II, where the short arrow represents the direc-
tion of the e.m.f., to the meshes (a) e, r 3 , r 4 , (b) e, ri, r 5 , (c) n f r,
r 3 , (d) r 2 , r 5 , r 4 , always keeping an arbitrarily chosen counter-
clockwise direction, we have,
(a) ri + r 3 i 8 + r 4 i 4 = e (5)
(b) ri + nil + r b i b = e (6)
(c) nil - r,i 2 - r 3 i 3 = (7)
(d) r 2 iz + r 6 i 8 - r 4 i 4 = (8)
There is one extra equation in each group as there are only
six unknown quantities, i, ii, i 2 , is, i 4 , is.
In calculating the resistance of more or less complex circuits
it is helpful to remember that current does not flow between
points of the same potential.
If, in Fig. 3, there is no difference of potential between points
B and C there will be no current in the branch r 2 .
PROBLEMS
Problem 1. If the resistivity (resistance of a cubic centimeter between
parallel faces at 0C.) of copper is 1.6 X 10~ 6 ohm, (a) show that the resist-
ance of an inch cube of copper is 0.63 X 10~ 6 ohm; (b) show that if the
temperature coefficient, a. = 0.004, the resistance of a centimeter cube at
20C. is 1.73 X 10~ 6 ohm; (c) show that __ B
the temperature coefficient per degree
Fahrenheit is 0.0022.
_
FIG. 4.
FIG. 5.
Problem 2. If a wire be connected across the terminals of a source of
constant e.m.f., a current will flow. Will this current increase, decrease, or
remain constant as time goes on, and why?
Problem 3. Deduce the equation for the equivalent resistance of three
resistances connected in parallel.
Problem 4. Find the line current 7, and the voltage across r 3 in the
circuit, shown in Fig. 4. E = 100 volts, r : = 1, r 2 = 2, r 3 = 3.
Problem 5. Let the outline of a cube, Fig. 5, consist of resistances, each
6 ELECTRICAL ENGINEERING
edge being 1 ohm. Prove that the total resistance between A and B is
^f 2 ohm; between A and C is % ohm; between A and D is % ohm.
Effects of Current in a Wire. When a current is set up in a
wire three effects may be noted, namely: (1) the wire gets warm,
(2) a compass needle placed near the wire is deflected, and (3)
when the voltage is high enough bits of paper may be attracted.
The amount of energy delivered through the wire does not bear
a relation to any one of these effects, but if the second and third
effects are multiplied together, or, as commonly expressed, if
the strength of the magnetic and electric fields are multiplied to-
gether the product is a value which is proportional to the amount
of energy transmitted through the wire per second, or to the
power. Thus we may write,
P = kei
where P is the power and k is a constant, k is unity when e,
which is proportional to the strength of the electric field is ex-
pressed in volts, i t which is proportional to the strength of the
magnetic field is expressed in amperes, and P is in watts.
The first effect, that is, the production of heat is due to con-
sumption of energy in the wire due to its resistance. The second
effect is due to the setting up of a magnetic field about the wire
by the current. The third effect is due to the setting up of an
electric or electro-static field in the region about the wire by the
difference of potential between the wire and other points in space.
Power. In a given circuit, then,
P = El = IE X / = PR
in which E is the total e.m.f., I the current, and R the total re-
sistance of the circuit.
This relation, known as JOULE'S law, is very important, as it
shows that the power is proportional to the square of the current
strength and to the first power of the resistance.
The heat developed by this power depends upon the duration of
the current, and is expressed in joules. Thus, heat energy =
Elt = I 2 Rt joules, where E is in volts, 7 in amperes, and t in
seconds (the current and voltage being assumed constant during
time t).
r*
In general, the energy converted to heat is W = I i 2 rdt.
Jti
Problem 6. Prove that if the current is represented by equation
i I sin ut
UNITS 7
T
the energy per cycle is W = 7 2 r -^ where T is the time of a complete cycle.
4
W Pr
The average power is then -7=- = -=-
j. z
(/I 2 /3 2 \
~o~ + ~o~) r when
i = /i sin co + / 3 sin (3J + a).
Heat Units. The practical heat units most frequently dealt
with are the British thermal unit (B.t.u.), and the large and small
calories (C. and c.).
One B.t.u. is the energy required to raise the temperature of 1
Ib. of water 1F.
1 B.t.u. = 1.055 kw. sec.
One large calorie is the energy required to raise the temperature
of 1 kg. of water 1C.
1 C. = 4.2 kw. sec.
One small calorie is the energy required to raise the temperature
of 1 gram of water 1C.
1 c. = 0.0042 kw. sec.
Problem 8. A 16-cp. lamp which consuntes 3 watts per cp. is immersed
in a quart of water at 20C. Assuming no loss of heat, (a) what will the
temperature of the water be after 2 min. ? (6) How long would it take to
evaporate the water?
Solution. (a) Temp, will be 20 + C. rise.
C.rise = ^ W ^ Xqt.inlkg.
2
kw. sec. = X 16 X 2 X 60 = 5.76
qt. per kg. = 1.057
f\ 7 A
.'. C. rise = -~- X 1.057 = 1.45.
Temp, after 2 min. = 21.45C.
(6) Time to evaporate = time to raise .to boiling + time required to
furnish latent heat of vaporization.
Time required to boil 1 qt. = time to raise 1 qt. 1 X (100 - 20)
= ~ X 80 = 110.3 min.
Time required to evaporate = calories required to evaporate -* calories per
min. supplied by lamp.
= 742 min.
.'. Total time required = 110.3 + 742 = 852.3 min. = 14 hr. 12 min.
8 ELECTRICAL ENGINEERING
Problem 9. Transform problem 8 into F. and B.t.u.
Problem 10. If electric energy costs lOc. per kw. hr., how much would
it cos,t to prepare a hot bath by electric means, if the bath required 50 gal.
of water raised in temperature by 50F. ?
Solution. Cost = kw. hr., X $0.10
_ kw. sec. _ kw. sec, to raise 1 gal. 1 X 50 X 50
3600 3600
1 gal. weighs approx. 8.4 Ib.
/. kw. sec. to raise 1 gal. 1 = 8.4 X 1.055 = 8.86
8.86 X 2500
' ' kw ' hr ' = 3600 " = 6 ' 15
Cost = 6.15 X 0.10 = $0.615.
Problem, 11. Four car heaters each take 4 amp. at 125 volts. Find the
cost per 10-hr, day at lOc per kw. hr., to operate them on a 500-volt circuit,
(a) when they are connected in series, (6) when they are connected in parallel.
Answer. In series, $2.00; in parallel, $32.00.
Problem 12. If the car contains 3000 cu. ft. and is insulated against loss
of heat, how much time is required for a rise in temperature of 20C. when
the heaters of problem 11 are connected (a) in series, (6) in parallel?
Answer. In series, 12 min. 44 sec.; in parallel, 48 sec.
NOTE. Specific heat of air at constant volume = 0.167.
CHAPTER II
FORM OF WORK
In order that students may gain the greatest possible advantage
from pursuing the course of study, it has been thought best to
include in the body of the book, at this point, a brief statement
of the procedure which the student should adopt in the working
out of the problems. He is urged to familiarize himself with the
method, and to follow it rigidly until, in so doing, he has thor-
oughly acquired the habit of careful and accurate work.
Object of Problems. Problems are almost universally con-
sidered to be indispensable in any engineering course. Their
function is similar in many respects to that of laboratory experi-
ments. They illustrate the theory. In this respect problems
may be divided into two groups, namely:
(a) Those in which the general equation is applied to a definite
concrete case, and
(b) Those in which the general equation is investigated for the
purpose of finding out the whole range of definite value which
may be obtained from one variable by assigning definite values
to one or more other variables.
As an illustration of the first group, we will take the following
example :
Problem 13. Ten arc lamps, in series, are used to light a certain building.
They require 6.6 amp., and the potential-difference (drop) across each lamp
is 80 volts. Current is supplied from a power house 2000 ft. distant, by
means of No. 6 B. & S. wire. If the energy is measured at the power house,
find the cost at lOc. per kw. hr. to light the lamps 8 hr. per day.
Solution. Cost per day = power X hr. X $0.10
Power,
P = PL + Pw
where
PL power required by lamps = nEI
where
n = number of lamps
and
PW power lost in the wire
Pw = I*R
9
10 ELECTRICAL ENGINEERING
where
R = total resistance of wire
R = resistance per 1000 ft. X 1 -j~
resistance per 1000 ft. of No. 6 wire = 0.4 ohms at 75F.
Then
R = 0.4 X ~ = 1.6 ohms.
PW = 6.6 2 X 1.6 = 70 watts
PL = 10 X 80 X 6.6 = 5280 watts
p = p L x P W = 5350 watts = 5.35 kw.
Cost = 5.35 X 8 X 0.10 = $4.30. Ans.
Such problems are typical of existing conditions. An engineer
continually meets them where he is trying to find what results are
being obtained from a given installation. In solving them,
accuracy is the prime consideration, and this is obtained by avoid-
ing short cuts and following through, step by step, a logical de-
velopment. These problems are of far less importance and inter-
est to the engineering student than problems of the second group.
+ 2000 r
r H
\ 1
^VAAWWNAA^-
t J>
1 1
250 ^50 Kw.
%'/////'//ti
FIG. 6.
As an illustration of these take the following example:
Problem 14. A load of 50 kw. at 250 volts is to be supplied by a power
house distant 2000 ft. from the load. If the line costs 20c. per Ib. of copper
laid, find and plot (a) efficiency of transmission against size of wire; (6) cost
of copper against size of wire; (c) efficiency of transmission against cost of
copper.
load 50,000
load + line loss = 50 , 00 + /*
50,000
. . Efficiency = 50)000
X resistance per 1000 ft. = 4 X r
X wt. per 1000 ft. = 4w.
FORM OF WORK
11
Tabulation :
Wire No. (B. & S.)
0000
00
1
4
8
12
16
r per 1000ft. 1
R = 4r
0.049
196
0.078
312
0.125
0.50
0.25
1.0
0.64
2.56
1.60
6.4
4.0
16.0
Wt. per 1000 ft = w
wt. = 4.... '...;.
Cost at $0.20
641
2,564
512.8
403
1,612
322 .4
253
1,012
202.4
126
504
100.8
50
200
40
20
80
16
7.9
31.6
6.32
40 00072 = 7 2 #
7,840
12,480
20,000
40,000
102,400
256,000
640,000
50,000 + 40,000/2 . .
Efficiency
57,840
0.865
62,480
0.8
70,000
0.715
90,000
0.55
152,400
0.33
306,000
0.16
690,000
0.072
The curves are plotted in Fig. 7.
600 100 1
90
400 80
70
300 |60
~ 1
i IB 50
J.I
200^40
30
100 20
10
OC
<
v
^
.-C09]
1
^.
\
\
^
-^
><
X
V
7
\
\
A
I
\
%!
\
/
\
^.
\
/
N
\
X
\
[
^
"^^.
*"
^
1 >
00 00 1 3 5 7 9 .11 13 15 17
Size of Wire
) 100 200 300 400 500
Cost. $
FIG. 7.
Summary. The curves show (1) efficiency of transmission
decreases as wire becomes smaller, at first slowly, then rapidly,
and then, for very small wires, slowly again; (2) the cost of wire
decreases as wire becomes smaller, at first very rapidly, then more
and more slowly; (3) efficiency increases with cost, rapidly at
first, for low efficiencies and costs, then more and more slowly.
1 From wire tables.
12 ELECTRICAL ENGINEERING
It is evident that this problem could be greatly extended so as
to include other variables, such as current density in the wire,
cost of lost energy, etc., and indeed it is characteristic of this
type of problem that there are always suggestive lines of investi-
gation which tend to stimulate the student's interest.
The work of solving the problem may be divided into a number
of parts, thus: (1) statement of problem, (2) diagram of circuit,
(3) analytical work, (4) tabulation of values, (5) plotting of
curves, (6) summary, or statement in words, of the results
obtained.
The statement of the problem should be concise. The dia-
gram should be an illustration of the statement, and should con-
tain the symbols to be used.
The analytical work should be carried out as far as possible
with symbols before the numerical values are substituted. In
the above example there is very little opportunity for the use of
symbols, owing to the shortness of the problem. In later prob-
lems this feature will be more apparent.
Tabulation should be arranged with care, and should be planned
so that columns can be conveniently added. As a rule, it is well
to assign along the horizontal various values of the independent
variable, and proceed, step by step, to the dependent variable.
As in this case, there may be different combinations of variables,
as number of wire, cost, and efficiency. This makes the tabula-
tion more complex, as it would be by any other procedure, but it
is still entirely clear. The plotting of curves is then carried out,
and this should be done neatly and preferably in ink.
The problem should then be completed with a brief statement
of the results obtained. It is not always easy to make students
take this last step, but they should be required to do so, and to
follow this general plan throughout, until they have formed the
habit of doing it and need no further compulsion.
There may be other ways of working these problems efficiently,
but it seems justifiable to urge teachers and students to adopt this
method in preference to any other to which they are accustomed.
It will insure uniformity and logical arrangement, will make cor-
recting easy, and will commend itself to the student as well as
the teacher.
In working problems of this nature there are other objects than
merely to illustrate and enforce the theory.
Great stress is laid on them, not only for the engineering knowl-
FORM OF WORK 13
edge which they contain, but because of their structure, which, it
is believed, strongly tends to develop those qualities most essen-
tial in an engineer. For instance, the mathematical develop-
ment calls for insight and understanding, the tabulation calls
for concentration of mind, the summation of results calls for
accuracy, and a study of the plotted curves calls for judgment.
At the same time, efficiency, the keynote of the engineer, would
be lacking if the problems were not done in the shortest and best
way consistent with obtaining the desired results, and it is obvious
that many hours will be wasted, both to student and instructor,
unless the work is done with order, accuracy and neatness.
CHAPTER III
MAGNETISM
FARADAY explained magnetic phenomena by assuming that
surrounding a magnet or a wire carrying current were lines of
force.
The stronger the magnet or current, the stronger is the magnetic
field, that is, the more lines of force per square centimeter.
The introduction, then, of a magnet into a space means the
establishing of a field of force.
To get quantitative ideas about field strength he made use of
the symbol H which was called the intensity of the field, or the
force on unit pole placed in the field. It seems an unfortunate
term since intensity and density are readily confused.
B, the density of the field, or the number of lines of force per
square centimeter, is proportional to H, and also to a quantity n,
the permeability or magnetic conductivity of the medium in
which the intensity, H, exists. Thus
B = fj.H.
= 1, it follows that the number of lines of
force per square centimeter is
numerically the same as the
intensity of the magnetic
field H.
H, the force per unit pole,
is expressed in dynes.
The force exerted on a pole
not of unit strength, but of
Jr IG. o. .
strength m, is
F = mH dynes,
where, of course, H is caused by other poles than m.
Consider, now, an isolated elementary pole of strength m, from
which n lines of force, per unit pole, protrude radially and uni-
formly in all directions (Fig. 8).
14
MAGNETISM
15
At a distance r from ra, no matter what the medium is provided
it is uniform, the density of the field is B T %, since the area
of a sphere of radius r is 47rr 2 . The force, H, on unit pole is then
_, , , . jr,
Thus the force on pole mi is r
B
nm
COULOMB, working in air, found experimentally that the force
between two poles of strength m and mi could be expressed by
p = k y-, and he would have found F = k ^ had he experi-
mented in a medium of permeability, /*.
Therefore, k may be written unity if n = 4ir. In other words,
if it is assumed, as is the case, that 4?r lines protrude from unit pole.
GAUSS came to the same conclusion from another point of view,
and the relation <j> = 4irm is called GAUSS'S theorem.
In words, GAUSS'S theorem states that from a pole of strength
m radiate outward 47rm lines of force, or the total outward flux,
$, from pole m is 4irm lines. 1
Cylindrical Poles. To find the intensity of the magnetic field
H at a point distant r from a uniform cylindrical pole of strength
m (Fig. 9). By GAUSS'S theorem the, flux <f> = 47rm, and # =
The area of a cylinder of radius r and length I is
, . _ 2m
at p, is
_ flux
area*
2irrl. Thus
2m ,
= -> and
\
1
t
I
FIG. 9.
FIG. 10.
Flat Poles. To find the intensity of the field, H , at a point dis-
tant d from one side of a flat pole of strength m (Fig. 10). As-
sume that the lines of force are perpendicular to the surface. Let
B = flux density at any distance. Then the flux coming from
one of the surfaces of the magnet is < = ^ = 2irm.
4
If the area of the pole face is S. then B = -> and H = FT-
S MO
1 For a more complete discussion of GAUSS'S theorem see "Advanced Course
in Electrical Engineering."
16 ELECTRICAL ENGINEERING
Magnets as Commonly Used in Meters. To find the magnetic
intensity between poles (Fig. 11). Let S be the area of a pole
face and d the distance between poles.
The density in the gap between the two pole faces is due to the
magnetic north pole, N, as well as to the south
pole, S.
If the lines of force flow outward from the
north pole, they flow inward from the south
pole. Thus a simple examination will show
that the fluxes add in the gap and cancel each
other in the outside region.
The total flux from N is 4irm and one-half of
this flux is assumed to be in the gap, the other half extending
outward.
The density in the gap due to N will then be
_ 4?rm _ 2-rrm
^ n " : ~2S~ ~S~'
Similarly, due to S,
and = # = 4?rm
M Bfj.
In all practical problems where magnets act in air only, /* is,
of course, unity.
Consider, now, the pull between the faces of a magnet as shown
in Fig. 11.
The flux density at the south pole due to the flux from the north
. . n 2irm 2irm , . ,
pole is B n = g . . H = -g- = force on unit pole at the sur-
face of the south pole.
Since the south pole has a strength m, the force on it is
therefore
' do)
Usually the density, B, in the gap is known.
Substituting the value of m from (9) into (10) gives
27r
~
4V
or the force in dynes per sq. cm. is F = ^
MAGNETISM 17
In air, where ju = 1,
B 2
F = - dynes per sq. cm. (11)
In Ib. per sq. in., the formula becomes
It is seen that if the pole strength, m, remains the same while the
faces of the magnet approach each other, the density, and thus
the force, is constant.
The work done is then Fd, where d is the distance between the
TT/ B * Sd
poles, or W = ~
Energy Density in a Field. The volume of space through
which the body is moved is Sd. The energy density, or joules
per cu. cm. of space between poles, is then :
1 B 2 B z
* 3 = S ergs = 8000* J0ules '
The conception of energy density is merely mentioned at this
point. Similarity of magnetic and eledtric fields will be shown
later on together with the development of theory and problems
in electro-statics.
Limits of Pole Intensity. In practice it is found that the limits
to which pole intensity -~- can be pushed are as given in the fol-
lowing table:
TABLE II. APPROXIMATE LIMITING VALUES OF
o
For wrought iron magnets, 1600 units of pole strength per cm.
For soft steel magnets, 1600 units of pole strength per cm.
For cobalt magnets, 1300 units of pole strength per cm.
For nickel magnets, 500 units of pole strength per cm.
For permanent steel magnets, 800 units of pole strength per cm.
The Magnetic Cycle. According to the molecular theory of
magnetism, magnetic bodies are composed of minute magnets
which attract and repel each other, and which are partly free
to turn under the influence of magnetizing forces. When strongly
magnetized, these molecular magnets are pointed in the direction
of the magnetic force. When the force is removed, they still
tend to point in the same direction, and thus the body exhibits
magnetization, which is called residual magnetism.
18 ELECTRICAL ENGINEERING
The magnetic state of a body is shown with reference to the
"magnetizing force" by a curve called the hysteresis loop (Fig.
12).
Magnetization of an iron bar is ordinarily accomplished by
sending current through a number of turns of wire wound around
the bar. The magnetization is thus produced by the ampere-
turns (A.T.). The number of lines of flux set up per unit area
enclosed by the turns will with a long bar be shown to be
r / ' M = B, where ju is the permeability of the bar and I is
its length. Since in air /-t = 1 and H = B, it follows that the
intensity of the magnetic field in a solenoid is:
QAirA.T.
The hysteresis loop is drawn with flux density, B (in lines per
square centimeter or per square inch), as ordinates and the mag-
netic field intensity, H (or frequently, for convenience, ampere-
TT7\
turns per inch length of magnetic circuit, j J , as abscissae.
The construction of the loop is as follows: Imagine a bar of
iron wound with many turns of insulated wire. If the iron has
no residual magnetism at the beginning, be-
fore current is sent through the wire, there
will be no magnetizing force and no flux, and
consequently the first or starting point on the
curve will be at a (Fig. 12). As more and
more current is sent through the wire, that
is, as the magnetizing force is increased pro-
portionally to the current, the flux or induc-
tion density, B, is increased, not according
to a simple law, but in such a way as to give
the characteristic curve (1) from a to 6.
If the magnetomotive force (m.m.f.) expressed in ampere-
turns is now decreased, the curve (1) is not retraced, but B
follows curve (2) from b to c. At c, H = 0, while B continues
to have a value represented by the line ac. This value of B
corresponds to the residual magnetism of the iron.
If, now, the current be reversed, so that H is given negative
values, B continues to decrease from c to d. At the point d,
B = 0, while H has the negative value ad. This value of H is
MAGNETISM 19
called the "coercive force" of the magnet. It is the magnetizing
force necessary to reduce the remanent magnetism, ac, to zero.
As H is further increased, negatively, B follows the curve de.
At 6, which corresponds to b with positive H, the current is again
reduced, and B follows curve (3) to /, which gives the value, of,
of negative remanent magnetism corresponding to ac for H = 0.
Thus, the point a is not reached again, but as H is now given
increasing positive values, the curve goes through g to b, complet-
ing the loop.
In obtaining a single loop, the points do not usually come into
such close agreement, due primarily to the fact that there is
always some remanent magnetism at starting, which prevents
the curve from beginning exactly at a. But in the case of many
uniform reversals of H, as occurs in electrical machinery, the
loop is retraced uniformly so long as the limiting values of H re-
main constant.
It will be later shown, in connection with the study of hysteresis
losses, that the area enclosed by the loop is proportional to the
work done on the magnet per cycle.
T>
Permeability. The ratio -p is called the permeability, and is
a measure of ease with which lines of flux are set up in a given
material. Permeability is denoted by the symbol /*. Numer-
ically, B = H in air (or vacuum) since /* = 1. In the magnetic
metals, particularly iron, steel, nickel and cobalt, ju undergoes
wide variation in value, with different values of H .
For a more complete discussion of the subject of magnetism the student is
referred particularly to EWINQ'S "Magnetic Induction in Iron and Other
Metals."
CHAPTER IV
N
PRINCIPLE OF THE ELECTRIC MOTOR
A wire carrying a current was discovered by OERSTED to be
surrounded by a magnetic field, which is strongest near the wire.
A small needle, placed in the field (Fig. 13), is directed along the
lines of force, but there is practically no tendency for it to move
toward the wire as the forces of attraction exerted on its poles
are equal and opposite. A long needle,
however, tends to move toward the wire
as there is a component of force on each
pole in the direction of the wire.
A wire carrying current, placed in a
field perpendicular to the lines of force
(Fig. 14), causes the flux to be distorted,
and this tends to force the wire in such
a direction that the lines shall again
take up their normal position. This is
the principle of the electric motor.
The electric motor consists (Fig. 15) of a number of wires
wound on a drum, and so placed in a magnetic field that the
current is caused to flow downward (toward the plane of the
paper) on, say, all the wires adjacent to the north pole, 1 and up-
ward on all the wires adjacent to the south pole. The wires on
FIG. 13.
N
N
FIG. 14.
FIG. 15.
the left, then, tend to move downward, and those on the right
upward, and thus rotation is produced.
1 In the diagram a cross, <8>, is used to represent down-flowing current and
a dot, O, up-flowing current in accordance with notation in common use.
20
PRINCIPLE OF THE ELECTRIC MOTOR
21
The current which, when flowing in a wire 1 cm. long placed
at right angles to a field having a density of 1 line per sq. cm.,
gives a force of 1 dyne is called the abampere.
The force, in dynes, is then
F = IIB
where / is the current in abamperes, I the length of wire in centi-
meters, and B the flux density of the field in lines per square centi-
meter. The force is due to the interaction
of flux and current.
If, however, the lines are not at right
angles to the wire, B must be replaced by
its component which is at right angles to
the wire. If the angle is a (Fig. 16), then
the force is F = IIB sin a, where B sin a is the component of flux
at right angles to the wire.
Problem 16. A copper wire carrying 10 amp. is placed in a magnetic field
of 10,000 lines per sq. cm.
What is the force in pounds on each centimeter of the wire (a) if it lies
perpendicular to the direction of the magnetic field, (6) if it lies parallel to
the field, (c) if it makes an angle, a, with the direction of the field?
N
FIG. 16.
(a)
Solution. F = IIB sin a
F, per cm. = IB sin a.
Sin a = sin 90 = 1
/ = 10 amp. = lab amp.
B = 10,000
.*. F, per cm. = 10,000 dynes.
10,000 dynes = == 10.2 grams
10
0.02245 Ib.
453.6
(6) Sin a = sin =
.'. F, per cm. = 0.
(c) For any angle, a,
F, per cm. = 10,000 sin a dynes
= 0.02245 sin a Ib.
Determinations of Magnetic Intensity.
Magnetic intensity at the center of a coil
(annulus) . Let a magnet pole, m, be placed
at the center of a coil (Fig. 17).' It will
send out lines in all directions, some of
which will strike an element of the coil, dl,
22
ELECTRICAL ENGINEERING
at right angles. Thus a force, dF, will be generated in the direc-
tion of the axis, as indicated, and its value will be
dF = IBdl = / dl
snce
The total force on the coil will be
- n dl
2irrlm
This will be the force, due to m, with which the coil will tend to
move along its own axis. It is obviously also the force on m due
to the coil. Thus if a unit pole (m = 1) replaces the pole of
strength m the magnetic field intensity at the center of the coil
is found. It is:
a _*a*,*L
r 2 r
Magnetic Intensity at Any Point along
of a Coil. The force dF will act
angles to the line joining m and
h
m
~
*
F
ji---"
I*
10. 18.
dF = IBdl = Idl
d
dF has components, dF cos a and dF sin a where sin a =
and
a
sin
The component of force which tends to move the coil in the
direction of its axis is dF sin . Call this component dFi.
Then
and
For
Jp
m
.
/ I. sin 3 .adZ =
27rm I sin 3
m = i p l = H
sin 3 a
PRINCIPLE OF THE ELECTRIC MOTOR
23
since the component dF cos a is balanced around the coil and
thus exerts no force.
Magnetic Intensity in the Center of a Long Coil. The force
at m, due to an element of the coil, dx (Fig. 19), is dF =
2irml sin 3 a
, where / is the current in abamperes, in the ele-
ment dx.
FIG. 19.
If the current per centimeter length of the coil is / c , then
, ,._ 2irml c dx sin 3 a
I = I c dx, and dF =
x / 1 \
But = cot a. Differentiating, dx = r ( =-^ ) da.
Substituting this value of dx in the formula,
2irml c r sin 3 a ,
dF = ^ da
r sm 2 a
= 2irm7 c sin ada.
Co. - ai
.*. F = I 27rm/ c sin ada = 4irl,
./a =" TT ai
m cos
or H = 47r/ c cos ai.
For very long coils, cos ai = 1, and
H = 47T/ C .
The relation between H and the ampere-turns of a coil may be
found as follows:
Let there be a current of I abamp. in the coil, and let n =
number of turns. Then nl = abamp.-turns. Abamp.-turns
nl
per cm. = -r- = J c , where I = length of coil in centimeters.
47m/ 1
I
Then H = 4ir7 c
When the current is in amperes,
1 NOTE. It should be noted that the above value of H
holds only
for infinitely long solenoids since it was derived on that assumption. For
practical purposes, according to the accuracy required, this value of H may be
24
ELECTRICAL ENGINEERING
H
j , whence, amp.-turns = 0.
SHI
(12)
i
If I is in inches, amp.-turns = Q.313HL
When the coil has an air core, H is numerically equal to B, and
amp.-turns = Q.SBl or = 0.313J5Z. 1
FIG. 20.
FIG. 21.
Application of Magnetic Formulae to Instruments. Let a rec-
tangular coil of height, a, and width, 6, be suspended in a magnetic
field of uniform density, B (Fig. 20). The two sides, a, are per-
pendicular to the flux, and therefore, with a current of / abamp.,
there will be a force on each wire of F a = IBl = IB a dynes.
FIG. 22.
This force will produce a torque around the axis, on each wire, of
T = IBa ~ dyne-cm. The total torque per turn = 2T a = I Bab,
i
and if there are n turns, T = InBdb, dyne-cm.
In practical instruments, T should be about 1 gram-cm.
Let a circular coil of radius r, as in Fig. 21, be suspended in the
field. To find the torque on any element dl. The useful part
used whenever it is desired to find the magnetic field intensity along the axis
of a solenoid, and not very near the ends, provided the length of the coil is
about 50 times its diameter.
It is necessary to observe this (always depending on the accuracy desired)
on account of the disturbing effects of the ends, as can be easily seen by
comparing the figures. (Fig. 22.)
1 NOTE. It can be proven that the density in the middle of such long coil
is uniform, thus if A is the area inside of the solenoid the total flux is AB,
area
PRINCIPLE OF THE ELECTRIC MOTOR 25
of dl is its component perpendicular to the lines of flux, = dl sin 6.
Then, dF = InB sin Bdl. This force acts with a lever arm = r
sin 6, and the torque is therefore dT = InB sin 6 X r sin 0dl. But
dl = rd0. Hence dT = InBr* sin 2 0d0, and T = I 7nr 2 sin 2
Jo
[01 T 27r
- - 7 sin 20
Zl 4 J
= InBr*ir = InB A y where A = area of the loop.
In practice, permanent magnets are generally used to produce
the flux.
CHAPTER V
DESIGN OF A LIFTING MAGNET
It has been shown that for a path of magnetic lines in air,
the following relation obtains: amp.-turns = 0.313 Bl", if inch
measurements are used. For an iron path, the necessary ampere-
turns are obtained from a curve of B vs. AT, where B is the
flux density. Such a curve called either magnetization or " satu-
ration " curve, is obtained experimentally from a sample of any
desired magnetic material. The curve thus obtained will be
approximately correct for that material, but variations are always
100 120
Ampere-Turns per Inch
FIG. 23.
juo
160
180
liable to occur due to either physical or chemical influences by
which any portion of the material is made to differ from the
sample used to derive the curve.
By testing many samples a typical curve is obtained for
any given material. In Fig. 23 is given a set of these satura-
tion curves of iron and steel as commonly employed in electrical
machinery.
26
DESIGN OF A LIFTING MAGNET
27
Let it now be required to make the calculations for the design
of a cast-iron electromagnet to lift a weight of 1000 Ib. through a
gap of 1.5 in. Let it be assumed
that the magnet core is of the
shape and dimensions given in
Fig. 24. Then the area of a pole
face is 10 X' 5 = 50 sq. in. The I
two pole faces have an area of
2 X 50 = 100 sq. in. Then the
weight to be lifted per square inch
of area is
<M
1000 Ibs,
FIG. 24.
Wt. per sq. in.
1000
100
= 10 Ib. = F.
But, from (11), F =
72,134,000
.*. B, the flux density in air, = V721,340,000 = 26,800 lines
per sq. in. From (12), the amp.-turns required for each gap =
0.313 HI in. = 0.313 X 26,800 X 1.5 = 12,600.
AT required for two gaps = 2 X 12,600 = 25,200.
AT required for the iron, from the magnetization curve for
cast iron = 32 per in. length of the flux path in the iron.
Length of mean path in the iron = 70 in.
.'. Total AT required for the iron = 70 X 32 = 2240, and total
AT for both air and iron = 25,200 -f 2240 = 27,440.
It is now necessary to arrange the winding so that the heat
developed by the current in the coil shall not cause an excessive
temperature in the coil. "
Assuming a permissible power loss of 0.4 watt per sq. in. of
exposed coil surface, an estimate can
be made of the proper amount of
space to be occupied by the coil.
Assuming, as a guess, a depth of wind-
ing of 3 in. and a coil length of 20 in.
(Fig. 25), the exposed surface of the
coil is 1270 sq. in. The total watts developed should then be
0.4 X 1270 = 508. Assuming, further, that the magnet is to be
designed for operation on a 100-volt circuit, the current is
508
loo = 5 - 08 amp -
Cross-section of Coil
28 ELECTRICAL ENGINEERING
27 440
The number of turns = -^ - = 5400, and the resistance of
O.Uo
., W 533 ,.,_ ,
the coil = yj = ,- oo\ 2 = 19.7 nm s.
The mean length of one turn may be estimated to be 42 in.
42
Then, total length of wire = 5400 X TS = 18,900 ft.
i si ,
19 7
Res. per 1000 ft. = = 1.04 ohms.
From tables, the nearest size wire is No. 10. Diameter of
No. 10, with double cotton covering, is 0.112 in. It must then
be found out if there will be room enough for the turns in the
space allowed.
Problem 16. Assume cast iron to cost l^c. per Ib. and copper 16c. per
Ib. Find the least cost of a magnet, of any desired shape, to lift 1000 Ib.
with a gap of 1.5 in.
Use 0.4 watt per sq. in. of coil surface as permissible power loss.
A
dx
CHAPTER VI
GENERATION OF ELECTROMOTIVE FORCE IN A
DYNAMO
FARADAY found in 1831 that the electromotive force produced
in a circuit was proportional to the rate of change of the lines
of force enclosed in the circuit. That is, he found that e k -57*
This discovery was really the foundation upon which electrical
engineering was built.
The truth of the relation may be seen by considering a rec-
tangular loop of wire carrying current I
moved a short distance dx in time dt, the
motion being at right angles to the direction
of the lines of force (Fig. 26). 1
The force on 1 cm. of wire, A, which is in
the field is F = IB. Thus the mechanical
work done in moving the loop from AB to
A'B' is Fdx = IBdx. F 7 G , 26<
As has been shown, the electrical work is
eldt and the mechanical and electrical work must be equal and
opposite.
dx
.'.IBdx = eldt or e = B -57- But Bdx is the change of
flux d<j> enclosed in the loop. Thus
j ," - "'''.-. e= ~ d it '
Consider now that the loop revolves in a uniform magnetic
field. When the loop encloses the entire field it may be said
to be in the zero position. Let it be assumed that in zero posi-
tion it encloses 100 lines. In position 1, displaced 10, it will
then enclose 98.5 lines. The loss of 1.5 lines from the loop has
resulted in a generated e.m.f. or, in general,
dt t% ti At
29
30
ELECTRICAL ENGINEERING
If the coil rotates at the rate of 1 r.p.s., the time required for it
to move 10 is : of 1 sec. = 0.0277 sec. = A.
Then
e =
1*5
0.0277 " 0.0277
= 54a6 volts.
This procedure may be followed and a tabulation made for
every 10, so as to obtain data from which to plot, point by point,
an e.m.f. wave, thus:
Angular position of coil
10
20
30, etc.
Flux enclosed, <
100
98.5
94.0
86 6
Change of flux, <f>z <i ...
1.5
4 5
7 4
A0
a
54.0
163.0
267
A
Plotted at angle
5
15
25
Values of e.m.f. are plotted at angles given in the last line, that
is, at the mid-angular divisions, since they represent average
values of e over each 10 of displacement of the loop.
Problem- 17. Assuming the field to be uniform, carry out the procedure
as just indicated for a complete rotation of the loop, and show by plotting
volts against angular displacement that the curve is a sine wave.
E.m.f. Waves in Fields that are not Uniform. Consider a
field between rounded poles of radius r,
distant 2r, in which a coil, of width 2r,
revolves (Fig. 27).
The field will be most dense in the
middle. The density will be assumed to
be inversely proportional to the distance
between poles. To find the density along
any line a&. The length, db = 4r 2r sin 6.
Then
B
4r 2r sin
where k is some constant i.e., proportionality factor.
This equation holds only for 6 from o to TT, in a revolution.
k
If B is taken from TT to 2ir. B = - A r-= :
4r + 2r sin 6
GENERATION OF ELECTROMOTIVE FORCE 31
When the coil moves a distance ds, there is a change d<{>, in
the amount of flux enclosed by the coil, per unit length of the
coil parallel to the shaft. d<fr is propor-
tional to the component of ds at right
angles to the direction of the flux, or to dx
(Fig. 28).
Thus, d<j> = 2Bdx is the change in flux per
centimeter length of coil, due to both con-
ductors. Then d<f> = 2Brd0 sin 0, since
ds = rd0, and dx ds sin 0.
Substituting,
kr sin 0d0 k sin 0d0
nax =
FIG. 28.
4r 2r sin
4 - 2 sin
Then,
_ d<j> __ JL^ /2k sin 0d0 \ k s
e " "" dt ~ dt U - 2 sin 07 " 2T^
fc sin
sn
Sin may be written sin 2irnt, where is expressed in radians
and 2irn denotes angular velocity. Then -j=r = 27rn, and n is
in revolutions per second, or is frequency jn a two-pole machine.
For machines of any number of poles -jj = 2wf where / =
frequency.
k sin
2 - sin
X2wf.
Let
Then
and
100 volts. This occurs when = ~
100 =
Substituting this value of k,
_ e max 2-n-f sin _ e max sin
= 27r/ X 2 - sin " 2 - sin '
Problem 18. Calculate and plot the e.m.f. for th$ above condition, for
one-half wave.
E.m.f. Wave when the Coil is Wound on an Iron Core. In
all these cases it is sufficiently correct to consider only the lengths
32
ELECTRICAL ENGINEERING
of the flux path in the air. By following the general procedure of
the preceding paragraph, the e.m.f. of a coil on the iron core is
found to be
(m 1) sin
where
m - sin
2mr
m = 7^-' or D
Problem 19. Calculate and plot for one-half wave, the e.m.f. for this
case, when e max = 100 and m = 1.1. (Fig. 29.)
Additional Problems for the Determination of E.m.f. Waves.
It is very good experience for the student to work out and plot
a number of these waves. For this purpose a few additional
problems are suggested.
Problem 20. Determine the e.m.f. of a coil wound on a wooden drum
when B max = 100 lines per sq. cm., speed = 1 r.p.s. and the dimensions of
the dynamo are as given in Fig. 30. Dimensions are given in centimeters.
Plot the wave, point by point, for each millimeter of distance across the pole
face.
_L
Problem 21. On an iron armature between rectangular poles as in Fig. 31,
let two coils, at right angles to each other (that is, in space quadrature), be
joined in series, so that their e.m.f. waves add. Plot the resultant e.m.f. per
centimeter length of the armature. Show that the resultant e.m.f. due to
the two coils is less than their sum.
Problem 22. Same as last, but for three coils spaced 120 apart.
Problem 23. Continue the development of the method of the last two
problems and finally obtain the average value of the e.m.f. of an armature
whose conductors are spaced uniformly around the periphery.
CHAPTER VII
FIG. 32.
INDUCTANCE
Inductance. When a circuit connected to a source of e.m.f., e t
is closed through a switch, S (Fig. 32), a current is established
in the coil, and sets up a magnetic flux
which links with the turns of the coil.
This flux produces a back, or counter
e.m.f. in each turn, e = -TJ-, or in the
N turns, e = N - Expressed in volts
this is
N_ d$
10 8 dt
Inductance is defined as the number of iijterlinkages of flux with
turns, per unit current, or, in symbols,
Expressed in practical units it is:
e =
(13)
L =
where i is the current in amperes.
Problem 24. Let a coil of 200 turns be supplied with various amounts of
current, and let the flux produced, when 1 amp. flows, be 1000 lines. Find
the inductance. Tabulating from Eq. (14):
i
i
2
4
10
tf). .
1000
2000
4000
10000
N
N<f>
200
0.2 X 10
200
0.4 X 10 6
200
0.8 X 10 6
200
2 X 10 fl
T
0.2 X 106
0.002
0.2 X 10 6
0.002
0.2 X 10 6
0.002
0.2 X 10
0.002
1 Students almost invariably have difficulties with inductance. This
problem is given solely for the purpose of impressing upon them the fact that
inductance is a "constant" of the circuit.
3 33
34 ELECTRICAL ENGINEERING
From the values of I/, thus obtained, it is seen that L is a
constant, and is independent of the current; it is a " circuit con-
stant" similar to resistance in a coil having a non-magnetic core.
Transposing Eq. (14),
"J&-
Differentiating with respect to time,
di _ N d<j>
L Jt"l0 8 ~dt
Substituting into (13),
which is the common expression for induced electromotive force,
or counter e.m.f. of self-induction. 1
In Fig. 33 is represented a circuit of
resistance, r, and inductance, L, which is
oj connected to a source of e.m.f., e. When
1 *- o
*' the switch, S, is closed, the e.m.f. has
Tfl _ qq
to overcome the resistance, r, and also
the counter e.m.f. of self-induction due to the setting up of flux
in the coil. Therefore we may write:
e = ir + L^ (15)
This equation is fundamental, and is general for circuits
possessing only resistance and inductance of constant value.
An algebraic relation between the impressed e.m.f. and the
current assuming in this case that the e.m.f. is kept constant
is found as follows:
e-ir
1 In an ironclad magnetic circuit the inductance is not a constant. It
depends upon the permeability.
The flux is not proportional to the current producing it but is a complicated
function thereof. In that case ^ = ~ (Li) = L ~ ' + i^.
INDUCTANCE 35
/. t = - ? log (e - ir) + C,
whence
rt
log (e - ir) = - + Ci,
and
" + Cl - r ^
e ir L = C 2 L
(16)
C 2 is determined from the nature of the problem.
The rate of energy supply or power equation corresponding
to (15) is obviously
ei = i z r -f Li
r
\
Jo
and the energy supplied by the generator is
T
eidt.
The energy dissipated in heat is
f*T
\ i*rdt
Jo
and the energy supplied and thus stored in the magnetic field is
JT j S*T
Li^di= ( Lidi = y 2 LP
where I is the value of the current at time T.
Starting and Stopping Current in an Inductive Circuit. Re-
ferring to equation (16) it is evident that for t = 0, i = when
starting the current and i = I for t = in stopping the current,
since energy cannot be altered in an infinitely short time and
therefore current cannot be established or changed in an infinitely
short time.
Thus when considering the starting of a current we have for
t = 0, i = 0.
Substituting these values in (16)
= - r [e - C 2 e] = -i [e - Cj,
36
whence,
and
ELECTRICAL ENGINEERING
(17)
This equation gives the value of the current at any instant after
the closing of the switch, S.
If the impressed e.m.f., e, is sud-
denly short-circuited by the closing of
the switch, S' (Fig. 34), then e = 0,
and, at the instant of closing, t = 0,
FIG. 34. * = I,
where Z is the current in the circuit
just before closing the switch.
Substituting these values into (16)
whence,
and
7 = [0 - CJ,
C, = - rZ,
(18)
This equation gives the current at any instant, t, as it is dying
away in the circuit after the e.m.f. has been suddenly removed.
The inductance of coils
varies with the size, shape
and number of turns. If a
given length of wire of defi-
nite size is to be made into a
coil, maximum inductance F IG< 35.
will very nearly be obtained
if the coil has the proportions given in Fig. 35. 1
the inductance, in this case, will be:
0.27 cm. 2
The value of
10Xc
,
henrys >
where the length of the coil is given in centimeters, and c is in
centimeters.
1 BROOKS and TURNER, "Inductance of Coils," Bulletin No. 53, Univ. of
Illinois Engineering Experiment Station.
INDUCTANCE 37
Problem 26. Find and plot current vs. time when the circuit is closed on
a coil of 1 km. of No. 15 B. and S. wire (diam., d.c.c., = 0.066 in.; r/1000' =
3.17 W ), designed for maximum inductance, e = 100 volts.
Problem 26. Find and plot the curve of dying away of the current when
the coil of problem 25 is short-circuited.
Problem 27. Find the average value of the inductance of the lifting
magnet previously designed (Chap. V), and determine how long it will
take for the current to rise to 90 per cent, of its permanent value.
CHAPTER VIII
ALTERNATING CURRENTS
It has been shown that the fundamental equation in an induc-
tive circuit where the resistance and inductance are constants and
not depending upon the current is :
This equation gives the relation between the particular values
of e.m.f. and current at any instant.
In the case previously discussed it was assumed that the im-
pressed e.m.f., 6, was constant.
In most engineering problems the e.m.f. is, however, not con-
stant but it varies from instant to
instant. Almost all electrical instal-
lations now use alternating current
rather than direct current. In this
case it will be seen that the e.m.f.
and current can almost always be as-
FIG. 36. j- L i
sumed to vary according to a simple
sine law.
In other words it can be assumed that the instantaneous value
of the current at any time, t, can be found from equation i = I m
sin ut (Fig. 36), where o> = 27r/ = angular velocity, and / =
frequency of alternation of the current = number of cycles, or
complete reversals, per second. I m = maximum value of cur-
rent. For 60 cycles, o> = 27r60 = 377.
.'. i = I m sin 377*.
Differentiating eqv: i = I m sin w
we get
di
- = 7 m w cos wt.
Substituting in (15),
e = r I m sin wt + L 7 m w cos orf
= I m (r sin wt + Lw cos o>0 (19)
38
ALTERNATING CURRENTS
39
Thus e is the sum of two component waves, one depending on
the sine of ut, and the other on the cosine.
Problem 28. Let I m
ponent waves of e.m.f.
1, r = 0.5, Leo =0.4. Find and plot the com-
Sin at
I m r sin ut = ir
Cos wt . .
w cos
L*
dt
e.
00
00
00
40
40
30
.5
.25
.866
.346
.596
60
0.866
0.4330
0.5
0.2
0.6330
90
1.0
.5
0.0
0.0
.5
120
0.866
0.433
-0.5
-0.2
0.233
150
0.5
0.25
-0.866
-0.346
-0.096
180
0.0
0.0
-1.0
-0.4
-0.4
These waves are shown plotted and combined in Fig. 37.
Problem 29. A similar set of waves should be obtained by each student
from values of I mi r, Leo, assigned at random.
By inspecting these waves it is seen that i lags behind e, that
FIG. 37.
is, it passes through zero later than e by about 40. This il-
lustrates one of the characteristic features of inductive circuits.
It should also be noted that ir is in time phase with i, and that
iLu is in time quadrature with i, being 90 in phase ahead of i.
The quantity Lo> is called reactance. It is measured in ohms,
and denoted by the letter X. Thus Lw = X, where L is the in-
ductance in henrys, w = 2ir/ is the angular velocity in radians per
second, and X is the reactance in ohms. X is not, like L, a
property of a coil or circuit, but depends on the frequency.
The average value of the e.m.f. generated in a coil of a dynamo,
depends only on the speed of rotation and the number of lines of
40 ELECTRICAL ENGINEERING
flux cut; that is, it depends on the average rate of cutting of the
lines of flux, by the conductors, and not on the distribution of
the lines under the poles. The effective value of e.m.f . does, how-
ever, depend on the distribution of the flux.
Frequency has been defined as number of cycles per second.
A two-pole generator, at 1 r.p.s. has the frequency,/ = 1.
A four-pole generator, at 1 r.p.s., has/ = 2.
fry
A p-pole generator, at N, r.p.s., has/ = -^N.
The coil, in position 1 (Fig. 38a), contains the whole flux.
The coil, in position 2, contains no flux. Thus, a change of
the whole flux takes place in a quar-
ter of a revolution.
If T is the time of 1 cycle, the
whole flux is therefore cut in the
T
time r -
4
The average rate of cutting is then
<|> 4$
FIG. 38. ~m" ~7p where & is the total flux.
T
Therefore, the average e.m.f. is -y-, where N is the number of
turns, and 2N is the number of conductors per circuit.
At 60 cycles,
In general,
a-.-L
* 60
.'. Average e.m.f. = 4N3>f = 4N$f X 10~ 8 volts.
In a four-pole machine (Fig. 386) all flux is cut in %
revolution. The average rate of cutting is therefore ~, w here
I i
T\ is the time of a revolution.
Average rate = -^ = 83>N g = ~^- = 4*/ where N s is the
number of revolutions per second, and f = - N s .
With N turns, average rate = 4/A r <l> = average e.m.f.
X 10~ 8 volts (20)
ALTERNATING CURRENTS 41
This equation is identical with that for a 2 pole machine. It
applies regardless of the number of poles as long as N is the
number of turns in series per circuit.
Average Value of a Sine Wave. The e.m.f. induced by rota-
tion of the armature conductors in the field is
Let <p = < m cos ut, be the flux enclosed at any instant. Then,.
-T = o>$ TO sin ut is the rate of change of the flux, and e =
volts.
sin ut.
In practical units,
sin
10 8
Since co = 2irf this may be written,
sin co/
10'
For maximum e.m.f.,
sin u* = 1, and E m = ^ s ' m volts (21)
To obtain the average value of e.m.f., integrate a half-wave
and divide by TT, that is, by the length of a half-wave.
Thus,
. = - ( "sin 6dd = -\ - cos 0| = - = 0.636.
irjo TT|_ Jo 7T
2 2
.'. The average value = - X E m . Multiplying (21) by -
7T 7T
Av. e = -
which agrees with the average value previously found (20).
Effective Value of a Sine Wave. Let i = I m sin 6.
If this current flows through a resistance r, it has been seen
that the heat developed at any instant is i 2 r.
Thus, the heat developed per cycle may be expressed as,
T 2 %
r I I 2 m sin 2 Ode.
Jo
By trigonometry,
sin 2 = y>, V<> cos 20.
42 ELECTRICAL ENGINEERING
Substituting,
The average value of energy flow or the rate at which energy
is being dissipated, or the power, is
rlmir = rim
27T 2
r/i
Thus, the rate of heat dissipation is --
The effective value of the current corresponds to a constant
or direct current which would give the same heat in the same
time if flowing through the same resistance.
. = , and i ett . = = 0.707 I m .
Similarly, the effective value of e.m.f. is obtained, and e e ff. =
0.707 E m , where E m is the maximum value of the sine wave of
electromotive force.
effective value . , . , -
I The ratio - is called the form
N average value
factor.
j With sine waves form factor (j(f) = QQ^Q =
^_ .
.
The equation for the effective value of the
I e.m.f. is obtained from (21) by multiplication
i
Fro. 39. '
m . m
- '"~ V210 ^0^^
This applies to a concentrated coil of N turns.
If, however, the turns are distributed over the periphery, as
in a direct-current armature, from Fig. 39 it is seen that coil 1
contains all the flux, while coil 2 contains the flux X cos 6.
Therefore the .effectiveness of coil 2 is N c cos 0, where N c
number of turns of the coil.
Let N = total number of turns. Then the turns per cm. of
N
armature periphery = ^ , where r = radius of armature.
ALTERNATING CURRENTS 43
N
Then the effectiveness of the turns per cm. is ~ cos 6. The
average effectiveness of the turns per cm. is
NcosB.. N
r+i
1 I N cot
,J l.r
N 2
The total effectiveness is therefore 2rr X -^ = -N.
Therefore, in a distributed winding, the turns are not so
effective as when they are concentrated.
Thus, for distributed winding,
4 44fAT<iv 2
*^*//> **m ^
CHAPTER IX
DIRECT -CURRENT GENERATORS
Homopokr Generators. These are also called by the names
" acyclic" and "unipolar." They are a small class of machines,
distinguished from the usual types of direct-current machinery
in that the conductors always move through the magnetic field
in a constant direction with respect to the
direction of the lines of flux.
Among the earliest of dynamos may be
mentioned one of this type known as
" FARADAY'S Disc Dynamo," in which a
copper disc was rotated between the poles
of a permanent magnet.
Current was collected by means of two
brushes making contact, respectively, with the rim and axle of
the disc (Fig. 40) . A more modern type of homopolar generator
is shown diagrammatically in Fig. 41. For the permanent mag-
FIG. 40.
I,
\
; i
!
f
\
<
i
N
E
3
r
p
1 F
^
-
1
Q
f
L_
^
, >
^
)
^
V
(
f.
~""\
J
\
_s
FIG. 41.
net is substituted a powerful electromagnet, and two sets of
brushes are used instead of one. By connecting these brushes
in series outside of the machine the total e.m.f. at the terminals
is doubled.
44
DIRECT-CURRENT GENERATORS
45
From the fundamental considerations developed in Chap. VI
it is evident that the voltage between each set of brushes is
e
10 8
where N = revolutions per second and 3> = total flux, since
there is only one conductor between the brushes. With any ar-
rangement which permits the use of additional sets of brushes,
as in Fig. 41, the voltage is increased in proportion to the number
of sets of brushes connected in series, and becomes
e =
10 8
FIG. 42.
where c is the number of conductors and is equal to the number
of sets of brushes in series. Fig. 42 illustrates the use of bar
conductors on the armature. Each conductor is connected to
two slip rings on which brushes bear. There are thus twice as
many slip rings as conductors. Since the conductors are in-
sulated, they may be put in series by properly connecting their
brushes outside of the machine.
Direct-current Machines with Commutators. On most
direct-current machines use is made of commutators. To
understand these machines a knowledge of the principle of wind-
ings on the armature is needed. In Fig. 43 a single coil is repre-
sented in a magnetic field. The ends of the coil are connected
to the segments of a two-part commutator. In the position
46
ELECTRICAL ENGINEERING
shown, the e.m.f. is maximum. As the coil moves in the field,
the segments move under the brushes and the e.m.f. at the
brushes, AB, during a half revolution, has the values of a half-
sine wave. When this e.m.f. reaches zero, the segments pass
from under the brushes. The same operation is then repeated
and gives a succession of half waves,
all in the same direction. If now
another loop is placed on the armature
at 90 to the first one, a new series of
half waves will be added at 90 to the
first series. By connecting these loops
in series, suitably joining to commuta-
tor segments and continuing to use
only two brushes, the e.m.fs. of the
loops are added together and produce
a resultant e.m.f. shown in heavy dots by the wave "d" (Fig.
44). This wave never reaches zero and is much more steady
than that produced by a single coil. By continuing this process,
all irregularities are virtually wiped out and there results a
smooth wave of constant
e.m.f. A simple example
of armature winding with
commutator and brushes
FIG. 44.
is shown in Fig. 45, for the purpose of illustrating the connec-
tion of coils in series.
Types of Direct-current Commutator Machines. Direct-
current machines are usually divided into groups according to
the method of exciting the field magnets, as follows:
1. Permanent Magnet Machines. These have no field windings,
but the field structure consists of hard-steel permanent magnets.
They constitute a small group, used chiefly for telephone sig-
nalling and gas engine ignition.
DIRECT-CURRENT GENERATORS
47
2. Separately Excited Machines. In these, the field winding
is supplied with current from an external source. The chief
advantage of this type is that it enables a steady field excitation
to be maintained at all times regardless of the fluctuations in
voltage at the brushes.
3. Shunt Machines. In this type the source of excitation of
the field is derived from the terminals of the machine itself.
The field circuit is connected in parallel with the external cir-
cuit and the field current varies as the voltage of the machine
changes.
4. Series Machines. The current in the armature is made to
flow also through the field windings; that is, the field and armature
coils are connected in series with the external circuit. Thus the
field excitation is proportional to the load current.
1 Permanent Magnet 2 Separate Excitation 3 Series
FIG. 46.
4 Siunt
5 Compound
5. Compound Machines. These are excited partly by a shunt
winding and partly by a series winding, each pole being provided
with both a shunt and series coil. The total field excitation
thus depends upon the voltage of the machine as well as on the
load current.
These five types are illustrated in Fig. 46. Other com-
binations are sometimes used in special cases. The performance
characteristics of these various types of generators differ greatly.
In general, the characteristic of a generator is a curve showing
the relation between terminal voltage and the load current, the
latter being the independent variable. These curves and
others of a similar nature should be thoroughly studied, especially
in the laboratory.
Armature Reaction. When a generator is delivering no cur-
rent the direction of the field flux is along the axis of the poles.
48
ELECTRICAL ENGINEERING
When current is flowing, however, the armature becomes an
electromagnet on its own account, and the field flux becomes the
resultant of that produced by the field windings and that due
to the armature winding.
Fig. 47a shows a bipolar dynamo with a ring armature. Arrows
show the direction of current and also of flux. Starting from
the negative brush, the current divides as it enters the armature,
one half winding around to the left, the other half pursuing a
similar path to the right, and both finally joining again to enter
the positive brush. It is to be noted that the flux set up by
these armature currents is, in general, in space quadrature to the
flux due to the field winding.
FIG. 47.
Fig. 476 shows an equivalent diagram representing a drum
armature. When once the principles of current action in the
armature are understood, it is simpler to make use of the rep-
resentation of Fig. 476 than of Fig. 47a. For clearness, the com-
mutator is omitted in the case of the drum, the position of the
brushes being indicated with reference to the armature itself.
It makes no difference how the end connections are made, so
far as the armature m.m.f. is concerned. In this case, since the
brushes are not shifted but are placed on the so-called neutral
axis midway between the poles, the armature magnetomotive
force is directed vertically upward, while the field magnetomo-
tive force is, as always, along the pole axis. The resultant
magnetomotive force is the vector sum of these two. Since the
armature m.m.f. acts at right angles to the field m.m.f., its effect
is said to be wholly cross-magnetizing.
When the brushes are shifted a the armature m.m.f., still
acting along the brush axis, may be resolved into components.
DIRECT-CURRENT GENERATORS
49
p c = p A cos a, the cross-magnetizing component, acting at
right angles to the field, and FD = FA sin a, the demagnetizing
component, acting directly in opposition to the field. The re-
sultant m.m.f., OR, Fig. 48, is then due to the m.m.fs. of the
field OF and the armature OA, the latter being composed of OC,
cross-magnetizing, and OD, demagnetizing.
Cross-magnetization is always present when the armature
carries current. It distorts the field and displaces the neutral
axis, necessitating thereby a shifting of the brushes. When the
brushes are shifted, demagnetization also enters in, weakening
directly the field strength. Under such conditions the resultant
flux takes up a general direction as indicated by the shading in
the air gap in the figure. The pole tips are unequally magnet-
ized, the leading tips being weakened and the trailing tips
strengthened.
A c
FIG. 48.
The actual direction of the resultant flux is not along OR but
along OR' ', a line of somewhat less deviation from OF. This is
because of the unequal reluctances of the paths along the direc-
tions of the component m.m.fs.
Consider, for example, a generator whose flux per pole enter-
ing the armature is < r , under conditions of normal operation,
that is, voltage and speed.
To generate this flux at no load would require F amp. -turns
on the field core if all the flux generated in the field passed through
the armature. Some flux, however, passes around the armature
without cutting its conductors. This is called leakage flux,
and amounts to 15 or 20 per cent, of the net flux, in ordinary
machines. To provide this leakage flux as well as the net flux,
</v, requires kF field amp.-turns, where fc is the leakage co-
efficient and may be taken as 1.15.
Let I c amp. now flow in each armature conductor, and let the
50 ELECTRICAL ENGINEERING
total number of conductors be C. Then total turns = -a and
turns per pole = ~~> where p = number of poles.
CI e
The total armature amp.-turns per pole = -~~'
Let the brushes be on the geometrical neutral, that is, midway
between the poles. Then, since the conductors are distributed
over the entire periphery, the effective armature A.T. per pole
9 r T r
= / L = ffk.
TT c 2p irp
The effect of these distorting ampere-turns has been shown
to be to weaken the flux in the leading pole tips and to strengthen
that in the trailing tips. The net result owing to unequal
saturation of the iron, is to reduce the actual amount of the flux.
In order to compensate for this reduction extra ampere-turns
must be placed upon the field core to the amount of about 40
per cent, of the armature cross-magnetizing ampere-turns.
Thus,
F c = > if there is no brushshift
irp
and the total field amp.-turns per pole are
F t = kF. + -^.
vp
When the brushes are shifted a, the cross-magnetizing amp.-
turns are F c = ~ X -j~ (Fig. 49).
. 180 - 2 I C C
Their effective value is k c r^ TT~
loU 4p
where
r*
Ur
2 cos a
cos dO, =
7T
where = IT 2a
The demagnetizing turns consist of a belt of conductors of
width 2a. The effective demagnetizing amp.-turns are then
, 2a I C C
DIRECT-C URRENT GENERA TORS 51
where
/+!
2 sin a
cos ede = ~2^-
These latter ampere-turns act in direct opposition to the field.
If there were no leakage of flux between field and armature, they
would be compensated by placing an equal number of additional
ampere-turns on the field. Owing to leakage this number must
be multiplied by k.
The total required field ampere-turns under the condition of
brush shift of a and I c amp. in the armature conductors is
then
(90 - a)I c C
/
(
0.4
in order that the flux entering the armature shall be <p r . With
constant generated voltage the terminal voltage falls off as
the load increases, due to the IR drop of the armature. <p r
must therefore be increased sufficiently to make up for the IR
drop.
CHARACTERISTICS OF DIRECT CURRENT GENERATORS
From the discussion given above it should be possible to calcu-
late the change of voltage with load in any of the different types,
provided that the saturation curve could be expressed in a simple
manner. This is unfortunately not possible, but it can be approx-
imated by FROELICH'S equation, which is :
km 2
where m is the excitation in ampere-turns and e the corresponding
voltage, k and ki are constants depending upon the shape of the
saturation curve which constants can be determined by substi-
F *
1 It is seen that jf = cot a. Thus, on the basis given above, the ratio
between the actual ampere-turns needed to compensate for the cross-mag-
netizing and demagnetizing ampere-turns is .34 cot a.
2 e is the induced e.m.f. due to the rotation of the armature conductors in
the magnetic field, m is the resultant m.m.f. of the amp.-turns on field
and armature referred to the field structure. At no-load, m is obviously
the field excitation alone.
52 ELECTRICAL ENGINEERING
tuting two known values of e and m from the actual saturation
curve.
Consider a compound wound generator. Let the terminal
voltage be e and the load current i. If the resistance of the shunt
p
field winding is r f the shunt field current is if = . l If each field
spool has t turns then the m.m.f. of the field winding, per pole,
/j
is m\ i/t = t. If the series winding has ti turns, per pole,
the m.m.f. per pole of the winding is mz = ii\. Let the demag-
netizing ampere-turns per pole of the armature with full-load
current as determined above be D; then the demagnetizing ampere-
turns, with load current i t is m s = ~ji, where 7 is full-load current.
Let the equivalent demagnetizing ampere-turns with full-load
current, due to the " cross magnetizing" ampere-turns be C;
C
then the demagnetizing effect of i amp. is w 4 = ~^i.
The total m.m.f., m , on each pole were there no leakage
is:
Due to the leakage between the field poles the equation is
obviously modified. Assuming 15 per cent, leakage:
w = 0.85(wi + 7/12) w 3 w 4 ,
ra then being the m.m.f. which causes flux to interlink with the
armature conductors.
If the saturation curve were plotted on the basis of these
ampere-turns the corresponding voltage could be obtained either
directly or through FROELICH'S equation. This is, however, not
the case but the saturation curve takes into consideration the
leakage, therefore, in order to use the saturation curve we have
to use a new value of m , namely:
ra 3
, + ,- -
Numerical Application. Let the no-load voltage e , the no-
load excitation, and the full-load current be taken as unity, then
mi = e
C~*lT
1 It is really , but ir is usually very small.
DIRECT-CURRENT GENERATORS
53
Let the full-load series excitation be 40 per cent, of the no-load
excitation, then
ra 2 = OAi
Let the demagnetizing ampere-turns of the armature with full-
load current be 10 per cent, of the no-load field excitation, then
ra 3 = 0.1 Oz, and let the equivalent demagnetizing ampere-turns
of the cross-ampere-turns with full-load current be 20 per cent.,
then m 4 = 0.20i.
(This relation between ra 3 and ra 4 corresponds to 11 brush
shift.)
1.2
1.0
,0.8
'0.6
0.4
0.2
0.2
0.4
0,6 0.8 LO
Current
FIG. 49.
1.2
1.4 1.6
Then
Referring now to FROELICH'S equation and assuming the sat-
uration curve to be such that for e = 1, m = 1; for e = 0.6, m =
0.5; then it is readily proven that k = 1.5, and ki = 0.5
1.5m
. . e = .. . ~ , or m =
' 1.5 - 0.5e
0.5 m v ~ 1.5- 0.5 e
= e + 0.05i, or e = 0.5 - 0.025z +
V(0.5 - 0.025t) 2 + 0.:
The voltage at the terminal of the machine is less than e by
the ir drop in the armature winding, brushes and series field.
If at full-load the drop is 3 per cent, then at any other load it is
0.03i.
54 ELECTRICAL ENGINEERING
Thus e the terminal voltage is
l = e - 0.03*' = 0.5 - 0.055i -f V(0.5 - 0.025^) 2 + 0.15*
The student should verify curves a and 6, Fig .,.49. Curve a
applies to the compound wound generator discussed above.
Curve b to a typical shunt generator in which the ratio between
the armature reaction and the no-load excitation is less than with
a compound wound generator, and in which the saturation at
normal voltage is usually higher.
The constants used for the shunt generator are:
k = 2.33, fcj = 1.33, mi = e, ra 2 = 0, w 3 = O.OSz, m 4 = O.lOi,
Ir = 3 per cent.
It is seen that as the resistance of the load is gradually decreased
the current increases up to a certain maximum value, in this
case 20 per cent, more than rated current; after that, the current
and voltage both decrease.
The student should study the effect of the saturation on the
shape of these curves.
CHAPTER X
A STUDY OF THE DESIGN OF A DIRECT-CURRENT
GENERATOR
All the underlying principles of the direct-current generator
may be studied to good advantage from the basis of a concrete
example. The example here chosen is an ordinary compound-
wound generator with the following specifications:
M.P. 12 - 500 - 375 - 250 volts, which means that it be-
longs to the general multipolar class (M.P.), has 12 poles, 500
kw. rated output, 375 r.p.m. at normal speed, and the voltage
is 250 at both no-load and full-load. The normal-load current
may also be given. It is
= 2000 amp.
Other data for this machine are: Armature external diameter =
64 in., from which is obtained what is called diameter per pole
64
= 12 = 5 - 33 in -
Armature internal diameter = 44 in.
Number of armature slots = 216.
Dimensions of slots, 0.465 in. wide by 1.3 in. deep.
Armature winding is of the multiple drum type.
Current in each effective conductor is
^ - , /.-??- 167 amp.' ,
Each effective conductor consists of two bars in parallel. Each
bar is 0.075 in. X 0.45 in., without insulation.
Area of each effective conductor = 2 X 0.075 X 0.45 =
0.0675 sq. in.
.'.Current density in conductor = Q 0675 = 24 ^ amp * per
sq. in.
55
56 ELECTRICAL ENGINEERING
With direct-current generators, current density in the arma-
ture conductors generally lies between 2000
and 3000 amp. per sq. in.
Number of effective conductors per slot = 4.
Number of conductor bars in each slot = 8.
Arrangement of conductors in slot is as
y , . T,.
FIG. 50. shown in Fig. 50.
Number of effective turns per pole,
_ conductors per slot X number of slots _ 4 X 216 _
conductors per turn X poles 2 X 12
Flux Calculation. There are now sufficient data to apply the
fundamental e.m.f. equation to the determination of the flux.
The equation is
* = %< volts.
Supplying numerical values,
4 X 37.5 X a X 36
"lO 5 ""
whence
a = 4,630,000.
< a here is the required flux per pole entering the armature at no-
load. Neglecting the effect of the small shunt field current flow-
ing in the armature, the generated voltage and terminal voltage
are the same at no-load.
At full-load, in order to maintain the same terminal voltage,
250, it would be necessary to generate a slightly higher voltage
to supply the drop in the armature, series field and brushes.
Assuming this drop to be 2^ per cent., the required flux entering
the armature at full-load is
0' a = 1.025 X 4,630,000 = 4,750,000.
The total flux which must be generated is made up of the
armature flux and that which leaks across from pole to pole with-
out passing through the armature. Assuming the leakage flux
to be 15 per cent., the total flux in the pole core at no-load will be
C ** 1.15 X 4,630,000= 5,320,000.
DESIGN OF A DIRECT-CURRENT GENERATOR 57
At full-load the total flux will be
0' c = 1.15 X 4,750,000 = 5,450,000.
The Magnetic Circuit. In order to produce this flux it is
necessary to employ the required number of ampere-turns per
pole of the field winding. These are determined as the sum of
the ampere-turns required for each part of the magnetic circuit
supplied by the windings on a single pole. The separate
parts are (1) the armature teeth, (2) the air gap, (3) the armature
core, (4) the pole core, (5) the yoke.
The relations between ampere-turns per inch length of the
magnetic path and flux density in lines per square inch are
given by the saturation curves for the various materials com-
. FIG. 51.
posing the magnetic circuit. To ascertain the ampere-turns it
is necessary to know the cross-sectional area and length of each
component part of the magnetic circuit. These are best deter-
mined with the help of a scale drawing showing the armature and
the field cores in their relative positions. Such a drawing is
reproduced in Fig. 51.
Here the mean flux paths are indicated by heavy dotted lines.
The cross-sectional areas through which they pass are ascertained
directly from the given dimensions, except in the cases of teeth
and gap.
Area of Flux Path through Teeth. Since the slot is of uniform
width, the tooth must be narrower at the base or "root" than
58 ELECTRICAL ENGINEERING
at the face. The ampere-turns required for the teeth may be
taken as the mean of the ampere-turns which would be required
if the teeth area throughout were that at their face, and if it
were that at their base. This is not the same as the ampere-
turns required for the mean area of the teeth.
Width of tooth at face = slot pitch slot width
TT X 64
216
- 0.465 = 0.465 in.
Width of teeth at face = 0.465 in. X number of teeth under one
pole = 0.465 X teeth per pole X P ? le *f C . X 1.08
pole pitch
216
= 0.465 X -jg- X 0.72 X 1.08 = 6.53 in.
The factor 1.08 is inserted to allow for fringing, that is,
the spreading of the flux to teeth not immediately under the
pole.
Teeth area at face = 6.53 X net armature length = 6.53 X
(gross armature length air duct width) X 0.9. This armature
has a gross length, parallel to the shaft, of 9 in.; the air ducts
are six in number, each % in. wide, making a total width of air
duct of 2.25 in. The factor 0.9 is commonly used to allow
for space lost between the laminations due to the presence
of Japan insulation or natural inequalities in the material.
Substituting these numerical values, the teeth area at face
is = 6.53 X (9 - 2.25) 0.9 = 6.53 X 6.07 = 39.6 sq. in.
Width of tooth at base = * A ~ 0.465 = 0.428 m.
216
Teeth area at base = 0.428 X -jg- X 0.72 X 1.08 X 6.07 =
36.4 sq. in.
Area of Flux Path through Gap. It might be assumed that a
mean area between that of the pole face and that of the teeth
should be taken for the gap. . Consideration, however, will show
that this will give too small a result.
The flux in the gap fills practically the whole of it, though
near the teeth the distribution is no longer uniform. A fairly
satisfactory approximation to the effective gap area is obtained
by taking one-fourth of the sum of three times the pole face area
plus the teeth area. Thus, gap area,
DESIGN OF A DIRECT-CURRENT GENERATOR 59
_3 X pole face area + teeth area
Pole face area = pole arc X pole length = 12.2 X 7.25 = 88.4
sq. in.
. A 3 X 88.4 + 39.6 _
. . AO = - -j - = 76 sq. in. approx.
Areas of Armature Core, Pole Core and Yoke. The armature
core section perpendicular to the flux path is taken radially,
and is the product of the radial distance, a, in Fig. 51 and
the net length of the core. Thus
A a = 8.7 in. X 6.07 in. = 52.8 sq. in.
The pole core section is circular in this machine, the area
being
A p = irr p * = TT X (4.4375) 2 = 62 sq. in.
The effect of any variation due to the pole shoe is very slight
and may be neglected.
The yoke section is taken radially as at 6, in the figure. Its
form is somewhat irregular.
In this instance the area is
A y = 83.5 sq. in.
Materials. The armature core is of soft sheet-iron laminations
of high permeability, the poles are of soft steel and the yoke is
of cast iron. Magnetization curves of these materials are given
in Fig. 20.
No-load and Full-load Saturation Curves. Having now de-
termined the fluxes, areas, lengths and materials, it is in order
to put these together in tables to show the flux densities,
ampere-turns per inch, and ampere-turns for each part of the
magnetic circuit, and finally the total ampere-turns. This is
done for both no-load and full-load. In the former case a
point is obtained on the no-load saturation curve, Fig. 52.
Other points on this curve are obtained by repeating the tabu-
lation process, starting with any desired values of voltage, such
as 80, 150, 200, 260, 280, determining the fluxes from the e.m.f.
equation, then flux densities and ampere-turns.
With the rest of the design the student should hand in both
curves with complete tabulation of points.
60
ELECTRICAL ENGINEERING
Tabulations for 250 volts, no-load and full-load, are given
in Tables III and IV.
In the case of full-load there must be added the ampere-
turns required to overcome the armature reaction, in order to
give the total required ampere-turns and the resultant point
on the full-load saturation curve.
TABLE III
Part
No-load. E = 250 volts
Flux (mgl.)
Area
B
A.T./in.
Length, in.
A.T.
Teeth
Gao
4.63
4.63
2.315
5.32
2.66
uired amp.-tur]
(face) 39. 6
(base) 36 . 4
76.0
54.5
62.0
83.5
as.
117,000
127,000
61,000
42,500
86,000
32,000
}
19,100
3
40
50
1.3
0.3125
7.0
12.0
12.0
429
5,970
21
480
600
7,500
Armature. .
Pole . . .
Yoke
Total req
TABLE IV
Part
Full-load. E = 250 volts
Flux (mgl.)
Area
B
A.T./in.
Length, in.
A.T.
(face) 39.6
(base) 36. 4
76.0
54.5
62.0
83.5
120,300
130,700
62,600
43,600
88,000
32,800
230 \
550 / 39
19,600
2.8
43
52
1.3
0.3125
7 .0
12.0
12.0
507
6,140
19.6
516
624
Teeth
Can
4.75
4.75
2.375
5.45
2.725
ns
Armature. .
Pole
Yoke
Amp.-tur
Amp.-tur
Total req
7,807
2,620
ns required tc
uired amp.-ti
> overcome armature r
jrns
eaction
10,427
Armature Reaction. " Armature reaction" means effective
ampere-turns per pole on the armature. The actual amp.-
turns per pole, in this case, are 167 X 36 = 6000.
Since the turns are distributed over the armature surface the
effective amp.-turns are
- X 6000 = 3820.
7T
DESIGN OF A DIRECT-CURRENT GENERATOR 61
If there were no shift to the brushes, these ampere-turns would
all be cross-magnetizing, or distorting. To compensate for
them, it is necessary to supply about 40 per cent, of their value
in additional ampere-turns on the field core.
It is assumed, however, that the brushes will be shifted 15,
giving a distorting belt of 180 - 30 = 150. To overcome the
distorting ampere-turns at full-load there will then be required
240
200
.120
40
7
2000
4000
6000 8000
Ampere-Turns
FIG. 52.
10000
12000
150
180
X 36 X 167 X k e X 0.4 = 1480,
where
+ 2
k c = ~ \ cos ed0 = x^ [1.9318] = 0.737.
The demagnetizing ampere-turns constitute a belt 30 wide.
To compensate for them would require their exact numerical
equivalent, were there perfect mutual induction between these
turns and the field. Owing to magnetic leakage there should be
added about 15 per cent, to the effective demagnetizing ampere-
62 ELECTRICAL ENGINEERING
turns. To compensate for these, therefore, will require
30
X 36 X 167 X fed X 1.15 = 1140 amp.-turns,
where k d = 0.99.
To overcome armature reaction at full-load will require 1480
+ 1140 = 2620 additional amp.-turns on the field core.
For any other load, keeping the same shift, the required
ampere-turns will be proportional to the load current.
No-load and full-load saturation curves are shown in Fig. 52.
The Shunt Field Winding. Under no-load conditions it is
evident that the shunt field current must supply the entire ex-
citation. In this machine, therefore, the shunt field m.m.f.
must consist of 7500 amp.-turns per pole when an e.m.f. of 250
volts is being generated.
Actually, each shunt spool is wound with 460 turns of No. 7
B. & S., D.C.C. wire. The field current is therefore
= 16.3 amp.
The shunt coil has an actual length of 6.25 in. As the di-
ameter of No. 7 wire is 0.16 in., including insulation, there will be
Q-TQ X 6.25 = 39 turns per layer of wire. There will be -^ =
11.8 layers, or practically 12 layers, giving a depth of winding
of 0.16 X 12 = 1.92 in.
The mean radius of the coil, allowing for spool thickness, is
then
Mean radius = - ' ~ - = 5.46 in.
.*. Mean length of turn = 2ir X 5.46 = 34.35 in.
Total length of wire on each shunt spool is
-r^- in. X 460 = 1316 ft.
Resistance of No. 7 wire at 65C. = 0.586 ohm per 1000 ft.
.'. Resistance of each shunt spool is
0.586 X 1.316 = 0.77 ohm.
The resistance of the entire shunt field is
r f = 0.77 X 12 = 9.24 ohm.
DESIGN OF A DIRECT-CURRENT GENERATOR 63
The voltage drop on the shunt field is
i/r f = 16.3 X 9.24 = 151 volts.
The voltage drop in the shunt field rheostat is
e rh . = 250 - 151 = 99 volts.
The Series Field Winding. Consider two cases: (1) the genera-
tor to be flat-compounded, (that is, the no-load and the full-
load voltages are equal, as specified), and (2) the generator to
be 5 per cent, over-compounded.
In the first case, it is evident that the shunt field ampere-
turns will remain the same at full-load as at no-load since the
same voltage, 250, is impressed on the shunt circuit.
But by Tables III and IV, it is seen that at full-load there will
be required 10,427 7500 = 2927 additional amp. -turns. These
must evidently be supplied by the series field m.m.f.
FIG. 53.
The actual winding consists of 2>^ turns per pole. Each turn
is made up of 5 strips of conductor in parallel, each strip being
3^ in. wide by 0.095 in. thick. The accomplishment of half
a turn is illustrated in Fig. 53 which represents the arrangement,
in plan, of the series field winding.
The series field current must then be
2927A.2 7 .
l * = ocx " = H70 amp.
2.5 turns
This means that with full-load current 2000 - 1170 = 830
amp. must be diverted from the series turns by a shunt con-
nected in parallel with them. This shunt is, in practice, usually
composed of German silver strips whose length is so adjusted by
test as to divert exactly the required amount of current.
In the second case, the full-load voltage with 5 per cent, over-
compounding is 1.05 X 250 = 262.5.
To obtain this voltage requires the addition of 11,700 7500 =
4200 amp.-turns to the no-load ampere-turns. This additional
64 ELECTRICAL ENGINEERING
excitation is not all supplied by the series field m.m.f., however,
since the shunt field current is affected by the increased terminal
voltage. The shunt field m.m.f. now consists of 1.05 X 7500 =
7875 amp.-turns.
Therefore, the series field m.m.f. must consist of
11,700 - 7875 = 3825 amp.-turns.
The current in the series winding is then
i t = o g = 1530 amp.
a.O
The current diverted through the shunt to the series field is
2000 - 1530 = 470 amp.
The shunt field current is
if = 17.1 amp.
Consideration of the saturation curves will show that this is
nearly the limit of over-compounding for this machine. If
full-load voltage of 275 were desired, it would be necessary to
add another half turn to each series coil.
The series field m.m.f. has been made to compensate for the
armature reaction and the ir drop (assumed 2% per cent.) in
the armature. So far as the field design is concerned, this is
satisfactory. These calculations are, however, only approxi-
mate and the actual values should now be determined from the
known data of the machine.
Armature Resistance. Being multiple wound, there are 12
paths in parallel in the armature. Each path includes 72 con-
ductors, or 36 turns. The length of a turn is twice the gross
length of the armature plus the end connections. The end
connections for one turn may be taken as 9 X diameter per pole
of the armature = 9 X 5.33 = 48 in.
Length of one turn is thus 2 X 9 in. + 48 in. = 66 in.
f\f\ \/ Q \
Length of one path of 36 turns = - - = 198 ft.
iZi
Since the area of each effective conductor section is 0.0675
sq. in., its resistance is found to be 0.142 ohm per 1000 ft. at
65C.
Resistance of one path is thus 0.142 X 0.198 = 0.02812 ohm.
Resistance of 12 paths in parallel is
"
= 0.00234 ohm.
DESIGN OF A DIRECT-CURRENT GENERATOR 65
The true armature resistance will be somewhat less than
this owing to the intermittent short-circuiting of coils by the
brushes, and its average value may be taken as
r a = 0.00226 ohm.
Voltage drop in the armature is
e a = r a l a = 0.00226 X 2017 = 4.55 volts.
Brush Resistance. There is always a drop in voltage at the
brushes due to the true brush resistance and also to the re-
sistance of the sliding contact between brushes and commutator.
This combined resistance has no definite value which may be
calculated, but it is found by experiment that the drop which
it causes amounts to 2 volts when the current density in the
brushes is 30 amp. per sq. in. or more, while for densities less
than 30, the drop is proportional to the current density. 30
amp. per sq. in. is about the usual current density in brushes.
Drop across brushes is thus 6b = 2 volts.
Series Field Resistance. Total thickness of series conductor =
0.095 in. X 5 strips = 0.475 -in. Area of series conductor =
0.475 X 3.125 = 1.485 sq. in. Mean radius of series turn,
allowing ^32 m - insulation between turns, is found to be 5.12 in.
mean radius. 3 turns + mean radius, 2 turns
Mean radius = - 2
(4.5 + 0.475 + 0.0313 + 0.233) + (4.5 + 0.475 + 0.0156)
2
5.24 + 5 R1 _.
2 = 5.12 m.
.*. Mean length of series turn = 2 X 5.12 X TT = 32.2 in.
,, , . . ,. 12 X 32.2 X 2.5
Length of series winding = r~ - = 80.5 ft. approx.
To this should be added about 5 ft. for connections between
coils, making the series winding 85.5 ft. long. Resistance
per 1000 ft. of series conductor is found to be 0.00645 ohm at
65C.
Series field circuit resistance is therefore
r a = 0.00645 X 0.0855 = 0.000552 ohm.
As it was found that only 1170 amp. go through the series
field coils at full-load, the voltage drop on the series field wind-
ing is
e a = r 8 i a = 0.00055 X 1170 = 0.645 volt.
66 ELECTRICAL ENGINEERING
Total voltage drop in the machine is therefore
e a + e b + e 8 = 4.45 + 2 + 0.645 = 7.095 volts.
or ' = 0.0284, or, approximately, 2.5 per cent, as assumed.
/oU
If the assumption of percentage drop is not considered to have
been sufficiently close, the magnetic calculations should be
repeated using the new percentage just found.
Commutator and Brushes. The size of the commutator is
determined chiefly by the brush requirements. The number of
commutator segments is 432, that is, one segment to each effective
turn on the armature.
The brushes rest perpendicularly on the commutator. There
are 12 studs of brushes, each stud holding 10 brushes. Each
brush has a cross-section of 1.25 in. X 0.75 in., giving a brush
area of 0.94 sq. in., or 9.4 sq. in. per stud.
As there are six positive and six negative studs, the area
of the positive (or negative) brushes is 6 X 9.4 = 56.3 sq. in.
Therefore the current density in the brushes at full-load is
2016.3
, a = 35.8 amp. per sq. in.
OD.o
The commutator length must exceed that of the brushes on
the stud, that is, it must exceed 10 X 1.25 + some space of
separation between adjacent brushes. In this case the com-
mutator length is 17.5 in.
The commutator diameter is influenced by the peripheral
speed. Being built up of numerous copper segments each
separated by sheets of mica, the commutator is usually mechan-
ically weaker than any other revolving part. It must not only
be protected from forces which would cause it to fly apart, but
there must be no force acting upon it which will be strong
enough to cause even slight warping of its surface. Good com-
mutation demands smooth, even contact between the segments
and the brushes at all times.
On the other hand, too small a diameter results in very narrow
segments, thin and wide brushes and then, in turn, a longer
commutator.
The commutator diameter for this machine is 39 in., which is
approximately 60 per cent, of the armature diameter. From
DESIGN OF A DIRECT-CURRENT GENERATOR 67
this it is found that the width of segment plus the mica in-
sulation is
7r39
432
= 0.284 in.
0.75
The brushes will therefore extend over Q * = 2.64 segments.
Flux Distribution Around the Armature. It is of interest at
this point to investigate the distribution of the flux around the
armature periphery on account of its bearing on the commuta-
tion and also in order to be able to determine the potential
difference between any two adjacent commutator segments.
This is best accomplished with the help of a diagram in which is
shown a pair of poles drawn to scale in relation to the armature,
developed along the horizontal line.
FIG. 54.
A curve abcde, Fig. 54, is first constructed to represent the
flux distribution around 360 electrical space degrees of the
armature periphery. This curve is based on the assumption of
flux density, being inversely proportional to the flux path in the
air. Thus, the density is uniform under the pole and is so
represented by the line ab. To determine the densities between
the poles, empirical mean flux paths to the teeth are drawn, and
the flux along each path is taken as inversely proportional to
its length. The curve cde will obviously be the reverse of
curve abc.
68 ELECTRICAL ENGINEERING
The second step is the construction of a curve of armature
magnetomotive force. This m.m.f . will act in the direction of an
axis midway between the poles (assuming brushes to be set on
the geometrical neutral).
Along this axis the m.m.f. will consist of all the armature
ampere-turns per pole. Acting through the next adjacent teeth
s, s, the m.m.f. will be diminished by the amount of armature
ampere-turns included between these teeth. These ampere-turns
may be plotted, tooth by tooth, in the manner thus indicated,
and the result will be a curve, fgh, in the form of successive steps
corresponding to the armature teeth. To construct the flux
curve of the armature reaction from the m.m.f. curve, reluctance
of the air paths alone need be considered. To be sure, the rest
of the flux path, especially that of the teeth, would have some
effect on the accuracy of the curves so obtained. But the error
would not be great, being anywhere from 2 per cent, to 8 per
cent, according to the position of the point on the curve. The
flux density for each tooth is therefore determined from the
formula:
3.19A.7 7 .
B - - z ,
where I is the length of the path in air.
1 This is plotted as curve, ijk, to the same scale as the curve of
the field flux density abode.
The actual densities along the periphery will vary from tooth
to slot, and, indeed, this variation is noticeable on many oscillo-
grams of alternator voltage. The ripples which occur in the flux
wave due to alternate teeth and slots would exist equally with
reference to the field flux, armature flux and resultant flux.
In order t6 avoid confusion the ripples have not been shown on
the armature density curve, but all the waves are plotted as
smooth lines.
A study of the resultant wave reveals the great distortion caused
by the armature current, the strengthening of the flux in the
pole tips at A, the weakening at B, and the shifting of the neutral
point, c, in the direction of rotation.
The student may well discuss fhe effect on the flux density
waves of giving a shift of the brushes.
Losses and Efficiency. The .efficiency of a machine is given
by the equation,
output
efficiency = , = ----
DESIGN OF A DIRECT-CURRENT GENERATOR 69
The full-load output of the generator has been given
P = El -*- 1000 = 500 kw.
as
The problem of the efficiency is then one of determining the
The losses of a generator may be considered under three
heads: (1) copper losses, due to heat developed by the currents in
the windings; (2) core losses, due to hysteresis and eddy cur-
rents set up by the changes of magnetic flux in the iron and, to a
slight extent, in the copper of the machine and (3) friction losses,
including that of the bearings, the brushes and windage.
Copper Losses. These consist of I 2 r loss in the armature,
the shunt field circuit including the rheostat, the series field
coils, and that of the brushes and commutator.
It is not sufficient to ascertain these losses for full-load only.
The quality of a generator is displayed by its performance at all
reasonable loads. The efficiency will in this case, therefore,
be calculated for loads from zero to 150 per cent, of full-load.
The armature copper loss is 7 a V a , where r a = 0.00226 ohm.
Shunt field copper loss is I/Ef, where E/ is the voltage impressed
on the field circuit, and is in this case 250 volts. // = 16.3
amp.
.'. Shunt field copper loss = 16.3 X 250 = 4075 watts.
The series field loss is I S E S , where I 8 is the line current =
I a If, and E a has been found to be 0.645 volts at full-load and
varies directly with // for other loads.
TABULATION OF COPPER LOSSES
Per
cent.
25
50
75
100
125
150
load
la
16
516
1,016
1,516
2,016
2,516
3,016
/a 2
256
266,000
1,037,000
2,300,000
4,075,000
6,330,000
9,100,000
/ 2 r
0.58
600
2,340
5,200
9,200
14,300
20,550
ItEf
4,075
4075
4,075
4,075
4,075
4,075
4,075
L
500
1,000
1,500
2,000
2,500
3,000
E,
0.161
0.322
0.484
0.645
0.806
0.968
I.E,
81
323
725
1,290
2,018
2,905
E b
0.6
1.2
1.8
2
2
2
I a E b
310
1,220
2,730
4,032
5,032
6,032
Total
loss
4,075
5,066
7,958
12,730
18,597
25,425
33,562
70 ELECTRICAL ENGINEERING
Brush loss = I a Eb, where Eb is the voltage drop in com-
mutator and brushes, being approximately proportional to cur-
rent density in the brushes up to a density of 30 amp. per sq.
in. and being 2 volts for higher current densities.
Core Loss. The hysteresis loss is principally in the armature
and is due to the reversal of direction of the flux in the metal as
the armature spins around. The amount of energy expended
in reversals of the magnetic molecules is proportional to the
frequency and approximately proportional to (flux density) 1 - 6 .
Thus,
Hysteresis loss = kfB 1 - 6 .
The exponent 1.6 was found experimentally, by STEINMETZ;
it holds with sufficient accuracy for the usual range of flux densities
obtained in electrical machinery.
In direct-current armatures hysteresis loss usually amounts to
about 2.8 watts per Ib. at/ = 60 and B = 64,500. Assuming this
value as standard, the armature core loss and teeth loss are ex-
pressed by the equation
W h = 2.8 X j4 X (54 KQQ) L8 X wt. of core or teeth in Ib.
For both core and teeth, / = 37.5.
B, in core = 43,600 at 250 volts, full-load. B, corresponding
to average amp.-turns required by the teeth = 126,000.
Weight of armature core = vol. X wt. of 1 cu. in.
= 0.28 X 6.05 X 7r(30J 2 - 22 2 ) = 2430 Ib.
Weight of teeth = 0.28 X 6.05 X k(32 2 - 3O7 2 ) -
216 X 1.3 X 0.465]
= 0.28 X 6.05 X [258 - 130] = 217 Ib.
Substituting these values, the total hysteresis loss in teeth
and core is
= 1.75 [0.676 1 -* X 2430 + L95 1 - 6 X 217] = 3340 watts.
The eddy current loss is due to the heating of the core by local
or eddy currents set up in the material of the core by the chang-
ing flux within it.
DESIGN OF A DIRECT-CURRENT GENERATOR 71
It is therefore an I 2 R loss, or -~-, where E is the e.m.f. set up,
which is expressed by the equation
where <j> = B X area = total flux.
From this it may be seen that the eddy current loss may be
written
We = k^B*.
The eddy current loss may be reduced as much as desired by
making the laminations of the armature core sufficiently thin.
A satisfactory value for this loss may be obtained by assuming
it equal to the hysteresis loss. In that case, W e = 3340 watts
and the total core loss is
W c = 2 X 3340 = 6680 watts.
Losses in pole faces and copper due to eddy currents are here too
small to consider.
There will also be slight changes in the values of the core
loss as the load changes, due to variation in magnetic densities,
especially in the teeth. This variation is also slight, however,
and will be neglected.
Friction Losses. Loss due to brush friction is based on a
coefficient of friction of 0.3, and a brush pressure of 1.2 Ib.
per sq. in. of brush surface. From this, the friction per sq.
in. is 0.3 X 1.2 = 0.36 Ib. Surface area of one brush = 1.25 X
0.75 = 0.9375 sq. in.
Total brush friction force is then,
F = 0.9375 X 10 X 12 X 0.36 = 40.5 Ib.
Power loss,
h w " - Hm x 746 watts -
where
r = radius of commutator in ft. = 1.625
and
n = speed in r.p.m. = 375.
Thus,
_ 2w X 1.625 X 375 X 40.5 X 746
Wb = 33,000 = 35 WattS '
1 This is the equation for induced e.m.f. in a transformer. It holds also
in this case as will appear later when the transformer is studied.
72
ELECTRICAL ENGINEERING
Bearing friction and windage, together, make a complicated
loss to determine with accuracy. This loss is, however, one
which may be assumed with quite sufficient accuracy from the
data obtained in practice. A fair assumption to make for
generators of this type is 1 per cent, of the rated output of the
generator. In this case then
W f = 500,000 X 0.01 = 5000 watts.
Summary of Losses, Output and Efficiency. The combined
losses of the generator for different per cent, loads is given in
Table V.
'60
20
50
75
% Load
FIG. 55.
100
125
150
TABLE V
The efficiency curve is shown plotted against per cent, load in Fig. 55
Per cent, load
25
50
75
100
125
150
Copper loss
4,076
5066
7 958
12 730
18 597
25 425
33 562
Core loss
6680
6 680
6 680
6 680
6 680
6 680
6 680
Friction loss ....
Total loss
8,500
19,256
8,500
20,246
8,500
23 138
8,500
27 910
8,500
33 777
8,500
40 605
8,500
48 742
Output
Input
19,256
125,000
145,246
250,000
273,138
375,000
402,910
500,000
533,777
625,000
665 605
750,000
798 742
Efficiency
86
915
93
936
939
939
Temperature Rise. The final limit to the output of the
generator is the permissible temperature rise. The effect of
DESIGN OF A DIRECT-CURRENT GENERATOR 73
temperature on copper is to increase its resistance to a slight
extent; the effect on iron is to increase its permeability. These
effects tend to offset each other so that, as tar as these two
materials go, it would be permissible to attain very high
temperatures. .
On the other hand, the insulation is the real limiting feature.
Of the many insulating materials, none possesses the com-
bination of qualities necessary in the ideal insulator for electrical
machinery. This material should be of high insulation strength,
strong mechanically, and its insulating and mechanical qualities
should not change under long-continued heating. Mica is
the best insulator in these respects, except that it is poor from
the mechanical standpoint. Asbestos is useful owing to its
heat-resisting qualities, but it is a rather poor insulator and its
mechanical possibilities are limited. Cotton tapes .and varnishes
do not withstand the high temperatures.
In attempting to extend the limit of output of machines of
a given size there are two lines along which lie the main pos-
sibilities of success.
Either some new insulating material, more satisfactory than
those at present in use, may be discovered or invented, or im-
provement in ventilation and heat radiation may be accom-
plished by alteration of the mechanical design.
Under existing conditions a temperature rise of 40C. above
that of the surrounding air is quite conservative. The tem-
perature which different parts of a machine will attain is hard
to predetermine accurately from the design. Practical studies
have afforded certain empirical constants which permit ap-
proximate determinations to be made, but in any case, practical
experience will greatly assist the designer in his attempts to keep
close to the limits.
For the present it will be sufficient to determine the watts
per square inch of surface of field spools and armature. For
rotating machinery 0.5 watt per sq. in. will correspond roughly
to a temperature rise of 40C.
The external surface of a field spool, only, should be taken,
and the same applies to the armature. These should, of course,
be calculated separately.
Problem 30. In the machine just studied, show by calculation, as indi-
cated above, that the temperature rise in the field and armature coils will
not be excessive.
CHAPTER XI
ELECTRICAL CONSTANTS OF A DIRECT-CURRENT
GENERATOR HAVING COMMUTATING POLES AND
COMPENSATING WINDING
As a typical generator of this more complex type will be taken
the following:
M.P. 6 - 1000 - 600 - 1200/1260 volts.
The generator is thus 5 per cent, over-compounded. Being
designed for comparatively high voltage, commutation becomes
a matter of special importance.
To insure proper neutralization of the armature reaction, there-
fore, special field windings are supplied, and these are so placed
as to counteract the armature m.m.f. in space as well as in
amount. That is, neutralization is accomplished by means of a
compensating winding placed in the pole faces symmetrically
with respect to the armature conductors under the pole arc, and
an auxiliary commutating pole inserted between the main poles,
where the armature magnetomotive force is the strongest, and
whose duty is not only to neutralize this magnetomotive force
about the neutral point in which the brushes are placed, but to
supply a flux which will be in proper direction to balance the
e.m.f. of self-induction of the commutated coil. With such an
arrangement the brushes are given no shift, and, consequently,
the armature m.m.f. is entirely cross-magnetizing.
The series field m.m.f. proper is thus relieved of every duty ex-
cept those of compensating for IR drop in the armature and over-
compounding. The circuit diagram of this machine is given in
Fig. 56.
General dimensions and specifications are as follows:
Armature outside diameter, 48 in.
Armature inside diameter, 28 in.
Armature gross length diameter 15.5 in.
Armature effective diameter, 11.7 in.
Armature ventilating ducts, 4% in. wide.
2% in. wide.
74
ELECTRICAL CONSTANTS
75
Slots, number and dimensions, 144; 0.44 in. X 1.53 in.
Effective conductors per slot, 6.
Effective armature conductor section, 0.55 in. X 0.09 in.
Armature winding, multiple drum.
Compensating Commutating
Series Winding Poles
Rheostat
Compensating Winding 1
FIG. 56.
Yoke section, rounded, 17 in. X 6.5 in.
Main pole core section, 14.5 in. X 14.5 in.
Main pole core length, including pole shoe, 14 in.
Main pole core length, allowed for field spool, 13 in.
Main pole arc, 17.5 in.
76
ELECTRICAL ENGINEERING
Commutating pole section, 13.5 in. X 2.25 in.
Commutating pole length, 14 in.
Main air gap length, 0.3125 in.
Air gap under Commutating pole, 0.5 in.
Shunt field winding; 2256 turns per spool of No. 15 B. & S.
triple cotton-covered wire.
Series field winding; 3 turns per spool. Each conductor built
up of 4 strips giving total section, 1.5 in. X 0.35 in.
Commutating pole winding; 5.5 turns of copper ribbon 12 in.
wide X 0.05 in. thick.
1400
1000
800
600
400
200 400 600 800
Ampere-Turns
FIG. 57.
1000
1200
Compensating (pole face) winding consists of 16 conductors per
pole contained in 8 holes in the pole face. Each hole has 2
conductors, one, a tube, the other a rod within the tube. Tube
outside diameter, 1% in., inside diameter 2 % 2 in.
Rod diameter, % in.
Commutator diameter, 30 in.
, Commutator length, 14 in.
Commutator segments, 432.
Segment width, 0.219 in.
ELECTRICAL CONSTANTS 77
Brushes per stud, 7.
Brush section dimensions, 1.25 in. X 0.875 in.
The armature flux at no-load is readily found to be 13.9
megalines per pole.
The no-load saturation curve is given in Fig. 57, having been
determined in exactly the same manner as that of the previous
machine, given in Fig. 52.
This curve shows that 7600 amp.-turns are required to give
normal voltage at no-load. At full-load, the shunt field m.m.f.
1260
will supply J2QO X 760 = 798 am P-- turns -
Assuming 2 per cent, voltage drop in armature and brushes, the
total e.m.f. which must be generated is 1.02 X 1260 = 1285
volts. From the saturation curve, this voltage requires 8750
amp.-turns. Therefore the series field m.m.f. must supply
8750-7980 = 770 net amp.-turns per pole.
To supply these, however, account must be taken of the un-
fortunate situation of the series field winding with respect to
magnetic leakage. Being placed close to the yoke, the leakage
factor should probably be 1.50 instead of 1.25 as used for the shunt
field calculation. This factor could, of course, be calculated,
but it is hardly desirable to introduce such a refinement when
the means of adjustment of the series field current render a reason-
able assumption entirely satisfactory. On the basis of a leakage
1 50
factor of 1.50, the series amp.-turns are T^ X 770 = 924.
924
The series field current is I 8 = -$- = 308 amp.
o
Current diverted around the series field is
I d = 793 - 308 = 485 amp.
The entire load current of 793 amp. passes through the 9 turns
per pole of the compensating winding, and the 5J^ turns of each
commutating pole.
SATURATION CURVE CALCULATION
No-load. E = 1200.
_ E x 1Q8
*" = 4/<
where
600 6
* = 60 X 2 = 30 '
78
ELECTRICAL ENGINEERING
6 X 144
t =
x 2 = 72 turns per pole.
1 200 ^ 1
= 13 > 880 > 000 = flux in teeth
307 X 72 X 4
and gap at no-load. The flux in the pole core is
fa = 1.25 X 13,880,000 = 17,340,000,
where 1.25 is the leakage factor. It is fairly large in this case, as is usual
when interpoles are present.
Tooth width at face =
Teeth per pole =
Teeth width (face)
Teeth width (base)
Teeth area (face)
Teeth area (base)
Gap area
Arm. core area
X48
144
- 0.44 = 1.05 - 0.44 = 0.61 in.
pole pitch
10.85 in.
X 10.85 = 10.15 in.
24 X X 1.08
25.5
= 17.75.
17.75 X 0.61 in.
44.92
48
10.85 X 11.7 = 127 sq. in.
10.15 X 11.7 = 118.8 sq. in.
3 X 17.5 X 14.5 + 127
= 222 sq. in.
44.92 - 28
X 11.7 = 99 sq. in.
Pole core area = 14.5 X 14.5 = 210 sq. in.
Yoke area = 17.5 X 6.5 X 0.95 = 108 sq. in.
No-load. E = 1200
Part
Mate-
rial
Flux
Area
B
AT fin.
Length
AT
Teeth (face)
Teeth (base) ....
Gap..
13.88
13.88
13.88
127
118.8
222
109,200
116,800
62,500
93 \
iso/ 136
19,600
1.53
0.3125
208
6,125
Arm
Sheet
Pole
iron . .
Steel
6.94
17.34
99
210
70,000
85,200
7
33.5
9.5
14
67
470
Yoke
Steel
8 67
108
80,300
30
24.3
730
Total
7,600
E = 600
Teeth (face)
54,600
3.9 1 .
Teeth (base) . . .
58,400
4.4 I 4 ' 15
6.35
Gao
3,063
Arm . . .
35,000
2.4
22.8
Pole
41,250
9
126
Yoke
40,150
8.8
215
Total
3,433
ELECTRICAL CONSTANTS
E = 900
79
Teeth (face) ....
82,000
11 Ol
Teeth (base) . . .
87,500
14 5/ 12 ' 75
19.5
Gap..
4,594
Arm . .
52,500
3 77
35 8
Pole
61 900
15 9
222 5
Yoke
60,200
15.1
367
Total
5 239
E = 1400
Teeth (face).
127,400
420 1
Teeth (base)
136,100
1,264 f 842
1,290
Gap.
7,150
Arm
81,600
10.8
102.7
Pole
96,200
79
1,107
Yoke ....
93,600
63.8
1,660
Total
11,310
FIG. 58.
Calculations of armature, shunt and series field windings, as
well as brush losses and friction loss are made in exactly the
same manner as in the preceding example. The difference in
location of the shunt and series windings is given in Fig. 58.
The division of the shunt into two coils per pole is made to
80 ELECTRICAL ENGINEERING
allow the necessary room for end-connections of the compensating
winding.
The calculation of the commutating pole winding is likewise
a matter of applying the old principle.
The conductor itself is of extreme dimensions, being a band
of sheet copper 1 ft. in width.
For the compensating winding the mean length of 1 turn
is found to be 2 X (length of pole parallel to shaft + 4 in.
(extension)) + 2 X mean span between poles, = 2 X (14.5 +
4) + 2 X 19 in. = 75 in.
Total length of winding = -^ X 8 X 6 = 300 ft.
Area of conductor section = 0.442 sq. in.
Resistance of winding = p- = -TQJ- X TTTIo = 0-0054 ohm.
Voltage drop in winding = 793 X 0.0054 = 4.28 volts.
Loss in winding = 4.28 X 793 = 3400 watts.
Voltage drops and losses at full-load in other parts of the
generator are as follows:
Voltage drop Loss
Armature ........... ........ ..... 14.75 11,700
Shunt field ........................ (1,008)
including rheostat ................ ...... 4,500
Series field ........................ 0.636 505
Compensating winding .............. 4 . 28 3,400
Commutating field ................. 1.19 945
Brushes (7 2 #) ..................... 2 1,590
Brushes (friction) .................. ...... 1,760
Hysteresis loss ..................... ....... 7,220
Eddy current ...................... 7,220
Friction and windage ............... ...... 10,000
Total voltage drop ............. 22 . 856
Per cent, voltage drop ...... .... 1 . 82
Total energy loss .... r ......... 48,840
output 1,000,000
Efficiency = = 1,048,840 =
Efficiencies for all loads are as follows:
Per cent, load 25 50 75 100 125 150
Percent, eff. 88.5 93.5 94.85 95.3 95.5 95.5
Fig. 59 shows the efficiency curve.
ELECTRICAL CONSTANTS
81
60
40
25 50
75 100
% Load
FIG. 59.
125 150
O Q O Ok
/O 000
FIG. 60.
82 ELECTRICAL ENGINEERING
EFFECT OF COMPENSATING WINDING AND
COMMUTATING POLES
To study the effect of these windings in neutralizing armature
reaction, it is best to construct a curve of magnetomotive forces
showing their distribution along the armature periphery. From
this and the curve of field flux density the resultant flux density
along the periphery is obtained. Such curves are given in Fig.
60. The armature ampere-turns and field flux density in the gap
are plotted to separate scales as was done in Fig. 54. The corn-
mutating pole and compensating winding ampere-turns are like-
wise plotted, but their direction is, of course, opposite to that of the
armature m.m.f. The resultant m.m.f. of these three is given
by the heavy irregular line. The average of this resultant m.m.f.
is seen to be very nearly zero, showing the effective compensation
of the armature reaction. It is also observable that the commu-
tating pole m.m.f. is made sufficiently strong to overbalance con-
siderably the armature m.m.f. in the neutral axis, thus creating
a resultant flux oppositely directed to the armature m.m.f.
The maximum armature m.m.f. which acts along the commu-
tating pole axis is 9564 amp.-turns. Opposing this is the m.m.f.
of the compensating winding which is 6344 amp.-turns, and the
m.m.f. of the commutating pole which is 4360 amp.-turns. Thus
the resultant amp.-turns amount to (6344 + 4360) 9564 =
1140. When the armature is in the less advantageous position
(that is, with a slot in the commutating pole axis), the resultant
amp.-turns are 10,704 - (11.5 X 797) = 1554.
The average resultant amp.-turns along the commutating pole
axis are therefore 1350.
These ampere-turns, acting through a gap of % m - produce a
flux density of
1350
B = 0.313 X 0.5 = 862 lines per Sq " in '
Commutation. If the field in the neutral axis were completely
neutralized, commutation would still be poor due to the reversal
of the current in the conductors during the period of commuta-
tion. Therefore, to balance the e.m.f. induced in the short-
circuited coil under the brush, an approximately equal e.m.f. is
created in the opposite direction in this coil by causing it to cut
through the flux due to the commutating pole.
Exact neutralization of the induced e.m.f. in the short-circuited
ELECTRICAL CONSTANTS 83
coil is practically impossible by this means. Current in the
conductors does not vary logarithmically as in an ordinary circuit
when the impressed e.m.f. is removed. If it did, the fundamental
equation, (15),
e = ir + L ^, where r and L are approximately constant,
would apply for the induced e.m.f. of the coil. But in this case,
r is by no means constant due to the varying brush surface on the
commutator segments. The value of r is therefore some func-
tion of the time. Putting r = / (Q, and considering the varia-
bility of L, due to change of permeability in the iron part of the
flux path, the induced e.m.f. would be expressed by the equation
e = 7(0 + I (Li)-
To solve this equation to a satisfactory degree of approxima-
tion, certain assumptions may be made. First, let it be assumed
that the current dies down in the coil
as a sine wave (Fig. 61). The in-
duced e.m.f. would then be maximum
when the coil axis passed through the
center of the brush.
If this maximum value were de-
termined, it could be made equal to F
the e.m.f. produced by rotation of the
coil through the field set up by the commutating pole. Other
values than the maximum could be left to care for themselves,
being of secondary importance. The maximum value of the
e.m.f. of self-induction is
where I is the current in the armature conductor at the moment
when commutation begins, and, in this case, is 133 amp. and X is
the reactance of the coil.
The second assumption is that L, the coil inductance, is
constant. Hence
X = 27r/ c L,
where f c = frequency of current during commutation.
f c = ^TTT where T c is the time of cummutation, since this time
"I c
evidently corresponds to one-half wave length. The time of
84 ELECTRICAL ENGINEERING
commutation is that time taken by the commutator to move a
distance equal to the thickness of a brush. In the machine under
consideration each brush covers four segments.
/. T c = -- X
segments covered by brush
r.p.s. total segments
1 4
= TO X 432 = ' 000952 sec '
and
2 X 0.000925 = 54 cycles per sec "
In calculating the coil inductance, L, it is not sufficient to consider
only the interlinkage of each coil with the flux which it produces.
Mutual induction is also present, the value of L desired being
therefore not strictly the self-inductance, but including that due
to the interlinkage of the flux produced by the current in all
6 turns with each single turn. In this machine the conductors
in each slot are all in parallel; thus N is 1 turn, composed of 2
conductors. It should be noted that of the 2 conductors com-
posing any turn, one of them lies in the lower half of its slot, while
the other lies in the upper half of its slot. The interlinkage of
each conductor with the total flux will not be the same in the 2
cases.
However, by considering the total flux as due to the 67 amp.-
turns of a slot acting through an effective magnetic conductance,
G, and surrounding each of the 6 conductors, the inductance
thus calculated will be correct, provided the proper value of G
is determined. Thus,
<t> = 67 X G,
where < is the equivalent flux surrounding all conductors in 1
slot. (The inductance due to end-connections must also be
ascertained, as is done later.)
The magnetic conductance per centimeter effective length of
the armature is calculated by means of the general formula,
area
x
Considering the magnetic circuit (Fig. 62), it is seen to consist
of 3 parallel paths in air, namely: that of section A and length
B through the conductors, that of section C and length B above
ELECTRICAL CONSTANTS
85
the conductors, and that of section F and length D, from the top
of 1 tooth to the top of the other. The common path through
the iron may be neglected as offering comparatively little
resistance.
To find the effective magnetic
conductance per centimeter
length across the section A , con-
sider an elementary section, dx,
at distance, x, from the bottom
of the conductors. The con-
ductance across this section is '
QAirdx
B
FIG. 62.
The amp. -turns acting in this conductance are T-, where / is
**
in amperes.
Therefore the flux set up through dx is
d<f>
QAirdx
~B~
X
2.47T/ x dx
AB '
This flux interlinks with only -r- conductors.
A.
Nd + = - L -AJ8 '
14.47T/ r\ ,
z 2 cfo
Thus,
and
14.47T/A
SB
is the interlinkages of the flux with all 6 conductors.
Thus, if the flux is considered to be due to 67 amp.-turns
acting through conductance, g, and this interlinks with each
conductor, the total number of interlinkages is
14.4rr7A
whence,
14.47T/A
~ 3B/X36 ~ 3B
The other two paths are entirely outside of the conductors,
86 ELECTRICAL ENGINEERING
and hence are acted on by all of the ampere-turns in the slot.
These conductances are then, respectively,
0.47T ^ and 0.47r y:-
Hi U
The total effective conductance is then
A . C . F
per cm. length of armature, and the flux per slot is
6 = ING X 2.54?
where I is the effective length of the armature, in inches. For
both slots and 1 complete turn the inductance is
<t>N 2 X 6X3.2 X 11.7in.rA C F
a ~ 10 8 7 10 8
Substituting numerical values for the slot and tooth dimensions,
A = 1.233 B = 0.44 C = 0.213
D = 1.047 E = 0.51 F = 0.607,
this inductance becomes
L, = 449 [0.934 + 0.417 + 0.58] X 10~ 8 = 867 X 10~ 8 henrys.
To this must be added the inductance of the end-connections.
The flux produced by the end-connections per ampere-turn per
inch of coil length may be taken as one-twentieth of that in the
slot.
It is, therefore, ^ X 3.2 X 1.931 = 0.309 lines.
zo
The coil divides as it passes out from the slot, so that only
3 conductors 3 conductors are grouped together. There-
fore there are 37 amp.-turns producing flux
around each conductor. If the length of the
end-connections for 1 turn is assumed as 8 X
diameter per pole, = 8 X 8 = 64 in., the flux
surrounding each turn is </> = 0.309 X 37 X
FIG. 63. 64 = 59.47 lines. The inductance is then
T <t> N 59.4 X 1 X 7
Le = TxW= 10* X 7 = 59.4 X 10- henrys.
The total inductance is thus
L = L, + L e = (867 + 59.4) X 10~ 8 = 926.4 X 10~ 8 henrys.
ELECTRICAL CONSTANTS 87
The reactance of the short-circuited coil is
X = 2irf c L = 6.28 X 540 X 926.4 X 10~ 8 = 0.0314 ohm,
and the maximum e.m.f. of self-induction is
E m = IX = 133 X 0.0314 = 4.17 volts.
To overcome this e.m.f. the short-circuited coil is made
to rotate in the field of the commutating pole. This field
has been found to have an average density around the neutral
axis of about 8620 lines per sq. in.
In this case, the commutated coil has only one turn. Thus
e.m.fs. are generated in one conductor under a " north" com-
mutating pole, and in the other conductor under a " south"
commutating pole. These e.m.fs. are similar, and together
make up the total e.m.f. generated in the coil by rotation in the
commutating field. The maximum value of this induced e.m.f.
corresponds to the rate of cutting the flux in the center under the
commutating pole.
Consider a small distance, dx, Fig. 64, at this
point. The flux through the area of width, dx,
and average length, 11.7 in., of the iron in field
and armature is FIG. 64.
d<f> = 8620 X 11.7 dx = 101,000te.
The speed of conductors at the armature periphery is irD X
r.p.s. = TT X 48 X 10 = 4807T in. per sec.
The time required for a conductor to go the distance dx, is
dT-
dl 480ir
,
* induced = 10^ - J^_ = L53 V ltS
J 4807T
per conductor, or 3.06 volts per coil.
This voltage opposes that due to self-induction, leaving
as a resultant,
4.17 - 3.06 = 1.11 volts
acting in the circuit.
Since experience has taught that two volts potential difference
can be taken care of by the resistance of the carbon brush no
difficulties from sparking need be anticipated.
CHAPTER XII
DIRECT-CURRENT GENERATORS IN PARALLEL
AND SERIES
Shunt generators operate in parallel without the slightest diffi-
culty. Generator No. 1 is first started and thrown on the line.
Generator No. 2 is then brought up to about normal speed, the
voltage is adjusted and the line switch is closed. Since genera-
tor and line voltage are the same, no-load is taken by generator
No. 2. By adjusting the field excitation of No. 2 the generator
takes the desired share of the load. As its load increases its
engine slows down, the governor opens and the speed is restored
to normal.
Series generators do not operate naturally in parallel. Assume,
for example, that two series generators are in parallel, each
taking its share of the total load. Suppose then that for some
. . reason the voltage of No. 2 (Fig. 65)
_ J j becomes slightly reduced. Its share of
the load will fall off proportionately
and, with this, its field excitation.
Falling off of the field excitation
further reduces the voltage and, con-
sequently, the load, the excitation,
and so on. The current is reduced to zero, then reversed in
direction in both the field and the armature coils. The rotation
of No. 2 remains the same, but the machine now acts as a series
motor driving its engine. In practice, the rush of current dur-
ing this period when the counter e.m.f. of generator No. 2 has
been destroyed is so great that the circuit is opened by its fuses
or circuit breakers. Series generators are not in common use,
but this principle of instability in parallel operation applies
equally to compound generators through their series field
windings.
With shunt generators there is no such instability. If the
voltage of No. 2 falls off, its current likewise is reduced. But the
effect of reduced current is to lessen the armature reaction, thus
88
DIRECT -C URRENT GENERA TORS 89
bringing up the voltage. The shunt field current is not affected
since it is derived from the bus bars.
Series generators and, more particularly, compound generators
may be made stable in parallel operation by the use of an
' 'equalizer bus." This consists of a very heavy copper connection
situated, as shown in Fig. 66, between the inner terminals of the
series field circuits of the two (or more) generators. If, now,
the voltage of No. 2 becomes reduced to a slight extent, current
will flow from the + brush
of No. 1 through the
equalizer and into the series
field coils of No. 2, main-
taining the strength of the
field of the latter.
If the two generators,
in normal operation, do F
not divide the load prop-
erly in the proportion of their respective ratings, this may
be . corrected by inserting resistance in the series field circuit
of that generator which takes too much of the load. The
effect of the equalizer is to put the series field coils always in
parallel. The voltage across these coils is therefore the drop
between the positive brushes and the positive bus. The re-
sistance of the equalizer is so low that its drop is negligible, so
that the drop across all the series field coils is the same. Putting
a shunt or diverter around one of the series field coils has no
effect on the distribution of the load on any particular generator,
as it affects all the series field currents alike, the proportions
remaining the same.
Direct-current Generators in Series. No inherent difficulty
is encountered in connecting direct-current generators in series.
Owing to the limited possibilities of constructing commutators
that will permit the generation of very high voltages, where
these are required in direct-current machines recourse is usually
had to series connection.
In electric railway work it is the general rule to employ both
series and parallel connection of the motors to give flexibility
in speed control.
The Three -wire System. Two generators in series afford the
simplest means of obtaining the three-wire system. This system,
invented by EDISON, was devised to enable the use of large
90
ELECTRICAL ENGINEERING
E
numbers of low voltage incandescent lamps without, at the same
time, entailing the use of a prohibitive amount of copper in the
distribution system. As seen in Fig. 67, the voltage of the
system is 2E t while that across any ele-
ment of the system is only E.
There are other ways by which power
may be supplied to such a system. Thus,
the source of power may be a single gen-
erator of voltage, 2E, across whose termi-
nals may be connected either a storage
battery, as in Fig. 68, or two small gener-
ators mounted on the same shaft, called a
FIG. 67. balancer, and shown in Fig. 69. In either
case the necessary condition is to have available some connec-
tion point the potential of which is intermediate between those
of the outer wires. The amount of current actually flowing in
FIG. 68.
FIG. 69.
either the battery or the balancer set is small in case the two
sides of the load are reasonably well balanced.
Another scheme consists in the use of the three-wire generator.
This is illustrated in Figs. 70 and 71. Fig. 70 shows a bi-polar
machine constructed by reversing the windings on two adjacent
FIG. 70.
FIG. 71.
poles of a four-pole generator. The potentials of the two
brushes on the horizontal axis are the same and are midway
between the potentials of the two other brushes. The object of
making the machine bi-polar is to give an intermediate inactive
DIRECT-CURRENT GENERATORS 91
belt along the commutator on which a brush may be placed
without causing disruptive sparking. A better scheme is 'that
of DOBROWOLSKY shown in Fig. 71. The armature is tapped at
two opposite points which are connected, through slip rings, to a
" choke" coil, which is simply an induction coil. This coil is
wound upon a laminated iron core, and therefore is of high in-
ductance. The e.m.f. impressed upon it is evidently alternating,
and therefore very little alternating current can flow through
the coil.
The middle point of the coil must always be at a potential
midway between those of the brushes. It may therefore be
connected to the middle wire of the system.
The disadvantage of using a battery is that some cells may be
called on to supply more energy than others. It then becomes
difficult to keep the battery uniformly charged, and deteriora-
tion results.
No such difficulty occurs with the use of balancers. They
may be small, inexpensive machines, which when running idle
take only a small current.
As an example of the use of balancers and the economy of the
three-wire system, consider the circuit illustrated in Fig. 72.
20
FIG. 72.
The load consists of 40 amp. on the upper branch and 30 amp.
on the lower. The system is therefore unbalanced. Currents
and directions of flow are indicated for each portion of the circuit.
The current in the middle or neutral wire varies, being 10 amp.
in some sections and in others. Let it be assumed that the
current required to run the balancer set is 1 amp. which would
be indicated, if shown in the figure, by an arrow pointing down-
ward in the balancer set. The current returning to the balancer
over the middle wire is 10 amp. This current divides equally,
5 amp. flowing upward in balancer A, combining with its down-
ward flowing 1 amp. to give 51=4 amp. in A, and 5 amp.
92 ELECTRICAL ENGINEERING
flowing downward in B, combining with its downward flowing 1
amp. to give 5+1=6 amp. in B. Current in A flows similarly
to that in the main generator. Thus A acts as a generator,
supplying 4 amp. to the load. Current in B flows in the opposite
direction; thus B acts as a motor and drives A. The difference
in current between that in B and that in A is 2 amp., which,
when multiplied by E, the voltage across B } gives 2E, the power
required to drive the balancer set.
If the generator voltage be assumed as 200 (that is, 2^ = 200),
then the generator output, or rating, if this be full-load, is
200 X 36 = 7.2 kw. The balancer, A, rating, as a generator,
is 100 X 4 = 0.4 kw. ; the balancer B, as a motor, receives input
= 100 X 6 = 0.6 kw. The line drop from the generator to the
load is (40 -+- 30) r = 70r, where r is the resistance of each of the
outer wires. The line loss, in transmission, is (40 2 -f-30 2 )r
= 2500r.
If the entire load were on the two-wire system, the current in
each wire would be 70 amp., the line drop, using the same size
wires, would be 140r, and the line loss would be (2X 70 2 )r
= 9800r. Comparing the two-wire system, using the same size
of outer wire,
drop, three-wire _ 70 _
drop, two-wire ~"~ 140
loss, three-wire _ 2500 _
loss, two- wire ~ 9800
The middle wire, carrying 10 amp., has no effect on the
total drop between the outer wires. It does have some effect
in slightly unbalancing the voltage of the two branches of the
system. Thus, assuming the voltage across the two machines
of the balancer to be exactly equal, which is very nearly true,
and taking this voltage as E, the voltage across each branch of
the load may be found. Across the upper branch it is,
E - 40r - lOr = E - 50r.
Across the lower branch the voltage is
E - 30r + lOr = E - 20r.
The amount of unbalancing of the voltage is therefore
(E - 50r) - (E - 20r) = 30r.
DIRECT-CURRENT GENERATORS 93
To get a concrete idea of the amount of this unbalancing, let
SO
the line drop, 70r, = 10 per cent. Then 30r = ^ X 0.1 =
0.043 = 4.3 per cent.
When the load consists of lamps it is necessary that the
two branches shall be sufficiently well balanced to prevent
excessive variation in voltage. This is usually very easily
accomplished.
The middle wire adds, directly, a small amount to the line
loss. In this instance, the loss in this wire is 10 2 r = lOOr.
The total loss in the system is therefore 2600r, and the ratio
2600
of losses of the two systems is QQnn = 0.265.
Where the percentage line drop or the percentage line loss is
specified, and must be the same with either system, the ad-
vantage of the three- wire system is in the saving in the cost of
copper. On that basis, let the calculations as already carried
out for the three-wire system be assumed as fulfilling the re-
quirements, that is,
Line drop = 70r.
Line loss (two-wire) = 2500r.
The two-wire system, to give equal line drop must be com-
posed of wires determined by the equation,
2 X 70 X r' = 70r,
where r' = resistance of one wire of the two-wire system, and
r, as before, is the resistance of one of the outer wires of the
three- wire system. Then
and each wire of the two-wire system will be twice as large as
each outer wire of the three- wire system. Assuming the middle
wire equal to the outer wire, the two-wire system will require
four-thirds as much copper as the three- wire system.
Since, however, the variation in voltage . is felt by all the
lamps on the two-wire system, while on the other system ap-
proximately one-half the variation in voltage is felt by each
branch, it is more reasonable to calculate on the basis of equal
percentage drop in the two systems.
94 ELECTRICAL ENGINEERING
Percentage drop, three-wire, = = 35 -
For equal percentage drop, therefore, the two-wire system
will require eight-thirds as much copper as the three-wire
system.
On the basis of equal power loss in the outer wires,
9800r' = 2500r,
' r ~ 9800 "
Adding the middle wire, equal to an outer wire, the two-wire
system will require n 055 _u n 12Y5 = 2.61 times as much copper
as the three- wire system.
Problem 31. What saving in copper does the three- wire system give
over the two-wire system, when the load is balanced, on the basis of (a)
equal percentage line drop, (&) equal line loss?
1. Middle wire equal to outer wire.
2. Middle wire one-half of outer wire.
3. Show that with a balanced load no current flows in the middle wire.
Problem 32. One hundred 60-watt tungsten lamps are to be supplied
with power at 3 per cent, line loss. The line length is 600 ft. Lamp
voltage is 120. The neutral wire is to be one-half the cross-section of
each outer wire.
Find the size of the required wires, and show that the weight of copper
is approximately 190 Ib. Show that on the two-wire system 610 Ib. would
be required.
Feeder
/ T f Trolley Wire
4
rf
/
E5
iiUj
Rail EEL
COJQ
FIG. 73.
Boosters. Generators are frequently connected in series for
the purpose of regulating the voltage and equalizing it along a
line in which there is considerable voltage drop.
Fig. 73 shows a simple arrangement of a street railway circuit
in which a booster is used. The generator, G, supplies power
DIRECT-CURRENT GENERATORS
95
to the system, including that delivered directly to the car and
that used in driving the motor M . The motor and booster form,
usually, a directly connected set. One terminal of the booster is
connected to the trolley wire at the station, the other is con-
nected through a heavy feeder to some distant point on the
trolley wire.
As an example of the effect of using a booster, consider the
following :
Problem 33. A trolley line 3 miles long is supplied with power by a
generator at 600 volts. The trolley wire is of No. 00 B. & S. wire, having a
resistance of 0.4 ohm per mile. The rail return has a resistance of 0.05
ohm per mile. A feeder, consisting of three No. 0000 B. & S. wires, of
0.087 ohm per mile, extends from the station to a point 2 miles distant,
where it connects with the trolley wire. The booster voltage is maintained
at 40. Find the voltage on the car as it proceeds from the distant end of
the line toward the station, assuming that the current taken is at all times
200 amp.
Solution. It will be of interest, first, to determine the voltage on the car
at the distant end when the booster is
disconnected.
Drop in the trolley wire is 200 X 0.4
X 3 = 240 volts.
Drop in rail = 200 X 0.05 X 3 = 30 v.
volts.
.'. Voltage on car without booster =
600 - 270 = 330.
This is to illustrate the necessity of doing something to improve the regu-
lation of the line. With the booster connected, the problem becomes one
for the application of KIRCHOFF'S laws.
The circuit is represented diagrammatically, in Fig. 74, for the case of the
car at the end of the line. Arrows indicate arbitrary directions of flow of
current. Let the voltage on the car be denoted by E.
By KIRCHOFF'S laws,
ii + 12 = 200
and
40 - ; 2 r 2 + iiri = 0,
whence, eliminating i\ between the equations,
200r 1 + 40
iz = :
74
Substituting values,
ri = 0.4 X 2 = 0.8 ohm
r 2 = 0.087 X 2 = 0.174
r 3
r
= 0.4
= 0.05 X 3 = 0.15
200 X 0.8 + 40
-^8Tol74-= 2
200 - 205 = - 5 amp.
96 ELECTRICAL ENGINEERING
The equation of the mesh composed of the generator, ri, r 3 , E and r is
600 = iiri + 200r 3 + E + 200r .
Substituting values and solving for E,
E = 600 + 4 - 80 - 30 = 494 volts.
Thus, there is a total drop of 106 volts instead of 270 volts without the
booster.
Now let the car be at the point, O, where the feeder joins the trolley wire.
Evidently the same equations hold, and z' 2 = 205 amp., ?i = 5 amp.
r is now 0.05 X 2 =0.1 ohm.
The mesh equation is now
600 = iiri + E + 200r .
/. E = 600 + 4 - 20 = 584 volts.
When the car is at a point 1 mile from the generator, the current and
voltage equations are :
^ + 12= 200
and
40 - i, (r, + + ti = 0.
Solving for i 2 gives
i = 123 amp.
and
ii = 200 - 123 = 77 amp.
The mesh equation of voltage is
600 = ii ^ + E + 200r ,
where r is now 0.05 ohm.
/. E = 600 - 31 - 10 = 559 volts.
Thus, it is seen that, by this simple connection of the booster to a point
chosen more or less at random, the voltage has been rendered much more
nearly uniform than it would be without the booster.
Problem 34. As a further study of the booster problem, consider that in
the above case the feeder is to be connected to the trolley wire at two points,
namely, at 1.5 and 2.5 miles from the generator.
Find the voltage on the car at each half-mile point, and plot against
distance from the generator.
CHAPTER XIII
DIRECT-CURRENT MOTORS
If two shunt generators connected in parallel supply power to a
certain load, as in Fig. 75, the division of the load between the
generators will depend upon their respective degrees of excitation.
By weakening the field of No. 1, it will take less of the load
until, by continued weakening, it takes none at all and finally
receives current from No. 2, thus being run as a motor.
With a change in direction of flow of current in the armature
comes a change in the direction in which the armature tends to
rotate due to its current, the direc-
tion of the field remaining constant
in shunt machines.
As a generator the rotational force
of the armature is counter to the
actual direction of rotation which is
, ., , . . . TT FIG. 75.
due to tne driving engine. However,
the actual direction of rotation does not change when the
machine ceases to act as a generator and becomes a motor.
With the series generator, reversal of the armature current
also reverses the field. To obtain a generator action from a series
motor, therefore, requires reversal of rotation.
It has been shown that, with generators, a forward shift of
the brushes increases the armature demagnetization.
With a shunt motor the armature currents are reversed, the
armature ampere-turns are reversed, and the effect of the arma-
ture, in shifting the resultant flux, is consequently reversed.
Therefore, the brushes of a motor require to be given a backward
shift. The effect of a backward shift on a motor, like the for-
ward shift on a generator, is to increase the armature demag-
netizing ampere-turns.
With direct-current motors, the impressed e.m.f. is the sum of
the counter e.m.f. and the ir drop.
Thus, the fundamental equation is
E = Ei + ir
7 97
98 ELECTRICAL ENGINEERING
where E = impressed e.m.f.,
Ei = counter e.m.f.,
i = current,
and
r = resistance of armature, brushes, etc.
The generator equation (20) also applies to the counter e.m.f.,
since the counter e.m.f. is the generated e.m.f. of the motor due
to the rotation of its armature conductors in the field.
where
, 4<
= 105'
Substituting this value of E i}
E = kf<f> + ir (24)
whence
f = frequency. To transform frequency to speed,
r.p.m. p
J ' 60 *\i*
where
p = number of poles.
For ordinary operation,
pi
f = 7' approximately.
There are three ways of changing the speed of a direct-current
motor: (1) by changing E, the impressed voltage; (2) by changing
<f> by means of a field rheostat ; (3) by changing </> by shifting the
brushes.
Shifting the brushes is not an effective means of speed regula-
tion since it introduces trouble from sparking at the brushes.
Types of Direct-current Motors. The principal types of
direct-current motors are known as shunt, series, cumulative-
compound, in which the series and shunt turns act in the same
direction, and differential-compound, in which the two field
m.m.fs. are arranged to oppose each other.
DIRECT-CURRENT MOTORS 99
Speed Characteristics of Direct-current Motors. These are
curves between speed and load, the latter being the independent
variable. To determine the effect of load upon speed, in the case
of shunt motors, it is seen from Eq. (25),
_ E - ir
* = ~
that an increased ir drop tends to reduce the speed. It has also
been shown that <f> is reduced by armature reaction, in pro-
portion, roughly, to the load. Therefore, for shunt motors, the
relation between the armature reaction and the ir drop will de-
termine whether the motor will speed up or slow down with an
increase of load. In general, if the magnetization of the field
extends above the knee of the saturation curve, the motor will
slow down, while below the knee the motor will speed up. Evi-
dently, a degree of magnetization might be obtained which would
result in practically constant speed.
The cumulative-compound motor slows down with increase
of load, since the effect of the series turns is to strengthen the
field.
The differential motor speeds up
with increasing load, due to the op-
position of the series and shunt field
m.m.fs.
The series motor speed is governed
almost entirely by its field, which is
nearly proportional to the load cur-
rent. At light loads, the speed be-
comes high and the operation of the motor is unstable. In Fig.
76 is shown a set of speed characteristic curves.
The student should be able to establish the general speed
equations and derive curves for each type of motor.
Power and Torque. Power input to the motor is obtained by
multiplying Eq. (24) by i, thus
Wi = Ei = Ej + i*r = kf<(>i + i 2 r.
In this, equation E# represents the output of the motor in
mechanical work, including bearing friction and windage; i 2 r is
the power lost as heat developed in the armature.
Expressed in horsepower, the output is
Load
FIG. 76.
100
ELECTRICAL ENGINEERING
Horsepower may also be expressed as
2irRnF
P' ~ 33,000
where R = radius of armature in feet,
n = revolutions per minute,
F = force in pounds on the armature conductors.
2wn = co = angular velocity,
and RF = T = torque.
Thus
and
746 33,000
33,000 Ed
27rnX746'
But output is also, by (24), kf<f>i.
. ^ 33,000 kfoi
= 2irn X 746 '
Also, since / =
T =
where p = number of poles,
0.0587 kp4>i =
120
33,000 kp<t>i
27rX 746X120
This expression may be reduced still further, since
Thus, for
where t = number of turns in series on the armature.
a motor of p poles and t turns in series,
T = 0.2348 tp<t>i X 10~ 8 ft.-lb.
Torque Characteristics. From the above equation of torque
it is possible to construct curves showing torque variation with
load current. It is necessary, however,
to be able to find the value of in
each case. With shunt motors </> is
nearly constant, and torque is therefore
nearly proportional to current. With
series motors <f> increases with i, and
torque therefore goes up as the square
of the current, approximately. l Fig. 77
gives a set of torque characteristics for
the four types of direct-current motor.
1 When the field core becomes saturated, increase of current does not
produce much increase of flux. Under heavy loads, therefore, the torque of
a series motor increases more nearly in direct proportion to the current.
Current
FIG. 77.
DIRECT-CURRENT MOTORS 101
Problem 35. Direct-current motors and generators being entirely
similar as respects fundamental equations, armature reaction, etc., it is
thought best to submit to the student the problem of the direct-current
shunt motor instead of presenting it here in detail. Let the generator
whose design was worked out in Chap. X be now considered as a shunt
motor. The series turns will then be disconnected. With 250 volts im-
pressed on the armature and maintaining constant shunt field amp. -turns
of 7500, let it be required to calculate the speed and plot its values against
those of the load current.
Choose current values of 0, 1000, 2000, 3000 amp. Assume a constant
brush shift of 15.
The fundamental speed characteristic, Eq. (20), has been found to be
E - ir
'--*r-
number of poles X r.p.m. pn
where 2X60 = 120'
E = impressed voltage,
r includes both armature and brush resistance.
where t = number of turns per pole on the armature and $ is the flux cutting
the armature conductors. For this last it is sufficiently exact to assume
$ at load amp. -turns at load
at no-load ~~ amp. -turns at no-load
(See armature reaction, Chap. X.)
Problem 36. The same problem as the preceding should now be worked
out, using (1) E = 270 volts, (2) E = 220 volts.
Question. What, in general, is the effect on shunt motors of increasing
or lowering the terminal voltage, as regards (a) speed, (6) torque, (c) output,
(d) efficiency?
Problem 37. Let the above motor be calculated as a differential-com-
pound machine, the series ampere-turns to be so adjusted as to give the
same field strength at full-load as at no-load.
Plot speed vs. armature current for impressed voltage E = 250.
Problem 38. Same as 37 only the motor is to be connected as cumulative
compound.
Problem 39. If, now, the entire field strength were determined by the
series turns, so that at full-load there should be 10,427 series amp.-turns, 1
calculate and plot the speed for variation of load.
Series field circuit resistance may be taken as 0.00134 ohm.
Problem 40. In problems 35, 37, 38 let the speed be maintained con-
stant by variation of the shunt field current. Let this speed be that of the
shunt motor at no-load (E = 250).
Plot curves between field current and load current.
Problem 41. Show how to obtain constant speed by shifting the brushes,
and work out numerically, as far as possible, the case of the shunt motor.
Plot a curve between degrees of brush shift and load current.
1 Same as required for the generator at full-load, Chap. X.
CHAPTER XIV
N
FIG. 78.
THEORY OF THE BALLISTIC GALVANOMETER
This particular type of galvanometer is of importance in
magnetic measurements, especially in the determination of the
hysteresis loop.
It consists, usually, of a coil of
fine wire wound upon a steel cylin-
* der, freely suspended between the
poles of a magnet as illustrated in
Fig. 78.
It has been shown that the force
exerted on a wire carrying current,
when placed in a field perpendicular to the lines of flux, is
F = Eli dynes,
where i is current in abamperes, I is length of wire in centimeters
and B is flux density in lines per square centimeter.
If the wire is one side of a rectangular loop, then the turning
couple of the loop is
C = 2pBli dyne-cm.
When the loop is displaced by an angle,
0, from the direction of the flux lines (Fig.
79), the couple is
C = 2pBli cos 6 = ABi cos 6,
where A = 2pl = area of the loop.
When the current is sent through the
loop, the action of the couple produced is
to turn the loop through an angle, 0. In order to oppose this
action, a spring is so attached to the loop as to introduce an op-
posing couple, k&, which balances the swing of the loop. Then
ke = ABi cos
and
K0
AB cos 6
where k is a constant of the spring.
102
Force
FIG. 79.
THEORY OF THE BALLISTIC GALVANOMETER 103
For small angles, 6 = sin 6. Substituting this,
k sin 6 k
cos
tan
Thus, the current in the loop is directly proportional to the
tangent of the angle of deflection; hence, the " tangent"
galvanometer.
For small angles, also, 6 = tan 6.
Thus, the galvanometer may be used as an ammeter to
measure directly the current, so long as the angle of deflection is
kept small.
In the ballistic galvanometer the moving part is designed
to have much inertia, so that its natural period of vibration
shall be long in comparison with the time of change of the flux
to be measured. Thus, a change of flux, produced in a sample
of iron under test by altering the number of ampere-turns on
the iron, will take place before the loop can move, that is, while
= and cos 0=1.
The couple on the loop is then
C = ABi,
which causes the loop to accelerate.
Therefore, ABi is the couple of angular acceleration, and
where /o = moment of inertia of the moving element, and o> =
angular velocity.
But idt is the quantity of electricity flowing in any time, dt.
Therefore the total quantity
... /o | , h <*>
J ldt = AB J d " = ~KB
(26)
AD
where a> is the final velocity attained.
The deflection is, however, limited by k0, the torsion of the
spring. The work done in overcoming this torsion is then
W = kOdd =
where is the maximum deflection.
= (k
104 ELECTRICAL ENGINEERING
Solving this equation,
I/O
t/o ^OA/
But by (26),
0)fl
Substituting this value of co ,
whence,
Q = ^^ (27)
In this equation all terms are constant except , the maximum
deflection of the loop.
If, now, the change of flux is d<f> in a time dt, the e.m.f . induced
in the coil surrounding this flux is
N_ d$ _ .
10 8 dt '' IT)
where N is the number of turns of the coil, r is the resistance
of the circuit, and i is the current set up in the circuit.
Then, transposing,
Nd<t>
idt = -
r X 10 8 '
The total quantity of electricity set flowing by the change
of flux is then
r X 10 8 L r X 10 8 '
whence, from (27),
Qr X 10 8 _ 6 r X 10 8 \/J7fc
N ABN
is the maximum value of the flux.
There is thus a direct relation between flux and maximum
deflection, and 0o is therefore a measure of the flux.
CHAPTER XV
VECTOR REPRESENTATION OF ALTERNATING-
CURRENT WAVES
In Chap. VIII the graphical relationships of the waves of
voltage and current in an alternating-current inductive circuit
have been developed, and the values and meaning of average
and effective values of a sine wave have been discussed.
The waves of Fig. 37 may also be represented as vector pro-
jections of their maximum values on the vertical axis, as shown
in Fig. 80. Since i = I m sin the length of ^ r
the current vector is taken as I m and the
value, i, at any instant, is the vertical pro-
jection of I m as it uniformly rotates, at speed
2-7T/ about the origin. The vectors all have FIQ ^
the same speed of rotation so that their re-
lations to each other are constant. Hence their position in
space at any desired instant may be chosen. Let that instant
be when = o, in Fig. 37. Then i = I m sin = o, and I m
must be laid off horizontally. rl m , the maximum value of the
e.m.f. consumed by the resistance, since it is in time-phase with
I m , is also laid off horizontally; xl m , the maximum value of the
e.m.f. consumed by the inductive reactance, x, is 90 ahead of
I m , and is therefore laid off vertically upward. Thus xl m
is positive maximum when I m is at zero, becoming positive.
rl m and xl m may now be added vectorially, giving ZI m or
E m which is the maximum value of e. E m is seen to be placed
at an angle a ahead of I m , such that tangent a = -y =
J- m '
This relation is also of fundamental importance. The numerical
value of E m is obtained by the relation
E m = Vl m 2 r* + I m V = I m Vr 2 + z 2
The quantity \/r* + x 2 is called the impedance and is denoted
by the letter z.
105
106 ELECTRICAL ENGINEERING
Problem 42. Draw the vectors of e.m.f. and current of problem 28,
Chap. VIII, and show that the angle of lag of current behind e.m.f. is
38 40'.
In the representation of waves by vectors, the vectors are not,
in reality, moved, but their relative positions in space are con-
sidered. Since no rotation is required, they may therefore be
drawn in length equal to their effective values, and this is the
common method of representation.
Also, since (7 m z) 2 = (I m r)* + U m x) 2 (28)
2 = r 2 _|_ 3.8 (29)
and the vector relationship holds for (29) as for (28). There
can be constructed what is known as the im-
pedance triangle, Fig. 81, in which a = r, b =
b x
x, c = z, and tan a = - =
Thus,
FIG ' 8L
x = z sin a.
Substituting these values in (19), e = I m (r sin 6 + x cos 6),
gives,
e = 7 m 2(sin B cos a + cos 6 sin a)
= I m z sin (8 + a),
or, substituting for 6, its equivalent, coZ,
e = I m z sin (ut + a), (30)
in which at is a variable angle depending on t, and a is a constant
angle determined by the relative values of x and r. Eq. (30)
shows that e, like i, is a sine wave quantity, but that there is a
constant angular or phase difference, a, between them, a is
called the angle of lead or lag, depending on whether it is posi-
tive or negative.
The relations indicated in Fig. 81 may also be expressed by the
notation of complex quantities.
Thus,
c = a -f jb.
The addition of the letter j to the equation simply means that
the quantity, 6, is to be drawn vertically upward. If it were
j, b would be drawn vertically downward. A dot is put under
ALTERNATING-CURRENT WAVES 107
c which means that c is dealt with as a vector quantity. Without
the dot, the scalar or numerical value of c, only, is meant.
Thus, _
c = \/a 2 + V.
Problem 43. Show graphically that 3-J3 is a vector of length 4.24,
which makes an angle of 45 with the horizontal axis. Show that a vector
of length 12, at angle 120, is represented by the expression, 6 + j 10.4.
Calculate and draw the following vectors : c = 3 j'2, c = 4 + j, c =
-2 + j3, c = -4 -J2.
j also means a rotation of 90 in the positive or counterclock-
wise direction. If the vector, a, is multiplied successively by j,
several times, its direction is shown as follows:
Vector Angle
....
90
J u
jja a,
180
270
iiiia = a. .
. . 360 =
Thus may be written,
whence
or, j is identical with i, used commonly in mathematics to denote
imaginary quantities. 1
If it is desired to rotate a through 30, we can write
a = a cos 30 -f j a sin 30.
To rotate correspondingly, ^~~ ' a 8in30
a = a cos a + j a sin a,
/ o i o\ FIG. 82.
= a (cos a -f j sin or).
Suppose a is first rotated 30, then 60 more. Then a = a (cos 30
+ j sin 30) (cos 60 -f- j sin 60).
Problem 44. Prove that this double rotation results in
a = ja
Consider the simple case of alternating current in an inductive
resistance, Fig. 83, where current, /, resistance, r, and reactance,
1 In electrical engineering j is used instead of i, because i is used to denote
current.
108
ELECTRICAL ENGINEERING
x, are known. / is chosen as the zero vector. Then I = i.
Frequently it is well to choose as the zero vector, or vector drawn
at 0, some known quantity. In order to determine the positions
of the vectors of electromotive force, etc., with respect to the zero
vector, there are two rules, previously brought out, which are
important to remember:
Rule I. The e.m.f. consumed by resistance is in time-phase
with the current, and in the same direction.
Rule II. The e.m.f. consumed by inductive reactance is in
time-phase 90 ahead of the current
By these rules may be drawn the vector diagram, Fig. 84, in
which the vector sum of ix and ir is iz, which is the total electro-
motive force consumed.
FIG. 83.
FIG. 84.
This electromotive force consumed, or vector E, numerically
equal to iz, is represented by the relation
E = ir + jix i(r + jz).
The impedance is thus expressed as r + jx, and it is a vector of
magnitude z = \/r 2 + x 2 , and the angle between the impedance
and the resistance is defined by the relations.
and
Z COS a
tan a = -
r
The e.m.f. consumed in the circuit is, in general,
E = IZ = (iji'} (r+jx).
The current may or may not be chosen as the zero vector. If it is
so chosen, / = i. If not, then / = i ji' } where i' is the wattless
component of the current.
The impedance is always z = r + jx.
Assigning positive or negative values to the wattless component
i f , we may write, in any case,
I = i+ ji'.
ALTERNATING-CURRENT WAVES
109
It should be remembered that a leading component requires a
+ sign, and a lagging component requires a sign.
Therefore, E = IZ = (i + ji') (r + jx)
= ir i'x + j(i'r + ix)
If the current is taken as the zero vector, then
E = i(r + jx)
In the general expression (31), an arbitrary zero line is chosen,
as in Fig. 85. In the simpler case (32), the direction of / is
chosen as the zero line, whence I i and i' = 0, and the vector
diagram becomes that of Fig. 86.
(31)
(32)
ix
FIG. 85.
%r
FIG. 86.
Problem 46. One ampere flows in a circuit of 1 ohm resistance and a vari-
able reactance. Plot curves of Ir, Ix, Iz drops and phase angle against x,
when x varies from to 5 ohms. Take / as the zero vector. Then 7 =
i = 1.
Solution.
Tabulating :
' tan a
X
0.5
1
2
3
4
5
ix
0.5
1
z
1
1.12
iz
1
1 12
x
5
r
-
a
26 35'
The blank spaces may be filled in by the student.
Consider the same case, Fig. 79, but with E known and /
unknown. E, then, may conveniently be chosen as the zero
vector, and
I = e _ = e e(r-jx)
z r + jx (
(33)
110
ELECTRICAL ENGINEERING
The last expression of (33) is obtained in accordance with a third
rule, as follows:
Rule HI. Never allow an equation to remain with a complex
denominator. Thus (33) becomes
where
g
(34)
(35)
FIG. 87.
g + jb F, is called the admittance-; g is called the conductance,
and b the susceptance of the circuit. The diagram of currents
may now be drawn to correspond with
Fig. 87, for e.m.fs., in which eg is the com-
ponent of the current in phase with e, that
is, it represents energy expended, and eb
is the component 90 behind e, called the
reactive or wattless component because
it does not represent any expenditure of
energy.
The power input to the circuit is then
Power input = e X eg = e z g,
and this is found equal to I 2 r.
The numerical value of the current = I = e\/g 2 + 6 2 .
Problem 46. Let E = e = 1; x = 1; r varies from to 10. Plot curves
of / vs. r, and I 2 r vs. r.
Calculate the maximum value of the power loss and find the value of the
resistance which gives the greatest dissipation of power.
Plot the 3 current waves, that is, the power current, eg., wattless cur-
rent, eb, and total current, ey, for the condition of maximum power loss.
Solution.
e 1 1
Vr 2 -f x 2 \/l + r 2 * -
r x 1
v = - z ', = --,; y = -
Tabulating :
r
2
4
6
8
10
Z
1
2.27
4.12
7
1
0.44
0.242
/
1
194
7r.. '.....
0.388
*
ALTERNATING-CURRENT WAVES 111
Power, W = Pr = - z r
*?-..
For maximum power -p- = 0. .'. x 2 r 2 = 0, and r 2 = x 2 . .'. r 2 = 1,
r = 1, and W = 1 X j-qjj = 0.5 watt.
To get current waves for maximum power loss, then
r = 1; x = 1; Z = 1.41.
e _ _e(r - j x) _er , ex _
where ^ = -^- 2 and 6 = ^*
The effective values of current are, therefore,
eg = 1 X K = 0.5, in phase with e,
jeb =-je~=-jlX^ = -jO.5, or 0.5, 90 behind e,
e(g +jb)=eY=^ = ^ - 0.707, lagging behind e, by an angle tan" 1
& V2
Maximum values are ^ TO = \/2^ = 1.41,
(Eg) m = 0.707; (J^6) w = 0.707; (EY) m = 1.
Circuit of Resistance in Series with an In-
ductive Impedance. The impressed e.m.f.,
E, of the circuit is known, also the resistance, E
r, and impedance, Zi = r l + jxi (Fig. 88). # I
is taken as the zero vector. Then, FlG 88
where
r , Xi
Q v~2> o = -- ^- 2 ; Z
and r = r -j- ri.
The drop across the impedance, Zi, is
#1 = 7 Zi = e(g + j6) (7*1 + ji
(s^i + n.
where a = 0n 60; ij 6 = gxi +
112 ELECTRICAL ENGINEERING
Problem 47. In the above circuit, Fig. 88, let E = 10, r = 1, n = 0.5,
Xl = 2.
Draw the vector diagram and waves of e, EI and /.
Circuit of Two Inductive Impedances in Parallel. Let E, Z\ and Z 2 be
known (Fig. 89). To determine 7, /i and /2.
FIG. 89.
We have:
/i = eY l} J 2 =
I -li + J, -a(Fi-+ 7,). '
Or,
/ 2 = e(flfi +j'6i)i
62) = e(G
where
G = 0i + g t , B = 61 + 62.
D
Tan a -^-> gives the phase relation of / and e.
Problem 48. In the circuit of Fig. 89, let E = 1, n = 1, x l = 0.5,
r 2 = 2, x 2 = 2.
Draw vector diagrams and waves of E, Eg, Eb, and /.
CHAPTER XVI
THE SYMBOLIC METHOD IN TRANSMISSION LINE
CALCULATION
KENNELLY AND STEINMETZ have introduced the so-called
symbolic method of representing electrical relations.
This method is neither vector analysis nor quaternions, but
is in many ways similar to both. It enables the use of simple
algebraic transformation when dealing with vector quantities of
the same rate of rotation or frequency. Thus, it is directly
applicable when, for instance, multiplying a current by an
impedance, since the resultant e.m.f. is of the same frequency as
the current. But when multiplying current and e.m.f., it is
applicable only after some modification, since the product
represents power, which is a vector of double frequency.
Addition. Let there be two vectors,
a\ + J&2, and bi -\- jb 2 , and let their sum be a vector C.
Then,
C = ai + ja 2 + 61 -f jb 2 = ai + bi + j(a 2 + 6 2 ) = ci + jc z ,
where
Ci = ai + bi and c 2 = a 2 + b z .
Multiplication. We have, evidently,
Oi + ja 2 )(6i + J6 2 ) = ai&i - a 2 b 2 + j(a^ -f bia z ) = di + jfa,
where
di = dibi a^bz', dz = Q>ib% + b\a,z.
In general, if i + ja z = bi + jb 2 then ai = 61 and a 2 = 62-
Power. At any instant,
p = ei,
where e and i are instantaneous values of voltage and current.
In the case of sine waves, where e = E m sin ui and i = I m sin
(* + ),
p = ei = E m l m sin cot sin (co + )
8 113
114
ELECTRICAL ENGINEERING
Problem 49. Plot waves of voltage and current, and by multiplying
their values at certain instants along the curves show that the resulting
power curve is a sine wave of double frequency.
Let E m = 1.4, Im = 0.7
(1) a = (2) a = 45 (3) a = 90.
Fig. 90 shows the curves plotted for the case of a. = 0. The
energy developed in the circuit, in any time dt is pdt. The total
energy during a cycle is then J] T
pdtj where T is the time of one
complete cycle. But this is the
- area enclosed by the power curve
and axis, shown shaded. As the
values of power are always posi-
FIG. 90. tive, the area represents energy
expended, or work done.
The student should show that when a is not 0, there is also
negative power, which represents energy returned to the source,
the total energy expended during a cycle being the difference
between the positive and negative areas enclosed by the power
curve and the axis.
Average Value of Power during a Period. -r-This will be,
p
m l m sin ut sin (coZ -f a)dt,
which, the student should show, is
cos a.
This may be written
E m I,
V2 V2
COS a = El cos a
(36)
where E and I are effective values as usual.
Thus the important result is found, that, in
case of sinusoidal current and voltage waves,
the average power is equal to the effective
value of the current times the effective value
of the voltage into the cosine of the phase
angle between the two. This is illustrated in Fig. 91, and it is
seen that when 7 is zero vector = i, P = El cos a = el = ei.
Similarly, when E is zero vector = e, P = Ei = ei.
FIG. 91.
METHOD IN TRANSMISSION LINE CALC ULA TION 1 15
Power is obtained by multiplying either quantity by the pro-
jection of the other upon it.
In general, if E makes an angle 7, and 7 an angle with the
zero axis,
where a = 7,
Therefore, 0> 7 .
P = EI cos a = #7 cos (0 - 7)
= EI (cos cos 7 + sin sin 7) (36)
t e
But cos
cos 7 ==
sin/3 = sin 7 =
Substituting these values in
P = ei + e'i'
which is the general expression for the average power.
Example. Let E = e + je'
i = i -f ji r .
Then by (37) P = ei + e'i'.
Suppose, however, that we carry out the
multiplication of the vectors. Thus,
EI = ( e +je')(i+ji f )
= ei e'i' + j(ei' + e'i)
The numerical value of this expression is
(36)
(37)
FIG. 92.
\(t- eY) 2 + (e'i- ei') 2
which is obviously not the same as (37), neither is its real com-
ponent the power, since it has a minus sign.
It has been shown in Fig. 90, that power is a quantity of double
frequency. It can therefore have no phase relationship with
E or 7. Hence, in the case of power or any double frequency
quantity, the operation of multiplying single frequency quanti-
ties is inadmissible.
On the other hand, it is known that the product
(i-ji>) (r+jx) = E
is quite correct, since the fundamental frequency only is involved.
116 ELECTRICAL ENGINEERING
The operation of obtaining the power from two vectors
E and I, is called " telescoping" the vectors. Thus, the prod-
uct of the "real" components is added to the product of the
"imaginary" components, without any change of sign due to
the presence of j.
Power Factor. In the expression for power (36) the term
cos a is called the power factor. The product El represents true
power only in certain special cases, particularly with direct
currents.
T r v j ^ j .- true power
Power factor may be defined as the ratio - '
apparent power
where apparent power = El.
El is also called the volt-amperes.
Since El cos a is the true power,
. El cos a
power factor = p.f. = ^7 = cos a.
Jiil
Transmission Line Calculation. The calculation of circuits
may now be continued to include the case which represents a
simple transmission line possessing concentrated resistance and
inductive reactance, being supplied with power at one end by a
generator, or source of alternating
current, and terminating at the other
end in any prescribed load. A cus-
tomer usually desires constant volt-
FlG 93 age, E, at the load.
In Fig. 93 is shown a generator
supplying power over a transmission line of impedance, Z = r
+ jx t at voltage E Q to a load, the current of which is i + ji'
at voltage E. E and i are in time phase ; E and i f are in time
quadrature.
(1) Let E be known, and be taken as the zero vector, = e.
Then, the voltage at the generator terminals,
Eo = e + IZ = e + (i + ji') (r + jx)
= e -f ir + ijx + ji'r i'x
= e + ir i'x + j(ix + i'r) = a -f- jb,
where
a = e -f ir i'x,
b = ix + i'r.
METHOD IN TRANSMISSION LINE CALCULATION 117
The power factor of the load, cos a = -=. = y
Generator volt-amp. = IE Q .
Power of Generator. P* Q = E I cos 7, where 7 is the angle
between E Q and 7.
Vector relationships are shown in Fig. 94: (a), for leading
and (6), for lagging current. In the first case
7 = a |3 is the angle between E Q and 7.
/. Cos 7 = cos (a |8) = cos a cos + sin a sin /3.
i i'
But cos a = j, and sin = y-'
a 6
Likewise, cos /5 = yr and sin = ^-
/. Cos 7 = yyr (m + i'6).
Substituting this value into the
equation for power of generator,
PQ = E I COS 7 = itt + *'6.
P could be more quickly obtained
by simply telescoping the vectors a
+ jb and i -f ji'.
Power factor at the generator
power PQ
Ei
Efficiency of transmission = ^5
-T
output
Apparent efficiency == ratio,
Regulation = -- ^ --
Having obtained the general expression for the various quan-
tities which enter in, we may now take a specific example of
transmission line calculation.
Problem 60. In Fig. 93, let E = 1, r = 0.1, x = 0.2, i = 1.
Let i' vary from 1 to +1.
Determine all the quantities, i.e., current, generator voltage, volt-
amperes, power factor, power, transmission efficiency, apparent efficiency
and regulation.
118
ELECTRICAL ENGINEERING
Tabulating:
1
j' '.
-1.0
-0.75
-0.5
-0.25
0.0
0.25
0.5
0.75
1.0
If
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
i'x
0.2
0.15
0.1
0.05 *
0.0
-0.05
-0.1
-0.15
-0.2
a
1.3
1.25
1.2
1.15
1.1
1.05
1.0
0.95
0.9
ix
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
i'r
-0.1
0.075
0.05
0.025
0.0
0.025
0.05
0.075
0.1
I
0.1
0. 125
0. 15
0. 175
0.2
0.225
0.25
0.275
0.3
a j .
1.69
1.56
1 .44
1.32
1 .21
1.10
1.0
0.90
0.81
6 2
0.01
0.0156
. 0225
. 0306
0.04
. 0506
0.0625
0.0756
0.09
a 2 + 6 2
1.7
1 . 5756
1.4625
1.3506
1.25
1.1506
1.0625
0.9756
0.9
-y/ 2 4. b 2 .. . .
1.31
1.25
1.21
1.16
.12
1.08
1.03
0.985
0.95
Eo
1.31
1.25*
1.21
1. 16
.12
1.08
1.03
0.985
0.95
i' 2
1.0
0.56
0.25
0.0625
.0
. 0625
0.25
0.56
1.0
i 2 + i' 2
2.0
1.56
1.25
1 .0625
.0
1.0625
1.25
1.56
2.0
V* 2 + i' 2 ..-.
1.41
1.25
1.12
1.03
.0
1.03
1.12
1.25
1.41
I
1.41
1.25
1.12
1.03
.0
1.03
1.12
1.25
1.41
E I
1.85
1.56
1.36
1.23
.12
1.11
1.15
1.23
1.34
i/I
0.707
0.8
0.895
0.94
.0
0.94
0.895
0.8
0.707
CM
1.3
1.25
1.2
1.15
.1
1.05
1.0
0.95
0.9
bi'.
0.1
0.0937
0.075
0.0437
0.0
0.0563
0.125
0.206
0.3
Po
1.2
1.156
1.125
1.106
1.1
1.106
1.125
1.156
1.2
i'/i
-1.0
-0.75
-0.5
-0.25
0.0
0.25
0.5
0.75
1.0
tan a
-1.0
-0.75
-0.5
-0.25
0.0
0.25
0.5
0.75
1.0
a
45
37
26 30'
14
14
26 30'
37
45
Cos a
0.707
0.8
0.895
0.94
1.0
0.94
0.895
0.8
0.707
P /EoI
0.65
0.742
0.827
0.9
0.982
0.999
0.978
0.941
0.895
Cos y
0.65
0.742
0.827
0.9
0.982
0.999
0.978
0.941
0.895
Fff Ei
tin. = p~" '
0.834
0.864
0.89
0.903
0.907
0.906
0.89
0.865
0.835
App. eff. =
Ei
0.54
0.64
0.735
0.813
0.894
0.9
0.87
0.813
0.74
Eol
Regulation =
Eo - E
0.31
0.25
0.21
0.16
0.12
0.08
0.03
-0.015
-0.05
E
Problem 51. Draw vector diagrams for the cases of i' 1, 0.5,
0, 0.5, 1, of problem 50 showing the relative positions of E , E and /. Also
plot the curves of regulation vs. power factor of generator and load.
The preceding example, like many others included in this book, is con-
structed on the basis of percentages. That is, by choosing e = 1 and i = 1,
whence p = ei = 1, the results obtained may be made to apply to any
case in which the constants, r and x, give the same percentage of ri and xi
0.1 = 10 per cent., ~ = 0.2 = 20 per cent.
drops. In this example,
If, now, 10 per cent, resistance drop and 20 per cent, reactance drop be
specified, let it be required to find, with varying power factor of the load,
the same quantities determined in problem 50, when v e = 2200 volts and
i = 300 amp.
All that is necessary, now, is to multiply those quantities representing
voltage by 2200, those representing current by 300, and those representing
watts, or volt-amperes, by 2200 X 300 = 660,000.
METHOD IN TRANSMISSION LINE CALCULATION 119
Thus, for i' = - 1 X 300 = - 300, E = 1.31 X 2200 = 2882, / =
1.41 X 300 = 423, E I = 1.85 X 660,000 = 1,221,000, P = 1.2 X 660,-
000 = 792,000.
The other quantities sought power factor of load and of generator,
efficiency, apparent efficiency and regulation are the same as already
calculated.
The advantages of problems on the percentage basis are thus quite
obvious.
Problem 62. A transmission line 1 mile long supplies power to a load of
100 kw. and 1000 volts at power factor of 0.8 and frequency of 60 cycles.
The line is composed of two parallel No. 000 B. & S. wires, 18 in. apart.
Find generator voltage, current, power factor, power output, line effi-
ciency, apparent efficiency, regulation, with the current both lagging and
leading.
The resistance of No. 000 B. & S. hard-drawn copper wire may be taken
as 0.06 ohm per 1000 ft. at 20C.
The reactance is 2wfL, where L is the inductance, in henrys, per centi-
meter length of wire. L may be calculated from the formula,
10 9
where D and r are, respectively, the distance between
centers of wires and the radius of the wire (Fig. 95).
D
FIG. 95
CHAPTER XVII
CONSTANT POTENTIAL-CONSTANT CURRENT
TRANSFORMATION
It is sometimes desirable that the current in a circuit shall
remain constant while the load varies. In series lighting circuits,
for example, the current through each lamp must be nearly
constant, while the number of lamps may vary from none at
all up to the most that the system can sustain. Generally,
however, it is desirable that the energy shall be supplied from
a source of constant potential, such as a constant potential
generator. Such a system is possible with a circuit arrangement
like that shown in Fig. 96. Here, a high resistance, r, is placed
in series with the lamps. When the lamps are comparatively
few, changing their number will not alter the total resistance of
the circuit very much, and the current will therefore be fairly
constant.
This arrangement is not, however, economical, as a large pro-
portion of the power developed is always lost in the resistance, r.
FIG. 96. FIG. 97.
We may, therefore, try substituting inductive reactance, x, for
r, and determine if this will give better results.
In this case, Fig. 97, let the generator voltage be unity, that
is e = 1. Let the largest value of the permissible current, in
the circuit also be unity, that is, / = 1, and let the resistance of
the lamps, that is, the number of the lamps, vary. We may
then find the maximum resistance of the lamps which may be
obtained without reducing the current lower than, say, 0.925,
which will be considered the minimum, permissible current.
The current is obviously a maximum when the resistance is
zero, that is, when no lamps are used.
e I
Then, x = j = j = 1 ohm.
120
CONSTANT CURRENT TRANSFORMATION 121
Let r be the resistance of the lamps, r is variable, depending on
the number of lamps in circuit at any time.
Let E be made the zero vector, = e
e e e (r - jx)
Then
where
I =
7 = e
Tabulating for varying r:
(38)
r
0.1
0.2
0.4
0.6
0.8
1
r 2 + X 2 .
1
1 01
1 04
1 16
1 36
1 64
2
a
0.099
0.192
0.345
0.441
0.488
0.5
b
-1
-0 99
-0 96
861
-0 735
-0 61
-0 5
a 2
o
0098
0369
119
195
238
25
6 2
1
0.98
0.921
0.741
0.54
0.373
0.25
a 2 + 6 2 ..
1
9898
9579
86
735
611
0.5
Va 2 + 6 2
7
1
1
0.994
0.994
0.978
0.978
0.927
0.927
0.857
0.857
0.782
0.782
0.707
0.707
It is evident from the calculation that the limit of current
is reached by a resistance of 0.4 ohm. This resistance could
evidently be obtained by directly substituting the value / = 0.925
into Eq. (38) and solving for r. However, it is frequently prefer-
able to carry out the tabulation, thus gaining the material for
plotting the curves. These curves are far more instructive than
the mere numerical answer.
In this case, where reactance has been used instead of resist-
ance in order to obtain (approx.) constant current, all the
energy is consumed in the lamps themselves since the reactance
(assuming zero resistance) does not consume any energy. Thus
the system is efficient. However, the power factor appears to be
very low.
Problem 63. Determine the power factor of the circuit and plot it
against the resistance of the lamps.
Altogether, it may be said that in practice this arrangement is cheap and
practical. A constant-current "tub" transformer has a higher power
factor, but is also more expensive. In this machine regulation is obtained
by altering the reactance in the circuit by means of the repulsion between
the primary and the secondary coils.
Problem 64. A constant-current system is supplied with power by a
122
ELECTRICAL ENGINEERING
generator at 2300 volts and 60 cycles. The line resistance is negligible.
Each lamp has 6 ohms resistance. Current must be maintained between
the limits of 7.2 and 6 amp.
Find the maximum number of lamps, both by the "resistance" and by
the "reactance" method of obtaining constant current. Plot and compare
curves of number of lamps vs. current for the 2 cases.
Solution. (a) Resistance method.
Let r = res. in series, TL = res. of lamps.
_ 2300
Then r = -=-^
r + -TL =
r L =
No. lamps =
2300
= 320 ohms.
= 383.3 ohms.
63.3 ohms.
383.3 - 320
~ = 10.5 = 10 lamps.
Lamps
2
4
6
8
10
12
TL
12
24
36
48
60
72
r + r L ..
320.0
332.0
344.0
356.0
368.0
380.0
392.0
I
7.2
6.93
6.68
6.46
6.25
6.05
5.87
(6) Reactance method. Neglecting the resistance of the reactance coil,
2300
rL 2 + 320 2
No. lamps
2300
6
383.3
210.8
= 320 ohms.
= 383.3 ohms.
/. T L = 210.8
35.1 = 35 lamps.
Lamps
10
20
30
35
40
T L
r, 2 .
A
60
QfiOO
120
14 400
180
Q9 CJOO
210
44 200
240
K7 700
z 2 -
102,500
102,500
102,500
102,500
102,500
102,500
r L *+x*
102,500
106,100
116,900
135,000
146,700
160,200
W+z 2
/
320
7.2
326
7.05
342
6.72
369
6.23
383
6.00
400
5.75
NOTE. In an example of this kind it may be as convenient to work
directly, without the use of complex quantities as has just been done.
With more complicated circuits, however, it may be far more convenient
and far safer to adhere strictly to the complex method.
There are many other schemes for obtaining constant current.
Some of these involve the use of condensers which are treated
in the next chapter.
This subject will be discussed further in Chap. XXI.
CHAPTER XVIII
CAPACITY AND CAPACITY REACTANCE
Two conducting surfaces, insulated from each other, are said
to possess electro-static capacity. Such an arrangement em-
bodied as a piece of apparatus is called an electrical condenser.
Condenser. When the plates of a condenser are connected
respectively to the positive and negative terminals of a direct-
current generator, the condenser becomes charged. That is,
when a switch, s, Fig. 98, is closed, completing the circuit con-
taining the generator and the condenser, ammeters A, placed in the
leads, will indicate a momentary current in
the direction of the arrows. No current, in
the ordinary sense, could pass between the t
plates. The phenomenon thus resembles
the piling up of electricity, as so much ma- F
terial, on one plate, the positive plate, since
it is connected to the positive terminal of the generator, and
the withdrawing of an equal amount of electricity from the other,
the negative plate. This quantity of electricity which seems to
have been transferred from one plate to the other is the charge
placed upon the condenser. The condenser is maintained in an
unstable state by the e.m.f. of the generator. If the generator
is disconnected, the condenser continues to remain charged so
long as its plates remain insulated from each other, but as soon as
electrical connection is made between them, the condenser dis-
charges itself by a rush of electricity from the positive to the
negative plate, as indicated by the flow of electricity or current
through the meters. If the condenser is charged to a difference
of potential which is excessive, the insulating dielectric breaks
down, allowing a discharge to take place between the plates.
This indicates that the dielectric is placed under a strain when the
condenser is charged. In fact, the dielectric behaves much like
an elastic medium compressed between plates. When the
pressure is removed the medium assumes its normal condition.
The plates act merely as carriers or distributors of the charge,
123
124 ELECTRICAL ENGINEERING
while its actual seat, as found out by FRANKLIN, is the surface of
the dielectric.
The capacity of any given condenser is determined by the
dimensions of its plates, their distance apart, and the nature of
the dielectric which separates them.
Capacity is not a property solely of apparatus arranged in the
form of a condenser, but any body may be said to possess
capacity for instance, a metallic sphere, insulated and isolated
in space. But this may also be considered as a limiting form
of condenser in which one plate is the surface of the sphere and
the other is a surrounding sphere of infinite radius. In this
case the strain in the dielectric may be represented by the lines of
force, or tubes of force, extending radially outward from the
surface of the sphere and terminating on the surface of the
imaginary sphere infinitely distant. This conception of lines,
or tubes of force, due to FARADAY, makes the direction of a line
or axis of a tube the direction of the force at any point, and the
number per square centimeter, or density, of lines or tubes be-
comes a measure of the force at the point.
FARADAY assumed that the number of tubes is the same
numerically as the charge per unit surface, and that the number
of lines emanating from a charge Q is $ = 4?rQ. Thus each tube
contains 4ir lines of force.
By connecting a sphere to one terminal of a battery, Fig. 99,
and connecting the other terminal to earth, assumed infinitely
distant, we establish a number of electric lines
/ of force extending outward from the sphere.
The number of lines established is 471-Q, 1 where
Q is the amount of the charge placed upon the
sphere.
If the sphere is in air, the practical limit to
^FIG 99 ^ ne num ^ er f nnes which it is possible to
establish is very closely 100 per sq. cm. of sur-
face. Thus, to produce 100 lines per sq. cm. requires a charge
100
Q = -7 = 8 absolute electro-static units per sq. cm.
To increase the number of lines established in any given
case, the difference of potential, or voltage, should be increased.
These lines are conceived to displace the ether, until by continu-
ally increasing the voltage, the crowding of them becomes so
1 See Advanced Course in Electrical Engineering.
\\ I /
"
CAPACITY AND CAPACITY REACTANCE 125
great that the dielectric breaks down. The ability of any dielec-
tric to withstand rupture under the strain of potential-difference
is called "dielectric strength."
With a parallel plate condenser, Fig. -f~
100, the lines or tubes are parallel, zip
except at the edges, where they bow z =jT
outward.
By definition, the* charge due to
current i during interval dt is:
dq = idt. (39)
The practical unit of charge or quantity, q, is the coulomb.
Another fundamental relation is that
q = Ce (40)
where C is the capacity, and e the e.m.f. or difference of potential.
This law, found experimentally, shows that the number of tubes
which can be set up in a condenser of capacity C depends directly
on the potential difference. In practical units, the charge in
coulombs is equal to the product of the capacity in farads and the
potential difference in volts. "Charge" is not a material quan-
tity, but may well be thought of as a measure of "tubes."
Substituting from (39) into (40), since
*-*,
and
dq = Cde,
which is called the charging current, or capacity current of the
condenser.
Assuming a sine wave of e.m.f., impressed on a condenser
then, e = E M sin wt (42)
i = CE M w cos orf. (43)
The capacity, C, is a constant of the circuit, that is, like resist-
ance and inductance, it is a quantity fixed by the mechanical
arrangement of the circuit.
Eq. (42) may be written:
i = CE m o> sin (cd + 90). (43a)
126 ELECTRICAL ENGINEERING
Comparing (42) and (43a) it is seen that the charging current
is 90 in time phase ahead of e. Also,
The effective value of the charging current is then
whence
E
X c .
/ 27T/C
The quantity x c is called capacity reactance, and its use in cir-
cuit calculations is similar to inductive reactance.
Expression of Condensive Impedance. It has been shown that
the charging current leads the impressed e.m.f. 90 in time.
Thus, if the charging current / is made zero vector, the im-
pressed e.m.f. is jkl where k is some constant and is obviously
x c . Thus E = jxj.
In an inductive circuit the current lags 90 behind the im-
pressed e.m.f.
Thus E = jxl.
Convention has settled that an inductive impedance is Z =
r + jx\ thus the condensive impedance is Z = r jx c where x c
as well as x is always a positive number.
Circuit Containing Resistance, Inductance and Capacity in
Series. To find the current. Let E, the impressed e.m.f., be
the zero vector. Then
/ = - -A- - =
r + jx - jx c
where
a =
b = -
^X X c )
(x - x c )
r (x - z c )-
To find the voltage E c across the condenser. We have:
EC = I(-jXc)
= e(a H- jb)( jx c ) = e( ajx c + bx c ).
Similarly, the voltage across the inductance is
E L = I X jx
= e(a + jb)jx = e(ajx bx).
CAPACITY AND CAPACITY REACTANCE 127
Problem 56. Let the constants of a circuit be r = 1 ohm, L = 0.0265
henry, C = 0.000265 farad, and let 100 volts be impressed on the circuit
at variable frequency. Find, and plot against
the frequency, 7, E c , EL, E r for frequencies from
to 100 cycles per sec. Jl /. i fo
Solution. We have:
7
*np- f c
e(a + jb); I = eV a 2 + b 2 .
E c = e( - ajx c + bx c ); E c =
E L = e(ajx - bx); E L =
Also,
E T = e(a + j'6)r; E r = er\/a 2
Tabulating:
FIG. 102.
/
20
40
50
55
60
65
70
100
2irf
125.6
251.2
314.0
345.2
376.8
408.0
440.0
628.0
X
3.33
5.65
8.33
9.15*'
10.0
10.8
11.65
16.66
x c
30.0
17.65
12.0
10.92
10.0
9.25
8.57
6.0
(x - x c )
- CO
-26.67
-12.0
-3.67
-1.77
0.0
1.55
3.08
10.66
(X - X C )2. . . .
+ oo2
712.0
144.0
13.5
3.14
0.0
2.4
9.5
114.0
r2 + ( x -x c )2..
oo2
713.0
145.0
14.5
4.14
1.0
3.4
10.5
115.0
a
0.0014
0.0069
0.069
0.242
1.0
0.294
0.095
0.0087
b
0.0374
0.0827
0.253
0.428
0.0
L0.455
-0.293
-0.0925
a*
0.000002
. 000048
0.0048
0.059
1.0
0.086
0.009
0.000076
62
0.0014
0.0068
0.064
0.183
0.0
0.207
0.086
. 0086
(I* + 62
0.0014
0.0069
0.0688
0.242
1.0
0.293
0.095
. 0087
Va 2 + 62. ...
0.0374
0.0828
0.262
0.492
1.0
0.54
0.308
0.093
7
3.74
8.28
26.2
49.2
100.0
54.0
30.8
9.3
EC
112.2
146.1
314.0
537.0
1000.0
500.0
264.0
55.8
EL
12.5
46.8
218.0
450.0
1000 .
583.0
370.0
155.0
E r
3.74
8.28
26.2
49.2
100.0
54.0
30.8
9.3
40 60
Frequency
FIG. 103.
128 ELECTRICAL ENGINEERING
Resonance. Curves of the form shown in Fig. 103 are
called resonance curves, and their maximum points of the de-
pendent variables are called resonance points. In this case, it
is said that 60 cycles is the frequency of resonance.
On examining the problem it is seen that resonance is attained
at that frequency for which x x c = 0, or when the effect of
inductance is just nullified by that of capacity. The circuit then
behaves as though it possessed resistance only.
CHAPTER XIX
PARALLEL CIRCUITS
Let 1 1 and 1 2 be any currents in the branches of a parallel
system, such that I\ = i\ + ji f \ and 7 2 = it + ji f z-
Laying off these vectors (Fig. 105), and adding them, gives
7 = 7i + 7 a = ii + la + j(i'i + *' 2 ). (44)
Let the impedances of the branches (Fig. 104) be Zi = ri jxi
and Z 2 = r 2 + j^2, respectively.
FIG. 104. FIG. 105.
To find the currents 7i, 7 2 , and 7. We have:
. _ e
where g\ =
r\
+
T
+ xi 2
is the conductance,
is the susceptance,
and FI = gi + jbi is the admittance of the first branch circuit.
eg i is the power component of I\.
cbi is the wattless component of I\.
Similarly,
_. ^2 JXz) , . ., v _ y
where _^^
r J
and
r 2
/=
(45)
129
130 ELECTRICAL ENGINEERING
To find the joint impedance, Z, of the branches,
e_ _e_ !_
~~ I ' eY '' = Y'
Example. In Fig. 104 let n = r z = 0, and Xi = ^- f ^> x- 2 = 2ir/L.
Then
-
= 0,
1
" z 2 27T/L'
Then, from (45),
7 = e (o + j(2*/C - g^) ) (46)
From this it is seen that the line current is in time quadrature with the
voltage.
If I = 0, then from (46) we have the relation
or
= 1.
that is to say, that if in a circuit such as is here considered the frequency
be varied, a value may be reached for which the line current will be reduced
to zero. In such a case the currents in the branches will be
7i = e(0
Both of these currents are in time quadrature with the voltage, but
/i is leading while 7 2 is lagging.
Thus, they are in time phase opposition to each other.
Problem 66. In the circuit of Fig. 104 let e = 100, n = r 2 = 1, L =
0.0265, C = 0.000265. Let the frequency vary, as in problem 54. Find
/ Ii, Jzj and plot them against the frequency.
Transmission Line Supplying Power to Parallel Loads. Let
a transmission line of impedance Z Q = r Q + j%o be used to supply
power to a load consisting of two impedances, Zi = 7*1 + jx\ and
2 = r 2 + jxz, which are in parallel. Besides the impedances,
let E the voltage at the receiving end be known.
Find 7, /i, 7 2 , E Q , P.F. of generator and of combined load,
regulation and efficiency of the line.
E is chosen as the zero vector = e.
PARALLEL CIRCUITS
131
Then
where
Zi ri+jxi rf + xi'
(0i + jbi),.
Similarly,
where
/2 =
FIG. 106.
And
where
Then
/ = /I + / 2
= m + jn,
= e (0i 4- 02 + j(&i + 62)
\ __ _/t i^ t \
= e + (w -f jn)(r + jx )
nx Q + j( nr o -
Eo = e +
= e +
= a
where
a Q = e -{- mr nx 0) 6 = nr
Power of generator, by telescoping E Q and I =P Q = a m + 6 n.
PO a w -f- 6 ?i
.'. P.F. of generator =
P.F. of combined load = -v = F'
EJ
- e
Regulation =
, . P em
Efficiency = ^- =
Problem 67. In the same circuit (Fig. 106), let E be known and E
unknown. Find all the quantities obtained in the last problem.
NOTE. In solving this problem the student is again urged to pay
particular attention to the form of his work. In order to add emphasis to
this matter these similar problems are here given, the one being worked out
and the other left for the student to do.
132 ELECTRICAL ENGINEERING
The numerical or scalar expressions are not put down. It is assumed
that they may always be obtained when needed by the simple process of
rationalizing a simple complex expression. By omitting them in the process,
confusion is eliminated.
Approximate Transmission Line Calculation. The two parallel
wires of a transmission line may be regarded as constituting the
plates of a condenser. When alternating e.m.f. is impressed
upon the line there will therefore flow a charging or capacity
current over the line, whether the distant end is open or closed.
Fig. 107 gives an approximate representation of such a line in
___
'
FIG. 107.
which the line capacity is replaced by two condensers, one at
each end, so proportioned that each shall take one-half of the
charging current. The charging current is taken as 2i 2 . ii is
always positive, whereas i', the wattless component of the load
current, is positive or negative depending on the load.
Then
EQ = e + I Z = o + j
where ao = e + ir Q i s x Q)
/o = / + jit = i + j(i' + 2i 2 ) = i + ju.
From these, the power, power factor, efficiency, etc., may be
determined. Expressions should be obtained by the student for
practice, as follows:
Power given by generator = P =
Apparent power at generator = E Q I Q =
p
P.F. at generator = cos <*o = jrj~
Efficiency of transmission = ,5- =
* o
Apparent efficiency
CHAPTER XX
DISTORTED WAVES. RESONANCE EFFECTS
So far, only current and voltage waves have been dealt with
which followed a sinusoidal variation with respect to the time
and had the same frequency or period. In the laboratory, re-
sults obtained are found not always to agree with those expected
from the theory. This is frequently due to the assumption in
theory of pure sine waves, whereas, in practice, a pure sine wave
is only approximately attainable, and the actual waves may
differ greatly from that form.
It can be proven that any curve representing changes occurring
with time can be resolved into a number of sine waves of differ-
ent frequency as long as the curve representing the changes
is a univalent function of time which it always is in electrical
problems.
It can also be proven that if the curves traced are symmetrical
above and below the axis no matter how distorted the sine
waves contain only the odd frequencies.
Thus assume as the simplest case that the current is distorted
in such a way that it can be represented by the first two terms
of the series, that is that:
i = Ii m sin wt -f J 3m sin
a).
It is seen that the frequency of the second component wave
is three times that of the first.
The first wave is called the
fundamental of the complex
wave, the second wave is
called the third harmonic.
The angle a denotes the per-
manent phase difference be-
tween the waves. Such a
combination of waves is seen ina
rIG. luo.
to be a distorted wave, as
shown in Fig. 108. To find the amount of heat such a wave
will develop in a circuit, that is, to find the effective value of
the complex wave.
133
134 ELECTRICAL ENGINEERING
Evidently the heat developed at any instant is proportional to
i 2 , and
i 2 = [Iim sin ut + 7 3m sin (3co -f a)] 2 .
The mean value of the heat developed during a cycle of the
wave will then be proportional to
1 C T
mean i 2 = I [I lm sin ut + h m sin (3co + a)] 2 dt. (47)
Thus, the effective value of the current is
\/mean^ 2 = 7 = \/ I t^im sin cot + 7 3w (3co + )] 2 cfr. (48)
The student should solve (47) and (48), and show that
1 2 7 2
. -Mm , -* 3m
mean z 2 T^+jT (49)
and / = \/Ii 2 + /a 2
where 7 im = maximum value of the fundamental current wave,
7 3m = maximum value of the third harmonic,
7i = -^7= = effective value of the fundamental,
7am
7s = j= effective value of the third harmonic.
V2
Also, in general, where there are any number of component
harmonic waves in a circuit,
/ = Vli 2 + h 2 + 7 6 2 + . . . (50)
Thus is found the important rule that the effective value
(ammeter reading) of any number of currents of different fre-
quencies is equal to the square root
w of the sum of the squares of the in-
dividual effective values.
15
NOTE. Eq. (50) holds for any combina-
tion of harmonics whatsoever. With alter-
nating-current machinery, we have to deal
only with odd harmonics, as the positive
and negative waves are always symmetrical
except during transient periods not considered in this volume.
Example. In Fig. 109 are represented three generators which
supply respectively 20 amp. at 60 cycles, 15 amp. at 25 cycles, and
10 amp. at 10 cycles. They all use a common wire for a part
DISTORTED WAVES. RESONANCE EFFECTS 135
of their circuits. Then the current which flows in the common
wire is
/ = V20 2 + 15 2 + I6 2 = 27 amp.
Problem 68. If still another generator is added to the above system and
it supplies 12 amp., direct current, to its load, using the common wire, find
the current, I, that will then be in the common wire and explain the result.
E.m.f. Which Causes Distorted Waves of Current. If the
current in any circuit is given by the equation
i = 1 1 sin at + 7 3 sin (3orf + a) (51)'
the question may naturally arise as to what kind of e.m.f. wave
will cause such a current to flow. Will the e.m.f. wave be more
or less distorted when the current is supplied to a circuit of re-
sistance and inductive reactance?
We have (Eq. 15),
, T di
e = ir+L dt'
Substituting from (51),
e = Ijr sin wt + 7 3 r sin (3co + a) + L(/io> cos co +
37 3 co cos (3co + a))
= 7 if sin ut + L/io) cos coZ + 7 3 r sin (3o>Z + a) +
3L7 3 o> cos (3coZ -f a)
lir sin ut + IiX cos ut + I s r sin (3orf + a) +
I 3 x 3 cos (3orf + a). (52)
Let - = tan /?; = tan /3 3 .
Then
r = Zi cos j8 = 2 3 cos /3 3 .
Substituting these values in (52),
e = /ii(cos j8 sin co + sin cos $)+ /3^s(cos 3 sin
[3orf + a] + sin & cos [3orf + a])
= JiZi sin (coi + 0) + 7 3 Z a sin (3w< + a + ft)
= Ei sin (co + 0) + Ei sin (3arf + + ft)- ( 53 )
Thus the amplitude, EI, of the fundamental voltage wave is
Zi times that of the current fundamental; the amplitude E 9 of
the triple frequency voltage wave is Z 3 times that of the cur-
rent triple frequency harmonic.
136 ELECTRICAL ENGINEERING
The difference between the multipliers, Zi and Z 3 , is due to
their respective reactances, x and x 3 , since r is the same in each.
But z 3 is 3x.
Therefore, it is seen that the triple frequency voltage wave
is greater in proportion to its fundamental than the triple fre-
quency current wave to its fundamental. In other words, the
voltage wave is more distorted.
Conversely, it may be said that when a distorted voltage is
impressed on a circuit, the effect of the inductive reactance is
to smooth out some of the distortion in the current wave.
Problem 69. Show that when the e.m.f.,
e = EI sin (at + E 3 sin (at + a),
is impressed on a circuit of resistance- only, the current flowing will have
the same amount of distortion as the voltage has.
Problem 60. Show that when the e.m.f. of problem 59 is impressed on
a circuit containing resistance and capacity, the effect of the capacity is to
increase the distortion of the current.
If the voltage (53) is measured by a voltmeter, what will the
reading be? From the development of (51) in respect to dis-
torted currents, since both currents and voltages are similar
in form it follows that the effective e.m.f. shown by a voltmeter
will be
E =
Problem 61. In Fig. 110 let E be the known impressed voltage, let
capacity = C farads, inductance = L henrys and resistance = r ohms.
Then
X e = ; XL = 27T/L.
Find the current, and the voltage drops across the inductive impedance
and the capacity, when the impressed voltage is composed of a fundamental
and a third harmonic. The fundamental component of current will be
where
b = -
Zo"
and
Zo 2 = r 2 + (XL x c )*,
and EI = e\ is the zero vector.
DISTORTED WAVES. RESONANCE EFFECTS 137
The voltage drop across the inductive impedance, z, due to /i, is
Eiz = I\Z = ei(g -{- jb)(r -\-JXL) = a + jb'
where
a = e\gr e\bx] b r = (br + gx}e\.
The voltage across the capacity reactance due to /i is
Eic = 7i(0 - jxc) = d +jf,
where
d = eibxc] f = e\gx c .
The third harmonic components of current and voltage are similarly deter-
mined, remembering that
x 3L = SZL,
_ _?5.
Problem 62. In the circuit of Fig. 110,
let r = 1, L = 0.0265, C = 0.000265, E =
100, E 3 = 30. Find and plot the current
waves /i, 7 3 and 7, as the fundamental p IG HQ
frequency is varied.
NOTE. Solve for frequencies of 15, 20, 25, 35, 50, 55, 60 ,65, 75, 100.
Solution. We have, first,
*T
Then
where
7 3 =
where
- # 3 2 = lOO 2 - 30 2 = 95.
95 95r-a;i) 95r
95(r-.ysi) _
r 2 + zi 2 r 2 +
1
= XiL ~
30
30r
+
+ji f 3,
= x 3L -
1 600 200
= 2./3L - ^77, = 0.167/a - 77 = 0.5/t - -jr-
27T/ 3 C
Waves of 7 lf 7 3 and / are plotted in Fig. 111.
It is seen that the maximum /i occurs when XIL
t * when
0.167/1 = -7-. or, at the frequency /i = 60, and it is I lm = -j = 95 amp.
200
Maximum 7 3 occurs when x 3 L = x 3 c, i-e-j when 0.5/i = -71 or, at the fre-
30
quency /i = 20 and it is I sm = ~r = 30 amp.
138 ELECTRICAL ENGINEERING
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DISTORTED WAVES. RESONANCE EFFECTS 139
40
Frequency
FIG. 111.
Fig. Ill gives a striking illustration of resonance, in that there are two
distinct resonance points.
Correspondingly more points of resonance would be produced if the
e.m.f. wave possessed more harmonics. The higher the harmonic, the
lower the frequency at which resonance occurs.
CHAPTER XXI
CONSTANT POTENTIAL CONSTANT -CURRENT TRANS-
FORMATION (Cont'd from Chapter XVII)
It has already been shown how a fairly constant current may
be obtained from a constant potential source by the use of re-
sistance and inductive reactance.
It is recollected that either the efficiency or the power factor is
poor, and that the range of fairly constant current is quite narrow.
While a far better control can be obtained by the introduc-
tion of a condenser in connection with the reactance, this latter
method has found little practical application because of the
rather high cost of condensers and their unsatisfactory operation
with the distorted current taken by arc lamps. It is, however,
probable that such system will be extensively used in the future
on account of the fact that series incandescent street lighting is
being used to an increasing extent.
In Fig. 112, let E be the constant voltage impressed on the
system, and let E be the variable voltage across that part of the
system in which the constant current, 7,
is to be maintained. Let the system be
composed of z\ = r\ + jx\, in series with
the parallel circuits, z 2 = r 2 + jx z and the
* 2 f [ f variable load impedance, z r + jx. E,
the voltage across the variable impedance,
FlG 112 will be taken as zero vector, = e, although
it might seem more logical to use E Q
which is known. (With E Q chosen, the calculations are some-
what more involved, yet, of course, perfectly possible, and the
student is advised to apply both methods and verify results.)
/ 2 = - = e (g 2 + j6 2 ), (55)
22
E = e + I l Z l = e(h+fl).
140
CONSTANT-CURRENT TRANSFORMATION
Numerically,
#o =
and
141
-
which is the value of the zero vector.
Substituting this value of e into (54), (55) and (56),
(58)
(60)
These equations give the currents in the three branches of the
circuit, but there is nothing about them which determines that
7 shall be constant. EQ, g% and 6 2 are constants, g and b are
variable, and g and b Q are, respectively, g + $2 and 6 + 62.
h and I are made up of combinations of g, #2, b, 6 2 and rj. and x\,
where r x and #1 are constants. I\ and 7 2 are essentially variable
because of the variation of z, the load impedance.
Problem 63. Let a circuit be chosen as in Fig. 113,
such that 7*1 = r-z =
Xl = - x, = 125 ohms
x = 0.
Let
Eo = 250 volts.
Find e, /i, 7 and the power factor, and plot
them against r for varying values of r from
to 500 ohms. FIG. 113.
x z
Solution.
Since
.'. (54) becomes
Similarly, (55) becomes
x = 0, 9 = -' t b = 0.
and (56) becomes
e I - 7
142 ELECTRICAL ENGINEERING
and
E = 250 = e +
whence
+ -<*+*>,
Substituting values for xi and z 2 ,
125 *" 1252
(57) then becomes
(58) becomes
(60) becomes
250
-lM- 2r -
r
/ = = 2 = constant for all values of r.
+ 12 2
62.5
This case is, of course, ideal in that it assumes absence of re-
sistance in both reactive branches of the circuit.
In Chap. XVII, it was found that with the system which used
only inductive reactance to obtain constant current, the power
factor was quite low.
To obtain an expression for the power factor in the present case,
h jl) = eh + jel,
,e
whence, by telescoping, the power is
_o n ^ m
e i x\\ e x\ e
= ( 1 -f- ) - =-
r \ xj x 2 r r
Substituting numerical values, P = 4r.
CONSTANT-CURRENT TRANSFORMATION
Volt-amp. = #0/1 = (eh + jel) (- - j -}
\T 3/2/
e% eH
143
Substituting values for h and I,
This equation reduces to
when Xi = x^ and, substituting numerical values,
1 = 4Vr 2 + I25 2 '
.'. Power factor =W^F~ =
Tabulating :
2 + 125 2
T
o
10
20
50
100
200
500
e
20
100
40
400
100
2,500
200
10,000
400
40,000
1,000
250,000
r 2 + 125 2 . . .
15,625
15,725
16,025
18,125
25,625
55,625
265,625
Vr 2 + 125 2
P.F
/i
7 2
125
2
125.2
0.080
2.00
16
126.5
0.158
2.02
32
134.5
0.372
2.15
8
160
0.625
2.56
1.6
236
0.847
3.78
3.2
515
0.97
8.25
8.0
Fig. 114 shows the curves of current, voltage, and power factor
for change of resistance of the load.
In this case, the power factor is seen to be very much better
than it was found to be where inductive reactance alone was used.
Problem 64. Let the load in the preceding problem be made up of both
r and x. Find the effect of reactance in the load and make a general study
of the conditions under varying power factor of the load.
144
ELECTRICAL ENGINEERING
This may be done as follows: Imagine the load to consist of any number
of lamps, each lamp possessing a certain resistance and a certain reactance.
Then the ratio of - will be constant.
1. Let x = 0.5r. The power factor of the load will then be
1
V/r 2 +0.25r2 Vl.25
0.895
2. Let x = r. The load power factor is then 0.707.
Supplying these values in turn to Eqs. 54, etc., as in the previous problem,
tabulations and curves may be obtained from which a report on the effect
of power-factor variation of the load may be made.
In order to bring the subject to a practical basis, the effects of
actual constants of the apparatus should be investigated. It
400 500
Resistance
FlG. 114.
will be found that the resistance of a suitable reactive coil for
such a case as developed above, would not need to be above
0.5 ohm, and that the resistance in the condenser circuit would
be very much less. Consequently the effects of resistance are
extremely small, and the case, as worked out, may be considered
as approximately attainable in practice.
Many schemes have been proposed for the attainment of con-
stant current from a constant potential source, in which more or
less elaborate combinations of reactances have been arranged. A
CONSTANT-CURRENT TRANSFORMATION 145
study of the possibilities of different schemes is profitable for the
student as it affords excellent practice in circuit calculation. 1
Power and Wattless Components of Volt-amperes. The
quantity El cos a is called the power component P of the volt-
amperes.
By a similar conception, El sin a is called the " wattless"
component P 1 of the volt-amperes.
Thus
El = V(EI cos a) 2 + (El sin) 2 -
Referring to Eq. (36)
El sin a = El sin (0 - 7)
= El (sin )8 cos 7 cos sin 7)
= EI ^]^ = i'e - e'i. (61)
1 For some of these developments see STEINMETZ, "Alternating-current
Phenomena," Chap. X.
10
CHAPTER XXII
THEORY AND USE OF THE WATTMETER
The most accurate way of obtaining results in the measure-
ment of alternating current, voltage or power is by the use of the
electro-dynamometer.
As generally employed the electro-dynamometer, invented by
SIEMENS is a combination of 2 coils, one movable and the other
fixed, whose planes are set at right angles to each other. When
current is sent through the coils, each sets up a magnetic field
in the region occupied by the other, thus causing forces which
tend to move the coils relatively to each other. The forces are
balanced by tension on a calibrated spring.
If the same current is sent through both coils, the scale, prop-
erly calibrated, measures the current. If the instrument is
placed in series in a circuit, the current measured is that of the
circuit, and the meter becomes an ammeter. If it is placed in
shunt to a given circuit, the current in the coils is proportional
to the voltage drop in the circuit, and the meter becomes a volt-
meter.
If, however, one coil is placed in series and the other in shunt
to a given circuit, the effect on the instrument is proportional to
the product of the amperes and the volts at any instant, and the
electro-dynamometer becomes a wattmeter and measures power. 1
For practical construction, the coil to be connected in series is
made of few turns of comparatively heavy wire, and is usually
the fixed coil, while the coil to be connected in shunt is made of
very many turns of fine wire and is movable.
Accurate results are obtained by the dynamometer because the
coils, while readings are made, are always kept in the fixed rela-
1 The flux set up by 1 coil (fixed) is proportional to the current flowing
in the circuit, while the flux set up by the other coil (movable) is proportional
to the voltage across the circuit. But the force at any instant acting on the
coils is proportional to the product of the fluxes set up by the coils, that is,
the force on the coils is proportional to the product, E X /, where
E = voltage across the circuit, and
/ = current in the circuit.
146
THEORY AND USE OF THE WATTMETER 147
tive position at right angles to each other, thus eliminating mutual
flux; 1 also because no iron is used in construction, and there are
no other materials which might cause variation in the results.
Accuracy must be obtained, however, by the correction of
certain errors in the readings. The error due to friction of the
movable coil is small. The error due to changes of resistance
by temperature is obviated in good
instruments by the use of resis-
tance which is not affected by
change of temperature. When
used as a wattmeter, the readings
of the dynamometer must be cor-
rected for error due to the manner
of connection in the circuit. This
correction is of great importance.
Let the wattmeter be connected
as in Fig. 115 (A), in which the
power consumed by the impedance,
Z = R + jX, is to be measured.
The current coil is represented
by the impedance z r + jx] the
voltage coil by the impedance z\ = <
7*1 -\-jx\. The load voltage is e,
which is chosen as the zero vector.
(B)
(c)
FIG. 115.
The meter should be first calibrated by direct current, so that
its reading for any given direct-current power is known.
Wattmeter Connections. Connection (A) is wrong, because
the current coil has to carry /i, the current taken by the voltage
coil, in addition to the load current /, thus causing the wattmeter
to indicate the power lost in the voltage coil, plus the load. There
will thus be a reading even at no-load. If JiVi, the power lost in
the voltage coil, is subtracted from the wattmeter reading this
error is eliminated. This error is usually negligible in connection
with circuits carrying large current at low voltage.
In the connection shown in Fig. 115 (B), the current coil
carries only the load current. The voltage coil, however, is so
connected as to include the drop in the current coil as well as
that across the load. Thus the wattmeter indicates the power
1 Mutual induction will be treated more fully in connection with the
study of the transformer. See Chap. XXVI.
148 ELECTRICAL ENGINEERING
lost in the current coil in addition to the load. This error may be
corrected by subtracting the Io 2 r power lost in the current coil.
Connection, (B), is best adapted to measurements at high voltage
and low current.
A third arrangement, Fig. 115, (C), is known as the compen-
sated wattmeter. In this there is wound a fine wire coil of the
same number of turns as the current coil directly upon the latter.
By its connection, it is seen that this coil, c, carries the current of
the voltmeter coil. It therefore supplies to the current coil just
enough back ampere-turns to neutralize those due to that excess
of current in the current coil. This arrangement causes the
wattmeter to read correctly so far as its connections are con-
cerned.
Readings of the dynamometer calibrated by direct current
must also be corrected for an error due to change in frequency.
In connection (A), Fig. 115, the load current is
Current in the voltage coil is
e
Current in the current coil is
/o = 7 + 1 1 = e(g Q + jb ).
\ These currents are plotted in Fig.
" e l!6. Usually, the angle of lag of /i, is
very small, even less than 1. This
slight lag may, however, give a large
error.
Tan 7 =
Since the coils are at right angles, the
torque at any instant on the movable coil is proportional to
the product of the currents.
Thus,
T = ki i 1}
where i Q and ii are instantaneous values of current in the two coils.
THEORY AND USE OF THE WATTMETER 149
Let
io = I 0m sin coJ, and ii = I lm sin (ut + a).
Then the average value of the torque through one-half cycle is
fc/
- I (/ Ow sin ut)(Ii m sin (co< -f a)dt
/i A .
I (si
.
sin 2 co< cos a + sin ut cos w sin a)dt
o/] Ah - cos2o; sin2co 1
I ~~2~~ ~ COS a "^ --- 2 Sln a
2fc/o/i["fa)< cosa __ cos a sin 2co^ sin a cos 2<on*
TT L 2 4 ~4 Jo
7r sin a , sin af| 2fc _
cos a -- - + -- = 7 /i cos a
cos a,
/o and 7i being, as usual, the effective values.
The wattmeter reading is then proportional to
/i/o cos a,
where a is the angle between J and /i.
But the true power is el cos ft where ft is the angle between
e and /.
In order that the wattmeter shall read true power it is therefore
necessary to correct each term in the reading. These corrections
are: (a) for the current in the voltage coil; (6) for the currerit in
the current coil; (c) for the angular displacement.
(a) Assuming the meter to be calibrated by direct current, the
current in the voltage coil at any frequency, /, will be less in the
ratio /---=> due to the inductance of the coil.
V ri 2 + Zi 2
The reading should therefore be corrected by the factor
in order to bring it proportional to e.
(b) The current in the current coil is too large in the ratio, -j
The correction factor is therefore,
L
/o
150 ELECTRICAL ENGINEERING
(c) The correction factor for angular displacement is evidently
^-^i where and a are as shown in Fig. 116. The complete
cos a
correction factor is then
21 IG 2 + B 2 cos ft
The constants of the wattmeter are assumed known, which per-
mits of obtaining all angles except the phase angle of the load.
Thus, a is known, but not 0. In order to obtain 0, a reading may
be taken of the wattmeter, voltmeter and ammeter as in the ex-
ample below. Then, roughly,
= ~- (62)
Substituting this value of cos into the correction factor, a
new value of W is obtained. Replacing the approximate W of
(62) by this new value, a new value of cos is obtained in which
the error is of the second magnitude. By repeating this process
any desired degree of precision may be obtained.
Example. To find cos /3, when by
reading of instruments the approxi-
mate power factor is found to be
= _ = 50
volt-amp. 100
pow^r Factor' 25 There must be a correction-factor
Fm. 117. curve of the dynamometer for varying
power factor.
From this curve, let the value of k be 0.99 f or P.F. = 0.5. Then
multiplying, 0.5 X 0.99 = 0.495 = power factor to second ap-
proximation. It is evident that a repetition of the process will
be hardly necessary in most practical cases.
Problem 66. With the wattmeter connected as above (115, A), determine
and discuss the correction factors: (1) with non-inductive load; (2) when the
power factor of the load is just equal to the power factor of the voltage
coil; (3) in the theoretical case when there is no self-induction in the voltage
coil.
Problem 66. By a process similar to that just given, find the correction
factors for wattmeters when connected according to (Fig. 115, B and C).
The errors actually obtaining in practice with good commercial
indicating wattmeters are quite small. Thus, at normal voltage,
THEORY AND USE OF THE WATTMETER 151
2000 cycles and power factor from 0.8 leading to 0.8 lagging,
the error is usually less than J4 per cent.
As the power factor is lowered the error becomes larger. At
normal voltage, 60 cycles, the error may be less than 0.2 per cent,
with the power factor down to 0.1.
If the impressed voltage is low, say 15 per cent, of normal, the
error may, however, be several per cent.
In operation there are also errors which enter
with the use of "current" and "potential" trans- j| g
formers. When these are used, the error is |^iovoit 5
practically negligible for power factors above f ' _j
0.8 except for small loads. With non-inductive-
load of, say 10 per cent, normal, the error may,
however, be several per cent.
Problem67. An uncompensated wattmeter (Fig. 115, A &ndB and Fig.
118) has a rating of 400 watts. At 100 volts the resistance of its voltage
coil is 2000 ohms. At 50 volts the resistance is 1000 ohms, at 10 volts it is
200 ohms. The inductance of the voltage coil is 0.007 henry. Resistance
of the current coil is 0.03 ohm. Inductance of the current coil is 0.0003
henry. Find the wattmeter reading, the actual watts and the correction
factor for all combinations of voltage, current and power factor, when
e = 100, 50.0, and 10.0, volts
7 = 4, and 0.4 amp.,
P.F. = 1, 0.1 lead, and 0.1 lag.
/ =60 cycles.
CHAPTER XXIII
SIMPLE PROBLEMS IN ELECTRO -STATICS
It is desirable at this point to introduce certain principles of
electro-statics. These should, of course, be more or less familiar
to every student who has had an adequate course in physics.
Potential. By definition, the potential at a point in an electric
field is equal to the work done per unit charge in bringing a
positive charge from a place of zero potential (usually infinity)
to the point.
Intensity. Also by definition, the intensity of the electric
field (lines per square centimeter in air) is numerically the same as
the force which that field exerts on unit charge.
FIG. 119.
Thus, if R is the intensity of the field at a distance r from a
point charge, Q (Fig. 119). See also Chap. XVIII.
R = _ _ = = Q
area of sphere of radius r 4.irr 2 r 2
Therefore the potential at p is
VP = - fRdr = - fR cos 6ds (63)
where ds is an element of the path of the unit charge and
dr ds cos 6.
The minus sign is used because work is done in bringing unit
positive charge against the charge Q which is also assumed posi-
tive. Thus, the repulsion between the charges must be overcome
152
SIMPLE PROBLEMS IN ELECTRO-STATICS 153
and work has to be supplied. This designation is, of course, a
matter of convention.
Substituting the value of R, just obtained,
V P = - 1 Rdr = - \ Qdr =
J- Jj 2 r_ p
Q
P Q
where p is the distance from Q to p.
Capacity of a Sphere. Suppose the charge, Q, to be on an iso-
lated sphere of radius, n. Then, at the surface of the sphere
p = ri, and the potential is Vi =
The capacity of a condenser is defined as the charge per unit
potential. Thus, C = y, where C is capacity, and V is the
potential of the charge Q.
Since, therefore, with an isolated sphere, Vi = i the capacity
of the sphere is
in cm.
Thus, the capacity of a sphere is numerically equal to its radius;
the value of the capacity expressed in farads, C is found by
dividing C in centimeters by the constant 9 X 10 11 .
Potential Gradient. The potential gradient, usually denoted
by G, or the rate at which the potential changes at a given point,
is of very great practical importance since it is a measure of the
electric stress to which the dielectric is subjected. The potential
gradient, G, and the electric field intensity, R, are the same nu-
merically. Thus, if the potential of a certain point falls at the
rate of 5 units of potential per cm., the actual number of lines
per sq. cm. at the point is also 5.
By definition,
Since,
dV = - Rdr
.
In a dielectric of specific inductive capacity, K, the intensity
as well as the potential gradient for a given charge is less than in
154 ELECTRICAL ENGINEERING
air. It is - times the intensity in air. Thus, in the case of a
sphere,
c v l Q
G== R== - -,-
The maximum possible value of G, or R, under ordinary con-
ditions in air, is not known exactly, but is in
the neighborhood of 30,000 volts per cm., or
100 electro-static units of potential.
Capacity of a Spherical Concentric Con-
denser. Consider 2 spherical concentric bodies
with charges plus and minus Q.
By (63), the potential difference is
r* = r
= - Rdx,
Jx = n
where r and n are ^radii respectively of the inner and outer sur-
faces of the condenser.
But
Q
-- r
t/ n
Therefore the capacity of the condenser is
C = Q -?^-in cm. (65)
ri -f- r
Potential gradient between concentric spheres.
Since
rfF
G = T->
dr
and
dV = - Rdr,
But
SIMPLE PROBLEMS IN ELECTRO-STATICS 155
rr
c =
r
At the surface of the smaller sphere, x = r, whence the gra-
dient is
7*1 V
= 7 (r, - r)'
The Capacity of a Concentric Cylinder. Let the charges be
Q per cm. of length of the cylinder (Fig. 121).
Then, by GAUSS' theorem, the flux emanating
from each centimeter of length = 4irQ.
Lines of flux are here assumed to extend radi-
ally, which they actually do.
At any distance, x } from the center of the
cylinder the intensity at a point is the total
number of lines divided by the area, or,
Thus, the potential difference is:
dx
+ 2Q log
r c^Q
I Rdx = I dx = 2Q[log r log rj
Jn -Jri
(66)
and the capacity is
n
C = =
in cm.
per cm. length of the concentric cylinder.
The gradient at any distance, x, from the center is
x x x " _ . r, . r, (68)
2 log - x log -
At the surface of the inner conductor, x = r.
a
r log -
& r
i and this is the greatest value of the gradient.
1 See "Advanced Course in Electrical Engineering."
156
ELECTRICAL ENGINEERING
In these formulae no account is taken of any effects due to the
ends of the concentric cylinder. For the special case of an outer
cylinder of radius ri = , C = 0.
Capacity of Two Parallel Plates so Large that the Effects of
Their Edges may be Neglected. The total flux set up by a
charge, Q, is IwQ (Fig. 122).
4x0
The intensity, R j->
where A is the area of one side of
the plate.
The potential difference is:
A
t +Q
1 f
/
^
1
1
X 1
^ -Q
1
V
JS
FIG
122.
-$>
*> ' -
where d is the distance between the plates.
The capacity
Q A K A
C = ~~ = ~ 4 j = - A ; in cm.
e 4ird 4?ra
(69)
(70)
where the dielectric has a specific capacity K.
The potential gradient, G y is a constant in the dielectric be-
tween the plates, since the flux lines are parallel.
Thus,
C - ^Q _ ^TrCe _ efc
~ dx = ~~A A = ~A
in which e is the difference of potential of the plates and k is a
constant, = 4irC.
Capacity of a Transmission Line. 1 The line is represented in
section in Fig. 123, with, r, the
radius, and, D, the distance be-
tween centers, of the wires A and
B. Let A be charged + Q, and"*"
B } Q. The flux lines emanat-
ing from A enter B. The inten-
sity at a point, p, due to the charge on A, is R A ; that due to
the charge on B is R B .
*|^r
I
FIG. 123
1 For more exact deduction see "Advanced Course in Electrical
Engineering."
SIMPLE PROBLEMS IN ELECTRO-STATICS 157
Then
r> --" v& ^vl
RA - ^z =
p
"
2ir(D-x) ~~ (D-x)
The intensity due to the two charges is the sum of R A and R B ,
since the direction of the lines of electro-static force from A , due
to a positive charge, is the same as that due to B, which has a
negative charge.
The potential difference is:
C r C T /I 1 \ dx
e = - \ Rdx = - 2Q I (- + ~ -)
JD-T JD-T \* D ~ x/
= 4Q log ^^ (71)
and the capacity is therefore
1
(72)
per cm. length of circuit not of wire.
This capacity is expressed in centimeters. If the line is in a
dielectric of specific inductive capacity, AC, the capacity in air as
determined above, must be multiplied by K.
To transform capacity, expressed in electro-static units in (72),
into electromagnetic units, the former should be multiplied by
2 where v is the velocity of light = 3 X 10 10 cm. per sec.
The practical electromagnetic unit of capacity is the farad.
C
Capacity in farads = -^ X 10 9 ,
where C is capacity expressed in electro-static units.
.'. Farads = electro-static units X
..Q// 9 \/
Thus, C/cm. of circuit, in farads, =
k
4 log ^~ X 9 X 10" (4 logio ^f- r ) X 9 X 10"
158 ELECTRICAL ENGINEERING
When connected to a source of alternating e.m.f., the effective
value of the charging current is I c = 2ir{CE, where E is the effect-
ive value of the line voltage.
The voltage is frequently taken from one
~E~ n " side of the line to neutral, that is, to the
E --- * -------- point of zero potential of the system (Fig.
| _ 124). When this voltage to neutral is used,
FIG. 124. the capacity to ground, or to neutral, is
twice as great as the capacity between lines.
This follows since I c = 2irfC n E nj where C n and E n are capacity
E
and voltage to neutral, and for single phase systems, E n = ^r-
z
For three-phase systems, E
k
2 X 9 X 10 11 X log
_ , farads per cm. of line, since in
r
using the neutral, the length of line is the transmission distance.
0074
Reducing values to practical units, C n /1000' = - ^ _
logic
is the capacity to neutral per 1000 ft. of line, in micro-farads.
7 C /1000' = 1Q n 6 is the charging current per 1000 ft. of line, in
amperes.
Capacity of a Three-phase Cable. Capacity to neutral per
1000 ft. of line is given in micro-farads by the formula
0.0074 i
Cn/1000'
V3a R* - a 2
logio
r R* + a 4 +
Such a cable is represented in section in
Fig. 125, where R is the radius of the sur-
rounding sheath, a is the distance from the
center, or neutral point, to the center of one FlG - 125>
of the wires and r is the radius of 1 wire.
Problem 68. (a) Prove that the greatest charge which may be put on a
ball of 10 cm. radius is 10,000 electro-static units. (Assume that the maxi-
1 This will be understood from later discussion of polyphase systems and
deduced in the volume dealing with advanced electrical engineering.
SIMPLE PROBLEMS IN ELECTRO-STATICS 159
mum gradient is 30,000 volts per cm. when air at atmospheric pressure
"breaks down" and a glow called corona appears around the wire.)
(6) Prove that the greatest surface charge, in coulombs per sq. cm., is ~ffp'
(c) Show that if the inside conductor of a concentric cable has a radius
of 1 cm., and the outside conductor is 2 cm. in radius, 0.0027 coulombs
must be put into 1 mile of cable to cause it to glow (corona). Show that
the potential difference between the 2 conductors is 20,800 volts.
Inductance of a Concentric Cable. The inductance is recol-
lected to be the interlinkages of the flux and turns per unit current.
In general, if the m.m.f . acting in a circuit is F then the flux
4iTrF X area of magnetic circuit
length of magnetic circuit
The interlinkage factor is the fraction of the total current en-
closed by the flux, and
is
L = 7 S flux X interlinkage factor.
(73)
Consider first the flux in the inside conductor due to the as-
sumed uniform distribution of the current
in it.
At a distance x from the center (Fig. 126),
TTX 2
the m.m.f. is ^ / where I is the total current.
?rr 2
The area enclosing the flux per centimeter
length of conductor is dx and the length of
the magnetic circuit is 2irx
x* ,dx_ _ x_
. Wi = 47T r , I 2wx - L r , a
TTX 2
This flux interlinks with ; of the total current; thus the
FIG. 126.
?rr
u
interlinkage factor is 5-
FIG. 127.
(Assuming that /* = 1) (74)
d Xl Between the conductors, the flux inter-
links with the whole current (Fig. 127).
Thus by a similar reasoning we get :
fr = 21 '*f
160 ELECTRICAL ENGINEERING
The current in the inner conductor interlinks with the entire
flux which is in the outer conductor but which is caused by the
difference in m.m.f. in the inner and outer conductors.
At distance x the m.m.f. is thus
__
"
R Q 2 - R 2 R, 2 - R 2
The interlinkage of this flux with the current in the inner con-
ductor is, of course, unity, thus
The inductance of the outer conductor should be added to give
the total inductance of the cable.
The m.m.f. is shown above to be
j. R<> 2 - x<> 2
R, 2 - R 2
(Ro 2 - x, 2 )
(Ro 2 - R 2 ) 2
2
'
1 R 2 + R 2 2Ro 2 R 2 Ro
2 R Q 2 - R 2 + (#o 2 - ^R 2 ) 2 g R
The total inductance L = LI + L 2 + L 3 + L 4 which is readily
proven to be
1 01 R , 2/V Ro 1 3^o 2 - R*
L = - 2 + 21og 7 + ^g^kg - - - R ^ _ R2 cm.
This inductance is expressed in the absolute system of units.
By dividing by 10 9 the inductance is expressed in henrys.
Problem 69. Prove that there is no flux outside of the sheath, the flux
set up there by the current in the sheath being exactly neutralized by the
flux set up in the same space by the oppositely directed current in the
inner conductor.
Inductance of a Transmisson Line. Let a transmission line
be represented as in Fig. 128 by 2 conductors, A and B, of
radius r. Let the distance between their centers be D. Each
conductor surrounds itself with flux lines, the directions of which
are indicated by arrows. The flux through any zone of width,
dx, between the conductors, due to the current in A, is
SIMPLE PROBLEMS IN ELECTRO-STATICS 161
where x is the distance of the zone from the center of A, and
F x is the m.m.f. due to A.
Similarly, the flux through dx, due to the current in B is
The flux due to both A and B is then
/ /
The inductance due to the / I /
interlinkages of the conductors f I j
with the flux between them is \ 1 \
then, since F x = I in this case, \\
V
, D-r 4 M . D - r ,
= 4/i log ^ ' cm. or j^ log - henries per cm.
To determine the total inductance per centimeter length of
circuit, that due to mterlinkage within the material of each con-
ductor must be added. This has been found (74) to be ^ for
2
each conductor. Therefore, the total inductance is
L (total) = 4;u log -- + n cm. per cm. of circuit (75)
In practical formulae, this becomes
L = 0.000015 + 0.00014 logio^-
in henrys per 1000 ft. of wire, not 1000 ft. of circuit.
Note that if the capacity between transmission lines is given
in farads and the inductance in henrys ~7ffi ^ on ^ verv
less than the velocity of light which is 3 X 10 10 cm. per sec. or
187,000 miles per second.
Problem 70. Explain the effect of increasing the size of the wire on the
inductance of a transmission line.
Similarly, explain the effect of increasing the distance between the wires.
11
CHAPTER XXIV
DISTRIBUTED INDUCTANCE AND CAPACITY
In the electric and magnetic problems dealt with so far it has
been assumed that the electro-static and magnetic fields propa-
gate with infinite velocity. In other words, it has been assumed
that the instantaneous values of the currents and e.m.fs. are the
same at all points of the circuit. This of course is practically
true except in very long transmission lines, since the propagation
of the electric and magnetic fields in a dielectric such as air is the
same as that of light, or very nearly 3 X 10 10 cm. per sec. or
187,000 miles per sec., and along a transmission line it is re-
tarded only a small percentage due to the fact that the current
is not confined to the surface of the conductor.
Assuming, however, that the transmission line is very long,
say 300 miles, then the time interval between, say, the maximum
, , value of the current at the beginning
dl : "1 r~^* an d the end of the line is evidently
;Hj20 sec -> corresponding in a 60-cycle
- system to approximately one-tenth of
one cycle, or, approximately, 36 in
FIG ' 129 ' time phase.
It is thus seen that in a long transmission line not only do the
instantaneous values of the currents and e.m.fs. vary from instant
to instant, but at a given instant the values of the currents and
e.m.fs. are different at different points of the line.
This problem has been treated very completely by many
authorities. The simplest solution appears to be that by STEIN-
METZ, 1 which is largely followed in the succeeding paragraphs.
Let Fig. 129 represent a long transmission line. Let r = re-
sistance per unit length of line, X Q = reactance per unit length
of line, </o = leakage conductance per unit length of line, 6 =
capacity susceptance per unit length of line, r = rj, = total
resistance of the line. Let dl be any small section of the line.
Then assuming sine wave of current when complex representa-
1 "Electric Discharges, Waves and Impulses."
162
DISTRIBUTED INDUCTANCE AND CAPACITY 163
tion can be used, the current entering this section is / + dl.
The current leaving the section is /.
Then / + dl - I = E(g Q + jbo)dl is the small difference in
current in passing through dl, or is the combined leakage and
capacity current across the section of width dl. This may be
written
dl = EY<41 (76)
Likewise, dE = 7(r + jx Q )dl is the e.m.f. consumed by the re-
sistance and reactance of the section dl, or
dE = IZodl (77)
(76) and (77) become, then,
dl
_
Differentiating these,
dE
~dl
d 2 E
(78)
dl
dl
dE
Substituting values of -rr and r from (78),
dE
jr
(79)
Thus, the second differentials of E and / are found to be pro-
portional to E and /, respectively. Since the two equations are
similar, their solutions are similar, differing only in integration
constants.
The equations (79) are of the form
whose solution is
y = Ae
164 ELECTRICAL ENGINEERING
The equation of the current then becomes
or if, for the sake of briefness, YoZ Q = v
I = Ac h + Be' 1 '
We also have, from (78),
Ti 1
Differentiating (81),
~ = Ave* -
dl
Substituting this into (82)
E = ~ [Ave lv -
-*
(81)
(82)
(83)
(84)
Representing the exponentials of (81) and (84) in series,
Substituting these into (81) and (84) gives:
72 7 ,2 72,,2
7 = A + Alv + A -- + . . . + B - Blv + B ~- -
= A + B + lv(A -B) + --(A +
i n v \
= (A + B) (l + ^) + (A - B)lv
and, similarly,
(85)
If I is made any length counting from the receiving end of the
line at I = 0, E = e, the receiving end voltage, and / = /i, the
load current, both of which may be known.
DISTRIBUTED INDUCTANCE AND CAPACITY 165
Substituting these values (85) becomes, for the receiving end,
/i = A + B
By substituting these values of I I and e in (85) we get finally:
(87)
where / and E are the values of current and voltage at any point, /,
along the line. The current and voltage at the generator are
found by substituting in (87)
7 ., 7V
1/2^2 & 1
~2~ ^T
where Z and Y are the values of impedance and admittance for
the entire line. The equations (87) become
, (88)
#o = e(.
YZ is found from the constants of the line, thus:
YZ = (g + jb) (r + jx) = gr + gjx + jbr - bx
= gr bx + j(gx + br).
Problem 71. Transmission Line Calculation. A 200-mile, three-phase
transmission line is composed of three No. 000 B. & S. wires, and runs at an
altitude of 1200 ft. where it may be assumed that the corona loss is 1 kw. per
wire per mile at a potential difference of 125,000 volts between the lines.
Let E = 125,000 volts at receiving end, between wires;
/ = 60 cycles; r = 64 ohms per wire;
x = 154 ohms per wire; g = 0.000038 per wire;
6 = 0.00107 per wire; D = 10 ft. = 304.5 cm. between wires.
166 ELECTRICAL ENGINEERING
Check the constants of the line and find :
Power per phase at the generator = P .
Current per phase at the generator = 7 .
Voltage per phase at the generator = E .
Volt-amp, per phase at the generator = E I .
r>
Power factor at the generator = ^ry
Power per phase of the load = P.
Volt-amp, per phase of the load = el.
Power factor of the load = -*
p
-*
P
Efficiency of transmission
Voltage regulation = (Eo e) -f- e.
Plot the voltage and current vectors for both ends of the line.
Solution. Resistance of No. 000 B. & S. wire, from tables,
= 0.0605o>/1000 ft. at 60F.
.'. r = 0.0605 X 5.28 X 200 = 64 ohms.
Inductance = 0.00014 logic ^-^ = 0.00014 lo glo 304 ' 5 Q 5 ~ ' 52 per 1000
ft., where r = radius = 0.5202 cm.
.'. L/1000' = 0.00014 logio 585 = 0.00014 X 2.767 = 0.0003874.
L = 0.0003874 X 5.28 X 200 = 0.409 henry.
X = 0.409 X 377 = 154.2 ohms.
C/1000' = ' r = 0.0074 X 0.361 = 0.002672 micro-farad
logio jr-
C = 0.002672 X 5.28 X 200 = 2.82 micro-farads.
6 = 2T/C = 377 X 2.82 X 10~ 6 = 0.00107
W corona loss per wire 200,000 nnnn oo,
9 ~ * ~ (voltage to neutral) = (72,250) * "
1 The voltage to neutral on a balanced three-phase system is the line
voltage divided by \/3 = 7-=^-
l.f O
The current supplied from the generator is found from (88) to be:
(V7\ I V 7\
1 + "27 + Y = (10 ~ J ' 50) I 1 + T) + e Y-
YZ = gr - bx +j(gx + 6r); Y = g -f- j6; e = 72,250.
/. -^ = - 0.081 +;0.037.
YZ
1 +-2~ - 0.919 + J0.037; eY = 2.741 + J77.3.
Substituting values,
I = (100 -j50) (0.919 +J0.037) +2.741 + J77.3
= 93.75 - J42.2 -f- 2.741 + j'77.3 = 96.49 + J35.1 = 102.5 amp.
DISTRIBUTED INDUCTANCE AND CAPACITY 167
The voltage at the generator terminals is obtained in a similar way, and is
o = e(l + ^r) + IZ = 72,250(0.919 +;0.037) + (100 - J50) (64
= 80,625 + .;14,928 = 81,200 volts.
The power per phase at the generator is, by "telescoping" E I ,
Po = e i + e V = 80,625 X 96.49 + 14,928 X 35.1 = 8200 kw.
The apparent power input to the line is
#0/0 = 81,200 X 102.5 = 8325 k.v.a. at the generator.
The power factor at the generator is
Po 8200
P.F.o =
8325
0.985.
The power supplied to the load is
p = e i = 72,250 X 100 = 7225 kw.
Load
FIG. 131.
The apparent power supplied to the load is
el = 72,250 X VlOO 2 + 50 2 = 72,250 X 111.9 = 8060 k.v.a.
The power factor of the load is
.,' ; r.*--ig-".
P 7225
Efficiency of transmission = p- = OOQA = 0.882.
168 ELECTRICAL ENGINEERING
81,200 - 72,250
Regulation = - 72 25Q = 12.4 per cent.
The vectors E , 7 , e and 7 are plotted to scale in Fig. 130.
Problem 72. Consider a circuit as shown in Fig. 131. Let the con-
stants be:
r = 0.01 x = 0.02
ri = 0.01 xi = 0.02
r = 0.01 x = 0.002
When the load voltage is e = 1, and the load current is 7 = 1+ Q.5j,
find the generator voltage, current, power factor, and the voltage and current
of the branch (r , So).
CHAPTER XXV
NOTES ON THE MATHEMATICS OF COMPLEX
QUANTITIES
This chapter is inserted in order that the common mathe-
matical operations shall be kept fresh in mind by review and
frequent practice. It is very desirable that the student shall
possess and retain facility in common, though not always fre-
quent, operations. For instance:
Solve Vo.008, using log tables.
Solve 6-- 216 , using log tables.
Differentiate y = ax n ; y = ae~ ax ;
y = sin x; y = cos x;
u
y
u
uv; y = -
Find the log, and differential, of 4 3j.
Find \/-3].
Representation of Complex Quantities. The general expres-
sion for a complex quantity is A = ai -f j2. The numerical
en
FIG. 132.
FIG. 133.
value, or modulus, of the complex is A = \/ai 2 + a 2 2 and the
vectorial angle is tan" 1 a = .
These various quantities may be represented as in Fig. 132.
Then a : = A cos a; a 2 = A sin a, whence A = A (cos a + j
sin Q) = Ae 3 ' a , the latter relation being proved later.
Addition of Two Complex Quantities.
Let C = A + B (Fig. 133).
169
170 ELECTRICAL ENGINEERING
Then
C = ai + ja 2 + 61 + j& 2
= ai + bi -f j(2 + W>
and
V (ai + 6]) 2 + (a s + 6 2 ) 2 .
Multiplication of Two Complex Quantities.
Let C = A XB.
C = (ai+ja 2 )(&i+j& 2 )'
and
C =
Division of Two Complex Quantities.
A
Let C = -
4- j2
__ a\l>\
and
Tan 7 =
Similar processes may be carried out when the complex
quantities are expressed in polar coordinates.
Multiplication.
C = AB = a(cos a + j sin a)6(cos + j sin 0)
= A(cos cy cos + j sin a cos + cos aj sin |S sin a sin ft)
= AB(cos (a + 0) + j sin(^ + ).
Involution and Evolution.
A 2 = A V' a = A 2 (cos 2 a + j sin 2a).
A n = A n (cos na + j sin na).
A^ = A^ (cos - + j sin -) (89)
\ n nl
MATHEMATICS OF COMPLEX QUANTITIES 171
Since cos a = cos (a + 2?rp) and sin a = sin (a + 2irp) where
p is any integer, the simple complex expression should be written :
A = A [cos (a + 27rp) + j sin (a + 2irp)],
where there is any question about the number of different
solutions.
In evaluating such expressions, a is in radians.
Sin X and cos X may also be written as series, 1 in which
cr
Sm # =
r 3
*
.
-f- FT
Cos x = 1 - ry + r^ -
(90)
Example. Calculate, from series expression, the value of
sin 2. Since the angle must be expressed in radians,
2X27T 7T
* = "360T = 90 radians '
Substituting this value into the series,
o _
sm J " 90 6 X 90 3
= 0.0349.
120 X 90 5
The Roots of a Complex Quantity. Using the more general
expression, Eq. (89) may be written:
2irp
cos
+ j sin
(91)
n n
1, 2, 3, 4, etc., and solve, continuing
To find the roots, put p
until repetition begins.
Example. Find \/l =
where
A = 1 = 1 + jo.
A = A (cos a + j sin a) = 1 ; a = l',n = 4; tan a = y =
Tabulating, and supplying values to (91)
p 01234
2wp
4
cos
-1
-i
sm
jO jl JO jl JO
v 7 A 1 j -1 -j 1
The roots are represented as vectors in Fig. 134.
1 Developed by MACLAURIN'S theorem.
-3
FIG. 134.
172 ELECTRICAL ENGINEERING
Exponential Representation of Complex Quantities. The ex-
ponent e" may be written as a series known as the exponential
series, developed from MACLAURIN'S theorem.
Thus,
u u 2 u 3
Let
u = jB.
Then,
* _ , , # , !! +
1 + I 2 + ' ' '
je P _ JP ^
-I- T~ I 1 '" O I O I 'I A
"(i-f+S) <
These two component series are seen to be those of the sine
and cosine (90). Hence (92) may be written:
j0 = cos 6 +j sin e (93)
Since A = A (cos a + j sin a),
substituting from (93),
A = Ae ja .
Thus, a third form of writing the complex quantity, A, has been
developed.
This last may be extended by letting A = e ao . Thus,
A o Va _. ao+jat
jtl e c C
in which the exponent is complex.
Differentiation of a Complex Number or Vector.
Let
A = Ae ja .
then
dA = Aje ia da + J a dA
' = e ja [Ajda + dA] (94)
Logarithm of a Complex Number or Vector.
Cdu
Iqgn-J
We have
log A
4= f
- J 4'
MATHEMATICS OF COMPLEX QUANTITIES 173
from (94),
rt>"Ajda . CdA^ r rdA
1 * A = J ~AfT + J "At* == J ^ + J T
= j(a + 27rp) + log A.
The logarithm of a vector has thus an infinite number of
values.
/
/
CHAPTER XXVI
THE TRANSFORMER
The alternating-current transformer is used to change electric
energy from one voltage to another. This is done by interlinking
two electric circuits having different numbers of turns with the
same magnetic alternating flux.
If the two circuits enclose exactly the same flux it is evident
that the voltages induced in the windings will be proportional to
the numbers of turns. If, however, as is the case, the flux is not
exactly the same for each circuit, the ratio is slightly affected and,
as will be shown later, the secondary voltage has a value differing
slightly from what the ratio of turns would demand.
When one circuit is connected to an alternating e.m.f., the
other circuit being open, a current flows in that circuit (Fig.
135). This current is called the no-load or the exciting current,
and may be assumed to consist of two components, one of which
Inf
FIG. 135. FIG. 136. FIG. 137.
supplies magnetism to the core and is called the wattless com-
ponent, while the other supplies power for hysteresis and eddy
current losses and is called the power component.
These component currents of the exciting current may be rep-
resented as flowing in a circuit of resistance and inductance in
parallel as in Fig. 136, where e is the e.m.f. which sets up these
currents. They may be represented vectorially, as in Fig. 137.
In the latter representation i m , in quadrature with e, produces
the flux 0, but no power; 4, in phase with e, supplies the core
loss. The exciting current, 7 o, lags behind e by an angle tan -1 -r-,
ih
It is not strictly correct to represent the core loss by a resist-
ance r, Fig. 136, with varying e, for part of the core loss is pro-
portional to e 1 - 6 and part to e 2 .
174
THE TRANSFORMER
175
Neither is it correct to assume that the magnetizing component
is proportional to the e.m.f., since the magnetization curve is not a
straight line. However, in most cases, the variation of e is
slight, and proportionality may be assumed without appreciable
error.
The Transformer Diagram. The relations of voltage, current
and flux which exist in a transformer under normal operation are
shown with great clearness by the aid of the transformer diagram.
<f> represents the flux that interlinks with the primary and second-
ary of the transformer; e t - is the e.m.f.
induced in the primary and secondary
windings (assuming the same number
of turns in each). This e.m.f. is 90
in time behind the flux, as is seen
from Fig. 139 and by the following
simple proof:
If
$ = <J> m sin (
then
N d<t> N
FIG. 139.
1 2 is the secondary or load current which in this particular
diagram is shown lagging behind the induced e.m.f. 7 2 r 2 and
7 2^2 are respectively the e.m.fs. consumed by the secondary
resistance and reactance, /2r 2 being in phase with 7 2 and 7 2 z 2 being
90 ahead of 7 2 . 7 2 z 2 is the e.m.f. consumed by the secondary
impedance, which subtracted vectorially from e t gives E 2 as the
secondary terminal voltage.
The primary current may be assumed to consist of three com-
ponent parts: the first I\, which corresponds to the secondary
current and is equal and opposite thereto; the second 7 m , which is
176 ELECTRICAL ENGINEERING
the magnetizing current producing the flux <f> and is in phase
with the flux; and the third Ih, which is the power loss current due
to the core loss and is in quadrature to the magnetizing com-
ponent, that is, in phase but opposite to the induced e.m.f. e t .
I m and I h combine in 7 00 which is the exciting current.
To overcome the induced e.m.f. e t - in the primary winding an
impressed e.m.f. e { is required.
To overcome the resistance and reactance drop in the primary
windings an e.m.f. I\z\ needs to be supplied. Thus the pri-
mary impressed e.m.f. E\ is the vector sum of these.
0i is evidently the angle between the primary current and
e.m.f.
6 2 is the angle between the secondary current and e.m.f.
The total primary current, /i = I'\ -f- 7 o.
In phase with /i is the voltage, I&1, consumed by the primary
resistance, r j; and at right angles ahead of /i is the voltage, I\Xi t
consumed by the self-inductance of the primary coil.
These two voltages combine to form /iz 1; the voltage consumed
in the primary of the transformer.
The total impressed primary voltage, EI, is the sum of IiZi
and d. The angle 0i is the phase angle between EI and /i.
The transformer diagram is obviously not suitable for accurate
calculation. For this purpose, another de-
gj . velopment will be made.
.4, j lf l Let there be two mutually inductive coils,
one of them, called the primary, having NI
L N turns, TI ohms resistance, and LI henrys in-
ductance, while the similar quantities of the
FIG. 140. other, or secondary coil, are N* y r 2 and L 2
respectively.
Then, in Fig. 140, if the secondary current 7 2 = 0, the primary
impressed voltage, ei = i l r l -f LI -p where e\ and i\ are instan-
taneous values of voltage and current, and Li is assumed con-
stant. If a secondary current flows, there will be induced in the
secondary an e.m.f. e t = - L 2 -J 2 - The secondary induced e.m.f.
perturn=-f -*?.
Nt N 2 dt
If it be assumed that there is no leakage, that is, that all the
magnetic flux links with both the primary and the secondary
THE TRANSFORMER 177
coils, then the induced e.m.f. in the primary due to 7 2 must be
N l T di*
~N~ 2 L ^t'
Then,
The sign of the last term changes from to + because the in-
duced e.m.f. must be overcome, or balanced, by an e.m.f. of the
opposite sign.
But
JV2 L 2 = \LA
for, from fundamental relations,
^1 = T7^r> and #i =
Substituting,
L! =
Similarly,
L 2 =
Then the ratio
Li _
L 2 ~
whence
(96) then becomes
*'r \/rr
F 2 ^ 2 Nf ^ i ^ 2>
&
61 = i\r\ -f~ LI -37 -f- \LiL 2 j.
at at
Let \/L]L 2 be denoted by 3f .
Then,
Similarly for secondary,
_i_ / ^ 2 4. M l = n (98)
since no e.m.f. is impressed on the secondary coil.
The constant, M, is called the coefficient of mutual induction,
and may be defined as the number of interlinkages of flux with
both coils of a mutually inductive circuit when unit current is
flowing in one of the coils.
12
178 ELECTRICAL ENGINEERING
Mutual inductance, like self-inductance, is measured in henrys.
It does not always follow that M is equal to \/LiL 2 . In fact,
that condition is attained only when no magnetic leakage exists,
which never occurs. If part of the flux set up by the primary
does not interlink with the secondary, that part constitutes the
primary leakage flux. Similarly, when current flows in the
secondary, some secondary leakage flux is set up. Whenever
there is leakage flux,
M
Eqs. (97) and (98) hold at all times provided the proper value of
M is supplied, and M is usually about 95 per cent, of v LiL 2 .
Equivalent Transformer Circuit. The differential equations
given above are not readily used, but, fortunately, STEINMETZ
has evolved a simple treatment involving a diagram of simple
series and multiple circuits, which, while not showing the physics
of the phenomenon, lends itself to very simple and quite accurate
treatment.
He represents the transformer by a circuit which is shown in
Fig. 141.
r l
%
r a
J a
t
^ t ri.
f
*oo|
Si
l&oo
M
1
I
1
i V
FIG. 141.
Let the secondary or load current be 7 2 = iz + ji't, and let
the secondary terminal voltage be e 2 , the zero vector.
Then the secondary induced e.m.f . E t = e* + hZ 2 = e z + itfz
The exciting current is
/oo = EiY Q = (i -f je'i) (goo + j&oo)
The primary current is
/i = /2 + /oo = ^2 + IO
The impressed voltage is
THE TRANSFORMER 179
From these values may be obtained:
power output = e^iz,
power input = &iii -f e\i'\ t
f . . , power input e\i\ + d\i\
power factor at primary terminals = rr- = ' T >
volt-amp. EJi
regulation =
2
62*2
efficiency
In using these equations r*i, r 2 , Xi and x 2 are positive, & o is
negative because the magnetizing circuit is necessarily inductive,
iz is negative for lagging, positive for leading current.
Transformers are rated on the basis of kilovolt-amperes, not
kilowatts.
Example of Transformer Calculation. Given a 2200 to 220-
volt, 60-cycle, 50-kv.a. transformer, in which r*i = 0.97, r 2 =
0.0097. Assume that on test 98.5 volts on the primary produces
full-load current in the short-
circuited secondary (142, a).
At no-load, with the normal
voltage (220) impressed on the
secondary, the primary circuit
being open, the watts input are
WQ = 1000, and the exciting
current / O o = 12.25 amp. (Fig.
142, 6). The percentage rl
drop in the primary is
22.7
(6)
FIG. 142.
22.7 X 0.97
-2200" = 0.01 == 1 per cent.
where 22.7 is the normal primary current.
In the secondary,
per cent, rl drop =
= .01 = 1 per cent.
The total impedance, calculated from the short-circuit test
(142, a), is
QO K
Z Mal = 2277
4.35 ohms.
180 ELECTRICAL ENGINEERING
Total per cent, impedance drop, referred to the primary
voltage, is
= - 0448 = 4 ' 48 P er cent '
<n
2200
Percentage total reactance drop is then V4.48 2 2 2 = 4 per cent.
total per cent, resistance drop = 1 + 1 = 2).
Therefore, assuming primary and secondary percentage react-
ances to be equal,
per cent. Xi = 2 per cent.; per cent. x z = 2 per cent.
Thus, on the percentage basis, or assuming e 2 = 1 and i = 1, then,
ri = 0.01 xi = 0.02
r 2 = 0.01 x z = 0.02.
The core loss of 1000 watts obtained on test is supplied at 220
volts by the component of no-load current, 4.
1000
4.55 amp.
The per cent. 4 of the secondary current is
per cent. i h = - = 0.02 = 2 per cent.
0.02.
The magnetizing component of the no-load current is obtained
from 7oo and the core-loss current. Thus,
V 12.25 2 - 4.55 2 = 11.35 amp.
11 35
The percentage i m = ' = 0.05 = 5 per cent.
= - 0.05.
Having obtained the above constants, values may now be tabu-
lated to find the effect of variation of the load current with
constant power factor.
Problem 72. Let power factors of 100 per cent., 80 per cent, lagging and
80 per cent, leading be assumed, and let the calculations be made for
secondary currents of 0, 0.5, and 1.
THE TRANSFORMER
181
Tabulating :
0.8 Lagging
P.F. = unity
0.8 Leading
12 ..
i't.
oo. .
e' iff o
t oo. .
ii. . . .
ei
t'm..
Pj. . .
em . .
e'li'i.
Pi...
P.F.i.
Eff...
Reg..
0.0
0.0
0.0
0.0
0.0
0.0
0.0
1.0
0.0
0.02
0.0
0.02
0.0
-0.05
-0.05
0.02
-0.05
0.054
0.0002
-0.001
1.0012
-0.0005
0.0004
-0.0001
1.001
0.0
0.02
0.0
0.02
0.054
0.37
0.0
0.001
0.000250
0.02045
0.0001
0.0505
0.0504
0.5
0.4
-0.3
0.004
-0.006
0.008
-0.003
1.01
0.005
0.0202
0.42045
-0.3504
0.547
0.0042
-0.007
1.0212
-0.0035
0.0084
0.01
1.021
0.4
0.49
-0.003
0.426
0.557
0.762
0.939
0.021
1.0
0.8
-0.6
0.008
-0.012
0.016
-0.006
1.02
0.01
0.0204
.0005
0.0209
0.0002
-0.051
-0.0508
0.8209
-0.6508
1.046
0.0082
-0.013
1.0295
-0.0065
0.0164
0.02
1.03
0.8
0.854
-0.013
0.841
1.087
0.772
0.951
0.03
0.0
0.0
0.0
0.0
0.0
0.0
0.0
1.0
0.0
0.02
0.0
0.02
0.0
-0.05
-0.05
0.02
-0.05
0.0538
0.0002
-0.001
1.0012
-0.0005
0.0004
-0.0001
1.002
0.0
0.02
0.000001
0.02
0.0538
0.372
0.0
0.002
0.5
0.5
0.0
0.005
0.0
0.01
0.0
1.005
0.01
0.0201
-0.0005
0.0206
0.0002
-0.0505
-0.05005
0.5206
-0.05005
0.522
0.0052
-0.0010
1.011
-0.0005
0.0104
1.012
0.5
0.5265
-0.001
0.525
0.5272
0.995
0.952
0.012
1.0
1.0
0.0
0.01
0.0
0.02
0.0
1.01
0.02
0.0202
-0.001
0.0212
0.0004
-0.0505
-0.0501
0.0
0.0
0.0
0.0
0.0
0.0
0.0
1.1
0.0
0.02
0.0
0.02
0.0
-0.05
-0.05
1.0212
0.0501
1.022
0.0102
-0.0010 -0
0.02
-0.05
0.0538
0.0002
001
1.021
-0.0005
0.0403
1.022
1.0
1.041
-0.002
1.039
1.0404
0.999
0.963
0.022
1.0012
-0.0005
0.0004
-0.0001
1.0013
0.0
0.02
0.000001
0.02
0.0538
0.372
0.0
0.0013
0.5
0.4
0.3
0.004
0.006
0.008
0.003
0.998
0.011
0.01996
-0. 00055 ;
0. 02051 !
0.00022
-0.0499
-0.0497
0.4205
0.2503
0.4965
0.0042
0.005
0.997
0.0025
0.0084
0.0219
0.9974
0.4
0.419
0.0055
0.424
0.494
0.859
0.942
-0.0026
1.0
0.8
0.6
0.012
0.016
0.006
0.996
0.022
0.01992
-0.0011
0.02102
0.00044
-0.0498
-0.0494
0.82102
0.55U6
0.988
0.011
0.0055
0.0168
0.0443
0.9932
0.8
0.815
0.0242
0.839
0.975
0.86
0.954
Problem 73. Write a discussion of the results obtained in problem 72 for
the three values of current and the three power factors.
Approximate Method of Determining the Regulation, Effi-
ciency and Power Factor of Transformers. Let 1 2 = iz -f- ji\ be
the secondary current.
Then the primary current is approximately
/i = 7 2 + 4 Jim = iz + ih + j(i r * im) = ii + ji'i-
In the secondary winding only the secondary current, I 2 , flows;
in the primary winding, the primary current. The average cur-
rent in the two windings considered as one is, then,
la = 12 + 0.54 +j (^2 ~ 0.54).
If the secondary voltage, referred to primary, is the zero vector
and is e 2 , then the primary voltage is
#1 = 2 + /Zo,
182 ELECTRICAL ENGINEERING
where Z = r + jx* is the sum of the impedances of the primary
and secondary referred to the primary.
Ei = 62 + [i % + 0.54 + j(*' - O- 5 *"*)! ( r + ^ o)
0.54r - *Vo + O.S^zo
and the real value of the primary voltage is, neglecting second
power of small terms, _ __
E l = vV + 2e 2 fero + 0.5ur - i'*x* + 0.5i m xo)
Regulation is
^~ 62 = ^1 _ l (100)
6 2 02
Let Ei and /i represent the primary
e.m.f., and current, as in Fig. 143.
Then the power input is
P =
cos B.
But
cos 6 = cos (a - j3) = cos a cos + sin a sin /3
:. p
x
The secondary output is 62^2-
The primary input is e\i\ + e'\i f \,
= e 2 i 2 + ^4 + I^TO, approximately.
The efficiency is
Output #2^2
input 2^2 + e^ik + I fro
, Similarly,
p i = - e 2 i' 2 + e 2 i m + h z x Q , approximately.
(101)
and cos 0i is the power factor of the primary.
Problem 74. (A) Determine the numerical values of the primary and
secondary resistances and reactances, the core-loss current, the magnetizing
current, the exciting current from the 1000-volt side, the core loss in watts,
THE TRANSFORMER
183
and the short-circuit impedance when taken from the 1000-volt side, for the
following, 1000-100 volt transformers. The primary and secondary re-
sistance drops are each 1 per cent. ; the primary and secondary reactance
drops are each 2 per cent.
The conductances, o, and susceptances, 6 o, are calculated at 1000 volts.
Rating in k.v.a.
10
20
40
80
160
320
0.0002
0.0006
0.0004
0.0012
0.0008
0.0024
0.0016
0.0048
0.0032
0.0096
0.0064
0.0192
Problem 74. (B) Find the power factor, regulation and efficiency of these
transformers by the approximate method, assuming unity power factor of
load.
Problem 74. (C) For any one transformer, plot the regulation and effi-
ciency vs. power factor, and find the points of per cent, regulation and
maximum efficiency.
(A) Solution for the 10 k.v.a., 1000-100 Volt Transformer. Since, with non-
inductive load (on the secondary) the primary voltage and current will be
nearly in phase, the approximate primary current is
10,000 k.v.a.
"
/1=
1000 volts
10am P'
rj drop = 1 per cent. = 0.01 X 1000 volts = 10 volts
10 volts
TI = T^~~ ' 1 ohm.
10 amp.
TZ\ " = 0.01 ohm.
Similarly,
Xi = 2 ohms.
X 2 = 0.02 ohm.
The core-loss current is
i h = e g 00 = 1000 X 0.0002 = 0.2 amp.
Magnetizing current is
i m = eb 00 = 1000 X 0.0006 = 0.6 amp., lagging,
.'."no-load current is
/oo = \/ih z + im 2 = \/0.04 + 0.36 = 0.632 amp.
The core loss is
eVoo = 1000 2 X 0.0002 = 200 watts.
The short-circuit impedance is
= \/2 2 + 4 2 = -\/2Q = 4.47 ohms.
(B) Solution E! = 1000 = V^ 2 + 2e 2 (i 2 r + 0.5^r + 0.5 i m x )
Tabulating for equations (100), (101), (102):
10.0 erfi 9820.0
Kw.
7 2
10.0
1.0
196.4
100.0
184
ELECTRICAL ENGINEERING
AiAi
U)
0.5 t
im
0.5 i
e 2
JSi,
e 2
Reg.
1.0
2.0
2.0
2.0
4.0
20.0
0.2
0.2
-0.6
-1.2
982.0
1.018
0.018
Vr
Eff.
tan <p
P.F. = cos
200.0
0.961
400.0
-589.2
-189.2
10,216.4
-0.01853
0.9998
(C) Solution. In finding the efficiency and regulation it makes no differ-
ence in the results whether the problem is solved on the percentage basis or
by supplying numerical values for any given machine.
The former method is more general in its application, and it will be used
here, percentage values being taken from the data of the 10-kw. transformer
and applied in formulae (100) and (101).
The percentage data then, are :
E l = 1, / 2 = 1, i h = 0.02, i m = -0.06, r = 0.02, x = 0.04
0.25 0.50 0.75 1.00 0.75 0.50 0.25
Lagging Leading
Power Factor
FlG. 144.
THE TRANSFORMER
185
000000000 05 000000
OOr- (OOOOOi-i O OOOOOO
III 1
*O 00 O
8 3 g
OOOOOOOi-n O OOO
III 1
OO
OOOO
O<N(N
OOO
OOOOOOOi-H O OOO
III 1
OOO
OOOOOOOT-H O OOOOOO
I I I I
SO O i i OO I-H i i 00 i-H C<1 tN CO
OOOOO5 O OO5OOOO5
i-HOOOO'-HT^I> 10 IOOOO5
IOIOO'-HO<MO I ^O Tt< T^rHrH
I>I>OOOOOOO5 O Ot^O
OOOOOOOOO rH OOOOOO
1 1
lOtNt^CllN lOrH CO
CD (M O T^ i t CO O N. l>t^Oi CO 1 ^
COrHOCOO^O T^ T^t^i-tC<Ji-l(M
OIOGOOOOOOO3 O OrfiOOiOOS
OOOOOOOOO I-H OOOOOO
1 1
W 00
rH<MGO *
00
iO
OOOOOOOOO 1-1 OOOOOO
I I
<M <N (N <N
CO i ^ O5 C^ O5 O5 O5 O5
OOi IOOOOOO5 O OOOOOO
ooooooooo i-l doodod
I I
S.3
4f
v 10 ? 10 .
;^o'o
bO
b
1
>>
I
o
0>
w
CHAPTER XXVII
HYSTERESIS AND EDDY CURRENT LOSSES
Hysteresis Loss. The hysteresis loop is interesting in that it
indicates by its area directly the work done on the electromagnet
per cycle of change of current.
The work done in an electric circuit has been shown to be
feidt. If T is the time of the cyclic variation of current then,
C T
W = I eidt, is the work performed during the cycle.
J Q
But the induced e.m.f. in a winding of N turns is
e = jgg -=r> where -^ is the rate of change of flux.
.'. W = I TT^ -j7 dt. But (j) = SB, where S is the cross-
sectional area of the magnetic circuit in square centimeters.
. W . . rx&Ma
Jo 10 8 dt C
Also the magnetizing force is:
QAiriN
tf = -_ ,
where i is given in amperes.
Thus,
A7 IH
iN =
0.47T'
and, substituting this value,
r SIH as si r
J 0.47T xW* ~di c ' WxT*h HdB '
But SI is the volume, 7, of the magnetic structure. Thus,
V C T
W = - HdB.
10 7 X 47rJ
But HdB is the area of the hysteresis loop corresponding to
maximum density, B, as seen from the loop. The work is given
in joules.
186
HYSTERESIS AND EDDY CURRENT LOSSES 187
STEINMETZ found that the hysteresis loss in watts could be
expressed (approximately) by the following equation:
W
10 7
where V is the volume and rj is a constant which depends upon
the quality of the iron. The equation shows that the loss is
proportional to the 1.6 power of the maximum density and
directly proportional to the frequency. In centimeter measure-
ments ri varies from 0.001 to 0.002 in ordinary sheet iron and
may be 10 times as great in tempered steel. In the best silicon
steel it is 0.0006, which corresponds to 0.54 watt per Ib. at 60
cycles and a density of 64,500 lines per sq. in. or 10,000 lines
per sq. cm.
Eddy Current Loss. Eddy currents differ in no way from other
currents, and the loss of power by them is therefore i 2 R or if E
is the e.m.f. causing the current and Z is the impedance of the
path, then,
and the loss is
It follows, then, that the loss is proportional to the square of the
e.m.f. or, what is equivalent, to the square of the maximum
density and to the square of the frequency, since the e.m.f.
itself is proportional to the frequency of flux variation and the
maximum density.
Even in the simplest cases it is difficult to calculate the loss
since the distribution of the flux and, therefore, the e.m.f. in
different parts of the material is often very complex.
Consider as an illustration the simple case of eddy current
loss in transformer steel. The cores are
built up of laminations in such a way that
the flux path is divided up into a number
of elements each having the section of the
edge of a lamination and following parallel,
or as nearly so as possible, to the sides of "^j^ 145
the laminations.
With the flux entering, as is shown in Fig. 145, currents will
flow as indicated by the dotted lines. The current flowing
188 ELECTRICAL ENGINEERING
through a section of area l\dx encloses a flux which is r 0, where
u
</> is the flux passing through the entire area of one lamination
(assuming uniform flux density). 1
The effective value of the e.m.f. induced is
4.44 X flux X turns X frequency _ \/27r2o;<ft/
10 8
The resistance of the path, neglecting that of the ends, is
2lp
lidx
. \/2ir2x<j>flidx
where p is the specific resistance of the material.
.'. t' 2 r in the elementary circuit is
x 2/p 4
p 2 h(dx)
and the total loss is
p ' 4**<l>*f*l0*(dx)
Jo IVWp
6 X 10 16 Z P
Since the volume is Hid, the loss per cm. 3 is
W TT^hd 1 TT 22
X
F 6 X 10 16 Z P Z X W ~ 6 X 10 16 / 2 p
But = 5 X W. /. </> 2 =
and
W_ Tr 2 BH 2 d 2 f*
V ~ 6 X 10 16 Z 2 p ~ 6 X 10"p Watts *
p for sheet iron is about ^ ohms.
1 For a more complete discussion see " Advanced Electrical Engineer-
ing."
CHAPTER XXVIII
WAVE DISTORTION IN TRANSFORMERS
If on a transformer containing no iron a sine wave of e.m.f.
were impressed at its terminals, the flux and the exciting current
would also follow sine waves.
With the introduction of iron, however, while the flux values
would still follow a sine wave, or very nearly so being distorted
only due to the ohmic drop of the distorted current the exciting
current wave would necessarily be considerably distorted.
Its shape is shown in Fig. 148, which is derived from the hyste-
resis loop given in Fig. 147.
Conversely, if by some arrangement the exciting current were
made to follow substantially a sine wave, the flux wave, and
therefore the wave of voltage across the transformer, would be
greatly distorted.
This distortion in current or e.m.f. waves is of considerable
importance in connection with the grouping of transformers in a
three-phase system, as will be seen later. At present, however,
only the condition in a single-phase transformer will be studied.
A representative hysteresis loop is shown in Fig. 147, which was
obtained from actual tests with a sine wave of impressed e.m.f.
The test data are recorded in Table VI.
If the effect of the ohmic drop be neglected, then the impressed
and counter, or induced, e.m.f. are the same numerically and
where N is the number of turns and < is the flux.
With a sine wave of flux </> = $> m sin co,
dt
-^ = 4> TO co cos at.
cos a)t = E m cos
The induced e.m.f. has its negative maximum when the flux
begins to rise, and lags behind the flux by 90 time degrees. Thus
189
190
ELECTRICAL ENGINEERING
the impressed e.m.f ., E, which is equal and opposite to the induced
e.m.f., leads the flux by 90 (neglecting the ir drop), Fig. 146.
If instead of being a sine wave the flux were distorted and yet
symmetrical, it would be represented by FOURIER'S series of odd
harmonics, thus:
= 3> lm sin at + <J> 3wi sin (3o>Z + a)
/. e . = N -ir = &i m w cos ut
The e.m.f. wave would be relatively
more distorted than the flux wave as
is evident from the coefficients of the
different trigonometric terms.
sn
cos
-fa)...
FIG. 147.
When a hysteresis loop is given, if either the flux wave or ex-
citing current wave is known, the other may be at once obtained.
For example, let the flux wave be assumed to be sinusoidal.
TABLE VI. HYSTERESIS LOOP DATA
Ord.
Abs. Aba.
0.0
0.5
-0.5
0.2
0.56
-0.43
0.4
0.63
-0.32
0.6
0.71
-0.18
0.8
0.82
0.08
0.9
0.9
0.35
1.0
1.0
1.0
EXCITING CURRENT DATA
Time
Flux
ioo
0.0
0.5
10
0.174
0.55
20
0.34
0.6
WAVE DISTORTION IN TRANSFORMERS 191
The exciting current data are obtained from the hysteresis loop
by reading off the current values corresponding to the flux values
which have been taken at uniform intervals along the flux wave.
Thus, at on the flux wave $ = 0. This value of <, on the
hysteresis loop, corresponds to i 00 = 0.5 amp. At 10 on the
flux wave, < = 0.174. This value on the loop corresponds to
z'oo = 0.55, etc. Data for the exciting current are given in Table
G/T. It should be noted that the flux
naximum and current maximum always
>ccur at the same instant. / *-
The phase relations and character-
istic current wave shape for a sine wave
of flux are shown in Fig. 148 The im F IG
pressed voltage wave leads the flux by
90. The scales to which the waves are plotted are quite in-
dependent of each other, and should be so chosen as to exhibit
the waves most clearly.
When the induced e.m.f. is not a sine wave, the flux wave is
also distorted. In this case the impressed e.m.f.
-N**.
~ * dt
Transposing,
edt
where N is the number of turns.
Hence
ft - <2 />2
I &- A^d0.
./ = i ^i
If ti is chosen as the time when <f> is zero, and tz is the time when
</> is maximum, then
AC-* ^ N N
!*- ioi^ = ioi
Ji = (i t/0
This equation shows that the maximum value of the magnetic
flux or flux density in which the electrical engineer is very much
interested, since it determines the magnetizing current and core
loss is proportional to a certain area of the e.m.f. wave, and it
remains to determine where this area is located.
When the flux is a maximum then - is zero; thus e is zero.
192
ELECTRICAL ENGINEERING
The value of tz is therefore easily ascertained, as is shown in Fig.
149.
The ordinate through ti must bisect the e.m.f. wave in order
that the flux wave be symmetrical, as
can also be seen by slight consideration,
since the flux wave must be symmetrical
above and below the zero line.
Thus, in finding the flux wave, the
first step is to bisect the area of the
e.m.f. half-wave, which gives the posi-
FIG. 149.
tion of ti and the zero of the flux wave.
Problem 76. From the following readings on a distorted e.m.f. wave
obtain and plot the flux and current waves.
NOTE. Choose a scale to give <b m 1.
t
ei
t
Ci
0.0
100
0.73
10
0.005
110
0.90
20
0.01
120
1.0
30
0.04
130
0.98
40
0.1
140
0.91
50
0.15
150
0.78
60
0.22
160
0.5
70
0.31
170
0.12
80
0.42
180
0.0
90
0.58
Solution. By bisecting the area of the e.m.f. half-wave it is
found that the zero of the flux wave will be at 120 in this ex-
ample. This is also the point of maximum e.m.f. Starting
from 120 and tabulating values proportional to the areas
enclosed for each 10 gives values proportional to the flux when
these areas are successively summed up. Thus at 120, flux = 0.
At 130, the area enclosed between 120 and 130 ordinates and
the curve and base line is proportional to the mean ordinate, say
2~~ ~ = 0.99. At 140, the mean ordinate between 130
and 140 is 0.95.
The area from 120 to 140 is proportional to 0.99 + 0.95 =
1.94. Thus, three points on the curve are obtained, namely,
0, 0.99, 1.94.
WAVE DISTORTION IN TRANSFORMERS 193
These values may conveniently be reduced by a factor to
bring the maximum of the flux wave to unity.
The tabulation is as follows :
t
120
130
140
150
160
170
180
\
1.00
0.98
91
78
50
12
Av
99
95
85
64
31
06
Area
0.0
0.99
1.94
2 79
3 43
3 74
3 8
263 X area
26
51
735
903
985
1 00
t
190
200
210
220
230 \
)
240
i
-0.005
-0.01
-0.04
-0.10
-0.15
-0.22
Av
-0.0025
-0.0075
-0.025
-0.07
-0.125
-0.185
Area
3.8
3.79
3.77
3.7
3.57
3.39
0.263 X area. .
1.00
0.997
0.992
0.975
0.94
0.893
t
250
260
270
280
290
300
i
-0.31
-0.42
-0.58
-0.73
-0.90
-1.00
Av
-0.265
-0.365
-0.50
-0.655
-0.815
-0.95
Area
3.12
2.76
2.26
1.6
0.785
-0.165
0.263 X area. .
0.822
0.727
0.595
0.421
0.206
0.0
1.0
0.8
0.6
0.4
0.2
-0.2
-0.4
-0.6
-0.8
-1.0
f
N
\ /
^
/
/\
/
ty
; *
y
\
^
S
/
V.
4
8
/
20
1
2(
0~*^s
/
.
/
^
X
Angular Displacement
FIG. 150.
13
194 ELECTRICAL ENGINEERING
The tabulation is carried out for values of e t from 120 to
300, values from 180 to 300 being the same as from to 120
but reversed in sign.
The flux wave is then plotted from to 120 by reversing the
sign of the values of flux obtained from 180 to 300. These
waves are shown in Fig. 150.
Problem 76. With three-phase systems, the exciting current of Y-con-
nected transformers resembles fairly closely a sine wave. 1 Assuming,
FIG. 151.
therefore, a sine wave of exciting current, determine the flux wave from the
hysteresis loop (Fig. 147), and from this find and plot c,-. These waves
are shown in Fig. 151, in which the characteristic form of the induced vol-
tage, 6i, is noteworthy.
Problem 77. Analyze, by FOURIER'S series, 2 the typical wave of exciting
current shown in Fig. 148, determining and plotting the fundamental and
third harmonic and, if sufficient time is available, also the fifth harmonic.
Dependence of Core Loss on the Shape of the E.M.F. Wave.
The core loss of a transformer, which is due to hysteresis and
eddy currents in the iron core and is equal to eVoo, depends on the
maximum value of the flux, since the greater the maximum flux
the greater the area enclosed by the hysteresis loop. In modern
transformers, hysteresis loss is about 70 per cent, and eddy current
loss about 30 per cent, of the core loss.
But $ m depends upon the area of the e.m.f. wave, as has been
illustrated in the problems, and hence on the average value of
the e.m.f.
Hysteresis loss is approximately proportional to the 1.6th
power of the maximum flux.
Thus, if a comparison is made of two e.m.f. waves of equal
effective value, but of different shape and average value, the
ratio
Hysteresis loss in wave A _ /av. e.m.f. of A\ 1>6
Hysteresis loss in wave B ~ \av. e.m.f. of B/
1 This is demonstrated on p. 228, Chap. XXXII.
8 See Chap. XXIX.
WAVE DISTORTION IN TRANSFORMERS 195
By definition,
Form factor (f .f.) =
effective e.m.f.
average e.m.f.
. Hysteresis loss in A _ rf.f. (B)"| L6
' 'Hysteresis loss in B Lf.f. (A)J
Therefore, the higher the form factor the less the core loss.
The form factor of a sine wave is 1.1. In general that of a flat-
top wave is less; of a peaked wave, more.
Wave A (Fig. 152) has maximum core loss.
Wave B has minimum core loss.
FIG. 152.
CHAPTER XXIX
DISTORTED WAVES
It is often necessary to express a distorted wave in the form
of an equation. This can readily be done since it has been
found that any periodic univalent curve can be expressed by a
series of terms involving a constant and sine and cosine terms.
That is,
y = a + ai cos + 2 cos 26 + + a n cos nB
+ 61 sin + 6 2 sin 20 + + 6 n sinr*0 (103)
represents any distorted wave in which for every value of
abscissa only one ordinate exists, provided that the abscissa
is so chosen that the curve repeats itself at a value of = 2?r,
i.e., the wave is periodic.
Obviously, if the distorted wave is given graphically it is
always possible to read off the ordinate corresponding to each
abscissa (Fig. 153).
2 7T
FIG. 153.
The problem then resolves itself into finding the coefficients
o, i, n, &o, &i, &in (103).
To do this a mathematical transformation has been worked
out involving convenient integrations and the fact that sines
and cosines have the same values at = as at = 2?r or any
multiple of 27r, that is, 2?rn, where n is an integer number.
To find a integrate Eq. (103) between and 2?r. Thus,
yds
r^
I
- I
ade +
rzr
I
cos BdB +
cos nOde -f I 61 sin 6d0 +
196
1
ri*
I a n -
b n sin n0d0.
DISTORTED WAVES 197
From what has been said above, all integrals except the first
must be zero. Thus
j ydO = I a dB = a (2ir - 0) = 27ra .
1 F*
.'. a = -^~ yds.
^
But I yds is the area of the curve during one complete
period and 2ir is the abscissa.
.'. a is the average value of all the ordinates, or the average
value of y.
To determine any other coefficient, for instance a, Eq. (103)
is multiplied by cos nB and integration is again carried out be-
tween limits and 2ir.
In this case it is also remembered that the integral over one
period of any product of sine and cosine terms is zero.
rr r2* r2*
y cos nBdB = a I cos nBdB + i J cos nB cos BdB
+ a n I cos 2 nBdB + fei I cos nB sin BdB-
+ fe n I cos nB sin n&dB.
All these integrals on the right-hand side must be zero with
the exception of
cos 2 nBdB,
and this integral, as is readily seen, is = TT.
I C 2v
.'. a n = - I y cos nBdB.
But J*y cos nBdB is the area, not of the original curve, but of
another curve which is obtained by multiplying each value of y
by the particular value, at phase angle B } of cos nB.
Since that area is divided by TT the integral must be just twice
the average of the instantaneous values of ?/, multiplied by cos nB.
.'. a n = 2X avg. of y cos nB between and 2ir.
In a similar way all values of fe are obtained so that,
6 n = 2X avg. of y sin nB from to 2ir.
2
.'. a = avg. (y)
198 ELECTRICAL ENGINEERING
ai =
2X
avg.
(y
cos
0)
a 2 =
2X
avg.
COS
20)0'
a 3 =
2X
avg.
(y
COS
30)1'
a n =
2X
avg.
(y
COS
I 2 '
n0)!o
2
61 =
2X
avg.
(y
sin
?)
62 ==
2X
avg.
(y
sin
20
)o
b n = 2X avg. (y sin n0)l
It should be noted that dividing the curve up, say every 10
from to 360, 37 readings are obtained. It is better then to use
36 and to take the average value of the values at and 360
instead of using both of them.
In a symmetrical wave only those harmonics can exist, which,
with an increase of the angle by 180 or TT, reverse the sign of the
function.
This is only the case when n is an odd number. Since, if n is
2, 4, 6, etc., then increasing the angle by TT means 27r, 4?r, 6V, etc.,
and the values of the sine and cosine are the same for a, (a + 2ir),
(a -f 4?r), etc., whereas if n = 1, 3, 5, etc., we get, TT, 3r, STT, in
which the sign of the function reverses.
If sin a is positive, then sin (a + TT) is negative.
If cos a is positive, cos (a + v) is negative, etc.
Thus, for symmetrical waves such as are given by alternators
under stable conditions, the trigonometric series becomes :
y = ai cos + a 3 cos 30 + a 5 cos 50 +. . . -f- 61 sin
+ 6 3 sin 30 + &5 sin 50 + ....
Obviously, in that case, it suffices to analyze one-half a wave
only. 1
Problem 78. Plot the wave,
e = Ei sin 6 + E 3 sin (36 + a),
for
E l = 1
E z = 0.5
a = 30,
and analyze the wave, proving that the analysis gives the original equation.
Show also that no 5th harmonic exists.
1 For a more complete discussion of this method of wave analysis see
STEINMETZ'S " Engineering Mathematics."
DISTORTED WAVES
199
Tabulating :
6
10
20
30
40
50
60
70
80
90
Ei sin
0.0
0.174
342
50
643
766
866
94
935
1
38 + a
Sin (30 + a)
tfasin (39 + a)...
30.0
0.5
0.25
25
60.0
0.866
0.433
607
90.0
1.0
0.5
842
120.0
0.866
0.433
933
150.0
0.5
0.25
893
180.0
0.0
0.0
766
210.0
-0.5
-0.25
616
240.0
-0.866
-0.433
507
270io
-1.0
-0.5
485
300.0
-0.866
-0.433
567
100.0
110.0
120.0
130.0
140.0
150.0
160.0
170.0
180.0
Ei sin
0.985
0.94
0.866
0.766
0.643
0.50
0.342
0.174
0.0
36 + a
330.0
360.0
30.0
60.0
90.0
120.0
150.0
180.0
210.0
Sin (30 + a)
-0.5
0.0
0.5
0.866
1.0
0.866
0.5
0.0
-0.5
Et sin (36 + a).. .
-0.25
0.0
0.25
0.433
0.5
0.433
0.25
0.0
-0.25
e
0.735
0.94
1.116
1.2
1.143
0.933
0.592
174
-0 25
Analysis. a must be zero because the wave is symmetrical above and
below the center line. The coefficients of the fundamental cosine and sine
waves are found from
ai = 2 X avg. e cos 0,
bi = 2 X avg. e sin 6.
e
Cos e
e
e cos
Sin $
e sin
1.0
0.25
0.25
0.0
0.0
10
0.985
0.607
0.598
0.174
0.1057
20
0.94
0.842
0.792
0.342
0.288
30
0.866
0.933
0.808
0.5
0.466
40
0.766
0.893
0.685
0.643
0.575
50
0.643
0.766
0.493
0.766
0.587
60
0.5
0.616
0.308
0.866
0.534
70
0.342
0.507
0.173
0.94
0.477
80
0.174
0.485
. 0844
0.985
0.478
90
0.0
0.5fc7
0.0
1.0
0.567
100
-0.174
0.735
-0.128
0.985
0.725
110
-0.342
0.94
-0.322
0.94
0.885
120
-0.5
1.116
-0.558
0.866
0.966
130
-0.643
1.2
-0.772
0.766
0.920
140
-0.766
1.143
-0.875
0.643
0.735
150
-0.866
0.933
-0.808
0.5
0.466
160
-0.94
0.592
-0.556
0.342
0.202
170
-0.985
0.174
-0.171
0.174
0.0303
180
-1.0
-0.25
-0.25
0.0
0.0
200
ELECTRICAL ENGINEERING
The sum of the 18 cosine readings, using the average of and 180 as
one, is - 0.2486 and the average value is - 0.0138.
Thus,
ai = 2 X avg. = - 0.0276.
Similarly, the sum of the sine readings is: 9.007
The average is 0.5004,
Thus,
61 = 2 X avg. = 1.0008.
The coefficients of the 3d harmonics are found from,
a 3 = 2 X avg. e cos 30,
6 3 = 2 X avg. e sin 36.
e
30
Cos 30
e cos 30
Sin 30
e sin 30
1.0
0.250
0.0
0.0
10
30
0.866
0.525
0.5
0.304
20
60
0.5
0.421
0.866
0.730
30
90
0.0
0.0
1.0
0.933
40
120
-0.5
-0.447
0.866
0.774
50
150
-0.866
-0.664
0.5
0.383
60
180
-1.0
-0.616
0.0
0.0
70
210
-0.866
-0.439
-0.5
-0.254
80
240
-0.5
-0.242
-0.866
-0.420
90
270
0.0
0.0
-0.1
-0.567
100
300
0.5
0.368
-0.866
-0.637
110
330
0.866
0.815
-0.5
-0.470
120
360
1.0
1.116
0.0
0.0
130
390
0.866
1.040
0.5
0.600
140
420
0.5
0.571
0.866
0.990
150
450
0.0
0.0
1.0
0.933
160
480
-0.5
-0.296
0.866
0.513
170
510
-0.866
-0.151
0.5
0.087
180
540
-1.0
0.250
0.0
0.0
The sum of the 18 cosine readings is 2.251. The average is 0.125.
.'. a 3 = 2 X avg. = 0.25.
The sum of the sine readings is 3.899. The average is 0.2165;
.*. 6 3 = 0.433.
The exercise of proving that no 5th harmonic exists is left for the student.
Summing up the values already obtained, the equation may be written :
e = - 0.0276 cos 6 + 0.25 cos 30 + 1.0008 sin - 0.433 sin 30
which is, approximately,
y = sin + 0.433 sin 30 + 0.25 cos 30.
DISTORTED WAVES
201
The second and third terms may be combined or added, being in quad-
rature, by the vectorial method in which
where
Thus,
A sin 6 + B cos = \/A 2 + B 2 sin (0 + a),
tan a = A
A.
0.433 sin 30 + 0.25 cos 30 = Vo.188 + 0.0625 sin (30 + a)
= 0.5 sin (30 + a), where a = tan" 1 Q-TOQ = 30.
The complete wave is, therefore,
e = sin -{- 0.5 sin (30 + 30).
The wave is shown plotted in Fig. 154, in which also the component
waves are indicated by the dotted lines.
FIG. 154.
CHAPTER XXX
MECHANICAL STRESSES IN TRANSFORMERS
It is recollected that a mechanical force is exerted on a con-
ductor carrying current if it is placed properly in a magnetic field,
the force being 1 dyne per cm. of conductor per abamp. in a field
intensity of 1 line per sq. cm. provided the field is at right angles
to the conductor.
Referring to Fig. 155, which represents the cross-section of a
transformer, it is evident that the main flux which interlinks with
both" the primary and the secondary windings and is confined to
the iron does not cut through any part of the windings carrying
current, but that the leakage flux more or less completely cuts the
windings and therefore is responsible for a force which tends to
warp the coils out of shape and thus to damage them. The deter-
mination of the mechanical stresses resolves itself therefore largely
into the calculation of the leakage
core flux or leakage inductance of the
transformer.
To calculate the leakage induc-
tance of the secondary coil, con-
sider this made up of the interlink-
. ages of flux with turns in the space
FIG. 155. occupied by the secondary coil
itself, plus the interlinkages of the
flux between the coils with all of the secondary turns. Similarly
with the primary.
Approximation of the Leakage Inductance of the Secondary.
- turns, where
In Fig. 155, a portion of the coil of depth x, has
a is the total depth of the coil, and N 2 is the total number of
secondary turns on 1 leg of the transformer.
The magnetomotive force of this part of the coil is
t r *r x
m.m.f x = I 2 N 2
a
The flux which this m.m.f. produces is
flux =
202
MECHANICAL STRESSES IN TRANSFORMERS 203
where p is the reluctance, and p = = - when we con-
area max
sider only the flux which passes through the small area of width
dx and length m. m is the length of a turn at distance x in Fig.
155. It is almost impossible to determine accurately the length
Z . It is the equivalent length of the lines of force which going
through section mdx return upon themselves. Part of these lines
can be readily traced. They go almost straight across the trans-
former windings of length h; then they spread apart, and the
equivalent length, as a result, is relatively short. Then, the
majority of the lines enter the iron and their reluctance is insig-
nificant. Some, however, enclose the winding that is outside of
the iron and these meet with considerable reluctance. Therefore,
it might be fairly conservative to assume Z , the equivalent length,
as I the height of the "window" of the transformer.
If mz is the mean length of a secondary turn, this may be sub-
stituted for m, thus
I
p '
Then the flux in any elementary band, dx, is
m 2 dx x
d<f>'
This flux interlinks with N z turns. Therefore, the interlink-
/> /y
ages with the flux = ^irl^Nz "**""" - N%, and the inductance due
to the interlinkages within the space occupied by the coil is
(104)
4
To determine the inductance due to the flux in the
gap between coils, consider Fig. 156 which shows a
section through one side of the coils. The current is
oppositely directed in the two coils, as indicated by
dots and crosses. On a 1:1 basis, the turns and
currents in the two coils are equal, and the figure may
be regarded as merely showing a section through a
single coil, of N z turns, or of Ar 2 /2*amp.-turns.
The area of the core of this imaginary coil will be 6m 3 , where
ra 3 is the mean circumference between the actual coils, and 6 is
204
ELECTRICAL ENGINEERING
the distance between them. The flux produced in this region by
the m.m.f., / 2 N 2 , is then
o X 4-n A r 2 2 iw 3 , due to
The number of interlinkages is
This represents an inductance of L" 2
the secondary coil, since half of the inductance is due to the pri-
mary and the other half due to the secondary.
The total secondary inductance of this coil is then
L 2 =
Lff
2 =
and the primary inductance is, similarly,
T 27rAVr 9 c
Li- j [2rai^
where c is the depth of the primary coil.
Since ^ =
N
former, referred to the primary is
L = ^y 1 -^!^ +
If two legs are in series, L (tote/)
or, if in parallel, L (tota0 = -
In practical units,
L = 32 X 10- 9 ~
2> the total inductance on 1 leg of the trans-
n. (105)
2L,
m 2
henrys ' (106)
where the dimensions are in inches.
The same reasoning may be applied
to a core-type transformer in which
the coils are differently arranged, for
example, as in Fig. 157.
Here are two secondary coils, with
the primary placed between them.
Consider the primary as if made up
of two equal coils, separated by a
dividing line shown dotted. The
calculation should then be made of the combined inductance of
the secondary, S', and one-half of the primary, which are grouped
MECHANICAL STRESSES IN TRANSFORMERS 205
together as A in the figure, and similarly, the secondary S" and
the other half of the primary grouped as B.
From (105), the inductance for A is,
U
and for B,
in which
]2 |~,w,/
i m
w'i = mean length of inside one-half primary turn
w"i = mean length of outside one-half primary turn
m' 2 = mean length of inside secondary turn
m"z = mean length of outside secondary turn
m 3 = mean length of inside gap
ra 4 = mean length of outside gap
w"i = 2rai
m 3 + w 4 = 2m.
If coils are symmetrical, mi
m 2 .
Supplying all of these values, the total inductance is
L = L'+L" = -^r-\mi~ + m 2 % + ra&lcm.,
I L 6 3 J
where Ni is the number of turns in half the primary coil. If TI
is the number of primary turns per leg of the core,
c a
- + m 2
If dimensions are in inches,
T 16 TYr c
Lj -, /-vn T
cm. per leg.
mb]
FIG. 158.
In a similar manner, shell-type transformers may be dealt with.
Such a transformer is shown in Fig. 158. In this, let m = mean
length of 1 turn, NI = number turns in half of a primary
206 ELECTRICAL ENGINEERING
coil, = one-quarter total primary turns. Using the same
reasoning as with core-type transformers, the inductance of a
unit combination, A, in the figure, is
[| + I + b] cm. (107)
Note that m ^ + + Z is the equivalent area, whence the
total inductance is
SL = j |g + F+&| cm.
In inch units,
128
Calculation of Stresses. Under ordinary conditions of load,
these would not be excessive, but for maximum current, as in the
case of short-circuit, or heavy transient currents from switching,
they may be very great. Calculation may properly be based on
the short-circuit current, remembering again that a wire 1 cm.
long, carrying 10 amp. (unit current), if placed perpendicular
to a field of 1 line per sq. cm., is repelled by a force of 1 dyne;
or the force in dynes = BI'l, where I' is expressed in absolute
values abamperes.
If the flux density in the gap, &, between coils, is B max then it
may be assumed that the average density of the flux leaking
D
through the coils themselves is ^, which is then the average
density of the flux passing through the coils of any section A,
Fig. 158, and the force per turn on any coil, will be
F t = -^ X 1\ X m dynes
B
max
98f grams '
where m is the mean length of the turn.
If /2 is in amperes,
B max z
F t = grams per turn.
Let the effective value of the short-circuit current be 7 2 , and
let the total secondary turns be T 2 , then the turns in a half coil
(Fig. 158) are -
MECHANICAL STRESSES IN TRANSFORMERS 207
The maximum value of the force will be
2 X 9810 v 4
= 8X9810 m grams ( 108 >
or in the case of the primary short-circuit current /
m V2ITmB m
8 X 9810
where / is the effective value of the primary short-circuit current
and T the total number of primary turns.
The leakage flux must, in the case of short-circuit, be the main
flux (neglecting the flux due to the voltage which is consumed
by the ohmic drop), if it is assumed that the generating station
is large and the voltage impressed upon the transformer is
normal even though the transformer is short-circuited. (See note.)
The maximum value of the flux between a group of coils is
obtained by multiplying the maximum value of the flux density
B m by the equivalent area as given in (107).
That is
The group contains in this case one-quarter of the turns and
ET
the voltage per group is -r where E is the effective value of the
impressed e.m.f.
The relation between the maximum value of the flux and the
voltage is given by the well-known relation
Substituting this in (109)
(110)
The average value of the force is obviously one-half of the
maximum value.
208 ELECTRICAL ENGINEERING
The force between the coils is proportional to the rating
assuming the same regulation.
NOTE. The actual flux enclosed by the secondary turns depends upon
the terminal voltage and the ir drop.
At short-circuit the secondary terminal voltage obviously is zero. Thus
if as a limiting case the ir drop is neglected the secondary winding encloses
no flux.
As long as it is assumed that the primary voltage is normal voltage and
that the ir drop is again neglected the primary coil encloses the same flux
during the short-circuit as it does at no-load. The path of the flux must
therefore be essentially different. In the latter case it traversed the two
windings and is therefore mainly in the iron, while in the former case it
must traverse only one winding the primary. Thus the flux must find
its way between the primary and secondary coils and is thus the so-called
leakage flux.
CHAPTER XXXI
GENERAL PRINCIPLES OF TRANSFORMER DESIGN
Type. Transformers may be classified as belonging either to
the "core type" or the " shell type."
Core-type transformers frequently have a single magnetic
circuit of rectangular form. On the two vertical sides of this
core are placed the windings, each side being provided with
half of the primary and half of the secondary coils, the low-
voltage coils usually being placed next to the core (Fig. 159).
Shell-type transformers usually have a multiple magnetic
circuit the coils being placed upon a central core, the outer limbs
of which extend around the coils, somewhat resembling a shell
(Fig. 160). As illustrated diagrammatically in the figures, it is
Secondary
Core
Primary
FIG. 159.
FIG. 160.
seen that the coils of the core-type transformer have the form
of a cylindrical shell, while those of the shell type are in the form
of discs. The former lend themselves readily to designs of great
mechanical strength, while the latter tend to be mechanically
weak.
The present tendency seems to be more and more toward
the core type, and it remains for the superiority of the shell
type to be demonstrated in any given case in order to justify its
existence at all.
Recently transformers having a multiple magnetic circuit have
been introduced. The coils are of the cylindrical form placed
around the central core. Thus, this is called the cruciform type.
14 209
210
ELECTRICAL ENGINEERING
An important consideration with respect to the choice of type
is the method of cooling the transformer. Core-type transform-
ers are usually immersed in oil in such a way as to provide free
circulation of the oil about all surfaces of the coils and core.
The oil then receives the heat and carries it to the outside case
which is frequently corrugated to present greater effective
surface to the outer air.
Shell-type transformers are cooled by the above method, but
more frequently this is augmented by the addition of coils of
pipe through which is forced a stream of cool water. These
coils are placed in the oil above the transformer.
The addition of the cooling water is essentially a feature of
large transformers, since they have less area of possible cooling
surface per unit volume than have smaller units.
A common form of the shell type is known as the air-blast
type. The method of cooling consists in forcing a continuous
blast of cool air up through the ducts with which the core is
provided, and between and around the coils.
Efficiency. Transformers are not designed to give the highest
possible efficiency as this would involve too great an expense in
materials and manufacture, but, rather, the highest practical
efficiency, so as to meet competition both in price and in quality.
Consequently, from results obtained in practice, it is easy to
construct a table of efficiencies which might reasonably be
expected of various sizes of transformers of moderate voltages,
say up to 10,000 volts. This table is as follows :
Efficiency
25 cycles
60 cycles
1
94.0
96.0
5
96.5
97.5
10
97.0
98.0
50
98.0
98.5
200
98.0
98.5
Knowing the approximate efficiency of the transformer which
is to be designed, the total losses are of course also known.
For example, let it be required to design a 10-kw., 60-cycle,
200 %oo- v lt core-type lighting transformer. The efficiency is
to be about 98 per cent. The losses are 2 per cent., or 0.02 X
10,000 = 200 watts.
GENERAL PRINCIPLES OF TRANSFORMER DESIGN 211
Losses. These losses are made up of the I 2 r loss in the copper
windings and the hysteresis and eddy current losses in the iron
core and windings.
Maximum efficiency is obtained at that load for which the
copper and iron losses are equal. It becomes a matter of choice
in design as to what ratio shall be given these losses or at what
load they shall be equal. Thus, for power purposes, the copper
and iron losses should be about equal at full-load, giving maxi-
mum efficiency at full-load. For lighting purposes, however,
owing to the peculiar conditions of operation, this is not generally
desirable. A lighting transformer carries full-load only for a
very small period during each 24 hr., while the rest of the time it
is operating practically at no-load. Thus the copper loss is quite
small even with a large value of 7 2 r, while the core loss is larger
since it is continuous through the whole day. It would be better,
therefore, to make the copper loss relatively greater than the
core loss, at full-load, and thus reduce the total losses for the
daily operation. Fairly good values to choose for these losses
are: copper loss = 60 per cent., core loss = 40 per cent, of the
total loss.
In the example considered,
copper loss = Pr = 200 X 0.60 = 120 watts,
core loss = 200 X 0.40 = 80 watts.
The core loss may be further divided between loss due to
hysteresis and loss due to eddy currents. The former is usually
larger because it depends on the magnetic quality of the iron or
steel used, whereas the latter depends largely on the degree of
thinness of the laminations of the core, and this may be carried
to any extent mechanically practical. Values of hysteresis and
eddy current losses when silicon steel laminations .014 in. thick
are used are:
hysteresis loss = 0.7 watt per Ib. at 60 cycles,
eddy current loss = 0.3 watt per Ib. at 60 cycles,
when the maximum induction density is 64,500 lines per sq. in.
(10,000 lines per sq. cm.).
Since 1 cu. in. of this material weighs 0.28 Ib., the loss per cu.
in. at 60 cycles and 64,500 lines per sq. in. is:
hysteresis loss per cu. in. = 0.28 X 0.7 = 0.196 watt,
eddy current loss per cu. in. = 0.28 X 0.3 = 0.084 watt,
total core loss per cu. in. = 0.28 watt.
212
ELECTRICAL ENGINEERING
Hysteresis loss for any frequency and density is given ap-
proximately by the equation,
hyst. loss = W h = 0.196 X ^ X
where V = volume of iron.
Similarly, eddy current loss is
/
V,
0.084 X
From these two equations and the core loss which is given,
the volume may be obtained for any value of B. Assuming, as
will later be done, that B = 70,000, in the example, *
80 ^ [0.196 X (1.086) 1 - 6 + 0.084(1. 086) 2 ] =
80
0.2205 + 0.099
= 250 cu. in.
And the hysteresis loss is W h = 0.196 X 1.125 X 250 = 55.2
watts, and the eddy current loss is W e = 0.099 X 250 = 24.8
watts.
5 10 15 20
Volume per Watt Hysteresis Loss Cu, In.
0.1 0.2 0.3 0.4 0.5 0.6
Watts per Cubic Inch
FIG. 161.
B and V. The relation between B and V is shown by the
following curve, Fig. 161, from which it is evident that values
of B should lie between 50,000 and 90,000.
GENERAL PRINCIPLES OF TRANSFORMER DESIGN 213
Hysteresis and Eddy Current Loss per Cubic Inch.
(!)/ = 60
B
50,000
60,000
70,000
80,000
90,000
100000
B
775
93
1 085
1 24
1 395
i fit)
64,500
/ B \i.e
Of\K
OQQ
1 19^
1 d.9
Iijt*
2f\f
\64,500/
TF OIQfi ( B V 6
0197
01 79 f\
099O
097ft
. to
00.^0
. uo
OAfyy
W k - 0.19G ^ 4
1 B y
OAfll
OftAC
11 ft
.^/o
1A
.o^o
IQr
.4UZ
241
\64,500/
TF n ns4 / B \ 2
.OU1
OfkKfJK
.oDO
OO797
. lo
Onoo
. O4
01 9QJ.
.O
01 AQQ
.41
Oonne
W. - 0.084 ^ 64j50Q j
TF
7
0.1775
0.2452
0.319
0.4074
. iDoy
0.5069
0.6045
(2) / = 25; = 0.417; = 0.174
W K
0.053
0.072
0.0917
0.116
0.143
0.1675
W,
W
0.0088
0.0618
0.0127
0.0847
0.0172
0.1089
0.0225
0.1385
0.0285
0.1715
0.0384
0.2059
v
As a matter of fact the usual limits are:
for 60 cycles, B lies between 60,000 and 75,000,
for 25 cycles, B lies between 80,000 and 90,000.
In the example, let B = 70,000, which will be taken as a trial
value.
From Fig. 162 the volume per watt loss by hysteresis is 4.55
cu. in. The total volume of iron is 4.55 X 55.2 watts = 250 cu. in.
Magnetizing Current. Having chosen a suitable value of B,
we can at once find out the required number of ampere-turns per
inch length of magnetic circuit, from the saturation curve, p.
Let
M o = ampere-turns per inch and
I = length of magnetic circuit.
Then total ampere-turns = M Q l = -\/2imt, where -\/2i m =
maximum value of magnetizing current and t = number of turns
on the primary.
Using the fundamental equation for e.m.f.,
E = 4.44 ft* X 10- 8 = 4A4ftBA X 10~ 8 ,
214 ELECTRICAL ENGINEERING
the magnetizing volt-amperes are
1-1 "
X 10 8
= vfBAMol X 10~ 8 = irfBMoV X 10~ 8 ,
since
4.44 = \/2ir, and V=Al.
The percentage magnetizing current is obtained by dividing by
El, thus,
T = 10 u Xkw.'
In the example, M is found to be 6.5. Therefore, substituting
known values into the equation,
i m 3.14 X 60 X 70,000 X 6.5 X 250
T : io" x 10 l5 '
or approximately 2 per cent.
This is a reasonable value. In practice, magnetizing currents
range from 2 to 8 per cent., being larger in smaller transformers
and at lower frequencies.
Number of Turns, Total Flux, Area, and Length of Magnetic
Circuit. Returning to the fundamental e.m.f. equation, it is seen
that turns and flux are both unknown. A practical limit in help-
ing to decide what value to assign to either one of these unknowns
is found from the fact that the number of turns should depend
upon the voltage. While it would not be safe to allow too great
a difference of potential to exist between adjacent turns, this
consideration is not the deciding feature. The choice of number
of turns is governed largely by cost considerations. From prac-
tice it is known that volts per turn should lie between 0.4 X \/kw.
and 0.6 X A/kw. in core-type transformers. The former value
is more suitable for distribution transformers when it is desirable
to keep down the core loss, while the latter is suitable for power
transformers. The value for shell type is from two to three
times as great.
In the example, it will be assumed that volts per turn = 0.5 X
= 1.56.
Then, turns on primary
GENERAL PRINCIPLES OF TRANSFORMER DESIGN 215
and flux
- 586 > 000 '
area
$ 586,000
= A = B =: "TpOO" == 8 ' 37 sq '
and length,
Resistance, Length of Mean Turn, Total Length and Size of
Windings. Returning now to the windings, it is possible at first
to calculate the primary resistance, since the copper loss and the
current are known.
In the example Pr =120 watts. This must be divided be-
tween primary and secondary, and half may be assigned to each,
as a reasonable approximation.
Thus primary
Also,
Pr = = 60 watts.
W 10,000
7l = Yi = " "2000T = 5
60
.'. #1 = ^ = 2.4 ohms.
&o
Knowing the resistance and number of turns, the size of wire
may be found when the mean length of one turn is estimated.
As a basis for this, the cross-sectional area, A, of the core is known,
and experience tells about how much space is necessary for insu-
lation between core and coils and for circulation of the cooling
oil between the coils. Also, since the heat generated in the in-
terior of the coils has to pass through the thickness of copper and
insulation, it will be unwise to make the coils too thick.
Practical thickness of insulation against voltage is given in
the following table.
TABLE VII
,. ,. Insulation
Volta thickness (mils)
110 40
440 50
1,000 70
2,300 100
6,600 180
16,000 260
216
ELECTRICAL ENGINEERING
For circulation of oil, space of not less than J m width should
be allowed. This width is governed by the height of the coils.
Thickness of the coils should hardly exceed 1 in., but may
reasonably be % in.
Applying this procedure to the example, it is found that with
an area, A = 8.37 sq. in. of iron, the gross area occupied by the
laminations will be about
If this area is in the form of a square, the side of the square
will be -\/9l3 = 3.05 in. Fig. 162 is
next drawn, showing the relative
positions of coils, core, insulation, etc.
In this case, the length of mean turn
of the secondary winding is
L 2 = 4 X 3.05 + 2^(0.25 +
0.04 + 0.375) = 16.4 in.
Since the secondary winding is
nearest the core, its features will be
discussed first, thus avoiding any
error in the final determination of
TI
,~\
t75-
^
V
-f
-1
1
25,
= 3.05 :=>
Core
^
1
Primary
FIG. 162.
the mean length of primary turn.
Total length of secondary is
16 4
X < 2 = -TH- X
12
ft.
In general,
100
2000
X 1280 = 64.
In practice, however, it is found convenient to put the two
primary coils in series and the two secondary coils in parallel to
obtain the 20: 1 ratio.
If this procedure is adopted, the voltage impressed on one
, 2000
primary coil is -y- = 1000, while the whole secondary voltage
of 100 will be across each of the secondary coils.
The secondary turns per coil will then be
100
1000
X 640 = 64
GENERAL PRINCIPLES OF TRANSFORMER DESIGN 217
and each coil will carry half of the total secondary current or
10,000
= 50 amp.
Total length of each secondary coil is then
L 2 X t, = ~ X 64 = 87.3 ft.
27?
Resistance per 1000 ft. of secondary coil = AQ i *
U.Uo7o
Since the coils are connected in parallel, the resistance of one
coil is 2R Z .
The resistance of a secondary coil is obtained from the fact
60
that the secondary copper loss per coil is -^ = 30 watts.
Thus, the resistance of each secondary coil is
30
= - 012 onm -
Resistance per 1000 ft. of conductor is n ' 7Q = 0.1375 ohm.
U.Uo/o
This corresponds to an area of 0.07 sq. in.
The conductor chosen must be of copper strip, of rectangular
cross-section. In using strip, the practical dimensional limits
are about 0.1 in. in thickness and 0.5 in. in width. These
dimensions give an area of 0.05 sq. in. If greater area is re-
quired, any number of strips may be wound in parallel. Each
strip is, however, insulated, usually with double cotton covering,
to prevent too great eddy current loss in the copper.
In the present case there will be two strips required, each of
0.1 X 0.35-in. section.
With insulation, the dimensions of the double conductor become
0.36 X 0.22 in.
It will be seen that the most practical arrangement of the
turns will be to have two layers deep and 32 turns per layer.
Then the thickness of the coil becomes 2 X 0.22 in. = 0.44 in.;
the length of the coil is 32 X 0.36 in. X 11.5 in.
The corrected mean length of turn is
L 2 = 4 X 3.05 + 27r (0.25 + 0.04 + 0.22) = 15.4 in.
I K A y f\A
Total length = - ^~ - 82 ft.
Resistance per 1000 ft. = ^ = 0.147 ohm.
1375
Corrected cross-section is X 0.07 sq. in. = 0.0655 sq. in.
218 ELECTRICAL ENGINEERING
Maintaining the same thickness, i.e., 0.1 in. the width of the
strip, with insulation, now becomes 0.332 in., and the coil length
is 32 X 0.332 = 10.6 in.
The mean length of the primary turn may now be found. It is
L l = 4 X 3.05 + 2T (0.25 + 0.04 + 0.44 + 0.04 +
0.25 + 0.1 + 0.375) = 12.2 + 2w X 1.495 = 21.6 in.
Total length of primary is then
/
21.6 X || = 21.6 X = 2304 ft.
r> . 24-
Resistance per 1000 ft. of primary is 2304 = 2394. =
ohms.
Referring to wire tables, this resistance is found to be nearly
that of No. 10 B. & S., which has resistance of 1.18 ohms per 1000
ft. at 65C.
If now it should be desirable to use copper strip for the
primary winding, the requisite area may be found by comparison
with that of No. 10 wire. Thus,
1 18
area = - X 0.00815 = 0.00922 sq. in.
In this case, however, it will be practical to use No. 10 wire.
Wire larger than No. 10 is not generally used, but smaller sizes
are preferable to rectangular strip.
Choosing then No. 10 wire the space which the 640 turns of
each coil will occupy must be determined.
With a layer 10.3 in. long there will be
10 3
Q IQ2 = 100 turns per layer,
and
640
100 =: 6 ' 4 layers '
Obviously, the best arrangement of these 640 turns will be to have
8 layers of 80 turns each, giving a coil length of 8.24 in., and coil
thickness of 0.824 in. The thickness will be slightly less, owing
to the bedding of the layers. Perfect bedding would give 0.824
X 0.866 = 0.714 in. The value of 0.75 in. originally assumed
may therefore conveniently be taken as correct.
The mean length of the primary turn is then LI = 21.6 in.
as previously calculated.
GENERAL PRINCIPLES OF TRANSFORMER DESIGN 219
21 6
Total length of primary = -^ X 1280 = 2300 ft. Total
primary resistance = 2300 X 1.18 = 2.72 ohms.
This resistance deviates considerably from the value of 2.4
ohms assumed, but not enough to warrant the choice of another
size of wire for the primary.
Having determined the core cross-section, and the coils, an
assembly sketch may be made, as shown in Fig. 163. Allowing
0.3 in. between the coils on the two legs, the size of the window is
FIG. 163.
found to be 4.24 in. wide by 10.75 in. high. The total core
height is 10.75 + 3.05 + 3.05 = 16.85 in.; total core width is
4.24 + 3.05 + 3.05 = 10.34 in.
The total volume of iron is length X net cross-section, or,
V = (2 X 16.85 -f 2 X 4.24) X 8.37
= 42.18 X 8.37 = 354 cu in.,
which does not compare very favorably with the first assumption
of 250 cu. in., but this is not very important since it should
be noted that I calculated from core loss and I calculated from
220 ELECTRICAL ENGINEERING
the space required for the copper windings will generally not be in
agreement. We must have sufficient space for the windings,
but I should not be any greater than necessary. Therefore,
unless we wish an entire recalculation of the design based on
altered assumptions it is sufficient to accept the new value of I
and the attendant new value of V. The mean length of the flux
path is,
I = 2 (10.75 + 4.24) + 27r X 1.5
= 29.98 + 9-44 = 39.42 in.,
as against 29.5 in. in the preliminary calculations.
Per Cent. Magnetizing Current and Core Loss. Applying the
new values of V and I, we may obtain new values for per cent.
magnetizing current and the core losses. Thus,
amp. turns = M Q X I = 6.5 X 39.42 = 256.
256
Max. exciting current = \/2i m = TQC = 0.2.
'
Per cent. i m = = = 0.028.
/i 5
Hysteresis loss, W h , will be in the ratio of the two volumes thus
far obtained, namely; 250 cu. in. and 354 cu. in.
354
Wh = 250 x 55<2 = 78 * 2 watts ' and similarlv the edd y
354
current loss is W e = X 24.8 = 35.1 watts, giving a total
core loss of 113.3 watts.
Efficiency. The approximate efficiency is then
input losses
77 = -- ; - >
input
where the losses are:
Primary copper loss = 5 2 X 2.72 =68 watts.
Secondary copper loss = 2 X 50 2 X 0.012 = 60 watts.
Core loss = 113.3 watts.
Total loss = 241.4 watts.
*'* * = ~ 10 ooo = * 976 = 97>6 per cent>
It is seen that the efficiency is very nearly that which was
assumed at the outset, so that the variations in values, even where
GENERAL PRINCIPLES OF TRANSFORMER DESIGN 221
they have been large, as with volumes and length, have not been
such as to produce any considerable effect.
TT copper loss .
However, the ratio of ^ ron | oss nas decreased, the former
now constituting only 53 per cent, of the total loss, instead of 60
per cent, as at first assumed.
There is, of course, nothing hard and fast about these relations
as it is impossible to say just how the transformer will be oper-
ated from day to day. If it is desired to approximate more
closely to the original assumptions of losses and efficiency, it will
be necessary to go back to the beginning and choose from among
the numerous variables, let us say, another value of turns and
another value of the flux density.
However, in the present instance, the values so far obtained
will be regarded as satisfactory, and then there remains only
to determine the regulation, heating and cost of material, to see
if these also will be satisfactory.
Regulation. In Chap. XXVI, p. 182, the regulation of a trans-
former was found to be
Reg. = ^ " 1>
where EI = V E 2 2 + 2E z (i 2 r Q + 0.54n> - i'&o + 0.5i m z )
approximately.
Of these quantities, EI = 2000 volts, assumed impressed on
the primary, i z = 7 2 = 5 amp. = the energy component of the
load at unity power factor (assumed) and referred to the primary
basis. On this assumption, the wattless component of the load
current, i'z = 0.
4 = energy component of the exciting current.
To obtain 4, we have: core loss = Eih = 113.3 watts.
1100
" ih = ^ = - 0566 amp *'
and
too = Vi^+4 2 = V(H41 2 + 0.0566 2 = 0.152 amp.
r = fli + # 2 = 2.72 + 0.009 X 400 = 2.72 + 3.6 = 6.32.
To determine X Q , the combined leakage reactance of primary
and secondary, we have
X Q = 27T/ (Li + 1/2) = 27T/I/0.
LI and L 2 may now be calculated by the help of equations,
p. 204, Chap. XXX, but each must be done separately since the
222 ELECTRICAL ENGINEERING
two primary coils are in series while the secondary coils are in
parallel. We have
Referring to Fig. 155, the constants in these equations are
readily evaluated. Thus, we have,
Ni = 640; N* = 64; I = 10.75,
mi = mean length of primary turn = 21.6 in.,
mz = mean length of secondary turn = 15.4 in.,
mz = mean length of gap between coils = 18 in.,
a = secondary coil thickness = 0.44 in.,
b = distance between coils = 0.39 in.,
c = primary coil thickness = 0.75 in.
Supplying these values,
fc - 64 X 10-9 41 X 1Q4 2L6X ' 75 18X ' 39
Ll ~
10.75 3
= 0.00244 [5.4 + 3.51] = 0.02175 henry
= 6.1 X 10-6 [2.26 + 3.51] = 0.0000352 henry,
where L 2 is the actual secondary inductance.
Referred to the primary, L 2 = 0.0000352 X 400 = 0.0141
henry.
Lo = Li + L 2 = 0.02175 + 0.0141 = 0.03588 henry
and
Zo =27r/Lo = 377 X 0.03588 = 13.53 ohms.
Supplying all the values into the formula for Ei, we have,
Ei = 2000 =
X 6.32+0.5 X 0.0566 X 6.32 + 0.5 X 0.141 X 13.53)
4 X 10 6 = # 2 2 + 2# 2 (31.6 + 0.179 + 0.95) =
V 2000
.'. E 2 = 1968 volts, and regulation 1 1 =
1.016 1 = 0.016 = 1.6 per cent, for full non-inductive load.
Heating. The total radiating surface of each primary coil is
found by calculation to be 388 sq. in. Therefore, the watts per
square inch which must be radiated from the primary coil are
. . 0875 .
388 sq. in.
GENERAL PRINCIPLES OF TRANSFORMER DESIGN 223
Similarly, the area of the secondary coil radiating surface is 340
sq. in. The watts per square inch that must be radiated are
-
The radiating surface of the core is about 400 sq. in. Therefore
watts per square inch that must be radiated are
__ 113.3 watts
' 400 sq. in.
Watts per square inch serve as an empirical guide by which it
may be determined satisfactorily whether the design is sufficiently
liberal to permit of dissipation of the heat without undue rise of
temperature of any part. In general a loss of 0.4 watt per sq. in.
of surface of the coils and core is quite satisfactory.
In designing the case, however, about 0.15 watts per sq. in.
only should be allowed. In the transformer, then, since the
entire loss in watts must be radiated from the case we should
need an area of
241.3 watts . . .,, ,, .,
fT-^= = 1610 sq. m., in contact with the oil.
U.lo
Weight and Cost of Material. The core volume has been
found to be 354 cu. in. At 0.28 Ib. per cu. in. the core weight is
354 X 0.28 = 99 Ib.
Cost of core at 3.5 c. per Ib. is
99 X 0.035 = $3.46.
The primary copper volume is length X section,
= 2300 X 12 X 0.00815 = 225 cu. in.
Secondary copper volume is
82 X 12 X 0.0655 X 2 = 129 cu. in.
Total volume of copper is 225 + 129 = 354 cu. in.
Weight of copper at 0.32 Ib. per cu. in. is
354 X 0.32 = 113.4 Ib.
Cost of copper at 16c. per Ib. is
113.4 X 0.16 = $18.15.
Total cost of iron and copper is $3.46 + $18.15 = $21.61
Of course, such a calculation of cost has comparative merit
only, as it does not include labor or such materials as insulation,
oil and case.
224
ELECTRICAL ENGINEERING
Summary of data of 10-kw., 60-cycle 2000-100-volt core-type
distributing transformer.
-
High side
Low side
Kilowatts
10
Frequency
60
Ratio of transformation
20:1
Volts
2 000
100
Arnp6r6s
5
100
Window dimensions, in
10% by 424
Total width of iron!! in
Total height of iron, in
10.34
16.85
Depth of lamination, in
Electrical
Number turns in series
3.05
1 280
64
Section of conductor
Amperes per square inch
0.00815
614
. 0655
763
Number of coils
2
2
Connection of coils. ...
Series
Parallel
Width of coil
75
44
Height of coil
8 24
10.6
Number turns per coil
640
64
Mean length of turn
Resistance of circuit at 65C
21.6
2 72
15.4
0.009
Magnetic
Total maximum flux
Effective core section, sq. in.
586,000
8 37
Effective core length, in
Core density
Effective core ampere-turns
39.42
70,000
180.5
Magnetizing current
Thermal
/'flloss
0.141
68
60
Radiating surface of coil
388
340
Watts per square inch
0875
0885
Core loss
113 3
Radiating surface of core
Watts per square inch
400
28
Total loss, full-load
241 3
External radiating surface
Watts per square inch
1,610
15
Efficiency and regulation
Per cent, core loss, full-load
Per cent, copper loss, full-load .
1.15
1 3
Per cent, efficiency, full load
97 6
GENERAL PRINCIPLES OF TRANSFORMER DESIGN 225
High side
Low side
Per cent, magnetizing current
Per cent, resistance
2.8
1 58
Per cent, reactance. . .
3 38
Per cent, regulation
1 6
Weight and cost
Copper, pounds
113 4
Iron, pounds
99
Pounds copper per kilowatt
11 34
Pounds iron per kilowatt .
9 9
Cost of copper at 16c
Cost of iron at 3.5c
Total cost.
$18.15
$3.46 .
$21 61
Cost per kilowatt
$2 16
Having now developed the general principles and procedure in
transformer design, it is desirable that the student should carry
through the calculations for some assigned machines. Trans-
formers of different capacity may be assigned to the students of a
section, each student being required to complete the calculations
for both 60 cycles and 25 cycles, tabulating the specifications and
making sketches to scale of the core and windings.
Many of the finer points in design are omitted here, since the
principles are the primary interest. For practical designing the
fact that experience is a factor of the greatest importance should
always be remembered by the student who is attempting to
master the practical aspects of the subject.
To assist in the further study of the principles of transformer
design, the following suggestive questions are added.
1. Find regulation at 80 per cent, power factor.
2. Why are less volts per turn used with lighting than with
power transformers?
3. If the core loss is too great, how may it be reduced?
4. What relation does per cent, exciting current have to core
loss, copper loss, efficiency and regulation?
5. How may per cent, exciting current be reduced?
15
CHAPTER XXXII
COMBINATIONS IN MULTIPHASE TRANSFORMER
SYSTEMS
When the primary of a transformer is connected to a source
of e.m.f . the following equation relating the impressed e.m.f . cur-
rent, resistance and inductance obviously obtains
e = ri + ^ (Li).
The drop, ri, is small, being perhaps 5 per cent, of 1 per cent.
of the normal voltage, if the exciting current is 5 per cent, of
normal current and the resistance drop at full-load is 1 per cent.
Thus the e.m.f. consumed by the transformer counter e.m.f. is
approximately
If the transformer were merely a coil having an air core, the
inductance would be constant, and the induced voltage would be
. di
If i were a sine wave of current, e would also be a sine wave
displaced 90 behind i. However, with iron cores, as with trans-
formers, the inductance is not constant, but is a function of the
current i. Hence
di . .
and if i is a sine wave of current, the counter e.m.f. of self-induc-
tion is no longer a sine wave. Similarly, if a sine wave e.m.f.
is impressed on the transformer, the current will not have the
sine shape, but will be made up of fundamental, 3d, 5th, etc.,
harmonics.
In Chap. XXVIII, the characteristic wave of exciting current
with a sine wave e.m.f. impressed was determined. On analyzing
this wave by FOURIER'S series as indicated in problem 78, it is
found to consist of a fundamental and triple with higher har-
226
MULTIPHASE TRANSFORMER SYSTEMS 227
monies of lesser amplitudes. The presence of the triple-fre-
quency wave is an important feature in the exciting current of
every iron-cored transformer operating with impressed sine wave
e.m.f.
If, however, the triple frequency wave is suppressed in some
way, as is often the case in three-phase systems, so that the ex-
citing current is of sine shape (neglecting small higher harmonics),
the induced voltage, and consequently the terminal voltage, will
not be of sine shape, but will have the characteristic form shown
on p. 194.
The transformer, of course, must generate its' own counter
e.m.f. or the induced voltage, and hence may be regarded as a
generator, electrical energy being supplied to it instead of me-
chanical energy.
So far the transformer has been dealt with as a single unit.
It is common practice, however, to group transformer units in
various ways so that they shall serve as group units in the trans-
mission and distribution of energy in systems other than the
single-phase system.
The Three-phase System. While two-phase, four-phase and
six-phase systems are used to some extent and under certain con-
ditions, yet the three-phase ^
system in its various forms
is far more important than all
of these. Its study forms a
basis for the development of
any multiphase theory. The
principles of two-phase and
three-phase working from the
standpoint of the alternator
are explained in Chap.
XXXV. In the present in-
stance, three-phase will be
dealt with in reference to the
transformer alone.
Consider three similar transformers, A, B and C, receiving
current from three sources of simple sine waves of e.m.f. (Fig. 164).
Let the voltage impressed on A be e t = EI sin 6, that on B,
e* = E l sin (0 + 120), that on C, e 3 = EI sin (6 + 240). In
each case the arrows in the figure indicate outgoing and return
wires.
FIG. 164.
228 ELECTRICAL ENGINEERING
Neglecting higher harmonics the currents flowing will be, re-
spectively
ii = /i sin (0 0)
i z = I, gin (6 + 120 - 0)
i z = /! sin (0 + 240 - 0).
If the transformers are so arranged
that the return currents shall flow
through the same wire, as in Fig. 165,
the value of the current in this return
FIG. 165. wire will be
i n = ii + iz + iz = Ii (sin (0 - 0) -f sin (0 + 120 - 0)
+ sin (6 + 240 - 0).
The student should prove that this current is zero.
He should prove, also, that if there is current of triple frequency
flowing in the lines, the triple frequency current in the fourth or
neutral wire will be three times that in any line.
Problem 79. Let the three line currents be given by the equations:
ii Ii sin (0 $0 + /a sin (30 <s) + It sin (50 $5) + . . . .
it = /i sin (0 + 120 - 00 + 7 3 sin [3(0 + 120) - 3 j
+ h sin [5 (0 + 120) - <fa] + . . .
iz = /i sin (0 + 240 - 00 + h sin [5 (0 + 240) - ( ] +
/ssin [5(0 +240) -] + . . .
Prove that the current in the neutral wire will be
in = 3[/ 3 sin (30 < 3 ) + 1 9 sin (90 < 9 ) + /i 6 sin (150 #12) + J. >
that is, the current in the neutral wire is three times the sum of all
odd harmonics which are multiples of three which are present in
any line wire, all other harmonics becoming zero in the neutral.
Voltage Waves in Three-phase, Four-wire System. The neu-
tral wire serves to make the system virtually three single-phase
systems instead of a three-phase system having peculiarities of
its own. Thus if a sine wave e.m.f. is impressed on each trans-
former, the e.m.f. between lines is the vector sum of any two
of these and is also a sine wave.
Since there is no current of fundamental frequency in the
neutral wire, there is no necessity of having the wire there. Its
absence will not be the occasion for any interruption in the circuit.
However, since the circuit of the higher harmonics has been inter-
rupted, it becomes of importance to study higher harmonic effects
in connection with three-phase and with any other systems.
Transformers so arranged with or without the neutral wire are
MULTIPHASE TRANSFORMER SYSTEMS 229
said to be Y-connected. If the circuits were unbalanced the
situation would be somewhat different, as will be discussed later.
For the present, however, it is sufficient to see that a three-phase
circuit may be composed of only three wires, each representing
the outgoing wire of one of the phases.
Three-phase, Y-connected Transformers. Let it be assumed
that the three, so-called, phase voltages are
OA = BI = E l sin B + Es sin (30 + a),
OB = 6 2 = E l sin (0 + 120) + E z sin (3[0 + 120] + a),
OC = e 3 = #1 sin (0 + 240) + E 3 sin (3[0 + 240] + a),
these voltages being represented vectorially in
Fig. 166. To find the line voltage AB. Evi-
dently this is
e AB = AO + OB = -
since it is taken in direction from A to B.
Directions from outward are taken as posi-
tive. Therefore
CAB = - EI sin - E z sin (30 + ) + E l sin (0 + 120) -f E 3
sin (3[0 + 120] + a)
= #i[sin cos 120 + cos sin 120 - sin 0]
+ Et[sm (30 + a) - sin (30 + a)].
The last term vanishes since sin (3[0 + 120] + a) = sin (30 -f
360 + a) = sin (30 + a) and e AB = 1.73 EI sin (0 + 150).
The student should prove this by performing the intermediate
operations.
Thus, it is seen that in a balanced three-phase Y-connected sys-
tem, a triple frequency e.m.f. cannot exist in the voltage between
the lines. The same will be shown to be true also for what is
called the A-connection. This does not mean, as stated, that
there can be no triple frequency e.m.fs. in the phase windings,
but simply that they cannot be between the lines.
Likewise, the other line voltages are:
e BC = - 1.73 EI sin (0 + 90)
CCA = 1-73 E! sin (0 + 30).
These voltages are represented in Fig. 167 (a) and (6), which
give two ways of representing the same thing. The line and
phase voltage relations may also be shown graphically, as in
230
ELECTRICAL ENGINEERING
167.
.e
Fig. 168. Here, as in the equation for e^ B) OB is combined with
OA reversed.
Carrying out the same process by which the assumed triple-
frequency voltages in the line were eliminated, it could also be
found that any higher harmonics which
were multiples of 3, as 9th, 15th, 21st,
etc., would vanish. Even harmonics are
of necessity absent if the waves are sym-
metrical. Thus there remain only the
5th, 7th, llth, 13th, 17th, 19th, etc.,
which could exist in the lines.
In the three-phase Y-connection, the triple frequency voltages
in the phases are all in time-phase with each other, and the phase
therefore acts like three circuits in parallel. Their extremities
could be joined without causing any triple frequency current to
flow. If the neutral point, 0, be connected to ground, the triple
frequency in the phases would cause all three transmission lines
to oscillate just as would be the case with a single-phase line one
side of which was grounded.
In this case the three lines cor-
respond to the ungrounded side
of the single-phase line.
Three-phase A-connected
Transformers. W hen three
transformers are so connected as
to form a closed circuit, there
are two facts in connection with
their operation which are of
great interest, namely: (1)
there can be no circulating cur-
rent of fundamental frequency
in the windings; and (2) there always flows in the windings a
current of triple frequency or an odd multiple of triple frequency.
In proof of the first fact let the phase voltages be assumed, as
before,
Ci = EI sin + E z sin (3 + a)
e 2 = E l sin (0 + 120) + # 3 sin [3(0 + 120) + a]
e 3 = Ei sin (0 + 240) + E* sin [3(0 + 240) + a)]
Adding the fundamental components,
E l [sin 6 + sin ( 0+ 120) + sin (0 + 240)]
FIG. 168.
MULTIPHASE TRANSFORMER SYSTEMS 231
FIG. 169.
= Ei [sin + sin cos 120 + cos sin 120 + sin cos 240 +
cos sin 240]
= Ei [sin + sin X (- 0.5) + cos X (0.866) + sin X
(- 0.5) + cos X (- 0.866)]
= EI [sin sin 0) = 0.
Thus, if there is no e.m.f. of fundamental frequency acting in
the closed winding, there can be no current of fundamental
frequency circulating in it.
In proof of the second fact, adding the triple-frequency com-
ponents gives
# 3 sin (30 + a) -f E z sin [3(0 + 120) + a] + #3 sin [3(0 +
240) + a) = SE's sin (30 + a).
Thus if the delta is open at one point (Fig. 169) the triple voltage
across the opening is three times the triple-frequency
voltage of one phase. Similarly, with a Y-connec-
tion with neutral point grounded, the triple-fre-
quency current flowing into the ground is three times
the triple-frequency current of one phase. When
the neutral is grounded, it is no longer necessary to regard the
system as three-phase, but it may be considered as three single
phases having a common return, just as with the three-phase four-
wire system already discussed.
The triple-frequency current is
then perfectly free to flow in each
line wire, returning by way of the
neutral, whereas without the neutral,
the triple-frequency current cannot
exist.
To find the relation between line current and phase current
in a A-connected system.
Let the direction of the phase currents be assumed as indicated
in Fig. 170 where
ii = 7i sin + 7 3 sin (30 + a)
iz = h sin (0 + 120) + 7 3 sin [3(0 + 120) + a]
i 3 = 1 1 sin (0 + 240) + 7 3 sin [3(0 + 240) -f a]
Taking the direction of the arrows as positive, the funda-
mental line current,
*A = ii ~ i* = 1 1 (sin - sin (0 + 120)) = 1.73/isin (0 - 30)
FIG. 170.
232 ELECTRICAL ENGINEERING
This relationship is similar to that of the voltages for Y-connec-
tion. Similarly, also, there can be no triple-frequency current in
the line.
Voltage Waves with Y-connected Transformers. In problem
76, was assumed a sine wave of exciting current. This is approx-
imately the case with the three-phase Y-connection since there
can be no triple-frequency current in the line or phase.
The phase voltage must then look like that of Fig. 151. The
line voltage as previously seen will
be a combination of two phase
voltages, one of which is reversed,
as in Fig. 171.
(The depression in the line volt-
age wave is not actually as deep as
would appear from using sine
waves of magnetizing current.)
If the generator develops a sine wave e.m.f . and the transformer
counter e.m.f. is much distorted, due to the hysteresis loop effect,
then the difference between these two waves must be taken up by
drops along the lines and in the apparatus.
Problem 80. Given three A-connected transformers. Make a picture
of the e.m.f. and compare it with that of a single-phase circuit. What is
the shape of the wave of phase current? What is the shape of the wave of
line current? Show also, by a sketch that the sum of two exciting current
waves of a three-phase A-connected system makes nearly a sine wave, the
triple frequency vanishing. How does the core loss in this system compare
with that of three single-phase circuits?
Problem 81. Given three Y-connected transformers. The line current
can contain no triple harmonics but only 1st, 5th, 7th, llth, etc., harmonics.
The phase voltage, however, has a large triple harmonic. The line voltage
is a combination of two-phase voltages. What is the ratio of the phase
voltage to the line voltage? Is it 58 per cent.? Evidently it is higher, as
the phase voltage contains also the triple-
frequency voltage. How does the core loss
of this system compare with that of the A-
connected system and with the single-phase?
The flux wave is flat, since there is no
triple-frequency current. Therefore the p IGi 172.
maximum value of flux is less, and the
core loss is less (by about 30 per cent.), than with sine waves of flux.
Moreover, the exciting current is less because the max. value of the flux
density is less.
With open delta connection what will the voltmeter read? Evidently
three times the triple-frequency voltage per phase. Why? With the
MULTIPHASE TRANSFORMER SYSTEMS 233
neutral wire connected in a Y-system as in Fig. 172, what would be the
current in the neutral?
As has been pointed out, this is no longer a real three-phase system, but
three single phases in which the neutral wire is common to all the phases.
Evidently the triple-frequency currents can flow in each phase, and since
they are all in time-phase with each other the neutral will carry three times
the triple harmonic current of each phase.
What then, will be the effect on the core loss of connecting in the neutral,
as compared with leaving it out?
These problems are stated in such a way as to form the basis for a fairly
complete discussion, on the part of the student, of the effects which would
be produced by the different ways of connecting the transformer.
Three-phase Transformers. A natural development in the
use of three single-phase transformers for three-phase work is the
substitution therefor of a single three-phase transformer.
(d)
Po Po
'
p o (e) Po
FIG. 173.
Let there be three cores of laminated iron, symmetrically placed,
connected by legs, each core having on it the windings of one
phase. This may be done as in Fig. 173, a, b. How, then, should
the sectional area of the core be calculated? This should evi-
dently be done in the regular way since each leg has its coil, and
must have its flux set up by the coil. The yoke, however, cor-
responds to a A-connection, and the flux in any leg of the yoke is
= = 58 per cent, of the flux in any core.
The yoke may also be formed as a Y-connection (Fig. 173, c, 6),
in which the flux in any branch of the Y is the same as that in
any leg.
234
ELECTRICAL ENGINEERING
In practice, however, it is common to employ a form such as
Fig. 173, d y in which there are three equal legs carrying the coils
and the yoke is straight across the top and bottom. The whole
core is built up of laminations, which, except for the diffi-
culty of placing the coils, could be of one piece. In Fig. 173, d,
the flux paths are outlined by dotted lines, and it is evident that
so far as the magnetic core is concerned, coils 1 and 3 are sym-
metrical with respect to each other, while coil 2 is unsymmetrical
with respect to 1 and 3.
Assume the reluctance of one leg to be p, and the reluctance
of one section of the yoke to be po.
Then there may be constructed an analo-
gous electric circuit, as in Fig. 173, e. This
gives the magnetic circuit which is supplied
with flux by the m.m.f. of coil 1, on leg 1.
It may be simplified to Fig. 174, in which
.
p' p p + 2p
FIG. 174.
p' = p(p
2p )
2(p + po)
The total reluctance of the magnetic circuit of coil 1, is thus
2p
(3p -f 2 Po )
^-
2P:
"2(p +
It is also evident that pi = p 3 .
For p 2 , the circuit may be considered as
made of two parallel paths, as in Fig. 175.
Here,
1 p ;
P2 = 2 (^P + 2p ). FIG. 175.
Thus the relative reluctances of the two circuits are
P2
Pi
PO
2p
In a good transformer, the ratio of height to width of the
window is from 4 to 8:1; the average is about 6:1.
.'. P is from 4 to 8 times as large as p .
MULTIPHASE TRANSFORMER SYSTEMS 235
Assuming p = 6p ,
P2 _ Tpo _ 7
Pi 8po 8
.'. p 2 has 87K per cent, as great a value as Pl , and the exciting
current in the middle coil is from 80 per cent, to 90 per cent, of
that in the outside coils.
Suppose it were necessary to have equal exciting current in all
the coils. This could be accomplished by reducing the section
of the middle leg.
But, in this case, the core loss would be unbalanced, for it is
approximately proportional to the square of the flux density
which would be increased in the middle leg.
Therefore, there must be some unbalancing. In practice all
parts are made of equal section, including the yoke. It is there-
fore easy to calculate the saving in material over three single-
phase transformers.
Question. Considering wave shapes as discussed above if the
coils are connected Y, will there be a triple-frequency voltage?
It has been shown that the triple-fre-
quency currents are in time-phase in the
different phases. Hence they produce
fluxes in time-phase with each other. These
then neutralize each other or pass around
through the air which makes them very
weak. The induced triple-frequency volt-
ages are therefore very small. Thus a three-phase Y-connected
transformer acts like three choking coils as far as the triple-fre-
quency current is concerned. Their
flux paths being largely in air, the
hysteresis loops are very thin, causing
small distortion. Therefore, core-
type three-phase transformers behave
much like three A-connected single-
phase transformers as regards triple-
frequency harmonics.
Shell-type Three-phase Trans-
formers. These could be made by
FIG. 177. placing three single-phase shell- type
transformers one on the other. In such a case, the leg with the
coil has a width, a, while the other legs have widths a/2 (Fig.
177).
FIG. 176.
! !
236
ELECTRICAL ENGINEERING
The combined intermediate sections or widths, 6, could be re-
duced to n~ a, since the fluxes differ by 30 in time-phase, in
adjacent intermediate sections. This is seen to be the case by
noting the dotted lines in the figure. Arrows indicate what may
be called the positive direction of the flux and these directions
are opposite in the adjacent intermediate sections.
If now the middle coil is reversed, the positive direction of the
flux in the middle transformer is reversed, and the flux phases in
intermediate adjacent sections have a 60 relation. The total
flux in these sections is therefore
exactly the same as that in the out-
side section, and required width of
section b is also that of the outside
section namely, a/2. Transformers
are therefore designed as in Fig. 178,
with all width dimensions a/2, except
the middle legs which have the
width, a.
To prove that in reversing the
middle coil the flux produced is in
amount the same flux as when the
coil is not reversed.
The flux produced by coil 1 is $ sin 0. Flux produced by coil
2, reversed, is - $ sin (0 + 120). The flux due to the two coils
is then
$ [- sin - sin (B + 120)]
= 3> [- sin - sin 6 cos 120 - cos 6 sin 120]
- $ [0.5 sin + 0.866 cos 0] = - & sin (0 + a)
where a. = 30.
Problem 82. Discuss the wave shapes of three-phase transformers.
Show that in the core type, it makes very little difference whether the neutral
is connected or not.
Show that, in the shell type, the waves are essentially the same as those
of three single-phase transformers.
Open Delta Transformer Connection. If one of three delta-
connected transformers is disabled it is possible to operate at
reduced output with the remaining two, connected as shown
in Fig. 179.
The following is a comparison of the use of two transformers
FIG. 178.
MULTIPHASE TRANSFORMER SYSTEMS 237
and three transformers when the power delivered is assumed
equal in the two cases.
A <
Current in transformer /, 7= /
V
Voltage across transformer. . . E p , E E
TJIT
Rating of each transformer. . . El
v 3
ETT
Rating of installation 3 /= = \/3EI 2EI
Ratio of transformer capacity = -~ in favor of the three
transformers. However, it may be cheaper, in a given initial
installation, to buy two large transformers r
than three small ones. i
Now, assume a three-transformer in- /
stallation in which one transformer has \
been disabled. How much should the
load be reduced to give normal operation - --
of the remaining two on open delta?
The line current must evidently be reduced in the ratio -p
v 3
since the line current and the phase current are now the same.
The output, which was -\/3EI, therefore becomes -\/3E 7= = El.
V 3
ETT
Therefore the ratio of outputs is x - = 0.58 or less than the
V3-M ___
ratio of transformer capacity which is 2/3 or 0.0SB.
b Two transformers are frequently
used both for three-phase and for
Teaser PLJr a combination of three-phase-two-
phase transformation being con-
Mam primary Mam secondary nected in a manner known as the
FIG. 180. T- or SCOTT connection.
T-connection of Transformers.
This connection is illustrated in Fig. 180. The connection is
used commonly in circuits with rotary converters, where a wire
may be brought out from the neutral, h', and connected to the
middle wire of a three-wire system on the direct-current side.
In this case the direct current flowing in the transformer wind-
238
ELECTRICAL ENGINEERING
ings has no magnetizing effect since it flows in opposite direc-
tion in the two halves of the transformer windings.
It consists of a so-called "main" transformer with a tap
brought out at the middle points of its windings and a " teaser"
transformer of 0.866 times as many turns, one terminal of which
is connected to the tap, d, of the main transformer. The three-
phase lines are brought to the terminals, a, b, c, which are at the
three vertices of an equilateral triangle. Thus, if the base, ac,
of the triangle has the length, Z, its height, bd, will be 0.866.
The center of this triangle will be at a point, h, called the neutral.
Rating of T-connected Transformers. Three-phase output
= -\/3EI, where E and / are line voltage and current, respect-
ively. Rated output of the two transformers
= El + 0.866#7 = 1.866#7.
.". Ratio of the output to the transformer
1 73
rating is -T = 0.925 & 92.5 per cent. This
means that for the same values of E and 7,
FIG. 181. three single transformers would need to have
only 92.5 per cent, of the kva. rating which
the T-connected transformers would have. Thus, the T-con-
nection is nearly as good. It may in some cases be cheaper, as
it involves only two transformers.
Two-phase Three-phase Transformation. Let two-phase
currents be led to the primaries,
while three-phases are taken
from the secondaries. Con-
sidered as 1:1 ratio of the main
transformers for convenience //
only, the teasers would be in the'
ratio 1 : 0.866. Neglecting excit-
ing current the two-phase input = 2EI, = three-phase output
= -7= 7 = 1.167.
FIG. 182.
The rating of a transformer may be taken as the average of the
input and output, and it is therefore,
Rating = % [2EI + 1.1QEI + (0.866# X 1.167)]
= M [2#7 + 1.1QEI + El].
= El + 1.08 El = 2.08 EL
MULTIPHASE TRANSFORMER SYSTEMS 239
2
The so-called cost efficiency is therefore ^-7^ = 0.96, that is.
^.Uo
the rating is 96 per cent, of that of two transformers for an
ordinary two-phase transformation, or for two single-phase trans-
formers. That is nearly as good as using three transformers for
the three-phase and, there being only two transformers, possible
economy is suggested.
The question arises as to how it is that, with such connections
the magnetization is uniform. If it is not uniform, there will
be complications due to over and under saturation in the different
parts of the cores. Therefore, the sum of the magnetomotive
forces due to the load current in the branches of the windings must
add up to zero, that is, the two-phase load ampere-turns in each
branch must be balanced by the correspond-
ing three-phase ampere-turns. Let oa, ob,
oc (Fig. 183), represent the three-phase AT,
in amount and direction. Let de represent
the two-phase AT 7 in the main transformer.
To obtain the three-phase projections on the
two-phase line, it is necessary to take the j? IG 183
components of oc and ob on the horizontal.
These equal de. The components, however, are in opposite
directions, but due to the fact that the ampere-turns from c to
6 are evidently all in the same direction, ob must be projected
backward to ob'. This causes the vertical components, b'd and
dc to be in opposition and they therefore cancel each other.
The proof of this by trigonometrical relations is as follows:
m.m.f. of primary = It sin 0, where t is the number of pri-
mary turns.
m.m.f. of od = ~- sin (0 + 210)
m.m.f. of oe = - sin (0 - 30).
Total m.m.f. = It sin + 1.16 ~ [sin (0 + 210) + sin (0 - 30)]
= It sin - 0.587* [sin (0 + 30) + sin (0 - 30)]
= It sin - 0.587* [sin cos 30 + cos sin 30
+ sin cos 30 - cos sin 30]
= It sin - 0.587* (1.73 sin 0) = 0.
This means that the load current does not increase the magneti-
zation of the transformer.
240 ELECTRICAL ENGINEERING
In the case of the teaser transformer, both the primary and the
secondary are in the same direction in space and time, that is,
they bear the same phase relation as with single-phase trans-
formers. Therefore, the ampere-turns relation is; secondary
A.T. = 0.866 X 1.16EI = El = primary ampere-turns. There-
fore as turns are proportional to voltage, the m.m.fs. are equal.
It is always possible to buy transformers of both 10:1 and 9:1
ratios from stock. For practical reasons 9:1 is used instead of
8.6: 1. With these ratios, connection can be made to nearly any
system in practical operation.
Auto-transformer's (also called compensators). Auto- trans-
formers are transformers with only one winding. The primary
voltage is applied to the coil terminals; the sec-
ondary voltage is obtained by connecting to taps
at any desired places of the winding.
The general connections of the single-phase
auto-transformers are as in Fig. 184. Let /i and
J 2 be the primary and secondary currents, respec-
tively.
Then
/i = current in ab.
/2 /i = current in be.
The rating of the section ab is h(Ei E z ). The rating of
section be is (7 2 Ii)E 2 . The rating of the auto- transformer is
the average of the sum, or
rating = ^ [/^ - IE 2 + I 2 E 2 - hEz]
= 1 A [IiEi + / 2 #2 - 2/i^J. (Ill)
Neglecting exciting current, as in any transformer, the volt-
"F I
age and current ratios are -^r = 7^ or IiEi = I 2 E 2 . Substi-
&2 1\
tuting for I 2 E 2 in (111), the rating becomes,
rating = % [2I 1 E 1 - 2/ 1 # 2 [ = I I [Ei - E t ].
The per cent, rating for a given current is
Ii(Ei - Ei) = E l - Ei
IjEi EI
Thus, if E 2 = 90 per cent, of E ly
Per cent, rating = - = 0.1, or 10 per cent. That is, it is
MULTIPHASE TRANSFORMER SYSTEMS 241
necessary to supply only 10 per cent, of the rating of an ordinary
transformer to effect this transformation which is obviously a
great gain in cost efficiency. If the voltage is to be reduced in the
ratio 2:1 the economy of using an auto-transformer instead of an
ordinary transformer is not so great. The saving is in this case
about one-half.
Problem 83. Show the advantage of using auto-transformers by plotting
a curve between per cent, rating of the auto-transformers and transformation
ratio.
Compensators for Two-phase Three-phase Transformation.
In Fig. 185, let the two-phase taps be cb and ef, and let the
three-phase taps be a, d, g, and let Ez and /2 be two-phase volt-
age and current respectively and E s and 7 3 be
corresponding values for three-phase. Neglecting
g
losses, \/3^3^3 = 2# 2 7 2 , is the power relation be- / \*\
tween input and output. ,//
Considering separate parts of the windings, cur- -^
rent in ab 7 3 ; voltage in ab = 0.866# 3 E 2 , FIG. 185.
since voltage in ac = 0.866# 3 ; rating of ab =
(0.866# 3 - # 2 )7 3 ; current in bh = 7 2 - 7 3 . In this case 7 2 >
7 3 , E 3 being > E 2 . Voltage in bh = E 2 - H 0.866 E 3 since he =
J^ac; rating of &ft = (7 2 - 7 3 )(# 2 - M 0.866^ 3 ); current in he
= 1 2 7s, since the resultant sum of two equal currents 120
apart is numerically equal to one of them. The three-phase
current in he is the sum of the currents of the phases hd and hg,
indicated by dotted lines.
Voltage in he = l /i 0.866 E 3 ;
rating of he = % 0.866 # 3 (7 2 - /) ;
current in de = 7 3 = current infg;
voltage of de = H (#3 - #2) = voltage of fg;
rating of de = or (#3 - #2) = rating of /gr;
current in ec = V(/2 - h cos 30) 2 + (/ 8 sin 30") = current in
c/, that is, it is 7 2 - the component of 7 3 , in phase with 7 2 + j X
the component of 7s normal to Iz>
p
Voltage of ec = -~ = voltage of cf.
Rating of ec = yV(/2 - I* cos 30) 2 + (/, sin 30) 2 = rating
of cf.
16
242 ELECTRICAL ENGINEERING
The combined rating, which is one-half the sum of the ratings
of all the parts, is
0.933# 3 /3 ~
0.5# 2 X V/2 2 - 1.73/ 2 /3 + / 3 2 .
An examination of this rather complicated expression will show
that the same ratio of cost efficiency holds with reference to the
T-connected transformers having primary and secondary wind-
ings, as holds for single-phase auto-transformers compared with
ordinary single-phase transformers.
Dissimilar Transformers in Series. Transformers may not be
indiscriminately connected in series with safety.
To connect two transformers of different design but proper
rated voltages in series is not always safe, since they may not
take their proper share of the total voltage. One may even burn
out at no-load due to excessive core loss. Suppose that their
normal exciting currents are different Since the same amount
of current must flow through each transformer (as they are in
series), this current will be insufficient to give the proper flux in
one of the transformers and will be more than necessary in the
other. Thus the voltages will not divide according to the rating
and the core loss will be low in one and excessive in the other.
Let the open circuit or exciting impedance of A, Fig. 186, be
r + jx = z,
that of , A
~\- JXi = Z\. B
The total impedance is then Z=z + Zi = .R-f jX. FIG. 186.
Then the exciting current of the two transformers in series is
, _ CQ _ impressed volts
" Z = R+jX
The voltage across A, is
Voltage across B is
V B = e
Neglecting the power component, we get as a fair approximation,
MULTIPHASE TRANSFORMER SYSTEMS 243
snce
' approximately,
A -
_
/.I
I m ,A
where I m ,B> and 7 mrA , are the normal magnetizing currents. Thus,
the respective voltage drops across A and B, when in series, will
be approximately inversely proportional to the normal magnetiz-
ing or exciting currents.
This assumes constant values of x and rti, which would not be
true if the resultant exciting current differed widely from the
normal values of exciting current of the two transformers.
If x is nearly equal to x\ 9 the above ratios would hold. Where
one transformer, however, is saturated, its reactance is greatly
diminished, which allows a greater current to flow in the circuit but
tends to equalize the voltages.
Dissimilar Transformers in Parallel. This is the usual mode
of connection and it offers no difficulty due to unequal exciting
currents. The question, here, is one of proper division of the
load. In giving orders for additional equipment, it is customary
to specify what the percentage reactance of the new transformers
shall be. With equal, or proportional, reactances there results
a proper division of the load. Consider the parallel connection
as shown in Fig. 187.
The two load currents are:
I A = i + Ji',
IB = i\ + fi'i-
The total load current,
I = I A + I B .
' '
Let
Z A = r + jx and Z B = TI +
FIG. 187.
be the impedances of A and B respectively.
Then, since the terminal voltages are the same on each, the
voltage drops in the transformers are equal, and are:
(i + ji')(r + jx) = (ii + ji'i)(ri + jxi).
244 ELECTRICAL ENGINEERING
Multiplying out,
ir + jix + ji'r - i'x = itfi + ji&i + ji'tfi - i'&i.
Here, the real components must be equal and the imaginary
components must be equal.
.". ir - i'x = itfi - i'&i,
and
ix + i'r = iiXi -j- zVi.
Neglecting resistances,
and
ix =
whence
/
i> X\ i> 3/1
z'i "~ a; ' ii "~ #
Thus, the load is divided in inverse proportion to the reactances.
The best method of connection is, as in Fig. 188. The student
is advised to explain why this is so. ' A ^c
ft P i
FIG. 188.
Three-phase Connection of Dissimilar Transformers. If
the three transformers are connected Y Y there will not be
symmetrical distribution of voltage. Consider the neutral
point, 0, Fig. 189, with reference to the transformers A and B.
With line voltage impressed on AB, the potential at may have
any intermediate value, just as with two single-phase trans-
formers in series, depending on the relative open circuit im-
pedances of the two transformers.
The point of junction of the three transformers may, for in-
stance, be displaced to 0'. The secondary Y-voltages would
have a similar relationship to each other. Dissimilarity may
consist merely in variation in the iron of two supposedly similar
transformers.
If the secondaries are connected in A, the primaries being
Y-connected this difficulty of unbalanced potentials is eliminated.
The induced voltage in each secondary will, of course, be
MULTIPHASE TRANSFORMER SYSTEMS 245
proportional to that of its primary, giving the closed A, ABC,
Fig. 190, when the primary circuit is balanced.
With unbalanced condition, if the A is left open at B, the vol-
tage vectors will not make a closed figure, but as shown by the
dotted lines, will leave an opening between B' and B". If the
A is then closed, the voltage B'B" will act in the A circuit, sending
a local current which will increase the
magnetization of the transformer whose
flux is below normal and decrease that
of the transformer whose flux is above
B'
FIG. 190.
FIG. 191.
normal. Thus the magnetization is brought back to normal
value, the local current in the A serving to anchor the neutral
point of the Y.
Three-phase transformer systems may be extended in a variety
of ways to cover cases where it is desirable to use six phases.
This practice finds application especially with rotary, or syn-
chronous, converters, and it will be discussed more fully under
that heading. Such combinations of transformers as permit
symmetrical grouping of voltages are illustrated by the double A,
double T, or double Y shown in Fig. 191.
i 2 3
(a)
CHAPTER XXXIII
ALTERNATORS
Fundamentally, direct-current and alternating-current genera-
tors are alike. An alternator becomes a direct-current generator
by adding a commutator. The essential principles of both
machines have been developed in Chaps. VI and VII.
In Fig. 192, a, is represented a simple alternator with a two-pole
field core magnetized with direct current from some independent
source, and an armature with
a single coil. As this armature
revolves there is generated in it
an e.m.f. which follows closely
a sine wave of time values. It
is apparent that the space on
the armature periphery is not
all utilized, and that another
coil could be put on in space
quadrature to the first. In
such a case, two similar e.m.f. waves would bfi produced but in
time quadrature with each other, or at 90 time-phase displace-
ment, that is, one wave would reach its maximum one-quarter
of a period later than the other (Fig. 192, 6).
Such an alternator is called a two-phase, or, sometimes, a
quarter-phase machine.
On the same principle, an armature may be supplied with three
coils, or groups of coils, spaced 120 apart, each group giving its
separate e.m.f. wave (Fig. 192, c).
In this way, any number of coils or groups of coils may be
wound on an armature, giving any desired number of phases.
In practice, however, the majority of alternators are three-
phase, and very seldom is one built for a greater number of
phases than three.
The voltages generated in the various phase windings may be
conveniently shown in their proper relations by vectors.
If in the two-phase case, the ends 1' and 2' (Fig. 193, 6), are
246
(6)
FIG. 192.
ALTERNATORS 247
joined together, the voltages of the two coils will be added
vectorially, so that a voltmeter placed across the terminals, 1, 2,
would read \/2 times the voltage of either coil taken separately,
since the two voltages are in time quadrature. Likewise by
connecting 1 and 2', the joint reading across 1' and 2 will obvi-
ously also be \/2 times the voltage of one phase.
With a three-phase machine it is not quite so apparent that the
voltage between the three collector rings is \/3 times the voltage
generated in one phase. At first sight it might be expected that
the resultant voltage should be the same as that generated in
each phase since the voltages are 120 apart.
C
FIG. 194.
Let, in Fig. 194, OA, the voltage of phase A, be represented by
e A = E m sin at.
Then OB, the voltage of phase B, is evidently
e B = E m sin (at + 120),
and
e c = # m sin (orf + 240).
The voltage between collector rings A and B is thus
SA e B = E m [sin at sin (co + 120)],
which, by simple trigonometric transformation becomes,
A _ B = V3#m sin (w< - 30).
Thus the numerical value of the potential difference between the
collector rings is \/3 times as great as the voltage generated in
each phase and the resultant voltage is displaced 30 from the
voltage generated in phase OA.
Problem 84. Prove that the voltage, between B and C is:
\/3E m sin (wt + 90),
and that the voltage between C and A is:
\/3E m sin (ut + 210).
It is seen, thus, that the voltages between the collector rings are also
120 apart.
248
ELECTRICAL ENGINEERING
It is interesting to note here that a single-phase machine might
be treated as a two-phase machine in which the two phases are
180 apart as is shown in Fig. 195.
Let the voltage generated in OA be E m .sin co. Then that
^ generated in OB is E m sin (coZ + 180). Thus
~B~~ o ~~5 the difference of potential between the collector
FIG. 195. rings at A and B is
(sin (at sin (' + 180) = 2E m sin ut.
The resultant potential difference is twice the voltage generated
in each phase, as should, of course, be the case.
This fact could have been developed also geometrically.
To find the potential difference between A and B } Fig. 194, we
should subtract OB from OA as shown in Fig. 196.
It is not necessary that the windings shall consist of separate
coils. A closed ring winding, or Gramme ring, may be tapped at
symmetrical points and these connected to slip rings, as in Fig.
197. Thus, if a voltmeter is connected across the slip rings (1,1),
the voltage of one phase is read. If connected across rings (2, 2),
the same value of voltage will be indicated, but it is evident that
the phase of this e.m.f. is displaced by 90 time degrees from that
of the first.
FIG. 196.
FIG. 197.
With the three-phase connection taps are brought out at
points 120 space degrees apart and led to slip rings. A volt-
meter, connected across any two rings, will read the voltage
of one coil, say coil a, Fig. 197. But this must also be the sum of
the voltages of the other two coils, since any one coil is in parallel
with the other two coils with respect to the external circuit.
The voltages in this case form a closed, so-called delta, A, and it
is evident that the phase voltage and line voltage, or voltage
between the collector rings, are equal.
In these diagrams only three collector rings and three lines are
shown. Yet in the discussion it has been assumed that one side
ALTERNATORS 249
of each winding is connected to a common point. It would seem,
therefore, that at least four collector rings and lines might be
necessary to form a complete system, in other words, that even a
balanced three-phase system would involve four wires as is shown
in Fig. 199. It is evident that if, with a balanced system, the
current in an ammeter placed at N is always zero, then no return
or fourth wire is necessary.
FIG. 198. FIG. 199.
Let the current in phase A be
ia = Im (sin a>t) and the current in phases
B and C be i b = I m sin (orf + 120) and i c = I m sin (at + 240).
The current in N is then
in = ia + ib + ic = Im sin 0=0.
Problem 86. Prove that no circulatory current of fundamental frequency
flows in the delta-connected generator.
Since with the Y-connected generator the transmission lines really form
extensions of the windings, it is evident that whatever current flows in the
line also flows in each winding.
With the delta-connected generator this is not so, because the line current
is the vector sum of the currents in the adjacent phases, as is shown in Fig.
198. The current in phase 1-2 may be considered the zero vector. Thus
the currents in the phases are:
in = Im sin (at,
i* = I m sin (at + 120),
iai = Im sin (tat + 240).
Then, since the sum of the currents flowing to a point is zero,
it + las - in = 0,
or it = iiz iza I m sin wt I m sin (wt + 120)
= Vzlm sin (cat - 30).
The line current is thus \/3 times as large as the current in the individual
phases.
Referring to Fig. 198, it is evident that
is + in *32 = and i\ + in isi = 0.
Problem 86. Prove that the currents in lines 1 and 3 are, respectively,
sin (tat + 90) and \/3/m sin (tat + 210).
The power given by a three-phase alternator is
P = \/3EI cos a,
250 ELECTRICAL ENGINEERING
whether the alternator is connected Y or A, where / is the effec-
tive value of the line current and E the effective value of the
voltage between the lines, and a is the angle of lead or lag of the
phase current in reference to the phase voltage, that is, cos a is
the power factor.
To prove this, consider a Y-connected generator.
Since I is the line current, it is also the current in each winding.
Since E is the line voltage, the voltage of each of the three
E
phases of the generator is =? Thus the power given by each of
V3
pr
the three phases of the generator is = cos a, and the total
V3
El
power, 3 -= cos a, = \/?>EI cos a.
v 3
Problem 87. Prove that this also applies in the case of a delta-connected
generator.
Voltage to Neutral. In Y-connected alternators the neutral
point is the center of the Y. On a three-phase distribution
system it is often advantageous to run a fourth wire from the
neutral.
The voltage between any of the other wires and the neutral
is the phase voltage, and is equal to the line voltage divided by
Vs.
In a A-connected alternator there is no actual neutral point.
However, for purposes of calculation, a neutral point is imagined
at the center of the delta, and the voltage to neutral
is then the phase voltage divided by \/3, or, since
the phase voltage and the line voltage are the same,
it is equal to the line voltage divided by \/3 as with
Y-connected alternator. The voltage to neutral is
thus independent of the manner of connecting the alternator
windings.
Rating of Alternators. As with transformers, alternators are
rated in kilovolt-amperes, not in kilowatts. This is because
the permissible output of an alternator depends on the current in
its windings, regardless of the phase relation between the current
and the voltage.
The nominal rating of an alternator may be designated as, for
example, A.T.B. 12-400-600-2300, where A signifies alternator,
T signifies three-phase, (S is for one or single-phase), B signifies a
ALTERNATORS 251
revolving field. If the armature is the revolving part, the third
letter is omitted.
12 signifies the number of poles.
400 signifies the rating in k.v.a.
600 signifies the speed in r.p.m.
2300 signifies the rated voltage.
Sometimes a subscript is added to the second letter; thus,
A TZ signifies a three-phase revolving armature alternator having
two slots per pole per phase on the armature.
If the above alternator is Y-connected, the phase voltage, or
2300
voltage to neutral, is -1= 1330 volts.
400 k.v.a. 400,000
The line and phase current is 23QQ = 3^1330 = 10
amp.
If delta-connected, the voltage to the imaginary neutral is
likewise 1330.
100
The line current is also 100, but the phase current is /= =
v 3
57.7 amp.
CHAPTER XXXIV
ARMATURE REACTION
The so-called armature reaction of a machine is a measure of the
m.m.f. of the armature. It is thus expressed in so many ampere-
turns, either on the whole circumference of the armature or,
more often, the m.m.f. on one pole of the armature. This latter
convention will be used in this book.
As will be seen, the m.m.f. of the armature current sometimes
acts against the m.m.f. of the field excitation, sometimes it
assists it, and often its effect is only to shift the flux.
In Fig. 201 the coil (1, 1) is in the position of zero, or mini-
mum, e.m.f., assuming the flux to be symmetrical in the field
system, or due to the field ampere-turns alone. The coil (2, 2)
is in the position of maximum e.m.f. This condition may be
assumed to hold for no-load. The coil (3., 3) is in an interme-
diate position. The current may or may not be in time-phase
FIG. 201.
FIG. 202.
with the e.m.f., but whatever its time-phase relation may be,
in spa6e, it is evident from Fig. 201 that the m.m.f. of the coil
is at right angles to the surface of the coil, and therefore at right
angles to the line which represents the position of the coil.
In Fig. 202 let the armature current lag behind the e.m.f. Its
m.m.f. is seen to be largely in opposition to that of the field which
causes the main flux, and this opposition increases the greater the
lag and becomes complete at 90 lag. Similarly a leading current
252
ARMATURE REACTION
253
umru
S"
FIG. 203.
increases the flux. The current, i, may be divided into two.
components, one of which, i", is entirely wattless and exactly
opposes the field flux, and the other, i 1 ', the watt component in
phase with e { , which merely distorts the field. The effect of
current in the armature is to weaken the resultant flux and to
displace its maximum position, if the current lags, as shown in
Fig. 203. The weakening is due to the wattless
component, the displacement or distortion to the
power component.
Thus the trailing pole-tip may even become
saturated, while the leading pole-tip is robbed of
a large part of its flux. The position of the coil
for maximum induced e.m.f. is shifted ahead, with
lagging current, and behind, with leading current.
These relationships are shown in Fig. 204, where
the induced e.m.f., e iy is taken as the zero vector.
6i is at right angles to the resultant flux, fa, and lags behind it.
The armature current, /, is taken at any angle and produces a
flux (f> a in time-phase with it. This flux vectorially subtracted
from fa, gives </, the field flux. In phase and 90 in time
ahead of the current are respectively, Ir and Ix, the e.m.f.
consumed by the armature resistance and that consumed by
the armature reactance. These com-
bine to make Iz, the impedance drop,
which, subtracted from e t , gives e the
terminal voltage. The phase angle of
the load is then 0.
In constructing the diagrams there is
some advantage in using ampere-turns
instead of fluxes, since then no compli-
cations arise from variable magnetic
reluctances. This has been done in
Fig. 207, Chap. XXXV.
At no-load, I = 0, e t = e and fy = fa.
In Fig. 204 the angle y is the angular
space displacement of the armature with respect to the field
poles, due to the load, that is, the angle between <f> f and fa,
or, 7 = /? 90, where is the angle between e t and </. a is
the angle between e t and e. Considered on the basis of experi-
mental data, e, /, r and x are known and e is chosen as the zero
vector.
I
FIG. 204.
254 ELECTRICAL ENGINEERING
Then the induced e.m.f.,
Ei = e + IZ = e + (i + jii)(r + jx)
e + ir + jix + j't'ir iiS
= (e + ir - iix) + j(ix + i = a + j6 (112)
where I = i + jii for leading current, and ii is negative for
lagging current. From the saturation (magnetization) curve of
the generator the number of ampere-turns needed to produce this
voltage is found. Let F r the resultant m.m.f. = C (a + jb)
or, rather, F r jC(a + jb) because, in space, as has been shown,
the m.m.f. is rotated 90 with respect to Ei.
Then,
F r = C(- b+ja). _
But the resultant m.m.f. is the vector sum of the field and
armature m.m.fs. That is,
F r = F f + F a = F f -f m (i + jii) (113)
Thus, C is the proportionality factor between the resultant
field ampere-turns and the volts, and m is that between the
armature ampere-turns and the current.
Examples If E t = 2500 volts and F r = 3000 amp.-turns,
c _ 3000 _
C "2500"
,. V2 X 1.5 X It 010 .
In a three-phase machine, m = j = 2.12, where
i = turns per phase on the armature. (This factor is discussed
more fully later.) The quantity, \2, enters in order to derive
the maximum value of the ampere-turns from
the effective value; 1.5 comes from the fact
that the resultant field of a three-phase system,
as an armature, or induction motor field, is 1.5
times that of a single-phase system. This is
shown as follows: Consider the components
of flux along the two axes (Fig. 205). The three
x-components in space are H, H cos 120, H cos 240. The
components along the i/-axis in space, are 0, H sin 120, H sin
240. The components of all the phases in time are H cos 8,
H cos (0 + 120) and H cos (0 + 240).
Hence the sum of components along the z-axis in time and
space is H cos + H cos (0 -f 120) cos 120 + H cos (0 + 240)
cos 240 = 1.5# cos B.
ARMATURE REACTION . 255
The sum of components along the 2/-axis in time and space, is
+ H cos (0 + 120) sin 120 + H cos (0 + 240) sin 240 =
1.5tf sin 0.
Transposing (113), the field m.m.f. is
F f = F r - m(i + jii)
= C(-b+ja) - m(i
= bC + jaC mi
= - 60 - mi + j(aC - mi^) (114)
(With lagging current, ii is negative.)
Numerically, F f = \( - bC - mi) 2 + (aC - mt\) 2 (115)
Problem 88. In a certain alternator let the reactance drop be 10 per cent.,
the resistance drop 2 per cent, and the armature reaction equal to no load
one-half the field ampere-turns. How many ampere-turns are required in
the field winding when the alternator is carrying full-load current at 80
per cent, power factor?
Since 'the drops are given in percentage, e will be taken = 1, and 7 = 1.
Then at 0.8 P.F.,
i = 0.8, ii = 0.6,
and a = e -\- ir i\x
= 1 + 0.016 + 0.06 = 1.076
and b = ix + i& = 0.8 - 0.12 = 0.68.
On the percentage bases, also, let
C = 1.
Then m = 0.5.
The ampere-turns required for the field will then be
F f = V(- 0.68 - 0.4) 2 + (1.076 + 0.3) 2 = 1.45.
For non-inductive load, ii = and i = 1. Then
Ei = Va* + b* = 1.023,
and F f = 1.186.
As a continuation and amplification of this problem, consider the follow-
ing:
Problem 89. A three-phase generator has 2 per cent, resistance and 10 per
cent, reactance. Its armature reaction is one-half the no-load field ampere-
turns. The magnetic reluctance is uniform all around the periphery and the
saturation curve is a straight line through the origin, at the point of
operation.
Plot the field excitation (a) against armature current, with variable non-
inductive load, up to high overloads; (b) at full-load current, but with vari-
able power factor leading and lagging; (c) with full-load power output and
variable power factor; (d) same as (a) but at 20 per cent, higher voltage; (e)
same as (6) but at 20 per cent, higher voltage; (/) same as (c) but at 20 per
cent, higher voltage; (g) same as (a) but at 80 per cent, of rated voltage; (/O
same as (b) but at 80 per cent, of rated voltage; (i) same as (c) but at 80 per
cent, of rated voltage.
256
ELECTRICAL ENGINEERING
By equation (115)
F f = V (- bC - mi) 2 + (aC - mil) 2
where
C = 1, m = 0.5
r = 0.02, x = 0.1
o = e + ir ii#,
5 = ix -f- iir
(o) e = 1, i = variable, ii = 0.
f (l+0.02i) 2 ~ =
- 0.6i) 2 +
= \/0.36i 2 + 1 + 0.04* + 0.0004t 2 = \/l + 0.04i + 0.3604^
Tabulating:
i
25
5
75
1
1 25
1 5
2
0.04i ....
i
0.0
0.0
0.01
0.0625
0.02
0.25
0.03
0.5625
0.04
1.0
0.05
1.56
0.06
2.25
0.08
4.0
0.3604i* .
"'
0.0
1.0
1.0
0.0225
1.0325
1.014
0.09
1.11
1.052
0.2027
1.2327
1.109
0.3604
1.4004
1.182
0.562
1.612
1.269
0.811
1.871
1.366
1.442
2.522
1.587
(d)
1.2; Ff = \/1.44 + QJ048i + 0.3604*2
048f
0.0
0.012
0.024
0.036
0.048
0.06
0.072
0.096
(F/)
1.44
1.4745
1.554
1.6787
1.8484
2.062
2.323
2.978
Ff
1 2
1 211
1 244
1 293
1 359
1 433
1 522
1 722
(g) e - 0.8; F f = -y/0.64 + 0.032i + 0.3604i2
0.032i...
0.0
0.008
0.016
0.024
0.032
0.04
0.048
0.064
(f/)i
0.64
0.6705
0.746
0.8667
1.0324
1.242
1.499
2.146
F/
0.8
0.819
0.863
0.93
1.014
1.115
1.222
1.463
(b) 1 = 1, P.F. = ~ = variable, e
- 0.6i - 0.02ii)2
+ 0.02i -
P.F
0.0
o o
0.25
25
0.5
5
0.75
75
1.0
1
0.75
75
0.5
5
0.25
25
0.0
ti
1
968
866
661
661
866
968
1
0.6i
0.02n . . .
*
*
0.0
0.02
-0.02
0004
0.15
0.01936
-0.16936
0288
0.3
0.01732
-0.3173
1008
0.45
0.01322
-0.4632
215
0.6
0.0
-0.6
36
0.45
-0.01322
-0.4368
191
0.3
-0.01732
-0.2827
08
0.15
-0.01936
-0.13064
017
0.0
-0.02
0.02
0004
0.02i ....
-0.6n..
t
1*
0.0
-0.6
0.4
16
0.005
-0.58
0.425
181
0.01
-0.52
0.49
24
0.015
-0.3965
0.6185
0000
0.02
0.0
1.02
0.015
0.3965
1.4115
2A-\
0.01
0.52
1.53
2 sue
0.005
0.58
1.585
0.0
0.6
1.6
2C
* + < 2 ...
Ff
0.1604
0.40
0.2098
0.4575
0.3408
0.584
0.598
0.772
1.4
1.182
2.191
1.48
2.43
1.56
2.532
1.59
2.5604
1.6
(e) e - 1.2, Ff = \/(- 0.6i - 0.02ii)2 + (1.2 + 0.02i - 0.6ii)2 - Vs* + <i 2
tl. .
6
625
OftQ
0010C
Inn
tl*
36
390
476
67
1400
2fi
3 A
30
3 OK
* + 1*...
Ff. .
0.3604
6
0.4188
646
0.5768
07EC
0.885
004.
1.852
IOfi
2.791
3.08
3.217
3.2504
(A)
e - 0.8,
Ff = V
- 0.6i -
0.02*02
+ (0.8 -f
0.02* -
0.6ti) 2 =
Vs 2 + t
2 2
fl
2
225
Ff... 2 . '.'.
0.04
O.C404
2
0.0508
0.0796
282
0.084
0.1848
43
0.176
0.391
Ococ
0.673
1.033
1.475
1.666
1.77
1.85
1.92
1.937
1.96
1.9604
(c) -l, P.F.
= variable, e = 1.
ARMATURE REACTION
257
Ff = (- 0.6i - 0.02ii) 2
0.02i - 6ii) 2 =
P.F
0.25
3. 87
0.5
1.73
0.75
0.883
1.0
0.0
0.75
0.883
0.5
-1 73
0.25
3 87
0.02n....
0.0774
-0 6774
0.0346
-0.6346
0.01766
-0.6177
0.0
-0 6
-0.01766
5823
-0.0346
5654
-0.0774
5226
s 2
-O.Gii. .
t
t 2 .
0.46
-2.322
-1.302
1 7
0.402
-1.04
-0.02
0004
0.382
-0.53
0.49
24
0.36
0.0
1.02
1 04
0.34
0.53
1.55
2 4
0.32
1.04
2.06
4 25
0.274
2.322
3.342
11 2
s 2 + * 2 . .
Ff
2.16
1 47
0.4024
634
0.622
788
1.4
1 181
2.74
1 65
4.57
2 14
11.474
3 38
(/) e = 1.2, F f = \/s 2 + (1.2 + 0.02t - 0.6ii) z =
1 1
-1.102
18
69
1 22
1 75
2 26
3 542
fl2
1 22
0325
476
1 49
3 07
5 11
12 6
S 2 + *1 2 ..
Ff
1.68
1 295
0.4345
659
0.858
925
1.85
1 36
3.41
1 846
5.43
2 33
12.874
3 59
(i) e = 0.8, F f -
+ (0.8 + 0.02i - 0.6ii) 2 -
tz ...
1 502
22
29
82
1 35
1 86
3 142
fz 2
S 2 + <2 2 ..
Ff
2.26
2.72
1.65
0.0485
0.4505
0.671
0.0841
0.4661
0.682
0.673
1.033
1.015
1.82
2.16
1.47
3.45
3.77
1.94
9.9
10.174
3.182
Curves showing the variations brought out by this problem are given in
Fig. 206.
2.4
0.25
0.50
0.25 0.50
Leading
0.75 1.0 1.25
Current
0.75 1.0 0.75
Power Factor
FlQ. 206.
1.50
0.50 0.25
Lagging
2.00
17
CHAPTER XXXV
CHARACTERISTICS OF ALTERNATORS WITH DEFINITE
POLES
In the preceding chapter it has been assumed that the magnetic
reluctance is uniform in the direction of the main field magneto-
motive force as well as in the transverse direction along the
armature surface. This condition exists practically in machines
with distributed field structures as in induction generators,
and, to a very fair degree, turbo-generators.
In engine-driven generators the condition of uniform magnetic
reluctance rarely exists, since such machines are usually built
with definite pole structures. In this type which includes the
majority of machines, the magnetic reluctance in the direction
of the field poles is almost constant for all m.m.fs., and therefore
the flux is proportional to the m.m.f. In the direction along the
surface of the armature the shift of flux is by no means pro-
portional to the m.m.f. but is largely limited 1 by the mechanical
construction, that is, the width of the poles, and it is always less
than proportional to the m.m.f.
Similarly, the self-induction of an armature coil is greater
when it is immediately under the poles than when it is midway
between them. Thus, considering that the armature current is
made up of two components, one, the power component which is
maximum when the coil is under the pole, and the other, the
wattless component, which is maximum when the coil is midway
between the poles, it might be assumed that the reactance is
greater for the power component than for the wattless component.
The general equation for the induced e.m.f. of such a machine
can therefore not be expressed as simply as:
E< = e + IZ = e + (i + ji,)(r + jx)
= e + ir - i& + j(ix + iir),
but must be written:
Et = e -f- IT i&i -f j(ix + z
258
CHARACTERISTICS OF ALTERNATORS 259
where x\ belongs to the wattless component and is about 0.6z,
and x is the reactance belonging to the power component of the
current.
In the synchronous impedance test, x if is obviously determined
since the current in that test lags nearly 90 degrees in time and
hence in space. 1 Similarly, the expression for the field excitation
is not
Ff = mi Cb + j(Ca mil),
but is
F f = _ m i - Cb + j(Ca -
where m is always smaller than mi since m determines the shift
due to the power component of the current. The relative
values of m and mi are not by any means fixed but vary
over a considerable range. It may, however, be assumed that
m = O.Swi.
The value of mi is determined from the winding data. For
instance, in a three-phase machine it is, \/2 X 1.5 X turns per
pole and phase.
The constants for a definite pole machine of the same general
dimensions as the generator previously calculated would thus be
C = 1, mi = 0.5, r = 0.02, x = 0.10,
m = 0.8 X 0.5 = 0.4, Xl = 0.6 X 0.10 = 0.06.
The angular space displacement of the armature with reference
to the field structure between no-load and any particular load is
of interest. Consider first a machine with round rotor.
FIG. 207.
Let e be the Terminal Voltage. At no-load the axes of the
field poles are in the directions of the field flux, that is, in direction
oF , in Fig. 207. With any load, 01, as shown in the figure, the
direction and the magnitude of the field excitation, the former
1 To calculate Xi from synchronous impedance test, assume i = 0. Then,
substituting in (115), F f = -C6 + j(Ca - wiii), where a = -to, 6 = if.
260
ELECTRICAL ENGINEERING
assumed the same as the direction of the field poles, is F/. Thus
the angular space displacement of the field structure in reference
to the armature is represented by the angle a, and tan a has
been shown to be
mi + Cb
tan a = 79 -- -'
Ca mi\
With " definite pole" machines this becomes,
mi + Cb
tan a =
79
Ca
where, of course, a and b are different from a and b in the "round
rotor" case.
As an illustration consider the same two generators, whose
constants have been given. To find a with varying power factor.
(1) For the round rotor type we have:
C = 1; m = 0.5; x = 0.1; r = 0.02; 6 = 1;
/ = t + ji' = 1
a = e + ir - i'x = 1 + 0.02* - 0.1*'
6 = ix + *'r = 0.1* + 0.02^'.
0.1* + 0.02?: / + 0.5* 0.6* + 0.02*' ^
= 1 + 0.02; - 0.1*' - 0.5*' "
Tabulating :
1 + 0.02* - 0.6*'
Leading current
Lagging current
P.F
0.0
0.0
1.0
0.0
0.02
0.02
0.0
-0.6
0.4
0.05
2 52'
0.25
0.25
0.968
0.15
0.01936
0.169
0.005
-0.5808
0.425
0.397
21 40'
0.5
0.5
0.866
0.3
0.0173
0.317
0.01
-0.519
0.491
0.645
32 50'
0.75
0.75
0.662
0.45-
0.01324
0.463
0.015
-0.3972
0.618
0.75
36 53'
1.0
1.0
0.0
0.6
0.0
0.6
0.02
0.0
1.02
0.588
30 27'
0.75
0.75
-0.662
0.45
-0.013
0.437
0.015
0.397
1.412
0.31
17 15'
0.5
0.5
-0.866
0.3
-0.017
0.283
0.01
0.519
1.529
0.185
10 30'
0.25
0.25
-0.968
0.15
-0.019
0.131
0.005
0.58
1.585
0.0826
443'
0.0
0.0
-1.0
0.0
-0.02
-0.02
0.0
0.6
1.6
-0.0125
-45'
'
0.6i
02i'
02i ...
- 0.6i'
(
tan a
(2) For machines with definite poles.
By (112),
E t = a + jb
where a = e + ir - i'x^ b = ix + i'r.
By (114), <
bC + mi
tan a =
~
aC
CHARACTERISTICS OF ALTERNATORS
261
= 0.4; x = 0.1; x l
In this case,
C = 1; mi = 0.5; m = 0.8
0.6 a = 0*)6
r = 0.02; e = 1; / = i + #' = 1.
a = 1 H- 0.02; - 0.06t'
6 = Q.li + 0.02z"
O.U* + 0.02^ + QAi 0.5i + 0.02z'
tan a =
s'
1 + 0.02^ - 0.06^ - 0.56^ 1 + 0.02z - 0.56^ ~ = t 1
Leading current
Lagging current
P F
0.0
0.0
0.0
1.0
0.02
0.02
0.0
-0.56
0.44
0.046
2 40'
0.25
0.25
0.125
0.968
0.019
0.144
0.005
-0.542
0.463
0.31
17 17'
0.50
0.50
0.25
0.865
0.017
0.267
0.01
-0.49
0.527
0.51
26 58'
0.75
0.75
0.375
0.662
0.013
0.388
0.015
-0.37
.0.644
0.60
31 6'
1.0
1.0
0.5
0.0
0.0
0.5
0.02
0.0
1.02
0.49
26 7'
0.75
0.75
0.375
-0.662
-0.013
0.362
0.015
0.37
1.39
0.26
14 35'
0.50
0.50
0.25
-0.865
-0.017
0.233
0.01
0.49
1.50
0.155
8 49'
0.25
0.25
0.125
-0.968
-0.019
0.106
0.005
0.542
1.55
0.068
3 53'
0.0
0.0
0.0
-1.0
-0.02
-0.02
0.0
0.56
1.56
-0.0127
-45'
5i
V
0.02^ . .
s'
0.02i
- O.SGi'
t r
tan a
The curves for both machines are plotted in Fig. 208.
0.25 0.5 0.75
Leading
0.75 0.5 0.25
Lagging
Power Factor
1.0
wer
FIG. 208.
Let it be required to solve the following problem:
Problem 90. An alternator has the rating:
A.T.B. 8-100-900-2300 volts Y.
262 ELECTRICAL ENGINEERING
The saturation curve is given by the following data :
A.T.
A.T.
e
400
250
3650
2300
1000
700
5000
2700
2000
1470
6000
2900
3000
2060
The normal gap density is 40,000 and the average gap length is 0.25 in.
The armature reaction is 1490 A. T. per pole.
The synchronous impedance test gives 1890 A.T. excitation with full-load
current at 0.25-in. average gap, and 1990 A.T. with 0.1875-in. average gap.
The armature resistance per phase is 0.69 ohm.
The weight of the revolving element is 800 lb.; the radius of gyration is
0.86ft. 1
First. Draw the saturation curves with 0.25-in. gap (data given above)
and also with 0.1875-in. gap.
Second. Determine m, m^ x, Xi both for round rotor and definite pole
machines.
Third. Draw the curve of field excitation, Ff, for varying non-inductive
load /, (compounding curve), for the two round rotor and the two definite
pole machines.
Fourth. Calculate and plot the terminal voltage as the non-inductive load
is increased and the field excitation kept at the normal no-load excitation.
Use 0.25-in. gap and 0.1875-in. gap with the two types of machines.
Solution. First. From the saturation curve data, the total A.T. at
2300 volts, for 0.25-in. gap = 3650.
Thus, 3650 = gap A.T. + iron A.T.
From Eq. (12), gap amp.-turns = 0.313 X flux density X gap length.
/.for 0.25-in. gap, gap A.T. = 40,000 X 0.313 X 0.25 = 3130; for 0.1875-in.
gap, gap A.T. = 40,000 X 0.313 X 0.1875 = 2347. The iron A.T. are the
same for both gaps.
To plot: (1) plot the given curve; (2) draw the straight line gap satura-
tion (magnetization curve) for 0.25 in., through and 3130 at 2300 volts;
(3) draw the straight line saturation for 0.1875 in. through and 2347 at
2300 volts; (4) add to (3) the difference between (1) and (2).
The curves are given in Fig. 209.
Second. To determine m, mi, x, XL
(a) Definite pole machine; air gap =0.25 in.
From p. 259, Chap. XXXV,
arm. reaction 1490 1490
arm. current
since
/ =
100,000
V3 X 2300
25
= 25 amp.
= 59.5,
Data for later calculation of hunting. See Chap. XXXVIII.
CHARACTERISTICS OF ALTERNATORS 263
Assuming
m
m = 0.8 X 59.5 = 47.6.
To determine o?i. Under short-circuit test e = and, in (112) therefore,
where / is the effective current.
And
F r = jCIr - CIxi.
3000
(3) (2)
2200
1800
1400
1000
200
7
r
17
7
z
(4)
1000 2000 3000 4000 5000
Ampere-Turns
FIG. 209.
Since the current lags nearly 90, / = - jii, and, neglecting r in compari-
son with x\j this last equation may be written. 1
F r = - Clxt = - Cixi + jdiXi = F f + F a
where ^o = fWiJti
/. F f = j(CiiXi + mil)
and Ff = diXi -\- miii
whence
(116)
Cii
1 When r is not neglected, Xi =
264 ELECTRICAL ENGINEERING
Supplying values:
(a) Definite pole machine, 0.25-in. gap
F f = 1890
i l m l = 25 X 59.5 = 1490,
" 1330
_ 1890 - 1490
*i = 2.74 X 25
and
5.84
* - 06 = 9 ' 5 '
(6) Round rotor machine, 0.25-in. gap.
m = 59.5, being the same as mi in definite pole machine.
x = 9.5, being the same as x in definite pole machine.
(c) Definite pole machine, 0.1875-in. gap.
wi and m are the same as for the 0.25-in. gap, since the armature ampere
turns are independent of the gap.
.*. mi = 59.5, m = 47.6.
1990 - 1490
since
9870
F f - 1990, and C = ~ = 2.16.
9.25
(d) Round rotor machine, 0.1875-in. gap.
m = 59.5
x = 15.4.
Third. Compounding curves, (/'/vs. 7). (a) Round rotor machine;
gap = 0.25 in. Non-inductive load. By (115),
F f = V(- bC - mi)* + (aC - mil) 2 ,
where
3650
= 1330 = 2 -' 742 1 m = 59 - 5 J i = variable;
ii = 0; x = 9.55; r =0.69; e = 1330
a = e + ir - i& = 1330 + 0.69i
b = ix + iir = 9.55i.
Whence,
F f = V(- 9.55i X 2.742 - 59.5i) 2 + ((1330 + 0.69i) 2.742) 2
= \13.3 X 10 6 + 13,800i + 7354i 2
= 85.75 Vi 2 + 1.875* + 1802 = 85.75 vT.
1 C is constant only on the assumption of a straight line magnetization
curve.
CHARACTERISTICS OF ALTERNATORS
Tabulating :
265
I
10
20
25
30
40
50
1.875*
o
18 75
37 ^
4fi S^
K(\
5 c fi
i 2
o
100
400
fioe
onn
/o. u
i Ann
93.6
t
1802
1921
2239
2474
97 KQ
1OUU
1477
2500
1 '*(!'
Vf
Ff
42.5
3650
43.8
3760
47.25
4050
49.6
42^0
52.5
4.crjc
58.9
Cftcrv
loyo
66.2
CA*rr
oo/o
(6) Round rotor; gap = 0.1875 in. Non-inductive load.
2870
C = j7j3Q =2.16; x = 15.38; 6 = 15.38*.
Other quantities are as in (a).
' Ff = V( - 15.38* X 2.16 - 59.5*) 2 + ((1330 -f 0.69*) 2.16) 2
92.7 Vi 2 + 0.995* + 960 = 92.7 \/t .
Tabulating :
i
10
20
25
30
40
50
0.995*
9.95
19 9
24 95
29 85
39 8
49 75
t
960
1070
1380
1610
1890
2600
3510
V7
31
32.7
37.1
40.1
43 5
51
59 2
Fj
2870
3030
3440
3720
4030
4730
5490
(c) Definite pole machine; gap = 0.25 in. Non-inductive load.
bC - mi) 2 + (aC -
where C = 2.742; mi = 59.5, m = 47.6; i = variable; ii = 0; e = 1330;
x = 9.55; b = ix = 9.55i; a = e + ir i&i = 1330 + 0.69i.
/. F f = V(- 9.55i X 2.742 - 47.6i) 2 + ((1330 + 0.69i) 2.742) 2
= 73.8 \/* 2 + 2.53i + 2430 = 73.
Tabulating :
i
10
20
25
30
40
50
2.53i
25.3
50.6
63.25
75.9
101.2
126.5
t
Vt
2430
49 3
2555
50.5
2881
53.6
3118
55.8
3406
58.3
4131
64.3
5057
71.0
Ff.
3650
3735
3960
4120
4310
4750
5250
(d) Definite pole machine; gap = 0.1875 in. Non-inductive load.
C = 2.16; X = 15.38; m = 47.6; 6 = 15.38*.
/. F f = V(- 15.38* X 2.16 - 47.6i) 2 + ((1330 + 0.69*) 2.16) 2
= 80.8 Vi 2 + 1.3H + 1265 = 80.8 \/T.
266 ELECTRICAL ENGINEERING
Tabulating:
i
10
20
25
30
40
50
1.3K
13.1
26.2
32.75
39.3
52.4
65.5
t
1265
1378
1691
1923
2204
2917
3831
VT
35.5
37.1
41.1
43.8
47.0
54.0
61.9
F f
2870
2995
3320
3540
3800
4360
5000
10 20 30 40 50
Armature Current
FIG. 210.
Compounding curves for all four machines are given in Fig. 210.
Fourth, (a) Round rotor machine ; gap =0.25 in.
At no-load, the induced voltage to neutral is e = 1330.
The field ampere-turns are E/ = 3650.
Eq. (115) may be written
Ft = V[- C(i x -Hif) - mi] z + [C(e + ir - iix) - mil] 2
or, since ii =
F f = V[ - Cix -
[C(
CHARACTERISTICS OF ALTERNATORS 267
Numerical values previously found are:
C = 2.742; x = 9.55; m = 59.5; r
.". F f =
0.69.
-2.742 X9.55z - 59.5z] 2 + [2.742(e + 0.69i)] 2
= \/7.54e 2 + 10.4ei + 7364i 2 = 3650.
Squaring both sides and reducing, gives
(Ff)* = e 2 + 1.38ei + 977i 2 = 1.763 X 10
whence
Tabulating :
e = - 0.69i 31.25 V1805 -
i
10
20
25
30
40
50
-0.691...
i
- 6.9
100
-13.8
400
-17.25
625
-20.7
900
-27.6
1600
-34.5
2500
1805 - i 2
1805
1705
1405
1180
905
205
-695
V1805 - i 2
31.25V
e
42.5
1330
1330
41.25
1290
1283
37.45
1170
1156
34.35
1073
1056
30.05
940
919
14.3
447
420
\/3e
2300
2220
2000
1830
1590
726
(6) Round rotor machine; gap = 0.1875 in.
The field ampere-turns are F f = 2870.
F f = V[- Cix -mi] 2 + [C(e + ir)] 2 ,
where
C = 2.16; x = 15.38; m = 59.5; r = 0.69.
F f =
- 2.16 X 15.38i - 59.5i] 2 + [2.16(fl + 0.69i)] 2
= 2870
whence
e
Tabulating :
= \4.68e 2 + 6.44ei
= - 0.69* 42.8 \/960 - i 2 ,
i
10
20
25
30
40
50
-0 69i
-6.9
-13.8
-17.25
-20.7
-27.6
-34.5
i 2
100
400
625
900
1600
2500
960 - i 2
$60
860
560
335
60
-640
-1540
V960 -i 2
42.8\/~
e
31
1330
1330
29.3
1260
1253
23.65
1013
999
18.3
783
766
7.74
331.5
311
\/3e
2300
2170
1730
1325
538
(c) Definite pole machine; gap = 0.25 in.
Ff = V[- Cix - mi] 2 + (C(e + ir)] 2
where
C = 2.742; x = 9.55; m = 47.6; r = 0.69.
_
F f = \/[- 2.742 X 9.55J - 47.6J] 2 + [2.742(e + 0.6&)] 2
= V7.54e 2 + 10.4 ei + 54641 7 " 2 = 3650
268
whence
ELECTRICAL ENGINEERING
e = -0.69i 26.9 V 2430 - i 2 .
Tabulating:
i
10
20
25
30
40
50
-0.69i
-6.9
-13.8
-17.25
-20.7
-27.6
-34.5
i*
100
400
625
900
1600
2500
2430 -i 2
V2430 - i 2
26.9\/~~
2430
49.25
1330
2330
48.25
1300
2030
45
1212
1805
42.5
1144
1530
39.1
1053
830
28.8
775
-70
e ....
1330
1293
1198
1127
1032
747
\/3e
2300
2240
2072
1950
1785
1293
(d) Definite pole machine; gap = 0.1875 in.
C = 2.16; x = 15.38; m = 47.6; r = 0.69
'[- 2.16 X 15.38i - 47.6i] 2 + [2.16(e + 0.69i)] 2
whence
= \/4.68e 2 + 6.44ei + 6532T 2 = 2870
= - 0.69* + 37.3 Vl265 - i z .
2400
10
20 80 40
Load Current, Amperes
FIG. 211.
50
CHARACTERISTICS OF ALTERNATORS
Tabulating :
269
i
10
20
25
30
40
50
-069i
-6.9
-13.8
17.25
-20 7
-27 6
-34 5
i 2
o
100
400
625
900
1600
2500
1265 - i 2
1265
1165
865
640
365
-335
-1235
A/1265 -i 2
37.3\/~~
e
A/3e
35.5
1330
1330
2300
34.1
1272
1265
2190
29.4
1097
1083
1875
25.3
944
927
1605
19.1
712
691
1196
The curves of terminal voltage with varying non-inductive load are shown
plotted in Fig. 211.
Problem 91. Determine all the above quantities and plot the curves for
the conditions of load power factor of 80 per cent., both lagging and leading.
Problem 92. Carry out the above investigation for the same alternator
when the phases are delta-connected.
CHAPTER XXXVI
Pole
APPROXIMATE DETERMINATION OF THE SELF-INDUC-
TION OR LOCAL MAGNETIC LEAKAGE REACTANCE
OF AN ALTERNATOR
Consider a single slot of an armature situated directly under-
neath a pole. When current flows in the winding, lines of flux
are set up, linking with the turns of wire. Thus, as in Fig. 212,
some lines will pass across the space of width, a, occupied by the
conductors; some will pass across the width, c, of the upper insu-
lation; some across width, d, which is
occupied by the wedge; and some will
pass across the face of the tooth, /,
and air gap, g, into the pole. Each
of these fluxes is set up by a mag-
netomotive force acting through a
reluctance, the values of both of
which may be determined in each
case. The reluctance, and conse-
quently the flux and the inductance
are, in the following, first determined
per centimeter length of effective iron
parallel to the shaft. The inductance of the end connections or
the parts of the coil which are outside of the iron is considered
separately.
1. Inductance, LI, Due to Flux through Section a. The
x
magnetomotive force is FI = nl -? where n is the total number of
a
turns in the slot and I is the current; - is any portion of the
depth of the coil, n - gives the number of turns included in any
distance x from the bottom of the coil. The flux set up by this
47rn7-
a b
m.m.f., FI, is d<f>i = - i where pi = -r- = reluctance of a
Pi dx
270
DETERMINATION OF THE SELF-INDUCTION 271
small path of length 6, in air, and of cross-section, dx sq. cm.
(dx X 1). The reluctance of the iron is neglected. Then,
x dx
- -=-
a o
The interlinkages or turns linking with this flux are n- Hence,
the flux-turns interlinkage per unit current, or the inductance
across the width, a, of the coil, is
1 C a n 2 x 2 a n 2
Li = j I 4ir-^I -^dx = g47ry, per cm. length of effective iron.
2. Inductance Due to the Flux through Section c. The mag-
netomotive force is F 2 = nl. p 2 =
The flux is
c
P2
All the n conductors link with this flux.
3. Inductance Due to the Flux across the Section d. This, by
a similar process, is
e
4. Inductance over the Face of the Tooth. The magneto-
motive force is F 4 = nl. The reluctance, p\ = -r- The flux
set up is
All the n conductors link with this flux.
5. Inductance of End-connections. The inductance of the
part of the winding that projects outside of the iron is almost
impossible to estimate accurately. It depends largely on the
mechanical design. If the end shields are some distance away
from the winding, a fair approximation is obtained by assuming
the flux per ampere-turn per centimeter of wire to be inversely
proportional to the square root of the pole pitch, or what is
equivalent, the square root of the armature diameter divided by
272 ELECTRICAL ENGINEERING
the number of poles. A fair approximation to the flux per ab-
solute ampere-conductor, per centimeter of wire is
' 5 = 13\JY) (
Y) (or 1.3 A when amperes are used),
where p is the number of poles, and D = armature diameter.
If the length of the end-connections of a coil, counting both
sides of the core, is 8 X > then the flux per turn, per ampere-
conductor, is
Thus, with a single-coil winding, where all conductors of a
coil are wound together, the flux per coil is
05 = 104n/ X \ >
Mp
where n is the number of effective conductors and / is the current
per conductor in absolute amperes.
The inductance, L 5 , will be due to only one-half of this flux
since each coil occupies two slots.
.-. L b = ^- = 52n<
I D
A/
\ p
The total inductance for a single slot exclusive of end-connec-
tions is then, per centimeter net length of armature iron,
and the total inductance, including end-connections for length,
I, of iron, is, in henrys,
a , c , d , / 13 ID
10 9 ,__.
For a three-phase alternator,
with one slot per pole per phase,
there is also to be added a term
due to the flux, 6 , in parallel
with 04, which passes from the
next adjacent half tooth, across
the gap (Fig. 213).
FIG. 213. The inductance due to this flux
will vary greatly, according to the
air gap, whose cross-section and length may be very different
from the values used in determining L 4 .
DETERMINATION OF THE SELF-INDUCTION 273
Problem 93. Calculate the reactance per phase of the following alter-
nator when the slots are under the poles.
A.T.B. 8-100-900-2300 v.
48 slots; 24-in. armature diameter; therefore, 2 slots per pole and phase ; 28
effective conductors per slot. Dimensions, referring to Fig. 214, are:
a = 1 . g = average gap under adjacent teeth = 0.25
6 = 0.75 g' = average gap under distant teeth = 0.5
d = 0.14 n = 28
e = 0.27 s = 2
/ = 0.10 p = 8
h = 0.85 I = 9
k = 0.82 D = 24
The wires are confined to the distance, a, of the slot.
Each slot has n effective conductors and there are s slots per
pole per phase. Let 0i be the field which crosses the conductors
due to the m.m.f. of the coil which is between the bottom of the
slot and the distance X.
Then the m.m.f. is
*,/,
where 7 is the current in amperes in the conductor.
The flux, in section dx per cm. depth of magnetic circuit
parallel to the shaft, is therefore,
4?rFi _ birsnxl
Pi Pi
But the reluctance pi of the path is
Pi = -r- neglecting the iron.
18
274 ELECTRICAL ENGINEERING
Thus, 4wsnxldx 4:wnlxdx
dd>i = r = T *
asb ab
M
This flux interlinks with - cs conductors. Therefore the
inductance of this part of the magnetic circuit, which is the
interlinkages of the turns and flux across the conductors per
unit current, is
^a
4.Trn 2 s T 7 a
Ix 2 dx = 47rn 2 s ^r-
3o
Consider next the inductance of the part of the magnetic
circuit which is above the coil proper.
The m.m.f. is that of all conductors, and is F z = snl.
The flux 02, per cm. length, is
Pz P2
In this particular case there are three magnetic paths in multi-
ple, the first, of reluctance, -j> the second, of reluctance, > and
(t 6
sb
the third, of reluctance, y
1 ;; 1 . 1 1 _d+f e_
* " 2 sb sh sb sb sh
and
d +f
Some flux crosses the two gaps from the teeth adjacent to the
coils and causes an inductance which is similarly determined.
Thus, the m.m.f. is
F 3 = snl.
4irsnl k
PS 2g
7 1 A T k
. . L 3 = j 4TrsnI pr- sn
L zg
DETERMINATION OF THE SELF-INDUCTION 275
Similarly, the flux which crosses the gaps from the more dis-
tant teeth causes an inductance,
L 4 = 47Tsn 2 -r
Thus, the total inductance per centimeter depth of magnetic
circuit covered by the iron is
T t 9 F a , d -f / , e , ks , ks~\
L = 4?rm2 L5 + V + 1 + *, + w\'
If I is the net length in centimeters of the iron of the armature
core, and p is the number of poles, then the inductance per phase
of the part of the electric circuit which is in the slots is
, 7 r a d + / . e . ks . ks~\
Lo = 4r.pl [36 + V + h + 2g + W\'
(The dimension of inductance and capacity in the absolute system
of units is centimeters.)
By extending the reasoning in the case of a single slot, the in-
ductance of the end-connections per phase is found to be
L 5 = 52sWp\ = 52s 2 ra 2
\ p
With bar winding, when the coils are split up, as shown in
Fig. 215, the inductance of the end-connections becomes L 5 =
77 9 /
13s 2 n 2 \/Dp, since the m.m.f. per end coil is ^-t and the inter-
linkages are ^- The inductance of the ma- / \
chine per phase is then
L = Lo + L 6 in cm., or
r LO ~r LS ,
L = ^ henrys.
If inch measurements are used, FIG. 215.
f . e . ks
and
L 5 = 83s 2 n 2 \/Dp for single coil winding, or
= 20.8s 2 n 2 \/T)p for split coils, and
the inductance in henrys is
L =
276
ELECTRICAL ENGINEERING
Applying these equations to the particular three-phase alter-
nators given above, and noting that the coils are not split up, we
get:
Lo = 32 X 2 X 28 2 X 8 X 0.9 [3^75 + ^ + 5^ +
(0.445 + 0.32 + 0.33 +
o^ 1 _ 2 + q : ^x_ 2] = 21)600j000cinj
3.27 + 1.64)
and
and
L 5 = 83 X 4 X 28 2 X Vl92 = 3,600,000 cm.
.'. L = 25 > 20 ^ 000 henrys = 0.0252 henrys.
x = 27r60 X 0.0252 = 9.5 ohms.
This is, then, the reactance for the slot under the pole, that is,
the reactance which should be used with the power component
of the current. The reactance be-
tween the poles is less and may be
taken as 0.6z or xi = 5.8 ohms.
It is very convenient in designing a
slot, to make it accommodate four
coils. As this is a very common
arrangement, the calculation of the
inductance of a single tooth armature
having four coils in the slot is also
made. The cross-section of the slot
is shown with dimensions in Fig. 216.
The procedure is practically the same as in the preceding case.
The flux through a small section, dx, of the space occupied by
the lower pair of coils is
x N
where the magnetomotive force is F x = -^I,N being the total
(i &
number of turns.
Thus, the flux-turns interlinkage per unit current or the induct-
ance through the lower pair of coils, LI, is
dx
ba
DETERMINATION OF THE SELF-INDUCTION 277
The inductance across the insulation, h, between the layers is
ik?
L2 = ^_4 _ h
b b
The inductance across the upper coils is
1 f 4 'T J1+ g
/JL -r
a
b
a _ SrrJV^a ^ TrN 2 a _ 7irN 2 a
o ~ 36 36 36
The inductance across the insulation, c, beneath the wedge, is
The inductance across the wedge is L 5 =
The inductance across the face of the tooth is
The inductance of end-connections is, as in the previous case,
L 7 = 527V 2 J
\ p
The total inductance per centimeter effective length of core is
Problem 94. A certain three-phase, 60-cycle alternator has one slot per
pole per phase. The dimensions in inches of slot, etc., are as' in Fig. 216,
where a = 0.45, 6 = 0.75, c = 0.14, d = 0.37, e = 0.85, / = 0.82, g = 0.15,
h = 0.1. There are twenty-four slots, thirty-two effective conductors, the
effective length of the armature core is 9 in., the armature diameter is 32
in. Show that the armature reactance is approximately 4 ohms.
CHAPTER XXXVII
FIG. 217.
ARMATURE REACTION IN MULTIPHASE MACHINES
With current in the armature of an alternator, two magneto-
motive forces exist, one, that of the field winding, and the other,
that of the armature winding.
Sometimes these add directly
but more often they are more or
less in opposition.
If the resultant field flux is in
the direction of the field poles,
Fig. 217, and the armature winding
is assumed concentrated in a coil
in position a-b, then the induced e.m.f. due to the rotation of the
coil in the field is
e { E m sin
and the current is
i = I m sin (6 + a),
where a is the angle of lead of the current in respect to the e.m.f.,
that is, tan a = -, where x and r are the total reactance and re-
sistance of the external and armature circuits, and EM and IM the
maximum values of the e.m.f. and current respectively.
Jf the armature coil has T turns, the m.m.f. of the armature is
obviously,
iT = JJTsin (B + a),
In the position shown the m.m.f. of the armature does not act
in line with the m.m.f. of the field winding, but its component in
the direction of the field is a' - &' or the total m.m.f. multiplied
by cos 0.
The component b - b' of the armature m.m.f. at right angles
to the field is, of course, the total m.m.f. multiplied by sin B
But this component does not increase or decrease the field, but
only distorts it.
278
ARMATURE REACTION 279
Let M be the component of the armature m.m.f. in the direc-
tion of the field m.m.f. Then
M = I m T sin (0 + a) cos
= I m T (sin B cos a + cos sin a) cos
= I m T(% sin 20 cos a + cos 2 sin a).
But
COS2 e = L-*.
/mT 7
' M = ~T~ t sin 2e cos a + sin a cos 2 + sin 1
7 T
= -y- [sin (20 + a) + sin a].
It is seen that the average value of the armature reaction in the
T rp
direction of the poles has a constant value which is -^- sin a,
2i
and superimposed upon this is a pulsating reaction, a m.m.f.
which pulsates at double frequency. The effect of the latter is
zero when considering the average effect over a cycle.
T rp
IT J-mJ-
. . M av . = y sm a,
But I m sin a is the maximum value of the wattless component
of the current (Fig. 218).
Thus the armature m.m.f. (or armature reac-
tion, as it is called), in the direction of the poles
corresponds to the wattless component of the FIG. 218.
current.
Thus, if the current is in time-phase with the induced e.m.f.
(in which case there is no wattless component), the armature
current neither magnetizes nor demagnetizes the field, but only
distorts the distribution of the flux.
If the armature current leads the induced e.m.f., then it is seen
that the armature reaction is positive. It helps the field m.m.f.
If the current lags, then a is negative and the armature reac-
tion opposes the field m.m.f.
In a three-phase machine the e.m.fs. of the different phases
may be expressed as
ei E m sin
e 2 = E m sin (0 + 120)
e 3 = E m sin (0 + 240).
280 ELECTRICAL ENGINEERING
Prove that the average armature reaction in the direction of
the poles is 1.57 m T sin a, and is not pulsating but steady.
NOTE. In specifications of alternators one item is usually called armature
reaction and the value given is ~ o~> in a single-phase machine, I m T in a
two-phase machine, and l.5I m T in a three-phase machine.
In this case, however, I m is the maximum value of the rated current, and
T is the effective number of turns per armature pole per phase.
Example. Find the so-called armature reaction in an 8-pole,
100-kw., 2300- volt, three-phase generator which has 224 armature
turns per phase and which is Y-connected.
Answer. The voltage per phase is
= 1330.
The full-load effective current is
100,000
^=- - = 25.1 amp.
\/3 X 2300
.'. I m = 25.1 \/2 = 35.5 amp.
The winding is practically concentrated so that all turns are
effective, thus
.'. M a = 1.5 X 35.5 X 28 = 1490 A.T.,
and this is the numerical value given to " armature reaction."
If the armature actually carried full-load current and the cur-
rent was lagging 90 time degrees behind the e.m.f., and hence was
90 space degrees displaced from the main field flux then the de-
magnetizing ampere-turns would be 1490.
If the current was leading then the armature current would
assist the field to the extent of 1490 A.T.
With a phase angle, say 30, the actual magnetizing or demag-
netizing ampere-turns would obviously be only 745.
In an n-phase machine the armature reaction is not pulsating
but has a constant value,
M Im
M a = - ^ - Sm a '
Consider any particular phase indexed m.
Its voltage is
e = E m sin (0 + - J ;
ARMATURE REACTION
281
its current is
; = 7 m
its m.m.f. is
M = iT = 7 ra T sin
cos
The total m.m.f. at any instant is, thus
But,
m = n
M = S
m = 1
in (20 +
sin
+
= sin (26 + a) cos
cos (20 + a) sin
The sum of all terms containing cos must
n
be zero, because
sides in a closed
the sum of the cosines of all
polygon is zero. Similarly the ..
are zero. Thus it fol-
sn
n
terms containing
lows that,
M a = M (since it is constant for all values of 0) = n
FIG. 219.
sin a.
FIG. 220.
Effect of Distributed Winding on the
Armature Reaction. Consider a single-
phase armature wound with a number of
coils as is shown in Fig. 220, 6, all of whose
coils are connected in series.
The effective value of the e.m.f. gener-
ated in coil A may be represented by.OA.
The e.m.f. in coil B is then represented by
AB, and so forth.
It is seen that in this case the resultant
e.m.f. is less than the algebraic sum of the
individual e.m.fs. of the coils. It is the
vector sum of the e.m.fs and is 2/7r times
the algebraic sum.
If the total winding has N turns, the
equivalent number of turns of a concen-
trated winding would be T = 2/ir N.
282 ELECTRICAL ENGINEERING
If instead of being distributed all around the periphery the
winding covered an arc of, say, 60, as is shown in Fig. 221, the
effectiveness would again, by a similar diagram, be found to be
the ratio of the chord to the arc. Thus, the chord is evidently
2 sin 30 and the arc --
/" A\. \
,. fc _ i-taW _ 8 f g;sft
_.. 7T
T
and
T = 0.955AT FIG. 221.
In general, if the winding covers a electrical degrees,
a
360 ir
Example. A completely distributed single-phase winding has
a = 180.
.'. * = 2 -
7T
Three-phase winding uniformly distributed. In this case, the
winding covers 60. Thus, k = 0.955.
CHAPTER XXXVIII
HUNTING
The periodic oscillation of synchronous machinery is a familiar
and oftentimes troublesome phenomenon, It manifests itself
principally by the swinging of the needles of meters connected in
the circuits. When the effect is cumulative, it continues to in-
crease until rupture occurs somewhere in the system. Often it
is not cumulative, and resembles simply the movement of any
vibrating body such as a pendulum.
The difficulty of visualizing hunting of a revolving machine
comes from the fact that the vibration is superposed on the steady
rotation of the moving part. It can be well imagined as similar
to the motion in space of a pendulum swinging east and west while
at the same time the earth, on which the pendulum is fixed, is in
rotation.
Hunting of electrical machines is possible because the position
of the armature core in the field structure at any moment is deter-
mined by the balance of mechanical and electromagnetic forces.
Assuming the mechanical force to be steady, as represented by the
shaft or belt in connection with the prime mover or load, the
electromagnetic force is variable owing to the highly elastic
property of the magnetic field. Under absolutely steady condi-
tions there would, of course, be no hunting. But such conditions
do not exist, and any variation of the electromagnetic forces
results in a change of speed as the machine re-establishes the
momentarily lost equilibrium. Hunting, or oscillating, is thus
started and continues as equilibrium is gradually restored in the
elastic medium of the field.
The mechanical force is not always steady. Steam engines,
and especially gas engines, are subject to pulsation of driving
torque. This may appear in the generator in the form of forced
electrical vibrations, especially where the machines are directly
connected. When the generator is free to oscillate in response to
any impulse, it does so at a definite rate called its natural period
in distinction to a forced period.
283
284
ELECTRICAL ENGINEERING
The natural period of a pendulum depends on its length and
mass, the length being the radius of gyration. Similarly the
natural period of an armature or revolving field structure depends
on its mass and radius of gyration.
To find the natural period of a machine, consider the motion of
a stretched spring as illustrated in Fig. 222. The
spring suspends a weight, and its motion is damped by
a piston working in a dash pot.
Let F = pulling down force in the spring, y = dis-
placement of the weight. Then, f t = ay = tension on
spring, where a = number of pounds, per unit length,
of the downward pull.
If the friction force due to the dash pot is assumed
proportional to the velocity, the force necessary to
overcome friction = // = k -T- (The power required
varies as the square of the velocity.)
The force required to overcome inertia = M or, where
Weight
Dash Pot
FIG. 222.
M = mass and a = acceleration,
or,
. .
= M
= ay
M
is the total force required to balance those acting in the system.
If the applied force is removed, or if F =
0, the equation becomes,
dt
= 0.
Applying this equation to an alternator, ^-^yy^r^nm^-^
the condition is as illustrated in Fig. 223.
The moment equation is
Pp = I _}- - _|_ a Q } FIG. 223.
where Fp = the applied moment, F being the force and p the lever
arm.
7 = moment of inertia,
6 = initial angular displacement,
)3 = moment of retarding force per unit angular velocity,
a = twisting moment per unit angular displacement.
dB
-J7 = angular velocity,
HUNTING 285
-77 = moment of angular velocity.
p = radius at which the force is applied.
la = moment of angular acceleration.
This is, by ordinary mechanics,
Md 2 s d ds
When the force is released, Fp = 0, and
The solution of this differential equation is
= Ae*V + Bt m J,
in which A, B t mi and niz are to be determined from known
conditions.
Let
dO
Then,
m 2 ! + 0m + a = 0,
arid
2_L^
m z -\- j m = jp>
whence,
_
- ~ 21
- 2l 4P ~ I ; m2 = " 27
If vp j is positive, then 5 is real, and the equation shows that
6 gradually decreases to zero without oscillation. If, however,
j3 2 a
jp -j is negative then the square root is imaginary and 6
reaches zero after a certain number of oscillations.
2 a
Thus, hunting can take place only when ^ j is negative.
Let then
Then
._.__.
5 = 7~4/ 2
/3
mi = - +
286 ELECTRICAL ENGINEERING
and
a
m * = ~ 27 ~ jd
or putting
Wi = - 7
m 2 = - 7 -
.'. = Ae-^e js
When t = T, the time of one period,
6T = 27T
Assume the case of suddenly throwing off full-load from the
alternator. Then 6 = .
At t = 0, the hunting has not yet begun.
"
The period,
T=^= , M ,
^ VW - |8 2
where /3 is the friction torque, and has little influence on the period
of hunting, but rather affects the amplitude.
We may assume = 0.
Then
where T 7 is in seconds.
The beats, or oscillations per second, are > or
oo
Beats per minute = *
7T \ x
The angular space position of the alternator armature with
reference to the field pole may be determined for any load (Fig.
224). Let this angle be assumed to be 20 for a two-pole machine
HUNTING 287
at full-load, or 10 for a four-pole machine. If 6 = mechanical
angle, and <f> = electrical angle,
8 = where v =
P
number of poles.
Torque,
7050 X kw. .
r.p.m.
If a = torque per unit
angular displacement,
T
57.3
where = angle in degrees for the load being considered.
Therefore,
= 24.25 X kw. X / X 10 s
where / = frequency, N = revolutions per minute and <t> is in
degrees.
Finally, the solution is,
Beats per minute, S m =
The number of beats per minute may be changed by changing
/ or </>, the former by the addition of a fly-wheel, the latter by
altering the gap. Bridges, or dampers, between the poles may
also be used to produce eddy currents for the purpose of damping
the oscillations.
Problem 96. Determine the periods of the 100-kw. alternator of the
previous problems, both as definite pole and as a round rotor machine, and
with long and short gaps.
Solution. The equation is
= 4^000 /kw. X/
in which the constants previously given are.
N = r.p.m. = 900; kw. = 100; / = 60;
TPr 2 800* X 0.86 2
1 = moment of inertia, = = ~ oo i A
p = 0.86 ft. = radius of gyration.
288
ELECTRICAL ENGINEERING
Supplying numerical values,
47,000 /100 X60
\18.4X d>
942 \ -o
tan /3 = tan (5
900 \18.4X<^
To find 0, in Fig. 224, <f> = - 90 + a.
Assuming non-inductive load,
Ei = e + Ir + jlx = a + jb.
F/ = be ml -f- ,/aC = d + jf.
f_ _b
tan 5 tan a _ d a af bd
'' 1 + tan 5 tan a = bf_ = ad + bf
+ ad
where
a = e + Ir = 1330 + 25 X 0.69 = 1347.25,
d = -bC - ml = -25zC- 25m,
/ = a <7 = 1347.25C.
Tabulating, for the four cases :
Definite pole
Round rotor
Gap, in.
0.25
0.1875
0.25
0.1875
X
8.15
14.1
8.15
14.1
C
2.75
2.18
2.75
2.18
m
47,5
47.5
59.4
59.4
f
3,700
2,940
3,700
2,940
b
204
353
204
353
d
-1,750
-1,960
-2,040
-2,250
of
4,980,000
3,960,000
4,980,000
3,960,000
bd
-357,000
-692,000
-416,000
-795,000
ad
-2,360,000
-2,640,000
-2,745,000
-3,030,000
bf
755,000
1,040,000
755,000
1,040,000
af - bd
5,337,000
4,652,000
5,391,000
4,755,000
ad + bf
-1,605,000
-1,600,000
-1,990,000
-1,990,000
tan/9
-3.32
-2.92
-2.71
-2.39
/3
106.75
109
110.25
112.7
tan a
0.1514
0.262
0.1514
0.262
a
8.6
14.67
8,6
14.67
ft - 90 + a
25.35
33.67
28.85
37.37
1/0
0.0394
0.0297
0.0346
0.0268
Vl/*
0.190
0.172
0.186
0.1635
s m
187
162
175
154
CHAPTER XXXIX
STUDY OF THE DESIGN CONSTANTS OF ALTERNATORS
Alternators differ primarily in respect to the number of phases,
and whether the armature or the field structure is the revolving
part. Secondarily, they differ in respect to the frequency,
voltage, output rating and speed.
In practice, the very great majority of alternators are of the
three-phase, revolving-field type. In frequency, they are gen-
erally of either the 25-cycle or 60-cycle type in America; 25- and
50-cycle in Europe. Voltage may be any desired value up to
about 13,000. In output rating alternators are built up to
30,000 kva.
The speed is limited by the prime mover and the frequency.
Maximum speed, for 60-cycle machines is 3600 r.p.m., corre-
sponding to the requirement of a bipolar field; for 25-cycles, the
maximum speed is 1500 r.p.m. The chief types of prime mover
used with alternators are the reciprocating engine, representing
moderate speeds, the water turbine representing low speeds," and
the steam turbine representing high speeds. Certain roughly
approximate constants have been obtained from experience
which may serve as guides in preliminary design. These are
given in Table IX.
TABLE IX. APPROXIMATE CONSTANTS OBTAINED FROM EXPERIENCE
Prime mover
Recip. engine
Water turbine
Steam turbine
Frequency. . .
25
5
3,200
2.5
6
2.5
60
3
1,800
2.5
6
2.5
25
13
8,500
2.5
6
2.5
60
5
3,200
2.5
6
2.5
25
20
13,000
2.5
6
2.5
60
7.5
4,800
2.5
6
2.5
Arm. dia. per pole
Arm. reac. per pole
No load A.T.-per pole
Arm. reac.
Regulation (approx.), per
cent
Sh. cir. cur. at load exc.
Full-load current
19
289
290 ELECTRICAL ENGINEERING
Using them as a basis, the design constants will be calculated
for the following alternator:
A.r.B-8-100-900-2300 volt.
General Constants. From the rating it is seen that the machine
is a three-phase, revolving-field, 8-pole, 100-kilo volt-amp., 900-
r.p.m., 2300-volt alternator, evidently to be driven by a recipro-
cating engine.
It is first necessary to decide whether the phase windings shall
be connected Y or A. Y-connection is, in general, suitable for
higher voltages and lower currents. Therefore Y-connection
will be assumed in this case. The phase winding voltage is then
The phase current = line current
Kva 100,000
^ r.p.m. ^, poles 900 v
Frequency = ^ Q X ^ = -QQ X 4 = 60 cycles.
Slot Dimensions. The development of the design now depends
on the determination of size and number of slots and the conduc-
tors *in the slot.
It has been found that for an n-phase machine, armature reac-
tion per pole = ~ I P t, where t = effective turns per pole and phase,
z
and Ip is the maximum value of the current in the windings.
For three-phase, therefore, by Table IX,
1800 amp.-turns = 1.5\/2 X 25.
. ' . t = 34.
This number serves as a good preliminary value. Actually, 28
turns per pole per phase were chosen. Conductors per pole and
phase are then 2 X 28 = 56. The number of slots per pole per
phase depends primarily on the armature circumference and the
slot pitch. With many slots, a smoother e.m.f. wave is generally
obtained. The number of slots is also usually greater in low
voltage machines, where the requirements of higher insulation are
not so severe. Practically, at least two slots per pole per phase
are used.
DESIGN CONSTANTS OF ALTERNATORS 291
From Table IX, the armature diameter per pole is found to be
3 in. Hence the diameter is 3 X 8 = 24 in. and the circum-
ference is TT X 24 = 75.5 in.
The slot pitch may be determined for different numbers of slots
per pole per phase, as follows:
Slots per pole per phase . .
1
O
3
4
Slots per pole
3
6
9
12
18
Slots
24
48
72
144
Ql/^j. .-.;.*. ^U / *"" \ : r . / ,Vioe;
31 A
blot pitcn 1 , , i in incnes
. 1*
.57
.05
. 785
0.524
About half of the slot pitch will be required for the tooth.
Considering, therefore, the insulation requirement given in Table
X, it is fairly apparent that a small number of slots per pole per
phase should be chosen. It will be assumed that.
there are 2 slots per pole per phase. Each slot
will then contain,
56
-j- = 28 conductors.
A very good arrangement of conductors in a slot
is that shown in Fig. 225, which permits of easy
insertion of the coils.
TABLE X
Voltage (phase)
110
440
, 1,000
2,300
6,600
16,000
Insulation, a
20 mils
25
35
50
90
130
The size of the conductor must next be determined. As a guide
for this, it may be taken as permissible to use current densities
up to 2500 amp. per sq. in. in low-voltage machines, and up to
1200 amp. per sq. in. in high- voltage machines. Assuming a
density of 2000 as reasonable the required area of conductor to
carry 25 amp. is
X
It is seldom good practice to use wire heavier than No. 10 B. & S.
As 0.0125 sq. in. corresponds nearly to No. 8, it will be preferable
292 ELECTRICAL ENGINEERING
to divide this area among several wires in parallel. The con-
ductor used consists of four No. 14 wires in parallel, having a
combined cross-section of 4 X 0.00323 = 0.01292 sq. in., and
giving a resultant current density of 1925 amp. per sq. in.
The arrangement of wires in the slot is similar to that of Fig.
225. There are four groups of 28 wires each, the wires being
placed four abreast and seven deep. Each layer of four wires is
insulated from those above and below it.
Width of copper in the slot is 8 X 0.064 in. = 0.512 in.
Width of insulation = 0.238 in.
Width of slot = 0.512 + 0.238 = 0.75 in.
Depth of copper in slot = 14 X 0.064 = 0.896
Depth of insulation = 0.59
Depth of wedge =0.2
Depth of slot = 1.686 = 1 *K6 in.
Width of tooth at face = slot pitch slot width = 1.57 0.75
= 0.82 in.
circumference at base
no. teeth
TT X (24 + 3.375)
Width of tooth at base = - ^ t ee -th ' " ' 75
48 - 0.75 = 1.038 in.
Flux Determination. The general equation for effective e.m.f.
per phase is
10 8
where
4.44 = 4 X 6ffeCtive e " m f = 4 X -^= = 4 X 1.11
average e.m.f.
4> a total flux per pole entering the armature at no-load,
t = total armature turns in series per phase, = 8 X 28 = 224,
/ = frequency = 60,
k = constant depending on the distribution of conductors on
the armature periphery.
If the conductors were concentrated in a single slot per pole
per phase, k would be 1. With a three-phase machine, these con-
ductors would never be spread out over the entire 180 electrical
space degrees of the pole pitch as in single-phase or direct-cur-
rent machines, but would be restricted to one-third of this
amount, or to 60, on account of the space required for the other
DESIGN CONSTANTS OF ALTERNATORS 293
phases. Where there are two slots per pole per phase the e.m.fs.
generated in the two slots add vectorially, as illustrated in Fig.
226, where E = EI + E 2 . k is then evidently equal to ^-^
For n slots per pole per phase,
1
o 60
2n sin 7:
2n
Thus, for
Supplying these values in the e.m.f. equation and solving for
flux,
2300 1
* -
V 0.966X60X224X4.44
= 2.3 megalines.
The flux leakage factor for this machine is 1.125.
.'. flux in the field at no-load is,
4> f = 2.3 X 1.125 = 2.59 megalines.
. Air Gap. An approximate average value for the gap length
may be obtained by reference to Table IX. In the table is found,
no-load A.T. per pole _
arm. reaction
Armature reaction = 1.5 X A/2 X 25 X 28 = 1490.
Substituting this value of armature reaction,
no-load A.T. per pole = 2.5 X 1490 = 3725.
These ampere-turns are mostly required for the gap. Assum-
ing 80 per cent, for this,
gap A.T. = 0.8 X 3725 = 2980.
Assuming, now, a gap flux density at no-load of 40,000 lines
per sq. in., and substituting in the equation,
A.T. (gap) = 0.313 B X l a ,
where l g is the length of one gap,
2980 = 0.313 X 40,000 X 1 ,
0.313 X 40,000
= ' 238
294 ELECTRICAL ENGINEERING
With this value for a guide, definite values may be chosen.
With alternators, it is usual to shape the pole pieces so that the
generated e.m.f. may more nearly approach the sine form. The
gaps chosen for this machine are :
gap length in center of pole = 0.1875 in.
gap length at edge (maximum) = 0.386 in.
average gap length, l g , = 0.2535 in.
Gap area,
flux 2.3 X 10 6 __ K
A = fluTdelisIty = 40,000 = 57 ' 5 Sq " m '
Armature Length. The main factor bearing on armature
length is flux density in the teeth. This in turn depends upon
gap area, pole pitch and pole arc.
The pole pitch at the armature surface is
The pole arc is usually about 0.6 X pole pitch.
In this machine, the ratio
P le arc _ 0*0
i '.L l~ U.OO.
pole pitch
Assuming pole-face area = air-gap area, length of pole piece
parallel to shaft is
A A7 K
= 11.5 in.
A g 57.5
0.53 X 9.43 5
The armature gross length may be slightly greater than this to
assist in the free balancing in the field. The gross length is there-
fore taken as 12 in. This length would justify the use of four
J^-in. ventilating ducts, one for every 3 in. The length of lami-
nations is therefore 12 in. 2 in. = 10 in. Assuming 10 per
cent, loss of length due to insulation between laminations, the
net armature length is
l a = 10 X 0.9 = 9 in.
The ratio,
effective length _9_
total length = 12 =
Teeth Flux Density. Allowing 10 per cent, extra for "fring-
ing" of the flux entering the armature from the pole face, the
average number 'of teeth under the pole is
fi^S X L1 - 07 X L1 " 3 ' 5
DESIGN CONSTANTS OF ALTERNATORS 295
This number varies from moment to moment according as a
slot or a tooth is in the center line of the pole. Teeth area at
armature face is then,
At = 3.5 X tooth width X effective length of armature.
In order that this area shall carry a flux density of about 90,000
lines per sq. in. in the tooth, 1 it may be calculated on that basis.
Thus, the flux entering the armature at no-load is 2.3 X 10 6 lines.
2.3 X 10 6
' At= 90,000 = 256 sq ' m "
From this value of area, the length obtained is,
256
3.5 X 0.821
8.9 in.
Thus, the length of 9 in. previously obtained is quite satisfac-
tory, giving, as it does, a slightly less teeth density at no-load,
but, as will be seen, approximately 90,000 at full non-inductive
load.
Armature Resistance. All data has now been obtained that is
necessary to calculate the resistance of the armature winding.
The length of the mean turn may be taken as twice the gross
length of the armature core plus nine times the diameter per
pole; or the length of the mean turn
= 2L + 9 D/pole
= 2 X 12 + 9 X 3 = 51 in.
-g- 4.25 ft.
Total length = turns per phase X mean turn.
= 8 X 28 X 4.25 = 954 ft.
2 9
Resistance of four No. 14 wires in parallel = -j- = 0.725 ohms
per 1000 ft.
/. R a per phase = X 0.725 = 0.69 ohm at 60C.
For 25-cycle alternators 110,000 lines per sq. in. is suitable.
296
ELECTRICAL ENGINEERING
Voltage drop per phase = IR a = 25 X 0.69 = 17.25 volts.
The full-load e.m.f. per phase = E + IR a approximately.
= 1330 H- 17.25 = 1347 volts
Magnetic Circuit Dimensions. Sufficient data is now at hand
to enable the making of a sketch which shall show approximately
how the available space may be utilized.
Fig. 227 represents such a section of the magnetic circuit. The
next step is to construct a table for the condition of no-load and
normal voltage, from which is obtained the total required number
of field ampere-turns. Some of the data in this table have already
been obtained, especially the required fluxes in the different parts.
FIG. 227.
The yoke is left out of consideration, its magnetic length being
very small in revolving field machines of few poles.
The armature and pole sectional areas are arbitrarily chosen to
give appropriate densities. The length of the pole depends upon
the space required by the field winding. The field values ob-
tained for this machine are given in Table XI.
Material of the armature core is standard sheet iron of 0.014 in.
thickness. The field core is built up of thick steel punchings.
DESIGN CONSTANTS OF ALTERNATORS 297
TABLE XI
Magnetic data. No-load, normal voltage
Part
Flux (mgl.)
Area
B
A.T. per in.
Length
A.T.
Teeth
2.3 (face)
(base)
2.3
1.15
2.59
26.0
32.8
57.5
28.2
27.5
89,000
70,000
40,000
41,000
94,200
15. 6\
7.0/ 11 ' 3
2.8
74.2
1.6875
0.2535
6.0
6.25
19
3,150
17
464
3,650
21
3,203
17
484
Gap
Arm
Pole
Total amp. -turns .
Teeth
Full
2. 34 (face)
(base)
2.34
1.17
2.64
-load, i
lormal v
90,500
71,200
40,700
41,700
96,000
oltage
^o}"-'
Gao
Arm
2.85
77.5
Pole
Total amp. -turns .
3,725
To the total required ampere-turns to
excite the field at full-load must be
added those necessary to compensate for
the armature reaction. The number
3725 is the resultant, F r , Fig. 228. The
total ampere-turns on the field core, F/,
must be equal to the vector difference of
F r and F a , where F a is the armature am-
pere-turns multiplied by the field leakage
factor. For full non-inductive load, ap-
proximately
1.125 r
FIG. 228.
F f =
Supplying values already obtained,
F f = \&725 2 + L125~>0490 2 = 4090 A.T.
For any other power factor, say 80 per cent., the required field
ampere-turns are approximated as illustrated by dotted lines in
Fig. 228. Thus,
F'f = MF r + 1.125fl. sin 2 + l.l25F a cos 2
= \3725 + 1677lxf<X6 2 + 1677 X 0.8 2 = "N/4732 2 + 1341 2
= 4920 A.T.
298 ELECTRICAL ENGINEERING
The Main Field Magnetomotive Force. The ampere-turns
which must be supplied to each field pole are:
for no-load, normal voltage, 3650
full-load, non-inductive, 4090
full-load, 80 per cent, lagging, 4920
maximum exciter voltage = 110.
The field winding may be taken as composed of copper strip,
edge wound. The depth of such a winding may vary from % in.
to 1 Y in. under ordinary circumstances, being usually deeper with
short poles.
The choice of the actual dimensions for a given case is largely
a matter of experience. The limiting factor is, of course, the
amount of heat that may be radiated.
In this machine, the field conductor is 0.625 in. wide by 0.0175
in. thick.
Length of winding space, exclusive of that required for pole
insulation = 5.5 in.
Turns in series per spool = 230.5
3650
Field current, no-load = oon - = 15.8 amp.
Field current, full-load, non-inductive = 17.8 amp.
Field current, full-load, 80 per cent, lagging = 21.3 amp.
Mean length of field turn = 2.72 ft.
Total length of field winding (8 spools) = 8 X 230.5 X 2.72
= 5020 ft.
Cross-section of conductor = 0.01095 sq. in.
Resistance, at 60C. = 4.3 ohms.
Excitation volts, no-load = 15.8 X 4.3 = 67.5
Excitation volts, full-load, non-inductive = 76.5
Excitation volts, full-load, 80 per cent, lagging = 91.5
Current density in the field winding:
no-load = 1434 amp. per sq. in.
full-load, non-inductive = 1625
full-load, 80 per cent, lagging = 1945
Losses and Efficiency. Full-load, non-inductive.
Armature copper loss per phase = PR a = 25 2 X 0.69 = 432.
Total copper loss in armature = 3 X 432 = 1296 watts.
Field copper loss = J/ 2 #/ = 17.8 2 X 4.3 = 1370 watts.
DESIGN CONSTANTS OF ALTERNATORS 299
The core losses have already been calculated for direct-current
machines. The hysteresis loss for alternators is determined on
the same basis. The eddy current loss will not, as in direct-
current machines, be equal to the hysteresis, but owing to the
greater degree of lamination, will be much less.
In this case it will be assumed that the eddy current loss is 50
per cent, of the hysteresis loss.
Weight of armature core = 806 Ib.
Weight of teeth = 180 Ib.
Hysteresis loss in core = 1130 watts
Hysteresis loss in teeth = 730 watts
Total hysteresis loss = 1800 watts
Total iron loss = 1.5 X I860 = 2790 watts
Friction and windage loss, assumed 1 per cent. = 1000 watts
Total loss, full-load, non-inductive = 6460 watts
Efficiency = = 0.939.
Temperature Rise. This is determined for the different parts
by the use of coefficients obtained in practice. For rotating
armature machines the radiation of 0.8 watt per sq. in. of surface
corresponds approximately to a temperature rise of 100C.
Thus, for a rise of 40, the radiating surface should be sufficient
to dissipate 0.3 watt per sq. in.
With rotating field cores, owing to the greater fanning action
a larger amount of energy is dissipated for the same temperature
rise. In some cases, particularly with turbo-alternators, there
are placed on the revolving structure fan blades which increase
the heat dissipation still more. In such cases, 2 watts per sq. in.
might correspond to a 100 temperature rise. The actual tem-
peratures which different parts of a given machine will attain
can only be estimated from experience, from the current and
flux densities and from a study of the particular structure with
relation to the ventilating action which it will produce.
In the machine under consideration, the copper loss in each
field winding is ~~o~ = 171 watts.
The area of the coil surface, including both the external and
the internal surfaces, is about 400 sq. in.
Watts per sq. in. radiated are therefore J = 0.427.
300 ELECTRICAL ENGINEERING
It is safe to assume that this will not cause a temperature rise
greater than 40C. The total loss in the armature is 4090 watts.
This is dissipated from a total area, including air ducts, and al-
lowing for extension of the end-connections, of about 8200 sq. in.
4090
Watts per sq. in. radiated are therefore, QOAA = 0.5, which is
entirely conservative.
Regulation. This may be determined directly from the satura-
tion (magnetization) curve. Thus, the field excitation for full
non-inductive load has been found to be 4090 amp.-turns. Re-
ferring to the curve for this machine, shown in Fig. 210, the no-
load voltage with this excitation is 2440.
2440 2300
.'. Regulation = - OQnA - = 0.061 = 6.1 per cent.
Regulation may also be determined by adding, vectorially,
the IR and IX drops to the full-load voltage, to obtain the no-
load voltage. The reactance has been seen to have two com-
ponent values representing that of the coils immediately under
the poles, and that of the coils between the poles. These have
been designated x and Xi, respectively, and their values for this
particular machine have been determined in connection with
problem 90. The theoretical determination of x has also been
carried out in connection with Fig. 210.
CHAPTER XL
SHORT-CIRCUIT OF ALTERNATORS
The short-circuiting of a direct-current generator is a very
serious event. The commutator usually "flashes over" and the
belts or shafts are dangerously strained. With alternators, ex-
cept with large turbo-machines, such short-circuit results in
practically no excessive stresses and of course not in any "fire-
works."
However, the phenomena of alternator short-circuits are of
great interest and importance. They involve the passage from one
steady state that of normal operation to another steady state
that of the permanent short-circuited condition. Between these
two steady states is what is called the transient period, during
which the system is thrown out of equilibrium. It is during the
transient period, especially its first p^rt, that difficulties some-
times occur, and consequently the interest of the student lies
chiefly here.
In any circuit of resistance and inductance, in which a con-
stant e.m.f. is acting, the current flowing at any instant after the
closing of the switch has a value expressed by the equation
i = 7(1 -~').
Similarly, when the e.m.f. is removed and the circuit is closed
upon itself, the current, and therefore also the flux, dies down
according to the equation
According to this latter equation the effect of resistance is to
damp out the current, while the inductance tends to maintain
it. A most striking illustration of these effects is afforded by the
experiment of ONNEs, 1 who withdrew a magnet from a closed coil
immersed in liquid helium. The temperature of the coil was so
low that the resistance became a negligible quantity, and the
current continued to flow for hours.
Communication No. 119 from the Physics Laboratory, Leiden.
301
302 ELECTRICAL ENGINEERING
Applying these equations to an armature under the condition
of short-circuit, the current could be found from known values
of r and L, were it not for the e.m.f. of rotation of the armature
in the resultant field. To find the current under actual condi-
tions requires first a knowledge of the flux at any instant, and
then the derivation of the electromotive force from the flux.
Thus, at the instant of short-circuit, it may be assumed that
the alternator has its full field flux. After the permanent short-
circuited condition has been reached, the field has fallen to only
a few per cent, of its normal value on account of the armature
reaction (that is, the armature reactive magnetomotive force),
which demagnetizes the field. During the transient period the
field is not much affected by the fluctuations of armature current,
these being balanced if the field-circuit resistance is low, as is
always the case, by mutual induction with the field circuit, the
field and armature ampere-turns acting in opposition to each
other. At the instant of short-circuit, therefore, the value of
the current produced depends almost entirely upon the resistance,
r, and the reactance, x, of the armature. Armature reaction, or
the demagnetizing effect of the armature current, has no appre-
ciable effect, at first, in cutting down the resultant flux. The
current may rise to, say, 20 times its normal value. To main-
tain this current would require an abnormally large field excita-
tion, many, many times as great, in fact, as that which actually
is available. Indeed, it might, without great error, be assumed
that in comparison the actual excitation is practically zero. In
that case, then, the main field flux is surrounded by a short-cir-
cuited winding, and it must therefore decrease in value according
to some exponential law, such as,
or, if time is expressed in radians
The final value of the flux is determined by what is known as
the synchronous impedance of -the armature which consists of its
resistance and the equivalent reactance of the armature magneto-
motive force. This fact fixes the values of r and XQ.
The ratio is not readily calculated. It depends not only
XQ
SHORT-CIRCUIT OF ALTERNATORS 303
upon almost all constants of the generator, such as the armature
reaction, armature resistance, field-circuit resistance, field wind-
ing, eddy currents in field poles, etc., but also upon the nature
of the short-circuit, whether single-phase or multiphase. Suffice
it, therefore, in this elementary treatise, to state the fact that in
almost all types of machines it is in the neighborhood of from
0.01 to 0.02. In other words, the field flux dies down very
slowly, requiring several cycles before it reaches a small value.
Since the speed, during the transient period, may be assumed
uniform, the induced e.m.f. will decrease according to the same
exponential as governs the flux.
If the initial value of the e.m.f. is
c = E m sin ut
and the final value is
e 2 = Ezm sin cot,
then, during the transient period, the e.m.f. is
e = Ei m e~ Lo sin ut -f- E 2m sin at,
that is, it is the sum of the final value and a transient term, the
latter being proportional to the instantaneous value of the flux.
Re- writing this equation in terms of the phase angle, 0,
--(0-00 , m?^
e = Ei m e xo sin + E 2m sin
in which 0i is the phase angle at the instant when short-circuit
occurs.
01 represents any time elapsing after the instant of short-
circuit.
At the moment of closing the switch,
E m sin = Eim sin + E Zm sin 0,
and
77T T7T | 77T
&m &lm ~T~ &2m-
Since the electromotive force in (117) acts through the armature
circuit of resistance, r, and reactance, x, the fundamental Eq.
(15) will evidently hold, and,
<8 }
e = Ei m *<> sin + E Zm sin = ir + x -
304 ELECTRICAL ENGINEERING
The solution of this is
- T 8 r r T 9 Elm - (0-0i) .
t = e * \ J e.x e xo sin 6dO
l_ 4-C
sn
_L0 rE lm - r ~ ffl r .
= e x exojex si
l_ x
Ezm r r -0 . , j/l i /-
+ - -J * x sm rf(9 + Ct
.C J
where
R fr r \
^ = ( --- ) = cot |8.
X \3 Xo/
Substituting
Z 2 = .R 2 + X 2 ,
and
tan /3i = -
and determining C from the condition that when = 0i, i = o
the final solution is given by
=
This equation may be greatly simplified by introducing certain
approximations, which, for practical considerations, do not injure
the value of the results obtained. Thus, in practice E^ m lies
between 2 per cent, and 10 per cent, of E m , .being smaller in
larger machines.
Neglecting E 2m in (118), and writing E m for E lm , (118) be-
comes
i = ^ ^ [ - r ^ (e ~ 9l) sin (B - 0) - e~ r * (0 ~ &l} sin (0! - 0)]
(119)
Equation (119) is convenient for fairly accurate work and should
be used for ordinary wave determinations. Nevertheless, rough
approximations may be made by further simplification. Thus,
assume = 90; then sin (0 0) = cos 0. Also, assume
Y
w = 1. Then (119) becomes
SHORT-CIRCUIT OF ALTERNATORS 305
These assumptions are more or less reasonable since, in practice,
/3 lies between 85 and 90, and, in concentrated field windings,
the reactance is much greater than the resistance.
The condition for maximum current is when 61 = o, and 6 = TT.
Then,
E m - r -*
.
+e
L,-i
xo J.
The value of is about 0.02 in all alternators, and ^r = 0.06.
XQ XQ
giving e~ - 06 = 1 approximately.
Therefore the maximum current at short-circuit is
E m -JL.
Continuing the evaluation, - is from 0.6 to 0.8.
x
jp
.' We = - - X 1.75 (approximately).
x
As an example, take an alternator which has 4 per cent, react-
ance. The greatest possible current that can be obtained on
short-circuit is then
imax = Q-QJ_ X 1.75 = 44 times normal current.
To illustrate the effects of short-circuit, three typical generators
are taken as examples, as follows:
Class A. Engine driven generators. Reactance, x = 12 per
cent., = 0.12, resistance, r = 1 per cent., = 0.01, short-circuit
current under normal no-load excitation, I s = 27, where / =
full-load current.
Class #. Turbo-generators. x = 0.02, r = 0.01, /, = 21.
Class C. Turbo-generators with external reactance, x =
0.06 (0.02 internal, 0.04 external), r = 0.01, I 8 = 21.
All three machines are taken on the percentage basis; with
E m = 1, I m = 1. All are single-phase generators, or, the short-
circuit may be regarded as that of one phase only, of a multiphase
generator.
Problem 96. From the above data calculate and plot the first few
cycles (2 to 4) of armature current, voltage and power.
The current may be determined from (119), the voltage from
(117) in which Ez m is neglected, and the power from the funda-
mental relation, p = ei, where instantaneous values are con-
sidered.
20
306
ELECTRICAL ENGINEERING
The following values are at once obtained:
r is taken equal to r; X Q = j- = ^
1 8 1
~ = 0.5; -- = 0.02
6 XQ
Class A
Class B
Class C
o /~ *. \ r
K . IT '0\
-^r= COt /3= ( ) j
0.0833 - 0.02
0.0633
0.5 - 0.02
0.48
0.1667 - 0.02
0.1467
2 =8^)3 =
0.998
0.9013
0.9894
=
86 20'
64 20'
81 40'
The only other constant factor remaining to be supplied is 0i,
the time-phase angle representing the instant of closing the
switch. 61 may be taken at any desired value, and it should be
considered what effects are produced with different values. For
convenience of calculation, and also to work under extreme con-
ditions the following values of 0i may be chosen.
Class A
Class B
Class C
01 = - 3 40'
-25 40'
-8 20'
86 20'
64 20'
81 40'
41 20'
19 20'
36 40'
For each value of 0i a set of three curves may be obtained, and
a comparative study will then be possible, both in regard to the
effect of closing the switch at a different point in the cycle and
with regard to the influence of the constants of the different types
of machine. In the present instance the curves for the engine
driven generator (class A) are produced under the condition 0i =
3 40'. The equations, with numerical values supplied, are:
i = 8.32[e-- 02( ' + - 064) sin (0 - 86 20') + -- 0833 C + - 064 >
= 8.32[e~ a: sin a + ~ v ] = 8.32[a + 6]
e = 6 -o.02(* + 0.064) gin = 6 -* sin
p = ei = 8.32[e-- 04( * + ' 064) (sin 2 cos 86 20'
- sin cos sin 86 20') + -- 1033 ^ +- 064 ) s in 0].
It is not necessary to evaluate the power equation since the
product ei may be taken for each angular position. The tabula-
tion is given for 360 from the instant of closing the switch. The
three curves are shown in Fig. 229 for something over two cycles.
Figs. 230-237 are for the other cases which have been taken.
SHORT-CIRCUIT OF ALTERNATORS
Tabulating: Case A. t = - 3 40'.
307
0-01
15
30
45
60
75
90
120
-3 40'
11 20'
26 20'
41 20'
56 20'
71 20'
86 20'
116 20'
sin
-0.064
0.1965
0.4436
0.6604
0.8323
0.9474
0.998
0.8962
a = 0-8620'
-90
-75
-60
-45
-30
-15
0.0
30
sin a
-1.0
-0.9659
-0.866
-0.707
-0.5
-0.2588
0.0
0.5
0(rad.)
-0.064
0.198
0.459
0.721
0.982
1.244
1.507
2.03
+ 0.064....
0.0
0.262
0.523
0.785
1.046
1.308
1.571
2.094
X
0.0
0.00524
0.01046
0.0157
0.02092
0.02615
0.03142
0.04188
r x
1.0
0.9947
0.9895
0.984
0.979
0.974
0.969
0.959
y
0.0
0.0218
0.0436
0.0654
0.0872
0.109
0.131
0.1745
b = e -
1.0
0.978
0.957
0.936
0.916
0.898
0.878
0.841
a
-1.0
-0.96
-0.857
-0.695
-0.49
-0.252
0.0
0.48
a + 6
O.Q
0.018
0.1
0.241
. 426
0.646
0.878
1.321
t
0.0
0.15
0.832
2.01
3.55
5.38
7.30
11.0
e
-0.064
0.1953
0.439
0.65
0.815
0.923
0.967
0.86
P
0.0
. 0293
0.365
1.308
2.89
4.97
7.05
9.45
0-01
150
180
210
240
270
300
330
360
146 20'
176 20'
206 20'
236 20'
266 20'
296 20'
326 20'
356 20'
sin
0.5544
0.064
-0.4436
-0.8323
-0.998
-0.8962
-0.5544
-0.064
a = 86 20'
60
90
120
150
180
210
240
270
sin a
0.866
1.0
0.866
0.5
0.0
-0.5
-0.866
-1.0
0(rad.)
2.55
3.08
3.6
4.125
4.65
5.17
5.7
6.22
+ 0.064
2.614
3.144
3.664
4.189
4.714
5.234
5.764
6.284
X
0.05228
0.06288
0.07328
0.08378
0.09428
0.10468
0.11528
0.12568
"
0.949
0.939
0.929
0.9195
0.91
0.902
0.891
0.882
y
0.2175
0.262
0.305
0.349
0.393
0.436
0.48
0.524
b -e~"
0.804
0.768
0.737
0.706
0.674
0.65
0.62
0.592
a
0.822
0.939
0.805
0.46
0.0
-0.451
-0.772
-0.882
a + b
1.626
1.707
1.542
1.166
0.674
0.199
0.152
0.29
i
13.53
14.2
12.85
9.7
5.6
1.66
-1.266
-2.415
e
0.525
0.060
-0.412
-0.765
-0.908
-0.809
-0.494
-0.0565
P
7.1
0.0851
-5.3
-7.42
-5.08
-1.343
0.625
1.364
Class A. 0! = 41 20' =0.718 radian
i = 8.32[e- - 02 ^-- 718 > sin(0 - 86 20') - -0-0833(0-0.7l8) s i n (_4 5 )]
= 8.32[ ~ z sin(0 - 86 20') + 0.707 e~ v ]
e = e~ x sin
0-01
30
60
90
120
150
180
210
41 20'
71 20'
101 20'
131 20'
161 20'
191 20'
221 20'
251 20'
"*
1
0.989
0.979
0.969
0.959
0.949
0.939
0.929
sin
0.6604
0.9474
0.9805
0.7509
0.3201
-0.1965
-0.6604
-0.9474
a"
-45
-15
15
45
75
105
135
165
sin a"
-0.707
-0.2588
0.2588
0.707
0.9659
0.9659
0.707
0.2588
c~ x sin a"
-0.707
-0.256
0.2535
0.685
0.9255
0.916
0.664
0.2405
707-w
0.707
0.677
0.648
0.621
0.595
0.568
0.543
0.521
i
0.0
3.5
7.5
10.87
12.67
12.35
10.05
6.34
e
0.6604
0.936
0.96
0.728
0.307
-0.1864
-0.62
-0.88
P
0.0
3.28
7.2
7.9
3.89
-2.3
-6.23
-5.58
308
ELECTRICAL ENGINEERING
10
FIG. 229.
FIG. 230.
SHORT-CIRCUIT OF ALTERNATORS
309
10
FIG. 231.
Class A.0! = 86 20' = 1.50 radians.
The equations (423) and (421) become:
i = 8.32-- 02 ^ - J - 5 ) sin (6 - 86 20') = 8.32e- x sin a,
e = -0-2(0 - 1.6) sin e = -x sin Q
Tabulating :
9 - 0i
30
60
90
120
150
180
210
86 20'
116 20'
146 20'
176 20'
206 20'
236 20'
266 20'
296 20'
sin
0.998
. 8962
0.5544
0.064
-0.4436
-0.8323
-0.998
-0.8962
a
0.0
30.0
60.0
90.0
120.0
150.0
180.0
210.0
sin a
0.0
0.5
0.866
1.0
0.866
0.5
0.0
-0.5
0(rad.)
1.5
2.03
2.55
3.07
3.60
4.12
4.64
5.17
- 1.5
0.0
0.53
1.05
1.57
2.10
2.62
3.14
3.67
X
0.0
0.0106
0.021
0.0314
0.042
0.0524
0.0628
0.0734
e~ x
1.0
0.989
0.979
0.969
0.959
0.949
0.939
0.929
e~ x sin a
0.0
0.4945
0.848
0.969
0.831
0.4745
0.0
-0.4645
I
0.0
4.11
7.05
8.06
6.92
3.94
0.0
-3.86
e
0.998
0.886
0.542
0.062
-0.425
-0.79
-0.937
-0.833
P
0.0
3.64
3.82
0.50
-2.94
-3.12
0.0
3.22
e - 0i
240
270
300
330
360
390
420
450
326 20'
356 20'
386 20'
416 20'
446 20'
476 20'
506 20'
536 20'
sin
-0.5544
-0.064
0.4436
0.8323
0.998
0.8962
0.5544
0.064
a
240.0
270.0
300.0
330.0
360.0
390.0
420.0
450.0
sin a
-0.866
-1.0
-0.866
-0.5
0.0
0.5
0.866
1.0
0(rad.)
5.69
6.21
6.74
7.26
7.78
8.30
8.83
9.35
- 1.5
4.19
4.71
5.24
5.76
6.28
6.80
7.33
7.85
X
. 0838
0.0942
0.1048
0.1152
0.1256
0.136
0.1466
0.157
e~ x
0.9194
0.91
0.902
0.89
0.881
0.872
0.864
0.855
t~ x sin a
-0.796
-0.91
-0.782
-0.445
0.0
0.436
0.749
0.855
t
-6.63
-7.57
-6.51
-3.74
0.0
3.63
6.24
7.11
e
-0.51
-0.0582
0.40
^0.741
0.880
0.782
0.479
0.0547
P
3.38
0.441
-2.61
-2.77
0.0
2.84
2.99
0.39
ELECTRICAL ENGINEERING
FIG. 232.
Class B.--e l = - 25 40' = - 0.445 radian
i = 45.1[e-- 02 <' + - 445 > sin(0 - 64 20') + 6 -0.5(0 +0.445) j
= 45.1[e- x sin (e - 64 20') + -*']
c -0.02? + 0.445) gi
sin 6
e - 61
210
4 20'
0.0756
0.
-60.0
-0.
-0.856
0.265
0.767
-4.015
0.0748
-0.30
34 20'
0.564
0.979
-30.0
-0.5
-0.4895
0.525
0.592
4.625
0.552
2.55
124 20'
0.8258
0.949
60.0
0.866
0.821
1.31
0.268
49.1
0.784
38.5
154 20'
0.4331
0.939
FIG. 233.
SHORT-CIRCUIT OF ALTERNATORS
311
Class B.Oi = 19 20' = 0.336 radian
t = 45.1[e-- 02 <* - - 336 > sin ($ - 6420') - e" - 5 ^ ~ - 336 > sin(-45)]
= 45.1[e- x sin (6 - 64 20') + 0.707 e~^]
- 0.336) = -
0-01
30
60
90
120
150
180
210
19 20'
49 20'
79 20'
109 20'
139 20'
169 20'
199 20'
229 20'
sin
0.3311
0.7585
0.9827
0.9436
0.6517
0.1851
- 0.3311
-0.7587
-x
1.0
0.989
0.979
0.969
0.959
0.949
0.939
0.929
t"" 1 sin a"
-0.707
-0.256
0.2535
0.685
0.9255
0.916
0.664
0.2405
0.7076-2"
0.707
0.542
0.419
0.322
0.2475
0.1895
0.147
0.1118
t
0.0
12.9
30.35
45.4
53.0
49.9
36.6
15.9
e
0.3311
0.75
0.962
0.914
0.625
0.1758
- 0.3108
-0.705
P
0.0
9.67
29.2
41.5
33.1
8.77
-11.39
-11.21
FIG. 234.
Class B.6i = 64 20'
t = 45.1e-- 02 ^ -
1.12 radian
2 ) sin (6 - 64 20')
= 45.1 e- x ,sin (e - 64 20') = 45.1 <-~ x sin a'
e = c -o.02(0-i.i2)
gn
= -x
sn
e - 0i
30
60
90
120
150
180
210
64 20'
94 20'
124 20'
154 20'
184 20'
214 20'
244 20'
274 20'
sin 9
0.9013
0.9971
0.8258
0.4331
-0.0756
-0.564
-0.9013
-0.9971
t~ x
1.0
0.989
0.979
0.969
0.959
0.949
0.939
0.929
t~ x sin o'
0.0
0.4945
0.848
0.969
0.831
0.4745
0.0
-0.4645
i
0.0
22.3
38.25
43.7
37.5
21.4
0.0
-20.95
e
0.9013
0.985
0.809
0.42
-0.0725
-0.535
-0.846
-0.925
P
0.0
22.0
30.95
18.75
-2.72
-11.45
0.0
+ 19.4
312
ELECTRICAL ENGINEERING
FIG. 235.
Class C.Bi = - 8 20' = - 0.145 radian
i = 16.5[e- - 02 ^ + - 145 > sin (0 - 81 40') + 6 ~ 0-1667(0 + o.i45)j
= 16.5[ e - * sin (0 - 81 40') + e~ V]
e = e- 0.02(0 + 0.145) sin e = - x sin 0>
0-01
30
60
90
120
150
180
210
-8 20'
21 40'
51 40'
81 40'
111 40'
141 40'
171 40'
201 40'
sin
-0.1449
0.3692
0.7844
0.9894
0.9293
0.6202
0.1449
-0.3692
-*
1.0
0.989
0.979
0.969
0.959
0.949
0.939
0.929
a"
-90.0
-60.0
-30.0
0.0
30.0
60.0
90.0
120.0
e-* sin a"
-1.0
-0.856
-0.49
0.0
0.48
0.821
0.939
0.804
y"
0.0
0.0883
0.175
0.262
0.35
0.437
0.523
0.611
t -v>
1.0
0.915
0.84
0.769
0.705
0.647
0.593
0.543
i
0.0
0.973
5.77
12.7
19.55
24.2
25.3
22.2
e
-0.1449
0.365
0.767
0.959
0.89
0.589
0.136
-0.3427
P
0.0
0.355
4.425
12.18
17.4
14.25
3.44
-7.61
FIG. 236.
SHORT-CIRCUIT OF ALTERNATORS
313
Class C.6i = 36 40' = 0.637 radian
t = 16.5[e-- 02 ('-- 637 >sin (0 - 8140')- 6-- 1667 ^-- 637 ) S in (- 45)]
= 16.5[e-* sin(0 - 81 40') + 0.707 c~v"]
e = c -o.02(*-o.637) sin e = -x sin ^
e - Oi
30
60
90
120
150
180
210
36 40'
66 40'
96 40'
126 40'
156 40'
186 40'
216 40'
246 40'
sin 6
0.5972
0.9182
0.9932
0.8021
0.3961
-0.1161
-0.5972
-0.9182
r 9
1.0
0.989
0.979
0.969
0.959
0.949
0.939
0.929
e~ x sin a"
-0.707
-0.256
0.254
0.685
0.926
0.916
0.664
0.241
Q.7Q7<T V "
0.707
0.647
0.594
0.544
0.498
0.457
0.42
0.384
i
0.0
6.45
14.0
20.25
23.5
22.65
17.9
10.3
e
0.5972
0.908
0.972
0.777
0.38
-0.1102
-0.5605
-0.853
P
0.0
5.85
13.6
15.74
8.93
-2.5
-10.05
-8.78
FIG. 237.
Class C.d 1 = 81 40' = 1.42 radians
i = 16.5 -o.02(0-l.42) sin ( _ gl o 40 , }
= 16.5 -* sin (6 - 81 40') = 16.5 e~ x sin a'
e = -o.02(0-i.42) sin e = e-*sme.
e - Oi
30
60
90
120
150
180
210
81 40'
111 40'
141 40'
171 40'
201 40'
231 40'
261 40'
291 40'
sin
0.9894
0.9293
0.6202
0.1449
-0.3692
-0.7844
-0.9894
-0.9293
-
1.0
0.989
0.979
0.969
0.959
0.949
0.939
0.929
~ x sin '
0.0
0.495
0.848
0.969
0.831
0.475
0.0
-0.465
i
0.0
8.16
14.0
16.0
13.7
7.84
0.0
-7.67
e
0.9894
0.919
0.607
0.1403
-0.354
-0.744
-0.928
-0.862
P
0.0
7.5
8.5
2.245
-4.85
-5.83
0.0
6.61
It is important, in connection with the study of short-circuits,
to determine how great will be the stress placed upon the shaft of
the alternator. From the present calculations of power (class A),
314 ELECTRICAL ENGINEERING
the maximum value obtained was found to be 9.5 times normal.
As an example, let the normal maximum output rating of the
machine be assumed as 10,000 kw. Then the maximum power
developed under short-circuit would be 9.5 X 10,000 = 95,000
kw.
A portion of this power will be supplied from the stored electro-
magnetic energy of the field, and the remainder must come from
the stored mechanical energy, or from the shaft. Before short-
circuiting, the stored electromagnetic energy amounts to %Li 2 ,
where L is the inductance of the field system and i is the field
current.
Since
L = i X 10 8 '
the energy is,
w f^ TAS" i ou l es -
Since it has been assumed that the flux at any instant is de-
termined by the equation
the energy given out during any period of time is
which may be determined from the known constants.
As an example, let
$ = 150 X 10 6 lines of flux per pole,
n = 300 turns per pole,
i = 100 amp. field current.
Then
300 X 150 X 10 6
L = - 1QO x 1Q8 = 4.5 henrys per pole.
If all the flux is destroyed the energy is
W = 4 X l ALi* = 4 X 0.5 X 4.5 X 10,000 = 90,000 joules,
or 90 kw. sec.
If this energy all disappears in Hs sec -> tne average power
during this short interval is
90 X 25 = 2250 kw.,
which is furnished by the destruction of the flux.
SHORT-CIRCUIT OF ALTERNATORS 315
The total heat developed is i 2 Rt, or
W = CpRdt = Ci 2 RdO,
where ti and 0i are used to designate the initial moment of short-
circuit, and t and any subsequent moment. If, for instance,
6 0i is made equal to 2?m, w is the heat generated in n cycles.
The complete expression for the heat developed is obtained as
follows. From (119),
w = fi*Rdd = R^f f) 'yV 2 "'^' sin' (0-0) _&-!(-*)
sin (0 - 0) sin (0! - 0) + e - 2 <-*> gin 2 (0! - p)]dO, where is
written for > a for - and i for + -
#o a; XQ x
Carrying out the integration, this becomes
sin 2(0 - 0) - q cos 2(0 - 0)
, 2<M(9 _ 9l)
4\ x Zl
p?
o:o 2 + 1
~' " > -
The maximum heat is produced when the short-circuit occurs
at such a time that sin (0i /3) = 1. The maximum heat
produced in n cycles is then:
W = PRdO = R ^ - + - ^
, \ a? Z/ L 4a 2
.. , ai 2 (1 - 2 ai )J approx. (120)
The average power developed during n cycles is
E I
Since the rated power of an alternator is m > the ratio
Power during short-circuit _ 2^n ^ ^ W
= *r av ,
rated power E m l m E m l m irn
2
1? T 1
On the percentage basis, ^ m = ^ or if E7 = 1, where effect-
ive values of voltage and current are used, the ratio becomes ^
316
ELECTRICAL ENGINEERING
Problem 97. Calculate and plot the ratio of average power, under the
condition of maximum heat (Eq. 120), to rated power, for values of n from
1 to 10, for the three classes of alternators.
Class A. From the previous calculation (page 315),
f) 2 = - 01 ( 8 - 32 ) 2 = - 692
0.692 0.11
a = 0.02,
a = 0.0833,
ai = 0.1033,
! 2 = 0.0107,
4a = 0.08,
2a = 0.1667,
2! = 0.2067,
- ax 2 = 1.0107,
47ra = 0.2515
4 = 1.048
27rai =0.65
2- - 0.2043.
Supplying these values (121) becomes
_ -1.408/
0.11 /I - -0-2515n
= n \ " ^
0.08
0.1667
- 0.2043(1 -
2.01254
n
1*375 -0.2515n
c
n
0.66 _
0.02246 _ . 65n
n
= a b c -}- d.
Tabulating :
n
1
2
3
4
5
6
8
10
2.01254
2ni occ/i
IrtAAOT
OATAQO
OKAO1 A
OA HOK1
000 SA O
OOK1 K'T
0=3 n
.UI^5O4
.UVoZf
. o7Uoo
. oUo!4
.4UZ51
. ooo4.<2
. ZOluf
0. 20125
1.375
1.375
0.6875
0.4583
0.3438
0.275
0.2297
0.1719
0.1375
n
0.66
n
0.66
0.33
0.22
0.165
0.132
0.11
0.0825
0.066
0.02246
Onoo/m
f\ /\i i OO
OAAT/lfi
OAAKftO
Of\f\A >ino
Of\f\O*7A
OAAOO1
OAAOO>t A
n
. UZZ4O
U.Ull^o
.UO749
. UUOD i
. UU44y 2
.UUo74
.UU2ol
. 002240
0.2515n
0.2515
0.503
0.7545
1.006
1.2575
1.509
2.012
2.515
e -o. 2 s, 8n
0.778
0.605
0.47
0.364
0.284
0.22
0.134
0.081
1.048n
1.048
2.096
3.144
4.192
5.24
6.288
8.384
10.48
-1.04 8n
0.35
0.124
0.043
0.015
0.0058
0.0016
0.0
0.0
0.65n
0.65
1.3
1.95
2.6
3.25
3.9
5.2
6.5
-0..5n
0.523
0.272
0.142
0.074
0.038
0.019
0.006
0.0012
b
1.07
0.416
0.227
0.125
0.078
0.0505
0.023
0.01115
C
0.231
0.041
0.00945
. 00247
0.000765
0.000176
0.0
0.0
d
0.01175
0.003055
0.001063
0.000416
0.0001707
0.000071
0.000017
0.0000027
Pa,.
0.7233
0.5523
0.4354
0.3761
0.3239
0.2848
. 2286
0.1901
Ratio
1.4466
1 . 1046
. 8708
0.7522
0.6478
. 5696
0.4572
. 3802
Class B.
a = 0.02,
a = 0.5,
i = 0.52,
a! 2 = 0.271,
= 0.01 (45.1) 2 = 20.34
20.34 = 3.24
2irn n
4 = 0.08,
2* = 1.0,
2i = 1.04,
i 2 = 1.271,
47ra = 0.2515.
4ira = 6.28.
= 3.267.
= 0.817.
SHORT-CIRCUIT OF ALTERNATORS
317
Supplying values (121) becomes
' m ' = "^TL (U)8~~ ~T
41.09 40.5 _ . 2515n 3.24 * 28n . 2.65 _ 3 . 267n
- 0.817(1 -
n
n
= a ' _ 5' - c '
Tabulating:
n
1
2
3
4
5
6
8
10
a'
41.09
20.54
13.7
10.27
8.22
6.85
5.135
4.109
40.5
n
40.5
20.25
13.5
10.125
8.1
6.75
5.063
4.05
3.24
3.24
1.62
1.08
0.81
0.648
0.54
0.405
0.324
n
2.65
2.65
1.325
0.883
0.663
0.53
0.442
0.331
0.265
n
c 0.2516n
0.778
0.605
0.47
0.364
0.284
0.22
0.134
0.081
c -6.28n
0.0016
0.0
0.0
0.0
0.0
0.0
0.0
0.0
g 3.267n
0.038
0.0017
0.0
0.0
0.0
0.0
0.0
0.0
6'
31.53
12.25
6.34
^.688
2.3
1.485
0.678
0.328
c'
0.00518
0.0
0.0
0.0
0.0
0.0
0.0
0.0
d'
0.101
0.00225
0.0
0.0
0.0
0.0
0.0
0.0
Pa,.
9.66
8.29
7.36
6.58
5.92
5.37
4.46
3.78
Ratio
19.32
16.58
14.72
13.16
11.84
10.74
8.92
7.56
Class C.
(jjf y\ 2
-~ j) - 0.01(16.5) 2 = 2.7225
2.7225 0.434
= 0.02,
= 0.1667,
ai = 0.1867,
i 2 = 0.035,
27m n
4a =0.08,
2a = 0.333,
2 ai = 0.3733,
1 + i 2 = 1.035,
47ra = 0.2515.
4 = 2.093.
27r ai = 1.173.
2ai =0.361.
1 +
Supplying values (121) becomes
0.434 r
L av * ~^
n
0.434 rl - e -o.25i6
0.08
6.565 5.42 _ . 2515n
c
n n
= a" - V - c" + d".
-1.173n
>i
318 ELECTRICAL ENGINEERING
Tabulating :
n
1
2
3
4
5
6
8
10
6.565
6Cf C
q ooq
I CK
11
i qi q
i 078
0090^
OfiCCK
n
.000
o . Zoo
. 1OO
l . olo
1 . \Jl O
, o^uo
. OODO
5.42
5.42
2.71
1.807
1.355
1.084
0.903
0.677
0.542
n
1.302
0K1
326
OOfiO
217
n iAq
1302
n
l.oUJ
. OOJL
. ^UU
W . \. 9
\J . XUO
U . JLOU
0.157
0.157
0.078
0.052
0.039
0.031
0.026
0.0195
0.0157
n
^-0.25157*
0.778
0.605
0.47
0.364
0.284
0.22
0.134
0.081
e -2.093n
0.124
0.015
0.0018
0.0
0.0
0.0
0.0
0.0
e -1.173n
0.308
0.097
0.029
0.0095
0.0025
0.0005
0.0
0.0
V
4.22
1.64
0.849
0.493
0.308
0.1988
0.0906
0.0439
c"
0.1615
0.00975
0.00078
0.0
0.0
0.0
0.0
0.0
d"
0.0483
0.00756
0.00151
0.0
0.0
0.0
0.0
0.0
Po,.
2.23
1.64
1.31
1.15
1.00
0.88
0.73
0.61
Ratio
4.46
3.28
2.62
2.30
2.00
1.76
1.46
1.22
20
18
16
14
12
10
xo.
As an illustration of the
power developed under short-
circuit, consider the genera-
tor of class A. The average
power developed in the first
cycle under the worst condi-
tion is found to be 0.7233,
where E m = 1, I m = 1, and
P av . = normal power output
123
45
Cycles
678
10
FIG. 238.
If, now, a 25-cycle machine
of 5000-kw. rating of this
type, is considered, the aver-
age power during the first
cycle is 10,000 X 0.7233 =
7,233 kw. The instantaneous
maximum of power has
already been found to be
95,000 kw.
Problem 98. Determine the
above relation for the machines
of classes B and C.
SHORT-CIRCUIT OF ALTERNATORS
319
Stresses on End -connections of the Armature Coils. When
end-connections run parallel for some distance, the forces exerted
on them are often very great at the instants of heavy current
during short-circuit. The force at any time may be determined
to a sufficient degree of approximation by multiplying the average
density of the flux through one conductor due to the other
conductor, by the current in that conductor.
Consider two similar conductors of radius, r, with a distance, d,
between centers. To find the average flux through conductor
B due to the current in conductor A . The flux through any ele-
ment, dx, of B is, per centimeter length of conductor,
2Idx
nut ~ j
2-n-X X
where I is in abamperes, and /* is taken as unity. The average
flux density is then:
h ' efo /, d + r
r
In general, the force exerted is BIl dynes, where I is length of
the wire in centimeters
VI
.76"
1
FIG. 239.
FIG. 240.
MM, t F /2 1 d + r j
1 nus, force per cm. = -y = log -j - - dynes.
t T d T
... , F P , d+r
[f 7 is in amperes, y == log
If dimensions are given in inches, the formula remains the same.
Example. Consider two adjacent conductors, as shown in Fig.
240. The area of each conductor is 0.2345 sq. in. The current
density is taken as 2000 amp. per sq. in. under normal conditions.
Therefore maximum normal current is
I m = \/2 X 0.2345 X 2000 = 664 amp,
d = 0.5, r = 0.1562, I = 20.
320 ' ELECTRICAL ENGINEERING
The maximum force under normal load is then
(664) 2 X 20 0.6562 366,000
F = 100 X 0.1562 log 03438 = 366 > 00 dynes = 44^000 =
0.822 Ib.
This shows that the force under normal conditions is very
slight. Under short-circuit the ratios of maximum current to
normal current for the three classes of machines considered, were,
respectively:
for class A, 14.2
for class B, 52.0
for class C, 25.3.
Thus, the maximum short-circuit forces are, for the three classes
under the dimensions assumed:
F A (max.) = 0.822 X 1O 2 = 166 Ib.
F B (max.) = 0.822 X 52^_ = 2220 Ib.
F c (max.) = 0.822 X 25.3 2 = 527 Ib.
Problem 99. Discuss the effects of changing the values of r and d on
the forces exerted on the end-connections.
In general, the effect of short-circuit as obtained in machines of class B,
was much decreased by the addition of external reactance, as exemplified in
class C. What change in the relative positions of the end-connections would
be necessary to reduce the force as obtained for class B to that of class C
machines?
Multiphase Short-circuits. The voltage of any phase, m, of a
multiphase, alternator in the steady period of operation is ex-
pressed by
_ . / . 27TW
e m = E m sin
Thus, for a three-phase generator, the voltages are
e\ = E m sin (ut + 0)
6 2 = E m sin (ut + 120)
6 3 = E m sin (co + 240)
where m has the values, 0, 1, and 2, respectively. (For two-phase
alternators, n must be taken as 4, not as 2, since the voltages
differ by 90, not by 180.)
The currents of a three-phase alternator are:
ii = I m sin (coZ + 0)
it = I m sin M + 120 + 0)
it = I m sin (co* + 240 + 0).
SHORT-CIRCUIT OF ALTERNATORS 321
The transient voltage, for example, of the second phase, is,
from (117) in which E m is substituted for Ei m , and E 2 is neglected,
as in the later calculations,
6 2 = E m sin(6 + 120)~^ ( ' 9l \
Equating this to i z r + x ~i as previously done for the single-
phase machine, the current during the transient period is found
to be
' sin (0 + 120 - ft - '
sin (0i + 120 -ft]-
A still shorter but less close approximation is made by con-
X
sidering - = 1, and = 90. The current is then
12 = cos
120 ) - f-% (e - 9 * cos (B + 120)].
In a polyphase generator, the current for any phase is given
approximately by
E m r --(o-ei) ( n . -.- /
im = [e * ; cos^i+-^- -e * cos+^- (122)
where n is the number of phases and m has the values 0, 1, 2,
. (n - 1).
Power developed in any phase, at any instant, is the product ei.
The whole power of a three-phase generator is, at any instant, the
sum of the three products, e&i, e 2 iz, e 3 i 3 , of the individual phases.
Problem 100. Perform the operation just indicated and prove that the
power of a three-phase generator is
sin- (-
This equation shows that power of a polyphase generator is
entirely independent of the time of closing of the switch. This
time may have any value assigned to 81, but the time at any in-
stant after the switch is closed is represented by B 0i, which is
independent of 0i.
This is quite different from the case of single-phase short-cir-
cuits in which the power, similarly determined, is
21
322 ELECTRICAL ENGINEERING
E
P (one-phase) =
cos Oi sin d-Q.5e~ 2 x<> (0 ~ ei} sin 20]
In this equation, enters independently of lt Cos 0i is, of
course, a constant.
From the power equations, the torque on the shaft at any
instant may be determined.
Problem 101. Show that the maximum power of a single-phase short-
circuit on a three-phase machine is two-thirds of that of a three-phase
short-circuit on the same machine and explain in words the basis for this
relationship.
Armature Reaction. For a three-phase generator in the steady
state of operation, the armature reactions of the three phases
taken separately have been found to be:
F A1 = i^T cos 0,
F A2 = *V
F A3 = i,T cos (e + y) = i 3 T cos (0 + I) ,
where T is the number of effective turns per phase and 0, + -=->
o
4rr
+ -o~ represent the angular space positions of the armature core
with respect to the field core. Substituting the values of i from
(122) the transient values of the armature reaction are:
ErnT[ -'-(6-9!) -^(0-0i) /I + COS 20\ "|
-- ' -- 2 -- ) \
---
Al = -- x cos 0! cos -
m - g - l 2w\ / 27T\
F A2 = e s v cos i + cos 6 + -
cos
SHORT-CIRCUIT OF ALTERNATORS 323
Adding these three equations, the total three-phase armature
reaction is:
cos fl _ 9i _
Problem 102. Prove that the armature reaction of a polyphase generator
is:
ET rjl r~ T 7*0 ~~\
n Jl/m-i I (6 0j) . . - ((? 81) I /'1OQ'\
2 x L
Problem 103. Plot single-phase and three-phase armature reaction
curves for the alternator for which waves of e, i, and p have been derived,
and discuss their characteristic differences.
Electromotive Force and Current Induced in the Field Windings.
Excessive voltage may be induced in the field windings and
cause breakdown of insulation. In general, the induced voltage
is proportional to ^7 It is, however, difficult to obtain a reliable
value of the voltage owing to the fact that the flux cannot pene-
trate uniformly into the magnet cores during the exceedingly
short time allowed by the rapidly changing current.
The induced field current may also be abnormally great. By
installing a circuit breaker in the exciter circuit, the rush of cur-
rent may cause the circuit to be opened, thus taking off the field
current from the short-circuited alternator.
Example. Let the normal field excitation be 18,000 amp.-
turns per pole, and the normal armature reaction be 9000 amp.-
turns. If the armature reactance is 10 per cent., the maximum
short-circuit current would be approximately seventeen times
normal current. The armature short-circuit amp.-turns are then
153,000. Assuming 20 per cent, leakage between armature and
field, the effective armature reaction is
0.8 X 153,000 = 122,000 amp.-turns on the field core.
The field current may then attain the value of
1 22 000
' X normal = 6.8 X normal current.
If the circuit breaker is set for twice normal current, it will
open the circuit.
CHAPTER XLI
SYNCHRONOUS MOTORS
When the ordinary alternator is supplied with electrical energy
and made to do mechanical work, it becomes a synchronous
motor. The name is meant to indicate its chief characteristic,
namely that of running in exact synchronism with the generator
which supplies it with energy. If the frequency of the generator
is 60 cycles per second, that of the motor its counter e.m.f. is
also 60 cycles. This condition is the result of the electromag-
netic relationship between the field and armature cores; the field
core changes its position in space by means of mechanical rota-
tion, the position of the magnetic field due to the armature
magnetomotive force changes in space because of the time-phase
relationships and alternation of the currents. The driving force
of the motor is maintained only by the existence of a constant
relationship between the field and armature m.m.f. The rate
of rotation of the armature m.m.f. is fixed by the frequency of
supply. The field has no fixed rate of rotation of its own and is
therefore free to accept that imposed by the armature.
The operation of the synchronous motor may be affected either
by changing its load or by altering its field excitation. These
may be called primary means of adjustment since they are ap-
plicable to any motor in operation. Since, however, the speed
cannot be changed, it becomes a matter of great interest and also
of importance to find out what is changed, and what peculiar
and valuable characteristics are associated with this hitherto un-
encountered characteristic of synchronous speed.
There are also secondary means of adjustment by which varia-
tion in the motor performance may be brought about. These
involve changes in the constants of the line or the motor circuit.
Thus, in the matter of design, it is important to study the effects
of different values of resistance and reactance of the armature.
In operation, with a constant generator terminal e.m.f., resistance
and reactance may be inserted or withdrawn from the line, thus
altering the total r and x of the circuit.
324
SYNCHRONOUS MOTORS
325
A thorough understanding of the effect of these constants is
essential from a practical as well as a theoretical point of view.
A motor which, for instance, operates perfectly satisfactorily on
one line may be entirely unstable and even unable to carry its
load or even a small fraction thereof on another line.
It will, for instance, be evident that
a high resistance line tends to make
the motor unstable unless the reac-
tance is also considerable. In synchro-
nous motor operation a fair amount of
line reactance is essential; in fact, the
very ability of the motor to carry load
depends upon the presence of reactance
in the motor circuit.
Let E be the e.m.f. counter generated
in the motor. The resultant flux will
then 'be 90 ahead of E. Assuming a current I, as shown in
Fig. 241, this current produces a m.m.f. in phase with itself
and which may be taken equal to it, by choosing a suitable
scale. The armature m.m.f. thus produced, when
added vectorially to the field m.m.f., will produce
the resultant m.m.f., which gives the resultant flux
<j> r . </ represents the direction of the field m.m.f.
In order to force the current through the im-
pedance of the armature, it is necessary to have an
e.m.f. equal to the impedance drop IZ. As shown
in the figure, IZ is the voltage which overcomes
the resistance and reactance of the armature. The impressed
voltage, E Q must be the vector sum of this IZ drop and the voltage
E, necessary to overcome the counter e.m.f. of the motor.
. . EQ = Lfj E.
The space relations indicated by Fig.
241 are illustrated by the sketch of a
two-pole machine in Fig. 242. The
vector relationship may also be con-
sidered from a somewhat different point of view, illustrated in
Fig. 243. Here there are two e.m.fs., E and EQ, acting in a
circuit of impedance Z.
If E Q is the generator terminal e.m.f., then z is the combined
326 ELECTRICAL ENGINEERING
impedance of the line and the motor. The counter e.m.f., E t
of the motor, is naturally in a direction to more or less oppose EQ.
The vector sum of E Q and E is E z which is the e.m.f. which
actually overcomes the impedance 2, of the circuit. The current
0*
7 lags behind E s by an angle a, such that tan a =
... f
The motor output is E X / cos p = P.
This is seen to be negative thus representing power supplied
to the machine or motor action.
The generator input is E Q X / cos q = P .
The power lost in the circuit is then P P = / 2 r.
If a is a large angle, representing large reactance, the motor
is more stable than if a is small.
If a is small, the projection of I on
E may even be positive, giving genera-
tor power instead of motor power in
which case the motor cannot carry me-
chanical load. Oftentimes poorly act-
ing synchronous motors may be greatly
benefited by increasing the angle a by
the insertion of self-inductance in the
line. For a given load on the motor/
the angle 7, between the field m.m.f.
FIG. 244. an d the resultant m.m.f. is almost con-
stant. OF/ evidently depends on both
the amount and the phase of the current. The counter e.m.f.,
E, on the other hand, is fairly constant for all loads. It de-
pends on the actual resultant flux in the air gap which is fairly
constant for all loads. For constant motor load, P oE X 0/0
(Fig. 244) and the locus of the ends of the current vectors will
be along the dotted lines HQ. The corresponding locus of field
flux vectors will be along F/Fo. If, however, the angle 7 is
assumed constant the two locii will be IF r and OF/, for vary-
ing field excitation. But this condition will correspond to a
variable load. If OF r is great with respect to 01 that is, if
the angle 7 is small the variation of both 7 and the motor
power is small for a considerable variation of the field flux
about the normal value. Plotting the armature current against
the field or the field current, gives the familiar " F-curves."
SYNCHRONOUS MOTORS
327
As E cannot be assumed constant, especially where r or x
is large, the condition of constant power output cannot be
shown by the above vector diagrams, since the power is not
represented by a constant projection of / on the horizontal.
Constant power input may, however, be assumed with constant
e Q impressed, and the power input is then proportional to the
projection of I on e , Fig. 245. Moreover, constant power
input, over a considerable range of current on both sides of the
minimum, is approximately constant power output, since the
difference is only Pr which is small and which may have small
variation. It is readily possible, therefore, to calculate E for
constant power input, since
\(e Q - IZ cos (0 + a)) 2 + (IZ sin (0 + a)) 2
cos (0 + a) + PZ 2
E
- 2e (ri - xi'),
and this may be determined for
varying /, since i is constant and
known. Thus for any input,
P,- = e i, and i' =
FIG. 245.
m*
FIG. 246.
Synchronous Motor Equations. Assuming e, the motor
counter e.m.f. to be the zero vector,
E = Eo cos ]8 JE Q sin jS.
By using the minus sign is introduced into the equation since
the true angle is 180 + /3.
.*. Iz = e Eo cos jE sin 0.
.
and
(124)
r + jx
328 ELECTRICAL ENGINEERING
Also,
/ = i + ji'
whence,
. e EQ cos g
and
- (e cos a EQ cos (a /?))
i' = -(-/o sin (a ft e sin cr)
(125)
These values are obtained by clearing the denominator of (124)
T X
of imaginaries and remembering that cos a = - and sin a. = -.
Mechanical, or motor power P = ie.
Hence the generated power P = ie + I 2 r.
Substituting the values in (125) above, mechanical power =
P = ~ z (E Q cos (a - ft - e cos a) (126)
If /? = and $0 = e, P = or there is no mechanical power.
Also when = 2a, P = 0.
To determine the maximum output (126) may be differentiated
with respect to and the result equated to zero. Thus,
dP e
= = -# sin(a - ft.
In this, sin (a ft must equal zero, since - E Q is not zero.
This gives a = 0.
Hence, the power is maximum f or <* = j8 and is zero f or /3 =
and = 2a.
If is negative, there is generator action, or the motor acts as
generator.
When E Q and e are unequal, the limits of /3 are somewhat
altered.
Problem 104. Given:
Section A, E = 1.1, e = 1
Section B, E = 1, e = 1
Section C, E = 0.9, e = 1
Assume the generator bus bars kept at constant voltage not constant
generator field excitation. The synchronous motor armature has 2 per
cent, resistance, 10 per cent, reactance.
SYNCHRONOUS MOTORS
329
1. An overhead line connecting the machine has 8 per cent, resistance and
20 per cent, reactance, all referred to motor. Constants will then be, r = 0.1,
x = 0.3, tan a. = 3, rated power = 1.0 = P.
2. An underground cable connecting the machine has a high resistance
of 18 per cent, and has negligible reactance. The constants will then be
r = 0.2, x = 0.1. Find for the two cases, power output, total current, and
power factor of the generator, and plot against /3 (Fig. 246).
3. Find the maximum output, for variable r, with (a) x = 0.1 (6) x = 0.2
and plot.
[(From Eq. 126,
Pmax. =~(Eo COS a)]
Solution of the first case. Section A.
# = 1.1; E = 1; r = 0.02 + 0.08 = 0.1;
x = 0.10 + 0.20 = 0.30; tan a = ^ = 3; a = 72:
E
Mech. power, P [E cos 1 (a - 0) - E cos a] watts
E
Z
Z = A/0.3 2 + O.I 2 = 0.316; f = 3.16; E cos a = 0.309.
r
5
10
20
30
40
50
60
a
a - ft
72.0
67.0
62.0
52.0
42.0
32.0
22.0
12.0
0.0
Cos (a - 0)
0.309
0.391
0.469
0.616
0.743
0.848
0.927
0.978
1.0
Eo cos (a 0)
0.34
0.43
0.516
0.677
0.818
0.933
1.02
1.075
1.1
E cos a
0.031
0.121
0.207
0.368
0.509
0.624
0.711
0.766
0.791
P
0.098
0.383
0.655
1.16
1.61
1.97
2.25
2.42
2.5
Current = J =
- 2EE cos ft
Cos/3
1.0
0.996
0.985
0.94
0.866
0.766
0.643
0.5
0.309
2 
