LIBRARY
UNIVERSITY OF CALIFORNIA.
Deceived Ut^fa/i ,189%
Accession No. 7 <3 3 Cla *s No.
Library of the
aid li
EXAMPLES IN
ELECTRICAL ENGINEERING
PHYSICAL AND ELECTRICAL ENGINEERING
LABORATORY MANUALS. Vol. I.
ELEMENTARY PHYSICS.
By JOHN HENDERSON, B.Sc. (Edin.), A.I.E.E., Lecturer
in Physics, Manchester Municipal Technical School.
Crown 8vo, 2s. 6d.
LONGMANS, GREEN, & CO.
LONDON, NEW YORK, AND BOMBAY.
EXAMPLES IN
ELECTRICAL ENGINEERING
SAMUEL JOYCE, A.I.E.E.
11
LECTURER TO THE SENIOR CLASSES IN ELECTRICAL ENGINEERING
MUNICIPAL TECHNICAL SCHOOL, MANCHESTER
LONGMANS, GREEN, & CO.
LONDON, NEW YORK, AND BOMBAY
1896
All rights reserved
Engineering
Library
PREFACE
MANY of the examples in the following pages have been col-
lected during the past few years to illustrate the author's lectures
to Advanced and Honours Students in Electrical Engineering,
though the majority are here published for the first time.
Originally intended as a collection of exercises, the explana-
tory matter forming the bulk of the text was, however, found
necessary to make the book more complete in itself, though it
is not intended to act as a full treatise on the subject. These
explanations, together with the tables at the end of the book,
will, it is hoped, be found very useful by draughtsmen and
'others engaged in electrical machine design.
The author's best thanks are due to such writers as have
been made use of, too numerous to mention by name ; and
also to two of his third-year students, Messrs. A. B. Mallinson
and W. K. Meldrum, for many carefully executed diagrams.
Lastly, and not the least, the author's thanks are due to
his friend Mr. E. S. Shoults, for considerable assistance in
checking examples.
S. JOYCE.
LATCHFORD HOUSE, GREENHEYS,
MANCHESTER,
July, 1896.
NOTE. The author will be much obliged if readers will kindly notify
any errors that may have escaped observation.
CONTENTS
CHAPTER PAGE
I. SIMPLE CIRCUITS i
II. DISTRIBUTION. MULTIPLE WIRE CIRCUITS 15
III. CIRCUITS CONTAINING MORE THAN ONE E.M.F. ... 27
IV. DYNAMOS AND MOTORS 35
V. ARMATURE WINDING 55
VI. THE MAGNETIC CIRCUIT 65
VII. MAGNET COIL WINDINGS 82
VHI. ARMATURE REACTIONS 86
IX. EFFICIENCY 94
X. VOLTMETERS AND THEIR RESISTANCE COILS 102
XL ELECTRIC TRACTION, RAILWAYS, ETC 106
XII. ALTERNATING CURRENT CIRCUIT 113
XIII. IMPEDANCE COILS AND TRANSFORMERS 145
XIV. EFFECTS OF CAPACITY . . 160
XV. REGULATING RESISTANCES 174
XVI. PRIME MOVER COSTS . 181
EXAMPLES , . . 186
ANSWERS 205
TABLES 209
INDEX 236
EXAMPLES IN
ELECTRICAL ENGINEERING.
CHAPTER I.
SIMPLE CIRCUITS.
The Simple Electric Circuit consists of a conducting path,
insulated so as to confine the electric flux as much as possible
to that path. In it the relationship known as Ohm's Law
holds good, viz.
If e be stated in volts, R in ohms, then C will be in amperes.
The value of e in the above expression must be considered as
being the algebraic sum of all the electro-motive forces (E.M.F.)
acting in the circuit, and will be called the effective or active
E.M.F.
The resistance of a simple circuit may be composed of a
number of separate resistances in series ; in which case R will
be the sum of all these other resistances, thus
R = ri + r^ 4- ; 3 + r,
and therefore C =
r-L 4- 1\ 4- r s + r,
or e = Cfi + O a + Cr 3 + O 4
and each of these quantities may be called a potential differ-
ence, or P.D., for short. The E.M.F. e is thus the sum of all
the P.D.'s, or is equal to
20
writing the P.D. Oj = v lt we have that
e = ^ + z/ 2 + v a 4- 4
The Rate of doing Work. The amount of work done
2 Examples in Electrical Engineering.
in a circuit is equal to the quantity, Q, of electricity passed
multiplied by the E.M.F. And since Q is the product of the
current and the time, then the work done, W, in the time t
will be
W = eCt
from which the rate of doing work, or power, or activity, is
'-?-?/:*
and as e = Oj + O 2 -f O 3 + O 4 , etc., we have
P = eC = CV a + CV 2 + CV 8 -f CV 4 , etc.
which gives the distribution of energy expenditure round the
circuit, and states that the whole rate of doing work in a
circuit is equal to the sum of all the activities in the separate
parts of the circuit.
The activity or power P is measured in watts, one watt
being the activity when the product eC is unity. Owing to
the values chosen for the volt and the ampere, the watt is
equivalent to the y^ part of the mechanical horse-power, or
H.P. =
Calculation of Resistance. The value of the resistance
of a circuit is determined by the nature of the material of
which that circuit is composed, by its temperature, and by the
dimensions of the circuit; the relationship between these
quantities being that
where L and a are respectively the length and sectional area
(average if not uniform) measured in terms of the same unit,
and p represents the resistivity at a certain temperature of
the material; that is to say, j> is the actual resistance in ohms
of the unit cube of the material.
The term resistance has come more into general use than
conductance, and represents the opposite idea, the relationship
between them being
Simple Circuits. 3
where m stands for the conductivity, that is the actual con-
ductance, in mhos of the unit cube.
Thus, the value of the resistance of a circuit can be expressed
in two ways, viz.
R = If = L = ohms
a am
and similarly the conductance may also be written
am a
K = -r- = T = mhos
L Lp
Thus, if the dimensions of a circuit, both electrical and
mechanical, be known, its resistance or conductance can at
once be found. And in cases where the whole resistance is
made up of a number of items, we can find the value of each
item r by inserting in the equation
r J*
a
the proper values for /, a, and p, the latter being possibly
different for each item of the circuit, either on account of
temperature or on account of difference of material.
The tables on page 209 will give all necessary data for these
values.
Different Kinds of Circuits. There are two main ways
in which lamps are connected to form a circuit : the series and
the parallel.
The series circuit is arranged thus
FIG. i.
4 Examples in Electrical Engineering.
and the lamps are generally arcs, each taking such and such
a current at a P.D., usually from 45 to 50 volts in the case
of continuous currents.
The current in the circuit is that taken by any lamp, and is
constant ; that is, fixed in value for that particular circuit. The
E.M.F. is that required by each lamp multiplied by the number
of lamps and added to the P.D. required for the dynamo and
leads.
The parallel circuit is arranged thus
^ .
c
>;
) C
)
o o o
f_
c
) c
) C
)
FIG. 2.
and the lamps are generally incandescents, arranged as shown
in Fig. 2, at x. These lamps may generally be described as
being so many 16 candle-power (c.p.), 60 watt, 100 volt lamps,
for example. They thus all take the same current, viz.
, . watts required by it
current in each lamp = -
volts
and the whole current is the sum of all the currents, or is
watts
n X
volts
Simple Circuits. 5
In some cases there are arc lamps arranged two in series as
shown at y, Fig. 2. These lamps will be 10 ampere lamps,
for example, and the current in that case will be 10 x half the
number of lamps.
Calculation of Resistance of Parallel Circuit. In
considering the value as to resistance or conductance of a
number of items in parallel, it is generally simplest to take
the sum of all the values of current in amperes through each
item of the parallel circuit due to a common P.D. For
example, What is the resistance of four lamps in parallel
having resistances respectively 100 ohms, 50 ohms, 25 ohms,
and 10 ohms? Consider the common P.D. to be, say, 100
volts, whence the currents respectively will be i ampere,
2 amperes, 4 amperes, and 10 amperes, making a total of
17 amperes, which current is due to a P.D. of TOO volts acting
upon the united resistances in parallel, which must conse-
quently have the value R = - = - = 5-88 ohms.
The direct calculation of conductance is simpler, since the
total conductance is obviously the sum of all the conduct-
ances. Thus, the respective conductances are 0*01, 0-02,
0*04, and 0*1 mho, or the total conductance is o'lj mho. Now,
conductance is the reciprocal of resistance, and resistance is
the reciprocal of conductance, or
R
But the total conductance is the sum of all the separate con-
ductances, or is K = ki + &J 4- 3 4- 4 , etc., which can be
replaced byK = -+--f- + I , etc. And as the resist-
r \ r -2 ?s 7*4
ance is the reciprocal of conductance, we have
K
6 Examples in Electrical Engineering.
which may be expressed in words : " The resistance of a
number of items in parallel is equal to the reciprocal of the
sum of the reciprocals of their separate resistances."
In the parallel or multiple-parallel system of distribution
the circuit is always composed of a number of items in series,
such as the resistances respectively of the dynamo, the con-
ducting wires, and the lamps; and as these latter generally
have to work at a fairly constant P.D., notwithstanding varia-
tions in the current, it becomes a question of great importance
to know what size the leads must be to prevent more than
a certain small loss of pressure in them. As a good com-
pound dynamo gives a fairly constant P.D. at its terminals
for all values of the current, its own resistance may be disre-
garded, and the terminal P.D. taken as the whole E.M.F.
of the outside circuit. A few typical cases of this kind will
render this clear.
i Example I. What must be the sectional area in square
inches of the copper leads to convey the current required by
200 lamps in parallel, all situated at the far end of the leads,
each lamp taking 60 watts at 100 volts? The distance from
the dynamo to the lamps is 100 yards, and the loss along the
leads is not to exceed 2 volts.
Solution. Since the loss in the leads may be 2 volts, and is
equal to the current multiplied by the resistance of the leads,
i.e. the P.D. required to send the current through the leads,
or O,, it is necessary first to find the total current. Each
lamp is a 60 watt, 100 volt, and therefore takes -~ =0-6
ampere ; and the whole current is consequently 0-6 X 200, or
= 120 amperes. Therefore the P.D. required by the leads is
1 2 or,, and must not exceed 2 volts ; or
r l -~^ = 0-0166 ohm
which is the resistance of a certain length of wire, viz. 2 x 100
yards. We can write this resistance also in terms of its length,
sectional area, and specific resistance, or
2 X TOO x 36 x 0-00000067
1 ~ a
Simple Circuits. 7
where a is the sectional area required in square inches,
and
2 x ioo X 36 x 0-00000067
a = . ' = 0-2895 sq. inch >
0-0166
which would be slightly larger than a y stranded cable.
Example II. If in the above question the resistance of
the dynamo be 0-013 ohm, what will be the total E.M.F. in the
whole circuit, the P.D. at the dynamo terminals, and the P.D.
at the lamps at full load and at half load, when the lamps are
supposed to vary one volt up or down, from ioo volts as an
average ?
Solution, Neglecting the slight difference made in the total
current when the P.D. at the lamp terminals varies slightly,
and the shunt current of the dynamo, we have that at full load
C = i2oa, loss in leads = 2 volts, P.D. at lamps is 99 volts,
and E.M.F. total must be equal to the sum of these plus the
P.D. required to send the current through the dynamo. And
this last is 120 X 0-013, or 1*56 volt. Whence E = 99 -j- 2
-j- 1-56 = 102*56 volts.
Again, at half load only ioo lamps are in circuit, and C =
60 amperes; whence the loss in leads is only 60 x 0*01666
= i volt, and the P.D. required for the dynamo resistance
is 60 x o'oi3 = 0-78 volt. The whole E.M.F. is then the
sum of these, viz. 1*78, plus the P.D. at the lamp terminals,
which is now ioo volts, and total E.M.F. = 10178 volts.
Also the P.D. at the dynamo terminals is in each case the
sum of the P.D.'s required by the leads and the lamps. At
full load this is 2 + 99 = 101 volts, and at half load i + ioo,
or, again, 101 volts. Thus the P.D. at the dynamo terminals
is to be kept constant, at 101 volts, for all loads.
Example III. What will be the diameter of the copper
rods to convey 5000 amperes to an aluminium furnace at a
distance of 7 yards from the dynamo terminals, if the loss
along the leads is to be only \ volt ?
Solution. The P.D. required by the leads is 0-125 volt,
and is equal to 5000^, whence
8 Examples in Electrical Engineering.
0-125
r, = 0*00002 c; ohm
5000
and the leads are 2 x 7 X 36 = 504 inches long, or have a
resistance of
0*000025
= 0*0000000496 ohm
per one inch. Therefore their sectional area will be
0*00000067
- = 1-3-e sq. inches
0*0000000496
and if circular bars, their diameter will be
d = 2 x \j - = 2 x A/4*3
7T
= 2 x 2-073 = 4 >J 46 inches
Example IY. What will be the loss of pressure at half
load and full load in the leads in a parallel circuit running
40 arc lamps arranged two in series, and taking 1 2 amperes
each, when the dynamo is kept at a terminal pressure of 102
volts, and is 50 yards from the lamps, the sectional area of
the leads being 0*3 sq. inch ?
Solution. At full load the current will be
40
C = 12 x = 240 amperes
and at half load C = 120 amperes. The leads are 50 X 2
X 36 = 3600 inches long, and therefore have a resistance
Lp __ 3600 x 0*00000067
1 ~ a 0*3
= 0*00804 ohm
Therefore the loss of pressure at full load will be 0*008 x 240
= 1*92 volt, and at half load 0*008 X 120 = 0-96 volt nearly.
In simple series circuits there is no variation in the loss of
pressure in the leads, etc., with variation of load since the
current is constant. It is, then, only necessary to keep the
loss of energy in the leads within a reasonable percentage of
the whole activity. Thus in the following examples :
Simple Circuits. 9
Example Y. A series circuit consists of 60 ten-ampere
arc lamps, spaced 100 yards apart, and taking an average of
49 volts each. What will be the diameter of the single copper
leads required in order that the energy waste in them will be
within 2 J per cent, of the energy required by the lamps ?
Solution. The expenditure of energy in the lamps is at the
rate of 60 X 49 X 10 = 29,400 watts, and 2 \ per cent, of this
is - 3 x 29,400 =735 watts, which is a loss of 73-5 volts, as
the current is 10 amperes, and consequently their resistance
is 7-35 ohms. The length of the leads is 60 x 100 X 36 =
216,000 inches long, and thus the resistance per inch is
7*35
-, -- = 0*000034 ohm
216,000
ri-U 1 0'00000067 .
Ihus the sectional area is - = 0-0197 sq. inch,
0-000034
whence the diameter will be
diameter = 2 x \/ - = 2 x Vo"oo626
7T
= 0*158 inch
or about No. 8 B.W.G.
Example YI. What will be the brake H.P. of engine
required to run the above lamps if the loss in the dynamo
itself is 150 volts, and only ^ of the power received is con-
verted into electricity ?
Solution. The activity in the lamps is 60 x 49 X 10 =
29400 watts, and the loss in the leads is 735 watts. The loss
in the dynamo is 10 x 150 = 1500 watts; whence the whole
activity is 31,635, or the electrical H.P. is
= 42-41
But this is only T ^ of the H.P. at the engine pulley, whence
the brake H.P. of engine must be
42-41 X ^ = 47'i3 H.P.
IO Examples in Electrical Engineering.
Example YII. What will be the E.M.F. required to send
a current of 10 amperes through a circuit consisting of a
dynamo whose resistance is 7 ohms, of i^ mile of leads of
No. 6 B.W.G. = 0*203 i ncn diameter, and 80 lamps each of
2-5 ohms resistance? What will be the H.P. of engine re-
quired if the dynamo turns -~ of the power received into
electricity? And what will be the cost of i Board of Trade
unit in the lamps if i mechanical H.P. cost twopence per
hour?
Solution. Total resistance of circuit is
7 + 80 x 2-5 + leads
and the resistance of the leads is
i '5 X 1760 X 36 x 0-00000067 = 4x7
0-203 X 0-203 X 22
= 1*96 ohm
or whole resistance is 208-96 ohms ;
whence the E.M.F. is 10 x 208-96 = 2089-6 volts
and the total activity is 2089-6 x 10 = 20,896 watts
and therefore the engine power must be
20,896 X^X^ = 31-12 H.P.
Thus the cost per hour's run will be 31- 12 H.P. at 2^., or
62-24</., which has to be paid for by the charge for the lamps,
or is the cost of 2ooX iox 10 watt-hours, or of 20 B.T.U.;
whence the cost
Insulation Resistance. The insulation resistance of a
circuit, i.e. the resistance between any conductor and earth, is
measured for convenience in megohms, one megohm being
1,000,000 ohms. As far as insulation goes, all parts of a
circuit are in parallel, and if R be the resistance of any one
part to earth, then the resistance of a circuit consisting of n
such parts, whether joined electrically in series or parallel, will
only be - of R. Fuse blocks, joints, fittings, switches, and
Simple Circuits. 1 1
connections generally are all sources of leakage, and usually
bring down the insulation resistance very considerably, in
spite of the fact that the resistance to earth of the cable
employed may be many thousands of megohms.
Example YIII. A circuit 500 yards long is made of cable
having an insulation resistance of 3000 megohms per mile,
and has 20 joints in it, each of 10,000 megohms resistance,
and 250 fittings of 800 megohms each. What is the total
insulation resistance ?
Solution. All the above items are in parallel, but as the
cable is only 500 yards long at 3000 meghoms per mile, the
resistance of this item will be
3000 X ^Q- = 10560 meghoms
The 20 joints in parallel will be X 10000 = 500 meghoms
collectively, and the 250 fittings will together be
"2^0 X 800 = 3-2 meghoms
The whole insulation resistance is consequently all these in
parallel, or will be
= 3*1 megohms
10,560 500 3-2
being practically that due to the fittings alone, and quite
independent of the quality of cable, the high insulation of
which must, however, be regarded as an indication of its
durability.
Example IX. A central station makes a rule that the
insulation resistance of any consumer's installation must not
be less than 0*2 megohm. If there are 600 consumers, and
the P.D. between the mains and earth is 100 volts, what will
be the total leakage current ?
Solution. The total insulation will -be -^ x o'2 = 333
ohms, whence the leakage current is -|--| = 0*3 ampere.
In the measurement of activity by means of an ammeter
12 Examples in Electrical Engineering.
and a voltmeter, connections may be made in two
ways
WORK
FIG. 3.
I
or
WORK
V
FIG.
Simple Circuits. 13
In Case I., if the voltmeter be not of the electro-static type,
the ammeter reads the current in the work added to the
current in the voltmeter when simultaneous readings are
taken; whilst the voltmeter reads the true P.D. at the termi-
nals of the work.
Let R = resistance of voltmeter.
r ammeter.
e = true P.D. at terminals of work.
C = ,, current in work.
Then the current in the voltmeter will be =r, and the ammeter
K.
reads C + whilst the voltmeter reads e.
K
( e \ e*
The product of these two is e\ C + - ) = ec + , and is
the apparent activity in the work. But the true activity is
obviously ec, and the difference is easily recognized as the
power lost in the voltmeter. Thus the measured power is the
true power plus the power required to actuate the voltmeter.
In Case II. similarly the measured power is the sum of the
true power plus the power required to work the ammeter. It
is thus necessary, in readings of this sort, to make corrections,
especially when the activity to be measured is comparable in
magnitude to the power required to work the instruments
used. For example
Example X. It is required to know how many watts a
certain incandescent lamp requires, connections being made
as in Case I. (Fig. 3). The ammeter indicates a current of
0-66 ampere, whilst the voltmeter reads 101 volts, and has a
resistance of 2000 ohms.
Solution. Since 101 volts is the P.D. at the terminals, the
current in the voltmeter is
Current in voltmeter = -^V = '55 ampere
and the reading of the ammeter is o66 ampere ; therefore
the current in the lamp is 0-66 0-0505 = 0-06095 ampere.
Thus the total activity is
Activity = 0-6095 x 101 = 61-5595 watts
14 Examples in Electrical Engineering.
Connections as in Case II. (Fig. 3) do not give such re-
liable information as those in Case I., for the reason that in
ascertaining the loss of volts in the ammeter its resistance
must be accurately known ; and as this is low and partly made
up of variable contact resistances at the terminals, there is
generally considerable uncertainty as to its exact value. An
example is, however, given.
Example XL What is the resistance of a lamp when the
ammeter reads 0*67 ampere and the voltmeter indicates 101
volts? Connections are made as in Case II. (Fig. 3), and the
resistance of the ammeter is 075 ohm.
Solution. The actual current in the lamp is given by the
ammeter as 0*67 ampere, and consequently the loss of volts
in the ammeter is 0-67 X 075 = 0*5025 ampere, and thus
the P.D. at the terminals of the lamp will be
101 0-5025 = 100*4975 volts
, 100*4975
from which its resistance is - = ico ohms
0*67
Note. Power-meters and ohm-meters, which are virtually
combinations of an ammeter with a voltmeter arranged so as
to give, in the one case, the product of volts and amperes, and
in the other case, the ratio of volts to amperes, can, in calibra-
tion, be corrected for errors due to the current in the voltmeter
part, or P.D., required by the ammeter coil, but must be then
connected up in such a way as to agree with such correction.
CHAPTER II.
DISTRIBUTION.
MULTIPLE WIRE CIRCUITS.
So far the examples given have only dealt with cases where
the lamps are all connected in a bank at the end of the leads,
remote from the generator. In a great many cases, however,
the load is more or less uniformly distributed along the length
of the circuit. If the arrangement be quite uniform, the case
is simple. Consider two parallel mains lead and return,
having lamps uniformly spaced between them, as in Fig. 4.
* >
^
r 2 ->
oooo
/
B
D F
FIG. 4.
Let AB represent the terminals of a dynamo supplying current
to a number of lamps, the nearest being situated at a distance
AC from the dynamo, and any lamp being at a distance CG
from its neighbour. Let the resistance of the leads from A
to the first lamp be t\ ohms lead and return, and the resistance
of the lead and return between any two lamps be r^ ohms.
Now, all the lamps obviously cannot be at the same P.D.,
1 6 Examples in Electrical Engineering.
since there will be a loss of pressure in between each lamp
and the next nearer the dynamo, of an amount equal to r<>
times the current flowing to all the lamps beyond the point
under consideration. Thus, if the current taken by each lamp
be C amperes, neglecting the slight variation of current due
to the fact that all lamps are not at the same P.D., the loss of
volts between CG = r. 2 x 3Q and similarly the loss between
GH and HJ, will be r X 2C, and' r 2 X C. There will also
be a loss of r^ x 4C volts between the terminals of the
dynamo and the nearest lamp. If e represents the P.D. at
the dynamo terminals, the P.D. at the terminals of the four
lamps, C, G, H, and J, will be respectively
e - ?\ X 4C
e - /-! X 4C - r. 2 X $C
e r^ x 4-C 7*2 X $0 r. 2 x 2C, and
e - ?\ x 4C - ?' 2 x sC - n X 2C - r 2 x C
In order that the lamps may not differ from one another by
more than quite a small amount in brightness, it is necessary
that the difference of P.D. between the nearest lamp C and
the farthest E may not be more than 2 per cent, or say 2 volts
on 100. Furthermore, it is obvious that this comparative uni-
formity between the nearest and farthest lamps will be quite
unaffected by the loss along AC. Thus, it is general to fix
the P.D. between the terminals of one lamp, viz. that nearest
the dynamo, and cause the others to date from that.
Adding up all the losses between the lamps, it will be seen
that the total loss is 7-3(3 C + 2C + C), or 6/- 2 C volts, which
is only half the loss which would have taken place had the
whole current of 4C amperes been flowing all along the leads
from C to E. From this follows a rule for ascertaining the
loss along leads having a uniformly distributed load.
The loss of pressure in volts along leads carrying current to
lamps uniformly distributed along their length is equivalent to
the product of the whole resistance of the leads and the average
current.
The above is the loss along that part of the leads bounded
by the nearest and farthest lamp, and is independent of the
Distribution, 1 7
loss along the part AC (Fig. 4), which is equal to ri times the
whole current. Thus, calling R x and R 3 the resistances re-
spectively of the parts of the leads in between the dynamo
terminals and the nearest lamp and in between the nearest
and furthest lamps, and C the total current, then the loss from
the dynamo to the nearest lamp is
and the loss along the leads between the lamps is
Uniformity of brightness in the lamps dictates that the
latter should not be more than, say, 2 volts on 100; but this
condition will not be affected by the loss RiC, provided the
middle or some other convenient lamp be maintained at a
certain agreed upon P.D., irrespective of the number of lamps
in circuit. Economy of cost is the only consideration which
will determine the loss along Rj, and this will be considered
later on.
It is usual to distinguish between these two parts of a circuit
by calling that part of a system of leads which carries the
same current throughout its whole length and is not tapped
anywhere, a feeder ; whilst that part of a system which is
tapped at intervals along its length, and consequently carries
a current which is not the same in all parts, is called a
distributor.
A circuit may consist of a number of distributors connected
together, and is then called a distributing network. Each
portion of such network will have to be supplied by its own
feeder to maintain the required P.D.
Fig. 5 will render this clear, where the double line of
conductors bounding the figure represents the distributing
network, which is connected to the station, or electricity works,
; S, by the feeders Sa, S, etc. Each point a, &, c, etc. is
called a feeding centre, and is kept at a fixed P.D. by
manipulation of the feeders in the station
i8
Examples in Electrical Engineering.
the lamps situated in between a and b will be partly supplied
by the feeder Sa and partly by S. The total current taken
off the distributors between a and b being called C amperes,
and being the same as that taken off the distributors between
any two other feeding centres, it follows that each feeder will
have to carry C amperes ; and if R, be the resistance of a
feeder, lead and return, the loss of pressure in the feeder will
f
FIG. 5.
be CRy, an amount which is not necessarily the same for all
feeders, as their lengths may be different, or they may not be
.at all times equally loaded.
As the load between b and c (Fig. 5) is equally divided, the
Distribution. 1 9
C
current in the distributors will be amperes at the feeding
centres, but gradually falling till some point is reached about
the middle of the distance be, say at x, where the current will
be zero. In order, then, to state the maximum difference of
P.D. between any two lamps, it is necessary to allow that the dis-
Q
tributor in between a and b has* a mean current of - amperes,
4
and a length equal to half the distance be, say bx. Calling
R cZ the resistance of distributor ab, lead and return, the
maximum difference of P.D. between a lamp at b and a lamp
at x will be
C* R CR
Loss of volts along distributors = - x = ^- d volts
4-2 o
The determination of the size of the distributors depends
upon the difference of P.D. allowable between any two
lamps, and has already been considered. But in determining
the size of the feeders other considerations are involved, such
as economy of cost of transmission. If the feeders be of very
small sectional area, they will cost little for material, but will
waste much in heating, and the whole yearly cost made up of
interest on cost of -cables and laying, and value of energy
wasted may be very large. On the other hand, if the feeder
be of large sectional area it will cost much, whilst the energy
wasted may be very small, but the total yearly cost may again
be great. It is therefore necessary to take some intermediate
size of leads which shall have the yearly cost as small as
possible. This was first pointed out by Sir William Thomson
(now Lord Kelvin), whose results may be expressed in the
following simple rule
The size of the feeders must be such that the interest for
one year on money lying idle in their cost and laying together
with depreciation, must be equal to the value of the energy
wasted in them during one year.
Professor Baily has pointed out a simple way of ascertaining
that size of cable which gives the desired equality. Construct
2O Examples in Electrical Engineering.
two curves of cost (per mile, say) of cable for one year,
thus
ct
Size of Gable
FIG. 6.
The curve b represents the cost of waste energy in the
cable, whilst a represents the interest, etc., on the cost of the
cable. The point y, where these two curves cut, shows where
the two items are equal and also the size of cable.
In calculating out the loss in the feeders it is necessary to
allow that no central station is on full load for all the twenty-
four hours. It thus comes about that the current taken in the
above calculation of waste in the cables must not be the full
current ; it should, in fact, be more nearly the
square
value when this can be ascertained.
A little consideration will at once show that strict adherence
to Kelvin's law is not always desirable, seeing that where
power is very cheap and materials and cost of laying very
great, the most economical size of conductor might be such
Distribution. 2 1
that the energy waste was excessive, and though of no con-
sequence as energy, yet producing a dangerous rise of tem-
perature on the conductors.
A good many safety regulations limit the rise in temperature
of the conductors to 150 Fahr. as a maximum, thus permitting
a possible difference between the air and the conductor of
about 75 Fahr., and that for a current twice as large as the
normal maximum. Since the heating is, other things being
equal, proportional to the square of the current, it follows that
the rise in temperature due to the maximum working current
must not be more than one-fourth part of the 75 Fahr., or
about 1 8 Fahr. From a large number of experiments on
conductors of all kinds, Mr. Kennelley has discovered a rule
connecting the diameter of the conductor and the current to
be carried consistent with the fact that the rise in temperature
shall be within this limit, viz.
C = 56o^or*=|^
thus indicating that the current density should decrease as the
diameter increases. As, however, most Fire Office rules fix
the current density at 1000 amperes per square inch, irre-
spective of the size of the conductor, it follows that though
for small sizes there is a very high degree of safety, neverthe-
less for conductors exceeding \ sq. inch sectional area the
allowable current would produce a greater rise in temperature
than that allowed by Kennelley's rule. Thus, when the
sectional area is - sq. inch, Kennelley's rule would allow
about 17 per cent, more current for safety than the Phoenix
Fire Office ; when the area is \ sq. inch, the Fire Office rule
would be equivalent to the safe temperature rise; whilst at
i sq. inch area, the Fire Office rules allow 50 per cent, more
current than is consistent with the maximum allowable rise in
temperature.
Example XII. The sectional area of the copper con-
ductors forming a system of distributors is \ sq. inch, and
they are loaded with one 5o-watt ico-volt lamp per yard
22 Examples in Electrical Engineering.
of street. What must be the distance between the feeding
centres, which are maintained at a P.D. of 100 volts, in order
that the maximum difference between any two lamps shall
not exceed 2 volts ?
Solution. Neglecting the small difference in the total current
made by the fact that ail lamps are not run at exactly the
same P.D., viz. 100 volts, each lamp takes 0*5 ampere, and
consequently the total current supplied by any feeder will be
amperes, where L is the distance in yards between the two
feeding centres. The resistance of distributor, lead and return,
in between the feeding centres will be 2L x resistance per
yard, or
2L X 36 X 4 X 0*00000067 = 0-00019296!; ohm
and the loss of volts along half of L will be
L
X
= 2
8
whence L 2 = - .. = 165,803
0-00009648
and L = 407 yards
Example XIII. What will be the maximum difference in
volts between two lamps if the distance between two feeding
centres is a quarter of a mile, the size of distributors is J sq.
inch sectional area, and current is taken off at the rate of
\ ampere per foot of street ?
Solution. The maximum current in distributors is
\ X i X 13,200 =110 amperes
and therefore the average value is 55 amperes. The resistance
of i mile (lead and return) of distributor of sq. inch sec-
tional area is
4 x 1320 x 12 x 0*00000067 = 0*04244 ohm
whence the loss of volts is 0-04244 X 55 = 2-334 volts, which
is consequently the maximum difference between two lamps.
Distribution. 23
Multiple Wire Systems. In some cases the distributors
are more than two in number, three or even five conductors
being used. This method was first proposed by Dr. John
Hopkinson. The connections are made as shown in Fig. 7,
taking the five-wire system as an example
T
A Q| R r.
6 \
ft I
H
\
rY
> <
> N t ^
f P
u
nT
> c
!
<? j
A '
>
u
Q| R
FIG. 7.
The five wires constituting the distributors are shown at AG,
BH, CK, DL, EM, the outermost conductors AC and EM
being considerably larger in sectional area than the three
others. The object of the system is to gain the advantages
of transmission of energy from the central station at a high
pressure, say 400 volts, and nevertheless provide for the use
of loo-volt lamps.
A three-wire system can do the same, using 2oo-volt lamps,
which are now coming in. To transmit the same amount of
energy at an E.M.F. of 400 volts, will only require one-
sixteenth the weight of copper for the same percentage loss
as on a loo-volt circuit. Lamps may be connected in various
ways, as at N, loo-volt lamps in parallel between the con-
ductors; as at O, two arc lamps being connected in series
between each pair of distributors ; and as at P, where 200-
volt lamps are connected between the middle and outside
wires.
24 Examples in Electrical Engineering.
The objects of the middle wires BH, CK, and DL, are
(1) To enable the equality of P.D. between the wires to be
maintained. This is done by introducing between each of
them in the station a dynamo or other device, maintained at
P.D., of 100 volts each.
(2) To prevent any want of equality in the numbers of the
lamps connected between the parallels causing fluctuations in
the P.D. between the parallels, by giving the excess current
another path back to the station than through the lamps on
the lightly loaded side.
When the whole load has been judiciously distributed
between the parallels this excess of current is not large, and
consequently the middle wires need not be so large as the
outermost. Indeed, if all the loads be equally balanced, then
the middle wires may be of so small a section as to practically
vanish ; but in practice this cannot be ensured, and the middle
wires are made of such a size as to carry an agreed-upon
percentage of the full load.
Like the simple parallel system, the distributors are fed by
feeders connected to large dynamos running at a P.D. of 400
volts, plus the economical loss of volts in the feeders, con-
veying the current to the distributors at feeding centres as
shown at QQ and RR.
Example XI Y. Compare two systems. No. i. A two-
wire system to supply current to a circle of distributors 4700
yards round, at the rate of one 5o-watt loo-volt lamp per foot
of street, and having ten feeding centres at equal distances
round the circle, the station being at an average distance of
800 yards from the distributors. There is not to be more
than 2 volts difference between any two lamps, and the pres-
sure at the station end of the feeders may be 10 per cent,
more than at the distributors.
No. 2. The same town to be supplied on the five-wire
system with the same conditions ; provision, of course, to be
made for the connection of the distributors to the potential
equalizers in the station.
Compare the weights of copper used in the two cases.
Distribution. 2 5
Solution. No. i. The distance between the feeding centres
is 470 yards, whence the average current in the distributors is
-^- X i '5 X = 176 amperes, say
and thus the resistance of the distributor must be
2x2
^ = 0*02273 ohm
for a 2-volt drop. And as the length of distributor from a
feeding centre to feeding centre is (lead and return)
2 x 47 X 36 = 33,840 inches
the resistance per one inch is
0*02270
= 0*0000006712 ohm
33840
whence the sectional area is i sq. inch.
The feeders have to carry a current of 4 x 176 = 704, say
705 amperes, and are 800 yards long, with 10 volts loss. Thus
the resistance is
- = 0*01418 ohm
70S
or
0*01418
_ 7 = 0*000000240 ohm per i inch
800 X 2 x 36
and the sectional area is
- f 2*72850. inches
246
Taking i cubic inch of copper as 0*32 lb., the total weight of
distributors will be
4700 x 36 X 0*32 X 2 _
~2~2lo~ = 48-34*
and the feeders
IOX 8oox 3 6.JLLX-7'8xoj = (ons
2240
Or the whole weight of copper in No. i = 27274 tons.
26 Examples in Electrical Engineering.
In No. 2, as the outside wires will be at 400 volts P.D., the
current will only have an average value of \ as much as in
No. i, or 44 amperes. Also the fall of P.D. may now be
8 volts, being only 2 per cent, as before. If the feeding
centres are at the same distance apart, the sectional area of
the distributors will thus be only y 1 ^ that of No. i, or y 1 ^ sq.
inch. But their length will be increased by 2 x 800 yards to
connect them to the potential equalizers in the station ; or the
whole weight of outside wires will be
( 47 00 + 800) X 2 X 3 6 X Q'32 X yV =
2240
and allowing \ the section for the three middle wires each,
their combined weights will be half as much as the outsides,
or 176 ton.
The feeders also carry \ the current, and with four times the
loss of volts they will likewise be yg the area of those in No. i,
or weight will be }- = 14*04 tons, or the total weight will
be 19*33 tons, as against 27274 tons for No. i.
CHAPTER III.
CIRCUITS CONTAINING MORE THAN ONE E.M.F,
IN some cases there is more than one E.M.F. in the circuit,
and the consideration of the magnitude of the phenomena set
up will depend upon whether all the E.M.F.'s are in one
direction, or whether some are opposed to the others. The
cases where all the E.M.F.'s are in one direction have already
been dealt with, and we will now 'deal with some cases where
the effective or active E.M.F. is the difference between an
applied E.M.F. and another E.M.F., due to the nature of the
circuit, acting against the applied E.M.F., and consequently
called Back E.M.F.
Let E = an E.M.F. applied to a circuit to (i) overcome the
B. E.M.F., and (2) leave, sufficient over to send
the necessary current through the resistance of the
circuit.
e = the B.E.M.F., such as that possessed by chemical
cells, running motors, arc lamps, etc.
e = the active E.M.F., which is equivalent to the differ-
ence between E and e, or = E e, and which is
also equal to the product of the resistance of the
circuit and the current passing.
Thus E = e + e, or e E e, and consequently the current,
which will be produced by E in a circuit of resistance R and
containing the B. E.M.F. e, will be
c-^" e - <
R -R
The resistance R may be composed of various items as before.
28
Examples in Electrical Engineering.
Secondary Batteries. In the case of charging a Secondary
Battery, the circuit must consist of a dynamo, the Secondary
Battery, and the necessary connecting cables. Each of these
items will have their own resistances, that of the Secondary
Battery being composed of a number of items in series, each
being the resistance of a single cell. In addition to this, each
cell will have an E.M.F., which must be overcome in charging,
and this E.M.F. is a quantity which changes with the state of
charge, being small when the charge is low, and rising to a
higher value as the charge increases. This rise of E.M.F. is
partly due to actual chemical change, and partly due to a
variable temporary increase of local resistance at the surface
of the plates. It is, however, customary to refer to it on the
whole as an increase of the B. E.M.F. The curve in Fig. 8
shows the values of this E.M.F., corresponding with different
charges.
H
U R
54
FIG. 8.
From this it will be observed that, though a cell may require
2-3 volts or so to overcome its B.E.M.F. when charging, it
Circuits containing more than one E.M.F. 29
cannot give more than a fraction more than 2 volts for even a
short time when discharging.
We may thus be required to state what E.M.F. a dynamo
must produce in order to charge so many secondary cells with
such and such a current ; or we may have to find what will
be the value of the current through the cells when charged
throughout with a constant E.M.F.
Example XY. What must be the E.M.F. to charge with
100 amperes a battery of 54 cells at an average B. E.M.F. of
2*3 volts each, and each having an internal resistance of
0*0004 ohm, when the resistance of the dynamo is 0*02 ohm
and that of the leads is 0*03 ohm ?
Solution. The total B. E.M.F. is
e = 54 x 2 -3 = 124-2 volts
and the active E.M.F. required for the resistance of the
circuit will be the product of current irito the sum of all
resistances, or
e = 1 00(0*02 + 0-03 + 54 X 0-0004)
= 100 (0*0716) = 7*16 volts
whence the applied E.M.F. must be
E = 124-2 + 7*16 = 131*36 volts
Example XYI. What will be the current through the
circuit at the beginning and end of charging in the case of a
battery of 53 cells, each of 0*0002 ohm internal resistance,
charged by a dynamo at an E.M.F. of 135 volts constant,
when the dynamo resistance is 0-015 onm > an d that of the
leads is 0*025 ohm, if the charging is continued for 7 hours,
the value of e per cell rising from 2*1 volts to 2-35 volts?
Solution. The B.E.M.F. at starting to charge will be
2'i x 53 = ni'3 volts, and at the end of 7 hours will be
2 '35 X 53 = 124-55 volts. The total resistance is
53 X 0-0002 + 0*015 ~T" 0*025 = 0*0506 ohm
the active E.M.F. at the start of charging will thus be
e = E - e = 135 - 111*3 = 23*7 volts
3O Examples in Electrical Engineering.
and at the end of charging
e = 135 - 124-55 = 10-45 volts
whence the current at starting will be
c - ^- , = 468 amperes
0*0506
and at end of run is
10*45
c = -~ = 206 amperes
0*0506
Example XYIL In the last example (XVI.), if the dynamo
has to be run so as to produce 135 volts always, and it is
desired that the current shall not exceed 200 amperes, what
will be the value of the necessary resistance to be inserted
(i) at starting, (2) at end of charging?
Solution. The active E.M.F. at starting
e - 237 volts
and at end of run is
e = 10-45
whence the total resistance at starting must be
2 37
= 0*1185 ohm
200 J
and at end of run
10*45
= 0*05225 ohm
200
But the resistance of the circuit is already 0*0506 ohm, whence
the added resistance must be
0*1185 0*0506 = 0*0679 onm at start
and 0*05225 0*0506 = 0*00165 ohm at end of charging.
Example XYIII. A battery of 55 cells has a discharging
E.M.F. of 2*05 volts per cell at start, and gradually falling to
1*9 volts each cell at end of 7 hours' run. The internal resist-
ance of each cell is 0*0004 ohm. The resistance of the leads
from the cells to the lamps is 0-005 onm - What number of
Circuits containing more than one E.M.F. 31
cells must be used to supply 400 5o-watt loo-volt lamps if
the P.D. at the lamp terminals is to be kept at as nearly 100
volts as possible all the time ?
Solution. The lamps are to take 100 volts, and will also
require X 400 = 200 amperes, neglecting the slight differ-
ence made by the P.D. at their terminals not always being
exactly 100 volts.
The loss of volts along the leads is
0*005 x 200 = i- volt
and the loss in the cells is
0-0004 X 200 = 0-08 volt per cell
or the available E.M.F. per cell is 2-05 0-08 = 1:97 volt at
start, or
1-90 0-08 = 1*82 volt at end of run
The E.M.F. required for leads and lamps is to be
100 + i = ioi volts
whence the number of cells required will be
IOI
r8 7
and
IOI
= 51-2, say 51 at start
1-82
= 55-4, or 55 cells at end
Motors. In running motors we are also dealing with cir-
cuits containing a B. E.M.F. A motor is a dynamo caused
to run by sending through its armature a current at sufficient
E.M.F. to overcome the B. E.M.F., and leave sufficient to
maintain that current through the resistance of the circuit.
We have, then, as before, that the current in the armature
will be
c- E "-
R
R being the resistance of the whole circuit, and e the B.E.M.F.
32 Examples in Electrical Engineering.
produced by the rotating armature. The part of the energy
which is supplied to the motor that is converted into mechani-
cal work is Ce, and of this a part is expended in friction of
various kinds in the motor itself; this part will be dealt with
later on, and at present will simply be expressed as a waste in
watts or horse-power. The difference between Ce and this
waste is the available power of the motor, and is usually
expressed as so many brake horse-power (B.H.P.),
Example XIX. A motor having a resistance of 0*025
ohm, and producing a B.E.M.F. of 90 volts at a given speed,
is connected by leads of 0-005 onm resistance to a dynamo
of resistance = 0*02 ohm. It is required to produce a B.H.P,
of 7 H.P. What must be the E.M.F. of the dynamo if the
waste in internal friction, etc., in the motor is 300 watts ?
Solution. The total mechanical power is
Ce = C X 90 watts
of which 300 watts are wasted in the motor, leaving 7 x 746
watts available at the brake. Therefore
C x 90 - 300 = 7 x 746
9 ^
= 6 1 -35 amperes
And this current is produced through the resistance of the
whole circuit by an active E.M.F
e = C (resistance of circuit)
or
e = 6l '35(' 2 5 4- 0-005 + ' 02 )
= 3' 6 75 volt s
whence the total E.M.F. must be
90 -f 3*0675 = 93*0675 volts
Example XX. In the above question, what will be the'
brake horse-power of the motor if the total E.M.F. of the
Circuits containing more than one E.M.F.- 33
dynamo is 95 volts, assuming the same speed and loss due
to friction, etc. ?
Solution. Here the active E.M.F. will be
e = 95 ~ 9 = 5 volts
and consequently the current will be
C = = 100 amperes
'5
therefore the total mechanical power is
100 x 90
and of this 300 watts is still the waste, leaving for brake
power
9000 300 = 8700 watts
or
8 7^ = 1 1 -6 H.P. at brake
746
Arc Lamps. In the case of arc lamps, the energy required
for the vaporization of the carbons is generally considered as
made up of the product of the current and an E.M.F. at the
crater, called a back E.M.F., the magnitude of which is about
39 volts for a continuous current arc ; but which appears to be
as low as 30 volts (virtual) for some alternating current arcs.
Thus, in an arc lamp, part of the E.M.F. supplied is con-
cerned in sending the current through the resistance of the
circuit, and the remainder is to overcome the back E.M.F.
Example XXI. An arc lamp has a B. E.M.F. of 39 volts,
and is connected to leads at P.D. of 50 volts. The resistance
of the lamp leads is o'n ohm, the resistance of the lamp coil
is 0-09 ohm, and the carbons have a resistance of 0*08 and
0*12 ohm respectively, whilst the arc itself has a resistance of
0*1 ohm. What must be the resistance included in the circuit
so that the lamp may take a current of 10 amperes?
Solution. The available or active E.M.F. is
e= 50 39 = ii volts
D
34 Examples in Electrical Engineering.
and since the current is to be 10 amperes, we have the total
resistance must be
e n
-^ = i' i ohm
C 10
But the resistance of the lamp itself is
o-ii -f 0-09 + o'o8 + 0*12 -f- o'i = 0-5 ohm
whence the additional resistance required must be
i' i 0*5 = 0*6 ohm
CHAPTER IV.
DYNAMOS AND MOTORS.
IN questions relating to the proportions of parts of dynamos
and motors we have to consider the relationship between the
armature and magnet windings, etc. The electro-motive force
generated in a revolving armature can be arrived at from the
data of its parts. Thus, the whole generated E.M.F. is
io c
for a Bipolar machine, whether drum or gramme wound,
where
N = whole number of c.g.s. magnetic lines which traverse
the armature core,
n = the revolutions per second of the armature,
and w the whole number of wires counting all round the
circumference of the armature,
io 8 = the number of c.g.s. units of E.M.F. , equivalent to
the practical unit, the volt.
The factor N stands for the rate of cutting lines in c.g.s.
lines per second.
The number of lines N is made up of the product of the
sectional area of the iron armature core and the number of
c.g.s. lines per unit of sectional area. It is very usual to call
the number of c.g.s. lines per square centimetre the value of
J3, or
/? = c.g.s. lines per sq. cm. of armature iron
For different types of armature there are different values of
ft most suitable. Thus, for gramme or cylinder armatures, ft
36 Examples in Electrical Engineering.
may range from 14,000 to 18,000 sq. cms., or even higher;
whilst, owing to the necessity of passing the lines through the
same size of air gap in the drum as in the cylinder type, we
must keep the value of ft for drums at about 9000 to 12,000
per sq. cm., though in some types of drum higher values can
be used with advantage.
The speed n is limited by the strength of the structure, on
the one hand, to not more than 3000 to 4000 feet per minute
(roughly, 1500 to 2000 cms. per second) circumferential
speed ; or, on the other hand, to far smaller values where
slow speeds are desirable for direct coupling, to slow-speed
engines, for example.
The number of wires all round, w, is determined by the
E.M.F. desired, and their size by the current to be carried.
As the commutator connects the armature to the outside
circuit in two parallels in Bipolar machines, the size of wire on
the armature is that to carry half the current from the machine.
No fixed rule can be stated for the relationship between the
current and the size of wire, as this will depend upon the size
of machine, the efficiency desired, and the rise in temperature
allowable. This part of the subject is considered in a later
chapter.
The whole E.M.F. E is the total volts generated by the
armature, and, of course, a part of this is required to circulate
the current in armature itself. Thus, it
r a = resistance of armature winding measured from brush
to brush,
and C = total current in amperes in the armature, then Cr n
volts will be lost in the armature, and the differ-
ence E Cr a is the terminal P.D. of the dynamo,
in the case of a magneto machine or a shunt
dynamo.
The resistance r a is obviously the resistance of J- the whole
length of wire upon it (since the two halves, and in parallel,
and each half is half the resistance of the whole), plus any
contact resistance at the brushes and in the leads to the
brushes.
,
I (UNIVERSITY \
Dynamos and Motors. 37
Leaving for the present the exact statement of the relation-
ship between the armature and magnet windings, we will take
a few simple examples.
Example XXII. What will be the sectional area of iron
required in the armatures respectively of two dynamos, one
ring wound and the other drum wound, each having 240 wires
all round, running at 1200 revolutions per minute, and giving
a total E.M.F. each of 120 volts?
Solution. In each case we have that the number of lines N
required will be
nw
and since n = ^ - =20 revolutions per second
XT 120 x io 8
N = = 2x00.000 lines total
240 X 20
Now, the ring-wound armature may have! ft = say 15,000,
whilst the drum may have ft only 10,000, say.
But N = fta
whence in the ring type
2,500,000 a
a = - - = i66f sq. cms.
15,000
and in the drum
2.SOO.OOO
a = ~ = 250 sq. cms.
10.000 * *
area of iron core.
Example XXIIL What is the total E.M.F. produced by
a dynamo, having an armature ring or gramme wound
with 204 wires all round, when ft = 16,000, area of cross-
section of iron core is 125 sq. cms., and speed is 25 revo-
lutions per second ?
Solution.
Na = 16^000 X 125 X 25 x 204 = IQ2 yolts
IO 8 IO 8
38 Examples in Electrical Engineering.
Example XXIY. : At what speed must the above armature
be run so as to give 102 volts terminal pressure when the
resistance of the armature is 0*02 ohm, the current is 100
amperes, and the value of ft remains at 16,000?
Solution. The loss of volts in armature resistance will be
O' ft = 100 x 0*02 = 2 volts
whence the total generated E.M.F. must be
E = 102 + 2 = 104 volts
thus the speed
Eio 8 104 x io 8
~- = 25 ' 49 revo!utlons P er sec "
or 1529*4 per minute.
Example XXY. What is the value of ft in an armature
core, drum wound, 25 cms. diameter, with 7 cm. hole in the
centre, and composed of 500 discs half a millimetre thick, if
wound with 180 wires all round, and run at 1200 revolutions
per minute, producing an E.M.F. of 175 volts at the terminals,
when the resistance of the winding is 0*032 ohm, and the
current is 100 amperes ?
Solution. The total generated E.M.F. is
E = 175 +
= 175 -f 100 X 0*032
= 178-2
and therefore
._ 178*2 x io 8
N = - s = 4,<Ko,ooo
20 x 180
But N = ft x area, and area is
a = ( 2 5 - 7) X 500 x 0*05 = 450 sq. cms.
and therefore
4,qt;o,ooo
ft = - yj - = 1 1,000 per sq. cm.
45
Dynamos and Motors. 39
Example XXYI. An armature, gramme wound, has a
core 36 cms. diameter, with 23 cm. hole, and is built np of
1000 discs, each 0*28 millimetre thick. There are 192 wires
all round, each large enough to carry a current of 60 amperes.
The speed is 781-2 revolutions per minute, and the resistance
of the armature is 0-039 ohm. What is the value of /?
required to produce a terminal P.D. of 150 volts at a current
of 120 amperes?
Solution. At 1 20 amperes the loss of volts in the armature
is
Cr a = 120 x 0-039 = 4'68 volts
whence the generated E.M.F.
E = 150 -f 4-68 = 154-68 volts
thus to produce this E.M.F.
N = I5 . 4 ' 68 X I0 " = 6,188,000 lines
I," X I92
and the sectional area of the iron core is
a = (36 23) X 1000 x 0*028 = 364 sq. cms.
whence
N 6,188,000
364
= 17,000 per sq. cm.
Magnetization of Field-magnets. So far we have been
considering cases where the exact value of the current in the
armature is given. In the majority of cases, however, the
armature has to carry a current which is greater than that in
the external circuit by the amount taken by the coils to
magnetize the field-magnets. This magnetization is brought
about in several ways, viz.
4O Examples in Electrical Engineering.
i . By current from some independent source called
separate excitation, thus
FIG. 9.
Here the data of the magnet-winding are of no particular
importance in the present connection.
The current in the armature is the same as the current in
the external circuit.
2. By a current from the armature of the machine itself
Dynamos and Motors.
caused to circulate round the magnet-winding before going to
the outside circuit, thus
FIG. 10.
where + T and T represent the terminals of the machine
which are to be connected to the external circuit.
This is called series winding; and here there is a loss of
volts in the resistance of the series coil, which has to be added
to the loss in the armature to get the terminal P.D.
In this case also the current in the armature is the same as
the current in the outside circuit.
We will call r m the resistance of the series coil.
42 Examples in Electrical Engineering.
3. Sometimes the magnet legs are excited by a current
derived from the terminals of the machine, called a shunt
winding, thus
FIG. ii.
In this case it is clear that the armature has in it a current
which is greater than that in the external circuit, by the amount
taken by the shunt coil. The total resistance of the circuit is
composed of the armature in series with the external circuit
Dynamos and Motors.
43
which is in parallel with the shunt coil. Calling r s = the
resistance of shunt coil, and r tt that of the armature, and the
current in the external circuit = C amperes due to a terminal
P.D. of e watts, we have that the current in the armat'ure
and the loss of volts in the armature will be
whence the total generated E.M.F.
(4) There may, for certain reasons, to be considered later
on, be a combination of shunt and series excitation, and it is
possible to connect up these two coils in two different ways ;
firstly, thus
FIG. 12.
called "long shunt." Compound wound.
In this case, calling e = terminal P.D. ; C = current in the
external circuit, etc., as before, we have that the current in
the shunt
c.=:
44 Examples in Electrical Engineering.
and the current in the series coil and armature will be both
the same, viz.
whence there will be a loss of volts in the series coil of
r m f C 4- 4 ) volts
' s
and in the armature of
r a ( C + - ) volts
The total loss in the machine will be
(^ + r m ) (^ C -f ^ J volts
and thus the total generated E.M.F.
Or, secondly, connections may be made thus
Cell
FIG. 13.
which is called " short shunt." Compound wound.
Here there will be a loss of volts in the series coil, due to
the main current C, or this loss will be
O,,, volts
Dynamos and Motors. 45
whilst the current in the armature is greater than C, the external
current, by the amount taken by the shunt coil.
Calling e the terminal P.D. as before, we have that the P.D.
at the terminals of the shunt coil will be
and thus the current in the armature will be
s
and the loss of volts in the armature is thus
whence the whole E.M.F.
There are other important differences between "long" and
short" shunt winding, which will have to be considered
later on.
Example XXYII. A shunt dynamo is to give a current )/
of 50 amperes at a terminal P.D. of 104 volts. The resistance
of the armature is 0-04 ohm, that of the shunt coil 52 ohms.
There are 210 wires all round, and the sectional area of the
core is 160 sq. cms. The speed is 1080 revolutions per
minute. If the winding is gramme type, what will be the value
of/3?
Solution. The current in the armature
C a = C + e s = 5 + ^ = 52 amperes
whence the loss of volts in the armature is
52 x 0-04 = 2-08 volts
or the total generated E.M.F.
E = 104 + 2-08 = 106-08 volts
46 Examples in Electrical Engineering.
to produce which E.M.F. the total number of lines
_ 106*08 x io 8 10,608,000,000
1080 0*18 x 210
f. - X 210
60
= 2.,8o6,ooo lines
whence the value of the magnetic density is
2,806,000
(3= i6 - = 17,530 per sq. cm.
Example XXYIII. At what speed must a series machine
run to give a current of io amperes at a terminal P.D. of 1000
volts, if the resistance of the armature is 2 ohms, and that of
the magnet coils r8 ohm, when the area of cross-section of
the armature core is~i5o sq. cms., and there are 17,500 c.g.s.
lines per sq. cm.? The winding is ring type, and there are
2100 wires all round.
Solution. The loss of volts in the armature is
io x 2 = 20 volts, and in the magnet coil is
io x 1*8 = 18 volts
or a total loss of 38 volts ; whence the generated E.M.F.
E = 1000 -f- 38 = 1038 volts
The total number of lines through the armature is
N = 150 x 17,500 = 2,625,000 lines
whence the speed
.
2100 x 2,625,000
or 1129*8 per minute.
= 1 8*83 revolutions per second
Example XXIX. A long shunt compound dynamo is run
so as to light 200 50- watt loo-volt lamps at the end of leads
of resistance of 0*008 ohm. The resistance of the armature
is 0*022 ohm, that of the series coil 0*015 ohm, and that of
the shunt is 60 ohms. If the speed be 1200 per minute, what
will be the sectional area of iron core required when /? = 17,085,
and there are 102 wires all round the armature?
Dynamos and ^lotors. 47
Solution, The current in the lamps is
C = 100 amperes
And there is a loss of 100 x 0*008, or 0*8 volts in the leads;
whence the terminal P.D. is 100*8 volts; and thus the shunt
current is
C , "^=,-68 ampere
60
and consequently the current in the series coil and armature
will be
C n = 101*68 amperes
the loss of volts in the armature and series coil will be
101-68 X (r a + r n )
or
101*68 x (0*022 + 0*015) = 3-76 volts
then for the whole generated E.M.F.
E = ioo'8 H- 376 = 104*56
to produce which the number of lines total will be
10.4^6,000,000
N = = 5,125,490
20 X 102
and as & = 17,085, the sectional area
q, 12=5,400
"= ,7,085 =3 s q- cms -
Example XXX. A short shunt compound dynamo is re-
quired to give a current of 200 amperes through 53 secondary
cells of 0*00025 ohm each internal resistance, and having a
B.E.M.F. when charging of 2*41 volts each cell. The leads
to the cells are 0*005 onm resistance ; the series coil is 0*008
ohm; the armature is 0*012 ohm; and the shunt resistance is
50 ohms. At what speed must she run when there are 144
wires all round the armature, the sectional area of the armature
core is noo sq. cms., and there are 17,100 c.g.s. lines per
sq. cm.?
Solution. The total B.E.M.F. of the cells is
c = 2*41 X 53 - 127*73 volts
48 Examples in Electrical Engineering.
there is a loss in their internal resistance of
200 X 53 X 0*00025 = 2*65 volts
the loss in the leads is
200 x 0*005 = I v lt
and that in the series coil is
200 x 0*008 = 1*6 volt
Thus the P.D. at the terminals of the shunt coil is
I2 7'73 + 2 ' 6 5 + i 4- i'6 = i3 2 '9 8 , say 133 volts
and due to which the current in the shunt coil is
C, = - 33 - - 2*66 amperes
0*50
Whence the current in the armature will be
C ft = 202*66 amperes
and the loss in the armature
202*66 x 0*012 = 2*43 volts
and thus the total generated E.M.F. is
E = 133 4- 2 '43 = J 35'43 volts
to produce which the speed must Le
It.cjAl.OOO.OOO
n = = 5 revolutions per second
noo X 17,100 x 144
or 300 per minute.
Pull on Conductors. The relationship between the power
required to drive an armature as a dynamo and the current
and E.M.F. produced, is that the E.M.F. depends upon the
speed only, as is shown from the formula we have already
used, viz.
Nnw
since for any given machine N and w are fixtures. The
current which the machine will give depends upon the resist-
ance of the circuit connected to it, and upon the power which
Dynamos and Motors. 49
the prime mover can put into it. There is a magnetic effect
produced by the current in the armature windings, which acts
so as to resist the motion. This will be understood by refer-
ence to Fig. ii.
The prime mover has to overcome the resistance to motion,
called a drag on the winding, by exerting a pull on the one
side of the belt attached to the dynamo. The magnitude of
this pull can be simply expressed as
P = CL(3 g x 0-0000002244 pounds per wire
where C = the current in amperes in any wire,
L = the length of a wire passing through the field in
cms.
fa = the strength of the magnetic field in the air gap
in c.g.s. lines per sq. cm.
Owing, however, to the want of uniformity in the strength
of the magnetic field round the armature, the actual pull on
any wire will depend upon its position in the air gap. When,
for instance, the wire under consideration is midway in be-
tween the polar horns, the pull will practically be zero, since
there the field is very small ; but when under a pole piece the
pull will be considerable, though gradually varying from a
smaller to a larger value as the wire moves round the polar
arc. The average value of the pull is that given above, and
the total pull on the belt, as far as production of electricity is
concerned, will be the product of this average pull and the
number of wires all round which can be considered to be
under the magnetic influence, or cutting lines. This number
is approximately equal
20 WO
w l = W-T- X i'i = 1*1-5---
360 180
for bipolar machines. Where w is the whole number of
wires counted all round, 6 is the angle of polar embrace, and
i 'i is a numeric to take in the stray field in between the
poles.
The angle 6 varies from 120 to 140 generally, though it is
sometimes outside these limits.
5O Examples in Electrical Engineering.
Thus the total pull will be
C T
P = LpgW^- X i 'i X o '0000002244
where C = the total current from the machine in amperes,
L = the length of any wire that is practically the length
of the bore of the magnets,
j3 a = density of lines c.g.s. per sq. cm. in the air gap,
w = total number of wires all round,
= the polar angle,
and the formula is only for a bipolar machine. Simplifying,
we have
P = CLfl a w6 X 0-0000000006856
From the above, the rate of doing work can be found by con-
sidering that this pull is a weight of so many pounds lifted
through a distance equivalent to the mean circumference of
the armature once in the time taken by one revolution ; thus
let
d = mean diameter in feet of the armature, i.e. diameter of
core plus the depth of winding.
n number of revolutions per second.
Then the above weight is lifted
Trd feet in ^ minutes
or the work done in one minute is
6oPmrd foot-pounds
whence the power is
,
horse-power
33,000
_ 63 x 22ndCLp g wO X 0*0000000006856
7 X 33,000
,
- horse-power
For a dynamo this is, of course, only that part of the power
required which is concerned in the production of electricity ;
Dynamos and Motors. 51
the power to overcome frictional losses of various kinds must
be added. But in the case of a motor the above gives the
mechanical power developed, part, however, of which is also-
lost in friction, etc.
In the case of motors the E.M.F. produced by the rotation
of the armature is opposed to the supplied E.M.F., and is
called a back E.M.F. e. But as far as the motor is concerned
this B.E.M.F. is one of the factors of the power, the other
being the current ; this B.E.M.F. has to be calculated exactly
as though we were dealing with a dynamo.
Example XXXI. At what speed will the armature of a
series machine run, and what will be the brake H.P. if a
current of 20 amperes be supplied to it at an E.M.F. of 400
volts? Its armature has a resistance of 0*25 ohm ; its magnet
winding 0-15 ohm. The loss in friction is 380 watts. There
are 900 wires all round, and the total magnetic flux through
the armature is two million lines.
Solution. Since the total loss of volts in the machine is
20 x (0-25 -f 0*15) = 8 volts
there will be
400 8 = 392 volts
available for overcoming the B.E.M.F. And the speed neces-
sary to produce this will be
io 8 392 x io 8
n = r; = ^ z
Nw 900 X 2 X icr
= 2 1 '7 7 revolutions per second
Thus the total rate of doing work mechanically is
392 x 20 = 7840 watts, of which the
loss in friction is 380 watts, leaving
7460 watts, or-^-
10 H.P. available at the brake.
Example XXXII. A machine run as a shunt dynamo^
with a current in the armature of 100 amperes, gives 100 volts
at its terminals at a speed of 1200 revolutions per minute.-
The resistance of the armature is 0*035 ohm. What will be
52 Examples in Electrical Engineering.
the speed if this machine be run as a motor off mains at 100
volts, and taking a current of 102 amperes total, when the
resistance of the shunt coil is 50 ohms, and the loss due to
friction, etc., is at the rate of 325 watts?
Solution. As a motor the shunt coil takes 2 amperes,
leaving 100 amperes for current through the armature. As
a dynamo there was a loss in the armature of 100 x 0*035
= 3*5 volts, whence the total generated E.M.F. must have
been 103*5 volts, at a speed of 1200 revolutions per minute.
As a motor there will also be the same loss of 3*5 volts in the
armature, leaving only 96*5 volts for overcoming B.E.M.F.,
which B.E.M.F. will be produced at a speed of
o6"t;
X 1200 = 1118 per minute
103-5
Since the strength of the magnetic field will be the same as
before, also the total mechanical rate of doing work will be
Cc = 100 X 96*5 = 9650 watts, of which the
loss in friction is 325 watts, leaving
9325 watts for external work, or
^~ = 12-5 H.P. at the pulley
Example XXXIII. A shunt machine running at 720 revo-
lutions per minute as a dynamo, gives 50 volts P.D. at its
terminals at a current of 200 amperes in the external circuit.
Its armature resistance is 0*0147 orim > an< 3 triat f i ts shunt
coil is 12*5 ohms. At what speed will it run as a motor, and
what current will it take at a P.D. of 50 volts to develop
12 brake H.P., when H.P. is lost in friction in the gearing,
and 0*352 H.P. in the machine?
Solution. The total rate of doing work by the motor must
be 12 + o'5 4- 0-352 = 12-852 H.P., or 12*852 x 746 = 9588
watts, which must be the product of the current in the arma-
ture and the B.E.M.F. produced. The shunt coil takes
^ = 4 amperes
and thus the total current will be C = C +4.
Dynamos and Motors. 53
As the resistance of the armature is 0*0147 onm > the l ss f
volts therein will be C rt x 0-0147 volts, and thus
= 50 -
and C rt must be = 9588, or
(50 - C n X o-oi47)C rt = 9588
or
- 0-OI47C, 2 + 5oC ft - 9588 = o
whence C rt = 204 amperes, at which current the loss in the
armature is
204 x 0-0147 = 3 vo lt s
leaving 50 3 = 47 volts for e
The total rate of doing work mechanically is
204 x 47 9588 watts
and the speed is
4.7
X 720 = 628-4 per minute
53
Example XXXI Y. A lo-ampere series machine which
gives 1000 volts total E.M.F. at a speed of 1000 revolutions
per minute, is supplied with a current of 10 amperes to run
as a motor, and has to lift a weight of 3 tons at the rate of
50 feet per minute. If there is a loss due to friction in the
gearing of ^ H.P., and 0-32 H.P. lost in the friction, etc., in
the motor, what will be the speed and P.D. of supply mains
required if the resistance of the armatureTis 3-5 ohms, and
that of the magnet coils 2-5 ohms?
Solution. To lift 3 tons 50 feet per minute is doing work
at the rate of
3 X.2240XJO = iQ . i8 H p
33,000
and, adding the losses, the total H.P. will be
10-18 -f 0-5 -f- 0-32 = ii H.P.
or in watts = 8206, which is done by a current of 10 amperes
against a B.E.M.F. of
liM = 820-6 volts
54 Examples in Electrical Engineering.
eh the speed will be
X 1000 = 820*6 revolutions per minute
to produce which the speed will be
820-6
1000
To send 10 amperes through the resistance of the machine
will require a P.U. of
io(r a -f- r m ) = 10 X 6 = 60 volts
whence the P.D. at the motor terminals will have to be
820*6 + 60 = 88 1 volts, say
Example XXXY. The length of one wire on the surface
of an armature is 20 cms., the mean diameter is 21 cms., the
speed is 1200 revolutions per minute, ancl the density of lines
in the air gap is 2291 per sq. cm. The angle of polar embrace
is 135, and there are 360 wires all round. If this machine
be used to give a total current of 50 amperes, what will be the
H.P. required to drive it if H.P. is lost in various ways?
Solution. Applying the formula on p. 50, we have
1 200 21
3917 ' 7 X ~6o~ X 7o^7 X 50 X 20 X 360 X 135 X 2291
H P = oo 30 47
io 15
= 6*01 H.P., to which must be added \ H.P. as above for
friction, etc., or total H.P. required = 6-51 H.P.
Example XXXYI. What will be the drag on one wire of
a dynamo armature which gives a current of 800 amperes, and
has an average magnetic field of 3500 c.g.s. lines per sq. cm.
of air gap ? The length of the polar bore is 80 cms., and the
machine is bipolar.
Solution.- Using the formula given on p. 49, we have pull
is
P = CL/30 X 0*0000002244 pounds
and as the current here is the current in each wire, it is
consequently 400 amperes, and
P = 400 X 80 x 3500 x 0*0000002244
= 25*1328 pounds
CHAPTER V.
ARMATURE WINDING.
IN considering the size of wire on a given armature to carry
a given current, we have to allow a certain temperature rise,
and consequently have to take into account all the sources
of heating. These are mainly threefold, viz.
(a) The C/R a loss due to the effect of the current in the
armature on the resistance of the armature ;
(b) The H loss due to molecular friction in the iron core,
brought about by the changing of the magnetic flux ;
(c) The F loss due to Foucault or eddy currents in the core,
and other moving parts of the armature.
The energy developed in the armature due to these effects has
to be got rid of at some reasonable temperature, and it is
customary to fix a limiting difference of temperature between
the surface of the armature and the outside air.
In calculating this temperature difference it is of course
necessary to bear in mind that some part of the heating is
got rid of in other ways than by convection and radiation from
the outside surface of the winding. Thus, part is lost by con-
duction through the spider to the spindle, the winding to the
commutator. Experience shows that this part so conducted
away may be about \ of the total loss.
Mr. Esson has shown that the surface of an armature re-
volving at a circumferential speed of 3000 feet per minute,
can dissipate energy at the rate of 0*00444 watts per sq. cm.
of its surface covered by winding for every i C. temperature
difference between that surface and the air. The correspond-
ing figure for a stationary surface, such as a field-magnet coil,
is 0*00281 watts per sq. cm. per i C. temperature difference.
56 Examples in Electrical Engineering.
A combination of these figures will give the rate of loss of
heat for any armature in terms of the circumferential speed.
Thus the increase in rate of cooling due to the speed of
movement through the air of 3000 feet per minute (1500 cms.
per second) is
0*00444 0*00281 = 0*00163 watts
per sq. cm. per i C., or
0*00163
= 0*000000543 watts
per sq. cm. extra per 1 C. difference of temperature per i
foot per minute of speed through the air or the loss for any
speed may be put down as
0*00281 -f- 0*0000005 43/ watts
per sq. cm. per i C, where f is the circumferential speed
in feet per minute.
For a speed of/= 3000 feet per minute, we have that the
total difference in temperature between the air and the
armature surface will be
W
T = x - degrees Centigrade
A X 0*0044
where W = that part (say, J) of the whole expenditure of
energy in the armature, which is considered to
be radiated from its surface, expressed in watts ;
and A = the surface area in sq. cms. of the armature.
Similarly, for a magnet coil we have
T = JL
A X 0*00281
where A is the outside surface of the magnet winding, not
including the cheeks of the bobbin, expressed in sq. cms.
The loss W is approximately f (C ft 2 R a + H + F), and of
these items the first is, of course, quite easily found, allowance
being made that, the temperature being high, we have to
employ a high value for p.
The hysteresis loss H is due to the fact that molecular
friction in the iron causes heating when the iron core as a
Armature Winding. 57
mass is rotated by the rotation of the armature, whilst the
molecules of iron are themselves stationary. The effect pro-
duced is the same as if the molecules were rotated whilst the
armature core remained stationary. In a bipolar field the
armature core has all the magnetized molecules thus rotated
once per revolution, and the value of the energy so expended
has been found by Professor Ewing for all conditions of
magnetization. The table on p. 214 gives these values ex-
pressed in ergs per cc. per cycle of change, one cycle corre-
sponding to one revolution of an armature in a bipolar field.
Now, as one watt-second is io 7 ergs, we have that the total
ergs expended in the armature per revolution multiplied by
the speed in revolutions per second and divided by io 7 will
give the rate of loss in hysteresis in watts. Calling the value
of the ergs per cc. per cycle ^, we have that the total ergs per
revolution will be ^V, when V is the bulk of the armature iron
in cc. If, then, n be the revolutions per second, we have the
hysteresis loss is
H = _- watts
io 7
The other loss, viz. that due to eddy currents, F, is almost
impossible of calculation in an armature. Only a small part
of it is in the laminated iron core, the major part being in
other conducting masses, such as the winding, end plates,
spider and spindle, etc. Due precautions being taken to
minimize it as much as possible, we shall assume in these
pages that it is equivalent to that part of the total loss which
is got rid of in other ways than by radiation directly from the
surface of the winding.
Therefore we may now put the value of W as
W = C ft 2 R ft + H
and write the temperature rise as being
T = C " 2R " + H d Centigrade
A X 0-00444
for a bipolar armature revolving at 3000 feet per minute cir-
cumferential speed.
OF THB
UNIVERSITY
58 Examples in Electrical Engineering.
In finding the area of the surface of the armature, we must
only take that part of the surface which is exposed to the air.
In the case of a drum armature this is easy, but in a ring or
gramme armature, some care is required in calculating the true
size of that part of the surface which is inside the cylinder.
In all cases the surface should be considered as smooth, no
allowance being made for corrugations due to the curved
surface of the wires ; for the coefficients taken are stated higher
to allow for such irregularity.
Example XXXYII. What will be the hysteresis loss in an
armature ring wound, 25 cms. long, 25 cms. diameter, with a
T 5-cm. hole (all dimensions being those of solid iron) ? The
speed is 1226 revolutions per minute, in a field of 4 million
c.g.s. lines the value of the ergs per cc. per cycle is 10,750
per cc. for this particular iron.
Solution. The sectional area of the armature is
(25 15) X 25 = 250 sq. cms.
whence the density of lines is
A \f T rfi
(5 = = 16,000 per sq. cm.
corresponding to which the ergs per cc. per cycle are given
as 10,750. The number of cycles per second will be the
revolutions per second, or
1226
" = -SQ- = 20 ' 43
And the number of cc. of iron in the core will be
V = 5X25X2oX7r = 7855 cc.
whence we have that
7855 x .075 X 20-43 =
I0 7
say, 173 watts.
Example XXXYIII. If the resistance of the armature in
the last example is 0*0212 ohm when hot, and it is used to
Armature Winding. 59
light 200 5o-watt loo-volt lamps at a terminal pressure of
102 volts, the resistance of the shunt coil being 51 ohms,
what will be the rise in temperature due to hysteresis and
CV a losses, if the eddy current loss is got rid of by conduc-
tion, etc. ? The surface of the armature is 3675 sq. cms., and
the circumferential speed is to be taken as 3000 feet per
minute.
Solution. The current taken by the lamps is
200 x 50 .
= 100 amperes
100
and that taken by the shunt coil is 2 amperes, whence the
total current in the armature is
C,,, = 1 02 amperes
and thus
Ci,V a = (io2) 2 0*0212 = 22o'5, say 221 watts
and the loss due to H has been found to be 173 watts, whence
the total watts
W = 221 + i?3 = 394 watts
and thus the temperature rise will be
.
0-00444 X 3675
Example XXXIX. Suppose the armature mentioned in
Example XXXVII. were replaced by a drum having as before
25 cms. length of iron in it, but having only a 7-cm. hole, and
run at a density of only 11,000 c.g.s. lines per sq. cm., corre-
sponding to which the value of ergs per cc. per cycle is 5900
(1) What will be the temperature rise with the same winding
as to size of wire and the same current at the same E.M.F.,
and what the speed ? and
(2) What would be the temperature rise and output if the
current and the speed are to be the same as when a gramme
armature ?
(3) Also, what would be the value of the current to make
the temperature rise the same and the speed the same as in
the case of the gramme ?
60 Examples in Electrical Engineering.
The surface area of the drum may be taken as 4000 sq. cms.,
and the variations in the speed are not supposed to affect the
rate of cooling.
Solution. (i) As a drum there will be a sectional area of
the armature of
(25 7)25 = 450 sq. cms.
whence at /? = 11,000 the total number of lines will be
N = fia = n,ooo x 450 = 4,950,000
The total generated E.M.F. was before equal to 102 volts for
the terminal P.D. plus the loss in the armature of
102 x 0*0212 = 2*16 volts, or 104*16 total
to produce which E.M.F. the drum will have to run at a speed
inversely as the number of lines, or speed will be
;/ = 20-43 X - - = 1 6*5 revolutions per second
the volume of the drum core will be
4 4
or
V = 2 5 X 25 x 22 x_g5 _ 7 X 7 X 22 xj5 =
4X7 4X7
Thus the hysteresis loss is
H = 5900 . *JI'3J_4 *2^S = 1 10 watts
I0 7
And since in question (i) the current is to be the same as
before, we have that the C a V n loss is 221 watts as before,
whence the rise in temperature will be
T = ^- - = 1 8-6 C.
4000 X 0*00444
(2) If the speed is to be the same as before, there will be a
greater E.M.F. total in the ratio of
Armature Winding. 61
whence with the same current the output will be
100 x 12674 = I2 >6?4 watts
as against only 10,200 for the ring winding.
But at this speed of 20*43 revolutions per second the
hysteresis loss will be
H = 5
or the total watts expended will be
136 4- 221 = 357 watts
and the consequent temperature rise will be
T = - 257 - = 20 C.
4000 X o '00444
(3) If, however, the temperature rise is to be the same as
for the ring armature at the same speed, the current may be
larger, such that the new C tt V tt loss added to the H loss will
make the heating such as to cause a temperature rise of
24-1 C. Thus the total watts will be
W = T X A X 0-00444
= 24*1 X 4000 x 0^00444
= 428 watts
of which 136 watts are for hysteresis, leaving 292 watts for
C V a . If the armature resistance be taken as 0*0212 ohm
as before, then
O'O2I2
/
and C a = V -^- = 117 amperes
Example XL. An armature working at a density of 8000
lines per sq. cm. and giving a certain E.M.F. is run at half
the speed with a density of 16,000 per sq. cm. : what will be
the ratio of the watts lost in hysteresis in the two cases ?
Solution. At the higher speed and ft = 8000, h is 2860
(p. 215) ergs per cc. per cycle; and at the lower speed and ft
16,000, h is 10,400 ergs. But as the cycles per second are
62 Examples in Electrical Engineering.
only half as many in the second case as in the first, the two
losses will be 2860 in the first case to 10 ^- = 5200 in the
second case, or as
2860
5^60 ='5499, say 0-55
or the loss in H in the high speed low density machine is
only 0*55 as great as in the slow speed high density.
Multipolar Machines. When the dynamo is not bipolar,
it may be so wound that either the current or E.M.F. is pro-
portionally increased. Thus, with what is called the parallel
winding, the coils are connected up as before to a commu-
tator, which then has as many pairs of brushes as there are
pairs of poles, unless the commutator is cross-connected, when
all the positive brushes can be replaced by one brush, and
similarly all the negative brushes by another brush. Using
the same symbols as before, and N still standing for the
number of lines entering the armature from or leaving the
armature towards any one pole piece, the E.M.F. will be
_
E = - g- as before
but the armature is now in zp parallels instead of 2 parallels,
where/ is the number of pairs of poles, and can consequently
carry / times as much current if the wire be the same size,
or
where Q is the current from the machine with / pairs of poles,
and C the current from a bipolar machine with the same value
for N, etc.
On the other hand, with the series winding, which is to be
preferred to the parallel for various reasons, the E,M.F is
increased in proportion to the number of pairs of poles, whilst
the current-carrying capacity of the armature is the same as
before, being only in 2 parallels whatever the number of poles.
Thus the E.M.F. will be
volts
Armature Winding. 63
the current being simply the same as in a bipolar machine
having the same size of wire.
The effect on the hysteresis loss in the core is the same in
both cases and may be found by taking the product of the
volume of the armature into h, n, and^ and dividing by io 7
hVnp
or H = c- watts
10*
for both series or parallel windings.
Virtually the effect of the multiplicity of poles is the same
as having p equal dynamos connected on the one hand in
parallel and on the other in series.
Example XLI. A six-pole dynamo has an armature core
composed of 600 discs of iron 0*25 mm. thick, and having
a radial depth of 7 cms. It is wound with wire which will
carry 50 amperes, and has 540 wires all round. If the density
of magnetic flux through the iron is 15,000 per sq. cm., and
the speed is 6 revolutions per second, what will be the E.M.F.
and current if connected up in the parallel method ?
Solution. The total value of lines is
N = 2 X 600 X 0-25 x o'i X 7 X 15,000
= 3,150,000 lines
and the whole E.M.F. will be
' E = 3,150,000x^^540 = io2yolts
As each wire will carry 50 amperes, the current, as an ordinary
bipolar machine, would be 100 amperes; but there are now
2/ = 6 parallels in the armature, and thus the current is
C = 6 x 50 = 300 amperes
Example XLIL What will be the hysteresis loss in the
above armature if the outside diameter of the core is 70 cms. ?
Solution.
Wnp
H ~
64 Examples in Electrical Engineering.
and h from the table on p. 216, is 9300. The volume of the
core is
V = 600 x 0-025 x 7 X - 7 - x 63 = 20,790 cc.
and thus
CHAPTER VI.
THE MAGNETIC CIRCUIT.
THIS can best be illustrated and explained by a comparison
with the electric circuit. The quantities corresponding to
conductivity and resistivity, E.M.F., and current, are called,
in the magnetic circuit, permeability and reluctivity, magneto-
motive force (M.M.F.), and magnetic flux respectively. Per-
meability has this difference from conductivity, that it has no
fixed value in the so-called magnetic materials (like the corre-
sponding value of electric conductivity, which has a fixed value
for a given temperature quite irrespective of the magnitude of
the current or electric flux), but varies with the variation of
the magnetic flux even when the temperature is kept con-
stant.
In the non-magnetic materials, however, the value of the
permeability is independent of the value of the magnetic flux,
even when this is confined to the same path, and so varies in
density. The use of the idea of current-density is not so usual
as to have caused the adoption of a symbol for its indication ;
but for purposes of comparison we shall make use of D, as
standing for the density in amperes per sq. cm. The corre-
sponding symbol very frequently used in magnetic measure is
/3, and stands for the number of c.g.s. lines of magnetic flux
per sq. cm. of cross-section.
It is very usual to confine ft to the magnetic density in iron
or the other magnetic metals, but for sake of uniformity we
shall not so restrict its use, but let it stand for the density of
magnetic lines in any material. It has been shown before
f
66 Examples in Electrical Engineering.
that we may write the relationship between the three quantities
<?, C, and R in the electric circuit in a simple way, viz.
C = , and R =
also
Adding now to our symbols the above-mentioned D, the
amperes per sq. cm., we shall have that
C = D
and
e = CR = DrtR = D/p =
;;/
or, putting again C for the total electric flux
a am
The corresponding quantities in the magnetic circuit we will -
call
ffi = magneto-motive force (M.M.F.) measured in terms of
a unit, to be afterwards defined.
N = /3a = the whole magnetic flux, generally represented in
the form /te, on account of the importance of knowing
the value of the density.
The quantity corresponding to resistance is the magnetic
reluctance, and may be represented by &, whose magnitude
will have to be stated in terms of the permeability, since there
is no unit in use called unit reluctivity. Thus
where //, stands for permeability in the same way as we have
used m to stand for conductivity, that is the actual conduct-
ance of the unit cube. We thus call //, the actual permeance
of the unit cube, measured in terms of its Value for some
particular material, viz. air, in which /x is always regarded as
unity.
The Magnetic Circuit. 67
Consider a solenoid of infinite length having a current of
C amperes passing through it. Let the turns or convolutions
of wire on it be w per cm. length. It is found that there will
be produced inside the solenoid a magnetic flux having a
density of Cw lines per sq. cm. of cross-section of core,
provided the core consist of air, or some other so-called non-
magnetic material. This Cw is commonly called H, and,
owing to the permeability of air, etc., being unity, it follows
that
4?r
H = Cw = the magnetizing force per unit length
and it is in this sense that H is generally used. The magnetic
materials are chiefly iron, steel, nickel, and cobalt, all others
being practically non-magnetic, or having a permeability of
unity ; some few, such as bismuth, have a permeability slightly
less than unity, but only to a quite insignificant extent. The
so-called magnetic materials have a permeability much greater
than unity, not constant, however, but depending upon the
density of magnetization ; that is, upon the magnitude of the
magnetic flux through a given cross-section. We must also
class a vacuum as non-magnetic, and having a value of ft = i.
Thus, if the core of our infinite solenoid be replaced by a
bar of iron, we have in the first place expelled all the air, but
the ether may be regarded as being still present, and we now
have the effect of two cores in parallel, the original core with
/* = i, and a new core of iron with //, much greater than unity.
The magnetizing force is still
Cw per cm. length
and is undisturbed, and consequently we get a magnetic flux
of density H through the original ether core, and a magnetic
flux through the new core of magnitude
I = Cw X /A!
when fji { is the value of the permeability of the iron core,
68 Examples in Electrical Engineering.
Adding these two magnetic fluxes together, we have
= I + H
the density of magnetic flux in the composite core.
From a purely physical point of view it is important to
distinguish between I and H ; but as we can never actually
separate them in practice, we invariably speak of (3, their sum,
and it is much more convenient to so consider the matter.
Though the idea of current density is sometimes used, we
are not so familiar with it as we are with actual current total ;
but a little consideration will show that the magnitude of the
E.M.F. required to produce a given current density D in a
conductor depends simply upon the material and length of
the conductor, or, more simply, for a given material of given
length at a given temperature the E.M.F. required is pro-
portional to the value of D, irrespective of the sectional area
of conductor. Fortunately for the ease of calculation in the
electric circuit, there is a very simple connection between
the E.M.F. required and the value of D, viz. one of direct
proportionality subject to the slight effect of temperature in
altering the resistance of the conductor.
Thus the E.M.F. required for a given value of D was stated
above as
D/ C/
e = or =
m am
Similarly, calling the value of total magnetic flux N = j3a,
we have that the M.M.F. required is
H = or = in physical measure
ap p
In the electric circuit the proportionality between e and D
renders calculation easy without the use of tables ; but in the
magnetic circuit the want of direct proportionality between
H and (3 renders the use of curves or tables absolutely
necessary for exact work. The exact value of this relation-
ship has been investigated by various observers, but we shall
confine ourselves to the measurements made by Professor
Ewing and the Drs. Hopkinson. A combination of their
The Magnetic Circuit. 69
results for good qualities of iron is shown in the tables given,
which have been read off from carefully drawn curves, having
a general appearance as shown in Fig. 14, which gives the
relationship between -
cut = H and ft
H 600
Ordinates stand for /?, the c.g.s. lines per sq. cm. in the
4-7T
core of iron, and abscissae for the magnetizing force Cw = H,
required for every centimetre length of the iron. It will be
noticed that when only feebly magnetized the iron has a much
greater ratio of (3 : H than when magnetized to a considerable
extent, though at very low values of j3 this is not so notice-
able. The three curves stand for this relationship for wrought
iron, cast iron, and for air only. It has already been pointed
70 Examples in Electrical Engineering.
out that the iron adds to the value in air an amount called I,
which is the difference between the ordinate of the air curve
and that of the iron curve. This addition is varying in value,
gradually increasing with the value of /?, till, as the curve in-
dicates, it is likely that at a very high value of H, this addition
will have reached a maximum, and will then be simply a
constant addition to the value for air alone. Tables carefully
read off from such curves as these are given on pp. 214-218.
It is not possible to represent the relationship between /3 and
H for the whole extent of the curves by a simple formula, and
thus reference to these tables is necessary for finding out the
magnetizing force required for any particular degree of mag-
netization. We can thus say that the value of H required to
magnetize a given piece of matter considered separate from
all other matter and complete in itself as being
But - is the same quantity as H in our curve, and thus we
N
may say that the quantity is some function of ft expressed
in terms of H. But H = Cw, which is the same thing as
saying -- times the ampere-turns per cm. length, and as we
really apply magnetizing forces by winding with wire to carry
a current, it is most convenient to do away with the numeric
- by making allowance for it in the tables, thus
A.7T IO
H = Cw, or Cw = H
10 4?r
Cw = H x 0795
or
ampere-turns per cm. length = H X 0*8 =/(/?)
nearly enough, since our knowledge of the exact values of ft
and H is not so exact as to require a distinction between 0795
The Magnetic Circuit. 71
and o'8. The convenience of this was first pointed out by
Mr. Esson, and has since been extensively made use of by
engineers. Thus, in the tables, we have put in values of fi
with the corresponding values of H and /(/?), this last being
the mode of writing function of ft. The permeability //, is also
shown, being readily found from
These tables are for the magnetic circuit what tables, showing
the relationship between D (amperes per sq. cm.) and e the
P.D. required in volts per cm. length of conductor, would be
for the electric circuit.
Using the symbols
H = the whole magnetizing force required by a magnetic
circuit, in physical measure called M.M.F.
we have
The ampere-turns total = o'8H = $&
and the ampere-turns re- | = Q . 8H m = magnetic RD<
quired by the part
Thus, remembering that it is just as necessary to have a com-
plete circuit for the circulation of the magnetic flux as it is
in the case of the electric flux, we have
We must either have the whole circuit of one material of
uniform quality and sectional area and of length /, in which
case ^t is the whole magnetizing force required ; or if / be
the length of a part only of the circuit having a definite
property, we must then regard
N/
0-8, =
#/x
the magnetic P.D. ampere-turns required by that part alone j
whilst then the whole M.M.F, will be the sum of all such
items as o>,
i Examples in Electrical Engineering.
Thus in a composite magnetic circuit
tfo/Xo
when the same total flux N circulates round the whole circuit.
The following comparison of the Electric and Magnetic
Circuits will serve to sum up the previous considerations :
ELECTRIC CIRCUIT.
C = Current or whole electric flux
in amperes.
D = = amperes per sq. cm. of
section, or current density.
m = conductivity in mhos of the
material of the circuit ; it
is the actual conductance
in mhos of the unit cube.
It is constant for a constant
temperature irrespective
of D.
u = = the resistivity of the ma-
m
terial or actual resistance
in ohms of the unit cube.
It is constant subject to
the constancy of m.
R = = = resistance in ohms
a am
of a circuit or part of a
circuit of length / and area
a made of material having
resistivity p,
MAGNETIC CIRCUIT.
N = Magnetic flux in c.g.s. lines
total.
N
ft = magnetic density or c.g.s.
a
lines per sq. cm.
ju = permeability of the material of
the circuit, and is the actual
permeance of the unit cube.
It is constant only for air
and the non-magnetic
materials having a value
of unity (l).
It is variable for the magnetic
materials, and dependent
upon the value of .
; = - = the reluctivity of the
M
material or actual reluct-
ance of the unit cube being
constant, and of value
unity (i) for air and the
non - magnetic materials ;
but varies with /* in the
magnetic materials, ac-
cording to the value of j8.
3& = = = reluctance of a cir-
a ap
cuit or part of a circuit of
length / and area a made
of material having a per-
meability ft,
The Magnetic Circuit.
73
ELECTRIC CIRCUIT.
i/p = the D.P.
in volts required to send
a current of C ampei - es
through a circuit or part
of a circuit.
can readily be calculated
from any of the above re-
lations.
E = the whole E.M.F. in the cir-
cuit, being the algebraic
sum of all the P.D.s round
the circuit, or the sum of
all such quantities as
Though these P.D.s may be
required distributed round
the circuit E may be gene-
rated at one part, viz. the
armature of a dynamo, for
example.
MAGNETIC CIRCUIT.
= The magnetic P.D. required
to send a magnetic flux
through a circuit or part
of a circuit. It is equal
N/
to o-SNE = 0-8.
ap
(o can only be calculated
for the non-magnetic ma-
terials where /j. = I . For the
magnetic materials refer-
ence has to be made to the
tables for the value of /*, etc.
By reference to the /()
column of the tables, the
value of o> in ampere-
turns per cm. length can
readily be found, and cal-
culation thereby much sim-
plified.
= the whole M.M.F., being the
algebraic sum of all the
magnetic P.D.'s round the
circuit, or being equal to
Though these items may be
required distributed round
the whole circuit, the whole
of fft may be produced at on e
part, viz. the magnet coils
of a dynamo, for example.
In the electric circuit it is quite easy to confine the electric
flux to a very definite path, because some materials (air, for
example) are practically of infinite resistivity. In the magnetic
circuit, however, air and the other so-called non-magnetic
materials have a permeability which is, at the very best,
only gfoVo as great as that of iron, the very best magnetic con-
ductor we have. There is thus no possibility of magnetically
insulating any circuit to anything like the perfect extent attain-
able in the electric circuit ; and we have consequently to allow
for some leakage of lines. The permeability of iron varies
74 Examples in Electrical Engineering.
from about 3000 to 30 or so at such values of ft as are con-
venient for general work, and thus the leakage of lines may
often be as great even as if we had no better insulator than
manganin, or such like, for the protection of the copper con-
ducting path in the electric circuit. This leakage of lines is
particularly the case in dynamos and such apparatus where the
magnetic circuit is composite in character, and when the mag-
netic P.D. between point and point may be very great. Thus,
if the permeance of the armature and air gaps in series be
represented by u, and that of the surrounding air space be
called w, these two permeances are in parallel, and are subjected
to nearly the same magnetic P.D. Consequently, if N be the
total magnetic lines through the armature core and air gaps,
the magnetic P.D. between the polar faces will be , and this
u
magnetic P.D. will produce a flux of - lines through the
surrounding air space. Now, these two fluxes have to flow in the
magnet limbs, which will consequently carry, not N lines, but
N 4-N^ = N
In their paper on dynamos in 1886 the Drs. Hopkinson chose
to represent this fact in the very simple way of stating the
lines N m in the magnet limbs as being v times the lines in the
armature core ; the value of v depending upon the relation-
ship of w to u. Thus, when w = u, v = 2, a value met with
in such multipolar machines as the old pattern of Ferranti
alternator. The value of w is, however, generally much less
than u in a well-proportioned dynamo, some measured values
of v being given in the following table :
Type of machine. Value of v.
Edison, Hopkinson, single magnet, drum '32
Siemens ,, ,, ,, ... ... '30
Phoenix over type ,, ,, gramme "32
Phoenix double magnet ,, '40
Manchester,, -49
Victoria Mordey, 4 pole, ring or gramme '40
Ferranti, 16 pole, disc ... 2 - oo
Manchester with polar supports 1*5 to 1*60
The Magnetic Circuit. 75
Professor Forbes, many years ago, showed how a sufficiently
exact approximation to the value of v for practical purposes
could be calculated from the drawing of the machine.
Thus, the same magnetic P.D. which is required between
the two polar faces to produce N lines in the armature and
air gaps, is also producing a magnetic flux through the very
imperfectly magnetically insulating air which surrounds the
dynamo, and which additional magnetic flux has to be ac-
commodated in the magnet limbs, thereby causing the value
of (3 in them to be greater, and making the value of the
magnetizing force for that part of the circuit greater than
would otherwise be the case.
We have, then, to consider the magnetic circuit of the
dynamo as composed of a number of items, each of which
may be different, either on account of material or the fact
that a greater flux of lines has to be accommodated. This
composite magnetic circuit consists of
(i) The armature, which is almost invariably made of a
good quality of wrought iron. The sectional area is easily
found, as has been indicated previously. The average length
of the path of magnetic lines can be found with sufficient
exactness by taking the mean semi-circumference in the case
of a bipolar machine, whether drum or gramme ; or for a
multipolar machine it is
mean circumference
where / is the number of pairs of poles.
(2) The iron to iron space, called the air gaps, though
mostly occupied by the winding, etc., on the armature. The
sectional area of this part is equivalent to the area of the
curved polar surface, increased by a fringe of 0*8 x the iron
to iron space. This increase is to take into consideration the
spreading of the lines round the edges, and was first pointed
out by the Drs. Hopkinson.
The length of this space is twice the iron to iron space.
(3) The magnet limbs, which may be wrought iron, cast iron,
or cast steel. Their sectional area is easily found from the
76 Examples in Electrical Engineering.
drawing, and may vary in different parts, as, for instance, when
the pole-pieces are of cast iron. The length is the same as
that of a line drawn slightly nearer the centre of the machine
than the centre of the limb, and bounded by the centre of the
polar face on the one hand, and by the yoke on the other,
taken twice over.
(4) The yoke, which is generally counted separately, either
on account of difference of material or difference in sectional
area. The area of section is easily found, whilst the length
is that of a curved line joining the extremities of the lines
representing the path through the magnet limbs.
Considerable experience and discretion is required to accu-
rately fix these lengths.
Using the following symbols to represent these various
measurements, viz.
4, A a , and /* a for the armature, through which a total flux of
N c.g.s. lines is passing, and in which, consequently, there
N
is a magnetic density of /3 a = -r-
**
/ g , A^ for the iron to iron space or air gaps. In these the
value of /A is i, and consequently need not appear; and
the magnetic flux is also N, as through the armature. The
N
density of lines in the gap is thus p ff =
A ff
l m , A m , p m for the magnet limbs, through which a greater
number of lines than N, viz. j/N, will flow, and conse-
Nv
quently the magnetic density is /? ?w = -r
/, Ay, and ^ for the yoke, in which again the total flux is Nv,
Nv
and the density p y = A -
A </
Thus the total M.M.F. in physical measure will be
_ W- , N/
And we have shown above that the ampere-turns can be
found by multiplying the physical measure by 0-8, whence
The Magnetic Circuit. 77
Ampere-turns ) N/ N/, Nv/,, t
^i i o o A ~r o o --T- -4- o o-r- + c
total 3 A,,u,, A,, A.,, a,,,
Or again, to save trouble in calculation, we can call the
N
o'S^ the value off (ft) in ampere-turns, and put
A/A
Ampere-turns total = / a /(&) + o-8/,& + l m f(ft m ) + //(&)
the 0-8 as a multiplier being retained in the case of the air
gaps, since for air we have not made out a table or curve for
reference, and consequently the 0-8 is not included there as it
is in the/(/?) for iron.
Example XLIII. How many ampere-turns will be required
to produce a magnetic flux of 300,000 lines in a wrought-iron
ring of 20 sq. cms. cross-sectional area, made of round bar
iron, and being 50 cms. outside diameter?
Solution. The value of
300,000
P** 20 - = i5,ooo
whence f(ft) = 22*88 ampere-turns per cm. length.
The length of the magnetic circuit is, nearly enough, the
mean circumference of the ring. From its sectional area of
20 sq. cms. the diameter of the bar is 5*046 cms., whence,
nearly enough, the mean diameter is 50 5*046 = 45 cms.,
say, and thus the circumference is
45 x TT = 141-35 cms.
whence
lf(ft) = 141 '35 X 22-88 = 3238 ampere-turns
Example XLIY. Neglecting the air path or leakage lines,
what will be the magnetizing force required to produce 3 million
lines through a mass of cast iron i metre long and 517 sq. cms.
sectional area ?
3 X i o 6
Solution. The value of (3 = * = 5802, and the magne-
tizing force in physical measure
78 Examples in Electrical Engineering.
corresponding to which value of /?, //, = 290, whence
TT s8oo x 100
H = = 2000
230
which is in physical measure, and will be converted into
ampere-turns by multiplying by = 0*8, or
Ampere-turns = o'8 x 2000 = 1600
Example XLY. What will be the addition to the ampere-
turns required for the iron ring in Example XLIII. if a saw
slot, i mm. thick, be made through the iron bar in a plane con-
taining the axis of the ring ? Neglect leakage lines.
Solution. There will be slightly less ampere-turns required
for the iron, in as far as the iron is less in length by i mm. or
o*i cm. As the magnetizing force is 22*88 per cm., we shall
now require 2-288 less magnetizing force as far as the iron is
concerned, viz. 2-288 ampere-turns less. But for the air gap,
where the whole 300,000 may be taken as uniformly distributed
over an area of 20 sq. cms., plus a fringe of 0*08 cm. all round,
that is over an area equal to that of a circle 5*046 + 0*16
= 5*206, say 5*2 cms. diameter, which has an area of 21*23
sq. cms.
Thus the density is
300,000
A = "27^3" = I4 ' 127 per sq * Cm>
and the ampere-turns required per cm. length will be
0*8 x 14,127 = 11,301. But as the gap is only 0*1 cm. long
the additional ampere-turns will be only 1130-1 ampere-turns
for air space.
Thus the difference will be an addition of
1130-1 2-28 = 1127*82 ampere-turns*
* Strictly speaking, owing to leakage lines caused by the increased
magnetic P.D. between the ends of the gap and points near to the gap,
there will be leakage of lines outside the area of gap taken, and these lines
will have to be accommodated in the iron part of the ring, thus making the
ampere-turns for the iron greater than before.
The Magnetic Circuit. 79
Example XLYI. What will be the ampere-turns required
to magnetize a dynamo which has a ring armature 25 cms.
diameter, with a i5-cm. hole, and built up of 1000 discs of
iron 0-25 mm. thick, if the total magnetic flux through it is
4 million lines? The diameter of the bore is 27 cms., its
length is 27 cms., and the angle of embrace is 135 degrees.
The length of the two magnet legs together is 124 cms., with
a sectional area of 415 sq. cms., that of the yoke being 54 cms.
and 729 sq. cms. respectively. Both magnet legs and yoke
are wrought iron, and the leakage coefficient v 1-4.
Solution. Using symbols previously mentioned, we have
A rt = (25 i5)(iooo X 0*25 X o'i) = 250 sq. cms.
and also
7T 25 + 15
4 = - X -- = IOTT - 31 5 cms. say
thus
4 X IO H
fc-isS- = l6 ' 000 g
whence the ampere-turns required for the armature will be
" = *J(P) = 4/(i6,ooo)
= 3i'5 X 39-84 - 1254-9
say 1255 ampere-turns for armature.
For the air gap we have
l g = 27 25 = 2 cms.
and as the angle of embrace is 135, the curved length of the
pole-piece is
,. X TT x 27 = 31 '8 cms.
The length parallel with the shaft is 27 cms., and thus the area
of the polar surface is
31-8 X 27 = 858-6 sq. cms.
to which must be added the fringe all round of o'8 times the
iron to iron space, or in this case of 0*8 cm. all round, or
0*8(2 x 27 + 2 x 31*8) = 94*08 sq. cms.
80 Examples in Electrical Engineering.
making
A <, = 858-6 -j- 94 = 9527, say, 953 sq. cms.
o 4 X io 6
p a - = 4200 say
953
and thus
whence the ampere-turns for gap =
(Og - o'8 x 2 x 4200 = 6720
Again, for the magnet legs we have the total number of line
Nv = i '4 X 4 X io 6 = 5-6 x io fi
and as A w = 415, we have
5-6 x io 6
corresponding to which /(/?) =10-9 nearly enough, and the
ampere-turns required will be
m m = mf(P) ==i24Xio'9 = i352 ampfere-turns say
Lastly, for the yoke there will be the same total flux,
5 '6 x io 6 , whence
for which
m = 2-12
and therefore ^ = 54x2-12 = 115 ampere-turns say. Thus
the whole magnetizing force will be in ampere-turns
Armature ... ... ... 1255
Air gaps ... ... ... 6720
Magnet legs ... ... ... 1352
Yoke ... ... ... 115
9442 ampere-turns total
Example XLYII. The length of magnetic circuit of a
transformer iron core is 50 cms. Its sectional area is 300
sq. cms., and the maximum value of the current allowed for
magnetization is 0*15 ampere. How many turns of wire must
The Magnetic Circuit. 81
the primary coil have if the total magnetic flux is 1,200,000
lines ?
Solution.
12 x io 5
Density, j3 = = 4000
for which /(/?) is i '24, and therefore the ampere-turns re-
quired
lf( = 5 X 1-24 - 62
whence, the current being 0-15, the total turns must be
62
w = = 413 say
0-15
CHAPTER VII.
MAGNET COIL WINDINGS.
WE have seen in the previous chapter how to ascertain the
magnetizing force required in ampere-turns, and will now pro-
ceed to show how the size of wire and number of turns or
convolutions may be found. There are two main cases
(1) That in which the current to be used is a fixture.
(2) That in which the P.D. available is fixed.
The first is simple, and can be easily disposed of. Let
ffi = the ampere-turns to be provided, and C the fixed current,
then the number of turns is, of course
The size of wire is then fixed, partly by the magnitude of the
current to be carried, and partly by the fall of potential allow-
able. Generally this last is so limited that the expenditure of
energy is so small as to make the heating negligible ; but in
any case the dimensions of the bobbin, or former, can be made
such as to satisfy Mr. Esson's rule, given on p. 56.
The coils which mostly come under this, the " series," class,
are those of ammeters, cut-outs, series-wound dynamos, series
coils on compound dynamos, and independent coils generally.
In case (2), which may be called the " shunt " class, we
have different conditions to take. A certain E.M.F., say E
volts, is to set up such a current that this current, multiplied
by the turns of wire, shall be equal to JW, the ampere-turns
required ; or
Cw = m
Magnet Coil Windings. 83
Since the whole resistance R will be
R = wr
where r is the resistance of the mean turn, we have
and
wr r
or
r= E
whence a simple rule :
Shunt Coil. The resistance of the mean turn is equal to the
P.D. at the terminals divided by the ampere-turns required ; or
is the ratio of the E.M.F. to M.M.F.
It is easy in most cases to fix or find the length of the mean
turn, say /, whence the sectional area of the wire will be
when p is the resistivity at the particular temperature to which
the coil is to be raised. The diameter of the wire, if circular
in section, is then
Or, putting in the previously found value for the resistance
of the mean turn, we have
when, of course, / and p must be in cms. or inches, according
as d is wanted in cms. or inches. Also p should have the
value corresponding to the temperature to which the coil is
allowed to rise.
To find the number of turns, w, we must again have recourse
84 Examples in Electrical Engineering.
to the formula of Mr. Esson, and make the total expenditure
of energy
E E
TT or -
R wr
such as to make no greater rise of temperature than may have
been previously agreed upon.
A little investigation will at once show that the number of
turns can be varied considerably without altering the value of
ffi when once the right size of wire has been found, provided
the length of the mean turn be not thereby altered.
Thus with only one turn the energy waste will be
E 2
r
and may be hopelessly large ; but with a suitable number of
turns, /, this waste will be reduced to
E 2
wr
or to times its first value. Since, however, the ampere-
w
turns are
Cw and C =
wr
we have in both cases that
E E
$&, = - x w =
rw r
whatever value w may have.
The number of turns it is possible to get on a coil is found
when the size of the coil is known, and the depth to which it
is possible to wind it. Thus let d^ stand for the diameter of
the covered wire (easily found for double cotton covering by
adding 0-02" to the diameter of the bare wire in inches). Let
L stand for the length of the bobbin or former which may be
Magnet Coil Windings. 85
occupied by wire, and D- the depth to which the winding may
be laid. Then there will be
D
each of
1J i
-r layers
#1
- turns
d
and
LD
W = ^
all quantities being, of course, expressed in the same units.
The value of ffll for dynamos, etc., is calculated as shown in
a previous chapter ; whilst fflt for ammeters, voltmeters, and
other small instruments, is usually found in the first instance by
experiment. Many ammeters and voltmeters of the magnetic
type are alike in size of parts whether ammeter or voltmeter,
being only different in the winding. Now, it is evident that,
with all the different ranges of reading required, it is not pos-
sible to provide exactly any particular number of ampere-turns
with any particular current or P.D. Thus all ammeters and
voltmeters of this class have some power of adjustment that
enables the nearest value of $$l to be made use of. In
ammeters it is not always possible to make the product
cw = the value of $t desired, and owing to the trade range in
sizes of wires for voltmeters, it is not always possible to find
that size in stock that will make the right value of resistance
of mean turn. Thus both ammeters and voltmeters are calcu-
lated for on the assumption that the value of J& is variable
within certain limits. In some instruments of the permanent
magnet class ^l ranges round 300 ampere-turns, whilst in some
of the electro-magnet type it may range round 600 ampere-
turns for ammeters, and about 400 ampere-turns for volt-
meters.
CHAPTER VIII.
ARMATURE REACTIONS.
So far we have only considered the simplest case of dynamo-
winding, and will now proceed to discuss the calculations for
different types of machines. In order to appreciate what it is
we have to allow for in a dynamo at full load, we shall have to
consider the effect of the current in the armature and the
magnetic circuit. In Fig. 15 is shown a section through a
FIG. 15.
dynamo, with the wire on the field magnets and armature
indicated by lines.
The direction in which the current flows in the field-magnet
winding is indicated by arrows. By applying Dr. Fleming's
rule for the direction of the E.M.F. in any wire on the
armature, we are able to mark these wires to show the direction
in which the current would flow if the brushes were con-
nected. Thus, on connecting the brushes, a number of things
will happen
Armature Reactions. 87
(1) Due to the resistance of the armature winding, we shall
get a P.D. at the brushes which will be less than the whole
generated E.M.F. by an amount which may be considerable at
full load, though hardly noticeable at light load.
Thus, calling the whole E.M.F. = E, the current in the
armature C a , and its resistance r a , we have that the value of the
P.D. at the brushes
e= E - C a r lt
(2) In addition to this loss of volts in the armature resist-
ance, we have to allow for the effect of the current in the
armature winding as a M.M.F. acting in the magnetic circuit.
Thus, with the brushes on the diameter of commutation, corre-
sponding with no load, as shown in Fig. 15, we see that the
armature winding and current produce a M.M.F., causing a
magnetic flux at right angles to the flux set up by the magnets,
and acting in a direction up the page, whilst the flux due to
the magnet winding acts from right to left, the resultant flux
through the armature will, of course, be a flux having a
direction somewhere in between these two directions, or
inclined to the horizontal line in the figure by an amount
which will depend upon the relative magnitudes of the two
M.M.F.'s.
Thus, if the armature current and number of windings be
very large, we shall have the resultant magnetic flux taking
a direction chiefly down the length of the page ; the dynamo
would then be a very bad one, and spark considerably. On
the other hand, if the M.M.F. due to the magnet-field winding
greatly predominates, we should have the resultant flux nearly
horizontal as before taking any current from the armature.
This state of affairs is that to be desired in all good dynamos
and motors where sparkless collection among other advantages
is wanted. In the best of machines, however, there is some
distortion, and it is this we have next to consider. When a
current is flowing in the armature the direction of the resultant
flux will be as shown in Fig. 16, where a dotted line carried
round the magnetic circuit indicates the path of the lines
of force. The brushes have been adjusted so as to bear upon
88
Examples in Electrical Engineering.
a diameter at right angles to the direction of flux through the
core ; that is upon the new diameter of commutation. Two
things will now be noticed
FIG. 16.
(1) The cross magnetic effect of the armature will now be
somewhat decreased, being now only that due to the wires on
the armature contained by the line ad or be; and
(2) The remainder of the winding on the armature, viz.
that contained by the line ab or de, is now a M.M.F. causing
a magnetic flux in a direction parallel to the flux caused by
the magnet winding, but in an opposite sense, or tending to
directly decrease its magnitude. Since the value of the whole
generated E.M.F. depends upon the value of the magnetic
flux through the armature, this E.M.F. will no longer be as
large as before a current was taken from the machine.
We must thus allow that the generated E.M.F. is now less
than E by an amount, say E,, and can write the terminal
P.D.
e=E- C a r a - E x
In a great many cases we are required to produce a machine
which will give a constant value of terminal P.D. e l at all loads,
and in order to do this it will be necessary to make allowance
for the reductions indicated in the above equation. In some
cases, however, it is desirable to make a correction also for
the loss along leads, at the distant end of which lamps are to
be kept at a constant P.D., and we may then add to the above
losses the loss along the leads, or Cr ( , making now
e = E - C a r a - Cr, - E,
Correction for these losses is commonly made by adding to
the M.M.F. of the magnet-winding by adding a series coil to
the shunt, thus compounding the machine. Now, such series
winding has the effect, if connected as "long shunt," of in-
creasing the effective resistance of the armature, and our
equation will now stand
e = E - C a r a - C a r m - Cr, - E,
where r m is the series-coil resistance. When, however, the
connections are made "short shunt," the series coil acts as
an increase in the resistance of the leads, and the resultant
reduction of terminal P.D. is slightly different, viz.
e = E - C a r a - Cr m - Cr t - E,
There is, however, a more important difference between the
two cases : whereas the P.D. at the terminals of the shunt
coil is practically constant in the case of "long shunt," it is
gradually increased from no load to full load with the " short
shunt" connections, by an amount equal to the loss in the
series coil, and thus the amount of series winding necessary
in the case of " short shunt " is not quite so great as in " long
shunt," when other things are equal.
We have, then, that the generated E.M.F.
E = e+ C a r a + C m r m + Cr, + E x
putting C m for the current in the series coil, which may be
either C or C a . Of these items we can have generally the
most accurate knowledge, save in the case of Ej. The exact
effect of the band of winding contained by the lines ab or dc
in Fig. 1 6, called the "demagnetizing band," is somewhat
difficult to predetermine. Its magnitude depends upon the
position of the brushes, and this depends upon the exact dis-
tortion produced by the cross magnetizing band. This distor-
tion can be found by plotting out on the drawing of the
9O Examples in Electrical Engineering.
machine the path of the lines for every value of the current
in the armature, and thus finding the diameter of commutation
for every value of the current. Considerable experience and
care are required for this, and it will be out of place to further
consider it here.
An example of the ascertainment of the value of the neces-
sary series winding, when all the above losses have been found,
will serve to sum up the matter.
A dynamo is required to give a terminal P.D. of e volts at
all values of current, from zero to C. amperes. In the first
place, at very small output the whole generated E.M.F. need
not be appreciably larger than e, and consequently a calcula-
tion must be made of the M.M.F. necessary to produce such
a value of N that the generated E.M.F. = ^, or
N = el0 *
nw
which will require a M.M.F. of $1 ampere-turns. The winding
to produce this must be the shunt coil. In the second place,
we have to allow that the losses at full load make the necessary
generated E.M.F. to be E volts, and thus necessitate a greater
value of magnetic flux, or
N, = -
TIW
to produce which the M.M.F. will be larger, and may be
called #fc.
Thus we have the additional M.M.F. required at full load
is
/&i J& ampere-turns
which must be supplied by the series coil in addition to that
required to balance the demagnetization. The series M.M.F.
is a product of the series turns into a current, which may be
that in the main circuit (short shunt), or may be that in the
armature (long shunt). In the latter event it will be noticed
that when compounding has been accurately carried out for all
the usual items, the P.D. at the terminals of the shunt coil is
a constant at all loads (except when compounding for loss in
Armature Reactions. 91
the leads). But in the short shunt combination the P.D. at the
shunt terminals will only be e volts at very small loads, and
will increase to e 4- Cr m volts at full load, thus increasing the
M.M.F. and making consequently less series winding neces-
sary. This alteration of M.M.F., due to the shunt coil, is of
very great importance in all cases where there is imperfect
compounding as any defect. Both over- or under-compound-
ing is much exaggerated.
A little consideration will show that it is hardly possible to
produce a perfectly constant P.D. at the terminals by compound
winding. For if we calculate the additional ampere-turns to
be supplied by the series coil in, say, two portions, viz. one for
half-load, and the other as a further addition for the other half-
load to full load, we shall find that the latter exceeds the
former. Thus, let the extra M.M.F. required for half-load be
J& ; this is due to an increase in the total flux necessary to
increase the generated E.M.F. by an amount equal to the sum
of a number of items, viz.
C a r a + C m r m + Cr t
which will be all nearly enough half the amounts for full load,
whilst the other item, E lf will be less than half, since it will
be due to a demagnetizing band of less turns than at full load,
and carrying only half the current. The items for the further
increase up to full load will all be higher than for the first
half-load, since the resistances will all be slightly higher, and
also the demagnetizing band will have increased in number
of turns as well as having been doubled in current ; or ^& 2 , the
increase of magneto-motive force for the second half of full
load, will be considerably greater than J&i. But both are
produced by the increasing current flowing in coils of fixed
number of turns, which will cause the additional ampere-turns
to be simply proportional to the current. And also the
relationship between the magnetic flux produced and the
ampere-turns producing it is such as to make the additional
effect of J& 2 actually less than ^ instead of greater.
It thus follows that if a dynamo is to give, say, e volts P.D.
at its terminals at a certain speed at no lead and also at full
92 Examples in Electrical Engineering.
load, it must give a P.D. which is greater than e at the inter-
mediate loads.
Consequently it is customary to fix upon two points, say
quarter-load and three-quarter-load, and so wind as to produce
the same P.D. at the terminals at these points, letting the
P.D. be slightly less than this at no load and full load, but
slightly higher at half-load.
In the above considerations it is sometimes necessary to ask
ourselves what effect in a composite magnetic circuit a given
increase or decrease of M.M.F. will have. It is very easy to
find what increase or decrease of M.M.F. a given increase or
decrease of magnetic flux will require ; but the converse is a
much more difficult problem, since the relationship between
all the items of the circuit is at once altered as soon as the
flux is altered.
ARMATURE REACTIONS IN A MOTOR.
FIG. 17.
If in the above diagram DD represent brushes bearing on
the commutator of a machine used as a dynamo, we shall have
the current flowing as shown by the arrow when the direction
of rotation is clock- wise, as indicated by the slope of the
brushes. In order to cause rotation in the clock-wise direction,
a pull will have to be supplied to the belt to overcome the drag
between the current in the wires under the pole-pieces and the
lines of force passing into and out of the armature. When,
Armature Reactions. 93
however, the diagram is taken to represent a motor supplied
with current to flow through the circuit in the direction as
shown in the figure, this drag will now be the moving force,
and the armature will revolve in a counter clock-wise direction,
to admit of which the brushes MM will have to be used
sloping in the opposite sense to the dynamo brushes, but
bearing upon the same point on the commutator, since the
nature of the magnetic forces acting in the magnetic circuit
is still unchanged. But as the armature is now revolving in
the opposite sense to what it was before the case, the E.M.F.
generated will also be opposite to that previously generated,
and will, consequently, be against the E.M.F. of the source
supplying the motor with power, which will thus have to
exceed that generated by the amount required for the resist-
ance of the armature coils.
It thus follows that, as in a dynamo so in a motor, the current
in the armature produces a demagnetizing effect upon the mag-
netic circuit. But whereas this demagnetizing effect is dis-
advantageous in a dynamo, it actually tends to aid constancy
of speed in a motor by reducing the generated E.M.F. for a
given speed somewhat in the proportion required by a constant
E.M.F. of supply.
The above diagram is due to Mr. Swinburne.
CHAPTER IX.
EFFICIENCY.
THE ratio of the useful expenditure of energy in a system to
the whole energy is called the " efficiency ; " but we have, in
the many and various cases to be considered, to sometimes
separate the parts of a system and consider them inde-
pendently, as will be seen in the examples taken.
As a simple case, we may take a single cell used to light a
single lamp. Let the total E.M.F. of the cell be E volts,
its internal resistance be r ohms, whilst that of the lamp is
R ohms. Then the current flowing through the circuit will
be
and the rate of doing work in the lamp is
E 2
C2R = ( RT7)* R
whilst the whole activity is
C 2 (R + r) -= ^Trf X (R + r)
the efficiency is then the ratio of that part of the energy con-
cerned in lighting the lamp to the whole activity, or is
(R +
_
(R + r) a
Or we may split our circuit up into items, and consider the
ratio of the useful energy in the lamps to that in the lamps
Efficiency. 95
plus that in the leads, thus getting the efficiency of distribution.
Or we may consider the ratio of the energy given out by the
cell to the whole energy produced by it, in which case we have
the electrical efficiency of the cell or other machine.
Or, again, we may include the energy required to run the
generator a dynamo, for example against the various sources
of friction, and thus state the efficiency as the ratio of the
energy delivered from the terminals to the energy put in at the
pulley, thus getting the mechanical efficiency, or commercial
efficiency.
In the case of motors, we can take the ratio of the power
at the pulley to the power supplied at the terminals, which is
again the commercial efficiency. On the other hand, we may
regard the motor as a machine for the conversion of electrical
energy into mechanical energy, whether useful or not, and so
obtain the electrical efficiency.
With secondary cells we may, on the one hand, regard the
cell as a store of quantity of electricity, and state the quantity
efficiency as the ratio of the coulombs got out to the coulombs
put in ; or we may consider the cell as a store of energy, and
state the energy efficiency as the ratio of the watt-hours got
out to the watt-hours put in.
Also, as a dynamo and engine will be required for charging,
we may have a combined efficiency of the whole apparatus,
called the plant efficiency, and being the ratio of the watts got
out to the power required to run the dynamo stated in watts.
I. Efficiency of Distribution and Transmission.
(a) In a simple parallel system, with all the lamps at the
ends of the leads, if r t = resistance of the leads, C the current
through the lamps, and e the P.D. at the lamps, we have
energy used in the lamps = eQ,
,, wasted in leads = CV ?
and the efficiency will be
eC e
cC + CV, - e 4- Cr t
which is the same thing as the ratio of the useful P.D. to the
total P.D.
g6 Examples in Electrical Engineering.
(b} Simple parallel system, with the lamps uniformly dis-
tributed along the length of the leads. Here the loss of volts
in the leads will be
C
C 2
making the energy waste in the leads = ;- z .
But the energy expended in the lamps will be
C x ( <? H r l \ as an average
where e is the P.D. at the far end; or may be put more con-
veniently in terms of the initial P.D., thus
C X (E -- r, J average
where the efficiency is
CE
(c) A parallel system with feeders. Let r f be the resistance
of the feeders, and r d be the resistance of the distributors
between two feeding centres. Then the total current will be,
C
say, C amperes in the feeders, or an average of - amperes in
4
the distributors.
The loss of energy in the feeders will then be C% whilst
that in the distributors will be
C 2
Tf*
If also E be the P.D. at the starting ends of the feeders, the
mean P.D. at the lamps will be
E - Cr, - -r d
4
whence the efficiency will be
CE
Efficiency. 97
Or, again, the ratio of average P.D. at lamps to P.D. at the
dynamo.
(d) Series circuit containing n items, each of r ohms resist-
ance, and having a constant current of C amperes. Calling r t
the resistance of the leads, the efficiency is
again the ratio of the P.D. at the lamps to the total P.D.
II. Efficiency of dynamos, motors, etc.
(a) Magneto-dynamo. Here the only loss is in the armature
and brush contacts, etc., the circuit being a simple series
system. Let e be the total P.D., and r a the resistance of the
armature, etc. ; then the output is eC, and the waste in the
armature CV rt , whence the
/-i
electrical efficiency = r . rT"
e\s -f- v 'm
which is also the ratio of the useful P.D. to the whole P.D.
From a theoretical point of view the efficiency of a magneto
may be higher than that of a dynamo since there is no waste
of energy to magnetize the field-magnets.
The other losses in the machine will consist of M the
mechanical friction, H the hysteresis loss, and F the eddy
currents. Calling all these W lt we have the
s-*l
mechanical efficiency =
eC + C 2 r tt + \V l
(b) Series dynamo. Calling as before e, C, and r a respectively
the terminal P.D. current and armature resistance, and r m the
resistance of the magnet coils, we have the
electrical efficiency = r .--; r .a7 r T" \
^ ~T~ ^ V a T I'm)
And again, if W l stands for all the other losses expressed in
watts, then the
mechanical efficiency
(c) Shunt dynamo. Using the same symbols as before,
H
98 Examples in Electrical Engineering.
and calling the resistance of the shunt coil r a we have the
output is <?C, but the current in the armature is greater than
/>
C by the amount taken by the shunt coil, viz. , the waste in
'i
the armature being consequently
e 1
Whilst the shunt-coil waste is , and the
r o
electrical efficiency =
And Wj having the same meaning as previously, the-
eC
mechanical efficiency =
W,
(d) Compound dynamo, short shunt. Here the output
is eC, as before. The current in r m is C amperes, but that in
r (l is C + shunt current, or
whence the various losses are C 2 r w in the series coil,
in the shunt coil, and
in the armature, making the electrical efficiency
eC
the mechanical efficiency being
eC
c/g-/^ H c
Efficiency. 99
(e) Compound dynamo, long shunt. The current in the
shunt is , and flows through both r m and ;, in addition to
the main-circuit current C ;
whence the
f-*
electrical efficiency =
and the
mechanical efficiency = 3
+ (r.+rj+w,
' S '
(f) A series motor. The electrical efficiency is the ratio of
that part of the energy supplied to the machine, which is
actually converted into mechanical work, whether useful or
waste, to the whole energy supplied. Calling E the P.D. of
supply, with C the current, and r a and r m having the values
above, we have that the back E.M.F. of rotation is
c = E - C(r B + r m )
and the whole rate of doing work mechanically is G ;
whence the
electrical efficiency = ~ = ~
The mechanical efficiency is
EC - Cfo + rj - W,
EC
or
EC
Since of the total rate of doing work mechanically the part W 1
being internal friction, etc,, is wasted ; M is direct mechanical
friction, H is molecular friction, but F is a load on the arma-
ture of exactly the same nature as a closed secondary on a
dynamotor armature.
ioo Examples in Electrical Engineering.
(g) Shunt motor. Calling E the P.D. at which a current
of C amperes is supplied, we have that of C amperes, the
p 1
shunt coil takes -, leaving the
E
current in the armature = C
r f
due to which the loss in r a is f C - J / volts
leaving the B.E.M.F = e = E-( C- E }r a
\ rj
whence the
electrical efficiency =
EC
and the mechanical efficiency
EC
(ti) Compound motor, short shunt. Calling EC the supply
as before, we have the shunt current is
r g
Whence the armature current is
r.
and the B.E.M.F.
_
e = &
r s
and therefore the electrical efficiency is
EC
whilst the mechanical efficiency
EC
Efficiency. IOI
(k) Compound motor, long shunt.
Here the shunt current is
and the B.E.M.F.
e=E-(c--J r m - (c - - J r a
and the electrical efficiency
EC
whilst the mechanical efficiency
EC
(/) Secondary battery. Calling the charging current C
amperes and the time of charging / hours, the corresponding
quantities for discharging being c and f lt we have the ampere
hours put in = ct and output = ^A, whence the
/,
quantity efficiency -
and may be as much as 100 per cent.
Considering, however, the fact that the B.E.M.F. of the cell
is higher on charging than the available E.M.F for discharging,
we have the energy put in is
Cfc
and that got out is C^ when e and e l are respectively the
E.M.F. to charge against a certain B.E.M.F. and resistance,
and the P.D. available for use at the terminals on discharging.
The
energy efficiency is = p V
(m) Plant efficiency. This must include all sources of loss
and give the ratio of
The useful work done in a given time
The energy supplied to the prime mover for the same time
CHAPTER X.
VOLTMETERS AND THEIR RESISTANCE COILS.
IN the following examples notice will be taken of the effect of
temperature on the readings of voltmeters of the magnetic class.
As was mentioned in the chapter dealing with the winding of
coils, there is no difference between an ammeter and a volt-
meter in point of shape, etc. There is, however, this im-
portant difference between them that, whilst the former has
its dial graduated in terms of the current passing through its
coil, the dial of the latter is marked in terms of the product
of the current through the coil into the resistance of that coil,
such resistance being taken at the temperature at which the
instrument was calibrated. Thus the readings of an ammeter
are independent of the value of its resistance, but those of a
voltmeter depend upon the resistance of its circuit, and will
only be correct when that resistance has the value it had when
the dial was marked. It is, therefore, the object of the instru-
ment-maker to so contrive matters that the resistance of the
voltmeter circuit shall vary as little as possible. Alteration in
temperature may be produced in two ways, viz. by external
means such as atmospheric variation, or variation due to
locality ; or by internal means such as the expenditure of
energy in the coil due to the current through it. We will call
the first of these the " temperature " error, and the second the
"heating" error. The employment of a material such as
manganin or constantan, will make both these errors quite
small, or even negligible as far as mere temperature variation
of resistance (T.V.R.) is concerned, though the actual heating
may be very large, and even dangerous. For the energy waste
Voltmeters and their Resistance Coils. 103
in a coil of a given size producing a given value of ampere-
turns will be directly proportional to the resistivity of the
material of the wire. When the resistivity is very high, it is
not easy to make an instrument coil of reasonable dimensions
without excessive heating and waste of energy, and owing to
the want of a material of low specific resistance and small or
negative T.V.R., we are obliged to make a compromise
between the two extremes, and wind that part of the instrument
which is to produce the magnetic effect with wire of small
resistivity in order to get large ampere-turns with small heating ;
and the rest with wire of small T.V.R., and having with
advantage a high resistivity. The two things to be aimed at
are, firstly, to get the requisite M.M.F. with small energy waste,
and then to get the necessary freedom from variation of
resistance by addition to the instrument resistance of a coil
having a constant, or fairly constant, value of resistance. This
is effected by making the working coil of the instrument of
copper wire to get the maximum M.M.F. with the least energy
waste, and by adding to the instrument so wound an extra
coil or resistance of, say platinoid or manganin, materials
having very small coefficients of T.V.R., and thus causing the
effective total resistance to remain nearer a constant then if
copper alone had been employed. The effect of so adding
to the resistance of the working coil is twofold ; firstly, the
current through the instrument due to a certain P.D. will be
decreased, and consequently the deflection for a given number
of volts will be less than before, or the total reading of the
instrument will be increased ; and, secondly, there is the ten-
dency to greater constancy in total resistance. The increase
of total reading power is obviously proportional to the increase
of resistance ; thus if the extra coil have a resistance equal to
that of the instrument's working coil, the instrument will read
twice as much ; or, more generally, if the extra coil be n times
the resistance of the working coil, the total reading power will
be (n -h i) times the original. The T.V.R. of the whole com-
bination is, on the other hand, inversely as the number of
times the total resistance is increased if the material of the
extra coil has no T.V.R. If the T.V.R. of the extra coil be
IO4 Examples in Electrical Engineering.
important, having a value c say, whilst the T.V.R. of the
working coil is b, and the extra coil has a resistance n times
that of the working coil then the combined T.V.R. will be
b+nc
a = ~~ per cent, per i C.
when b and c are stated in values per cent, per i C. The
plus sign is mostly the case, but when carbon or manganin
is employed for the extra resistance, then c is negative, and
by adjusting n so that
nc = I)
we shall obtain a system of perfectly constant resistance.
It thus follows that the value of an indication on a volt-
meter in volts is found by multiplying the reading as marked
on the dial by the ratio of the present resistance total to the
total resistance when calibrated, or
present resistance
volts = deflection x -r-=-
resistance when calibrated
The deflection being of course marked by a number which is
equal to the product of the resistance of the instrument when
calibrated into the current which a given P.D. produces
through the instrument.
The value of the present resistance depends upon the
temperature and the T.V.R.. values for which are given on
A, p. 209.
Thus let R T be the resistance at which the instrument was
calibrated (temperature, say T C.), and let R, be the resistance
at some other temperature / C., which may be either higher
or lower than T C. Let a be the T.V.R. per cent, per i C.
Then, nearly enough for practical purposes
when the difference t r may be zero, positive when t is
greater than T, or negative when t is less than r.
Similarly, the value of any indication on the voltmeter will
be always proportional to its resistance at the time of taking
Voltmeters and their Resistance Coils. 105
a reading, and if we call the value per one division k volts,
when the instrument is at the correct temperature we shall
then have that the value of one division at any other tempera-
ture will be
TOO
CHAPTER XL
ELECTRIC TRACTION, RAILWAYS, ETC.
THE application of the electric motor to tramways and other
tractive work renders it necessary to ascertain the horse-power
required for any particular case, and the data necessary for
this purpose may now be discussed. In order to draw a car
along a line, the mechanical pull required will depend upon
the following things
(1) The weight of the car.
(2) The condition of the line.
(3) The inclination of the line.
By the condition of the line is meant its smoothness and
freedom from dirt or other obstruction. In the best possible
cases of railway work the pull required to keep a ton weight
of car in motion on a level line may be less than 10 Ibs., but
in some cases, even with grooved rails, the pull may be 15 to
20 Ibs. in good cases, and even 40 Ibs. or more in cases where
dust or road grit has collected in the rails. Curves also cause
a local increase of resistance to traction.
In order to allow for the inclination of the line we have to
take into consideration the effect of gravity, which effect may
be against us in going up an incline, but which is for us in
going down an incline. It is seldom that the inclination is so
much as to render it necessary to distinguish between the
length of the actual distance gone and the true horizontal
distance, and consequently we may write the pull due to
gravity as being
2240W
Electric Traction, Railways, etc. 107
where n is the number of feet along the line for a rise or fall
of i foot, this pull being 4- or according as the inclination
is up or down. We shall thus have the pull required to move
a car of W tons weight along a line of i in n inclination with
a resistance to traction of x Ibs. per ton as being in pounds
taking the -f sign when going up hill and the when coming
down hill.
Again, the rate of doing work is proportional to the speed
of running, and we have the horse-power required will be the
pull multiplied by the distance gone through in feet per minute
and divided by 33,000. Thus, if S = the speed in miles per
hour, we have the horse-power is
H.P. = 88PS
33OOO
and this is the useful mechanical rate of doing work, and to it
must be added that required to run and excite the motor and
that required to transmit the power of the motor to the axle
of the car.
In general the conditions are so disadvantageous to the high-
speed motor, that in practice it is found that for every horse-
power put into the motor-terminals we may get out from 60
to 80 per cent, as useful work to run the car ; the lower figure
is frequently exceeded, but the higher can only be attained
under the best conditions, and with large motors. Calling
this factor , we have that the electrical power put into the
motors must be
or, in full
100 / 88SP
y
H.P. required =
V 33 )
which is converted into watts by multiplying by 746. Now,
this power is a product of volts and amperes, and it is simply
io8 Examples in Electrical Engineering.
a matter of convenience as to what the E.M.F. may be. The
motor is in any case arranged with its gearing so that a given
current through its armature means a given pull for traction,
and the value of this pull per ampere can usually be stated
or fixed. From this and the actual pull required we can
readily find the current required to move the car, and can add
that to overcome internal friction, etc. From the current and
the speed of running we can next find the useful rate of doing
work, and thence the B. E.M.F. of the motor. In the case
of shunt motors this B. E.M.F. will be practically constant,
being in any case proportional to the speed, and the cars will
run at a nearly constant maximum speed both up and down
hill, and on the level if supplied with a constant E.M.F. A
knowledge of the electrical data will then enable us to ascer-
tain the necessary increase of volts on account of internal
losses, and to state the exact value of the P.D. of the mains,
In some cases the tractive pull per ampere is not a constant,
since with series motors the drag on the wires is increased by
an increase of current both on account of that increase itself
and also on account of the consequent increase of magnetic
flux through the armature. When series motors are run off
mains at a constant P.D., we shall have their speeds varying
by an amount which is greater than in the case of shunt
motors, having a practically constant excitation. Thus, at
places on the line taking a heavy load, the extra current in
the magnet coils will cause the speed to go down, since other-
wise the B. E.M.F. of the motor would soon exceed that of
the supply mains, were that possible. The tractive force being
thus increased in a proportion more nearly as the square of
the current, makes the series motor particularly suitable for
overcoming steep gradients and other heavy loads with less
heating and waste of energy than would be the case with a
shunt motor, for the series motor will automatically reduce its
speed up hill, and by increasing its field strength take less
increase of current to produce the necessary tractive pull.
When getting up speed the car has to store an amount of
energy, equivalent in foot pounds to half its mass, multiplied
by the square of its velocity in feet per second. The extra
Electric Traction, Railways, etc. 109
pull required to do this will depend upon the time taken to
attain the speed, and we can write the work to be stored in
foot pounds, as
W x 2240 / S_x 5280 \' 2 = K
2g \ 3600 /
When g is taken as 32, we can simplify this to
K = WS 2 X 75 '3 foot pounds
where W = tons weight of car and S the speed to be got up in
miles per hour.
Now, if this store of energy is to be acquired in ;/z, minutes,
we must remember that the car in gradually getting up the
speed of S miles per hour will go through a distance of
Sx 5280
since its average speed during the time will be - if the
acceleration be uniform. And through this distance a pull
over and above that required to overcome the tractive resist-
ance will be required of
K
-j pounds
or the extra pull during the time of m, minutes, to get up
speed of S miles per hour will be
WS 2 X 75-3 WS
- ' D = x 1711 pounds
448^ m
Or if speed is to be attained in / seconds, the extra pull is
1027 WS
~ -- pounds
If also / be the pull per ampere, the current required will
then be
n .
amperes
1 10 Examples in Electrical Engineering.
to which must be added the extra current to get up speed,
or
r 102 7 WS
1 = -- lt> am P eres
The value of/ the pull per ampere may be nearly a constant
for all values of current in the case of a shunt motor ; but in
the case of a series motor it has more nearly the value
where n speed per minute and a and b are constants, de-
pending upon the size of the machine.
In questions of electric traction involving a knowledge of the
value of the pull produced by a given current in the armature,
or of finding the value of the current required to give a
certain pull, we must be careful to distinguish between the
shunt, compound, and series types of motors.
N
FIG. 18.
Shunt Motor. When the E.M.F of .supply is constant, we
may regard the magnetizing force, acting upon the magnetic
Electric Traction , Raihvays, etc. in
circuit as practically constant, subject to the fact that distance
from the generating station will mean a lowering of the P.D.
at the terminals of the shunt coil by the ohmic loss in the
leads. The pull or torque is always proportional to the
product of current and lines total ; and these last should be
constant in a shunt motor with constant P.D. at the terminals
were it not for the fact that armature reactions produce a
demagnetizing effect, and the curve of relationship between N
and C, the current through the machine will be of the nature
of the curve shown in Fig. 18, and consequently the pull per
ampere will be decreasing slightly as the current increases.
Thus at times of very large current, as at starting, the actual
pull may not be very large, and the maximum pull or torque
may never be great.
Compound Motor. Taking next the case of a motor com-
pounded for ordinary work, that is for constant speed with
varying load, we find that the effect of the compounding series
coil is to still further increase the demagnetizing effect of the
current, with the result that the curve of relationship between
N and C will be still more drooping at high values of C than
that shown in Fig. 18, with the effect of a still further
reduced pull under similar conditions.
But when the series-magnet winding is connected so as to
magnetize the circuit, we may have this effect made, either to
give a constant value of N for all values of C (in which case
there will always be a constant pull per ampere, whatever value
the current may have), or to actually produce an increase of
lines with an increase of current, making the curve of N C
rise for an increase of C, and thus giving a pull per ampere
increasing with the current. It will not be necessary to inves-
tigate this action in combination with the shunt coil, but to go
on to consider the effect of a series coil alone.
Series Motor, Here the value of N is constantly increasing
with the current, and would have a curve of relationship
similar to the /2H curve for a composite magnetic circuit, were
it not for the fact of demagnetization causing a tendency to
drop at high values of C. Taking the shape of NC, as shown
in Fig. 19, we may also plot a curve, showing the relationship
Xu\BR A Vy*X.
OF \
( UNIVERSITY J
112
Examples in Electrical Engineering.
between C and the product CN, as shown in the upper curve.
The pull exerted is directly proportional to the ordinate of the
second curve, and can be found from the nature of the winding
C
FIG. 19.
and gearing employed. Calling the total pull for tractive
effect,/ = 0CN x io~ 6 , where a is a constant depending upon
the type of machine, we have from a knowledge of the value of
p in pounds that the product
and a reference to the curve will give the value of C and N
corresponding with this product.
It is also clear that the pull will reach a maximum value,
and the performance of the motor is thereby limited.
CHAPTER XII.
ALTERNATING CURRENT CIRCUIT.
WHEN the current in a circuit changes periodically in direction,
passing through a set of cyclic changes of value, it is termed
an alternating current, and, owing to the disturbing influence of
the medium which surrounds the circuit, the usual relationship
known as Ohm's Law requires modification before it can
completely represent the true state of affairs. The manner of
change of value of the current from time to time is different
in different types of machinery, but is very frequently of a
nature simply expressed as a sine function of the time. Such
a current can be represented in all possible values by the
curve in Fig. 20, where ordinates above and below the line Of
represent the instantaneous values of the current in amperes
corresponding with any instant in time marked along Of.
Such instantaneous values can be represented by the ex-
pression
Q = C m sin *?t
where Q is the value at the instant in time /, C m the maximum
value to which the current ever attains, and T is the time in
seconds taken to go through one complete cycle, or change
from zero rising to a positive maximum, declining to zero,
reversing and rising to a negative maximum, again declining to
zero. The curve can be drawn from a table of natural sines
by dividing the horizontal line into 360 parts corresponding to
the 360 degrees in a complete revolution, and marking points
at a distance above or below the horizontal equal to C m times
the natural sine of the angle. The circular measure of 360
i
1 1 4 Examples in Electrical Engineering.
degrees is 2tr, and it thus follows that when t has such values
as to make - x 2?r equivalent to convenient angles, the curve
may be easily drawn in. When / = r the angle = 2?r = 360.
FIG. 20.
The time T is called the time of one complete cycle or period,
and ranges with different makers from -^ second to T ~ second,
though smaller values were used in past times. It has a very
general value in English practice of either -~ or ~Q second,
the value ^^ being American. It is thus customary to
describe the circuit as one having, for example, 100 periods
per second j or to say it has a frequency or periodicity of so
and so per second. Calling the frequency p periods per
second, we may put instead of the above expression
C t - C
sn
where / stands for -
Alternating Current Circuit. 115
Now, it will be quite evident that in some cases the rapidity
with which the current changes in value will materially affect
the conditions of working ; and it will be necessary to investi-
gate as fully as possible the effect of any such disturbance for
the different conditions possible. There are four ways in
which an electric current can become manifest to us
(1) Chemical.
(2) Thermal.
(3) Magnetic.
(4) Electrostatic.
And we will examine the effect produced for each of these
considered separately.
I. The Chemical Effect. With continuous or uni-directed
currents the chemical change produced is always proportional
to the strength of the current and the time during which it
flows. It is also dependent upon the direction of flow, being
of the nature of a solution where the current enters, and a
deposition where the current leaves the circuit. Thus, if a
current of one ampere be passed for the space of time of /
seconds through a solution of copper sulphate, it will dissolve
from the plate of copper at which it enters the solution, and
deposit upon the plate at which it leaves the solution, an
amount of copper which may be expressed as
/ x 0*00032959 gramme
Thus, during the time from o to A in Fig. 20, we have the
current flowing all in one direction; and if such current be
passed through a solution of copper sulphate, it will cause a
deposit of copper on one plate and a solution of copper from
the other plate, which will be in amount equal to the product
of the average strength of the current into the time, viz.
- seconds, and into the numeric used above (the electro-
chemical equivalent of copper).
But if the circuit be not broken at the instant when the
current becomes zero, the contrary effect will be produced, and
there will now be a re-solution 'of copper from the plate which
before received the deposit, and a deposition on the plate from
Ii6 Examples in Electrical Engineering.
which the copper was dissolved ; and this reverse effect will be
of exactly the same magnitude as the previous effect if the
current be maintained for the next or second half-period.
It thus becomes apparent that an alternating current cannot
produce any chemical effect ; for what is done during any half-
period is exactly undone during the next.
When, however, the current is " rectified " by a commutating
device, timed so as to exactly reverse the connections at the
instant the current changes in direction, we shall get a total
chemical effect which will be proportional to the product of
the time and the average value of the current, or is
C,,/o'ooo32959 grammes of copper
Now, it can be shown, by carefully measuring the area of
the half-wave from o to A, that its average height is 0*6366
times C m the maximum height, and we thus have that the
amount of copper deposited by an alternating current which
has been rectified is
0-6366 x C m X 0-00032959 x /grammes
and knowledge of the electro-chemical equivalent of any other
material will similarly give the sum-total of the effect pro-
duced, as, for instance, in charging a secondary battery.
II. Thermal Effect. This is, at any instant, proportional
to the rate of expenditure of energy, and is independent of the
direction of the current ; thus, it is proportional to
In order to find an expression for the average effect so pro-
duced, it is necessary to draw a curve which represents the
quantity Q 2 R, and to find the average height of such a curve.
Thus, if C m = 10 amperes, and R be i ohm, this curve of
energy expenditure will have zero values coinciding with the
zero values of current, and will rise to a maximum of 100 at
the time when the current is maximum. If now the length
corresponding to half a wave be divided up into 18 parts, and
the area of each such slice be taken, all such areas being added,
it will be found to amount to 899*6, when the width of each
slice is called unity, and its height taken as that of its middle
Alternating Current Circuit. 117
value as marked on the scale of the diagram. The average
height is consequently
899-6
IT = 49 ' 95
which would have come out = 50 exactly, if we had been able
to estimate values more exactly. Now, this 50 is the average
rate of expenditure of energy in the circuit, and consequently
is of the form
but we know R to be unity, whence it follows that the C 2 must
be = 50, or the effect produced is the same as would be the
case had a current of magnitude
4/50 = 7*071 amperes
been maintained constant for the corresponding time. But
this current is
0*7071 x the maximum
and is also the
>v/(mean square)
It is called the " virtual " value, and is really the value about
which we should know.
As far, then, as working in the circuit in heating it, the
alternating current which follows a sine law has an effect which
is the same as would be produced by a continuous current
having a magnitude of
0-7071 x C M
and lasting for the same length of time.
From this value the maximum can be found by multiplying
the virtual by U. T ^ ?T , or 1-414, or
max. C = C virtual x i'4 T 4
Before going on to consider the magnetic effect in detail, it
will be as well to point out that an electro-dynamometer will
be affected by the value of the square of the current at any
instant as with continuous currents, and will thus read a
quantity which is this Vrnean square, or virtual value.
1 1 8 Examples in Electrical Engineering.
It follows from this that the rate of expenditure of energy
in heating a circuit will be proportional to the product of the
resistance in ohms into the square of the electro-dynamometer
reading.
And also, when the current has been rectified, the electro-
dynamometer will still read the virtual value, or
0*7071 x maximum
But it has been shown that the chemical effect produced by
a rectified current is only proportional to
0-63 6 6 X maximum
whence it follows that the dynamometer will indicate a value
which is
, or I'll times too much
0-6366'
III. The Magnetic Effect. This may take place under
circumstances which may be summed up in the following
order, beginning with the condition where no magnetic effect
can be produced :
(a] No magnetic effect possible ; circuit non-inductive.
(F) Magnetic effect set up in a medium which has a uni-
formly constant permeability say air where /A = i and which
is also non-conducting.
(c) Medium having constant /x = i, but being conducting.
(d) Medium having variable value of /x, as in a closed iron
circuit. Non-conducting, i.e. laminated.
(<?) Medium with ^ variable but conducting ; closed circuit.
(/) Medium with /* variable, but of composite character, as
in the case of an open-iron circuit. Non-conducting.
() Medium with variable /x, composite and conducting.
For the sake of simplicity, capacity is supposed to be absent
in all the above cases.
(a) Non-inductive circuit. Here the only effect which is
possible has already been considered in the section on the
thermal effect. It will be as well, however, to complete that
case by considering the value of the E.M.F. in the circuit.
Alternating Current Circuit. 119
In order to cause a current of Q amperes to flow in a circuit
having a resistance of R ohms, we know that an E.M.F. of
E; = QR volts
will be necessary, and this being the only E.M.F. in the cir-
cuit, will necessarily be always proportional to the current, and
is then said to be in phase with the current, having its zero
values and the positive and negative signs at the times when
the current has these values and signs.
When this is the case, we can represent the rate of expendi-
ture of energy in the circuit by the product of the readings of
an electro-dynamometer and an electro-static voltmeter, both
of which instruments give the virtual or A/mean square values.
Thus
watts = C y 2 R = C E U
where C tf and E u stand for the virtual values, that is are 0*7071
times the maximum.
(b) Magnetic effect when the medium surrounding the cir-
cuit is, say air, having /x, = i and constant, and also being
non-conducting.
Consider three rings superposed, the middle one carrying
an alternating current. These rings are represented in section
in Fig. 21.
The middle ring has a current flowing in it of a kind repre-
sented in Fig. 20, and consequently rising in a certain direction
which we will call positive, when it is as indicated by the arrow
in Fig. 21, where the middle ring has then a polarity of N face
at the top and S face at the under side. Imagine the current
growing from o upwards ; as it grows lines of force will increase
in number, threading through the upper ring so as to enter its
section inwards from all points round it. An effect will conse-
quently be produced which will be of the same kind as if the
upper ring were being dropped upon the middle ring when
that is producing a constant magnetic flux. By Fleming's rule
it is easy to see that there will be an E.M.F. set up in the
upper ring, having a direction contrary to that in the middle
ring, or as shown by the arrows; and this E.M.F. will be
greatest where the rate of change of lines is greatest, and will
120 Examples in Electrical Engineering.
be least, or zero, when the rate of change of lines is zero.
Thus, it will be greatest at the moment the current in the
FIG. 21.
middle ring starts, and will be zero at the moment when that
current has attained the maximum value.
Alternating Current Circuit. 121
Again, when the maximum has been passed and the current
in the middle ring is now decreasing, there will be produced
in the upper ring an effect of the same kind as if that ring
were being withdrawn upwards from the field produced by
a constant current in the middle ring, and having a direction
still shown by the arrows in the middle ring. Fleming's rule
shows that such effect will be to generate in the upper ring an
E.M.F. contrary in direction to that indicated in Fig. 20, and
consequently similar in direction to that in the middle ring.
Also the magnitude of such E.M.F. will be least when the rate
of change of lines is least, and greatest when such change is
greatest; that is, it will be least at the moment when the
current in the middle ring is a maximum, and greatest when
the current there is on the point of change from positive to
negative.
A similar investigation will show that when the current in
the middle ring is reversed in direction, we shall get an
exactly similar state of affairs only in an opposite sense. And
to some scale or other we can represent the relationship
between the current in the middle ring and the E.M.F. set
up in the top ring, as also that in the lower ring (which will
obviously be of the same order), as is shown in Fig. 22,
where C represents the value and direction of the current in
the middle ring, and the dotted line marked e represents the
value (to some arbitrary scale) and direction of the E.M.F.
induced in the other rings. In consequence of its opposing
character, this E.M.F. is called a back E.M.F. ; and just as it
is induced in the upper and lower rings by the changing
current in the middle ring, so and much more, therefore, will
a similar E.M.F. be induced by the middle ring in itself
self-induced, and therefore called the
B. E.M.F. of self-induction
Leaving for the present the determination of the exact value
of such E.M.F., we have next to consider the conditions under
which such a current as C in Fig. 22 can flow in a circuit
having this magnetic property. It is quite clear that an
122
Examples in Electrical Engineering.
E.M.F. will be required at any instant freely acting in the
circuit in such a direction as to do two things
(1) Nullify any B.E.M.F. which may be present; and
(2) Leave sufficient over to send the current C against the
ohmic resistance of R ohms.
This last quantity, RQ, has been drawn lin above C in
Fig. 22, and all that remains to be done is to add, for all
instants in time, RQ to an amount equal and opposite to e to
fcV
^K^H
i y
\
/'
A
'
1
/ ^"
s
i / !
-
\
-
j
.-
/
/
j
P
1
/ \
/
/
s
\ /
::r j u '
^
s
\
1
/ i <
}/ [S
\
4 /^
I /
\
^
1
/j
< /
i
I
/
\
\
\
/ y
\
i
i 7.
.'
i \
i
//
\
H S-
i
"^
5 ~
. . /\ - -jyr .
\
*
/
r
/
jf '
,
1 \
i
L I
/
y
2 - -j-r-
>
/
S
i
/
4 / -
//
\
\
i
'/
/ f
\
1
\ \
/
,
^
}\
8 t- ^ 7
/
\
t
\
^ ^
^
\
\
/^
/
I
\
/
f *
X S
"^^
^
/
j
/
/
2o ^
'
i E
FIG. 22.
get the value of E, the E.M.F. which must be supplied to or
impressed upon the circuit. This has been done after the
manner indicated by Dr. Barfield in I886, 1 and the curve E
so produced is marked in full line. It will now be
noticed that this E curve does not coincide in phase with
the C curve which it produces ; in fact, the C curve lags after
E by an amount AB, which we will call the angle X. A
1 "Industries," vol. i. p. 17.
Alternating Current Circuit. 123
little consideration will show that this lag X will depend upon
the relative values of RC and e, since E partakes most in
character and phase with that one which is of greatest value.
Thus, when e is zero, as in a non-inductive circuit, E = RC,
and there is no lag. But when RC is very small and e very
large, then E is equal and opposite to e, and there is a lag of
practically X = 90.
Now, if we regard RC as a sine curve, it is clear that the
opposite of e must be a cosine curve ; and as E is the sum of
these two, we have that
E = RC, (i sin / + llw cos /
when RC, (l is the maximum value of RC and e 1)W the maximum
value of ej. Now, this can be shown to be equivalent to
E- V(R 2 C 2 TO + f lm ) sin (7* +
where X is an angle whose tangent is
Calling RC,,, = 'e the active E.M.F., we may write shortly that
the impressed E.M.F.
E = *Jr + 2 in magnitude
and has a phase-position which is A degrees earlier than the
current, X being such an angle that
e B.E.M.F.
tan X = - = : ^-TT-ET
e active E.M.F.
Thus, when the circuit is of such a nature that a very powerful
magnetic field can be set up by a small current, and the resist-
ance is also small, there is a considerable lag of the current
after the E.M.F., and such E.M.F. is very much greater than
that necessary to merely cause the current to flow in the ohmic
resistance of the circuit. But again, should the resistance of
the circuit be very large, so that the active E.M.F. RC
required is large compared to the B.E.M.F. set up by the
124 Examples in Electrical Engineering.
changing magnetism, then the current does not lag much, and
E is only slightly larger than that required for RC.
This can also be graphically illustrated by remembering that
the E.M.F. to be supplied, E, is the resultant of the other two
E.M.F.'s which are present, viz. e and e, which two are also at
right angles. Thus we can construct a figure which shall show
this resultant E.M.F., as is shown in Fig. 23, which shows the
FIG. 23.
magnitude of the sum of e and e, and also the nature of its
position. Considering the phase-position of e to be repre-
sented by the vertical line, and remembering that the current
C is in phase with e, it is clear from the three triangles in
Fig. 23 that the difference in phase, as indicated by the
departure from the vertical of line E = \f e 1 4- 2 , is least of
all when is small compared with e, is very marked when e
is comparable to e in value, whilst, however large may
become, the sum E = J(e 2 + e 2 ) can never be more inclined
to e than the horizontal position, or can never differ in phase
from e by more than 90.
Further investigation will also show that the rate of expen-
diture of energy in the circuit will depend upon the lag ; being
greatest when there is no lag, and least when the lag is large.
Alternating Current Circuit.
125
If the instantaneous values of the product of E and C be
taken for the three possible cases, viz. no lag, small lag, and
large lag, we shall get the waves of watts, as represented in
Figs. 24, 25, 26, where the areas enclosed by the dotted lines
E
1
w c
^
06 12
,, - f
S"
^
64 &
V
\
/v
\
/
\N
48 5
/
/
A
'/
\
\
.
3~
/
\ N
16 2
6 1
o o
a i
i6 e
24 a
M 4
40 5
48 6
66 7
6* 8
72 9
80 |0
8B II
96 12
104
,
/
.
v
\
,
\
.
-.
J
-
-
N V
-
-
J
I
. "7
_
-
-
-
^
\
~
Z
. (2.
\ v
/'
\
2
\
/,
N \
/
/
\
\
y
\'
//
V
/
x s
/,
X
^
^
^>
FIG. 24.
W will represent the work done. It will be noticed, in the case
of no lag, that all the work is positive that is done in the
circuit, and it can be shown that the average height of the
watts area is then
E V C V = watts
the voltmeter reading multiplied by the ammeter reading.
But in the case of small lag there is part of the area of work
marked on the lower part of the zero line and called negative ;
this must be understood as representing the case when the
circuit is not receiving energy, but is giving it back to the
engine, the resultant work being the difference between the +
126 Examples in Electrical Engineering.
and areas, an amount which can be shown to be equal as
an average value to
E t ,Q, x cos X = watts
Lastly, in the case of maximum lag = 90 the -f and
W RC
E
^
\
GO 12
\
,
'.
ICO 10
/
\
90 9
^
\
>
t
-
-
~
/
70 7
1
/
: s
/
>^
50 5
/
/
\
\
/
v
\
I
\
/
50 3
.
20 2
/
2
f
\
\
.
!
/
i
Q
\
\
/
v
\
j
/
y
20 2
.
j
s
/
\
V
/
/
4
J
50 5
\j
j
/
60 6
/
N
/
70 7
t
/
/
00 8
SO 9
N
S
K30 10
<
\
v^
^
^^
II
\
/
12
13
\
/
14
--,
/
.C.RC
FIG. 25.
areas are equal, and consequently the work done nil.
can also be represented as
watts = E B C y X cos X
This
where X = 90 and cos X = o. Such a condition is, of course,
not absolutely attainable, though in some cases the lag is very
nearly 90.
The coefficient cos X has been called by Dr. Fleming the
power-factor, and though it cannot be directly measured, it is
of great importance. As shown above, its value can be found
Alternating Current Circuit.
127
from a knowledge of the ratio of the back E.M.F. to the active
E.M.F., or of
e e
tan A =-, whence cos A =
>
The active E.M.F. is easily found for any circuit whose resist-
W Ec
45090
42585
40080
37575
35070
325 65
30060
27555
2505010
225 45 .3
20040 8
175357
150306
125255
100 20 4
75 "5 3
50 10 2
25 5 1
O
25 5 '
50 w 2
75 15 3
100 204
125 25 5
15030 6
175557
200408
225 45 9
2505010
275 55
30060
32565
35070
37575
40080
42585
450 90
" ?
\
/'
7
/
/
\
/
\
/
\
/
,-"
v,
t '
\
1
/
A
/
^
{
/
\
^
^
/
\
\
,
/
/
's.
\
I
\
i
/]
\
\
/
/
3
^
/
/
A
\
\
1
j t
*
\
^
A
\
~7 .
/
/
*
/
\
\
^
/
\
\
1
/
,'
1
\
/
'
\
\
S
:
;
^
^
}
/
I
\
/
"t
.
/
\
s
/
FIG. 26.
ance and current are known ; and we will now proceed to ascer-
tain .
The magnitude of the B.E.M.F. set tip in a circuit of
constant //, = i.
Consider an infinite solenoid having no magnetic core, and
being
d cms. = mean diameter, i.e. d is the diameter to the middle
of the depth of winding.
w = number of convolutions per cm. of length of coil.
128 Examples in Electrical Engineering.
The value of the magnetic density set up inside this solenoid
will be
IO
when C is the instantaneous value of the current in amperes.
The total number of lines in the plane section of the core at
right angles to the axis will be
__ 47T
N = Cw x -
10 4
and this will alternate in value and direction with the current,
having a magnitude at any instant which will be
47T d^TT . 27T
N ( . = C m w x -- sin /
10 4 r
Putting values to the quantities of
w = i turn per cm. length
d = 2 cms., and
C m = i ampere
then the maximum value of N will be
47T 2
-^ X i X i X i = 3*948 lines
and a curve can be drawn coinciding in phase with C, and
having a maximum value of 3*948, or
N, = 3*948 sin /
Now, the E.M.F. which a changing magnetic field can set
up is, in volts, equal to the number of lines gained or lost per
second multiplied by the number of turns or convolutions so
gaining or losing lines, divided by io 8 . As before shown, the
lines are changing at the greatest rate at the instant when they
are zero that is to say, the slope of the lines' curve is then a
maximum.
Alternating Current Circuit.
129
Let the periodic time T = j second, then the number of
lines corresponding to the instant in time 0-000027 second
after the zero value, will be
+N
N
(2 X VI
1-
Too"
= 3*948 sin 0-017452
X 0*000027
130 Examples in Electrical Engineering.
which angle is in circular measure, and must be multiplied by
iSc-
to get degrees, or
N, = 3-948 sin ( 0-017452 x - 1 ^
\ 7T
= 3-948 sin i
But sin i is 0*01745, whence the value of N f is
N; = 3-948 X 0*01745 = 0-0689 nne
In Fig. 27 is given a magnified part of the curve of lines
showing these values, which are taken for a very small part
of the curve, in order to approximate the more nearly to the
absolute slope at the instant. This slope is called, in the
language of the differential calculus
-TT- when </N and dt
at
stand for the infinitesimal values of the change in lines corre-
sponding to the change in time.
In our case the slope
^N _ 0-0689
dt ~ 0*000027
and we have that the value of the E.M.F. set up by this rate
of change of magnetism is
~dT u
= B VOUS
I0 8
whence, putting values as above, we have for the E.M.F. per
i cm. of length of infinite coil
0*0680
e = i X - ^r T 10 = 0-0000248 volt
0*000027
which is the maximum value of the B.E.M.F. curve, and, as
has been previously shown, has a negative value in accordance
with Lenz's Law.
Now the value
</N 0-0689
dt 0000027
. - 2480
Alternating Current Circuit. 131
is the maximum rate of change of lines, and can be found
from the maximum value of N = 3*948 by multiplying by -
01-628-32. Thus
maximum = 3*948 x 628 32 = 2480
and we can write the maximum value of the B.E.M.F. set up
in a coil, subject to a magnetic field changing after the manner
of a sine curve as
putting p =- - periods per second.
And the virtual value of this, as it is a sine curve, can be
found by multiplying 0-7071, or
0*7071 X
= '
_ i 443__m^ virtual volts per i cm. of length
In ascertaining the value of for any circuit, it is necessary to
know N, and this is generally the real difficulty. So far we
have considered an infinite solenoid ; but in practice we have
quite finite circuits to handle, and it is sometimes very difficult
to accurately state their dimensions. If we next consider a
solenoid of finite length, it will be at once perceived that the
conditions are now no longer so simple, though we shall have
the same value for the E.M.F., set up in the middle cm. length
of the coil, provided the ends be fairly remote. Thus in a
Finite Solenoid without magnetic core, we have that the
B.E.M.F. per cm. length, at and near the middle is
I0 8
132 Examples in Electrical Engineering.
but near the ends, owing to the straying of lines, the value will
not be so great. The coil will, in fact, act as though it had
a length shorter than its actual length by an amount which is
found to be very nearly equal to half the mean diameter. Thus
in Fig. 28 is represented in section a coil having a total length
FIG. 28.
/ cms., a mean diameter of d cms., and let us say of w turns
per cm. length. If this coil were part of an infinite solenoid of
similar construction, it would have a B.E.M.F. per cm. of its
length of
10
-^-^ volts
or
4-443 C m
io
volts virtual
per cm. length, or a value of
io c
total. But as it is finite, its B.E.M.F. will approximately be
io
. ~
( / -- )
\ 2 /
volts virtual
which comes to saying that the effective number of lines set
Alternating Current Circuit. 133
miy u,
sectional area, but is only
up in this coil is not uniformly C M w per sq. cm. of cross-
Cll ,, x f
per sq. cm. average. Nevertheless the whole M.M.F. in
ampere-turns is at the maximum crest
jft = C^-wl
10
whence it follows that part only of this is concerned in causing
the flux through the cylindrical air core, the rest obviously
being required to send the lines through the surrounding air
space.
Now, the reluctance of the air core can be expressed as
K ;:--= l = 4/ -
e tf/A
.
* 4^ 2
' nw ~
and to send
through this reluctance will require a M.M.F. in ampere-turns
of
-- X N x K
47T
or
IO ^"7T_ 4?T 2 4/ N
- X C m w -, X ,., ampere-turns
4?r 4 10 / d"jr
being unity ; which may be simplified to
/ j ampere-turns
thus leaving C m w- ampere-turns to overcome the reluctance of
134 Examples in Electrical Engineering.
the air space surrounding the coil : and as the same number
of lines, viz.
C,,,- lines
4 10 /
traverse this air space, it follows that the reluctance of it
will be
C w-
_ ampere-turns __ "* 2 _
0-8 x lines " """ ~d
10 d-ir^ ATT 2
X C,,,--w 7
477-4 10 /
i 2 73/
= ,/ j _ ,v = reluctance of the air space
This expression is not often required in considering the
behaviour of an air-core coil, but will be required when deal-
ing with a composite core.
We have next to consider the effect produced when the
medium surrounding the coil, including the coil itself, is more
or less conducting.
(c) Air core, /x constant and unity, but medium conducting.
These conditions will be fulfilled by the coils in Fig. 28, if
some or all of them be individually joined so as to make
complete electric circuits.
We have seen that a given impressed E.M.F.
where ej stands for the equal and opposite of the B.E.M.F.,
which is produced by the current C. It follows from this that
the current which a given E.M.F. impressed on a circuit will
produce must be sufficiently large to provide the necessary
maximum number of lines to produce such a value of e as will
leave so much E.M.F. over as is required to send C amperes
against the ohmic resistance R. In circuits as generally met
with in transformers, impedance coils, and such like, the value
of e is usually small compared to e, and it is now necessary to
Alternating Current Circuit.
135
consider the effect which will be produced by variations in
the magnitude of e for a given value of E. Thus, let
E = loo virtual volts
e - CR, and R be, say i, whilst C for some cause or other
is made to range from i to 20
We have, then, that
and will be as shown in the following table :
100
I
99'995
100
2
99*979
100
3
99 '954
IOO
4
99-919
100
99-874
IOO
6
99-819
IOO
7
99754
IOO
8
99-679
IOO
10
99-498
IOO
15
98-868
IOO
20
97*979
which shows that e may range from i per cent, to 10 per cent,
of E, without affecting the value of d produced by more than
\ per cent., and may even be 20 per cent, of E, with only
a trifle more than a corresponding reduction in e, of 2 per cent.
It consequently follows that the number of lines produced
by an alternating P.D. applied to a coil will depend almost
entirely upon the P.D., and will be almost independent of the
nature of the core or medium surrounding the coil, even when
that medium is of such a nature as to make the current
required to magnetize so large as to represent a P.D. required
for the ohmic resistance of as much as 20 per cent, of the
impressed P.D. In other words, this comes to saying that the
constancy of the magnetic field produced is, as it were, the
sole concern of the impressed E.M.F. Should anything arise
tending to undo the magnetic effect first set up, the result will
be such an adjustment of the current with respect to the
E.M.F. as to make the magnetic flux produced, and conse-
136 Examples in Electrical Engineering.
quently the B.E.M.F. still the same as at first, subject only to
the restrictions indicated in the table on p. 135.
This important fact was first pointed out by the author in a
course of lectures in the summer term at the Municipal
Technical School, Manchester, in 1892, and was afterwards
confirmed by some experiments mentioned by Prof. S. P.
Thomson in a paper before the Physical Society in 1894.
Now, when the rings in Fig. 21 are considered to be closed,
the E.M.F. set up in them can cause a current to flow, which
current will obviously disturb the magnetic field set up by the
middle coil unless, as indicated above, the current in that ring
is altered in such a manner as to exactly balance such dis-
turbance, and still leave sufficient to produce the original mag-
netic flux. Foucault currents or eddies set up in the mass of
the copper coils, in the iron core, and in any other conducting
body within the influence of the magnetic field are of this
nature, as is also such a state as current flowing in one of the
closed rings, which may thus represent a loaded secondary
coil. To exactly balance the effect produced, there will have
to flow in the middle ring a current which will be magnetically
equal and opposite to these " secondary " currents. Such a
current may be called the balancing current C 6 , and when the
current to be balanced does not lag after its own E.M.F. will
come in opposite phase to the induced B.E.M.F., as is indi-
cated in Fig. 29, where C m stands for the current required to
produce the magnetic flux before any "secondary" load came on,
whilst C 6 is such addition to C w as will nullify the " secondary"
load. The curve marked C p is the sum of C w and C 6 , and
may be taken as the current produced by the given E.M.F.,
acting in a circuit containing a given B.E.M.F. and secondary
load.
The full consideration of such an effect will be deferred
till the question of regulation in a transformer presents itself.
Briefly we may sum up by saying that the effect of a
conducting property in the system will only be to cause an
alteration in magnitude and phase of the current, and affecting
only in a very small degree the value of the magnetic dis-
turbance set up.
f CTNIVERSIT
\rcmt. Br U^S*
Alternating Current Circuit
And also that the number of lines produced in a given coil
by a fixed alternating P.D. will be practically the same,
FIG. 29.
whether that coil has a core of air or of the best possible
laminated iron.
(d) We have now to consider the case for a coil surrounded
by a medium, such as iron, having a high but variable value
of //,, and for simplicity so well laminated as to be non-
conducting.
In iron of even the very best and softest quality it is found
that the relationship between the magnetizing force and the
flux produced is not of a simple nature ; and further, that
when the magnetizing force changes periodically, the flux
changes periodically, but after a different law. In Fig. 30 is
shown this relationship, the enclosed area being a measure of
the work required to overcome the molecular friction in turning
the particles of iron through a complete circle. Professor
Ewing has published tests of a large number of samples of iron,
and from his curves tables have been read off, showing the
138
Examples in Electrical Engineering.
values of H required for numbers of points round each cycle
for different maximum values of /?.
Corresponding with these a table is given, showing the waste
/Q
P
^
/
3600
^.
^-
OA/V)
^,
^
1
3200
^
o/V\n
/
/
PAftO
/
1
2600
/
^
/
1
POOO
/
>
1300
/
/
1600
f
/
1400
i
1200
t
/
1000
1
/
pnrv
/
600
/
/
/
i
/
o
-
HH
I
/
i
/
1
1
ICOO
I
1
1400
i
/
I
/
/
/
/
f
1
^
/
/
/
?
i
s^
'
a/V)
/
^
3AOO
_.
^^
'
4000
4200
^
=:
,
s-\
/3
/
FIG.
in friction in ergs per cubic centimetre per cycle, called h as
before, for dynamo calculations.
Now, the nature of the motion of the iron particles will
depend upon the change from moment to moment of the
Alternating Current Circidt. 139
E.M.F. acting in the circuit, and when this is of the simple
sine curve nature, it is found that the value of j3 also changes
after the manner of a sine curve. This is shown by the fact
that the shape of the secondary curve of E.M.F. is practically
a sine curve. Working backwards it is then easy, as was first
pointed out by Mr. Evershed, to plot the corresponding values
of current required when the curve of lines has been drawn.
The cycle of magnetism in Fig. 30 is supposed to be that
found when the iron has settled down to a steady state, and it
will thus be seen that the manner of change is such as would
be met with in going round the figure in a counter-clockwise
direction. It thus becomes apparent that, owing to hysteresis,
when (3 is zero, but becoming of positive magnitude, then C
has to be of quite a considerable value, and does not increase
nearly in accordance with the rise of /3 from o to maximum.
Also that when the maximum /? has been reached (which
always coincides with the maximum value of the current), the
current declines in value much more quickly than ft having
even to become of opposite sign before /? has again become
zero. The reverse process takes place for the negative half
wave of ft and a complete wave is shown in Fig. 31, where the
current curve C is drawn to a larger scale than ft for con-
venience. It will be noticed
(1) That the value of C is very small, owing to the high
value of fji, which is thousands of times what it would be in air.
(2) That the C curve does not coincide with ft though, of
course, e will have its customary position or lag 90 after ft
Consequently, when the B.E.M.F. is the major item in deter-
mining E, E will still come in nearly opposite phase to e, and
consequently C will not lag so much after E as would have
been the case with a similar air core.
This decreased lag is of great importance, as, together with
the altered shape of the current curve, it determines the supply
of energy which is wasted in hysteresis, in addition to that
necessary for C 2 R.
Now, we can easily draw in such a curve as that shown, and
know its maximum value. But as indicated on an electro-
dynamometer we shall of course get its virtual value, and this
140 Examples in Electrical Engineering.
will no longer be 07071 x maximum value. Thus, in such
cases as impedance coils made with a closed iron core, and
where the current is made fairly large by making the value of
FIG. 31.
w small, this must be remembered. Owing mainly to the
irregular shape of the C curve in such a circuit it is general
to make impedance coils with an open iron circuit, the ad-
vantage of which will be indicated in the section devoted to
the consideration of the open iron circuit.
(e) Consideration of the closed iron circuit when conducting,
or having eddy current in the iron core, or a secondary load.
The nature of the disturbance set up by Foucault currents, or
a closed secondary will be of precisely the same character as
in the other case considered ; it is only necessary to point out
that, as in general the magnitude of C m , as we will call the
current required to produce the flux in the iron, is very small,
and lags little after E, whilst the current to balance the
Foucault losses has a magnitude which is also small, but which
is practically in phase with the impressed E.M.F., their sum
is consequently larger than either, and lags less than C m , whilst
when there is a secondary load to balance there is practically
Alternating Current Circuit. 141
no lag of the current after E, for even a small value of the
necessary balancing current.
(f) Medium composite in character, as, for instance, in the
case of an open iron circuit impedance coil. Here, as in the
composite circuit of a dynamo, calculation has to be made for
the several items. Thus for the straight iron core of length /
and sectional area a, to be magnetized to a total of N lines,
we have with a given winding of coil surrounding it, that
the M.M.F. required will be
Co// =
a reference to the tables being necessary to find the values of
C w corresponding to various values of /?.
In addition to this there will be required a current to send
the same lines through the air space. This may be found by
considering the reluctance of the air space, as given in the
formula
M I>2 73^
d(2l - d)
where / and d now must be considered to have the values for
the iron core. To force the lines through this space additional
ampere-turns will be required of magnitude
or
Now, that part of the current C, which is required for the
iron core, will be, as indicated, a saw-tooth curve like that in
Fig. 31, will lag somewhat after E, but not nearly to 90, and
will also be small in magnitude compared to Q, which will be
a sine curve, will be large, and will lag practically 90 after E.
The total magnetizing current will consequently lag much, will
partake mostly of the sine curve shape of Q and will, on the
whole, be very large compared to the case of a closed iron
circuit. An example will render this clear. Let the iron core
142 Examples in Electrical Engineering.
(laminated perfectly) be 30 cms. long and 5 cms. in diameter.
Let the value of (3 be 5000 total, as an average of maximum
value throughout the whole length. Let there be 4 turns per
cm. length of core. Then
C#// = {f (13) = C X 4 X 30 = 30 x 1-41 = 0-35 ampere
maximum value of the saw-tooth shaped curve for the iron
core. The total number of lines produced will be a maximum
of
^ X ^ X 2 2
N = 5000 x - - - = 98,200
And the reluctance of the air space will be
1-273x30 1-273 X3Q_
Whence the ampere-turns required will be
dwt = 0-8 x 0-1388 X 98,200
0-8 x 0-1388 x 98,200
and Cj = - - = 90-86 amperes
4 X 3
maximum value.
Now, the 0-35 amperes required for the iron core will lag
after E considerably less than 90; but the 90*86 amperes, which
is a sine curve, will lag 90, and as its magnitude is so large
compared to that for the iron, we shall get the real magnetizing
current required, being practically a sine curve and lagging all
but 90 after E, whilst its maximum value will only be slightly
less than 91 amperes. Indeed, we may say that the iron
core has nullified the reluctance of the core, and all we have
to consider is the outer air space. Had there been no iron
core the reluctance of the air core would have been
and the current required would have been nearly 1000 amperes,
It is for this reason, amongst others, that open iron circuit
impedance coils are generally used instead of those with a
closed iron core. The advantages are
Alternating Current Circuit. 143
(1) Current curve is practically a sine curve.
(2) Easy shape to wind and adjust.
(3) Small quantity of iron required.
The Hedgehog transformer of Mr. Swinburne is an example
designed to obtain a large all-day efficiency. This will be
more fully dealt with in the section on transformers.
1
9
8
7
6
5
*s
V .4
f:
i
/
^
I
^
II
/t
?*
/
, -
~~-
~~~~
I
^-
^
/
^
^
^\
<A
/
^
^
/
/
/
/
/
/
/
/
/
/
/
jFractton& of ful S>econ,(ta,ry L octet,
FIG. 32.
(g) When the medium of an open-iron circuit coil is conduct-
ing, as when imperfectly laminated or when containing a
closed secondary circuit, we have the same balancing currents
required. Only here such currents are no longer the only ones
of any magnitude, and their effects in determining the total
current are considerably reduced by the already large mag-
netizing current. It thus follows that even at very large
144 Examples in Electrical Engineering.
values of the secondary load there is still a very considerable
lag of C after impressed E.M.F. In some curves, showing the
relationship between the power factor and various loads for
closed-iron and open-iron circuit transformers, published by
Dr. Fleming in 1892, this great difference is very apparent.
For convenience of reference they are here reproduced in
Fig. 32, where the relationship is shown for two closed-iron
transformers, viz. the Westinghouse and Kapp ; and for one
open-iron transformer, the Hedgehog.
CHAPTER XIII.
IMPEDANCE COILS AND TRANSFORMERS.
Impedance Coils, or choking coils, as they are sometimes
called, are pieces of apparatus designed to make use of the
property of self-induced E.M.F., by introducing such an E.M.F.
into a circuit acting so as to decrease the active E.M.F. with-
out at the same time absorbing any power except that inci-
dental to the properties of the materials employed. Thus, we
may have a pair of mains at an alternating P.D. of 100 (virtual)
volts, and it is desired to connect to these a lamp taking a
current of 10 amperes at a P.D. of 40 volts. With a continuous
current circuit one way is to introduce a resistance in series
with the lamp of 6 ohms, but then the energy wasted in this
resistance will be at the rate of 600 watts. Or a number of
secondary cells may be put in circuit, having together an
E.M.F. acting against that of the mains of about 60 volts.
In this case there will be a storage of the energy not required
by the lamp, only a small part being wasted in heating, and
other losses peculiar to secondary batteries. With the alterna-
ting current it is equally possible to introduce a dead resist-
ance of 6 ohms, and waste the same 600 watts ; but we can
introduce a similar B. E.M.F., as in the case of the secondary
battery, the problem being to know how much ? Remembering,
then, that this B. E.M.F. is to be produced by the passage of
the current through an inductive circuit, we have that
or
e = V(E 2 -"?), or d = V(E 2 -?)
146 Examples in Electrical Engineering.
and inserting values in this last, we have
e 1 = >v/(ioo X 100 40 X 40)
= ^(8400) = 91-65
This value is, however, slightly too large, since the very coil
which is to produce it will have a resistance, and so absorb
a small E.M.F. in becoming heated. In fact the impedance
coil is also a resistance coil, and the effect of its own resistance
should be allowed for as a resistance in series with the external
circuit This resistance of the coil is usually very small, com-
pared to the resistance of the rest of the circuit, in order that
the power wasted may be small.
Calling *i the active E.M.F. required by the coil itself, we
then have
e 1= VJE'-'^T^?}
and all these values may be in either virtual or maximum
values, as we please, provided they are all either the one or
the other.
Transformers. An unloaded transformer, that is with
open secondary circuit, is simply a case of an impedance
coil designed, as a rule, to take the minimum current and
produce a B. E.M.F. differing from E the impressed volts by
as small an amount as possible. In as far as, then, as it is
unloaded, it can be dealt with in precisely the same manner
as the impedance coils discussed above.
In allowing for the effects of eddy currents it has been
pointed out that the disturbing effect produced by them had
to be balanced (automatically) by the increase of the main
current by an amount such as would produce ampere-turns
magnetizing force sufficient for the purpose of overcoming the
demagnetizing effect due to the eddy currents. A little con-
sideration will show that the current taken from the secondary
coil of a transformer will be of an exactly similar nature
and will similarly have to be balanced by a corresponding
increase of current in the primary. The consequence of this
is that the current in the primary coil is not only variable in
magnitude but is variable in phase.
Impedance Coils and Transformers. 147
As with impedance coils, there are two cases : the closed iron
circuit and the open iron circuit.
Consider a simple type of closed iron circuit transformer
wound so as to have the same number of turns on the primary,
as on the secondary.
The current required on open secondary circuit will be of a
kind represented in Fig. 31, including the two items
(1) True magnetizing current ; and,
(2) Balance for eddy currents.
The sum of these two will be a curve partaking most largely
of the nature of the true magnetizing current, since the current
to balance eddies is small and also lags due to self-induction
in the core.
The phase position of this no-load current, which we will
call C , will be after the primary E.M.F. by an amount less
than 90 on account of the hysteresis, but will be very small
in magnitude.
Of the two items of C the magnetic effect of the balance
for eddy currents is nullified by the eddy currents, thus leaving
only C m , the true magnetizing current to be looked upon as
causing so many ampere-turns to act upon the magnetic cir-
cuit, setting up an alternating magnetic field in the core which
produces the E.M.F. in the secondary as an E.M.F. coinciding
with the B.E.M.F. in the primary, and as we have already seen
this will have a phase position approximately 180 after the
primary E.M.F.
So far as the secondary is on open circuit, we get no effect
due to its presence except in so far as that presence has pro-
bably increased the eddy current loss somewhat. But when
the secondary is connected, as is most usual, to anon-inductive
circuit, we get a current flowing in it which is in phase with
its own E.M.F., but in contrary phase to the primary E.M.F.
This current will, of course, mean so many ampere-turns acting
upon the magnetic circuit, and these would produce a very
serious disturbance were it not for their effect upon the momen-
tary value of the B.E.M.F., causing the primary current to
increase in magnitude and lag less. This increase is of
exactly such a nature as would be produced by an additional
148 Examples in Electrical Engineering.
current in the primary acting in contrary phase to the secon-
dary current, and of such a magnitude as to produce ampere-
turns exactly balancing the magnetic disturbance set up by
the secondary current.
The real new primary current will, of course, be the sum of
this balancing current C 6 , and the original no-load current C ,
which, it must be remembered, is itself the sum of two items
previously stated. In the case of a transformer without iron
core and with no eddy current waste, there will then be only
two currents to consider, viz., the current C w to set up the
necessary magnetic flux, a current lagging almost 90 after the
primary E.M.F.; and the balance for secondary load, a
current which is in phase with the primary E.M.F., and thus
90 apart from C OT . Calling the latter Q, we shall have that
the current at any time in the primary coil will be
C X
and its phase position will be after the primary E.M.F. by an
angle whose tangent is approximately W 5 , which shows that
^6
when C 6 is large compared to C Hl the lag of C^ after the
primary E.M.F. may be almost nothing. In the case of the
closed iron circuit type this is most marked ; for even at as
small a value of C 5 as corresponds to only one-tenth of the full
secondary load, we find the ratio of C m to C 6 so small as to
make no appreciable lag of C p after primary E.M.F.
But in the case of the no iron core or open iron circuit
transformer, cm. is so large as to prevent the ratio of C m to C 4
ever becoming very small, even with a value of C 6 corre-
sponding to full secondary load. In the case of the open iron
core we have C^ being composed of the sum of several items,
viz.:
(*) C^f to magnetize the iron core : this is small in magni-
tude, is saw-toothed in shape, and lags less than 90 after the
primary E.M.F.
( 2 ) C wa to magnetize the air space : this is large, is prac-
tically sinusoidal, and lags almost 90 after the primary E.M.F.
(3) C& to balance the effects of secondary load and any
Impedance Coils and Transformers. 149
eddy current waste there may be in the whole apparatus :
this is practically sinusoidal, and lags hardly at all after the
primary E.M.F.
The sum of all these is mainly a sine curve, even when C& is
quite small ; and its phase position is mainly determined by
the relationship of C MMt to Q. But as C 6 never attains such a
size compared to C mtt as to make
Q
"H 22 = < o'3 <> r so
^6
it follows that there is always considerable lag of C p after the
primary E.M.F.
This is strikingly shown in the curves reproduced from Dr.
Fleming's paper on Transformers, on p. 143, Fig. 32.
In transformer design care must be taken that the loss of
volts in resistance is very small \ that the loss of power in
hysteresis and eddy currents is small; and that, above all,
these last are small, inasmuch as they are a constant load and
so very materially affect the efficiency when taken over a
period of time during which the load varies considerably.
The various losses may be thus stated
(i) Drop of volts in the primary coil, causing the B. E.M.F.
e, which is a measure of the magnetic flux set up, to differ from
E the impressed E.M.F. This loss is such as to cause the
E.M.F. producing power of the transformer to be smaller
than E, by the amount CpRp taken in its proper phase, and
making the ratio of ~ slightly less than unity. It is, in the
closed iron circuit type, very small at no load, since, then, C p
is very small ; and it only reaches a value of importance when
the secondary is on full load. But in the open iron circuit
type there is not nearly so large a variation in the value of C p
for various secondary loads, and so not as great an alteration
in the primary loss of volts.
(2) The drop of volts due to ohmic resistance of the
secondary, viz. QR,, is always proportional to the secondary
load in all types, and can be prearranged to have any value
desired.
150 Examples in Electrical Engineering.
(3) A drop in the secondary due to the slightly inductive
character of the winding. This will increase with the load
being such as to make the effective E.M.F. in the secondary
less than
where w s and w p stand for the number of secondary and
primary turns, in as far as the secondary coil acts in the same
manner as an impedance coil in its own circuit. For this
reason the number of windings w, should be kept small.
(4) There is also a drop in E.M.F. on full load, due to
leakage of magnetic lines. It has already been shown that a
fully loaded transformer has two oppositely acting magnetizing
forces in the circuit, and if these be not evenly distributed
round the magnetic circuit the consequence is evident polarity,
followed by inequality of the magnetic flux through the coils.
In a good closed iron type with the primary and secondary
coils well intermingled along the magnetic circuit this drop
can be kept to as low as ~ per cent.
(5) The loss due to eddy currents in the core ; and, in the
case of open iron circuit types, due to eddy currents in the
winding.
Mr. Alexander Siemens has pointed out how this loss may
be arrived at in the case of a core composed of fine iron
wires. The following formula is adapted from his paper on
the subject, and is the value of the watts lost in eddy currents
per cubic centimetre of iron
6-6//3V 2
F = - - - watts per cc. of iron
io 12
where d stands for diameter of wire in cms.
p periods c/> per second.
ft ,, c.g.s. lines per sq. cm. of iron section.
Mr. Evershed has also shown how to express the loss when
the iron core is composed of thin iron plates, viz. :
F = -^-rr watts per cc. of iron
11
io
where / is the thickness of plate in cms.
Impedance Coils and Transformers. 151
(6) The hysteresis loss due to molecular friction in the iron.
This may be very accurately found in the way shown for the
armatures of dynamos by means of the table on p. 2 14. Thus
H = -y = hysteresis loss in watts total
Calling the total induction in the iron of a transformer N
c.g.s. lines, and using the symbols already mentioned, we
have, neglecting the slight difference at no load between
and E,,
E,, X io 8
4 '443 X p
and N = fta, ft being quite small, ranging from as low as 3000
to 6000, and seldom exceeding the latter, whence
E,, x io 8
7 ^ = 4^rxTx^
which gives an important product of the two unknown quan-
tities in terms of known, or easily fixed values. Only ex-
perience can fix the relationship between the primary turns,
and the sectional area of the core. On the one hand, if w p
be large we shall have, with a given weight of copper, a high
resistance and large copper loss ; but the iron losses H and F
will be small. If, however, a be made large, the copper losses
may be very small, but the iron losses will be large. The
relationship between these two losses, copper and iron, will
depend greatly upon the average output or load-factor. When
the transformer is to be on full load always, then the highest
efficiency will be attained when the copper losses are equal to
the iron losses. But in general the conditions of supply render
it impossible that the transformer can be on full load for more
than a very small part of the 24 hours, whilst for a very large
part it may be practically unloaded. The all-day efficiency
can then be made higher by making the constant iron-losses
smaller at the expense of large copper losses, which will only
be of importance in themselves for the short times of high and
full load. In this, of course, due regard must be had for a
reasonable constancy in available secondary P.D.
152 Examples in Electrical Engineering.
Example. A transformer for 40 amperes and 100 volts
secondary output. In order that the efficiency at full load
may be, say 95 per cent, it follows that the input must exceed
the output by about 200 watts, which is to cover both the
copper losses and the iron losses. If these be about equal,
we must allow for something like 79 watts in hysteresis, from
which, if the value of /? be fixed, we can find the volume of
iron. Thus from
we have
HX_1 7
h f
and if ft be 6000, h 1820. Putting these values and 79 for
H, we have, if/ = 80 en per second.
7
V =
If the primary E.M.F. E^ = 2000 volts virtual, we have
from
io
where N is the maximum induction, that
2000 x io 8
= 562 ' 7 ' OC
and as ft = 6000, we can put
w p a = 93,800
where a is the sectional area of the iron core in sq. cms.
Taking this to be 155 sq. cms., we have from the previously
found volume that the average length of the iron core must
be
V 5425
- = ^ - = 35 cms.
i55
If the core is composed of stampings of the shape and size
Impedance Coils and Transformers.
153
shown below, we shall require a pile of them 22*2 cms. of iron,
not counting lamination.
21
MA }**
- ^<
7
X-
*
FIG. 33.
The total volume of iron in the above plates is
(21 x 14- 7 X 7) 22-2 = 5439 cc.
but as the magnetic induction cannot be taken as filling the
iron into the corners, the amount under magnetic influence
will be more nearly that previously found, viz. 5425 cc.
The sectional area is
7 X 22-2 = 155-4, say 155 sq. cms.
as was desired.
Of the spaces for winding we will allow that owing to
superior insulation the space occupied by the primary coils
will be i4'5 sq. cms. out of the total of 24*5 sq. cms., and that
the secondary takes 8 sq. cms., leaving 2 sq. cms. for an insu-
lating partition between the two coils. From
w p a = 93,800
putting a =-. 155, we have
w p 606 say
154 Examples in Electrical Engineering.
which number of wires has to go into the space of 14*5 sq.
cms., or each wire will take up a room of
= ' 2 392 sq. cm. area
whence the diameter of the covered wire, if round, will be
d = ^0-02392 = 0*154 cm. covered wire
and if the covering increases the bar diameter by 0-07 cm.,
the diameter of the bare wire will be
0*154 0*07 = 0*084 cm.
which is practically equivalent to a No. 21 B.W.G., which is
0*032" diameter, and has a resistance of about u ohms per
1000 feet warm.
Allowing 2 -8 cms. for lamination, the average length of one
turn on the primary coil will be 72 cms. if the primary be
wound inside the secondary.
72 cms. = 28-5 inches say
whence the total length of the primary wire will be
606 x 28*5 inches = 17271
or 1440 feet. Whence the resistance
RP = 1*44 X n = 15*4 ohms say
Again, the mean length of the secondary turn will be 76 cms.,
or 30 inches, and
i 606
*=- 2Q v>* = 20 3i say
The space to be filled by these 31 turns is 8 sq. cms., whence
each wire will occupy a room of
o
= 0*258 sq. cms.
and if the wire be square the size will be ^0*258 - 0*507
cm. square covered, which will admit of a bare wire of 0*437
cm. side of square, about equivalent to B.W.G. 6J, and having
a sectional area of 0*1909 sq. cm., and a resistance of about
0*2 8 W per 1000 feet warm.
Impedance Coils and Transformers. 155
The total length of secondary wire will be
30X31 = 78 feet say
12
whence
R, = 0*28 x o'oyS = 0*022 ohm.
If, then, the maximum secondary current be 40 amperes,
there will be a full load ohmic drop of
40 x o - o22 = o*S8 volts
and a full load secondary copper loss of
(4o) 2 x 0*022 = 35-2 watts, say 36
Also, at full load the current in the primary coil will be
C s
C, = ^ + C M
where C Hl is to include the magnetizing current giving the loss
in hysteresis and also the current required to balance the eddy
current loss in the core. In as far as these are losses they can
easily be found, as shown below ; but as they are very small
compared to the secondary balancing current, they hardly at
all affect the value % of the primary copper loss. Thus, for
example, corresponding to ft = 6000 every centimetre length
of the iron core will require 1*624 ampere turns, or the maximum
magnetizing ampere turns on the primary will be
35 x 1-624 = 56*84
which must be produced by a current of
, = 0*0937 amperes (maximum value)
But this current is not a sine curve, and its virtual value will
be greater than 0*707 times the maximum, or the virtual value
will be, say 0*075 amperes.
Furthermore it is not in phase with the secondary balance,
which has a virtual value at full load of
4
= 2 amperes
Thus, including the hysteresis and eddy current losses, the
I $6 Examples in Electrical Engineering.
full load primary current cannot be greater than 2*1 amperes
virtual, if so much, and therefore the primary full load copper
loss is
(2*1)- x 14 = 62 watts say
Using the formula on p. 150 for the eddy current loss, and
taking the thickness of the stampings as 0*03 cms., we have
the loss per cc. is
80 x 80 x 6000 x 6000 x 0*03 x 0-03
r, - = 0'0020
I0 11
whence the total eddy current waste will be
5425 x 0*002 = 10*85 watts
say ii watts.
Thus, the total output will be 40 x 100 watts, and the input
will be 40 x ioo -f 79 -f 62 4- 36 + n = 4188, which is
within the prescribed limit.
If kept on full load the efficiency will be
4000
100 X 4188 = 95 ' 5 P ercent
But when it is remembered that during a day of 24 hours a
transformer on a lighting circuit is very variably loaded and
may be practically on open secondary circuit for several hours,
during which time the iron losses are constantly to be supplied,
it will be apparent that the all-day efficiency or ratio of
Units paid for during 24 hours
Units put in during 24 hours
will be a very different value to the full-load efficiency.
Consider, then, the case of the above transformer as on a
circuit where it is
On
No load.
A load.
i load.
load.
} load.
Full 1 ad.
For
8hrs.
4 hrs.
4 hrs.
4 hrs.
3 hrs.
I hr.
We shall have the following table of outputs and inputs for
Impedance Coils and Transformers.
157
24 hours, the total full load copper loss being 98 watts, a loss
which will vary practically inversely as the square of the load ;
and the iron loss equal to 90 watts, whatever the load.
Load. , Hours.
Output
watt-hours.
Input = watt-hours.
Output + iron loss + copper loss. Total.
8
O
o + 8 x 90 + o = 720
T'.
4
4 X 400 = 1600
1600 + 4 X 90 + 4 x 0-98 = 1964
1
4
4 X 1000 = 4000
4000 + 4 x 90 + 4 x 6'i =4385
i
4
4 X 1334 = 5336
533 6 + 4 X 90 + 4 X 10-8 = $739
i
3
3 X 2000 = 6000
6000 + 3x90 + 3x25 = 6345
I
i
i x 4000 = 4000
4000 + i x 90 + i x 98 = 4188
Output 20936
Input 23341
_ . ioo X 20936
And all-day efficiency = -
23341
= 89-6 per cent..
which is a very different thing to the 95*5 per cent, found
above for the full-load efficiency, being even less than the
efficiency if the transformer were kept on quarter load always,
which would be over 91 per cent.
Example XLYIII. A circuit has a B.E.M.F. of 1600 volts
and a resistance of 20 ohms. What E.M.F. alternating will
be required to send through it a current of 10 amperes?
Solution. The relationship between the E.M.F. required E
the B.E.M.F. e and active E.M.F. e - RC, is that
whence since
E = V(* 2 + 2 )
e - RC = 20 x 10 = 200
E = V(i6oo) 2 4- (200)" = 1612 -45 volts
Example XLIX. In the above circuit what will be the
current if the alternating E.M.F. supplied is 2000 volts, and
what will be the rate of doing work ?
Solution
e =
158 Examples in Electrical Engineering.
or
e - V(20oo)- (1600)- = 1200 volts
but -
20
and the rate of doing work is C 2 R = 72,000 watts or 96-48 HP.
Example L. The P.D. at the terminals of a house con-
nected to an alternating system of distribution is 100 volts. It
is required to run 2 arc lamps, taking 34 volts each, in series
off these terminals. If the lamps can be regarded as simple
resistances of 3^4 ohms each, what will have to be the value of
the B.E.M.F. of the choking coil to be used in series with
them?
Solution. Since each lamp takes 34 volts, and is equivalent
to a resistance of 3*4 ohms, they are taking a current of
10 amperes, and when in series will require 68 volts, which is
the value of R.C., the active E.M.F. required. Thus, since
E = 100 we have
. ? = ^/((ioo) 2 - (68) 2 )
= v^ioooo - 4624) = A/(5376)
= 73'3 2 volts B.E.M.F.
Example LI. In the last question, what is the saving
effected, by using the choking coil instead of a non-inductive
resistance, if the choking coil consumes energy at the rate of
30 watts ?
Solution. The total expenditure with a non-inductive
resistance would be CE = 1000 watts, and with the choking
coil it is only
C 2 R + 30
or
100 x 6-8 -f- 30 = 710 watts
whence the saving is
1000 710 = 290 watts
Impedance Coils and Transformers. 159
Example LIL What will be the angle of lag of the
current in Example L. ?
Solution, the lag is such that tan A = -
tan A = 7 -|| 2 = 1-078
68
whence A = 47 about.
CHAPTER XIV.
EFFECTS OF CAPACITY.
THE properties of a condenser may be easily studied by refer-
ence to the water analogy suggested by Dr. Fleming, as a
mechanical representation of the electrical conditions.
D
M,
M 2
P
FIG. 34-
In Fig. 34 is represented a system of tubes containing water,
divided at D by a flexible diaphragm of indiarubber, and
having in each leg a water-meter, Mj and M 2 . A tap and
pump are also placed in one leg ; the tap to represent a switch
Effects of Capacity. 161
in an electric circuit, and the pump corresponding to the
electro-motive force.
First of all suppose the diaphragm removed. On working
the pump a current of water will flow round the circuit in one
direction or another, according to the rotation of the pump.
This current will be indicated on the water-meters, and will be
at the rate of so many gallons per minute corresponding to
an electrical effect of so many coulombs per second. This
flow of water is constant in direction and in quantity, so long
as the pump be rotated in one direction at the same rate.
If, however, the pump be replaced by a simple piston re-
ciprocating in one of the tubes, we shall have the current of
water constantly varying in direction, according as the piston
is moving up or down ; this reciprocating motion of the piston
corresponds to an alternating E.M.F., and the alternating flow
produced by it will not make any permanent record on the
meters, since as much flows through in one sense as flows back
in the other sense. The force urging the piston on its course
will vary from point to point of the stroke, according as the
mass of water is moving with or against it, and according to
the resistance to the flow due to friction against the sides of
the tube. This friction corresponds to resistance and the
momentum of the water to the B.E.M.F. of self-induction.
But though the meters record nothing, they may be replaced
by suitable mechanism indicating the amplitude of the recipro-
cating motion, which is a measure of the work-doing property
of the moving mass of water.
When, however, the diaphragm is replaced we have a very
different state of affairs. Formerly there was no obstruction
to the motion of the water right round the circuit ; but now,
on rotating the pump, water will, let us say, be forced up the
right side, causing the elastic diaphragm to swell out, as shown
by the dotted line in the figure. Meter No. 2 will record the
quantity of water required to do this as a quantity of water
flowing into the diaphragm ; but meter No. i will also record
it as a quantity of water flowing out of the diaphragm. For
however long the pump be worked the diaphragm will only
become distended to an extent equal to the force exerted by
M
1 62 Examples in Electrical Engineering.
the pump, supposing, of course, that its strength is such as not
to admit of rupture by any force the pump can exert. Thus,
there will be a flux of water on the one hand into the dia-
phragm, and on the other hand out of the diaphragm for a
short time only; and the rate of flow will have been larger
at the start w r hen the elastic force exerted by the diaphragm
was small, than at the close when the elastic force due to
stretching is equal, and opposite to the force exerted by the
pump, thus leaving no available force for urging water along
the circuit.
Whilst still keeping the pump in rotation if the tap be closed,
the stretched condition of the diaphragm may be preserved
indefinitely, although the pump be afterwards stopped.
A careful examination will disclose the fact that the only
difference now existing in the apparatus is this stretched con-
dition of the diaphragm : not a store of water, for as much
obviously flowed in as flowed out, but a power to do work by
relaxing a store of energy.
The above conditions correspond to that of the continuous-
current circuit, and the flux of water which lasts for a short
time only after the starting of the pump is called the " dis-
placement current."
When the pump is now replaced by a reciprocating piston,
we have a representation of the alternating-current circuit.
Now, the diaphragm will be stretched first in one direction and
then in the other, and the water-meters will, as before, be
affected by a quantity of water surging to and fro in the tubes.
The rate at which this quantity of water will be flowing in any
one direction will depend upon the force of the piston and
upon the elastic force of the diaphragm. When unstretched,
the water will flow into the stretching diaphragm at a great
rate, even though the force due to the piston may be small ;
but as the diaphragm becomes distended, the rate of flow into
it will be small, and ultimately reach a zero value as the elastic
force of the diaphragm equals that of the piston at the end of
its stroke. As the pressure on the piston which caused it to
move forwards is relaxed, allowing it to return to its starting
position, the elastic force of the diaphragm causes water to flow
Effects of Capacity.
163
back, slowly at first, but reaching a maximum rate as the
piston reaches its starting-point (the middle of its total stroke),
and then again declining to zero as the piston moves on its
opposite stroke, the diaphragm now becoming distended in
the opposite way. This can better be seen by reference to
Fig- 35 j where P is the reciprocating piston, D the diaphragm,
B
B,
r
- TJ- ,.-'
/
(
/
(
i
\ \
P /\.
D ic,
>
\
(
/
n Xx --.
\
FIG. 35-
and ABC represent the course of the piston. As it starts
from B the diaphragm is unstretched as at B 15 and the rate at
which water will flow into it is greatest ; as the piston reaches
C the diaphragm is distended to Q, and the rate of flow is a
minimum, having declined from a maximum, but being always
in the same direction as the force applied to the piston. As
the piston, displaced thus far by a force, moves back to B by
a removal of that force, the water now flows out or back
towards the piston, not away from it as heretofore, till the
piston reaches B, when the diaphragm is now as at B x the rate
of flow, however, being now a maximum, but declining to zero
again, though still towards the piston, as the piston moves
towards A, due to a reverse force to that which urged it from
B to C. The diaphragm is now as at A x exerting a force away
from the piston, and as the latter returns, due to the gradual
reduction of the outside force acting on it, the water now flows
away from the piston at a slow rate at first, but reaching the
maximum again as the piston returns to B, when the diaphragm
is once more unstretched.
Calling the outside force which urges the piston along as
equivalent to the alternating E.M.F. E, and the current of
water which flows into the diaphragm as the current C, there
164
Examples in Electrical Engineering.
being no resistance to the flow and no magnetic effect set up,
we thus have the following order of effects
E.M.F. - E.
Current - C.
Zero but
rising to a
+ maximum
then declining
though still +
to zero
reversing and
rising to a
maximum
and declining
to zero again.
A + maximum
but declining to
zero
then reversing and
rising again
to a maximum
declining whilst
still to
zero
reversing and rising
to a + maximum again.
and so on, which can best be indicated by plotting the curves
of E.M.F. and current on the assumption that the variations
of strength of E.M.F. are sinusoidal, as is done in Fig. 36,
where E is the E.M.F. supplied and C the current, the circuit
FIG. 36.
being supposed to be unable to set up any magnetic effect and
to offer no resistance to the passage of the current, a condition
of affairs which would correspond, on the water analogy, to a
massless fluid moving in frictionless tubes.
Effects of Capacity. 165
Now, the degree to which the diaphragm can be distended
by any force will depend upon its size and upon the material
of which it is made ; or its capacity depends upon its con-
struction. Similarly the electric condenser has a capacity which
is proportional to its area, inversely proportional to its dielectric
thickness and also proportional to the property of the material
of which it is made.
Thus the quantity of water required to distend the diaphragm
so that its elastic force is equal and opposite to the unit force
acting on the piston will be a measure of the capacity of
the system. Also, in the electric circuit we may say that
the quantity of electricity in coulombs required to strain the
dielectric to an extent equal and opposite to the E.M.F. of
i volt is a measure of the capacity of the condenser; and
when that quantity is unity (i coulomb) the condenser is con-
sidered to have unit capacity, or a capacity of i farad. Thus
i farad is that capacity which requires i coulomb to pro-
duce a P.D. of i volt between its two coatings.
But this quantity Q coulombs must, in an alternating-current
circuit, be passed into, or out of, the condenser each quarter
period, and therefore the average current flowing into, or out
of, the condenser will be
C = T= r
4 4
where K is the capacity in farads and E M is the maximum
value of the E.M.F. supplied to the condenser.
Or, putting / = -, we have the average current
and thus the virtual current and maximum current will be
C. - 4'44^KE, (l
and C m = 27r/KE, w
But if E be stated as the virtual E.M.F., then
C, = 27T/KE,
1 66 Examples in Electrical Engineering.
And we have seen that the E.M.F. supplied to the condenser
reaches its maximum value, 90 degrees, later than the maximum
value of the current due to it, the condenser itself exerting
an E.M.F. in the circuit, always equal and opposite to the
supplied E.M.F. It is necessary to make a very careful dis-
tinction between these two E.M.F.s; that supplied to charge
the condenser is of course that of the dynamo, or other gene-
rator, and is called, as before, the impressed E.M.F. ; that
exerted by the condenser is of the nature of a back E.M.F.,
and may be called the B. E.M.F. of the condenser, K.
Thus, if a condenser be in a circuit having an E.M.F. of E ( ,
volts, it will take a current
C, = 27T/KE, amperes
and will exert a B.E.M.F. of
With a perfect condenser having plates of some material offer-
ing no resistance to the flow of the charge into them, and with
a dielectric of perfect insulating properties, this B. E.M.F. K V
will be exactly equal and opposite to E w the impressed E.M.F.
But since some E.M.F. is required to maintain the momentary
charging current, and there is a small actual leakage from one
set of plates to the other (in both cases causing heating), it
follows that the impressed E.M.F. E, must slightly exceed the
B. E.M.F. K P , the difference of phase being also slightly less
than 90 degrees.
Now, since the supplied E.M.F. lags 90 after the current in
the condenser, and the condenser B. E.M.F. is equal and oppo-
site, it follows that the current in a condenser lags 90 after its
own B. E.M.F. Therefore the effect of capacity in a circuit is
to oppose the effect of self-induction, as may be easily seen by
reference to the following diagram (Fig. 37), where the current
through a circuit is marked as C amperes, the circuit being
able to set up a magnetic field of such magnitude as to cause
a B. E.M.F. of self-induction of c volts, and also possessing a
capacity such as will give K volts, which E.M.F. may now be
Effects of Capacity. 167
called a forward E.M.F., in contradistinction to the B.E.M.F.
of self-induction e. To maintain the current C amperes through
\
\
\
\
FIG. 37.
the circuit, the impressed E.M.F. E will thus have to be suffi-
cient at any moment to (a) send C amperes through R ohms,
(b) to nullify e, and (<r) to nullify K ; and when = K, it follows
that E = e = RC simply, or the capacity effect has counter-
acted the magnetic effect.
Now, we have seen that
-/-x8
and also that
10 10
C
(see p. 132)
* 2:r/>K 4 44/lv
whence it follows, that when the capacity is such as to make
the circuit non-inductive in spite of a magnetic effect
~~70 8 ~ ~ = 27T/K
whence
C,,io 8
i\. =
1 68 Examples in Electrical Engineering.
or if N and C be maximum values, then
C m io* i
= N w X ~^~ 2
or
K = ?r x ^
Now, the capacity of a condenser in farads is
A
K = i
1-131 x io 13 x d
where A = the area of one set of coatings in sq. cms.
d = the distance between one set of coatings and the
other in cms.
and / = that property of the dielectric placed in between
the two coatings, which is called its specific in-
ductive capacity.
The values of / for different materials are :
Air i
Mica 5
Flint Glass 6*51013
Glass 3
Ebonite ... 2*5 to 3-1
Sulphur 2-8 to 3-8
Paraffin wax ... ... ... ... ... about 2
Paper baked in paraffin oil about 2' 7 to 3
In constructing a condenser, the conductors are composed of
thin sheets of metal, usually tinfoil, piled upon each other with
some dielectric in between. In order to obtain the large
surface required for even a small capacity, the sheets are made
of a reasonable size, and interleaved with each other, as shown
in section in Fig. 38, where the lines show the metal sheets,
and the spaces the dielectric. It will be noticed that five
alternate plates are connected together by the terminal block
T 1} and are interleaved with the other six, these last being
also all connected together by the terminal block T 2 . In
such an arrangement use is made of each side of each
plate, and thus the surface of such a condenser is found by
the products of half the number plates into the surface of
Effects of Capacity. 169
one plate counting both sides. If, consequently, each plate
measures 24 cms. by 48 cms., the area per plate will be 2304
T -
T
FIG. 38.
sq. cms. If now the dielectric be of oiled paper, 0-015 cm.
thick, each piece of paper and plate of tinfoil will be, say
100
0*017 thick, or there will be = 5882 plates, or 2940 say
connected to one terminal per metre thick, and taking the
specific inductive capacity as 3, the capacity per metre thick
will be
2940 X 2304
K = 3 x - 13 = 0-000119 farad
1*131 X io 13 X 0*015
or to make a condenser of i farad capacity would take a thick-
ness of about 8400 metres, which gives some idea of the great
size of a condenser to have so large a capacity.
Also, if used on a circuit working at 2000 virtual volts at
100 c/) per second, the current taken will be
C, = 27T/KE,
= 628 X i X 2000 = 1,256,000 amperes
to which should be added such current as may leak from one
set of plates to the other, a current which will probably be
almost in phase with the E.M.F. supplied, and consequently
lagging 90 after the condenser current.
Use of Condensers for modifying the effect of self-indue-
170 Examples in Electrical Engineering.
tion. We have already seen that the nature of the B.E.M.F.,
set up by a condenser, enables it to oppose the E.M.F. of self-
induction when the condenser is placed in series with the
circuit. We may thus say that capacity in series with an in-
ductive circuit renders that circuit non-inductive by nullifying
the E.M.F. of magnetic action, and thus leaving the circuit in
a condition which can be described fully by saying its resist-
ance is R ohms, it requires simply RC volts, but can, never-
theless, exert a magnetic effect. In practice, one of the most
interesting cases is that of the shunt circuit of a watt-meter for
measurement of alternating power; this circuit must be able
to produce a magnetic field, and must also be perfectly non-
inductive with its current consequently in phase with the P.D.
at its terminals.
Condensers are, however, sometimes used in parallel with
a circuit the properties of which they are desired to modify.
The circuit to which a condenser is a shunt is, however, un-
altered in any way, the main effect is in the generator and
leads connecting it to the circuit and its shunting condenser.
Thus, if we have a circuit such as the primary of an un-
loaded open-iron circuit transformer, we have seen that
the current taken by it will be large, but will lag almost
90 after the E.M.F. supplying it. This current will not
necessarily mean any great waste of energy in the transformer,
but the loss in the connecting leads and dynamo armature
will be a considerable item. If now a condenser be connected
across the primary terminals, the dynamo will have to supply
through the leads a current such as will be determined by the
size of the condenser, but coming 90 earlier than the E.M.F.,
and consequently opposed to the transformer current. The
current, therefore, which flows in the leads and armature will
be the sum of these two, and may be almost zero when the
condenser current is equal to the transformer current, being
only so large as will, in combination with the E.M.F., repre-
sent the waste in the transformer and condenser.
We have seen that the value of the current flowing into a
condenser is
C tt = 27T/KE,
Effects of Capacity. 171
whence in this case we must make the capacity such as will
take this current C tf equal to that required by the unloaded
transformer at a P.D. of E fl volts, or
K- C "x 1
-E, X 2,r>
where C K stands for the so-called " magnetizing " current of the
open-iron circuit transformer.
The use of condensers for the above purpose was proposed
by Mr. Swinburne, in conjunction with his " hedgehog " trans-
former, but very few are in use.
It is clear from the above that a condenser in parallel with
a circuit tends to make the whole combination non-inductive,
rather by rendering the apparent resistance larger, than by
opposing any B.E.M.F. in the circuit.
In systems using concentric cables, the capacity of the cable
considered as two conducting masses intimately associated but
separated by the insulating covering on the inner conductor, is
a shunt to the circuit supplied by those leads, and so may
modify the effect of self-induction in the circuit. Messrs.
Cowan and Still have published the values of the capacity of
a few cables, viz.
For yf concentric cables the capacity per mile is
British Insulated Wire Co. (paper insulation) ... 0-31 microfarad
Glover and Co. (vulcanized rubber) ... ... 0*615
(diatrine) ... ... ... 0-315 ,,
This capacity will take a current coming 90 earlier than the
impressed E.M.F., which current must be added to the current
supplied through the cable to the load at the far end. When
the load current is lagging much, due to self-induction, the
total current will be, as shown above, the difference between
the two, and therefore may be zero.
Example LIII. A circuit supplied from an alternator takes
a current of 10 amperes at a pressure of 100 volts, its resistance
being, however, only 5 ohms. Find the lag of the current and
the capacity in microfarads of the condenser which must be
used in series with the circuit in order that the current may be
2O I
K = ^>-^ x -= 0-000367 farad
86'6 27Tioo
172 Examples in Electrical Engineering.
20 amperes, and not lag at all. There are 100 periods per
second.
Solution. The active E.M.F. will be 50 volts, whence the
B.E.M.F. e is /y/ioooo 2500 = 86'6 volts, whence there
will be a lag of current after the E.M.F. such that tan A = -
5
or X = 60. In order to reduce this lag to zero, and thus make
the current 20 amperes, the capacity will have to be
20
X
27TIOO
or 367 microfarads, a capacity which could only be made in
a size of about 25 cubic feet.
Example LIY. An unloaded open-iron circuit transformer
is shunted by a condenser of 30 microfarads capacity, and
appears to take a negligibly small current. It is supplied at
2000 volts off a circuit, having 80 periods per second. What
is the true magnetizing current ?
Solution. It is clear that the condenser current is equal to
the magnetizing current as, being in opposite phase, it has just
nullified it. Now, the current taken by such a condenser will
be
^4
2?r^KE = x 80 x 0*00003 x 2000 - 30*17 amperes
which is, therefore, the true magnetizing current.
Example LY. A concentric main six miles long, having
a capacity of 07 microfarad per mile, is supplying energy to
an inductive load, taking 10 amperes and lagging almost 90.
If the P.D. is 1000 volts, what is the indicated current where
there are 100 c/5 per second?
Solution. The current taken by the main on account of
capacity will be
C = 27T/KE = X ioo X 4'2 X io- x 1000 = 2-64 amperes
which current will be almost in opposite phase to that in the
load, whence the apparent current will be the difference or
10 2*64 = 7*36 amperes
Effects of Capacity. 173
Example LYI. A concentric cable is 12 miles long, and
is connected to an alternating P.D. of 10,000 volts. If the
capacity is 1*5 microfarad per mile, what will be the current
supplied, the circuit working at 80 CO per second ?
Solution. Total capacity is 12x1*5=18 microfarads,
whence the current taken is
C = 27rK/E = x o'ooooiS x 80 x 10,000 = 90-5 amperes
Example LYII. Two condensers of \ microfarad capacity
each are connected to an alternating P.D. of 50 volts at 225 CO
per second. What will be the current taken
(a) when the two condensers are in parallel ?
(b) when in series ?
Solution. In parallel the total capacity will be f micro-
farad, since virtually the surface of the condenser is now double
that of either. Thus the current will be
44
- X 0-000000667 X 50 x 225 = 0-047 ampere
But when in series the effect is the same as making the thick-
ness of the dielectric twice as much while the surface remains
the same. The capacity is then only -| microfarad, and thus
the current will be a quarter of that in parallel, or 0*0117
ampere.
CHAPTER XV.
REGULATING RESISTANCES.
IN order to produce desired variations in the manner of working
of various types of electrical machinery, it is sometimes neces-
sary to employ resistance coils connected in circuit with the
machine to be regulated. For this and other purposes it may
be necessary to provide resistances calculated to effect the
following
(1) To replace some device cut out of circuit;
(2) To regulate and vary the speed of a motor.
(i) The value of a resistance to replace a bank of lamps, an
arc lamp, a motor, or other device which requires when working
a P.D. of E volts, and takes a current of C amperes, is simply
R = P ohms
This resistance must, of course, be made of such shape, size,
and material, that it will be able to carry the current for the
length of time it is to be in circuit without overheating. Prac-
tice varies considerably in this respect, the rise in temperature
allowed ranging from quite a few degrees above the surrounding
air to a temperature quite sufficient to discolour paper. No
rule can be given for the size of wire or current density, as a
great deal depends upon the spacing of the wires, or the degree
to which the convolutions of the spirals are separated from one
another. With a fair amount of spacing, however, a current
ranging from \ to of that required to melt the wire employed
according to the table of fusing currents determined by Mr.
Preece, may be taken as reasonable. The materials usually
Regulating Resistances. 175
employed are German silver, platinoid, iron, and manganin, the
latter having the advantage of constancy of resistance in spite
of variation of temperature.
(2) The resistance employed to vary and regulate the speed
of a motor is usually combined with a set of resistances and
switch contacts to enable the motor to be gradually introduced
into the circuit, so that excessive current through the motor on
the one hand, and sudden demand on the generator on the
other hand may be avoided.
In all motors the tendency to rotate is due to a pull on the
armature conductors by the poles of the magnets, the pull
depending directly upon the strength of the current and the
strength of the magnetic field. Thus, whatever be the magni-
tude of the current in the armature, there will be little or no
pull on its conductors if the strength of the magnetic field be
small. It thus follows that in starting a motor the first thing
to see to is production of a strong magnetic field. In the case
of a shunt motor, the switch for introducing it into circuit is,
therefore, arranged so as to, first of all, make circuit with the
shunt coil ; then the armature is introduced into the circuit in
series with a considerable resistance, in order that the current
may be only a small fraction of that at full load. In this way
the two essentials to rotation are produced without any sudden
strain either on the motor or generator. In the event of there
being no load on the pulley the motor now starts, and will
soon reach a speed near the maximum possible ; but if there
be a load on the pulley, the torque due to the small armature
current may not be sufficient to produce rotation. In this case
the switch has to be moved over several contacts, cutting out
of circuit a resistance at each step till the current through the
armature is sufficient to produce rotation. Regulation of speed
is effected in either of two ways, or sometimes by both com-
bined, viz. :
(a) Reduction of the E.M.F. available for the armature by
resistance.
(b) Reduction of the strength of the magnetic field by intro-
duction of resistance into the shunt circuit.
(c) A combination of the two previous methods.
176 Examples in Electrical Engineering.
Fig. 39 will illustrate this. The dynamo D has one
terminal connected directly to the one brush of the motor M,
to which is also connected one end of its shunt coil as usual.
The other terminal of the dynamo is connected to a contact
arm A, which may be brought into contact with either of the
contact blocks i, 2, 3, 4, 5, 6, or 7, or may be off any of them,
when circuit will then be broken. The other brush of the
motor is disconnected from the other end of the shunt coil, and
is connected to one end of the series of resistance coils placed
between the contact blocks ; the shunt coil is connected to
some part of the resistance coils, as shown at B. If now the
contact arm be upon block marked a, there will be no circuit,
SJvunt
FIG. 39.
but when contact is made with block i, the armature and shunt
coil will then be in circuit, the armature current, however, being
quite small as the coils between B and block i have a resis-
tance which is large relatively to the armature, but not so large
as to make much reduction in the shunt current. Thus there
will be produced a considerable field strength, and also some
armature current. The resistance now in circuit is such as to
prevent the motor running at anything like full speed since the
loss of pressure in the resistance coils will very considerably
reduce the E.M.F. available for overcoming the full B.E.M.F.
of rotation ; but on moving the arm A over to block 3 or 4,
the resistance in series with the armature is considerably
Regulating Resistances. 177
reduced, and the shunt is now directly connected to the dynamo,
this position of the switch being that intended for the slowest
speed of running, since the shunt is now strongest, and the
armature has still considerable external resistance. It is from
this point onwards that the double regulation comes in ; for as
the arm A is moved onwards towards block 7, the shunt is
being made gradually weaker, necessitating an increase of speed
in the armature to give the required B.E.M.F., whilst at the
same time there is less and less drop of pressure in external
resistance till the arm A is on block 7, when the armature then
has the full P.D. of the dynamo, and will, consequently, run
fastest.
In calculating the magnitude of the resistance coils to be
inserted in a switch of the above type it is as well to bear in
mind that the current taken by the motor depends mainly upon
the demand made upon it at the pulley, and may reach the
maximum value even at a slow speed when probably the
demand for work may be large. Thus all the coils between
blocks 4 and 7 may have to carry the full current, and must
consequently be of such a size as to prevent overheating. The
coils between contacts i and 4 must also be made of such
size as will make them quite safe at any current they may have,
the largest value of course being such as the full P.D. can
produce through the whole external resistance with the arma-
ture of the motor stationary ; these coils, though not intended
generally to be left in circuit for any length of time, are
frequently so left, and should therefore not be so slender as to
risk fusion. As to the value in ohms, a good deal depends
upon the range of regulation required, bearing in mind that in
any case such regulation is very wasteful of power since any
not used by the motor is dissipated in heating the coils.
Let the value of the resistance between blocks i and 4 be
called R ohms and that between 4 and 7 = r ohms, the resist-
ance of the armature being R a and that of the shunt coil R 4 .
For the present we will leave out of consideration the effect of
the resistance in the shunt coil, and deal only with that in the
armature.
Calling the dynamo P.D. E volts and the maximum value
N
178 Examples in Electrical Engineering.
of the motor armature current C tt amperes, it is first of all
necessary that R -f- r shall be so large that
C = ^ . = the starting current
may be less than C fl , or certainly not greater than it. Next
we will suppose that the switch has reached the contact block
4, and that the motor is so loaded that it is taking a current of
C amperes. Under these circumstances there is a loss of
pressure in the external resistance and in the armature, neglect-
ing the shunt current, of
C(R tt + r) volts
leaving
E - C a (R rt -t* *) * i
the speed being nearly proportional to the B.E.M.F. e. Thus,
by making C a (R a -f r) any desired proportion of E, the speed
can be thereby reduced in the same proportion.
Next consider that the arm A has reached the block 7, then
the value of e will be simply
E - C a R tt
and the speed will be normal. But with the switch connections
as shown in Fig. 39, p. 176, the shunt coil has now a resistance
inserted in it, the effect of which is to reduce the magnetizing
force, and therefore the magnetic field, but not in the same
proportion, since the magnetic circuit is composite, and conse-
quently the various items of it are altering in relative impor-
tance. Thus, whereas the air gap was, at full excitation, a
certain large proportion of the whole circuit, it is now relatively
much more important, since, due to reduction of magnetic
density, the permeability of all the iron items has increased.
The exact effect of such reduction of magnetizing force is not
capable of direct calculation, since the law connecting j3 and /x
is not easily stated in symbols; and, moreover, the greater
effect of armature reactions at reduced excitation renders the
problem a very difficult one. We can, however, form an
approximate estimate of such effect by remembering that,
initially, the reduction of lines will be always less than the
Regulating Resistances. 179
reduction of magnetizing force. But, counting armature re-
actions as more important at low inductions, we shall be very
near the truth in assuming that the field falls off directly as the
magnetizing force.
Thus in the above case the magnetizing force is proportional
to the shunt current, which is as a maximum
and as a minimum
r JL-
w STF?
If the speed be n at the point when the switch arm is on
block 4, when the shunt gets the full P.D., and the armature
has the resistance r in series, then this will be the slowest
speed if the motor be so loaded as to take the full armature
current C a . But when all the resistance is out of the armature
circuit and inserted in the shunt, the speed will have increased
from its former value to
E - C a R R,
;/1 = E
and whence
a
= < E - X
F r p j. *',
E-C.R. + -1
from which it will be seen that in any case the shunt effect is
small. But in the simpler case, where the shunt coil is always
at the full P.D. of the circuit, then
E - CR
n ~ E - C a (R rt + r)
and
r=
1 So Examples in Electrical Engineering.
In all cases ;/ x stands for the maximum speed, and n for the
slowest speed. The difference is not considerable between the
two cases, the resistance r required to vary the speed from i to
2 being about 8 per cent, less when the shunt is connected as
shown in Fig. 39, than when the shunt always has the full
P.D. at its terminals.
Example LYIII. A shunt motor having R, = 50 ohms,
E = 100 volts, C a = 100 amperes, and R a = 0*02 ohm, is re-
quired to have a speed ranging from 600 to 1200. Find the
value of the regulating resistances required.
(a) When the shunt always has the full P.D. 100 volts;
(b) When the shunt has the P.D. at its terminals varied by
the same resistance as in the armature circuit.
Solution. =. 2, whence in case (a) the resistance to be
inserted will be
2 TOO -f 2OO -j-4 I 06
; = o's-? ohm
200 200
and in case (b)
sooo 100 4000
r = - = - 5 = 0*485 ohm
100 2 4- 10000 10098
CHAPTER XVI.
PRIME MOVER COSTS.
IN most forms of heat engine the consumption of fuel may be
divided up into two items, one to run the engine idle, and the
other to supply the output. The latter is generally represented
by a regular constant amount for every actual brake horse-
power hour, whilst the former is also generally nearly a con-
stant, bearing some relationship to the difference between full
load consumption and idle consumption. The actual con-
sumption for any output is of course the sum of these two.
Calling
B = the maximum brake H.P. of the engine;
b consumption per actual B.H.P. hour;
a = a factor such that a(bty gives the idle or no load con-
sumption,
we have the actual consumption as being
b(aR + x)
where x = the load on the brake in horse-power.
The values of these constants will depend greatly upon the
size of the engine, and may be taken to have the following
values for steam and gas engines of the usual sizes employed
in electric lighting.
FOR STEAM ENGINES. FOR GAS ENGINES.
a = 0*3 to 0*4. a 0*16 to 0*28.
l> 17 to 3 Ibs. of coal per hour. b = 12 to 18 cubic feet of gas per
hour,
1 82 Examples in Electrical Engineering.
The smaller the engine in each case the higher the value to
be taken.
If, now, we call z pence the price per unit, either pound of
coal or cubic foot of gas, of fuel, we can write the cost of one
hour as being
zb(a% -f x)
And the cost of i B.T.U. will be found by putting for x the
actual B.H.P. required, with the particular gearing and dynamo
to give an output of i B.T.U. ; the cost of i B.T.U. thus
varying from a somewhat large value at small load to lower
values at full load.
Example LIX. What will be the cost of fuel for u B.T.U.
from a dynamo of 90 per cent, commercial efficiency, when
run at full load of ioo a at 100", by a gas engine in which a =
0*2 and b = 15, the price of gas being 2S. $d. per 1000 cubic
feet ? The gas engine has a maximum power of 15 B.H.P.
Solution. The H.B.P. required will be
1000 100
746 XIOX ~ 9 o ='4 74 = *
whence the cost of fuel per hour will be
zb(afc -f x)
where z is
27
1000
or
penny
= 7ooo X T5(o ' 2 X X 5 + I474
x 266-1 = 7 -185 pence
1000
>which is the cost of 10 B.T.U., whence the fuel cost per
B.T.U. = 07185 pence.
Example LX. If with the above engine the average load
is only 3 B.T.U. when the commercial efficiency of the dynamo
is only 80 per cent., find the fuel cost of i B.T.U.
Prime Mover Costs. 183
Solution. Here the B.H.P. required is
1000 100
74 6*3X 8o - = 5-*3 = *
and the cost of fuel per hour is
1000 X '5 (' 2X '5 +5 ''3) = 3-33 Pence
which is the cost of 3 units or rn pence per B.T.U.
Example LXI. Find the cost of running the above
dynamo by a steam engine having the following particulars.
a 0-4 and b = 2-5, B = 15 when the cost of coal is 0-043
pence per Ib.
Solution. (i) on full load of 10 B.T.U. the B.H.P as before
will be 14*74 whence cost of fuel will be
0-043 X 2-5 (0-4 x 15 + 1474) = 2-23 pence
or only 0-223 pence per B.T.U.
(2) But on load of 3 B.T.U, the B.H.P. required = 5-23 as
before and the cost of i hour will be
0-043 X 2-5 (0-4 x 15 + 5' 2 3) = I<2 7 pence
or only 0-402 pence per B.T.U.
Example LXII. A large gas engine having a = o'i6, b -
12, and of 200 B.H.P., is used to run a dynamo which has a
commercial efficiency of 50 per cent, at an output of 7 B/T.U.
per hour, and 95 per cent, at 140 B.T.U. per hour. The price
of gas is 2S. 6d. per 1000; find the cost of gas per B.T.U. at
an output of 7 and 140 B.T.U. per hour.
Solution. (i) At 7 B.T.U. per hour the horse-power re-
quired is
7 X looo ioo _
~746^ C 50 "
whence the cost of gas per hour will be
X 12(0-16 x 200 + 18-7) = 18-28 pence
IOOO
184 Examples in Electrical Engineering.
or
18-28
= 2-61 pence per i B.T.U.
(2) At 140 B.T.U. horse-power required is
140 x 1000 100
746 " C "95" =
and cost per hour is then
30
IOOO
or
X I2(o'i6 x 200 -f 198) = 82-8 pence
o .o
= o'59 pence per i B.T.U.
140
It will be seen from the foregoing examples that the cost of
fuel will depend very much upon the extent to which the
machinery can be loaded. Large engines and dynamos run to
supply only a small fraction of their full power will be very
wasteful, and so run up the cost of generation. To the cost of
the fuel must also be added the cost of attendance, lubrication,
etc., all of which items become still more costly as the load
goes down. Thus, though at times of full load the cost of fuel
may be but a fraction of a penny per B.T.U., the average cost
is much increased by the, to a large extent, unavoidable waste
at times of light load. The following example will illustrate
this.
Example LXIII. The average pressure generated in a
central station is 430 volts, and the system is a five-wire net-
work maintained at 400 volts between the outers. At full load
there are 10,000 30- watt lamps in use. The dynamos have an
average efficiency of conversion of 92 per cent. The station
is on full load for 2 hours, on half load for 8 hours, on ^ load
for 6 hours, and runs idle for the rest of the 24 hours. One
mechanical brake H.P. costs 1*75 pence per hour for wages,
coal, oil, etc., at full load; costs 2*5 pence at half load, and
6 pence at -^ load, whilst at no load the cost of running is
Prime Mover Costs.
185
i$s. per hour. What should be the price charged per B.T.U.,
in order that there may be a profit of 10 per cent. ?
Time
hrs.
Output Watts
watts. generated.
Brake H.P.
required.
Total cost for
time in pence.
B.T.U.
paid for.
2
300,000 $$ x 300,000
= 3 22 >5oo
W x w
= 469*95
@ 175 pence
per hour
= 1644-8
6OO
8
150,000 ^ x 150,000
= 161,250
W x ifl^
= 2 34'97
@2'5^. perhr.
= 4699'4
1200
6
30,000 3 2 5 2 5
46-995
@ 6d. per hr.
= 1691-82
1 80
8
o
@ 15*. hour
= H40
O
T<~ito1 ^^cf
Tntnl unite
9475 pence j sold 1980
whence the cost per unit is ^P = 478 pence, and to pay
10 per cent, the price must be 5*258 pence.
EXAMPLES.
CHAPTER I.
1. The suspension of a moving-coil galvanometer is made of platinoid.
It is 3| inches long, 0*0005 mcn thick, and 0*004 inch wide. If platinoid
will be 20 times the resistance of copper, what is the resistance of this
suspension ?
2. What relationship will the resistance of the above suspension have
to the resistance of the coil, which is I inch wide and 2\ inches long,
being wound with 400 turns of copper wire 0*002 inch diameter ?
3. What will be the E.M.F. required to send a current of 10 amperes
through a circuit consisting of a dynamo of resistance of 7 ohms, i| mile
of leads of No. 6 B.W.G., and 80 lamps each of 2^ ohms resistance?
State the loss in the dynamo and in the leads, and give the horse-power of
engine required, if the dynamo converts T 9 S of the power received into elec-
tricity. Also if the lamps are paid for at so much per B.T.U. in the
lamps, what will be the cost of I B.T.U. if a mechanical horse-power costs
twopence per hour ?
4. What size of leads will be required to keep 400 6o-watt loo-volt
lamps at a P.D. of 100 volts, if the dynamo is 170 feet from the lamps,
has a resistance of o'Oig ohm, and produces a total E.M.F. of 106 volts?
5. How many secondary cells would be required to run the lamps in
the above question when each cell has a resistance of 0*00007 ohm, and
an E.M.F. of 2*05 volts at the start, but falling to 1-92 volt at the end of
discharge ? The lamps are connected to the cells by the same leads as
above.
6. In the above question the numbers of cells do not come out exactly,
but as whole numbers have to be used, find what compensating resistances
would be necessary to make the current exactly right.
7. A series dynamo has a resistance of 10 ohms, and sends a current of
12 amperes through 40 arc lamps taking 48 volts each, and equal to 1000
candle-power, all connected in series on a circuit of a total length of mile
Examples. 1 87
of No. 9 B. W.G. What will be the horse-power of engine required if the
dynamo has a frictional loss of f horse-power ?
8. What will be the horse-power of a dynamo to run 40 incandescent
lamps in series each of 1000 candle-power, and taking 65 amperes each
at 48 volts, if they are on a circuit \ mile long of No. 3 B.W.G., and the
resistance of the dynamo is 1*25 ohm, with a friction loss of 3! horse-
power?
9. A dynamo has an armature resistance of 0*042 ohm, and is required
to run 300 i6-candle-power 5o-watt lamps at the end of leads which are
' OI 333 onm resistance. The total current is 150 amperes. What will
be the reading of a voltmeter connected to the terminals of the dynamo,
and what percentage is this P.D. of the whole generated E.M.F. ?
10. What will be the respective resistances of two bars, one of copper
and one of aluminium, but both the same weight and the same length ?
11. A pair of conductors, I square inch sectional area, conveys 400
amperes to a distance of f of a mile from the dynamo. If the load at the
far end requires a P.D. 410 volts, find the P.D. required at the dynamo.
12. In a secondary battery the lead lugs connecting cell to cell are \ x I
inch in section, and 10 inches long each. If there are 54 cells joined in
series, find the resistance of these connections.
13. If the cells in the above question have 15 plates each cell, each
plate being 10 inches square and a quarter of an inch distant from the
next, find the resistance of the electrolyte in each cell if the resistance
of one cubic inch of the solution is 3 ohms.
14. Neglecting the E.M.F. of the cells in the above questions, find the
E.M.F. required to send a current of 100 amperes through the mere resist-
ances of the connections and electrolyte if all the cells are in series.
15. The connections on a switch-board are made of rod brass, and have
an aggregate length of 27 feet. They are to carry 150 amperes, and the
drop of pressure over all is not to exceed half a volt. Find the sectional
area of the brass.
16. A standard manganin resistance of o'ooi ohm is 20 inches long. It
is made of sheet metal T ' s inch thick. How wide will it be ?
17. An arc-light circuit consists of 60 lamps in series, each taking 10
amperes. Each lamp has the positive carbon 18 inches long and \ inch
diameter, and the negative carbon 12 inches long and / 6 inch diameter.
Neglecting the effect of^heating, find the loss of energy in the carbons
alone. *\
1 8. If the arc lamps in the above question be evenly spaced along a
circuit at a distance of 30 yards apart, and the cable employed is T 7 g S. W.G.,
what will be the total E.M.F. required if each lamp takes 45 volts in
addition to the drop in the carbons ?
V
1 88 Examples in Electrical Engineering.
19. What will be the nearest cable of strands of S.W.G. copper to
make leads to convey 600 amperes a distance of 30 yards so that there
shall not be more than i '26 volt lost in the lead and return ?
20. A motor taking current of 100 amperes at a P.D. of 2000 volts is
run from a dynamo 8 miles away. If the loss in the leads is not to be
more than 10 per cent., what will be the nearest cable of S.W.G. wire, and
what will have to be the P.D. at the dynamo?
21. If the power in the last question be transmitted at 20,000 volts,
what will then be the size of the bare solid conductor employed with only
5 per cent, loss of pressure ?
22. Find the horse-power of engine required to run 1000 i6-candle-
power lamps taking 100 volts and 0*6 ampere each. There is a loss in
the leads of i volt, and the dynamo gives out T 9 5 of the power received
by it.
23. If all the above lamps are replaced by arcs in series, taking 10
amperes and 50 volts each of 900 candle-power, so as to give twice the
total light, and if the loss in the leads be 4 volts, and the dynamo gives out
85 per cent, of the power received by it, find the horse-power of engine
required, and the number of lamps.
24. In the calibration of an ammeter to read 2000 amperes, what will
be the smallest power required, if a number of secondary cells be used in
parallel, each cell giving 2 volts ?
25. A street 4 miles long is to be supplied with 50 arc lamps per mile,
each taking 10 amperes and 45 volts. The current is supplied by 4
dynamos, one to each mile of street, the leads being respectively i, 2, 3,
and 4 miles, lead and return, of -^ square inch sectional area. If the
mechanical efficiency of each dynamo be 90 per cent., and its resistance
10 ohms, find the E.M.F. of each, and the power required to drive each.
26. What size of leads will be required to send current to a number of
98-volt lamps from a dynamo giving 200 amperes at 100 volts ?
27. A current is required to be sent a distance of 50 yards to 400 lamps,
taking o'6 ampere at 100 volts each, from a dynamo at P.D. of 101 volts.
What will be the size of leads required ?
28. How many incandescent, and how many arc lamps, will be required
to give together 20,000 candle-power for 20,000 watts? Each arc gives
2000 candle-power for 500 watts, and each incandescent 16 candle-power
for 60 watts.
CHAPTER II.
29. The distance between the feeding centres of a two-wire system of
distributors is 400 yards, and current is required at the rate of I ampere
per yard. What will be the size of the distributors if no two lamps are to
Examples. 1 89
differ by more than 2 volts? And what will be the size of the feeders if
the loss along them is not to exceed 10 volts, and the station is an average
of 500 yards distant from any feeding centre ?
30. Two parallel mains running round a district are of \ square inch
sectional area, and are connected to the station by feeders J square inch
sectional area at points 400 yards apart. Each feeder has to convey 400
amperes to the mains, and there are 5 pairs, of lengths respectively of 100,
200, 300, 400, and 500 yards. If the P.D. at each feeding centre is main-
tained at 204 volts, what will be the P.D. in the station at the ends of each
pair of feeders ?
31. In the above circuit, what is the maximum difference in volts
between any two lamps at full load ?
32. The feeding centres in a two-wire system are 500 yards aparU
What must be the sectional area of the distributors, so that there may not
be more than 2 volts difference between any two lamps if the full load is at
the rate of I ampere per yard of street ?
33. The distributors in a street 1200 feet long are connected to feeders
midway. The load is equal to one 48-watt loo-volt lamp per yard of
street, and the feeding centre is maintained at 101 volts. Find the size
of the distributors when the furthest lamps are at a P.D. of 99*5 volts at
full load.
34. A system of distributors is \ square inch sectional area, and the load
is one 5o-watt loo-volt lamp per yard of street. If the feeding centres
are maintained at 101 volts, find the distance between them so that the
drop in the distributors shall not be more than i| volt.
35. A street 503 yards long is supplied with current at the rate of
I ampere per yard from a 200- volt circuit. What will be the size of leads
if the dynamo is connected at the middle of the street, and the difference
between any two lamps is not to be more than 3*5 volts ? Also, if the loss in
the feeders is 7 volts, what will be the horse-power required at the dynamo
terminals ?
36. In a five- wire system the outers are I square inch sectional area. What
will be the maximum distance allowable between the feeding centres if no
two lamps are to differ by more than \\ volt? There are an average of
four o'6-ampere lamps in series per yard of street.
37. If the distance between the feeding centres be 400 yards on a three- 1
wire circuit running two 2OO-volt lamps in series per foot of street, each
lamp taking 0*3 ampere, what must be the size of the outers if no two
lamps are to differ by more than 2 volts ?
38. In a three-wire system running 200- volt lamps, two in series between
the outers, the feeding centres are 500 yards apart and supply a maximum
current of 500 amperes each. If the outers are square inch sectional area,
Examples in Electrical Engineering.
what is the P.D. between the terminals of the poorest lamp when the
feeding centres are kept at 410 volts ?
39. A district is to be supplied with current at the rate of I ampere per
yard of street, the system is two-wire with loo-volt lamps, and the feeding
centres 600 yards apart. What will be the current supplied by each
feeder, and what size must the distributors be in order that no two lamps
may differ by more than I volt ?
40. If the feeders in the last question are 700 yards long on an average,
and the loss along them is 10 volts, find the weight of copper per mile of
street.
41. A district takes \ ampere per yard of street, using 2OO-volt lamps
on a two-wire system. If the feeding centres are 600 yards apart, what
current will they supply, and what will be the size of the distributors in
order that no two lamps may differ by more than 2 volts ?
42. If the feeders in the above case be 700 yards long, find the weight
of copper per mile of street with 20 volts lost in feeders.
43. A district is supplied with two loo-volt lamps in series on a three-
wire system at the rate of \ ampere per yard of street. The middle wire
is \ the section of the outers, and the feeders are 600 yards apart. If no
two lamps are to differ more than I volt, find the size of the outers. Also
find the weight of copper per mile of street, if there be a loss of 20 volts
in the feeders, which have an average length of 700 yards.
44. If \ ampere per yard of street be supplied on a three- wire system
with two 2OO-volt lamps in series, with feeders 600 yards apart, find the
size of the distributors so that no two lamps shall differ more than 2
volts. And if 40 volts be lost in the feeders which are an average of
700 yards long, and the middle wire be \ the area of the outers, find the
weight of copper per mile of street.
45. If the system be three-wire, with the outers i square inch sectional
area, and the middle wire \ square inch area, the feeders being 600 yards
apart, what may be the maximum load in order that no two lamps may
differ by more than 2 volts? Each lamp is for 200 volts, and takes 0*3
ampere.
46. A current is required along a street 2000 feet long at the rate of
I ampere per lo feet. Find the sectional area of conductors, so that there
shall be only 2 volts difference between any two lamps. 1000 feet of
copper i square inch section is O'OoS ohm
. (a) When current is supplied at one end of the street.
(b) When current is supplied midway. (City Guilds Examination.}
Examples. 191
CHAPTER III.
47. A secondary battery of 54 cells has a resistance per cell of 0*0002
ohm and an E.M.F. of from 2 volts at start to 1-85 volt at the end of a
run. What will be the current sent through a bank of lamps of 2 ohms
resistance 70 yards away, connected by leads of 0-02 ohm resistance ?
48. Find the speed at start and end of charging for a dynamo to send
150 amperes through the above cells. The dynamo produces T \j volt per
revolution per minute, is separately excited, and has armature resistance
cro2i ohm. In charging each cell has B. E.M.F. ranging from 2*2 to 2*4
volts, and the battery is 20 yards away, and connected by leads of 0*0004
ohm per yard.
49. What resistance should be introduced into the circuit on starting to
charge in order to keep the current 150 amperes, though the dynamo runs
at the higher speed ?
50. Find the E.M.F. required to charge 108 cells in two parallel with
200 amperes each cell, if the B. E.M.F. is 2*3 volts each. Each cell has
resistance 0*001 ohm, that of the dynamo being 0*005 ohm > and leads
ox> I ohm.
51. Find the E.M.F. required to charge 57 cells in series, each having
B. E.M.F. of 2*3 volts and 0*0005 onm resistance. The dynamo = 0*03
ohm, and leads 0^04 ohm, and current 100 amperes.
52. What will be the electrical efficiency of the above cells when dis-
charging at 100 amperes through lamps on the above leads, each cell
giving an average E.M.F. of 1*9 volt? Also state the P.D. at the lamp
terminals.
53. If a dynamo gives 100 volts at 1000 revolutions, at what speed
must it run to charge 54 cells at 50 amperes, each cell having B. E.M.F.
2*2 volts and 0*005 ohm internal resistance when the leads = 0*01 ohm ?
54. If the above cells are discharged at 50 amperes, the E.M.F. of each
being 2 volts at start, falling to 1*8 volts at the end of run, what per-
centage of the power put in is utilized ?
CHAPTERS IV. TO IX,
55. What will be the brake horse-power of a motor series wound having
armature resistance) 0^025 ohm, series coil 0*025 ohm, when running at
1 200 revolutions per minute with a current of 50 amperes supplied at an
E.M.F. of loo volts? The electrical loss in the machine is \ of the
friction losses.
56. In the above motor what will be the value of the total flux through
1 92 Examples in Electrical Engineering.
the armature if there be 360 wires all round ? Also find the density in
the air-gap, if the polar angle is 135 degrees, the bore is 22 cm. diameter,
and length 20 cm.
57. In the above motor, what will be the pull in pounds on any one
wire ?
58. A motor is supplied with 150 amperes at 100 volts. The magnets
are shunt wound to 40 ohms, and the armature resistance O'OI ohm. If
I horse-power is lost in various frictions, what will be the brake horse-
power, and what the mechanical efficiency ?
59. If the above motor armature has 120 wires all round, find the total
lines, and the density in the air-gap', if the polar angle is 140 degrees, the
diameter and length of bore respectively 30 cm. and 25 cm., and the speed
1 200 per minute.
60. A machine separately excited gives 100 volts P.D. at 15 revolutions
per second when run as a dynamo. The armature resistance is 0*015
ohm, the watts lost in eddy currents = 120, in hysteresis = 180, and in
mechanical friction = 150. At what E.M.F. and current will it run, at
the same speed as an unloaded motor ?
61. A magneto machine gives 50 volts as a dynamo at no load. The
armature resistance is I ohm, and 125 watts are lost in various frictions.
Find the E.M.F. and current required to run it as an unloaded motor at
the same speed.
62. A shunt dynamo having armature resistance o'O2 ohm, shunt resist-
ance 40 ohms, when run on open circuit at 20 revolutions per second,
gives P.D. of 100 volts. If it be separately excited at 100 volts, and run
as a motor at the same speed, to work a crane lifting 3 tons, at the rate of
30 feet per minute, what will be the current taken, and E.M.F. required,
if the crane has a friction loss at the rate of 200 pounds per ton lifted, and
the mechanical friction of the motor is equal to 150 watts, the hysteresis
loss 130 watts, and eddy current loss 170 watts? Assume that there is no
armature reaction.
63. What is the commercial efficiency of the above combined crane and
motor ?
64. The E.M.F. produced by a separately excited machine when run
as a dynamo is proportional to the speed, and is 100 volts at 1000 revolu-
tions per minute. The armature resistance is 0*04 ohm. The loss due
to hysteresis is proportional to the speed, and is 120 watts at 1000 per
minute. The eddy current loss, however, is 80 watts at 1000 revolutions
per minute, and is proportional to the square of the speed. The mechani-
cal-friction loss varies directly as the speed, and is 1 10 watts at 1000 revo-
lutions. If the above machine be supplied with current at 100 volts, at
what speed will it run, and what current will it take (a) unloaded, (l>)
loaded to 2 brake horse-power?
Examples. 193
65. A machine has armature resistance of 0*005 ohm, and shunt resist-
ance 25 ohms, and is separately excited at 100 volts. If run at 600 per
minute as a dynamo, it gives 100 volts on open circuit, the eddy current
loss being 250 watts, the mechanical friction 300 watts, and hysteresis loss
400 watts. When supplied at 100 volts, what will be the speed and
current required to run as an unloaded motor ?
66. A series dynamo gives a current of 12 amperes, and is required to
run a series motor of I brake horse-power. The dynamo has 360 wires all
round, and a resistance of 1*3 ohm, with one million lines through the
armature. The resistance of the motor is also 1*3 ohm, and \ horse-power
is lost in various internal frictions. At what speed must the dynamo run,
and what will be the P.D. at the motor terminals, the electrical efficiency
of the dynamo, and the electrical and mechanical efficiency of the motor ?
67. A series motor produces a B.E.M.F. of 94 volts, and is supplied
with loo amperes at a P.D. of 100 volts. If horse-power is lost in
various frictions, what will be the brake horse-power, and the electrical
and mechanical efficiency ?
68. A motor has an armature resistance of 3 ohms, with 3*6 million
lines through it. The shunt coil is 750 ohms, and J a horse-power is lost
in various frictions. If the armature has 720 wires all round, and the
machine is supplied with 15 amperes at 550 volts, find the speed, the brake
horse-power, and the electrical and mechanical efficiency ?
69. A series motor has 240 wires all round the armature, with a resist-
ance of O'o8 ohm, and 22 million lines through it. The magnet winding
is 0*12 ohm, and the machine is to give ipo brake horse-power when
taking a current of 150 amperes. If n be the revolutions per second,
and the mechanical friction loss is ioo#, the hysteresis loss Son, and the
eddy current loss is 3 2 , find the E.M.F., the speed, and electrical and
mechanical efficiency.
70. A series machine has 140 wires all round the armature, which is
222 '2 square centimetres sectional area, with a density of 18,000, and a
speed of 1200 per minute. If the P.D. at the terminals for a current of
loo amperes is 106 volts, what would be the electrical efficiency as a
dynamo, and also as a motor at the same speed and current ? Also, if the
friction loss is horse-power, in both cases, state the brake horse-power
as a motor and mechanical efficiency, both as dynamo and motor.
71. A series dynamo, when giving a current of 10 amperes, has a P.D.
at its terminals of 1000 volts at a speed of 1040 per minute. The armature
resistance is 1*8 ohm and magnet resistance 2*2 ohms. At what speed
will it run as a motor if supplied with 10 amperes at a P.D. of iooo volts ?
The internal friction losses are horse-power. Also give electrical
efficiency as a dynamo, the brake horse-power and electrical and me-
chanical efficiency as a motor,
OF THR
CTNIVERSITY-
194 Examples in Electrical Engineering.
72. What would be the E.M.F. required to run the above motor at
1040 revolutions per minute ? and what would then be its brake horse-
power and electrical and mechanical efficiency, the internal friction being
400 watts ?
73. An armature running at 20 revolutions per second has 96 wires all
round, and 4 million lines through it; its resistance is I '05 ohm, and the
shunt magnet coil 20 ohms. If 200 amperes be passed through the
armature, what E.M.F. will be required to run it as a motor? Also if
the hysteresis and eddy current losses are together equal to 300 watts, and
a horse-power is lost in mechanical friction, what will be the brake horse-
power, and the electrical and mechanical efficiency ?
74. A crane is to lift I ton at the rate of 500 feet per minute, and has a
friction loss of 5 horse-power. It is run by a motor having 6 million
lines through the armature, which has 400 wires all round, and a resistance
of 0*05 ohm. There is a loss of f of a horse-power in mechanical friction,
and 500 watts in iron losses. The magnets are series wound to o - o8 ohm.
If it is supplied at 600 volts, what will be the current taken, the speed,
the brake horse-power, the electrical and mechanical efficiency of the
motor, and the plant efficiency ?
75. A motor with ring armature i6J inches mean diameter and 120
wires all round runs in a field of 3000 lines per square centimetre at 800
revolutions per minute. The length of the bore is 16^ inches, its diameter
being 17 inches, and the curved length of the polar face 22 inches. The
resistance of the armature is 0*024 ohm, the shunt coil being 48 ohms.
Half a horse-power is lost in mechanical friction, and 400 watts in iron
losses. If the armature be supplied with a current of 200 amperes, what
will be the pull on one side of the belt on a pulley I foot mean diameter ?
Also give the brake horse-power and the electrical and mechanical
efficiency.
76. A mechanical contrivance requires 100 horse-power, and wastes 15
horse-power in friction. If run by a series motor having armature resist-
ance 0*06 ohm, magnet coil 0*08 ohm, 360 wires all round the armature
and 18 million lines through it, ij horse-power lost in friction, 600 watts
in iron, and supplied with a P.D. of 700 volts, find the speed, the current
taken, the electrical and mechanical efficiency of the motor, and the plant
efficiency.
77. A compound dynamo to give P.D. of 105 volts and a current of 90
amperes is to have 92 per cent, electrical efficiency. If N be about
3 million lines, the speed noo per minute, and half the loss is to be in
the armature, the rest in the magnets divided as f in the shunt and in
the series coil, find the wires all round, and the resistances of the armature,
shunt, and series coils.
78. A shunt motor has the shunt resistance 750 ohms, armature resist-
ance 2 ohms, and N 4*2 million. If supplied with 15 amperes at 500
Examples. 195
volts, find the brake horse-power, the speed, and the electrical and me-
chanical efficiency, if there be 720 wires all round the armature, and the
friction losses are equivalent to 300 watts.
79. The magnetic circuit of a dynamo has reluctance = 0^0025, and an
average of 5 million C.G.S. lines through it. Find the size of wire for a
shunt winding, if the average length of one turn is 30 inches, and the P.D.
at the terminals is 400 volts.
80. In the above question, what number of turns will be required on the
shunt coil so that the energy waste shall not exceed 600 watts ?
81. A dynamo takes 8000 ampere turns, and produces a terminal P.D.
of 100 volts. If the average length of one turn on the shunt coil is 2 feet,
find the size of wire required.
82. An armature running at 18 revolutions per second, has 288 wires all
round, and a resistance of 0*025 ohm. Find N so as to give 70 amperes
at 100 volts.
83. The electrical efficiency of a dynamo is 98 per cent., and mechanical
efficiency 93 per cent. If the output be 200 units, what power is wasted
in resistance, and what power in mechanical and other frictions ?
84. A drum armature is 3 feet in diameter, with a lo-inch hole and 3*8
feet length of iron parallel to the shaft. It runs at 400 revolutions per
minute, and is worked at such a density that the hysteresis loss is 5000
ergs per cubic centimetre per cycle. Find the hysteresis loss in watts.
85. If the resistance of the above armature be 0*005 ohm, the current
600 amperes at P.D. of 440 volts, shunt resistance no ohms, eddy current
loss 2 horse-power, and mechanical friction 2j horse-power, find the
mechanical efficiency.
86. A shunt dynamo has electrical efficiency 95 per cent., and gives an
output of 20 units. If there is an equal loss in the shunt and in the
armature, and the P.D. is 100 volts, find the resistances of the armature
and shunt coils.
87. If the mechanical efficiency of the above dynamo is T 9 , what power
must be wasted in other ways than in resistance ?
88. A shunt dynamo has magnet coil 33 ohms resistance, and armature
o'oo) ohm, and gives current to 500 6o-watt loo-volt lamps, with a
terminal P.D. of 101 volts. Find the electrical efficiency; and if the
mechanical efficiency is 81 per cent., what horse-power of engine will be
required, and what power is lost otherwise than electrically?
89. Two series machines, each having armature resistance 0*025 ohm,
and magnet resistance 0*025, also N is 5 million lines, and the current
100 amperes. They are connected together by leads of 0*1 ohm resist-
ance. Each machine has a frictional loss of 600 watts* which may be
considered to be independent of the speed. If one machine be run as a
196 Examples in Electrical Engineering.
dynamo by an engine of 15 brake horse-power, at what speed will they
run, and what will be the brake horse-power of the motor when the
current is 100 amperes ? Also state the ratio of the power given out by
the motor to the power received by the dynamo.
90. In the last question, what brake horse-power engine would be
required to run the dynamo if the motor be used so as to give out 15 brake
horse-power, the current and losses being the same as before ? Also state
the speeds.
91. A series motor is to give 50 brake horse -power at a speed of 600
revolutions per minute. Its armature has 720 wires all round, is f ohm
resistance, and has 12 million lines through it. If the field coils have a
resistance of 0*833 ohm, find the E.M.F. required to supply the above
power through leads 2\ ohms resistance, the friction losses being 2 horse-
power.
92. How many ampere-turns will be required on each limb on a
Manchester-type machine having N = 4/2 X io 6 , A = 280, / 35, A g 1620,
/,, 2-1, A; 587, l p 35, A m 294, /, 26, v - 1-5 ? The pole-pieces are of cast
iron, and magnet cores wrought iron.
93. A shunt dynamo to give 310 volts has the length of the average
turn on the magnet coils I metre, and is of the single-magnet type. It also
has the following dimensions : l a 70, A a 1400, N 21*6 millions, l g 2*4, A,,
4000, /, 240, A m 2700. If v = I '4, and the magnets are wrought iron
wound with 2000 turns of wire, what will be the shunt resistance at a
temperature of 35 C. ?
94. A single-magnet bipolar dynamo has the pole-pieces of cast iron
and yoke of wrought iron, on which is placed the magnet winding. The
armature resistance is 0-024 olim hot, with 144 wires all round, and
N = 7 million lines at no load. It is to give 200 amperes at a P.D. of
120 volts, and has the following dimensions: l a 40, A a 450, l g 2*5, A g 2300,
IP 80, A^ 1300, /, 76, A m 700, with v I '43. Its shunt coil has 4000 turns,
each of 45 inches average length, and the series coil is to be o'Oi ohm
resistance hot. If the brushes are lead forward 6 wires on the armature
at full load, find the windings of the series coil, the size of the shunt wire,
and the electrical efficiency at full load. The shunt coil is connected
"long shunt."
95. An armature, gramme wound, for a bipolar-field is required to
carry 100 amperes at a P.D. of 100 volts. It has the following dimensions :
N = 4 millions, n 20*43, w I2 *> eac ^ being 80 cms. average length and
0*24 sq. cms. sectional area. The iron stampings are 25 cms. diameter
with 15-cm. hole, and make altogether 25 cms. length of iron parallel to
the spindle. If the over-all surface is 3675 sq. cms., what will be the rise
in temperature above the air on account of copper and hysteresis losses,
neglecting the heating due to eddy currents ? Take i cc, of copper to be
o '000002 ohm resistance.
Examples. 197
96. A drum armature for bipolar-field is 25 cms. diameter with 7-cm.
hole, and has 25 cms. length of iron parallel to the shaft. It runs at
20-43 revolutions per second in a field of 4-95 X io 6 lines, and is wound
with 102 wires all round, each of 0*3 sq. cm. sectional area. Each
convolution has an average length of 135 cms. If p = 0-000002, and there
be the same rise in temperature as in the gramme armature above, find the
maximum current allowable, neglecting the eddy current loss.
97. The magnetic system of the gramme armature in question 95 above
has the following dimensions : 4 31, A a 250, l g 2, A^ 950, l m 170, A m 415,
all the iron parts being charcoal iron. The leakage factor is I "4. Find
the size of shunt wire required if the average turn is 3 feet long, and the
P.D. at the terminals is 100 volts, the maximum temperature of the shunt
coil being 35 C.
98. Find the size of shunt wire for a dynamo having the following
dimensions : Armature ring wound, w = 252, l a 37, A a 137, diameter of
core iof ins., bore n| ins., and length of 9 ins. Magnet limbs are of
cast iron, 9 ins. X 7 ins. area and 6^ ins. apart, with no polar projections,
l m 100. Speed 1000 per minute to produce 97 volts on open circuit. One
turn on the shunt coil is 40 ins. long, and the coils rise to a temperature
of4OC. Leakage factor I ^.
99. If the armature resistance in the above machine is 0*07 ohm, and
the shunt coil 42 ohms, find the number of series turns to keep the P.D. of
97 volts at 70 amperes. Allow the series coil to be 0*04 ohm resistance,
and the lead of the brushes 20 degrees. The connections are to be " long
shunt."
100. A dynamo is to generate 118 volts, and has 17,080 lines per square
centimetre in the armature, which has an area of 480 sq. cms., and runs
at 720 revolutions. The average length of I coil on the (ring wound)
armature is 46 ins., and the size of conductor 0*072 sq. in. If the
machine be bipolar, and gives a current of 200 amperes, find the P.D. at
the terminals at 35 C.
101. A dynamo has 200 wires all round the armature, which has a
resistance of 0*03 ohm, an area of 250 sq. cms. at a density of 17,000,
and is to run 200 half- ampere i6-candle-power lamps at the end of leads
of i sq. in. area and 20 yards long. At what speed must she run to keep
the P.D. at the lamps at 100 volts?
102. A shunt dynamo gives a terminal P.D. of 216 volts at a current of
50 amperes. Its armature has 7 million lines through it, and 160 wires all
round, each coil being 4 feet long, and bipolar ring wound. It runs at
1 200 revolutions, and the shunt resistance is 108 ohms. Find the size of
wire on the armature at a temperature of 35 C.
198 Examples in Electrical Engineering.
CHAPTER X.
103. A voltmeter coil has a core inch diameter, and can be wound to
a depth of \ inch. It is i'2 inch long, and requires about 300 ampere-
turns to produce a deflection to the full extent of the scale. What size of
wire will be required to make the instrument read 120 volts total, and
what will be the energy wasted in the coil if wound with copper wire
covered with silk to a radial depth of crooi inch ?
104. A certain type of instrument wound as an ammeter to read 20
amperes total has 40 turns of wire on its coil. The length of the mean
turn is 6 inches, and the diameter of the wire 0*2 inch. What is the waste
of energy at full reading ?
105. If the space for winding in the above instrument is 2 inches by
\\ inch, find the size of wire and total waste of energy to make it read
as a voltmeter to 20 volts with the coil wound with copper, and having an
extra coil to increase the reading up to 120 volts wound with manganin.
106. How much will it cost a central station to keep an indication of
105 volts on each of two voltmeters for a year of 8000 hours, if the cost
of a B.T.U. is 2f pence, and interest on the first cost of instrument is to be
at the rate of 5 per cent, per annum ?
(1) A magnetic instrument costing 10, and having a resistance of
2100 ohms.
(2) A hot-wire instrument costing 7, and having a resistance when in
use of 470 ohms.
107. An instrument is wound with copper wire, and has a resistance of
18557 ohms at o C. Its dial is divided into 120 divisions, and 37^5
milli-amperes gives a deflection of 75 divisions. What will be the constant
of this instrument as a voltmeter at 20 C., and what resistance should the
extra coil of manganin have to increase the range to 2400 volts ? Also
state the T.V.R. with the extra coil in circuit if manganin is of constant
resistance.
108. A voltmeter is wound with copper, and has a resistance of 2000
ohms at the temperature at which it was calibrated to read 120 volts total.
It is first connected to the terminals of a dynamo in a hot engine-room,
where its resistance rises to 2200 ohms. It is then taken outside, where
the temperature is so low as to make its resistance only 1800 volts, and
connected to the end of the circuit, but on the dynamo side of the load.
The leads have a resistance of 0*02 ohm, and the lamps take loo amperes.
If the true P.D. at the dynamo is 102 volts, what will the instrument read
in the two cases ?
109. A voltmeter of 1500 ohms resistance is connected to the terminals
Examples. 1 99
of a lamp. An ammeter connected in the main circuit reads O'6 ampere.
If the voltmeter indicates 150 volts, what is the resistance of the lamp?
no. An instrument reads 10 milli-amperes for 100 divisions deflection
at 15 C, being wound with copper to a resistance of 100 ohms. What
will it read as a voltmeter at 25 C., when in circuit with a manganin
extra coil of 2900 ohms resistance ?
in. A voltmeter to read 100 volts requires 1000 ampere-turns. If the
average turn of wire be 3 inches long, find the size of wire required.
112. Find the diameter of copper wire for a voltmeter requiring from
300 to 400 ampere-turns, reading 60 volts total, if the mean diameter of
the winding is f inch.
113. A voltmeter requires 750 ampere-turns, and reads 120 volts total.
If the mean turn is 6 inches long, find the diameter of copper wire
required.
1 14. If only one-fourth part of the resistance of the above instrument
is to be upon the working coil, the rest being of manganin on an extra
coil, find the diameter of the copper wire required for the instrument and
the total resistance, if the waste is 8 watts.
CHAPTER XL
115. A train weighs 120 tons, and is to go at the rate of 25 miles per
hour. The resistance to motion on the level is 25 Ibs. per ton; the
gearing is such that I ampere gives a pull of 10 Ibs. The line is
divided into four stages, viz. level, up an incline of I in 480, up incline
of I in 348, and down incline of I in 960. Find the current required for
each stage, going and returning, and also state the horse-power.
116. The resistance to motion on the level is 30 Ibs. per ton, and the
train weighs 100 tons. I ampere gives a tractive force of 20 Ibs. Find the
current, the E.M.F., and horse-power required to run both ways on the line
in the previous question.
117. If the E.M.F. of supply is 500 volts, find the current required to
move a train of 200 tons weight at a speed of 50 miles per hour both on
the level and up and down an incline of I in 200. The resistance to motion
on the level is 15 Ibs. per ton.
1 1 8. In the last question find the total current to get up full speed in
five minutes.
119. Find the current required to run a car of 12 tons weight at 10 miles
per hour on the level and up and down inclines of I in 200 and I in 80, if
the resistance to motion on the level is 25 Ibs. per ton, and the E.M.F. of
supply is IQO volts,
2OO Examples in Electrical Engineering.
1 20. For the above car find the extra current required to get up full
speed in half a minute.
121. A car weighing 10 tons runs up an incline of I in 100. The motor
is geared to give a pull of 10 Ibs. per ampere. If the resistance to motion
on the level be 30 Ibs. per ton, find the current required. Also give the
horse-power for a speed of 4 miles per hour.
122. A tram-car, weighing 6 tons empty, starts empty, on a level line,
picking up 20 persons halfway along. On reaching an incline up I in
50, 10 more persons get on. Fifteen persons get off when the car is going
down I in 100. What will be the current required on the level both
empty and loaded ; on incline up I in 50 loaded ; and down I in 100
loaded as above? The tractive force required is 25 Ibs. per ton on the
level, and I ampere gives a pull of 8 Ibs. Also state the horse-power
required for the various conditions for a uniform speed of 6 miles per
hour. Take 20 persons to one ton.
123. A series dynamo has the following relationship between N in the
armature and the current C :
10 12 13 14 16 18
N 0-90 1-5 1-85 2-i . 2-3 2-44 2-45 2-4}. 2-26 r8
where N is in millions of lines. The armature has 500 wires all round,
and a resistance of 077 ohm. The pull exerted as a motor is
4XC X NX io~ 6 Ibs. If supplied with an E.M.F. of 300 volts, find
the current required as a motor to haul a car 3 tons in weight on level, and
up and down incline of I in 150. Also give the speed of the car and the
horse- power required. The resistance to traction on the level is 25 Ibs.
per ton.
124. A series dynamo of 0^15 ohm resistance is supplied with current at
440 volts, and has the following relationship between C and N :
C 10 26 50 75 ioo 120 140 150 160 180 I 200 I 220 240 250
N I 4 | 8 | 12 | 14-8 | 167 | 177 | 18-1 | 18-2 | 18-1 | 17-5 | i6'6 | 15-3 | 13-4 | 12
the induction being in millions of lines. If the tractive force exerted by it
is O'5CN X 10 ~ 6 Ibs., find the C required to run a train of 50 tons weight
on level line, taking 20 Ibs. per ton ; the maximum speed in miles per hour,
and the revolutions per second of the motor.
125. With the above motor and train, give the C required, the maximum
speed of train and of motor to go up an incline of I in 200.
126. For the same motor and train, find the maximum speed of train, the
C required, and the speed of the motor to go down an incline of f in 200.
Examples.
201
127. Again, with the same train, find the shortest time in which the
maximum speed can be attained on the level line.
128. A shunt motor, working at 100 volts, has 200 wires all round the
armature and the following relationship between the current in the arma-
ture C and the magnetic flux N :
C
N
N being given in millions of lines. The armature resistance is O'I2 ohm,
and that of the shunt 50 ohms. The pull exerted is O-8CN X io~ 6 Ibs.
Find the motor speed, the current required, and the speed of the car to run
on the level if the car weighs 4 tons and resistance to motion is 25 Ibs.
per ton.
129. With the same motor and car, find the current required, speed of
car and motor to go up incline of I in 200.
130. For what current will the speed of the above motor be slowest ?
6
10
20
30 4
So
60
70 80
86
2-485
2-4825
2*475
2*46
2*4375
2-4023
2-3525
2-2823
2-17
2-0075
CHAPTERS XII. TO XIV.
131. The maximum value of the alternating E.M.F. in a circuit is 2820
volts, and the resistance is 100 ohms, being non-inductive. What will be
the average value of the current ; and if the terminals of the circuit be con-
nected to an electrostatic voltmeter, what will it indicate ?
132. A transformer has a primary resistance 12 ohms, a secondary
resistance 0*02 ohm. It is supplied with a virtual P.D. of 2010 volts at
the primary. What will be the P.D. at the secondary terminals when
running 50 lamps taking O'6 ampere each, the ratio of transformation being
20 to I, and the magnetizing current negligible ?
133. A circuit has 10 ohms resistance and a B. E.M.F. of 20 volts : what
will be the current through it if connected to an alternating P.D. of 29
volts as indicated on a hot-wire voltmeter ?
134. What will be the horse-power supplied to a circuit having a
B.E.M.F. of loo volts virtual, and a resistance of I ohm, when connected
to a dynamo producing a maximum E.M.F. of 200 volts alternating?
135. A circuit has a resistance of 0-015 ohm > and takes a current of 50
amperes when connected to an alternating P.D. of 60 volts. Find the
B.E.M.F.
V 136. An alternating P.D. of 100 volts virtual is supplying a current of
12 amperes to two 40- volt lamps connected by a switch with two choking
coils, in such a manner that when the switch is on contact No. I, the two
lamps in series are in series with coil x ; but when the switch is on contact
2O2 Examples in Electrical Engineering.
No. 2, only one lamp is in circuit, and is in series, with the two coils x and
y in series. Find the B.E.M.F. which the coils x and y must have.
137. What must be the maximum magnetic flux in the cores of the
coils x and y in the previous question, if they have each 200 convolutions
of wire, and negligible resistance, there being 80 complete cycles per
second ?
138. A transformer is to supply current to one hundred 6o-watt loo-volt
lamps, and the secondary resistance is 0*107 ohm. The primary has 20
times as many convolutions as the secondary, and its resistance is 6 ohms.
The magnetizing current is 0*8 ampere, and lags practically 90 after the
primary E.M.F. What must be the P.D. at the primary terminals to give
the lamps 100 volts?
139. A transformer is working off mains kept at a P.D. of 1015 volts.
The resistance of its primary coil is 13 ohms, and that of the secondary
o*li ohm. If the ratio of transformation is 10 to I, what will be the P.D.
at the secondary terminals, when the secondary current is 20 amperes,
and the primary magnetizing current is 0-5 ampere lagging 90 ?
140. With the above transformer, what should the primary terminal P.D.
be to make the secondary P.D. exactly 100 volts ?
141. A transformer with 9-6 to I ratio of transformation is to give
loo amperes at a secondary P.D. of 205 volts. The secondary resistance
is 0*035 ohm, and that of the primary 3 ohms. It is connected to the
station by leads having a total resistance of 4 ohms. Find the P.D. at
the station end, at zero load, half-load, and full load, when the magne-
tizing current may be neglected.
142. A transformer has an output of 10 units at full load, with a loss of
o'l unit in the iron, and 0*1 unit in the copper. What will be the all-day
efficiency if run on a load factor of?
Hours 10 4 6
Load
143. A transformer has to convert from 1000 volts to 200 volts, giving
a secondary current of 60 amperes. The primary resistance is 075 ohm,
the secondary being o - O2i ohm. What will be the efficiency at g, , J,
and full load, if there be an iron loss of 120 watts?
144. If the above transformer be used on a load factor thus
Hours
16 *
Load j o i
what will be its all-day efficiency ?
Examples. 203
145. The primary coil of a closed iron circuit transformer is to work at
P.D. of 2050 volts. The secondary is to give a maximum of 50 amperes,
at a P.D. of 101 volts. The secondary resistance is O'oiS ohm, and the
primary 8 ohms. If the iron loss is 140 watts, what will be the efficiency
of conversion at full load, 5, and ^ load ? And if employed on a load
diagram as under, what will be the all-day efficiency ?
Hours I i I I I i I 2 I 4 I 2 I 14
Load i 0-9 075 0-5
0-4 0-2 o
146. A transformer has a secondary current of 400 amperes, the resistance
of the secondary 0*0015 ohm, and the primary 0*8 ohm, with 20 times as
many turns as the secondary. If the iron loss is two-thirds of a horse-power,
and the magnetizing current is 5 amperes lagging nearly 90 after the
primary E.M.F., what will be the efficiency at full, , ^ and 5 ' load?
147. A number of open iron circuit transformers are supplied by a
concentric cable 5 miles long, and having a capacity of 2 microfarads per
mile. The total magnetizing current is 20 amperes lagging practically 90-
The cable has a resistance of 4 ohms, and the P.D. at the transformers is
to be 2020 volts. Find the P.D. in the station and the capacity of the
shunting condenser required to reduce the current in the cable to practically
zero. The dynamo is 140 CO 'per second.
CHAPTER XV.
148. A series motor which requires 50 amperes at 400 volts has a
resistance of O'8 ohm. It is to have a starting and regulating resistance
such that the starting current cannot exceed 30 amperes, and the speed may
range from 200 to 800. Find the values of the resistances required, and
state the maximum power for each speed with a current of 50 amperes.
149. A shunt motor having R s 35, R 0*05, and running at 800 when
supplied at 120 volts, is required to range in speed from 100 to 800. The
full-load current is 100 amperes. What will be the value of the regulating
resistance required, arranged so that the shunt is always at a P.D. of 120
volts ? Also, if there be a friction loss of 800 watts proportional to the
speed, what will be the brake horse-power at slow and high speeds with
full current in each case?
150. A shunt motor is to have a regulating resistance arranged so as to
insert resistance in the shunt circuit as it is removed from the armature
circuit. R s = 25, R a = O'O2, and the full-load current is 250 amperes at
loo volts. The speed is to range from 200 to 600. Find the value of the
resistance.
204 Examples in Electrical Engineering.
CHAPTER XVI.
151. A gas-engine consumes when on full load three times as much gas
as when running idle. At full load it gives 5 horse-power at the pulley,
and runs a dynamo having R* 50 ohms and R rt 0*3 ohm, giving 100 volts.
Find the average cost of I B.T.U. for a run of 12 hours, if on no load
2 hours, \ load 3 hours, % load 6 hours, and full load of 40 sixty-watt
lamps for I hour. The dynamo has a friction loss of 300 watts, and each
brake horse-power on the engine takes 18 cubic feet of gas costing 3 shillings
per 1000.
152. A steam-engine has a no-load loss equivalent to T 3 5 its maximum
brake power, which is 200 horse, for each of which the consumption of
coal is i '8 Ib. per hour, costing 15 shillings per ton. The friction load
attached to it is 20 horse-power, and useful load 25 amperes at 400 volts
for 5 hours. Find the average cost of I B T.U. during the run.
ANSWERS.
1. 23-45 ohms.
2. ^gR of coil.
3. 2089-6 volts, 7 volts in dynamo, 19-6 volts in leads, 31-12 H.P.,
3-112 pence.
4- 0-4557 sq. in.
5- 53 or 54.
6. 0-000916 ohm at start, and 0-00555 ohm at finish.
7- 33-69 H.P.
8. 179 H.P.
9. 102 volts ; 94 per cent.
10. 2 to- I.
JI - 435 '472 volts.
12. 0-018 ohm.
13. 0-000535 ohm.
14. 4-7 volts.
15. 0*272 sq. in.
16. 5-76 ins.
17. 1746 to 3492 watts.
1 8. 2894 or 37 volts.
19. f} S.W.G.
20. fi S.W.G. ; 2185-6 volts.
21. About No. 13 S.W.G.
22. 90-26 H.P.
23. 28-5 H.P., 36 lamps.
24. 5-3 H.P.
25. E.M.F. 2364, 2378, 2392, and 2406 volts; power 35-2, 35-4, 35-6,
35'8 H.P.
26. 0-804 sq. in.
27. 0-578 sq. in.
28. 8 arcs, 272 incandescents.
29. 0-4824 sq. in., 0*9648 sq. in.
30. 207-86, 211-72, 215-58, 219-44, 223-29 volts.
31. 3-86 volts.
32. 0-7545 sq. in.
206 Examples in Electrical Engineering.
33. 0-3087 sq. in.
34. 498 yds.
35. 0-432 sq. in., 138-7 H.P.
36. 1 290 yds.
37. 0-22 sq. in.
38. 202-73 volts.
39. 600 amps., 2*17 sq. ins.
40. 83-3 tons.
41. 300 amps., 0*5427 sq. in.
42. 20-7 tons.
43. 0-5427 sq. in., 21-93 tons.
44. 0*144 S< 1 i n -> 5'69 tons.
45. I -82 amps, per yard of street.
46. 1*6 sq. in., 0*4 sq. in.
47. 53'i amps, to 49-1 amps, at end of run.
48. 1259-7 and 1367-7 per min. at end.
49. 0-072 ohm.
50. 141 volts.
51. 140-95 volts.
52. 93*6 per cent., 101 '45 volts.
53. 1328 per min.
54. 77-2 per cent.
55. 6-03 H.P.
56. 2291 per sq. cm.
57. 0-2566 Ibs.
58. 18-47 H. P., 91-8 per cent.
59. 4,105,208; 4032.
60. 100-0675 volts ; 4-5 amps.
6 1. 56-25 volts and 2-5 amps.
62. 101*08 volts, 56*62 amps.
63. 79 per cent.
64. 998-76 per min. and 3-1 amps. ; 992*752 per min. and 18-119 amps.
65. 599715 and 9*5 amps.
66. 30*05 per sec. ; 90*18 volts, 83*3 per cent., 82*7 per cent., and
68 -9 per cent.
67. 12*1 H.P., 94 per cent., and 90*27 per cent.
68. 19-568 per sec., 9*18 H.P., 87*7 per cent., and 83*06 per cent.
69. 540*576 volts, 9*67 per sec., 94*4 per cent., and 92-0 per cent.
70. 94*64 per cent, dynamo, 94*91 per cent, motor; 14*3 H. P., 90-6
per cent, as dynamo and 90*6 per cent, as motor.
71. 960 per min., 96*15 per cent, as dynamo, 12*35 H.P., 96 per cent.,
and 92*2 per cent.
72. 1080 volts, 13*4 H.P., 96*29 per cent., and 92*59 per cent.
73. 79*8 volts, 19-68 H.P., 94*3 per cent., and 90*2 per cent.
74. 53-84 amps., 24*7 per sec., 38*93 H.P., 98*83 per cent., 89*9 per
cent., and 78*3 per cent.
Answers. 207
75- 367'35 Iks., 28-5 H.P., 94-6 per cent., and 91-3 per cent.
76. 10-56 per sec., 111-07 amps., 97-7 percent., 95-94 per cent., and
81-549 per cent.
77. 200 wires ; R, 39-8 ohms ; R rt 0-0467, R )M 0*0156 ohm.
78. 8-65 H.P., 15*6 per sec., 90*0 per cent., and 86'o per cent.
79. 0-00054 sq. in. area.
80. 6650.
81. 0-0419 in. diameter.
82. 1,962,770.
83. 5-49 H.P. in res., and 14-7 H.P. in frictions.
84. 2337.
85. 96-6 per cent.
86. 0*0125 ohm, and 18-9 ohms.
87. 1-56 H.P.
88. 96-3 per cent., 50*1 H.P., 7-88 H.P.
89. Dynamo 21*18 per sec., motor 17*18 per sec., 10*71 H.P., 71-4
per cent.
90. 19*29 H.P., dynamo 27*58 per sec., motor 23*58 per sec.
91. 1043*5 vol ts.
92. 7500.
93. 46 ohms.
94. Shunt wire 0*066 in. diameter, series 30 convolutions, 92*8 per cent.
95. 23*08 C.
96. 148*9 amps.
97. 0*057 in. diameter.
98. No. 15 S.W.G.
99. 96 convolutions.
100. 115*24 volts.
101. 730*2 per min.
102. 0*0081 sq. in.
103. 0*00224 in. diameter ; 2*14 watts.
104. 2*2 watts.
105. 0*0143 in. diameter, 11*5 watts.
106. (a) i nearly, (b) 2 los. od.
107. 38.000 ohms ; 0*0199 per cent.
108. 92*72 in station ; 1 1 I'll volts outside.
109. 300 ohms.
no. 30*0388 volts total.
in. 0*005 in. diameter.
112. 0*0031 in. diameter.
113. 0*0056 in. diameter.
114. 0*01 12 in. diameter ; 1800 ohms.
115. Level, 300 amps., 200 H.P. up I in 480, 356 amps., 237*3 H.P. ;
up i in 384, 370 amps., 246*6 H.P. ; down 1 10960,272 amps., 181*3 H.P. ;
up i in 960, 328 amps., 218*6 H.P. ; down I in 384, 230 amps., 153*3
H.P., down I in 480, 244 amps., 162*6 H.P.
208 Examples in Electrical Engineering.
116. Level 150 amps., 80 H.P. : up i in 480, 173-3 amps., 92-4 H.P. ;
up I in 384, I79'i amps., 95*52 H.P. ; down I in 960, 138*3 amps., 7376
H.P. ; up i in 960, i6i'6 amps., 86'i6 H.P. ; down I in 384, 120*8 amps.,
64-4 H.P. ; down I in 480, 126*6 amps. ; 67-52 H.P. The E.M.F. in all
cases being 397*8 volts.
117. 596*8 amps., 1042 amps., 151-1 amps.
118. 1732 amps.
119. Level 59*68 amps., up i in 200, 86*53 amps., down i in 200, 32 8
amps., up I in 80, 126*5 amps., down i in 80, put on brake.
1 20. 82 amps.
121. 52-4 amps., 5-589 H.P.
122. Level, empty 1875 amps., loaded 21*875 amps., 2-4 H.P. and 2'8
H.P., up i in 50, 65-43 amps., and 8*376 H.P. ; down i in 100, 2*193 amps.,
and 0-28 H.P.
123. Level 8*7 amps., 1504 ft. per min., 3*41 H.P. ; up i in 150, 12*3
amps., 1317 ft. per min., 478 H.P. ; down i in 150, 2*7 amps., 1426 ft.
per min., and 1*077 H.P.
124. 115 amps. ; 24-4 miles per hour, 6*71 revolutions per sec.
125. 177 amps., 23-56 miles per hour, 6*5 revolutions per sec.
126. 64 amps., 31*45 miles per hour, 8*69 revolutions per sec.
127. 3*05 mins.
128. 54*2 amps., 2165 ft. per min., 19-6 revolutions per sec.
129. 77 amps., 2273 ft - P er min., 20-53 revolutions per sec.
130. About 42 amps.
131. 17*76 amps., 1993 volts.
132. 99 volts.
133. 2-1 amps.
134. 13-4 H.P.
135. 60 volts.
136. x 60 volts, y 31*65 volts.
137. * 84,375, .727,630.
138. 2020*5 VOltS.
139. 99-24 volts.
140. IO22*2 VOltS.
141. 1968, 2021-55 and 2075*1 volts.
142. 95*51 per cent.
143. 92-4, 95-7, 97'3 and 97'5 P er cent -
144. 93*6 per cent.
H5- 95'5 93'9> 7 8>0 and all-day 86 -7 per cent.
H 6 - 97'3 8 > 94'7 88-41, and 79-36 per cent.
147. 2O2O volts, 1*25 microfarad.
148. I3!j ohms starting, 5-4 ohms regulating, 6 H.P. slow, and 24*1 H.P.
full speed.
149. 1-9 H.P. and 14*3 H.P.
150. 0*252 ohm.
151. 2*21 pence.
152. i -35 pence.
( 209 )
SPECIFIC RESISTANCE OF VARIOUS MATERIALS AT o C., WITH
TEMPERATURE VARIATIONS.
Temperature
B 5,c
Material.
Ohms per
cubic inch.
Ohms per
cubic centimetre.
variation of
resistance per cent,
per degree
S|.s
SPli.S
U 3,0
Centigrade.
> 3
"* O.O
Copper
0*00000063
0-0000016
0*388 increase
0.320
Iron
0-00000405
O-OOOOIO4
Q'453
0-280
Aluminium
O-OOOOOII4
0-0000029
0*390
0-092
Mercury
0-00003666
0-0000943
0*072
0*489
Lead
0-00000780
O'OOOO2OO
0*387
0*410
Platinum ...
0-00000353
0-0000089
0*247
0-828
Platinoid
0-000014
0-000034
0*021
German silver
O-OOOOO82
O'OOOO2I
0*044
0*30
Manganin
O-OOOOlS
0-000044
0*002 decrease
Brass
O-OOOOO28
O-OOOOO72
0*290
Platinum-silver ...
0*0000095
0-000024
0*031 increase
Carbon (arc lamp) j
0-0017 to
0*0034
0*0044 to
0-0086
>o*o52 decrease
Sulphuric acid
solution i* i sp. g.
3-33
8'45
2-2
I'2S P .g.
2-62
6*66
2-2
VALUES OF THE SPECIFIC RESISTANCE OF COMMERCIAL COPPER AT
VARIOUS TEMPERATURES.
Temperature degrees.
Resistance of one cubic
Resistance of one cubic
Cent.
Fahr.
centimetre in ohms.
inch in ohms.
32
0*00000l6o
0-00000063
5
41
0-OOOOOI63
0-00000064
10
5
0-OOOOOI67
0*00000065
15
O*COOOOl69
0-00000067
20
68
o 00000172
o 00000068
25
77
0-OOOOOI76
0*00000069
30
86
O'OOOOOI79
0*0000007I
35
95
0*00000l82
O*OOOOOO72
40
104
0-00000186
0*00000073
45
H3
0-OOOOOI89
0-00000074
5o
122
0-00000193
O*OOOOOO76
55
131
o 00000196
0*00000077
60
140
0*00000200
0-00000078
65
H9
O*OOOOO2O3
O-OOOOO079
70
158
0-00000207
O'OOOOOOSl
75
I6 7
O*OOOOO2IO
O-OOOOOO82
80
176
0*00000214
0-00000084
85
185 0*00000217
0*00000085
90
I 94
O*OOOOO22I
0*00000087
95
203
0*00000224
o*oooooo88
100
212
0-00000228
0*00000090
2io Examples in Electrical Engineering.
VALUES OF H CORRESPONDING TO VARIOUS VALUES OF ft FOR CYCLIC
CHANGES IN ft AS A SINE-FUNCTION OF THE TIME, AND HAVING
MAXIMA OF 1000, 2000, etc., to 7000.
The values are given for each hundred ft for a complete half-period.
Values for the other half-period can be easily put in by reversing the signs.
The hysteresis loss h in ergs per cc. per cycle corresponding with these
values of the cyclic j8H curve are given in the table of values of j8 and H
on pp. 214-218.
H
ft
H
ft
H
ft
H
0'49
600
072
900
0-61
400
0'20
100
0'55
700
075
800
0-42
300
-0-28
2OO
300
0'59
0-62
800
900
0-80
700
600
0-23
0*07
200
100
0^36
0-42
400
0-66
IOOO
0-83
500
0*07
-0'49
500
0-69
o
0*62
IIOO
0-88
1900
0-90
! 900
- 0-28
100
0-65
1200
0-90
1800
072
800
-0'34
200
0*67
1300
0*92
1700
0-55
700
-0-38
300
0-69
1400
0*95
1600
0-40
600
-0-42
400
072
1500
0-97
1500
O'26
500
0-46
500
074
1600
I '00
1400
0-16
400
O*5O
6OO
076
1700
1-03
1300
0*05
300
- 0'53
700
078
1800
1-05
1200
0-04
200
- 0-56
800
0-80
I9OO
I -08
IIOO
0-14
IOO
- 0'59
900
0-83
2000
ri2
IOOO
-0-21
0'62
1000
0-85
0-83
IOOD
09
29OO
I -06
1400
-0-45
100
0-85
1700
ii
2800
0-80
I3OO
-0-49
2OO
0-87
1800
13
2700
o'6i
1200
-o*S3
300
0-88
1900
'14
26CO
o - 46
IIOO
-0-56
400
0-90
2OOO
16
2500
0'33
IOOO
o'6o
500
0-91
0-92
2100
2200
18
20
2400
2300
0-21
O'll
900
800
-0-63
-0-66
700
0-94
23OO
22
2200
O'O
700
-0-68
800
0-96
2400
24
2IOO
-0-08
600
-071
900
1000
o'97
0-99
25OO
2600
26
27
2000
I9OO
-0-15
0*22
500
400
-074
076
IIOO
I -01
2700
2 9
I800
-0-27
300
-079
1200
I '02
2800
'SI
1700
-0-32
200
-0-81
1300
1-04
2900
33
1600
-0-37
IOO
-0-82
1400
I -06
3000
'35
1500
0-41
-0-83
1500
1-07
Cyclic fiH Values.
21 I
ft
H
ft
H
ft
o'gi
2100
20
3900
100
0-92
2200
22
3800
200
0-94
2300
2 3
3700
3 00
0-95
2400
' 2 5
3600
4OO
0*96
25OO
26
35oo
500
0-98
2600
28
3400
600
0-99
2700
30
3300
700
I '00
2800
31
3200
800
I -01
2900
'33
3100
9OO
I'O2
3OOO
'35
3000
1000
1-03
3100
'37
2900
IIOO
1-05
3200
38
2800
I2OO
1-07
330
40
2700
1300
1-09
3400
42
2600
1400
ITO
350
'44
2500
1500
I'll
3600
46
2400
1600
I'M
3700
48
2300
I7OO
1-14
3800
"So
2 2OO
I800
1-15
3900
'53
2100
1900
1-17
4000
56
2000
2OOO
1-18
H
ft
H
I'25
1900
-0-55
I -00
1800
- 0-58
0-80
1700
0*60
o - 65
1600
0-62
0-52
1500
0-65
1400
0*67
0-27
1300
070
0-17
1200
- 072
0-08
IIOO
-074
O'OO
1000
076
- 0-07
900
- 078
- 0-14
800
- 079
- 0'20
700
- 0-81
- 0-25
600
-0-83
- 0-31
500
- 0-84
- 0-36
400
- 0-85
- 0*40
300
-0-86
- O'44
200
-0-88
- 0-47
100
0-90
- 0-51
-0-91
0-98
2600
1-27
4900
i'34
2400
-0-68
0'99
2700
1-29
4800
i -08
2300
070
'00
2800
I' 3 I
4700
0-90
2200
072
oi
2900
I'33
4600
074
2IOO
-074
'O2
3OOO
I'35
4500
0-60
2000
-075
'02 I
3IOO
I'37
4400
0*46
1900
- 077
03 |
3200
I- 3 8
4300
o'34
I800
-078
04
3300
I-40
4200
O'22
1700
-0-80
'OS
3400
I- 4 2
4100
OT2
I6OO
-0-81
06 !
3500
I'44
4000
O'OO
1500
-0-82
07
3600
1*46
3900
-0-08
1400
-0-83
08
3700
I- 4 8
3800
-0-17
1300
- 0-84
TO
3800
r50
3700
-0-24
1200
-0-85
Tl
3900
i'S3
3600
0-30
IIOO
-0-86
T2
4000
i-*55
35oo
-0-36
1000
-0-88
*3
4IOO
i'57
3400
- 0-40
9OO
-0-89
'H
4200
i'59
3300
-0-44
800
- 0-90
"IS
4300
r6i
3200
-0-47
700
0-91
T7
4400
1-63
3100
o'5o
600
0-92
18
4500
1-65
3000
-0-54
500
-0-93
'20
4600
1-67
2900
-0'57
400
-0-94
21
23
4700
4800
170
172
2800
2700
-0-59
O'62
300
200
-0-95
0-96
2 4
%
4900
5000
'"74
176
2600
2500
0*64
-0-66
100
-0-97
- 0-98
212 Examples in Electrical Engineering.
CYCLIC )8H VALUES.
ft
H
ft
H
ft
H
H
I -06
3100
1-42
59oo
I-72
2900
-0-75
100
1-07
3200 ;
i'43
5800
1-40
2800
076
200
08
3300
i'45
5700
1-16
2700
-077
300
09
3400
1-46
5600
0-94
26OO
-079
400
10
3500
1-48
55oo
0-78
2500
-0'80
500 |
II
3600
1-50
5400 j
0*62
2400
-0-81
600
12
3700 !
1-52
5300 !
0-47
2300
-0-82
700
'13
3800
i'54
5200
0-37
2200
-0-84
800
14
3900 1
1-56
5100
0-27
2IOO
-0-85
coo
15
4000
1-58
5000
0-17
2OOO
-0-86
1000
1-16
4100
i '60
4900
0-08
1900
-0-87
IIOO
1-17
4200
1-62
4800
O'OO
1800
-0-88
1200
1-18
4300
1-64
4700
- O'lO
1700
-0-89
1300
1-20
4400
1-66
4600
-0-17
I6OO
0*90
14OO
1*21
4500
1-68
4500
0-24
I5OO
0-91
1500
I '22
4600
170
4400
-0-28
1400
- 0-92
1600
1-23
4700
1-72
4300
-0-33
1300
-o'93
I7OO
I'24
4800
1-74
4200
-0-37
1200
-0-94
I800
I-2 5
4900
1-76
4100
0-42
IIOO
-o'95
1900
1-26
5000
1-78
4000
0-46
IOOO
0-96
2000
I*27
5100
i -80
3900
-0-49
900
-0-97
2100
1-28
5200
1*3
3800
-0-52
800
-0-98
22OO
1-29
5300
1-85
3700
0-56
700
-0-99
2300
r30
5400
1-88
3600
-0-59
600
O'lOO
2400
I-3I
5500
1-91
1 3500 ;
0-62
500
O'lOI
2500
I-32
5600
2-00
34 oo
0-64
400
0-102
2600
i'34
5700
2'10
3300
0-67
303
O-IO3
2700
i '35
5800
2-20
3200
0-69
2OO
O-IO4
2800
1-36
5900
2-3C
3100
-0-72
100
- 0-105
2900
1-38
6000
2-40
3000
-0-74
o
o*io6
3000
i '40
i
I"I2O
1800
I'296
3600
536
5400
1-890
100
I-I28
1900
I-308
; 3700
552
5500
1-910
2OO
1-136
2OOO
I-320
i 3800
568
5600
1-940
3 00
I-I44
2100
I'332
3900
584
5700
1-960
4 00
I-I52
22OO
i '344
4000
600
5800
1-980
500
600
1*160
1-170
2300
1 2400
&
4100
4200
620
640
5900
6000
2-010
2*050
700
1-180
! 2500
1-380
4300
660
6100
2-080
800
1-190
2600
i '394
4400
680
6200
2'IIO
900
1-200
! 27OO
1-408
4500
700
6300
2^140
IOOO
I'2IO
2800
1-422
4600
718
6400
2- 1 80
IIOO
1-220
2900
1-436
4700
736
6500
2"2IO
I2OO
I-230
3000
i '450
4800
754
6600
2-250
1300
I-24O
3100
1-464
4900
772
6700
2*290
1400
I-250
32OO
1-478
5000
790
6800
2*340
1500
I-26O
3300
1-492
5100
810
6900
2-380
1600
I-272
3400
1-506
5200
830
7000
2-440
1700
1-284
3500
1-520
5300
860
6900
1-960
Cyclic ftH Values.
213
ft
H
ft
H
ft
H
ft
H
6800
i-6co
5000
0-460
33
- 0-830
1600
- 0-972
6700
1-380
4900
~ 0-49 2
3200
0-840
1500
0-980
6600
1-190
4800
-0-524 ; 3100
0-850
1400
- 0-988
6500
I'OOO
4700
- 0-556 1
3000
- 0-860
1300
- 0-996
6400
0-840
4600
-0-588
2900
- 0-870
1200
"004
6300
0-720
4500
0-620
2800
- 0-880
I IOO
-012
6200
0-590
4400
- 0-644
2700
- 0-890
1000
-O2O
6100
0-440
4300
- 0-668 '
2600
0-900
900
- -030
6000
0-320
4200 0-692
2500
0-910
800
-040
5900
O'2OO
4100
- 0-716
2400
0-918 700
'050
5800
O'lIO
3000 0-740
2300
0-926 600
'060
5700
o-ooo
3900
-0752
2 2OO
- 0'934
500
'O7O
5600
O'lOO
3800
-0-764
2100
0-942 400
- -080
55
0-200
3700
- 0-776
2 COO
- 0-950 300
1*090
5400
0-252
3600 -0-780
I9OO
- 0-956
200
I 'IOC
53
- 0-304
3500 1 0-810
I800
0-962 I ioo
I'lIO
5200
- 0-356
3400 0-820
1700
-0-968 ! o
I'I20
5100
- 0-403
1
2t4 Examples in Electrical Engineering.
ANNEALED CHARCOAL IRON. VALUES OF MAGNETIC PROPERTIES.
ft .
Density in
c.g.s. lines per
sq. cm. area.
H
in c.g.s. per
cm. length.
ft
Ampere-turns
per cm.
length.
M
Permea-
bility.
Hysteresis
loss in ergs
per cc. per
cycle.
&P
Product of
density and
ampere-turns
per cm. length.
100
0-17
0-I40
5 88
5
14
200
0-32
0-260
625
10
52
3 00
0-44
0-356
685
17
107
400
0*54
0-432
740
25
172
500
600
0-60
0-65
0-476
0-520
$
37
5
238
3 I2
700
800
0-69
0-76
O*552
0-584
1014
1052
65
80
386
467
900
0-78
0-624
H54
97
5 6l
1000
0-83
0-664
1204
115
664
1 100
0*87
0-696
1264
134
765
I2OO
0-91
0-728
1318
155
873
1300
0-94
0-752
1382
177
978
1400
0-97
0776
1443
200
1086
1500
0-99
0-796
I 5 I 5
222
1194
1600
02
0-816
1568
245
1305
1700
04
0-836
1634
270
1421
I800
07
0-856
1682
295
1540
1900
09
0-872
1743
32O
1656
2000
II
0-888
1802
345
1776
2IOO
' 1 3
0-904
1858
372
1898
2200
15
0-922
1913
400
2024
2300
17
0-940
1966
430
2l62
2400
19
0-958
2117
460
2304
2500
'22
0-976
2049
490
2440
2600
2 4
0-992
2096
520
2579
2700
26
I '01
2142
550
2727
2800
2 9
1-03
2170
580
2889
2900
'3 1
1-05
2214
610
3045
3000
'34
1-07
2238
640
3216
3100
36
i '09
2279
670
3379
32OO
38
10
2318
700
3520
3300
40
'12
2357
735
3696
3400
42
14
2394
770
3862
3500
'44
16
2430
805
4065
3600
'47
18
2448
840
4233
3700
'49
20
2486
877
4440
3800
'Si
21
2516
915
4590
3900
'53
'22
2548
952
4758
Annealed Charcoal Iron. 215
ANNEALED CHARCOAL IRON. VALUES OF MAGNETIC PROPERTIES.
/?
Density in
c.g.s. lines per
sq. cm. area.
H
in c.g.s. per
cm. length.
/ft
Ampere-turns
per cm.
length.
M
Permea-
bility.
h
Hysteresis
loss in ergs
per cc. per
cycle.
Product of
density and
ampere-turns
per cm. length.
4000
i'SS
24
2580
990
4960
4100
i'57
26
26ll
IO22
5166
42OO
i'59
27
2641
1055
5346
4300
1-61
29
2670
1092
55 6 7
4400
1-63
3
2699
1130
578i
4500
1-65
32
2726
1170
5940
4600
4700
1-67
1-69
4
2780
1210
1252
6H5
6392
4800
172
3 8
2790
1295
6604
4900
174
'39
28l6
1337
6811
5000
176
'41
2841
1380
7040
5100
178
42
2865
1420
7242
5200
i -80
'44
2889
1460
7488
5300
1-82
46
2912
1505
7738
5400
1-85
48
2918
1550
7992
5500
T RR
50
2 9 2S
1590
8250
5600
1-91
'53
2931
1630
8556
5700
1-94
'55
2 93 8
1677
8835
5800
i '97
58
2944
1725
9140
5900
2 '00
60
2950
1772
9445
6000
2-03
62
2955
1820
9744
6lOO
2'06
65
2961
1867
10065
6200
2'10
68
2955
1915
10416
6300
2'13
71
2952
1962
10773
6400
2-17
74
2949
2OIO
1 1 1 10
6500
2*21
77
2941
2060
ii55
66OO
2-25
80
2938
2110
11880
6700
6800
2-28
2'32
11
2933
2930
2l6o
2210
12261
12662
6900
2- 3 6
89
2923
226O
13041
7000
2-4I
i '93
2310
13496
7100
2-46
1-97
2886
2362
13987
7200
2-5I
2-01
2868
2415
14457
7300
2- 5 6
2-05
2851
2470
H965
7400
2-62
2'10
2824
2S 2 5
^S 10
7500
2-6 7
2-14
2808
2587
16050
7600
273
2-18
2783
2650
16598
7700
278
2-23
2769
2705
17171
7800
2-84
2-27
2746
2760
17721
7900
2-90
232
2723
2810
18328
Sooo
2- 9 6
2'37
2703
2860
18944
8100
3'01
2-4I
2691
2939
19521
8200
3'06
2'45
2679
301.8
20090
8300
3'12
2-50
2660
3097
20750
8400
3'i8
2'5S
2641
3176
21420
216 Examples in Electrical Engineering.
ANNEALED CHARCOAL IRON. VALUES OF MAGNETIC PROPERTIES.
Density in
c.g.s. lines per
sq. cm. area.
H
in c.g s. per
cm. length.
Ampere-turns
per cm.
length.
M
Permea-
bility.
h
Hysteresis
loss in ergs
per cc. per
cycle.
Product of
density and
ampere-turns
per cm. length.
8500
3' 2 5
2-60
2615
3255
22100
8600
3*3*
2*65
2598
3334
22790
8700
3*37
270
2581
3413
23490
8800 3-43
275
2565
3492
24200
8900 3-50
2'80
2542
357i
24720
9OCO 3-56
2-85
2528
25650
9100 3-62
2-90
2513
3730
26390
9200 3-68
2'95
2500
3810
27140
9300 376
3'01
2473
3890
27993
9400
3-85
3'08
2441
3970
28952
9500
3-92
3-14
2423
4050
29830
9600
4-00
3 -20
2400
4130
30730
9700
4-08
2377
4210
9800
4-16
3'35
2355
4290
32830
9900
4-27
3 '42
2318
4370
33858
IOOCO
4'37
3-50
2288
445
35000
IOIOO
4-48
3'59
2254
4535
36259
10200
4*60
3-68
2217
4620
37536
10300
471
377
2187
4705
38831
IO4OO
4-82
3-86
2157
4790
40147
10500
4-96
3-96
2117
4875
41580
10600
5'7
4-06
2090
4960
43036
IO7OO
5'2O
4-16
2057
5045
445 i 2
I0800
5'33
4-27
2026
46116
10900
4'37
1995
5215
47633
IIOOO
5-60
4-48
1963
5300
49280
1 1 100
573
4'59
1937
5390
50949
II2OO
5-87
470
1908
5480
52640
II3OO
6-03
4-83
1873
5570
54579
II400
6'2O
4-96
1838
5650
56544
JI500
6 '40
5-12
1797
5750
58880
II600
6-60
5-28
1757
5840
61248
II700
6-85
1707
5930
64116
IISOO
7-10
5-68
1662
6020
67024
II900
7-36
5'88
1617
6110
69972
I2OOO
7'6o
6-08
1579
6200
72960
I2IOO
7-90
6-32
1531
6300
76472
12200
8-20
6-56
I 4 8 7
6400
80032
12300
8-50
6-80
1437
6500
83640
12400
8-90
7-12
1393
6600
88288
12500
9'2O
I35 1
6700
92000
12600
12700
9-60
I0'0
8-00
1312
1270
6800
6900
96768
101600
12800
10-4
8-32
1231
7000
106496
12900
10-8
8-64
1192
7100
111456
Annealed Charcoal Iron. 217
ANNEALED CHARCOAL IRON. VALUES OF MAGNETIC PROPERTIES.
a
Density in
c.g.s. lines per
sq. cm. area.
H
in c.g.s. per
cm. length.
Ampere-turns
per c.m.
length.
Permea-
bility.
h
Hysteresis
loss in ergs
per cc. per
cycle.
ft/ft
Product of
density and
ampere-turns
per cm. length.
I3OOO
11-3
9-04
II 5 I
72OO
II7520
I3IOO
11-7
9-36
III5
7300
I226l6
I32OO
I2'2
976
1082
7400
128820
13300
I2'7
lO'I
1047
7500
134330
13400
I3-2
ID'S
1015
7600
140700
I35
137
I0'9
986
7700
I47I50
13600
I4-2
1 1 -3
958
7800
153680
13700
14-8
11-8
926
7900
I6I600
13800
15-4
12-3
8 9 6
8000
169740
13900
16-1
12-8
86 3
8lOO
177920
14000
16-9
13-5
839
8200
189000
I4IOO
I7'8
14-2
793
8310
2OO22O
I42OO
187
14-9
759
8420
211580
14300
197
157
725
8530
224510
14400
20-7
16-5
695
8640
237600
14500
14600
21-8
23-0
18-4
665
636
8750
8860
252300
278640
14700
24-2
19*3
608
8970
283710
14800
25-5
20-4
580
9080
301920
14900
26-8
21-4
555
9190
318860
15000
28-6
22-8
524
9300
340000
I5IOO
30-0
24-0
503
9410
362400
15200
31-5
25-2
482
9520
383040
15300
33' i
26-4
461
9630
403920
15400
35'o
28-0
440
9740
431200
15500
37'i
29-6
419
9850
458800
15600
39*2
3 J '3
398
9960
488280
15700
41*5
33*2
379
10070
521240
15800
43 '8
361
I0l8o
553000
15900
46-8
37'4
34 1
IO29O
594660
16000
48-8
39-0
321
10400
624000
I6IOO
53*2
42-5
303
I05IO
684250
16200
567
45'3
285
10620
733860
16300
60-7
48-5
270
10730
790550
16400
65-0
52-0
256
10840
852800
16500
69-9
55'9
238
10950
922340
16600
60- 1
221
Ilo6o
997660
16700
16800
87-0
64*5
69-6
207
193
III70
II280
I077I50
1169280
16900
93-0
74*4
II390
1257360
I7OOO
lOO'O
80-0
170
II500
1360000
I7IOO
107-0
85-5
159
Il62O
1462050
17200
115-0
92-0
149
II740
1582400
17300
123-0
98-4
141
II860
1702320
17400
131-0 \ 104-0 133
11980
1809600
2i8 Examples in Electrical Engineering.
ANNEALED CHARCOAL IRON. VALUES OF MAGNETIC PROPERTIES.
ft
Density in
c.g.s. lines per
sq. cm. area.
H
in c.g.s. per
cm. length.
Ampere-turns
per cm.
length.
M
Permea-
bility.
*
Hysteresis
loss in ergo
per cc. per
cycle.
ft/ft
Product of
density and
ampere-turns
per cm. length.
17500
140
112
125
I2IOO
I96OOOO
17600
149
119
117
I 22 2O 2094400
17700
158 126
III
12340
2230200
17800 167
133 106 12460 2367400
17900 176
140
ioi 12580 2506000
18000 186
148
96 I27OO 2664000
18100 196
156
90 12820
2823600
l8200 211
1 68
86
12940
3057600
18300 226
180
80 13060
3294000
18400 244
195
71 13180 3588000
18500 260
208
69 13300
3848000
18600
276
220
67
13420
4O92OOO
18700
293
234
65
13540
4375800
18800
3 11
248
64
13660
4662400
18900
330
264
60
13780
4989600
19000
350
280
54
13900
5320000
I9IOO
372
297
I4O2O
5672700
I92OO
396
316
48
I4I40
6067200
19300
422
337
45
14260
6504100
19400
448
358
43
14380
6945200
19500
476
380
14500
7410000
19600
505
404
39
14620
7918400
19700
537
429
36
14740
8451300
19800
575
460
34
14860
9I08000
19900
609
487
32
14980
9691300
20000
650
520
3i
I5IOO
IO4OOOOO
^HY
THB
Magnetic Properties of Soft Grey Cast Iron.
CVERSITY
MAGNETIC PROPERTIES OF SOFT GREY CAST IRON.
. ft
Density in c.g.s. lines
per sq. cm.
H
c.g.s. per cm.
length.
fft
Ampere-turns per cm.
length.
M
Permeability.
3000
i-5
I'2
2000
3IOO
1-9
1*5
1695
3200
2-3
1*8
1391
3300
27
2-2
I2 4 I
3400
3 - i
2 '5
1096
3500
3'5
2'8
1009
3600
3*9
3'i
923
3700
4*3
3'4
860
3800
47
3-8
808
3900
4' i
764
4OOO
5*5
4'4
727
4100
5'9
47
695
42OO
6-3
5'
666
4300
67
5 '4
641
4400
7-1
57
619
4500
7-6
6-1
592
4600
8'!
6-5
568
4700
87
7-0
540
4800
9'3
7'4
516
4900
I0'0
8-0
490
5OOO
107
8-6
467
5100
11-6
9*3
439
5200
12-5
I0'0
416
5300
i3'5
io'8
392
5400
H'5
11-6
372
5500
12-6
348
5600
17-1
137
327
5700
18-5 V
14-8
37
5800
2O'O
i6'o
290
5900
21-6
17-3
274
6OOO
23'2
18-5
258
6100
2 4 -8
19-8
246
62OO
26-5
21'2
234
6300
28-2
22'S
223
6400
30-0
24-0
213
6500
3 2-0
25'6
203
6600
6700
36;4
29-I
6800
307
178
6900
40-8
3 2-6
174
7000
43'2
162
220 Examples in Electrical Engineering.
MAGNETIC PROPERTIES OF SOFT GREY CAST IRON.
. ft
H
fft
Density in c.g.s. lines
per sq. cm.
c.g.s. per cm.
length.
Ampere-turns per cm.
length.
M
Permeability.
7100
45 '9
367
155
7200
48-6
38-9
I 4 8
7300
Si-7
41-4
141
7400
54'9
43 '9
135
7500
58-6
46-8
129
7600
62-3
49'8
123
7700
67-2
537
117
7800
70*2
S6-i
III
7900
743
59'4
1 06
8000
78-5
62-8
102
8100
827
66-1
9 8
8200
86-9
69-5
94
8300
91 'i 72-8
9i
8400
95 '4 76'3
89
8500
99'8 79'8
86
8600
104-2 83-3
83
8700
108-9
87-1
80
8800
113-6 90-8
77
8900
118-6 94-8
74
9000
123-7 98-9
72
9100
129-5 103-6
70
92OO
135-6 108-4
68
9300
141-9 113-5
66
9400
148-3 118-6
64
9500
154-8 123-8
61
9600
161-4 129*1
59
9700
168-0 134-4
57
9800
1747 1397
55
9900
181-5 H5'2
54
IOOOO
188-5 1 5'%
53
IOIOO
195-8 156-6
5i
10200
203-2 162-5
49
10300
210-8 168-6
48
10400
219-0 i75'2
47
10500
228-0 182-4
45
IO6OO
238-0 190-4
44
10700
248-0 198-4
43
IO8OO
259-0 207-2
42
10900
272-0 217-6
40
IIOOO
288-0
230-4
39
Details of Conductors.
221
DETAILS OF CONDUCTORS.
S.W.G.
Diameter
of each
wire.
Diameter
of the
strand.
Area.
Resistance at
60 F.
Weight.
Per looo
yards.
Inch.
Inch.
Square inches.
Per looo yards.
Ohms.
Ibs.
7/0
0'500
...
0-196349
0*1246
2270
6/0
0*464
0-169093
0*1446
1955
S/o
OH32
...
0-146574
0*1669
1694
40
0-400
0*125663
0*1946
1454
3/0
0-372
...
0-108686
0*225I
1256
2/0
0-348
0-095114
0-2572
1099
I/O
0-324
...
0*082447
0*2967
953
0-300
...
0*070685
0*3461
817
2
0-276
...
0*05982
0-4089
691
3
0-252
...
0*04987
0-4905
576
4
0-232
...
0*04227
0-5787
488
5
0'2I2
...
0*03529
0-6931
405
6
O-I92
...
0*0289
0*8450
334
7
0-I76
...
0*0243
1*0056
281
8
o'i6o
...
O*O2OI
1*2168
232
9
0-144
...
0*0163
1-5022
188
10
0-128
...
O-OI28
1*9012
148
ii
O'ii6
...
0-0105
2'3I50
122
12
0-104
0-0085
2*8800
9 8
13
0-092
...
0*0066
3-6803
7 6
H
0*080
0-0050
4-8673
58
15
0-072
...
0-0040 6-0089
47
16
0-064
...
0-0032 7-6049
37
17
0-056
...
0-0024
9'933 2
28
18
0-048
...
0*0018
i3"5!98
21
19
0*040
...
0*0012
19*4697
I4-5
20
0-036
...
0*0010
24'0354
11*7
21
0-032
...
0-0008
30-422
9'3
22
0-028
...
0-0006
39-729
7*1
3/25
0-020
0-042
0-00096
25'955
ii
3/24
0*022
O'O22
0-00116
2i*454
I3-5
3/23
0-024
0-024
0-00138
18-026
16
3/22
0-028
O-O28
0-00188
13*243
22
3/21
0-032
0-032
0*00246
10-144
28
3/20
0-036
0-036
0*00311
8-01 18
36
3/19
0-040
O-O4O
0*00384
6-4899
44
3/iS
0-048
0-048
0*0055
4-5066
64
222 Examples in Electrical Engineering.
DETAILS OF CONDUCTORS.
S.W.G.
Diameter
of each
wire.
Diameter
of the
strand.
|
A Resistance at
Area - 60 F.
Weight.
Inch.
Inch.
Square inches.
Per looo yards.
Per 1000
yards.
Ohms.
Ibs.
7/25
0*020
0'060
O'OO22
II*I24
25-5
7/24
O*O22
O'O66
0*0027
9*I952
3^5
7/23
0-024
0-072
0*0032
7-7256
37
7/22
0-028
0-084
0*0043
5-6757
5i
7/21*
0-030
0*090
0*0050
4"9445
58
7/21
7/20
0-032
0-033
0*096
0-099
0*0057
0*0064
4:346o
4*0864
66
75
7/20
0*036
O'loS
0*0072
3-4336
84
7/19
0-040
0-120
0*0089 2*7813
104
7/18
0-048
0-144
0*0129 I '93 I 4
149
7/17
0-056
0-168
0*0175 1*4190
203
7/16
0*064
0-192
0*0229 1*0864
266
7/15
0*072
0-216
0*0290 0-8584
336
7/H
0-080
0-240
0-0358
0*6953
415
7/13
0*092
0-276
0-0474
0-5257
549
7/12
0-104
0-312
0-0606
0*4114
701
7/i i
0-116
0-348
0-0754
0*3307
872
7/10
0-128
0-384
0-0918
0*2716
1062
7/9
0-144
0-432
0-1162
0*2146
1343
7/8
0-160
0-480
0>I 435
0*1752
1660
7/6
0-192
0-576
0-2067
0*1207
2390
19/24
0*022
O'HO
0-0073
3-3877
85
19/23
O'O24
0-120
0-0087
2-8463
101*5
19/22
0-028
0-I40
0-0119
2-0910
138
19/21
0-032
0-160
0-0156
1-6011
180
19/20
0'036
0-180
0-0197
1-2650
228
19/19
0-040
O'2OO
0-0244
1-0247
282
19/18
6*048
0*240
0-0351
0*7115
406
19/17
0*056
0-280
0-0478
0-5228
553
19/16
0-064
0-320
0-0624
0-4002
722
i9/*5
0-072
0-360
0-0790
0-3162
914
19/14
0-080
0-400
0-0976
0*2561
1128
I9A3
0*092
0-460
0-1290
0*1937
1491
19/12
O'IO4
0-520
0-1649
Q'^S
1906
19/11
0-116
0-580
0-2052
0*1218
2372
19/10
0*128
0-640
0-2498
0*1000
2888
19/9
0-144
O'72O
0-3162
0*07906
3655
19/8
0-160
0-800
0-3904
0*06406
45i3
19/7
0-176
0-880
0-4724
0*05292
546i
Details of Conductors.
DETAILS OF CONDUCTORS.
223
S.W.G.
Diameter
of each
wire.
Diameter
of the
strand.
Area.
Resistance at
60 F.
Weight.
Inch.
Inch.
Square inches.
Per 1000 yards.
Per 1000
yards.
Ohms.
Ibs.
37/24
0-022
0*154
0*0143
I-7396
I6 5
37/23
O-O24
O'l68
0-OI7I
1*4647
I 9 8
37/22
0-028
0*196
0-0233
1-0737
270
37/21
0-032
0-224
0-0304
0*8222
352
37/20
0-036
0*252
0*0386
0-6496
446
37/19
0-040
0-280
0*0476
0-5262
550
37/i8
0-048
0-336
0-0686
0-3654
793
37/17
0-056
0-392
0-0934
0*2684
1080
37/i6
0-064
0-448
0-1220
0*2055
1410
37/15
0-072
0-504
0-1544
0*1624
1785
37/14
0-080
0-560
0-1906
0-I3I5
2203
37/13
0-092
0-644
0*2521
0-09947
2914
37/12
0*104
0-728
0*3221
0-07783
3723
37/n
0-116
0*812
0*4008
0*06265
4633
37/iQ
0-128
0-896
0*4880
0-05138
5641
37/9
0-144
I -008
0*6176
0*04060
7140
37/8
0-160
I'I20
0-7625
0*03288
8815
61/24
0-022
0-198
0*02378
l '$S l
275
61/23
0-024
0*216
0*02831
0*8865
327
61/22
0-028
0*252
0*03854
0-6583
446
61/21
0-032
0-288
0*0503
0-4987
572
61/20
0-036
0-324
0*0637
0-3940
736
61/19
0-040
0-360
0*0786
0*3191
909
61/18
0-048
0-432
0*1132
O*22l6
1309
61/17
61/16
0-056
0-064
0-504
0-576
0-1541
0*2013
0*l628
0*1246
1781
2327
6i/i5
0-072
0-648
0*2548
0*09850
2945
61/14
0*080
0*720
0-3I45
0-07979
3636
61/13
0-092
0-828
0*4160
0*06033
4809
61/12
0-104
0*936
0-5316
O*O472I
6i45
61/11
0-116
1*044
0*6614
0*03795
7646
61/10
0-128
I*I52
0-8053
0*03Il6
9309
91/18
0-048
0-528
0-1692
0*14857
1956
91/17
0-056
0*616
0*2302
9*10915
2661
91/16
0*064
0*704
0*3007
0-08357
3476
9I/I5
0-072
0*792
0-3806
0*06603
4400
91/14
0-080
0*880
0-4699
0*05348
5432
9i/i3
0-092
I*OI2
0-6215
0*04044
7185
91/12
0*104
1*144
0*7942
0*03164
9181
91/11 0-116
1*276
0*9881
0-02543
11422
224
Examples in Electrical Engineering.
SQUARES.
|
1
2 \ 3
4
5
6
7
8
9
1 23 4
5
6 7
8
i
i
1
ro
1*000
*020 I*040
1-061
1-082
103
124
i-i45
1-166
1-188
2 4
6 8
10
3 15
17 19
*i
1*210
232 I -254
1-277]
1-300
"323
346
1-369
1-392
1*416
2 5
7 9
ii
4 16
18 21
2
1*440
464
1*488
i -513! i -538
563
588
1*613
1-638
1-664
2 5
7 10
12
5 i7
20 22
'3
1-690
* 7 l6
17421769
1796
823
850
1*877
1*904
1*932
! C
8 ii
13
6 18
22 24
'4
I-960
* 9 88
2-0162-0452-074
103
132
2*161
2-190
2-220
3 6
9 2
14
7 20
23 26
'5
2*250
280
2-310
2-341
2*372
"403
"434
2-465
2-496
2-528
3 6
9 2
15
19 22
2528
6
2*560
592
2*624
2-657
2*690
723
756
2789
2-822
2-856
! 7
10 3
16
20 23
26 30
7 2-890
924
2*958
2-9933*028
063
098
3-I33J3-I68
3^04
I 7
10 4
17
21 24!28 31
8 (3-240
*2 7 6
3-312
3 '349; 3 "386
"423
460
3-4973-534
3-572
\- 7
ii 5
18
22 26 30 33
9 13-610
2'0 I4-000
648
"040
3-686
4*080
3-725J3764
4*121 4*162
803
202
8423-881
"2444-285
3*9203*960
4 -32614 -368
i 8
i- 8
12 16
12 16
19
20
23 27
25 29
3i 35
33 37
2-1
2*2
4-410
4-840
4-452
J-88 4
4-494
4*928
4'537
4-973
4-580
5*018
623
063
666
108
47094752
5*1535-198
3796
5-244
1- 9
I 9
13 17
13 18
21
22
26 30
27 31
34 39
3640
2*3
5-200
5'336 5'382
5 "429; 5 "476
523
570
5*6175*6645712
5 9
14 19
23
28 33 38 42
5*760
5-8o8
5^56
5 '90S 5 '954
003
052
6*101 6*1506*200
5 io
15 20
24
29 34 39 44
2'5
6*250
6-300
6-350
6*401
6-452
503
'554
6-605
6-656
6*708
5 1
15 20
25
31 36
41 46
2-6
6*760
6-812
6*864
6-917
6*970
023
076
7*129
7*182
7-236
5 ii
16 21
26
32 37
42 48
27
7*290
7-4537-508
56:
7-6187-673
7728
7*784
5 ii
16 22
27
33 38
44 49
2-8
7-840
7-896
7-952
8 -009: 8 -066
8-1808-237
8*294
8-352
6 ii
17 23
28
34 4
46 51
2-9
8-4IO
8-468
8*526
8-5858-644
8703
87628*821
8*880
8*940
6 12
18 24
29
35 4147 53
3-0
g-OOO
9-060
9-120
9'i8ij9*242
9-303
9*3649*425
9*486
9-548
6 12
18 24
3
37 43149 55
3-1
9"6io
9-672
9734
97979-860
9-923
9-986
10*05
10*11
10*18
6 13
i
19 25
2 3
31
3
38 44
4 5
50 57
5 6
3 '2
10-24
10-30
10-37
10-43 1'5
10-56
10*63
10-69
1076
10*82
i
2 <
3
4 5
5 6
3 '3
10-89
10*96
II'02
11-0911-16
11*22
11-29
11-36
11-42
11-49
i
2 ;
3
4 5| 5 6
3'4
11-56
11-63
1170
117611*83
11*90
11*97
12-0412-11 12-18
i
2 <
3
4566
3'5
12*25
12-32
12*39
12*46
12*53
1 2 -60
12-67
1274 12-82
12*89
i
2 '
4
4566
3-6
12-96
13-03
13*10
13-18
i3'25
IS'S 2
13-40
13-47
13-54
13-62
i
2 3
4
4567
37
13*69
1376
13-84
14*06
14*14 14-21 14-29
14-36
i
2 *:
4
5 5 6 7
14*44
I4'52
H-59
! 4 *82
14-90114-98
15*05
I5-I3
i 23
4
5 5| 6 7
3'9
IS' 2
15-29
15-37
15-4415-52
I5-60
i5-68!i S - 7 6
15-84
15-92
I 2
2 3
4
5667
4-0
16*00
16*08
16*16
16-24 16*32
I6* 4
16*48
16-56
16-65
16-73
I 2
2 *:
4
5667
16-8
16*89
16-97
17-06
17-14
17*2
17-31
17-39
17-47
I7-56
I 2
2 3
4
5677
4-2
17*64
1772
17-81
17-8917*98
1 8*06
i8*is|i8*2 3
18-32
18*40
I 2
3 3
4
5678
4'3
18*4
18-58
18-66
1875 18*84
1 8*9
19*01 19*10
19-18
19*27
I 2
3 3
4
5678
4'4
19*36
i9'45
I 9'5^
19*62
,197
19*8
19*89
19-98
20-07
20*16
I 2
3 4
5
5678
4-5
20*2
20*34
20*42
20*52
20 '6
20*7
2079
20-88
20-98
21-07
I 2
3 4
5
5678
4'6
21*1
21 -2 q
21 '34
21-44
21 '5
21*6
21*72
21 *8l
21*90
22-00
I 2
3 4
5
6 7\ 7 8
47
22*0
22-18
22-28
22*3722*4
22-5
22-66
2275
22*85
22*94
I 2
3 4
5
6 7
8 9
4*8
23*0
23-14
23-23
23-3323-4
23 "5
23-62
2372
23-81
23*91
I 2
3 4
5
6 7
8 9
4'9
24*0
24*11
24*21
24*3024*4
24 '5
24 *6c
24*70
24-80
24*90
I 2
3 4
5
6 7
8 9
25*00
25*10
25-20
25 "3C
>25'4
25-5
25*60
2570
25-81
25-91
I 2
3 4
5
6 7
8 9
5'i
5' 2
26*0
27-0
26-11
27*14
26.21
27^
26*3226*4
27 -3527 '4
26-5
27'5
26*6312673
27-672777
26 '8c
27-88
26*94
27-98
I 2
I 2
3 4
3 4
5
5
6 7
6 7
8 9
8 9
5'3
28-0
28*20
28-30
28*4128-5
28-6
287328*84
28-94
29*05
I 2
3 4
5
6 7
9 10
5'4
2 9 -I
29-27
29-38
29-48 29-5
297
29*81
29*92
30*03
30-I4
1234
6
7 8
9 10
Squares.
SQUARES.
225
1
2
3
4
5
6
7
8
9
1 2
3 4
5
6 7
8 9
5*5
30-25
30-36
30*47
30-58
30-69
o-8c
0-91
31-02
3i*i4
31-25
I 2
3 4
6
7 8
9 10
c*o
31-36
?i"47
31-58317031-81
1-92
2-04
32-15132-2632-38
I 2
3 5
6
7 8
9 10
57
5'8
I' 9
6-0
32-49132-60
33 '64i3376
34'8i|34'93!
36-0036-12!
3272 32 -83132-95
33' 8 733*99|34*
35 '5i35*i 6 35*28
36-2436-3636-48
3-06
6'6o
3-1833-20,133-41
4 > 3434 < 46|34*57;
5-52135-643576;
67236-84^6-97
33-52
34 '69
35-88
37*09
I 2
I 2
I 2
I 2
3 5
4 5
4 5
4 5
6
6
6
6
7 8
7 8
7 8
7 9
9 10
9 JI
IO II
10 II
6-1
37*2i
37 *33|37 '45; 37*58!
3770
7-82
7-95
38-07
38-19
38-32
I 2
4 5
6
7 9
10 II
6-2
38 "44138 "56^38 "69138 '81538 "94
9-06
9 -I9i39-3ii39 "44 39*5ili 3
4 5
6
6 9
IO II
6-4
39-6939-82
40-9641-09
39-9440-0740-20
41-2241-3441-47
0-32
i -60
o -45:40-58 4070.40-83 i 3
i 73'4i -86 41 -99:42-121 1 3
4 5
4 5
6
6
8 9
8 9
10 II
10 12
6-5
42-2542-38
42-5142-644277
2-90
3 -03 43 -1643 -30 43-43, i 3
4 5
7
8 910 12
6-6
43'5643'69
43-82
43-9644-09
44-22
44 -36 44-49 44-62 4476! i 3
4 5
7
8 9 ii 12
67
6-8
44 '8945-02
46-2446-38
45-1645-29145-43
46 -51 46 '65146 79
6^2
57045-8345-97146-1011 3
4 5
4 5
7
7
8 9!n 12
8 lOJII 12
6-9
47-61 14775
47 -89J48 -02)48 -16
8-30
8 -44 48 -58 48 72J48-86 i 3
4 6
7
8 10 ii 13
49-00
49-14
49-28149-4249-56
970
9-8449-9850-13
50-27 i 3
4 6
7
8 xolii 13
7*1
50-4150-55
50-69
50-84
50-98
I-I2
51-27151-41
5^55
5i7o i 3
4 6
7
9 ioUi 13
7*2
7*3
51-8451-98
53-29153*44
52-I3 52-27J52'42
53-58 5373 S3**
2*56
4'02
527152-85
54*17154*32
53-0053-14
54-4654-61
t 3
i 3
4 6
4 6
7
7
9 io!i2 13
9 10 12 13
7*4
7'5
54*76
56'25
54*9i
56-40
55-0655-20
56-5556*70
56*85
5'50
7-00
55-65|55-8o
57-1557-30
55*9556-io
57-46157-61
i 3
2 3
4 6
5 6
7
8
9 10 12 13
9 II.I2 14
7-6
77
5776
59*29
57*9 r
59 '44
58-06
59-60
58-22
5975
58-37
59-9I
58*52
58-68
a "22
58-83
60-37
58-98
60-53
59* I 4
60-68
2 3
2 3
S 6
5 6
8
8
9 ii 12 14
9 ii 12 14
7-8 60-84
7-9 162-41
6i"oo
62-57
61-15161-31
6273:62-88
2
6304
6I-62
63-20
617861-94
63-3663-52
62-09
63-68
62-25
63*84
2 3
2 3
5 6
5 6
8
8
9 ii 13 14
ID 11)13 14
8-0 64-00:64-16
64-3264-48
64-64
6 4 -8o
64-96
65-12
65-29
2 3
5 6
8
10 II
I 3 J 4
8-2
65-616577
67-24,67-40
68-8969-06
65'93!66-io
67-576773
69-2269-39
66-26
67-90
69-56
66-42
68-06
6972
66-59
68-23
69-89
6675
68-39
70-06
66-91
68-56
70-22
67-08
6872
70-39
2 3
2 3
2 3
5 7
5 7
5 7
8
8
8
10 II
10 12
10 12
J 3 *5
i3 IS
13 i5
8-4
70-567073
70-90
71-06171-23
71-46
71*57
7174
71-91
72-08
2 3
5 7
8
10 12
H 15
72 -25 172 -42
72-59
727672-93
73 "1C
73*27
73'44
7362
73*79
2 3
5 7
9
10 12
H 15
8-6
73-9674*i3
74*30
74*48
74-65
74-8
75 "oo
75*i7
75*34
75*52
2 3
5 7
9
10 12
14 16
87
75-697S-86
76-04
76*21
76-39
76-5
7674
76-91
77-09
77-26
2 4
5 7
9
I 12
14 16
8-8
77-4477-62
7779
77*97
7 8-i
78-3
78-50
78-68
78-85
79*03
2 4
5 7
9
I 12
14 16
8-9
79-21179-39
79*57
7974179-92
Jo*i
80-28 80-46
80-64
80-82
9 4
5 7
9
i 1314 16
9-0
Si'oo
8ri8
81-36
81-548172
81-90
82-08
82-26
82*45
82-63
2 4
5 7
9
I 13
14 16
9-1
82-81
82-99
83*I7
83*3683-5;
837
8 3 ' 91
84-09
84-27
84-46
2 4
5 7
9
I 13
15 16
9*2
9'3
84-64
86-49
84-82
86-68
85*01
86-86
87-0587-24
87-4
37-61
g*
86-12
87-98
86-30
88-17
2 4
2 4
6 7
6 7
9
9
I 13
I 13
15 17
*5 17
9'4
88-36
88-55
8874
88-92
8 9 -i
89-3
89-49
89-68
89-87
90-06
2 4
6 8
9
I 13
i5 17
9'5
90-25
90-44
90-63
90*82
91-0
91-2
91*39
91*58
9178
91-97
2 4
6 8
10
I 13
15 17
9*6
97
9-8
92-16
94-05
96-04
92*3 q
94-28
96-24
92-54
94-48
96-42
9274
94 "6/
96-6'
92-9
94*8
96-8
93*i
95 -of.
97*o
93*3293*5i
95-26)95-45
97-2297-42
9370
95 "65
97-61
93*90
95*84
97-81
2 4
2 4
2 4
6 8
6 8
6 8
10
10
10
12 14
12 14
12 14
IS 17
16 18
16 18
9'9
98-01
98-21
98-41
98-60
>98'8
99 -oc
99-2099-40
99-60
99-80
i 2 4
6 8
10
12 I^
16 18
I II
226 Examples in Electrical Engineering.
RECIPROCALS OF NUMBERS FROM 1000 TO 9999.
1
2
3
4
5
6
7
8
9
1 2
3 4
5
6 7
8
10
O'OOIOOOO
9901
9804
9709
9615
9524
9434
9346
9259
9174
918
27 36
45
55 64
7382
II
12
o '0009091
0-0008333
9009
8264
8929 8850
8197 8130
8772
8065
8696
8000
8621
7937
8547 8475 8403
7874 7813 7752
8 15
6 *2
23 30
19 26
38
32
45 5361 68
38 45^1 58
13
0-0007692
7634
7576 75i9
7463
7407
7353
7299 7246 7194
5 "
l6 22
27
33 38 44 49
14
0-0007143
7092
7042 6993
6944
6897
6849
680367576711
5 10
14 19
24
29 33 38 43
o '0006667
6623
6579 6536
6494
6452
6410
6369 6329 6289
4 8
13 17
21
2 5 29 33 38
i6
17
0*0006250
0*0005882
6211
5848
61736135
5814 5780
6098
5747
6061
6024
5682
5988 5952
56505618
5917
5587
4 7
3 6
II 15
10 13
18
16
22 26
20 23
29 33
26 29
18
0-0005556
5525
5495 5464
5435
545
5376
5348 5319
5291
3 6
9 12
15
17 20
23 26
19
0-0005263
5236
5208 5181
5128
5102
5076,5051
5025
3 5
8 ii
13
16 18
21 24
20
0-0005000
4975
49504926,4902
4878
4854
4831
4808
4785
2 5
7 10
12
14 17
I 9 21
21
22
23
0-0004762
0-0004545
0-0004348
4739
4525
4329
4717 4695 4673
450544844464
431042924274
4651
4444
4255
4630
4425
4237
46084587
44054386
42194202
4566
4367
4184
2 4
2 4
2 4
I 1
5 7
II
10
13 15
12 14
II 13
17 20
16 18
14 16
24
25
0-0004167
0-0004000
4149
3984
4132
3968
45
3953
4098
3937
4082
3922
4065
3906
40494032
38913876
4016
3861
2 3
2 3
H
8
10 12
9 ii
13 IS
12 14
26
0-0003846
3831
38173802
3788
3774
3759
3745373 1
3717
I 3
4 6
7
8 10
II 13
27
28
0-0003704
0-0003571
3690
3559
3676 3663
35463534
365
352i
3636
359
3623
3497
36io|3597
348413472
3584
3460
i 3
X 2
4 5
4 5
7
6
8 9
7 9
II 12
10 II
29
0-0003448
3436
3425 3413
34oi
339
3367133^6
3344
X 2
3 5
6
7 8
9 10
30
0*0003333
3322
3311 3300
3289
3279
3268 3257J3247
3236
I 2
3 4
5
6 7
9 10
31
32
0-0003226
0-0003125
3215
35
3205 3195
3106 3096
3185
3086
3175
3077
3 l6 5
3067
3I553I45
30583049
3135
3040
I 2
I 2
3 4
3 4
5
5
a'?
8 9
8 9
33
34
0-0003030
0-0^02941
3021
2933
3012 3003
2924 2915
2994
2907
2985
2899
2976
2890
29672959
2882 2874
2950
2865
I 2
I 2
3 4
3 3
4
4
5 6
5 6
7 8
7 8
35
o '0002857
2849
2841
2833
2825
2817
2809
2801
2793
2786
X 2
2 3
4
5 6
6 7
36
0-0002778
2770
2762 2755
2747
2740
2732
2725
2717
2710
X 2
2
4
5 5
6 7
37
0-0002703
2695
26882681
2674
2667
2660
2653
2646
2639
2
4
4 5
6 6
38
39
40
0-0002632
0-0002564
0-0002500
2558
2494
26182611
24882481
2604
2538
2475
2597
2532
2469
2591
2525
2463
2584
25 J 9
2457
2577
2513
245 1
2571
2506
2445
X
I
X
2
2
2
3
3
3
4 5
4 4
4 4
5 f
5 6
5 5
41
42
43
0-0002439
0-0002381
0-0002326
2433
2375
2320
24272421
23702364
2315 2309
2415
2358
2304
2410
2353
2299
2404 2398 2392 2387
23471234223362331
2294 2288 228312278
X
X
X
2
2
2
3
3
3
3 4
3 4
3 4
5 5
4 5
4 5
44
0-0002273
2268
22622257
2252
2247
2242 2237
22322227
X
2
3
3 4
4 5
45
0-0002222
2217
22122208
2203
2198
2193
2188
2183
2179
o
X
2
3 3
4 4
46
O-O002I74
2169
2165
2160
2155
2151
2146
2141
2137
2132
o
I
2
3 3
4 4
47
0-0002128
2123
2119
2114
2110
2105
2101
2096
2092
2088
o
I
2
3 3
4 4
48
0-0002083
207-
2075
2070
2066
2062
2058
2053
2049
2045
o
I
2
3 3
3 4
49
0*0002041
2037
2033 2028
2O2/
2O2O
20l6
2012
2008
2004
I
2
2 2
3 4
50
0-0002000
1,96
1992
1988
198^
I 9 80
1976
1972
1969
1965
I
2
2 2
3 4
5*
0-OOOI96l
1957
1953
1949
1946
1942
1938
1934
i93i
1927
I
I 2
2
2 3
3 3
52
53
o "000192;:
0*000l887
^
1916
1880
1912
1876
I908
1873
1905
l86 9
1901
1866
1898
1862
1894
1859
1855
I
I
I I
I I
2
2
2 3
2 2
3 3
3 3
54
o '0001852 1848
1845
1842
1838
1835
1832
1828
1825
1821
X
X I
2
2 2
3 3
N.B. Three zeros follow the decimal point in the reciprocal of any four figure whole number
except the number 1000.
NOTE. Numbers in difference columns to be subtracted, not added.
Reciprocals of Numbers from 1000 to 9999.
RECIPROCALS OF NUMBERS FROM 1000 TO 9999.
227
55
56
%
g
61
62
63
64
65
66
67
63
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
5
86
87
88
89
90
9i
92
93
94
95
96
97
98
99
1
3
3
1808
1776
1745
1715
1686
1658
1631
1605
1580
4
1805
1773
1742
1712
1684
1656
1629
1603
1577
1553
*5 2 9
1506
1484
1462
1441
1420
1401
1381
1362
*344
1326
1309
1292
1276
1259
1244
1229
5
6
7
1795
1764
1733
1704
1675
1647
1621
1595
1570
8
9
1 2
3 4
5
6 7
8 9
0-0001818
o '0001786
0*0001754
0*0001724
0-0001695
0-0001667
0-0001639
0-0001613
0-0001587
0-0001563
0-0001538
0-0001515
0-0001493
0-0001471
0-0001449
0*0001429
0-0001408
0-0001389
0-0001370
0-0001351
0-0001333
0*0001316
0-0001299
0-0001282
0-0001266
0-0001250
0-0001235
0*0001220
0-OOOI205
0-OOOII90
0-OOOII76
o '0001163
0*0001 i4c
0*0001136
0*0001124
0*0001111
0*0001099
0*0001087
0*0001075
0*0001064
0-000105;
0*0001042
0-0001031
0*0001020
O'OOOIOIO
1815
1783
1751
1721
1692
1664
1637
1610
1585
1560
1536
1513
1490
1468
1447
1427
1406
1387
1368
1350
1332
1314
1297
1230
1264
124^
'233
1218
1812
1779
1748
1718
1689
1661
1634
1608
1582
1558
1534
rS 11
1488
1466
1445
1425
1404
'SfS
1366
1348
1330
13"
1295
1279
1263
1247
1232
1217
1202
1188
1174
1160
1147
H34
II2I
1109
I0 9 6
1085
802
770
739
709
1681
1653
1626
1600
1575
155
I5 2 7
1504
1481
1460
H39
1418
1399
1379
1361
1342
1325
1307
1290
1274
1258
1242
1227
1799
I7 67
1736
1706
1678
1650
1623
1597
1572
1548
1524
1502
1479
1458
1437
1416
1397
1377
1359
1340
1323
1305
1289
1272
1256
1241
1225
I2II
II 9 6
Il82
1168
"55
1142
1129
1116
1104
1092
1792
1761
1730
1701
1672
1645
1618
1592
1567
1543
1520
1497
1475
1453
1433
1412
1393
1374
1355
1337
1319
1302
1285
1269
1253
1238
1222
1208
"93
1179
1166
1152
"39
1126
1114
IIOI
1089
1078
1066
J 55
1044
1033
1022
1012
1002
1789
1757
1727
1698
1669
1642
1616
1590
1565
i54i
1517
1495
1473
I45 1
I43 1
1410
i39i
1372
1353
1335
1318
1300
1284
1267
1252
1236
1221
1206
1192
II 7 8
1164
" S o
H38
1125
III2
1 100
1088
1076
1065
I0 5^
1043
1032
IO2I
IOII
1001
O I
o
o
o
I I
2
2
2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
I 2
3 3
3 3
2 3
2 3
2 3
2 3
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
I 2
X 2
X 2
I I
J-OO-D
1531
1508
I 4 86
1474
1443
I 4 22
1403
1383
I3 6 4
1346
1328
I3II
1294
1277
I26l
1245
1230
1215
1200
1186
1172
H59
H45
H33
1 120
1107
IO95
I08 3
1072
1000
1049
1038
IO28
1017
1007
1 54
1522
1499
1477
1456
1435
1414
1395
1376
1357
1339
1321
I 34
1287
1271
1255
1239
122^
1209
"95
1181
1167
"53
1140
1127
1115
no;
1091
1079
1067
1056
1045
1034
1024
1013
1003
o o
O
O
o o
O
o o
O
O
o c
I
I
I
I I
I I
O I
O I
I
I
2
I
I
I
I
I
I
I
I
I I
I I
O
O
I
I
I
I
"99
1198
1183
1170
1156
H43
1130^
1117
1105
1093
1081
1070
105^
1047
1036
1026
1015
1005
1189
1175
1161
n 4 8
"35
1 122
IIIO
1098
1086
1074
1063
1052
IO4I
IO3O
1019
1009
1171
"57
1144
1131
1119
1106
1094
1082
1071
I0 59
1048
1037
1027
1016
1006
O
o o
O
o o
O
o o
I
I
I
1
o o
1068
1057
1046
1035
1025
1014
1004
1U ^
1062
1050
1040
IO29
ioi
1008
I
o
I I
I I
I I
I X
N.B. Three zeros follow the decimal point in the reciprocal ol any four-figure whole number
except the number 1000.
NOTE. Numbers in difference columns to be subtracted, not added.
228 Examples in Electrical Engineering.
NATURAL TANGENTS.
1
2
.30
4
5
6
7
.go
.90
oooo
0017
0035
0052
0070
0087
0105
0122
0140
oi57
I
0175
0192
0209
0227
0244
0262
o>79
0297
0314
0332
2
0349
0367
0384
0402
0419
0437
0454
0472
0489
0507
3
0524
0542
0559
0577
0594
0612
0629
0647
0664
0682
4
0699
0717
0734
0752
0769
0787
0805
0822
0840
0857
5
0875
0892
0910
0928
0945
0963
0981
0998
1016
1033
6
1051
1069
1086
1104
1122
"39
H57
"75
1192
1210
7
1228
1246
1263
1281
1299
I3U
1334
i35 2
1370
1388
8
1405
1423
1441
H59
1477
H95
1512
1530
1548
1566
9
1584
1602
1620
1638
1655
1673
1691
1709
1727
1745
10
1763
1781
1799
1817
1835
1853
1871
1890
1908
1926
ii
1944
1962
1980
1998
20l6
2035
2053
2071
2089
2107
12
2126
2144
2162
2180
2199
2217
2235
2254
2272
2290
13
2309
2327
2345
2364
2382
2401
2419
2435
2456
2475
14
2493
2512
253
2549
2568
2586
2605
2623
2642
2661
15
2679
2698
2717
2736
2754
2773
2792
2811
2830
2849
16
2867
2886
2905
2924
2943
2962
2981
3000
3019
3038
17
3057
3076
3096
3H5
3134
3153
3172
3191
3211
3230
18
3249
3269
3288
3307
3327
3346
3365
3385
3404
3424
19
3443
3463
3482
3502
3522
3541
35 6 i
358i
3600
3620
20
3640
3659
3679
3699
3719
3739
3759
3779
3799
3819
21
3839
3859
3879
3899
3919
3939
3959
3979
4000
4020
22
4040
4061
4081
4101
4122
4142
4163
4183
4204
4224
23
4245
4265
4286
4307
4327
4348
4369
4390
44"
443 i
24
25
'4452
4663
4473
4684
4494
4706
4515
4727
4536
4748
4557
4770
4578
479i
4599
4813
4621
4834
4642
4856
26
4877
4899
4921
4942
4964
4986
5008
5029
5051
573
27
5095
5"7
5139
5161
5^4
5206
5228
5250
5272
5295
28
5317
5340
5362
5384
5407
5430
5452
5475
5498
5520
29
5543
5566
5589
5612
5635
5658
5681
5704
5727
5750
3
5774
5797
5820
5844
5^7
5890
59H
5938
596i
5985
31
6009
6032
6056
6080
6lO4
6128
6152
6176
6200
6224
32
6249
6273
6297
6322
6346
6371
6395
6420
6445
6469
33
6494
6519
6544
6569
6594
6619
6644
6669
6694
6720
34
6745
6771
6796
6822
68 47
6873
6899
6924
6950
6976
35
7002
7028
7054
7080
7107
7133
7159
7186
7212
7239
36
7265
7292
7319
7346
7373
7400
7427
7454
748i
75o8
37
7536
7563
7590
7618
7646
7673
7701
7729
7757
7785
3*
7813
7841
7869
7898
7926
7954
7983
8012
8040
8069
39
8098
8127
8156
8185
8214
8243
8273
8302
8332
8361
40
8391
8421
8451
8481
8511
8541
857i
8601
8632
8662
4i
8693
8724
8754
8785
8816
8847
8878
8910
8941
8972
42
9004
9036
9067
9099
9i3i
9163
9195
9228
9260
9293
43
9325
93 S8
9391
9424
9457
949
9523
9556
9590
9623
44
9657
9691
9725
9759
9793
9827
9861
9896
993
9965
Natural Tangents.
NATURAL TANGENTS.
229
1
.go
. 3 o
4
5
6
7
8
. 9 o
45
I'OOOO
0035
0070
0105
0141
0176
0212
0247
0283
0319
46
i '0355
0392
0428
0464
0501
0538
0575
0612
0649
0686
47
i -0724
0761
0799
0837
0875
0913
0951
0990
1028
1067
*r/
4.8
i'iio6
1145
1184
1224
1263
1303
1343
1383
1423
1463
*T*J
49
1-1504
1544
1585
1626
1667
1708
1750
1792
1833
1875
50
1-1918
1960
2002
2045
2088
2131
2174
2218
2261
2305
CI
2349
2393
2437
2482
2527
2572
26l7
2662
2708
2753
*3
C2
'2799
2846
2892
2938
2985
3032
3079
3127
3175
3222
J
53
3270
33 J 9
3367
34i6
3465
35H
3564
3613
3663
3713
54
3764
38i4
3865
3916
3968
4019
4071
4124
4176
4229
JT^
55
4281
4335
4388
4442
4496
4550
4605
4659
4715
4770
56
4826
4882
4938
4994
5051
5108
5166
5224
5282
5340
J
57
'5399
5458
5517
5577
5637
5697
5757
5818
5880
594i
58
6003
6066
6128
6191
6255
6319
6383
6447
6512
6577
3
CO
6643
6709
6775
6842
6909
6977
7045
7H3
7182
7251
60
7321
739i
7461
7532
7603
7675
7747
7820
7893
7966
61
8040
8115
8190
8265
8341
8418
8495
8572
8650
8728
62
8807
8887
8907
9047
9128
9210
9292
9375
9458
9542
63
1-9626
97"
9797
9883
9970
0057
0145
0233
0323
0413
64
2-0503
0594
0686
0778
0872
0965
1060
"55
1251
1348
65
2-1445
1543
1642
1742
1842
1943
2045
2148
2251
2355
66
2*2460
2566
2673
2781
2889
2998
3109
3220
3332
3445
67
2-3559
3 6 73
3789
3906
4023
4142
4262
4383
454
4627
68
2-4751
4876
5002
5129
5257
5386
5517
5 6 49
5782
59i6
69
2-6051
6187
6325
6464
6746
6889
7034
7179
7326
70
2-7475
7625
7776
7929
8083
8239
397
8556
8716
8878
7i
2-9042
9208
9375
9544
97H
9887
0061
0237
0415
0595
72
3-0777
0961
1146
1334
1524
1716
1910
2106
2305
2506
73
3-2709
2914
3122
3332
3544
3759
3977
4197
4420
4646
74
3-4874
5105
5339
5576
5816
6059
6305
6554
6806
7062
75
37321
7533
7848
8118
8391
8667
8947
9232
9520
9812
76
4-0108
0408
0713
1022
1335
l6 53
1976
2303
2635
2972
77
4'33i5
3662
4015
4374
4737
5107
5483
5864
6252
6646
78
47046
7453
7867
8288
8716
9152
9594
0045
0504
0970
79
5-I446
1929
2422
2924
3435
3955
4486
5026
5573
6140
80
5-67I3
7297
7894
8502
9124
9758
0405
1066
1742
2432
Si
6-3138
3859
4596
5350
6122
6912
7920
8548
9395
0264
82
7'"54
2066
3002
3962
4947
5958
6996
8062
9158
0285
83
8-1443
2636
3863
5126
6427
7769
9152
0579
2052
3572
84
9-5I44
9-677
9-845
10-02
10-20
10-39
10-58
10-78
10-99
11-20
85
H'43
n-66
11-91
I2"l6
12-43
12-71
13-00
13-30
13-62
13-95
86
14-30
14-67
15*06
15H6
15-89
16-35
16-83
I7-34
17-89
18-46
7
19-08
19-74
20-45
21'20
22-02
22-90
23-86
24-90
26-03
27-27
88
28-64
30-14
31-82
33^9
35-80
38-19
40-92
44-07
47-74
52-08
89
57-29
63-66
71-62
81-85
95-49
114-6
143-2
191*0
286-5
573-0
230
Examples in Electrical Engineering.
NATURAL SINES.
1
2
.30
4
5
6
r
.go
.90
0000
0017
0035
0052
0070
0087
OIO5 OI22
0140
0157
I
0175
0192
0209
0227
0244
0262
0279
0297
0314
0332
2
0349
0366
0384
0401
0419
0436
0454
9471
0488
0506
3
0523
0541
0558
0576
0593
0610
0628 0645
0663
0680
4
0698
0715
0732
0750
0767
0785
0802
0819
0837
0854
5
0872
0889
0906
0924
0941
0958
0976
0993
IOII
1028
6
1045
1063
1080
1097
IIJ 5
1132
1149
1167
1184
1201
7
1219
1236
1253
1271
1288
1305
1323
1340
1357
1374
8
1392
1409
1426
1444
1461
1478
1495
*5*3
1530
1547
9
1564
1582
1599
1616
1633
1650
1668
1685
I7O2
1719
10
1736
1754
1771
1788
1805
1822
1840
1857
1874
I8 9 X
ii
1908
1925
1942
1959
1977
1994
201 1
2028
2045
2062
12
2079
2096
2113
2130
2147
2164
2181
2158
2215
2232
13
2250
2267
2284
2300
2317
2334
2351
2368
2385
2402
14
2419
2436
2453
2470
2487
2504
2521
2538
2554
2571
15
2588
2605
2622
2639
2656
2672
2689
2706
2723
2740
16
2756
2773
2790
2807
2823
2840
2857
2874
2890
2907
17
2924
2940
2957
2974
2990
3007
3024
3040
3057
3074
18
3090
3107
3123
3140
3156
3173
3190
3206
3223
3239
19
3256
3272
3289
3305
3322
3338
3355
337i
3387
3404
20
3420
3437
3453
3469
3486
3502
3535
3551
3567
21
3584
3600
3616
3633
3649
3665
3681
3697
37H
3730
22
3746
3762
3778
3795
3811
3827
3843
3859
3875
3891
23
3907
3923
3939
3955
397i
3987
4003
4019
4035
4051
24
4067
4083
4099
4H5
4131
4H7
4163
4179
4195
4210
25
4226
4242
4258
4274
4289
4305
4321
4337
4352
4368
26
4384
4399
4415
4431
4446
4462
4478
4493
4509
4524
27
4540
4555
4571
4586
4602
4617
4633
4648
4664
4679
28
4695
4710
4726
4741
4756
4772
4787
4802
4818
4*j3
29
4848
4863
4879
4894
4909
4924
4939
4955
4970
3
5000
5015
5030
5045
5060
5075
5090
5105
5120
5135
31
5150
5165
5180
5195
5210
5225
5240
5255
5270
5284
32
5299
53H
5329
5344
5358
5373
5388
5402
5417
5432
33
34
5446
5592
5606
5476
5621
5490
5635
5505
5650
5519
5664
5534
5678
5548
5693
5563
5707
5577
35
5736
5750
5764
5779
5793
5807
5821
5835
5850
5864
36
37
5878
6018
5892
6032
5906
6046
5920
6060
5934
6074
5948
6088
5962
6101
5976
6115
5990
6l29
6004
6i43
38
6157
6170
6184
6198
6211
6225
6239
6252
6266
6280
39
6293
6307
6320
6 334
6347
6361
6374
6388
6401
6414
40
6428
6441
6455
6468
6481
6494
6508
6521
6534
6547
4*
6561
6574
6587
6600
6613
6626
6639
6652
6665
6678
42
6691
6704
6717
6730
6743
6756
6769
6782
6794
6807
43
6820
6833
6845
6858
6871
6884
6896
6909
6921
6934
44
6947
6959
6972
6984
6997
7009
7022
7034
7046
7059
Natural Sines.
NATURAL SINES.
231
1
2
. 3 o
.40
6
6
7
.QO
.90
45
7071
7083
7096
7 108
7120
7133
7H5
7157
7169
7l8l
46
7193
7206
7218
7230
7242
7254
7266
7278
7290
7302
47
73H
7325
7337
7349
7361
7373
7385
7396
7408
7420
48
7431
7443
7455
7466
7478
7490
7501
7513
7524
7536
49
7547
7556
7570
758i
7593
7604
7627
7638
7649
50
7660
7672
7683
7694
7705
7716
7727
7738
7749
7760
5i
7771
7782
7793
7804
7815
7826
7837
7848
7859
7869
5 2
7880
7891
7902
7912
7923
7934
7944
7955
7965
7976
53
7986
7997
8007
8018
8028
8039
8049
8059
8070
8080
54
8090
8100
8111
8121
8131
8141
8151
8161
8171
8181
55
8192
8202
8211
8221
8231
8241
8251
8261
8271
8281
56
8290
8300
8310
8320
8329
8339
8348
8358
8368
8377
57
8387
8396
8406
8415
8425
8434
8443
8453
8462
8471
58
8480
8490
8499
8508
8517
8526
8536
8545
8554
8563
59
8572
8581
8590
8599
8607
8616
8625
8634
8643
8652
60
8660
8669
8678
8686
8695
8704
8712
8721
8729
8738
61
8746
8755
8763
8771
8780
8788
8796
8805
8813
8821
62
8829
8838
8846
8854
8862
8870
8878
8886
8894
8902
63
8910
8918
8926
8934
8942
8949
8957
8965
8973
8980
64
8988
8996
9003
9011
9018
9026
9033
9041
9048
9056
65
9063
9070
9078
9085
9092
9100
9107
9114
9121
9128
66
9135
9H3
9150
9157
9164
8171
9178
9184
9191
9198
67
9205
9212
9219
9225
9232
9239
9245
9252
9259
9265
68
9272
9278
928-;
9291
9298
9304
93"
9317
9323
9330
69
9336
9342
9348
9354
9361
9367
9373
9379
9385
9391
70
9397
9403
9409
9415
9421
9426
9432
9438
9444
9449
71
9455
9461
9466
9472
9478
9483
9489
9494
9500
9505
72
95"
95 * 6
9521
9527
9532
9537
9542
9548
9553
9558
73
95 6 3
9568
9573
9578
9583
9588
9593
9598
9603
9608
74
9613
9617
9622
9627
9632
9636
9641
9646
9650
9655
75
9659
9664
9668
9673
9677
9681
9686
9690
9694
9699
76
9703
9707
97"
9715
9720
9724
9728
9732
9736
9740
77
9744 1 9745
9751
9755
9759
9763
9767
9770
9774
9778
78
978i
97*5
9789
9792
9796
9799
9803
9806
9810
9816
9820
9823
9826
9829
9833
9836
9839
9842
9845
80
9848
9851
9854
9857
9860
9863
9866
9869
9871
9874
Si
9877
9880
9882
9885
9888
9890
9893
9895
9898
9900
82
9903
9905
9907
9910
9912
9914
9917
9919
9921
9923
83
9925
9928
9930
9932
9934
9936
9938
9940
9942
9943
84
9945
9947
9949
9951
9952
9954
9956
995?
9959
9960
85
9962
9963
9965
9966
9968
9969
9971
9972
9973
9974
86
9976
9977
9978
9979
9980
9981
9982
9983
9984
998;
87
9986
9987
9988
9989
9990
9990
9991
9992
9.93
9993
88
9994
9995
9995
9996
9996
9997
9997
9997
9998
9998
89
9998
9999
9999
9999
9999
rooo
I OOO
rooo
rooo
roco
nearly.
nearly.
nearly.
nearly-
nearly.
232 Examples in Electrical Engineering.
LOGARITHMS.
1
Q
8
4
5
e 7
8 1
13
3 4
5
e 7
8 e
10 OOOO
0043
0086 0128 017:
0212
0253
0294
03340374
48
12 17
21
25 29
33 37
|
II
0414
0453
0492 0531 0569
060 7
0645 0682
07190755
48
II 15
19
23 26
3 34
12
0792
0828
0864 0899 0934
0969
1004 1038
1072 1106
3 7
10 14
17
21 24
28 31
13
"39
H73
1206 1239,1271
1303
1335^367
1399' 1430
10 13
16
19 23
26 29
14
1461
1492
1523 155311584
I6l 4
164411673
17031732
3 6
9 12
15
18 21
24 27
15
1761
1790
1818 1847
1875
1903
1931
1959
1987 2014
36
8 ii
14
17 2O 22 25
16
17
2041
2304
2068
2330
2095 2122
2355 2380
2148
240-
2175
2430
2201 2227
2455(2480
22532279
2504 2529
3 5
2 5
8 ii
7 10
13
12
16 18
15 17
21 24
20 22
18
2553
2577
2601 2625
2648
2695 2718
27422765
2 5
7 9
12
14 16
19 21
19
2788
2810
2833 2856
2878
2900
2923,2945
2967 2989
2 4
7 9
II
13 16
18 20
20
3010
3032
3054 3075
3096
3 II8
3139 3160
3181,3201
2 4
6 8
II
13 15
17 19
21
22
23
24
3222
3f4
3617
3802
3243
3444
3636
3820
3263 3284
3464 3483
3655 3674
38383856
3304
3502
3692
3874
3324
3522
37"
3892
3345 3365
35413560
3729 3747
3909 3927
3385 3404
3579i3598
3766(3784
39453962
2 4
2 4
2 4
4
6 8
6 8
6 7
5 7
10
10
9
9
12 14
12 14
II 13
II 12
16 18
IS 17
IS 17
14 16
25
3979
3997
4014 4031
4048
4065
4082
4099
41164133
3
5 7
9
10 12
14 15
26
4150
4166
4183 42OO
4216
4232
4249
4265
4281
4298
3
5 7
8
10 ii 13 15
27
43 J 4
4330
43464362
4378
4393
4409
4425
4440
4456
3
5 6
8
9 J 3 14
28
4472
4487
45024518
4533
4548
4564
4579
4594
4609
3
5 6
8
9 ii 12 14
29
4624
4639
46544669
4683
4698
4713
4728
4742
4757
3
4 6
7
9 1012 13
30
477i
4786
4800
4814
4829
4843
4857
4871
4886
4900
3
4 6
7
9 10 ii 13
31
4914
4928
4942
4955
4969
4983
4997
5011
5024
5038
3
4 6
7
8 10
II 12
32
5051
5065
50795092
5105
5 JI 9
5132
5145
5159
5172
3
4 5
7
8 9 ii 12
33
5185
5198
5211
5224
5237
5250
5263
5276
52^9
5302
3
4
6
8 9'io 12
34 5315
532853405353
5366
5378
5391
5403
54i6
5428
3
4
6
8 9|io ii
35
544i
5453 6465 5478
5490
5502
55M
5527
5539
555i
2
4
6
7 9 ! io ii
36
5563
5575
5599
5611
5623
5635
5647
5658
5670
2
4
6
7 8
10 II
37
5682569457055717
5729
5740
5752
5763
5775
5786
2
3
6
7 8
9 10
38
5798 5809
5821
5832
5843
5855
5866
5877
5888
5899
2
3
6
7 8
9 10
39
5911 5922
5933 ! 5944
5955
5966
5977
5988
5999
6010
2
3
5
7 8
9 10
40
6021
6031
604216053
6064
6075
6085
6096
6107
6117
2
3 4
5
6 8
9 10
4i
6128613861496160
6170
6180
6191
6201 6212
6222
2
3 4
5
6 7
8 9
42
43
44
45
6232 6243
63356345
64356444
65326542
6253
6355
6454
655 1
6263
6 A 5
6464
6561
6274
6375
6474
657i
6284
6385
6484
6580
6294
6395
6493
6590
6304 6314 6325
6405 6415,6425
6503 6513:6522
6599 660916618
2
2
2
2
3 4
3 4
3 4
3 4
5
5
5
5
6 7
67*
6 7
8 9
8 9
8 9
8 9
46
6628J6637 6646 6656 6665
6675
6684 6693 6702 6712
2
3 4
5
6 7
7 8
47
6721 6730 6739 6749)6758
6767
6776 6785:6794:6803
2
3 4
5
5678
48
49
6812
6902
6821 683016839:6848
6911 6920 6928 6937
6857
6946
68666875
6955^964
688416893
697256981
2
2
3 4
3 4
4
4
5 6
5 6
7 8
7 8
50
6 99 c
699817007
7Oi6 i 7O2.i
7033
7042 7050 705917067
2
3 3
4
5 6j 7 8
51
7076
70847093
7101
7110
7118
7126
7135
7143
7152
2
3 3
4
5678
52
7160716817177
71857193
7202
72107218:722617235
2
2 3
4
5 6{ 7 7
53
724317251:7259
7267 7275
7284
7292173001730817316
2
2 3
4
5 6! 6 7
54
73247332:7340
7348 7356
1
7364
73727380 7388 7396
2
2 ^
4
5 6J6 7
Logarithms.
LOGARITHMS.
233
1
2
3
4
5
6
7
8
9
12
3 4
5
6 7
8 9
55
7404
7412
7419
7427
7435
7443
7451
7459
7466
7474
I 2
2 3
4
5 5
6 7
56
7482
7490
7497
7505
7513
7520
7528
7536
7543
755i
2
2 3
4
5 5
6 7
%
7559
7634
7566
7642
7574
7649
7582
7657
7589
7664
^
7679
7612
7686
7619
7694
7627
7701
2
2 3
2 ;
4
4
5 5
4 5
6 7
6 7
59
7709
7716
7723
773i
7738
7745
7752
7760
7767
7774
2 '
4
4 5
6 7
60
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
2 <
4
4 5
6 6
61
7853
7860
7868
7875
7882
7889
7896
7903
7910
7917
2
4
4 5
6 6
62
7924
793 1
7938
7945
7952
7959
7966
7973
7980
7987
2 <
3
4
6 6
63
7993
8000
8007
8014
8j2I
8028
8035
8041
8048
8055
2 <
3
4
5 6
64
8062
8069
8075
8082
8089
8096
5 102
8109
8116
8122
2 ^
3
4
5 6
65
8129
8136
8142
8149
8156
1162
8169
8176
8182
8189
2
3
4
5 6
66
8i95
8202
8209
8215
8222
8228
8235
82 4 !
8248
8254
2
3
4
5 6
67
8261
8267
8274
8280
8287
8293
8299
8306
8312
83*9
2 <
3
4
5 6
68
83 2 5
833i
8338
8344
835i
8357
8363
8370
8376
8382
2 ;
3
4 4
5 6
69
8388
8395
8401
8407
8414
8420
5426
8432
8439
8445
2
3
4 4
5 6
70
845i
8457
8463
8470
8476
8482
8488
8494
8500
8506
2
3
4 4
5 6
7 1
72
8513
8573
8519
8579
8525
8585
8531
859i
8537
8597
US!
ss
si 55
8615
8561
8621
8567
8627
2
2
3
3
4 4
4 4
5 5
5 S
73
8633
8630
8645
8651
8657
8663
8669
8675
8681
8686
2
3
4 4
5 5
74
8692
8698
8704
8710
8716
8722
8727
8733
8739
8745
2
3
4 4
5 5
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
2
3
3 4
5 5
76
8808
8814
8820
8825
8831
8837
8842
8848
8854
8859
2
3
3 4
5 5
77
8865
8871
8876
8882
8887
8893
8899
8904
8910
8915
2
3
3 4
4 5
78
8921
8927
8932
8938
8943
8949
8954
8960
8965
8971
2
3
3 4
4 5
79
8976
8982
8987
8993
8998
9004
9009
9 OI 5
9020
9025
2
3
3 4
4 5
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
2
3
3 4
4 5
81
9085
9090
9096
9101
9106
9112
9117
9122
9128
9133
2 2
3
3 4
4 5
82
9*3 8
9143
9149
9154
9159
9165
9170
9175
9180
9186
2 2
3
3 4
4 5
83
9 I 9 I
9196
9201
9206
9212
9217
9222
9227
9232
9238
2
3
3 4
4 5
84
9213
9248
9253
9258
9263
9269
9274
9279
9284
9289
3
3 4
4 5
85
9294
9299
9304
9309
9315
9320
9325
9330
9335
934
3
3 4
4 5
86
9345
9350
9355
936o
9365
9370
9375
9380
9385
9390
I
3
3 4
4 5
87
9395
9400
945
9410
9415
9420
9425
9430
9435
9440
2
3 3
4 4
88
9445
9450
9455
9460
9465
9469
9474
9479
9484
9489
2
3 3
4 4
89
9404
9499
9504
9509
9513
95i8
9523
9528
9533
9538
o
2
3 3
4 4
90
9542
9547
9552
9557
9562
9566
957i
9576
958i
9586
2
3 3
4 4
9 1
959
9595
9600
9605
9609
9614
9619
9624
9628
9633
2
3 3
4 4
92
9638
9643
9647
9652
9657
9661
9666
9671
9675
9680
2
3 3
4 4
93
9685
9689
9694
9699
9703
9708
9713
9717
9722
9727
2
3 3
4 4
94
973 1
9736
974i
9745
975
9754
9759
9768
9773
2
3 3
4 4
95
9777
9782
9786
9791
9795
9800
9805
9809
9814
98U
o
2
3 3
4 4
96
9823
9827
9832
9836
Q8 4 i
9845
9850
9854
9859
9863
2
3 3
4 4
97
9868
9872
9877
9881
9886
9890
9894
9899
9903
9908
2
3 3
4 4
98
9912
9917
9921
9926
9930
9934
9939
9943
9948
9952
2
3 3
4 4
99
9956
9961
9965
9969
9974
9978
9983 99-7
9991
9996
2
3 3
3 4
234
Examples in Electrical Engineering.
ANTILOGARITHMS.
1
2
3
4
5
6
7
8
9
1 2
34
5
7 ! 8 9>
oo
1000
1002
1005
1007
1009
1012
1014
1016
1019
102 1
O
I I
I
1222-
oi
1023
IO26
1028
1030
r 33
1035
1038
1040
1042 1045
o o
I 2: 2 2
"02
1047
1050
1052
1054
1057
1059
1062 1064
106711069
I 2
2 2
*3
1072
1074
1076
1079
1081
1084
10861089
1091
1094
o
I 2
2 2
04
1096
1099
1 102
1104
1107
1109
III2 III4
1117
III9
o
2 2
2 3
'OS
1 122
1125
1127
1130
1132
"35
"38II40
U43
n 4 6
2 2
2 2
06
II 4 8
5i
"53
1156
1150
1161
1164^167
1169
1172
2 2
2 2
07
"75
1178
1180
1183
1186
1189
1191 iig/
1197
1199
o
2 2
2 2
08
1202
1205
1208
I2II
1213
1216
1219
1222
1225
1227
2 2
2 3
09
1230
1233
1236
1239
1242
1245
1247
1250
1253
1256
2 2
2 3
10
1259
1262
1265
1268
I2 7 !
1274
1276
1279
1282
1285
2 2
2 3
II
1288
1291
129^
1297
I 3 00
1303
1306
1309
1312
1315
o
2
2 2
2 3
*I2
1318
1321
1324
1327
1330
1334
1337
1340
1343
1346
o
2
2 2
2 3
;i3
1352
1355
1358
1361
1365
1368
1371
1374
1377
o
2
2 2
3 3
1380
1384
1387
1390
1393
1396
1400
1403
1406
1409
2
2 2
3 3
15
1413
1416
1419
1422
1426
1429
I43 2
J 435
1439
1442
2
2 2
3 3
16
1445
1449
1452
X 4S5
1459
1462
1466
1469
1472
1476
o
2
2 2
3 3
17
J 479
1483
1486
1489
1493
1496
1500
1507
1510
2
2 2
3 3
18
1514
1517
1521
1524
1528
I53 1
1535
S
1542
1545
2
2 2
3 3
19
1549
1552
1556
1560
1563
1567
1570
1574
1578
1581
o
2
2 2
3 3
20
1589
1592
IS96
1000
1603
1607
1611
1614
1618
2
2 2
3 3
*2X
1622
1626
1629
1633
l6 37
1641
1644
1648
1652
1656
2
2
2 2
3 3
22
1660
1663
1667
1671
1675
1679
1683
1687
1690
1694
o
2
2
2 2
3 3
23
1698
1702
1706
1710
1714
1718
1722
1726
1730
1734
o
2
2
2 2
3 4
24
1738
1742
1746
1750
I7S4
1758
1762
1766
1770
1774
o
2
2
2 2
3 4
25
1778
I7 82
1786
1791
1795
1799
1803
1807
1811
1816
2
2
2 2
3 4
26
1820
1824
1828
1832
1837
1841
1845
1849
1854
1858
o
2
2
3 3
3 4
2 7
1862
1866
1871
1875
1879
1884
1888
1892
1897
1901
o
2
2
3 3
3 4
28
1905
1910
1914
1919
1923
1928
1932
1936
1941
1945
2
2
3 3
4 4
2 9
1950
1954
1959
1963
I 9 68
1972
1977
1982
1986
1991
2
2
3 3
4 4
3
1995
2000
2004
2009
2014
2018
2023
2028
2032
2037
o
2
2
3 3
4 4
*3*
2042
2046
2051
2056
2061
2065
2070
2075
2080
2084
o
2
2
3 3
4 4
32
2089
2094
2099
2104
2109
2113
2118
2123
2128
2133
2
2
3 3
4 4
"33
2138
2143
2148
2153
2I 5 8
2163
2168
2173
2178
2183
2
2
3 3
4 4
'34
2188
2193
2198
2203
2208
22I-:
2218
2223
2228
2234
2 2
3
3 4
4 5
*35
2239
2244
2249
2254
2259
2265
2270
2275
2280
2286
2 2
3
3 4
4 S
36
22911
22 9 6
2301
2307
2312
2317
2323
2328
2333
2339
2 2
3
3 4
4 S
'37
2344
2350
2355
2360
2 3 66
2371
2377
2382
2388
2393
2 2
3
3 4
4 5
38
2399
2404
2410
2415
2 4 2I
2427
2432
2438
2443
2449
2 2
3
3 4
4 5
"39
2455
2460
2466
2472
2477
2483
2489
2495
2500
2506
2 2
3
3 4
5 S
40
2512
2518
2523
2529
2535
2541
2547
2553
2559
2564
2 2
3
4 4
5 5
42
2570
2630
2 Ff
2636
2582
2642
2588
2649
2594
2655
2600
266l
2606
2667
2612
2673
2618
2679
2624
2685
2 2
2 2
3
3
4 4
4 4
If
43
'44
'45
2692
2754
2818
2698
2761
2825
2704
2767
2831
2710
2773
2838
2716
2 7 80
2844
2723
2 7 86
28 5 I
2729
2793
2858
2735
2799
2864
2742
2805
2871
2748
2812
2877
2 3
2 3
2 3
3
3
3
4 4
4 4
4 5
1 t
46
2884
2891
2897
2904
2 9 II
917
2924
2931 2938
2944
2 3
3
4 S
5 6
*47
2951
2958
2965
2972
2979
985
2992
2999 300 6
3 OI 3
2 3
3
4 5
5 6
48
'49
3020
3090
3027
3097
3034 3041
3I05J3H2
3048
055
3126
3062
3133
30693076
31413148
3083
3155
2 3
2 3
4
4
4 5
4 S
6 6
6 6
A ntilogarithms.
ANTILOGARITHMS.
235
1
Q
3
4
5
6
7
8
9
12
34
5
6 7
8 9
'50
3162
3170
3177
3184
3192
3i99
3206
3 2 i4
3221
3228
i i
2 3
4
4 5
6 7
'51
3236
3243
3251
3 2 58
3266
3273
3281
3289
3296
334
I 2
2 3
4
5 5
6 7
'52
33ii
3319
3327
3334
3342
335
3357
3365
3373
338i
I 2
2 3
4
5 5
6 7
"53
3388
3396
3404
3412
3420
3428
3436
3443
3451
3459
I 2
2 3
4
5 6
6 7
'54
3467
3475
3483
349i
3499
35*
35i6
3524
3532,
3540
I 2
2 3
4
5 6
6 7
"55
3548
3556
3565
3573
358i
3589
3597
3606
3614
3622
I 2
2 3
4
5 6
7 7
56
3631
3639
3648
3656
3664
3673
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INDEX
Activity, or rate of doing work (P), I
measurement of, by ammeter and voltmeter, 12
Alternating-current circuit, 113
shape of wave, 1 14
value of B.E.M.F. in finite solenoid, 131
- chemical effect, 115
- thermal effect, 116
virtual value of, 117
magnetic effect of, 118-144
B.E.M.F. of self-induction, 120, 127
- power in circuit, 125
- lag of, 123
lag of, with small B.E.M.F., 125
medium B.E.M.F., 126
large B.E.M.F., 127
Ammeter, winding of coil of, 85
Ampere-turns magnetizing force, 70
required for armature, 77
a j r gapSf 77
magnet limbs, 77
yoke, 77
compounding, 90
by transformer iron, 155
voltmeters and ammeters, 85
Amperes required by tramcar, 109
Arc lamps, 33
Area of armature iron, 75
air gaps, 75
magnet limb, 75
Armature, reactions in, 86
wire, pull on, 48
total pull on, 49
power to drive, 50
winding of, 55
losses in, 55
temperature rise of, 56
hysteresis loss, calculation of, 57
eddy-current loss, calculation of, 57
cross magnetizing wires on, 88
demagnetizing wires on, 88
Index. 237
0, the magnetic density, 65, 68
B.E.M.F. of secondary cell, 28
self-induction, 120
, magnitude of, 127
Capacity in alternating-current circuit, 160
in parallel with circuit, 170
in series with circuit, 169
, current due to, 165
, E.M.F. *due to, 167
of concentric cables, 171
size of I farad condenser, 169
, specific inductive, values of, 168
Circuit, simple, I
, series, 3
, parallel, 4
, multiple wire, 15
magnetic, compared with electric, 72
Coil winding, series class, 82
, shunt class, 83
Condenser, use of, in alternating-current circuit, 169
, size of i farad, 169
Conductivity (m), 2
Conductance, 2
Cross magnetizing wires on armature, 88
Demagnetizing wires on armature, 88
Distribution, 15
Distributor, definition of, 17
Drop of volts in distributor, 16
in feeder, 16
in armature, series, 36
, shunt, 43
, compound, long shunt, 44
short shunt, 45
in transformer primary, 149
secondary, 149
due to B.E.M.F. of secondary, 150-
due to magnetic leakage, 150
Dynamo, formula for E.M.F. of, 35
-, P.D. at terminals, 36
, separately excited, 40
, series, 41
, shunt, 42
, compound, long shunt, 43
, short shunt, 44
, calculation of amount of series winding, 86-92-.
Economical size of feeders, 19
Eddy-current loss in armature, 57
in transformer, 150
, current to balance, 136
Efficiency of distribution and transmission, 95
magneto machine, 97
Index.
Efficiency of series dynamo, 97
shunt dynamo, 97
compound dynamo, short shunt, 98.
, long shunt, 99
series motor, 99
shunt motor, 100
compound motor, short shunt, 100
, long shunt, 101
secondary battery, 101
combined plant, 101
transformer, all-day, 156
Electro-motive force, E.M.F. defined, i
E.M.F. active (e), 27
back (c), 27
total (E), 27
Feeder, definition of, 17
-, economical size of, 19
, working current in, 19
Field-magnet winding, series, 39-82
, shunt, 83
, compound, 90
Five-wire system, 23
Gap, air or iron to iron space, area of, 75
Gas-engine, cost of gas consumed by, 181
H, strength of magnetic field, or magnetizing force, 67
Heating of armature, 56
voltmeter, 102
Hysteresis loss in transformer, 151
in armature, 57
, effect of on shape of current curve in alternating-current
circuit, 137
in cyclic change of magnetization, 138
Impedance coils, 145
, why open iron circuit type, 140
Insulation, electric resistance of, 10
, magnetic, 73
Impressed E.M.F. alternating, how found, 122
Kelvin's law of economy, 19
Kennelley's safety rule, 21
Leakage of magnetic flux, 73
coefficient (j/), 74
drop in transformer, 150
Length, effective, of solenoid, 132
of magnetic path through armature, 75
air gaps, 75
magnet limbs, 75
yoke, 75
Index. 239
Magnet coil, series, size of wire for, 82
, shunt, size of wire for, 83
Magnetic properties of iron, 69
circuit compared with electric, 72
of dynamo, analysis of, 75
of finite solenoid, 131
Magnetizing force (H) physical measure, 68
(ffil) ampere-turns, 70
Motors, 31
, armature reactions in, 92
, compound winding of, 93
, B.E.M.F. of, 51
Ohm's law for continuous currents, I
Ohmmeter, connections of, 14
Open iron circuit transformer, 143
Periodicity or frequency, 1 14
Permeability (/*), 66
Potential difference (P.D.) definition of, I
Power to drive armature, 50
Power or activity, (P), 2
Power factor (<) in alternating-current circuit, 126
Powermeter, connections of, 14
Powermeter for alternating current circuit, 1 70
Prime mover costs, 181
Pull on armature wire, 48
Quantity- efficiency of secondary battery, 101
Railway, electric, 106
tractive force required, 106
power required, 107
pull to get up speed, 109
E.M.F. required, 107
current, 109
shunt motor on, 1 10
compound motor on, 1 1 1
series motor on, 112
Rate of doing work, or activity, I
Regulation of speed of motor, 1 74
Reluctance magnetic (2&), 72
of air space surrounding coil, 131
Resistance to start a motor, 174
in series with motor, 1 76
in shunt coil of motor, 177
calculation of series circuit, 2
parallel circuit, 5
of insulation, 10
measurement of by ammeter and voltmeter, 12
Resistivity (p) definition of, 2
Safety rules, 21
Secondary battery, E.M.F. curves, 28
240 Index.
Solenoid, infinite, magnetic properties of, 67
, finite, effective length of, 131
Specific inductive capacity, values of, 168
Steam engine, cost of fuel in, 181
Temperature rise, Esson's rules, 55
Transformer, balancing of secondary output in closed iron circuit type, 14 1
in open iron type, 143
, relationship between power factor and output, 143
, design of, 148
, eddy-current loss in, 150
, hysteresis loss in, 151
, all-day efficiency of, 156
Voltmeter, size of wire for coil of, 85
, resistance of, 102
, extra coil for, 104
, effects of temperature on, 102
, heating error in, 102
, temperature error in, 102
, constant of, at any temperature, 105
Work, rate of doing, I
Watt, defined, 2
THE END.
PRINTED BY WILLIAM CLOWES AND SONS, LIMITED, LONDON AND BECCLES.
UNIVERSITY OF CALIFORNIA LIBRARY
BERKELEY
Return to desk from which borrowed.
This book is DUE on the last date stamped below.
ENGINEERING LIBRARY
OCT 8 1948
JUL 5 1950 f<
LD 21-100m-9,'47(A5702sl6)476
VB 24105