 

1‘

.9
I
' I
.
Q
.. W
... \.
. 1‘.
...].
Y
a _
.it
. w
.,.‘.\ u
fan

I u‘

.. ;L
T

‘r | I ‘.

”Walt.“

.l .

W .v L.

.0
v. i, I
!
..¢ 1
... x ..

. ... .....u. . ...
J‘Aﬂuvia? ¢./oc{w.u¥

cg. . . hv..!.u5..va .D’VV rlhA..,

 

13.43“

. $‘f

. h . l‘ V
- .- r ' : A
. ' "' M“? II
‘ a \v ‘I N

'L‘.

, _ 'l ‘1' ‘ ‘
in, . '
1 o ~‘2 ‘ )Liu‘:\‘:'

u V {‘1' tr.
.0; ' ‘J
J l

r\,'
‘2?”

. . 3.!-
J

"-3? (“X f.

‘t‘ fv'iw :’é_ ‘ a‘

t“: .:-| “"1 '('§.II~

“1333313 "#113?"

£514- iﬂm 3115‘ A “...?

U v

.4 ‘révﬁﬁft‘ﬂ;
-' ,- .g‘v’yﬁt'M-Et‘» ' .

. '- 7.;
.'.- 3.)"
. ' _-n"
‘96:”, “3%.
. ,‘n' .
. 5‘ .
-. - .I -_
I- s
— "- \' l‘ 9‘
...;.I

( -_ m

|

,y‘ ‘ .\ "
~ . ’fv.
‘ " .v .33.. '

*0

'v '1” -": I n - f .. ‘uu '. '- - - 'a - ,. . ...! -‘ ‘fk ' I
_, 3L5: 14a, - ,' .. _ ._ ' . .‘ _ . . . { . . . . -. . . . . ..
r‘ cl‘e'J"£L.‘I"-$}r"¢;$f “‘ - . ‘ . ’ . . ' . ' ‘ .. " _ .. W»: . ' - " a '3' N

4 - '3?“ "1.”? ' '. . -. '-: .~ '4 " , ~ .2 ' .

1

< . K
.-r‘ .)‘- E1 - -.:.

. . 1.. . _ . ' : a .' t n
I . ‘ . ‘ '1: “' 3 . I ' ":5. . l ' r I ‘1‘“ ‘ H.” E '1 ; ' | -' I. ' T. : a}? ‘* - ELX' ’3‘ S ‘- ' ‘ V
o' ' ‘ 3‘ . l " ‘ I c l.' ‘ ,.‘ I i ' ‘f' K‘ ". 3' ' 4.". .5", ‘ " I 1 A“: "H.“ ‘ . - 3’ ‘4‘ ’
i a: d - ..n 0‘ P. 3 _ .. . .1" . ‘1 ‘. . .' :'_‘ ‘ ' ‘ ‘ is» ? .% ‘4’1?£"-', . .' l 1 ‘
N .‘ ‘;‘. ~. - , ' J, . . .‘. .. -_ I'. ." _ , y . I‘- ‘1’ V»

\N’
g. ,
r‘V.
7047‘ '~* .4 "3-,
' x ' ,3 ‘ .

2*...
{#54 ' 3‘ .r .-..

‘L~.IJAII.':‘. '. 1- “If?" ~ ‘ .
‘K "“‘.\*~ .‘A‘: .‘
:: ‘ " .l
v ‘t ‘1. }7§;:’f}1:.,.' .

‘5‘ .r.!";f$‘~" ' “a .a“ “

l3 1 5". .h ' . "‘ ‘ n

‘ I I' 1 _ u‘
..z. 3.1,.“ . .;

,| M; ' 4 9

1 _ V 570-3 _ '. T

. ﬁ.‘1#: :6“? 43.. .53.?
. ‘ '

.U r" , 9‘ , -
f". ‘.‘ “ $IIJ “'1‘: . ..f
, "."r ‘ - 1' .1 .

x .. ' ' .1..- Q

_~
_'.“ v4}
. ,.

.,c
I.

. 5“ ¢_.
.. ‘_.'.
.'I;"“.| -

~71:
I.\

r‘ .‘A’ { 5
, -'t-;'~«.._ .01" ? -. .
".- ' .Q ‘ .

-... ' ’7 4‘ ‘

. . - ' ‘- é '. I"! T.’ “- 01'
. r'-“‘.'-,.._'.',- ~. .3.
" t 7 14.33"“. 34"”? 1' .f -,' " ""545’19. V
‘. l ‘ V ‘ . 1‘ ‘ ' ,‘ .. '\ . ’1 ~‘ . .
{n \ ‘ 3 k‘ﬁpx. ¢ " z' - 3‘ “Ln-E. 3» ~
‘4 It ‘

~ it.» - a
. raw ‘-‘-9".‘v'6.". 7"
H ’- 'c ”51.5%.".«r »

.v
1‘ *

l‘ l ..
’ .‘k w"

.. :o' .
. .
t"

I ‘ .P‘
‘ ; $31.5
$ “it :11731r’
I. 1

Th ' . .5 ,

‘ I [_—v‘.£“?‘ ."‘ '_ '1 . .. 1‘" '7‘. ..‘L if
,‘nu‘f r " 'at‘.“ 5‘2“. [#31)! [ﬁxj‘nu'J‘L'
_.?r..fr‘-‘.T§-)‘ . ..9ieerw..ggf,>.é,..;' r

s
-_ . .’~‘ ”("4 -
f ”1:" x. 5' h } ff“ . ' ‘
J‘s :- . I .)._ r. _. ‘ A .

,. 15:» "i“.
~ an”. ' '
"tit-ﬁght; f3.
" $4376.?
. \- ‘

bl ..
.

.
l

- O
,

‘3:

\. V ,
. ‘.. '. h .' 1
“1‘ H 6.6! ‘VI; 7‘ l. y
‘ ' 21‘ '. ., .‘K‘ 3 ‘ _ \
.f - ....I ‘ _' . - J ‘ -
3".~ 35 ' ‘.
‘\ g i [‘3‘ . l
i L,'., ‘
'. \

«1
&ﬂ%§.
"“E‘ J“ Thy-i“ -'- “
.1, . .4 w?“ .... . .1
all; ". ‘.' I " ' .
'V'I 7’ .
.‘ . .1 . at? "73):; .
If: ’u-I’l-I‘yf v.24 ?’i.‘~“¢£ﬂv '
.Q , ".-. ‘ '. :- »_ . ~ .-. .... . .-r :- “... ‘1'.“
' ; .. .» r~ .. ...-:4.» ' y a «xx-5%? .. 3?.
u 1. ...-:~.'_“ .‘ ".- ... ..‘." y . ,‘ 1 4. I _ A” _ ‘n Y r‘ 'ﬁ{’p _. ‘ I

(X. , . '# - .
. J ', _ s . . .‘ \' .. ‘ " ‘ ‘
"MAM . . ' -' - ' -’- ..s. . 1-” .4 'x.
... ‘H. s_ ‘ . |' l . ,9, u . J 4‘.
1‘9‘ _‘ ‘ i . .' Y . ... “ .‘ ' .’ ‘ R b '1 ‘\ ,‘ ._ .
.j‘.» x . .. ‘l - ' . "p ' ‘ . "".- g ‘ . ‘ ' ' ' . ‘ _ .‘V‘ “ .
, ,. _ . ,. , , .

.7111?
‘3; 1xa~fﬁu§i;7*1
':.‘.~: . " ..N'L' 11.4“. x“. "
3‘31"" «.u’u'is'xr3‘1511“.“?.3575 . M n
l .. “- ,. #« ﬁ'r' A \‘l 43‘ _ 1., ‘\
7; V" i" I " l. mﬁ‘d‘!“"iﬂﬁw‘ '.\
I 5“. ..u,:{., . Q§3'|' ,;1¥Y : .v‘rl’dl'»; 1‘. . b
“‘. - (‘ It .).-)3‘ I' ’1‘ +~“f‘v* n
. ﬂ war“)? ,3. . .. --— ... ~
1 1“. ~‘. ,
P v;

u
n
.

. ‘L: "~."{."'-'. *1
lﬁéﬁtrﬁw -

u ' . 3 RNMSk tir'i'. ‘” ": fl I. )

- .,. rte “Jim 3&1“ diff.- ' '. '; f,

w 1.. ‘i’; ' . , ‘ - ‘ ‘ °' :1 "@Mmr-t"
(8‘12). an ' '3 . ‘f I“... - ...“ $L #31"

. 0S1 .~, . ‘ '
.. ' ' . 1 ' .
‘o _ v ‘9. ‘1. .. u
o . t7 .' 1‘ : ' 'K ‘

g .
.
‘ .

:V

V I ".‘§‘ﬁ._

. . r

' "Kw“

n . -' -' 7"." " - iz. '- . '
"-used4"..«zbewg’y' ' ' ' '

x}, it '4' ‘
gm. 3.;
.x.
. ‘1

¢ »— ....-
as.“ aw“:
4 4
AW:

.
4

v
i

* ‘ ""r'f -

 

A STUEY O? TF3 COﬂCATEBlTIOH O? IRTUCTICH ﬁOTORS.
BY

CYARLEB 511433; .

written as a 3.3. THESIS
At
MICHIGAN STATE COLLEGE

1930

THESIS

To Professor Foltz,and members of the Electrical
Engineering department faculty.to whom I am.1ndebted
for their oo-operation 1n the obttining of the data,
this thesis Is dedicated.

Contents.
Part 1 _
Introduction and general theory of the Induction Motor.

Chapter. Page
The Electrical and Eagnetio Circuits. ------- 1

11 The Revolving Field. ----------------..--- 6
111 Slip.Current and Voltage Relatione,and

Torque --------------~-------------------- 7

1? Lesson ------------------------------ 11

Part 11

Hethode of Finding data for the construction of
operating characteristic curves for induction motors.

Chapter Page
1 The Constant Magnetizing Current Circle

Diagram,Theory and application to the Set
under Consideration. ----- ------ --------- - 12
11 The brake Test. -------------~------------- 24
111 Separation of Core Loss from Friction and
windage. ------~-~--- ------------- -------- 27

Part 111

A Method of Speed Control.
Page
Chap.
1 The Concatenation of M-8 and M—9 30

A STUDY OF THE COHCATENATIOH 0F IHIUCTION MOTORS.

PART 1

----..-------

Chapter I
Introduction and General Theory of the Induction Rotor.

In the commercializing of alternating current as a means of
distributing electrical cnergy,probably no one factor has been
of greater importance than the invention and deve10pment of the
multi-phasc induction motor.lt was invented by hr.ﬁikola Tesla.
in 1888,a1though the first successful motor was designed most
probably by hr.C.E.L.Broun at the Cerlikon works in Switzerland.
Its simplicity of construction,ruggedness.and freedom from
trouble has made it very pepuler,especially for service WhiOhlﬁwacs

only one or two speeds.

The Electrical and :agnetic Circuits.

The induction motor is essentially a transformer in which the
secondary is free to revolve.Therc is also an air gap in the mag-
netic circuit of the induction motor.Due to the air gap.a much
greater magnetizing current is required in the induction motor
than in the transformer.This may be seen from the following

equations:

there, d is the desired flux
w =tho magnetizing force 8 .4 II n I

R =the reluctance - --l---—
u a

Then for the same flux in the magnetic circuit of the motor
as in the core of a similiar transformer,the following differ-
ences will exist.The reluctance of the magnetic path of the
motor will be equal to the sum of the reluctancee of the air
gap and the iron path in series.and,as the permeability of
the air gap is one and that of the iron may be several hund-
red.the reluctance of the air gap may easily enough be as
great or greater than that of the core of either the motor
or the transformer.1f the flux is to remain the same.then
an increase in the denominator must be accompanied by an in-
crease in the numerator F.As all the quantities comprising F
are fixed except I.then I must increase.The magnetizing com-
‘ponent of the current is 90°out of phase with the power com-
ponent and so the power factor is very low.Thue the air gap
in the induction motor is made as small as practica¥fThe
:rclatively large magnetizing component at no load and light
‘load causes a low power factor ,which gives rise to one of
the greatest objections to induction motors running under
'theee conditions.The vector sum of the power component and
‘the magnetizing component is often called the exciting current
and the magnetizing component simply thn magnetizing current.
ffhe ratio of the magnetizing current to the diameter of the
circle of current loci for any given motor.is called the
Ileakage factor,and is denoted by a. Thus if the magnetizing

current of a certain motor is 5 amps.,and the diameter of the

current loci circle‘to the same scale)is 100 amps..then the
leakage factor will be .05. This is an important ratio because
of its relation to many of the physical démgnsions of the
notor.It has been proven by experiment that the leakage factor
varies directly as the air gap and inversely as the pole
pitch.In experimenting with the air gap it was found that the
diameter of the current loci circle remained practically the
same regardless of the air gap,while the magnetizing current
increased in the same ratio as the gap.The leakage factor also
gives a means 2? estimating the maximum power factor which may

be expected from a given motor.This is shown by figure 1
e

d

*9

 

FIGI

 

i

“rom trigonometry.the cos9,or the maximum power factor will

1 *' ‘a

fine maximum power factor occuring where the primary current
()‘P is a tangent to the circlc.In the design of a motor these
«different relations are an aid in determining the correct and

Inost economical proportions.

The electrical circuit of the induction motor consists of
two windings.(l) The stator or primary winding and!2) the
rotor or secondary winding.

The stator winding must fulfill three requirements.(l)its
conductors must be of sufficient crossection to carry both
the load current and the exciting cureent.(2)it must have a
sufficient number of turns to set up the required magnetic
field.(3)the combination of the number of turns and the mag-
netic field must be such that a counter electro-motive force
will be generated which is always slightly less than the
impressed voltage.If all these conditions are satisfied.then
the motor stator winding should perform its function properly.

The rotor carries the load current.and as it has no elect-
rical connection to either the stator or the line.it must
receive this load current by transformer action from the
stator.There are two common types of rotor windings (l) the
squirrel cage and(2)the wound rotor.

In the squirrel cage type the conductors are placed in slots
in the outer edge of the rotor.the ends being welded to a
short circuiting ring.Tuus the phases cannot be cpened and
the same rotor may be used in machines having a different
number of poles in the stator.In fact the stator will induce
tan.opposite pole on the rotor for each one of its own.

The wound rotor has a winding placed on it which has the
same number of poles as the stator in which it is going to
'be placed.1t has many advantages as will be discussed later

‘under torque.rue to the extra amounthf labor required in

‘winding,its first cost is much greater than that of the

squirrel cage type.The later is more rushed,mahins a better

motor for all around use.
chapter III

The Revolving ?ield.

..----“--

The Operation of the induction motor depends entirely open
the revolving magnetic field.”rom a sine wave representation
of e three phase alternating current,it can readily be seen
that the resultant magnetic field at any instant will be zero.
Bce.if one of the waves be reversee(or in other words if one
of the coils in a three phase winding be reversed) the result-
ant will be seen to be one save of magnetizing current.$his

is shown by fig.£.

 

 

I r ’ D
f- ’ﬁL}: .:_
.',-' _.'"
.'«' .‘ k
j.
7313. P

Then if these three coils are shown in the stator in the

different time positions as follows;

' u 8" a [cor ' E . '
mm Leg ’l —— B:l;ﬂ:.5,‘C‘-.5

.. ﬂ..866;3:.866; c = o

 

 

 

 

 

 

 

 

 

 

 

 

 

I71]; 32.5; €175.

 

 

 

73m s‘r/a/v o F FI£L D.

 

 

The coils A,B.and C in the stator slots of figure thiee are
carrying current as represented by the waves A.B,and C of fig.
2 respectively.1t will be seen that the amplitude of each of
the waves is transferred to the chart at the designated points.
The crossectioned area represents the flux in both quantity
and sign.The arrows on the crest of the waves indicate that
the flux wave is moving to the left.In this way it produces
the revolving field.From a study of fig.2,it is apparent thet
each pair of poles will represent 360 electrical degrees.or
one complete revolution or cycle.Then if the frequency in
cycles per. second be multiplied by 60 to change it to cycles
per. minuite and this be divided by the number of pairs of
poles.the speed of the motor will be obtained.Thus;

f x-peere-efbpeles 60
(3) Speed in r.p.m. - ................. -----
pairs of poles

r.p.m. 1 pairs of poles

(4) Transposing, f = ------- ................. ---
60
2 x f x 60
( 5) or . 1:0. p“68: -------- --------- -
r.p.m.

In effect the flux wave travels as many electrical degrees
in a ten pole motor as in a two pole.but it travels a smaller
portion of’a revolution in passing 360 electrical degrees.
thus the speed depends cpon the number of poles a given point

9
would pass in going completly around the stator.

Chapter III.
Slip.Current and Voltage helations,and Torque.

When the rotor is at-standstill:the frequency of the current
in its coils is the same as that of the stator.The rotor now
bears the same relation to the stator in this respect.as the
secondary coil of a transformer does to the primary. The speed
at which the stator flux rotates is called synchronous Speed.
As the rotor starts to turn.the difference between its speed
and that of the rotating field becomes less.thus the frequency
in the rotor becomes less.until at synchronous speed it would
be zero. This relation will be a straight line function.

The slip.which is the ratidbf the difference between the rotor
and synchronous speeds,and the synchronous speed thus;
Syn. speed - Rotor speed
(6) Slip 3 ~~----~--------~----~---~--
Syn. speed
will vary in like manner. Then the frequency at any rotor speed
will be.

(7) f2: sfl .where fl is the stator frequency and s is the
slip expressed as a decimal. The resistance of the rotor re-
mains practically constant.

The reactance will be ZIIfL,where L is the inductance in
Henries.L remains oonstant.and so the reactance will vary as
the frequency. From this we obtain;

(8) x2'811.where x113 the reactance at standstill.

From the following equation for any induced voltage;

(9) a. = 4.44 n f 51:21:11; 10-8

it becomes apparent thet.assuming constant flux.the induced
rotor voltage will vary directly as the frequency.ussuming
e ratio of transformation from the primary to secondary of
cne.the voltage induced in the rotor at any instant may be

expressed by the following equation.
(10) E2 = Els .where E1 is the stator voltage.

The current in the rotor,12,will then be.
(11) 13 a ---§§----
22
e E2
°r ' I}? 335295“

From this equation it can be seen that with a small increase
in slip near synchronous speed.the current will be greatly
increased.and so within the rated capacity of the motor the
slip will be comparetively small.Take for example the following
small 110 volt motor;

Voltage induced/phase =70
Resistance of rotor/$18 1 ohm

Reactance of " " l "

a slip = 1 and 4

701.01
For 1 p slip. 13 8 ~----- ------- = .7 amp.
(‘1 § eOOOI)B '
_ 70 x e04
For 4 7:3 Blip 12 = ------- ----- a = 2.8 amps.
( 1 a .0016)

The later would be about full load for the motor.

r
The power factor in the rotor will be --§-.or in full
Z:
t tien cos9 ' r2 (12)
no a 2 ...-4......-
' (r2 + 3131*

The torque in any motor is proportional to 511% .or to
ﬂlzg. Applying this to alternating currents:

r on ﬁre ooaez (13)

From equation (12).it is seen that the maximum reactance
is reached when 8' 1.or in other words at standstill.Therefore
at this time the power factor will be the smallest obtainable
at any possible condition. According to equation (13).the
torque will be decreased materially.“ many conditions arise
which require 0. large starting torque.some means must be used
to supply this demand.

From equation (12),it is evident that if r2 be increased,
the power factor at starting will be increased. In the squirrel
cage type rotor.the rotor bars are rigidly fastened to the
short oircuiting rings,and so the phases cannot be opened
and resistance inserted.However in the wound rotor type the
phases are left apes and either brought out through slip
rings to external rheostats.or resistances are placed in the
rotor controlled by an automatic device which shorts the
resistances out as soon as the rotor comes up to a set speed.

The ratio between r and X which will give max. torque at

variable slip may be found by differentiating the general

expression for torque .

| 32 r2 ¢1
Toe ----~ ------ ~ ,and finding, the maximum by setting
r3+ s‘Xé

the result equal to zero.and solving for r in terms of X.

10
Differentiating with respect to s and setting the result
equal to sero.(replacing ¢iand Ez.which are constant .by K)
di'.‘ _[r§+ s 2X§)Kr2- Krzstzsxé)
'32” : ""7115; £5333 """""" " "
Discarding the denominator and collecting;

r: + sax; = 2 32X?
2 -
and extracting the squire root,
1'2 3 0X2 (14)
Then the power factor for maximum torque will be a constant,

thus;
substituting eq. (14) in (1?)

r
008 Qfor me torqua = -?:g--§2 2:;- o..-
+~r2

‘; 2 e707 1".
2% t 0)

it standstill the slip is unity. Substituting the full

notation for torque as used in the general equation thus;

Ezr2¢i ‘
Tex ----2~- (Sub. the notation for IPand c0892?
r ,

Now if the equation s: l for standstill and r2 therefore
I xz,be substituted in this equation,
VlEz

T ca ------ .which is the maximum torque at starting.
222

11
Chapter 1v

Losses in Induction motors.

The following in an induction motor must be found in order

to determine the efficiency;

(1) Copper loss in the primary. - lirl

(2) ” ' " “ ” secondary - 1% r2

(3) Iron loss in the magnetic circuit.due to hysteresis
and eddy currents.

(4) Windage and friction.

Whether the motor windings are connected delta or wye.the
copper loss will be 7 3/2 123 (l7).where R is the resistance
between the terminals of the three phase winding,and I is the
line current. Equation (17) applies to both the stator and
rotor.but of course in the squirrel cage winding the phases
cannot be cpened and so the equivalent resistance is used.

If the rotor of a motor is Open circuited and the line
voltage applied to the stator.the wattmeters will read the

sum.of the iron loss and the Izr loss in the stator.Then,
Iron loss = Total loss - IPr (18)

Then if the rotor is shorted and the motor started,the
difference between the total loss and that in eq.(18) will
give the windage and friction loss.(except for a small I?r
‘loss in the secondary.which in this case is negligible)

 

) Infallpll‘Lw Ii I I
ll.. 1 i

2&3? 11

12

Methods of Finding Data for the Construction of Operating

Characteristic Curves For the Induction Motor.

Chapter 1

The Constant Magnetizing Current Circle Eiagram.

In this discussion the data taken from the two motors H98.

and l99.which constitute the concatenation set under con-

sideration;will be used.The data taken from the name plates

of the machines is as follows;

H-B
Hodel.lo. 69 A 415
Type ' HT948 C-lO-lBOO
For: CL
Volts 220
awe VOItS 170
Alps. 25e9
Sec.Amps. 28.
Rae ' U“
Speed Fe Le 1725

Both motors 3¢.60 cycle -

MP9

73 a 586

3T596 6-10-1200
CL

220

165

27.8

30.

m .

1145

Built by General Electric

The first test run on the motors was the open circuit test.

This gives the windage;friction and core loss.The data taken

was as follows.

M-B
I 12 I W "32
Meter 1 3 1
.Ho.6188 8889 22012 3443 1118
Scale 0 ~10 _ O - 1.5
I: 1 2
t.4 7.15 8.7 2277 .515
A1.
Used 8.42

31-? E2-3 E1.3
15759 21873 21555
0 . 300
1
220 215 219
218.3

s-s

Sane meters and constants.

11 12 13 W1 32
’06 8.25 8.8 “057‘ e595
AVe 8.88

1"

51-3
220

13

E2-3 El-s
220 220
220

The next test taken was the short circuit test.This is taken

at reduced voltage.It gives the data for the computation of

short circuit current and the capper )osses.Reduced voltage

practically eliminates the iron losses by reducing’¢1.is the

motor does not run there is no friction and windage loss.

M-B

‘I

I I I d W
H.‘.r 1 2 6 1 2
55.10029 22010 21518 207 208
I’ 5 5 1 25 26

3.8 3.8 18e2 .26 -.09
Aye 18e9
. . “#9
(same meters)
4.00 See 19.5 “e115 e32
‘Ve 19e5

E1-5

3228

1

46.2

51.75

E2-3 31-2
2551 18023
1 1
45.2 45.2

45.2
50- 51.4
51.05

The data taken will now be worked-up into shape for dimension

~ing the circle diagrams.
M-B

After averaging the values in the short circuit data,the

following figures are found.

14

e26 - e09
II- 18.9 Amps. w ' ---------- x 8.5 II425 watts “‘ 46.2
.001

The ratio between line voltage and the voltage applied is.

220
---- 34.775 .and as the short circuit current at full line

vggtige will vary as this ratio,then ISS will equal.
4.775 x 18.9 = 90.25 amps.
The power varies as 12,therefore the power at short circuit
will be. 425 x 4.7753- 9575 watts.Th8 power factor is;
425
008959 ' --:-------------- 3 .2815
Six 46.2 x 18.9
The resistance between terminals-.28 ohms.
Then the 123 1088 in the primary end-eeeendary at short circuit
is. 3/2”:§90.252- 3414 watts.?rom this the 12381088 is found to
be 9675 - 5414 = 6261 hatts.
The following values are obtained from the open circuit data,
I3 8.42 amps. W'(515 - 277) x 2 8 476 watts EI218.3 v.
Then cosOO¢8 --1-33§--.----- . .1415 or p.:.-14.95
39: 8.4? x 218.3
Note: All wattmeter readings were checked for sign in case of
open.ctrcuit by adding a mechanical load,and in case of short
circuit by removing one wattmeter.
8-9

0n short circuit the following average values were used.

I‘ 19.5 6' (320 - 115)2.5-501 watts E-5§.o5~7.
220

Then the voltage ratio!I ------ 84.52.8nd the full voltage
51.05

staph-19.6 x 4.32-84.25 Amps.
The short circuit watts = 501 x 4,392: 954.3 wattg.

15
The power factor.cos€88 =--3§91------- = .291
321 51 I 19e5
The resistance between terminals is O .418 ohms(primary)

Then the copper loss in the primary is.
3/2 x .418 x 84.252 = 4454 satts,and the secondary copper loss

is 9340 - 4454 = 4886 watts.

On open circuit the values were as follows,

1' 8.88 amps. w- (595 - 570)2=450 watts 3:220 volts.
'9 450
° 3%: 220 x 8.88

In giving the proceedure for the construction of the circle
diagrams.the data taken on.n-9 will be used.that of the other
being constructed in the same manner.

The open circuit current of 8.88 amps. is first drawn in.
Ibis vector makes an angle with E.whose cos is .153.From a
table this angle is 82.3 degrees.The easier way is to resolve
the vector into its horizontal and vertical components and in
that way locate the point.It is assumed in this diagram that
the magnetizing current is constant,and.as the friction and
'windage in any given machine depend only on the speed.these
'will also be practically constant.These losses may then be
represented by a straight line drawn from the end of the vector
‘perpendicular to the E vector.The short circuit current is
then drawn 111.88 in the diagram.at an angle whose cos is .291
land in the same manner as the open circuit vector.The locus of
7a11.currents will then be a semi-circle whose diameter lies
on.the no load loss line ov.the points 0 and a being on the

circumference.This determines the circle.The primary current

16

will be a minimum when equal to mo,and all other points of its
extremity will lie on the circumference of the circle.the
corresponding secondary current vector being drawn from o to
the same pointe.The secondary copper loss is assumed to be
serc at c.and so a line drawn from c to Q will be the limit of
the secondary cepper losses.The line as is the in phase compon-
ent of the 84.24 amps.shcrt circuit current.howerer as the
product of this component and voltage gives the p0wer.and this
voltage is assumed constant.then if the line at be ldyed off in
watte.the total length being the total short circuit loss,it
will be a true measure in all current ccmponents.Thus if the
9340 watts be divided equally among the number of spaces on
the grep! paper.it will form a true scale of power in watts.
The Circle diagram of which the following is a cepy.was
drawn np'to a much larger scale and the recorded values taken
from that.giving much greater accuracy.The point C is the
division between the primary and secondary losses.and a line
00 will divide the primary and secondary losses at any load.
If a quadrant be drawn from the vector E to the x axis.and
the distance out to its intersection with E be diiided into
100 parts.then from trigonometty.a perpendicular dropped from
the intersection of the circumference of the quadrant and any
primary current vector.will indicate the power factor directly

on the scale.

17

The loss line cs is new extended to the line HT.intersecting
at the point 3.1 perpendicular is then erected at the point
B.end from a point 81 on this line .a line drawn,paralle1 to
HT,and intersecting the lineoa at H.Then if this line be divided
into 100 parts.the point of intersection of a line drawn from
point H.through any current value on the circle,and this scale.
will designate the efficiency.The scale reads 0 at E.and 100 at
BtThis method of finding the efficiency is derived from eimiliar
triangles thus;

FL:EH::881:RN1
and HL:PL: :nlnmnl

Hultiplying these two equations ;
FLzPL: :Hllmll

or Total loss: inputz: 215:100

Thus Elm represents per cent loss,and consequently ﬁn is
per cent efficiency.

If’a perpendicular be erected o.tangent to the circle.and
from some point on this line.another line be drawn parallel to
eG.thie line may be divided into 100 equal parts and used as
a slip scale.The point of intersection of the slip line and
the secondary current will give the 3% slip.This proof is based
on eimiliar triangles as in the proof in case of efficiency.

The large circle diagram was drawn th the scale of 5 amps .
to the inch in the case of current .the power coming out to

be 8000 watts to the inch.Thus in the case of full load,as is

18

illustrated in the accompanying diagram.the output PF must be
equal to 7460 watts.which on the large diagram would be 5.73
inch spaces.(is stated before all illtstrations refer to n-s)

The line H? is the primary current for full load and is equal

to 26.75 ampe.At the same point the secondary current is 22.00
amps..the slip 5.5%,the eff.85.7%,and the power factor 83%.

A series of points were then chosen along the circumference.

and the data obtained used in the plotting of performance curves.
The following are the data sheets;

2+9
1, I. Con Slip ﬁmyn. sort. H.P Dutput

10.25 5.8 .49 .957 99.055 74.7 2.01
14.00 8.75 .715 2.00 98.00 84.4 4.56
15.75 14.00 .80 5.19 96.81 85.57 6.9
25.75 22.00 .85 5.5 94.7 85.7 10.00
50.00 25.00 .829 67.00 95.00 84.8 11.65
56.25 51.25 .815 7.55 92.55 82.7 15.12
44.25 58.5 .78 10.47 89.55 79.00 14.75
55.00 45.4 .75 10.95 89.07 74.00 15.5
54.5 57.00 .55 21.55 78.44 54.2 15.9
59.5 51.5 .568 25.56 75.44 57.8 12.00

19
9.8
15.5
18.25
24.25
29.00
55.5
42.5
51.00
58.0
67.0
80.4

In

3.8 -

8.75
13.75
19.75
24.75
30.75
37.2
45.2
52.0
60.0
72.5

0080'
.52
.74
.82
.855
.85
.64
.82
.78
.74
.65
.50

558
clip
1.125
1.875
3.1
4.97
6.3
8.2
10.55
15.15
16.7
21.5
38.00

I

5833

98.875
98.125

96.9
95.03
93.7
91.8
89.65
86.85
83.3
78.5
62.00

5.7:.

71.41
83.5
86.00
86.5
85.6
83.75
81.25
77.5
73.00
66.5
51.7

H.P.Output

2.09
4.79
7.46

10.00

12.12

14.25

16.17

17.7

18.00

17.15

11.45

19

20

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L. J 2 as
.k 6.3. .. ...qu 29204.: s.QZ~W.J>€.§
, J «nonzﬁ
3.6:...ng \ 7.3.
1 r .
_ Q h
3 2. it“ A
.3 encounkH ... w “No a 15‘
.5 6. 5.5.4512 - - .. I - -..- 2.25:8
.4.
\ a 53
2 ... .
o
1‘
m

EQEQSQ 35.50

 

 

21

 

 

 

 

 

 

 

   

 

 

 

 

 

 

H
... 4 1 2
ﬁréetiaiemmgsgﬁ o r
1 0‘ WNﬁuoﬁd . ..1
3 :3. 4%L. I -..m
A
.16 rQM!
1 51.55 ..m ..m o
Wwabﬁ ..
vans.
“NW“... 4N MN llIlIlill
.2 A 5:65 A d
T 8.\
S .0 0w. .2
$08
:
m.

®-\>\ tttwmQQ wqomtu

 

 

mm

 

 

22

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

kblkbo mu{ m m.

o . \ 3 f .

D an \
Q\ \\A\\\“‘ Q\

\ \ H

_ ‘ QN
0N K\ \

- \ \\ QM
QM. _ H\\

1 \\ 2.
Q9. \

. 9h

... / \

_ f \.\.\v \ \_ QW
Q9 _ \WV 7 \

J \VA/ R
h \
o W \. \ \

XJVJW , \ \IW \ Q‘
a u NNHHMLH 7 jax 1.

_ 37L!!! mum in

. ]/ uv

.
Q6 . f/ .

_ a . 1h 52

H 52 _

 

 

 

 

 

 

 

 

 

0...: Ww> ESQ «QEQXTQKQNQ

mm
.2» 0
SS»

 

 

35

 

 

 

 

 

 

 

 

 

 

 

 

 

6 Q .3 3 m: m m. o
l\ ﬂ¥\
\ \
\\
Q N O N.
\\
on \\\\\ \ on
Q? \\ Qu.
a... \ \\: , H eh
Qm \.\\,\a \. A 8 0%
QR \ It: \ 4.. \\ 9h
- // \ :\.\\ _ x ..
QQ ,rfiufful I! \\.q\.\w\\ \\ Om
jjJ 4” Lft Lrlﬂlumkii\\.\
0 Q l/ [ cm.
[ .2“... [
03 3Q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ﬁst 9&1th NUEQXKQkImo.

imw
56 mm

 

 

24
Chapter II.

The Brake Test.

Only enough points were taken in the brake run to serve as
a check: on the circle diagram.and to offer a comparison bet'een
the two methods. The following is the data together with the

computed values of efficiency,power factor.and output.

599
‘11, wt 5 7 HP 3322:. )3 2. 2.
20.55 5100 218 50 lb-ft 6.7 82. 81.
26.00 7500 217 40 7 8.75 85.6 81.
50.25 9400 214 50 7 10.55 85.7 85.2
42.5 12400 212 70 7 15.6 81.5 79.5

In the above computation8,the following formulae were need;

it x 6.2832 x 1200 x % syn.

33000 x 100

8.2.

 

(The “,3 syn. was taken from the circle diagram)
HePe x 100

“Eff. : .. ”--.-..----”

7“ x 1e3‘
W5

P.F. . to... «ace-......

3 3311p

The same points of outpu) on the circle diagram compare as !

feline; ‘
I, )3 Eff. '73 2.17.
3GP. Bk. COD Bk. GOD. Bke COD.
6.7 20.55 18.8 82. 86. 81.00 80.00
8.73 26. 25. 85.5 85.7 81. 83.

10.55 30.25 27.00 83.7 85.00 83.8 83.
13.6 42.5 38.00 81.6 88.8 79.5 82.00

25

HEB
Brake test data and computed results;

1p 8. 5 7 7'3 Eff. 52 93 5.17.
16.15 4650 215 16 84. 5.44 05.5
27.5 9100 211 50 84. 10.25 ' 85.
45.5 15200 206 44 84.5 14.95 81.7

The only difference in formulas is that of HP,

7 x 6.2832 x 1800 x % syn.

HP . ----e-“cm----“--mce----~-

33000 x 100

Placing the comparative values of the two different methods

in adjacent columns.

ID 5522. 532.5.
HP Bk. COD. Bk. COD. Bk. ODD.
5.44 16.l5= 15.5 87. 86. 77.5 77.
10.25 27.6 25.5 84. 85. 85. 85.

14.95 45.3 38.00 84.5 84. 81.7 83.

Chapter III 27

Separation of Core loss from Friction and Windage.

The resistances of the different windings will first be
given in tabulated form;

The resistance of the stator windings was found by the IR dr0p
method.The following are the averages of a number of readings.
(These are the same as were used in the computations for the

circle diagrams.

H59 M98
.209 ohms/ phase .14 ohms/phase.

The resistances of the rotors were found by the wheatstone
bridge method.The bridge was set up on the motor side of the
set.and facing the motors.the phases were inga position denoted
by the following 761w.The bars denote the slip rings.Brass strips

were slipped under the brushes.thue assuring perfect contact.

/ / /
x

Z Y
Mh9 M-B
Ir! X93 sz X?! 192 Y-Z
.51 .5 .5 .45 .45 .45

is denoted in the 761w.these resistances are between terminals.
Lsmmaing that the rotor is wye connected,the resistance of the
phases will be one half the terminal values. The data for the
test with the secondary open ,and then with the secondary closed

will next be given.leading to the separation of the losses.

28

£59

Secondary closed-Motor running-Ho load.
I (average) at E

8.88 e45.KeWe 220
Secondary cpen - not running.
I ‘8') Wt E

Beg? e367 KeWe 220

898
Secondary closed-Motor running-ho load.
1 mo " wt 5 ‘
8.42 .476 218.3

Secondary open— not running.

w

1 (av) Wt
8.73 .337 220

Separation in 3-9

The core loss in 2-9 will be equal to the difference between
the watts loss at open secondary and the 19R loss at that current,
that is ﬁt - 12R . The current as seen above is 8.97 amps.
ant the resistance /0 is .209 ohms. The core loss will then be.

50' 307 - (3/2 x .209 x 8.972)
' 307 - 25.3 8 281.7 watts core loss.
0n open circuit the losses consist of core 1088.122 loes,and

windage and friction.This is.0f course with the motor running.

29

. 0..
The core 1053 has been found,and the l“n loss is found as

before.using the closed secondary test current.
Thus; aindage and 2riction 9 it - core lose - 1’3
or 35,: a 450 - 201.7 - (8.582x 5/2 x .209)

450-381.? - 24.67

143.63 watts Triction and windage.
Separation in 2-8
The core loss equation is;

no: 557 - (3/2 x .14 x 0.753)
. 557 - 15.95

3 321.05 watts core loss.
0 ’J’
The wiﬂdﬁﬁO and friction loss equation is;

3“,: . 475 - 521.05 - (5/2 x .14 x 8.422)

3 476 - 321e05 " 14e85

= 140.1 watts ~indsge and Friction.

?abu131133 these results;

Core 1055 hindage and friction.

5+9 281.7 watts. 143.63 watts
2-8 321.05 140.1 .atts

Changing tLBBG values to 7.3.

Core Loss 31ndage and Friction.
2-9 .577 hp .1025 hp
U~8 .43 hp .1827 hp

P ART 111 30

A Kethod of Speed Control.
Concatenation.

inppoee that c certain sir pole.wounﬁ rotor.60 cycle induction
motor ie loaded to such an extent that its slip beconee 20;.
Then the rotor current frequency will be .2 of 60 or 12 cycles.
ﬂow it this low frequency be fed to the stator of a second motor,
say a 4 pole machine.thgr3§eeﬁ of the eeooni machine would be
only 12/60 or .3 its normal upeel.Thuo if the motor were deeigneé
for GO cyclee.ite once: would be only .2 x 1803,0r 363 r.p.m.

Of course it would be difficult to take a common motor ani load
it to 23; slip,ne it would be producing about double its rated

output.zot only that.it woulﬂ require conetont loading of both

motors when feeoine one machine from the other.which coniition

would not mret proctical requirements.

Koo suppose that the rotor etefte of the two motors are coupled
together.ond neither motor loeﬁei.bnt fed in the some manner es
in the illustrationﬁg33ve.ln other words the motors are run in
ooecaae or concatenation.

The formulas for concatenation epeeﬂe will first be derived
for general ceeee enﬁ then applied to the concatenation oet
oomooeei of n—o eni 3-8.

ne—will Toke es an example two motore i eni 2.: having the
greater number of polee.Tten let A,the motor with the greater
number of polee.be fed free the line.ooﬁ B be fei from the rotor
of i.the rotors beiny ccuﬁﬁei togetner.Let the rotor of i feed

the stator of B in ouch a manner that the airection of rotation

22

 

 

 

 

 

 

 

 

 

 

 

 

 

 

mm
.2; ”
«new»

kblkba ml m.

o _ \ 3 L9 e .

h «H \.
E \\Xl\t 2

_ \N“ on
0N

_ \ \\ QM ,
QM. _ \ \

_ .\\\\
3w \ 0%
on \L a.
on _ / Vex“. \\ 3

_ \vx .

N. . \

_ .

;\VA/. .. N R
a... «Ax . N

\ ..\.

47..., WK. \ a
O“ " \w/rulfyu Jar—l ... .{ ;ﬁ 1!! .7ia\\=\ x? ..

_ +33! trrfi 1x2 .

,Il . w

2. m[ t em

. [/1 . /

H 2: _ 2 e2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

of: mnimbo moémtmektwm

 

23

 

 

 

 

 

 

 

 

 

 

 

 

 

kboﬁ .3 or .
o 3 .3 ~< W o m. o
..H \.
Q\ \m\\ Qx
\ . H\.
\ \
\\

QN \ ON
on \\\\\\\ \\\\\ on
3 \\ 3.
RV HF . a.
3 x \xx- .x on

\\ \. -..\ .

/ \ \\ Km

x .3. n.
2 \ III]. \ 2. mum &
xx\ .... 1N. ..

(If //I \r.\

0Q -.HHUII. H \%\ \ .\\ 8
jjﬁmﬂ- loerxH-m \\

l/ a.

on IILIJ «Q
[r]! .
.23. [I x

02 1! Q“:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

went 9%; ESQ MUETZKQKEMQ

 

 

Only

24
Chapter II.

“---

Tho Brake Test.

enough points were taken in tho brake run to serve on

a check on tho circlc diagram,and to offer a comparison between

the two methods. The following is the data together with the

computed values of cfficicncy.power factor,and output.

In
20.55
25.00
50.25
42.5

599
5. E T HP ﬁerr. % 2.5.
5100 210 50 1b-ft 5.7 82. 81.
7500 217 40 7 8.75 85.5 81.
9400 214 50 7 10.55 83.7 05.2
12400 212 70 7 13.5 81.6 79.5

In the above computatione,tbc following formulae core need;

HoP.
(The

ﬁkEtf.

P.F.

T x 6.2832 x 1200 x % syn.

--“----”m--.--.m - .'

33000 x 100

% syn. was taken from the circle diagram)
HOP. x 100

W‘ x 103‘

W:

3 xExIp

The some points of output on the circle diagram compare an !

fallout; 1
H.P. Bk. c.D Bk. 0.0. at. 0.D.
6.1 20.36 18.8 82. 86. 81.00 80.00
8.73 26. 26. 86.6 86.7 81. 83.
10.66 30.26 27.00 83.7 86.00 88.8 83.

13.6

42.5 38.00 81.6 82.8 79.5 82.00

85

398
We test data and computed results;

Ip wt 3 r 35 Eff. 52 3% 2.3.
15.15 4550 215 15 at. 5.44 00.5
27.5 9100 211 30 84. 10.25 85.
45.3 13200 205 44 84.5 14.95 81.7

The only difference in formulas is that of HP.

1' x 6.2832 x 1800 x 3% syn.

HP . Q“---”m----”--Q..---..~-

33000 x 100

Placing the comparative values of the two different methods

in adjecent columns.

11, ﬁéEff. ‘ 33 P.F.
HP 3:. GOP.- Bk. COD. Bk. GOD.
5.44 15.15. 15.3 87. 85. 77.5 77.
10.85 27.5 23.3 84. 85. 85. 85.

14.95 45.3 38.00 84.5 84. 81.7 83.

Chapter 111 37

Separation of Core Ross from Friction and Windage.

The resistances of the different windings will first be
given in tabulated form;

The resistance of the stator windings was found by the IE drop
nethcd.The following are the averages of a number of readings.
(These are the same as were used in the computations for the

circle diagrams.

H99 “#8
.209 ohms/ phase .16 ohms/phase.

The resistances of the rotors were found by the wheatstone
bridge method.The bridge was set up on the motor side of the
eet.and facing the notors.the phases were inla position denoted
by the following vilthhe bars denote the slip rings.Brass strips

were slipped under the brushes.thus assuring perfect contact.

///

Z Y X
H99 H98
I-Y X- Z Y- Z X-Y X- Z ' Y- 3
.51 .5 .5 .45 .45 .45

is denoted in the v61w3these resistances are between terminals.
[Assuming that the rotor is wyc connected.the resistance of the
phases will be one half the terminal values. The data for the
test with the secondary open ,and then with the secondary closed

will next be given,leading to the separation of the losses.

88

E-O

Secondary closed-Hotor running-Ho load.

I (average) wt E

8.88 .451K.W. 220

Secondary cpen - Bot running.

I ‘8') Wt E
8. 97 em? KeWe 220

398
Secondary closed-Motor running~N0 load.
I (IVO ' Wt E~
8.42 .476 218.3

Secondary cpen- not running.
1 (av) 8t E
8.73 .537 220

Separation in 3-9

The core loss in 3-9 will be equal to the difference between
the watts loss at Open secondary and the 153 less at that current,
that is ﬁt - 12R . The current as seen above is 8.97 amps.
ant the resistance /0 is .209 ohms. The core loss will then be.

36‘ 307 - (3/2 x .209 x 8.972)
3 307 - 25.3 8 281.? watts core 1088.
On open circuit the losses consist of core 1035.123 less,and

windage and friction.This is.of course with the motor running.

29

va
The core loss has been found,snd the I“n loss is found as

before.using the closed secondary test current.
Thus; aindage and 8riction 9 it - core loss - ICE

or 3,,f s 450 - 281.7 - (8.882x 3/2 x .209)

450-:81.7 - 24.67

143.63 watts Triction and windage.

Separation in H-B
The core loss equation is;

To“ 337 - (3/2 x .14 x 0.732)
- 337 - 15.95

3 321.03 watts core loss.
a (J)
The rindage and friction loss equation is;

:w ; a 475 - 321.05 - (3/2 x .14 x 8.422)

= 476 - 521e05 " 14085

140.1 watts «indage and Friction.

Tabulating these results;

Core loss windﬁge and friction.

£99 £81.? watts. 143.63 watts
K-a 321.05 140.1 .atts

Changing tusse values to 8.3.

Core Loss Lindnge and ?riction.
3-9 .377 hp .1923 hp
Y-8 .43 hp .18?7 hp

P83? 111

A Kethod of Speed Control.
Concatenation.

Suppose that a certain six pole.wound rotor.60 cycle induction
motor is loaded to such an extent that its slip becomes 20;.
Then the rotor current frequency will be .2 of 60 or 12 cycles.
new if this low frequency be fed to the stator of a second motor,
say a 4 pole machinc.th5r§peed of the second machine would be
only 12/60 or .8 its normal speel.TLuo if the motor were designed
for 00 cycles.its shoe: would be only .2 x 1803,0r 363 r.p.m.
0! course it would be difficult to take a common motor and load
it to 51} 8113.88 it would be producing about double its rated
output.hot only thet.it would require constant loading of both
motors when feedins one meetine from the other,which condition
would not mret proctical requirements.

how suppose that the rotor shafts of the two motors are coupled
together,end neither motor loadei.but fed in the same manner as
in the illustrationﬁggevs.ln other words the motors are run in
cascade or concatenation.

The formulas for concatenation speeds will first be derived
for general cases and then applied to the concatenation set
composed of M—9 and 3-8.

he—eill take as an example two meters 1 and 2.5 having the
greater number of poles.$h5n let s.the motor with the greater
number of poles.he fed fron the line,and B be fed from the rotor
of A.tne rotors being cuhiod together.Let the rotor of 3 feed

the stator of B in such a manner that the direction of rotation

31

of tne magnetic fields of the two stators in the oeme.Tnen let
X 3 Ike concatenation opccd.
Hith the motors running in ceccede,the number of alternations
of the current in the rotor of e (per. min.) will be.
( synchronous once‘ of A - x) x no. of poles in the stator of t,
which is eq.(1).
to the epeed varies as the no. of polee.the speed of the field
of B will be;
no.polee A
( 3n. speed of A - X ) --—----------- (2)
no. poles 3
But at no load there is no 1039 in the rotor of B.end coneequent
~13 there con he a some? to be no slip.Then the field spee‘ of B
will be the concatenation rpce‘ of the set.
no.pclee A

or, X ‘ (eyn. speed of i-X) -~----------- (3)
no. poles B

Th a method of connection is celled cumnulative concatenation.
gubetituting the constants of 2-9 and n—8 in the equation (3}.
6
x = ( 1200 - x ) --------
4
5/2 x a 1830
1 3 720 r.p.m..the creed of the set in cumguletive
concetenetion.1t may be seen that this is the spec? of a motor
having the same number of poles as the sum of these two meters

or ten poles.

Now suppose that motor B is fed from the line and motor A is
fed from the rotor of B .in such a way that the stator fields
rotate Oppositely.Then the connection is said to be diffcdbtiel
concatenation.es the rotor in motor B is now rotating against

the stator field.the number of alternations per minute will be;

(1800-? X)4
And the concatenation speed will be.
4
(1800 + x) «mm . x
6
Then, 1/3 x . 1200
X = 3600 r.p.m..the speed in diffractial concatenation.

This is seen to be the speed of a motor of the number of poles
equal to the difference of the number of poles of the two motors.
or two poles.

From the above diccussion and formulas it can be seen that if
the motors have the same number of poles,thc diffractial concat-
enation speed will be O.and so the set has only 2 possible speeds.
that of cummulative concatenation and that of either motor.Somc-
times one stator is provided with all the coils.but in this case
it is more efficient to regroup. the coils instead of using
concatenation as a method of speed control.

In the case of cummulstive cescade.the frequency of the rotor

current in A is ~139§335239---- x 60 t 24 cycles .or 40$ of the

stator frequency.Then 405 is transmitted to the other machine

sand GOﬁ is changed to mechanical power in A.

Concatenation Test late.

53

In oummulative concatenation,§brake test gave the following

data;

In
15.9
15.5
13.5
23.4
25.

a
218
sis
214

30.5

n
1000
2100
5600
4750
6020

6450

5‘

6000
6180
6885
8625
9400

,
ace 1

11100

.1;
1’9
'3 0‘

‘03:.

16.6
34.

52.25

55.1
54.
57.9

lb-ft

T Speed
0 723
14 732
28 721
40 718‘
48 710
52 695

2.2.
in , out
0

2.818 1.9
4.82 5.8
6.37 5.44
8.08 6.5
8.625 6.79

The breaking point 15 at about 55 lb-ft and as

5 Eff.
o
67.4
78.75
85.4
80.4
78.8

amps.

Using the same points and values of output.the following

data was obtianed for the power transfer to HAS,

15

6.9

8.48
12.1
16.25
19.62
22.25

E8

83.
81.
79.
75.
72.
68.

'37
’u 8

140
750
1360
1750
2225
2400

arequcncy

24.25
24.5
24.75
25.
25.25
25.5

How,if from the total input.the losses and friction and windage

together with the power transfer to H98.be subtracted.the rest

will be the power delivered th the shaft by “.9. Assuming that

the no load losses'are constant and equal to 450 watts.and that

the capper losses are the sum of the losses in the primary and

secondary.the following data is computed;

(10

(2)

Total Total Power Be load 1359 1253 Total 1- HP HP

HP Input Trans. Loss H-Q Hé9 iTrans. M99 H98
1.9 2100 750 450 155 54 1599 701 .94 .95
5.5 5500 1550 " 217 109 2235 1554 1.85 1.97
5.44 4750 1750 " 542 197 2759 2011 2.7 2.74
5.5 5020 2225 a 425 255 5255 2554 5.54 2.95
5.79 5450 2400 " 555 550 5795 2555 5.49 5.50

From an inspection of the H.E. output values for 2-9 and H98
it can easily be seen that the output does not divide as it was
explained that it should in the theory. The stator of M98 is fed
with current of about 25 cycles.which greatly reduces the ind-
uctive reactance. Thus the power factor of this motor is greatly
in excess of that of M99.Although fed with current of much lower
voltage than HA9.the impedance of its windings is much less and
so a large current flows at high power factor.The two motors are
of the same rating.and naturally their windings would be more
evenly balanced than as if one were a large motor and the other
a small one.As the rotor circuit of M99 and the stator circuit
of use are in series,the increase of impedance tends to hinder
the current in these circuits from rising very rapidly.as would
be the case if the motors were Operating independently.This tends
to favor M—8.but not m-9.0ther factors which aid use are(l);The
voltage being low.the iron losses are much less;(2) The same may
be said of the frequency;(3)The speed of “-8 is much less in
preportion to that of H~9(§elative to their normal speeds).that

the power loss in its cooling fan is small.

35

As a reference.and as an aid in gaining a better idea of the

division of load.the outputs have been plotted against torque

as follows;

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

7725-79“)

7
,4

6

5' ///
If /

5“"
4
3 z// ,/ffj::"“’
/ MI. «‘17,?
a A, 1:,» ”TWO
/ / ’:
I
o” In )5" .o .25“ 3o 35' 40 45 .52

 

 

36

Differential Concatenation.

---”---- --

The following no load data was taken with the motors
connected in differential concatenation;
1p Ep Wt ﬁjp.f. Speed
27.2 208 4550 46.6 3500
Do load test was run with the motors connected in this manner.
Full load current was already flowing to the machines.and the
input was well up toward the limit reached when loaded in cumm-
ulative. The set became very unstable when approaching the
speedhf a two pole motor.in fact if the resistance here cut out
of the rotor circuit as quickly as it is ordinarily done.the
set would stop. The resistance in he rotor circuit made a
difference of about 500 rpm. in the speed of the set.that is;
Speed with Resistance in. Resistance Out.
3000 rpm. 3500 rpm.
There are several reasons why the no load losses should be
so large. The great increase in the speed increases the power
loss in the cooling fans. This power loss raises as the cube
of the speed.thus it becomes 8 times as great in the 4 pole
motor and 87 times as great in the 6 pole. The frequency in
the rotor of the motor in which the rotor is revolving Opposit-
ely to the field.becomes excessive. 55 this is the rotor of
M-8.the frequency is three times as great,and this also feeds
the stator of I99. An increase in frequency is accompanied
by an increase in iron losses. Then again the motors are

carrying full load current.which means a largo Ik' loss.

37

Genera1 Discussion.

rho theory that has been deve10ped and applied to these motors
in the foregoing pages has checked quite closely with the results
of the brake tests on the machines. Probably the greatest variat-
ion was in the case of the division of load when in cummulative
concatenation. The theory as stated may be found in Karapetoff's
"Experimental Electrical Engineering: vol.2.page 384. He gives
no proof.but merely gives it as a statement. As before stated,
the motors in this case being of the same rating,and differing
by only two poles.the windings will be more nearly balanced.
which may account for the more nearly equal division of load.

The tests with the set running in differential concatenation
prose that this is not a practical method of obtaining 3600 rpm.
as the set consumes its full capacity in losses.and so is unable
to carry any outside load. Furthermore.the set is not alWays
self starting when oonnectei in this way,although without load
it will start in most rotor positions. Even if it did afford a
small amount of power,the time required to bring it up to speei
would make it too inefficient for modern shop practice. The
study of the set is excellent,however.in that it brings out

many points of interest to those dealing with induction motors
I

which would otherwise remain obscure.

QM 33E ONLY

6d?“
at

in!

I. I'.’ *l'.‘ 0“!

 

 

_ n u
C

.

A

a

.

7

a

V

a

T

w

..-

V

T.

..

u. .

W .1
r
_.1 . A.“
7

an ,

i E . u
_
... .

‘ x-‘ir'ox —. we};
' ‘JM’thi'V I Ly“ A‘v' .‘ ‘V .
' «x -. :c 1..“ We"

   
     

J I. . o
. " J. i "
i r 5 .‘V‘. '
’fl- ‘ .5 . r
f .
‘1' In ,I." I u q.
. . ‘."'I <' .‘ _ '_
I “‘ . .t .‘ .K .
. .
..».\ a 9
'. t I - I‘.r'
a c - .
s ‘ -‘ .o -l
f - _ .
. . ‘ o
" g: ' ‘.
"'i ,x. .‘ ‘t :
.
I _ ~,‘ . -
. ' I
v' .\ - A
. N :v r
I ‘. V .
\' 1
."H‘ v,
, ‘ .
1| ‘ , ,-
, . , u '
. , r
,
. J 'A l
O. I I
‘C
r ' .. ' . '._ L-
k ";
P .06 '.
x l | .
‘ ' l - I
v.‘ '
, . . ,-
,1.
(v I . '\
'_ _ f) '
'4 ‘. I

._ ‘
7‘ I
Q. ‘
.5. ‘
n
‘ \
- ._ .
' \
k " u
v
' 1
~.
.I ‘
l
.
.
n
I I
.
.
u
i
' \
i
.
.' V
v
I
.
.
‘
i '.
I
.
. u
I
.
u ‘ ‘
. I
l
' n
. .
I

 

. l

'."Jq‘

‘ - , \ _, _' .'-.‘"\-'"--'

., . .. - .-- - ~. -. -- H. t. ~ . -ttini z.‘.\

L . . ~’ .. ' . .'~ ".'.' p i“ '_ ' -" ~35 \ "‘V-IC‘. e“ l‘xuly
. ' ' ' -r '

 

M'Tﬂiﬁtiﬂﬁilﬂiiﬁljuﬁiﬂiﬂjfﬂiﬂkitiiﬁﬂiﬂmﬁr