THS

 

 

Siam '3? A PHASE SPECIAL TRANSFORMER

THESIS FOR DEGREE CF 3. S.

3.:‘7ARSHALL G. HOUGHT’JN

1928

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LIBRAfiY
Michigan State
University

 

 

 

 

 

PLACE IN RETURN BOX to remove this checkout from your record.
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DATE DUE

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6/01 cJClRC/DateDue.p65-p.15

 

DESIGN AND CONSTRUCTIDH

OF A
13 KW. SPECIAL TRANSPORTER

A Thesis Submitted
to
The Faculty
of

Elohigan State College

BY

_ 1

Marshall George Hougnton.

Caniiiate for the Degree of

Bachelor of Science

June,1926.

Introduction.

The Interconnected Star Connection.

Voltage Relet

Design.

Result 3.

Bibliography.

ions in a Rotary Convertor.

Introduction.

An explaination of the situation is the first step in
establishing an understanding and unless one knows what is
to be done there is not much use in going ahead with any kind
of work. This introduction will show the situation as it was
and will outline the steps taken giving the reasons for this
method of attack.

The power panel board in the laboratory at Michigan State
College has thereon a 270 volt D.C. circuit and it was desired
that there be a source of 110 volts D.C. This would he made
possible by an Edison three wire system and inasmuch as an
inverted convertor was available,the most systematic way of
completing this setup was to load the convertor with a star
connected transformer and bring out the neutral for the third
wire of the Edison system. Upon investigating the character-
istics of simple star transformers for this job,there'was a
set in the lab,it was seen that when a heavy D.C. load to
neutral was thrown on that the unbalanced D.C. flowing through
the windings would saturate the transformer cores and cause
a current to be drawn from the rotary which had a strong
lagging component. The lagging current weakened the field of
the rotary and it would ’3; 3:11 {brace . This situation was met
by using interconnected star primaries on the transformers;
the reason being that with this connection the unbalanced
D.C. flows through the windings in such a way that their
m.m.f‘s neutralize each other and do not affect the operation
of the rotary.

It was also desirous to have 110 and 280 volts A.C. as

a source of power in the laboratory to be used alone or in

20

parallel with other machines. The solution to this part of
the problem was to put a 110 and 883 volt secondary on the
transformer. It was not necessary to bring this neutral
out,neither was it necessary to keep it free from triple
harmonic voltages so the simple star connection was used
because of its requiring less wire, being cheaper,and
filled the requirenents as well as was necessary.

This thesis deals with the design,construction,and

I Q.
‘1 - .‘t '3 ~ " " “1-"...- J .-','.'-, o
C31!VU.L..~.L‘V 4¢L|-L'~..,~J»J.I5,

testing of a transfo mar with the follofing

Interconnected star primary with neutral tap brought out,

110 and 220 volt secondary using the simple star connection.

k_._. ..

-_ ~¢ ....—.—...—.~.

This transformer is to use the A.C. output of an inverted.
convertor as a source of power.

It is readily seen that one must be familiar with the
voltage and operating characteristics of a rotary convertor
in order to get the best results in this thesis so a short
summary of the convertor has been included.

The author wishes to acknowledge his indebtedness to

Professors Foltz and Cory for their suggestions and to

Professor Naeter for his assistance throughout the were.

m'!

the Interconnected Star Connection.

In the case under consideration there is an inverted

rotary convertor already in

V

m

talled and a three wire D.C.
circuit is desirable. The three wire arraingement is ob-
tained by connecting .he neutral wire of the sysien to the
neutral point in the primarys of the transformers. In such
cases the connections should be arrainged so that the direct
current in each transformer divides into two branches of
equal ampere turns. This is to prevent any undirectional
magnetic flux in the trausforusr cores which,if added to
that which is already there due to the 60 cycle current,
would tend to saturate the cor s to a point past which it
is not advisable to operate. This in turn would cause ex—

cessive heating and xeiting current,except where the un-

([1

balanced current is comparitively shall. The inter-connected
star arraingement eliminates the flux distortion due to

the unbalanced D.C. in the neutral. Two separate inter-

.4...

C) '..-' (A

{I

H

connected windin s are used for each oer of ihi

(

The unbalanced neutral current flowing in this system may
be compared in its action to the affect of a magnetizing
current in a transformer. The affect of the main trans-
former currents in the windings is balanced with regard
to the flux in the circuit,and the flux in the nagn:tic
circuit depends upon the magnetizing current. When a
direct current is passed through the transformer its
effect on the transformer iron varies as the magnetizing
current unless the fluxes produced by it neutralize one
another. If a distributed winding is used, the D.C. flows

in opposite directions artund the two h lvs. cf sch

m

winding and neutralizes the flux distortion.

When it comes to a decision between simple and inter—
connected star transformers it is only a question of
balancing the increased iron loss of the straight star
against the increased copper loss,and the greater cost
of the inter-connected star. E.G.Reed in his book on
Transformer Practice states that the straight star is
much simpler and would be permissable to use for transfor-
mers of small capacity where the direct current in the
neutral is not more than 30 percent ( 10 percent per
transformer) of the rated transformer current. The
current in the neutral in this special case may run as
high as 45 amperes and the average will be 30 to 35 amps
while the rated current of the transformer will be in
the neighborhood of EC amps. It is readily seen that the
interconnected star connection will be necessary, the

-4.
en...

(.1

average percentage being about 30 gar

This topic is also discussed,with especial emphasis
laid on the part the transformer plays in the operation
of the rotary , under the next heading - Rotary Convertors.

Voltage Relations.

The connection diagram and vector voltage relations
for the interconnected star are on the accompanying sheet.
The voltage Eco equals VE’EOG ,and the voltage EEC equals

Vg’Eoc- Therefore Eoe equals one-third EEC. In other words
each half of the secondary winding of each transformer has
a voltage of the interconnected star voltage between the
lines B and C or the voltage EEC . For example,the voltage

E00 between the neutral point and one of the lines is 30

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The Rotary Converter.

It is frequently necessary to obtain direct current
when the available supply is alternating. Such,for instance,
is the case with many electric railways,arc lighting systems
etc. Changing from direct current to alternating current is
also,but less frequently,required. The second is the case in
the laboratory at Michigan State College where there is a D.
C. supply of 220 volts and an inverted convertor available.
This work of conversion from D.C. to A.C.,or vice verse, is
done to great advantage by means of the rotary convertor.
When used to change A.C. into D.C. it is called a direct con-
vertor and when the reverse is true, namely a conversion from
D.C. to A.C.,it is called an inverted convertor.

A rotary is a motor and generator combined into one be-
cause it receives and transmits electrical energy. If it is
driven by mechanical power it may give out electrical power
either in the form of direct current or alternating current
or even both at the same time. It follows,then, that the
convertor can work in one of four ways which are ;direct
current generator, direct current motor, synchronous motor,
and alternator. In any case its performance becomes that of
one of these machines or a combination of them, and may be
studied in that light.

There are two main objects in the construction of this
thesis transformer,the first of which is to obtain a neutral
for an Edison three wire system and the second is to get
suitable output voltages so that this convertor may be used

as a source of power in parallel with other machines in the

laboratory,with the power house supply, or alone.

When an inverted convertor is operated in parallel
with another source of A.C. its speed will be fixed by the
alternator because they must operate in synchronism. If,
however, the convertor is the only source of A.C.,the speed
will be affected by a change in field strength in the con-
vertor and the machine will act as a shunt wound motor.
It follows that an inductive load on the A.C. side of an
inverted convertor,when operated alone, would tend to
weaken the field and cause the machine to speed up. For this
reason an inverted convertor should have overspeed protective
devices.

The design in this thesis has a direct bearing on what
has been stated in the previous paragraph. If a simple
star transformer is connected on the A.C. output of the
rotary to derive the neutral for an Edison three wire system
and a heavy D.C. load to neutral is applied,it will cause
considerable unbalanced D.C. to flow through the transformers.
This unbalanced D.C. will saturate the core of the trans-
former to such an extent that the magnetizing A.C. will
increase very materially and reduce the power factor on the
A.C. side. The lagging current component decreases the
field flux and the convertor speed increases. It follows
that if the D.C. load were thrown on to neutral suddenly
the convertor would tend to race to such an extent that the
overspeed devices would operate and take the machine off
the line.

The use of interconnected star transformers eliminates
this by dividing the primary into two parts and putting half

of the primary on the next core and connecting those on the

same core in such a manner that they are "bucking" each

7.

.)

other as far as the D.C. is cn-ernei.Then the mmf's set up

0
L

by the D.C. are neutralized,thus keeping the flux density
in the transformer cores the same and this in turn does not
give the rotary any chance to race because of its field

becoming weakened.

It is evident that the voltage and current ratios of the
unit taken as a whole are important . These are explained
to some detail in the succeeding paragraphs.
Voltage Ratios.
Symbols used; f % frequency.
b = flux per pole cutting the conductors.
f t - total number of turns between brushes.
The fundimental equation of a D.C. generator has been
found to be -

E: 4 f b t volts.
10

If the armature is tapped at two symmetrical points,for
each pair of poles,and the tape brought out to slip rings,
then the voltage across the slip rings according to the
fundimental equation for A.C. generators is;

E __ 4.44 f b t effective volts
10‘H

 

The previous equation holds only for concentrated Windings.

For distributed windings it is ;

E 2'.§ x gL44 fAb t effective volts.
1r 103

 

and by substituting for 4.44 its equal 2 the maximum
voltage is
Eafi'xgg fbt _ 4fbt volts
TIF— 10‘3 ~ 7.58—
which is the same as for direct current.
In polyphase machines it is more convenient to use one-
half the D.C. voltage and then use phase voltage,or voltage

to neutral for the A.C.

The D.C. voltage to neutral then is;
En : §_ - 2 f b t volts.
3 — 108
Single phase voltage to neutral then is;
Ep E
2 _ 2??
and in a three phase machine te slip ring voltage is;
{den = \[3— x 2fb__t_ volts.
10:3
In an N phase machine; the maximum voltage between rings,
E5. = 23 En x sin :1 z E xsin ‘l_1__ volts.
N ' N
The effective ring voltage is;
Eeff =2 EnpSin 1 = E Sin ll VOltSo
N 2 N
A three phase convertor is the one dealt with in this

thesis ,therefore the following will be true;

En:= E _ 220 =_llO volts.
8 - 2
and Em :BBO sin 1L. =_191 volts. ( phase)
3
and Eeff'= 220 sin 1%_: 135 volts. ( phase)

The preceeding discussion only applies to a condition
under which the A.C. voltage follows the sine law.Steinmetz
shows mathematically that if the A.C. voltage wave is flat
the D.C. voltage will be lower,and conversly if the A.C.
voltage wave is peaked the D.C. will be higher. This

is found to be the case in tests on convertors.

Design of the Transformers. 10'

GIVEN: Core stampings for the transformers,with dimensions
( as shown on the accompanying drawing. The loss due

to hysteresis and eddy currents in this iron is
approxima ely 1.36 watts per pound as determined
by tests on similiar iron in the department,This
figure is apt to vary but will be sufficiently
accurate for the computations.

REQUIRED: Complete design of a three phase,interconnected
star transformer with rating as 4 KVA per phase.

Note; All computations are for one phase.

** ** **
Volume of Iron;
(The core stampings are 6" deep per phase)

Vblume =(Extericr volume) - (Window volume) also

- volume cut off the corners.

9.25 x 6.75 x 6) - (5.75 x 3.25 x 6) -6.

(
=-856.5 cubic inches.

Weight of iron;
Weight=(Weight per cu.in.) x (Volume)
= .86 x 856.5

==66.75 lbs per phase.

Iron LOss (per Phase)
Given ; loss is 1.36 watts per pound.
Loss ::Lose per lb. x height.

= 1.56 x 66.75

H

91.3 watts loss per phase.

11.
Preliminary Design.

Terms Used; Ep:= Primary Voltage.

: frequency.

f
N :' Number of turns.
A = Area of core.

B

=“ Flux density. -. ( 60,000 lines)

 

** **
Number Of primary turns; (.85 is lamination factor)
Np: En X 108
f‘A B 4.44
lee/{E‘ £108

 

60 (6 x 1.75 x .85) x 4.44 x 60,000
= 54.8

Actual Number of primary turns;
(Due to phase relationship of half the coil)
Np» 3 x 54.8/V3 63.4

Np =use 64.

Effective Primary Turns.
It is the effective turns that produce the flux,

and all relations must take this into account.
Effective turns :rfgix 64/2 55.4

Number of secondary Turns;

N8: Np x ES/Ep

;.55.4 x 110/155
st’45 for 110 volts.

N8: 90 for 220 volts.

Preliminary Design.
Current Densities.

Use current density of 1500 amperes per square inch
for first approximation and assume the full—lead copper
loss equal to the iron loss which will give highest
efficiency at full load.

Primary Current;

Ip :L(Eatts ratsng plus Losses
x Voltage

2 3(4000+182) Ni? x 1:55

= 53.8 amps.

Wire size;

Area of wire _ amperes/ amperes per sq.in.
‘2 00359 sq. in.

From wire tables use 3 # 6‘s in parallel which

will give the necessary area and be easy to handle.
Area of 8 #6's in parallel = .0412 sq.in.

Current density then is '

’

53,8/.0412 : 1305 amps per sq.in.

Resistance of E #6‘s in parallel ;
R =(.3951)3/s(.3951)

:.1975 ohms per 1000 feet.

*1! *1! **

Preliminary Design. 13.

Losses.

Resistances.
Resistance of a coil = (Number of turns)x(Mean length

of turn in feet)x(Resis.per ft)

By inspecting the assembly drawing it can be seen
that the mean lengths per turn are as follows;
Primary == 1.5 feet.
Secondary-=1.8 " (110 volt sec.)

" = 2.0 " (220 part of sec.)

Rp: 64 x 1.5 x .1975/1000
:=.01952 ohms.

R8==45 x 1.8 x .1975/1000
=.Ol628 ohms for the 110 volt secondary.

The 820 volt part of the secondary is made of single
wire while the 110 part is double. It follows that the
resistance of the 280 volt circuit is 1.5 times that
of the partial 220 volt coil,there bning the same

number of turns in each.

Rs:;220 volt total
: 1.5 x 45 x 2 x .3951/1000
: .0534 ohms.

Currents.

1p: 53.8 from previous work.

Is: 3(4000)/110 x(3 63 amps (110 volts)

Is 3(4ooo)/zeo x(3 31.5 amps (220 volts).

14.
Preliminary Design.

12R Losses.
IZRp = (53.8)3x(.01953)
= 56.5 watts.
IBRB = (63.0)2x(.01628) (no volts)
: 64.6 watts.
1338 :— (31.5)3x(.0534) (220 volts)

= 5'5. 0 Watt s.

It is seen that the largest possible computed losses
are when the output is at 110 volts. By figuring through

for the worst case the limits will be found.

Total IBR losses per phase ; (using worst case)
Loss : 56.5 + 64.6

2 121.1 watts per phase.

Grand total losses per phase;
Total lossz' 12R -+ Iron loss.
:.121.1'f91.3
=-212.4 watts.

There are approximately 250 square inches of radiating

surface per phase on the finished transformer.

Watts radiated per square inch;
= 218.4/350
: .848 watts

.85 watts per square inch is very high and can not be
tunlerated for an air cooled transformer. Design data sdes

it ‘is not advisable to go above .5 watts per sq.in.

 

COIL . DATA.

 

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L " J

.7_A11 coils are 2%" wide at the face.

Dimensions of the average turn.

Primary Coils.

B is 2%". L is 6%".

110 Secondarv Coils.

s is s anil/lS" L is 9 and 5/16".

38C Secondary Coils.

B is 3 and 5/16”. L is 3 and 7/8".

 

~'“ -

 

on
-8.

 

15.
Preliminary Design.

It is also seen that the copper loss is 30 % higher
than the iron loss; i.e. 121.1 watts copper loss to 91.3
watts iron loss. For highest efficiency at full load the

copper loss should equal the iron loss at that point.

RESULTS of the preliminary design.
1. Losses are too high making radiation per sq.in.
too high.

2. Copper losses are out of proportion to iron losses.

Increase the size of copper and out down the copper
losses to a value near that of the Iron loss(at full

load Operating point.)
FINAL DESIGN.

WIRE SIZE.

Approximate check on size of wire needed;

Watts radiated per sq.in.==(I2R-rIron loss)/ Area.

.5 =-(53.s8

xR + 91.3)/ 250
R ==.18 ohms per 1000 feet.
By using the wire tables it is seen that 2 #4's in

parallel will apply all the correctinn needed.

NEW current density;
: 53.8 / .0657

=—820 amps per sq.in.

Resistance of 2 #4's in parallel;

R“(-24RR\8/9l DAQ:\ '- “CA: -L ItAAa-n n.

16.
Final Design.

Coil Data.

One-sixteenth of an lnG.Will be allowed for clearence
on all dimensions where clearence is necessary.

Sizes of coils are shown on the drawings included and
in this thesis and will illustrate questions which may
come up in the computations. Knowing the size wire and the
turns per coil there is no difficulty in getting the sizes

of the coils.

Primary Coils.
64 turns per phase, double #4.

Use 4 coils of 16 turns each(for convenience)

Average Length per coil; 0f one turn - the average one.
L=(2x2.5+2 x6.75) / 12
=-l.5l ft.
Resistance per coil.
lR=(Length in feet)x(resistance per foot)
=(1s x 1.51) x ( .1245/1000)
:.00301 ohms.

110 Volt Secondary coil.
46 turns per phase, double #4.
Use 2 coils , 23 turns each.
Length of average turn;

'L = (2 x 4.06 2 x 9.313) / 12

H

2.23 ft.
Resistance per coil;
R =;(23 x 2.23) x (.1245/1000)
= .00638 ohms

17.
FINAL DESIG .

220 volt secondary.
44 turns per phase,single #4.
Use 2 coils with 22 turns per coil.
Length of average turn;
L=(2 x 5.313 + 2 x 8.825) /12
= 2.09 ft.
Resistance per coil;
R::(22 x 2.09) x (.2485/1000)
=.0114 ohms. ‘
s: *t **
New Losses .
Total Resistances. (Per phase)
4 Primary Coils -
Total Resistance: 4 x .00301
==.01204 ohms
2 - 110 volt secondary coils;
Total Resistance = 2 x .00638
= .01276 ohms.
2 - 220 volt secondary coils;
Total Resistance : 2 x .0114
1.0228 ohms.
Then include the 110 volt secondary with which
this coil is in series;
Rt:1.5 x .0228
2.0342 ohms for total 220 volt circuit.

FINAL DESIGN.

18R losses.
13Rp==(53.8)3x(.01804) -= 34.85 watts.
I2R3:=(63.O)8x(.01276) .. 50.5 n (110 volts)
I2R3:'(3l.5)3x(.03420) :: 33.95 " (220 volts)

Now using the case which will give the highest losses
so as to be on the safe side ,it is evident that the 110

volt output is the one, the total losses are computed.

Total 12R loss per phase.
Loss : 34.85-t50.6
-=85.45 watts.

Grand total loss .(Taking iron loss into account)
LOSS =r85.45 +-91.3
==176.75 watts.

Watts radiated per square inch of surface;
=.l76.75 / 250
r: .706 watts.

The loss of .706 watts per square inch is still large
but inasmuch as this transformer is not for continual use,
being used only part of each day in the lab, it was not
thought to be necessary or economical to use more copper
in order to bring the losses below .5 watts per square

inch.

The transformer will then be constructed .

Wire Necessary to Build
the Transformer.
Primary - 4 coils per phase of double #4.
Average length per turn = 1.51 ft.
Number of turns : 16
Length per coil==l6 x 1.51 :r 24.2 ft.
4 coils per phase gives:.4 x 24.2: 96.8 ft.
Double Wire gives=-2 x 96.8 = 193.6 ft.
Total Primary,3 phases ::3 x 193.6==580.8 ft.

110 Volt Secondary. 2 coils per phase , double #4.
From similiar computations;
L==23 x 2.23 x 2 x 2 x 3
Total Length: 615.6 ft.
220 Volt Secondary. 2 coils per phase,sing1e #4.
1.:22 x 2.09 x 2 x 3

Total Length<=274.8 ft.

GRAND TOTAL LENGTH. (Allowing 30 feet for connections)
Length =580.8r615.6+ 274.8+ 3o
==1501.8 ft.

Weight of wire.

#4 Weight per 1000 feet 125.4 lbs.

Total weight;
: (1,501.2 x 126.4)/1000
: 189.6 lbs.

ORDER 190 lbs of #4 wire. Double cotton covered.

** ** **

to
0

RESULTS.

The first trouble experienced was in winding the coils.
A form.was made to wind them.on but due to the extreme
stiffnecs of the wire and the fact that no machines were
available to shape the coils it was not possible to get
them quite down to the dimensions and form as shown on the
sketch. This would be possible if a winding and shaping
machine were available and the transformer coils ,as
designed,would be all right. The handwound coils were
thicker than was expected and if they were made according
to the previous rating, 12 KW for both 110 and 220 volts
output, they would be too thick for the window of the
transformer iron. This difficulty was overcome by winding
the 110 volt secondary with single instead of double wire.
It is evident that this reduced the area of copper by one-
half so that the same conditions will hold if the 110 volt
rating is reduced to 6KW. The author was fairly liberal
in his values of current density and watts radiation per
square inch of surface and is of the opinion that 9 KW
could be drawn from the 110 volt secondary without over-
heating or otherwise injuring the transformer providing
the primary D C neutral and 220 volt secondary were not
called upon to serve at the same time.

The transformers were constructed as described in the

final design except for the change previously described.

TESTING THE UNIT. 21.

The transformers were tested by what might be called
the shop method. The first half of the test consisted in
reading the watts input to the transformer with no load
on the secondary and rated voltage applied to the primary.
The current in each lead was also read and recorded,this
being the exciting current of the transformer. The input
in this case,no lead being on the secondary,may be taken
to be entirely hysteresis ani eddy current losses and are
present in the same amount as long as the transformer has
rated voltage applied to it.

The values obtained in this first half are as follows:
Iron Loss and Exciting Current.

heter#21573 6183 8392 22010 1314 410

Constant l. 1. 1. 1. l. l.
. i/ I
135 2.3 2.25 2.5 282 30 50 cycles.

To get the total iron loss add the two wattmeter
readings which gives 31? watts for this value. It must
be remembered that the iron is over ten years old and
is of the soft steel variety which was used at that
time in transformers.

Bearing the previous point in mind the 312 watts
iron losses for a 12,000 watt transformer is not at
all out of prop:rtion,it being 2.6% of the total

output.

22.
The exciting current is found by taxing the aVerage 1

value of current which flows in the three leads. This
is 2.37 mmperes. The full-load primary current being
53.8 amperes it follows that the exciting current is
4.4% of the full—load current. This value is a very reason-
able one and is to be expected in a transformer of this
capacity.

The current in the three leads of the transformer do ,
not balance but this can be expected. Even a new induction
motor upon test dn the factory will very seldom have

current balance.

The second part of the test is the determination of
the copper losses at full load.

This is done by short—circuiting the secondary with
ammeters in the leads and impressing a low voltage at
rated frequency on the primary and raising this value
until rated current flows in the secondary. The watts
input are sensibly all copper loss because the voltage
is so low that the iron losses will be negligable.

It does not mess any difference which side is the input
side on this test but it was easier to use the 820 volt
terminals as the input side and short-circuit the 135
volt leads through the amne are. This was done with the
following results:

Copper Losses.

I Io I W W;

deter # lasts BlSTB 11§8 181i 4-10~
Con st ant 13 l 10 10 10
2.7 30 2.9 40 O

** **

‘0.)

3.
In order to find the full-load copper losses add the

wattmeter readings and multiply by the constant 10 which

came in because of using 53 - 5 curr n9 transfer :rs on
the input side in order to reduce the current to a value

which could be use: in the Wattneters available.

This gives the full—load cepper loss as éCC tatts
or 3.33% of the full-load output ,which value is very
reasonable and to be expected on a small transformer such
as this one is.

It is well to remember that this transformer was
designed primarily to get results with a safe radiation per
square inch and not for the highestOperating efficiency.
The efficiency obtained is comparable to that of other

small transformers of its size found on the market.

{0
”is
o

EFFICIENCY.

The efficiency of the unit can be calculated from the
data already t ran by using standard shop methods as

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follows:
The 312 watts iron loss are pres nt in the same quantity
as long as the transformer is supplied at rated voltage
and frequenC'r and do not change appreciably with the
load applied. This ds one element of the total loss and
the other is the copper losses.
It is evident that the c0pper losses change with the
load it readily foll we that the change is preportional
to the load change;i.e. with full-load copper losses equal

urn.- ,.-—-‘

s will be 30 watts,etc. {

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(D

400 watts, % load copper lo
Manifestly the total losses at any point will be the

sum of the 312 watts iron loss and the 00pper loss for

that point. Full—load losses are 312 plus 420 or 712 watts.

The efficiency can than be calculated from the formula,

%Eff is 100(Watts load)/(Watts load)plus(Losses at that

point). Calculation be this method shows:

Load 133 Loss. Total Loss. Efficiency.
t 100 watts 412 watts 83.0
t 200 n 512 n 92.1
t 300 " 81” " 93.8
1 400 n 712 " 94.5
1: 500 n 812 " .4.9
1% 500 n 912 n‘ 95.2

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MICHIGAN AGRICULTURAL COLLEGE

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SPEED REGULATION OF THE CONVERTOR.

One of the reasons for choosing the interconnected
star connection for the transformer was because it gave
so much better speed regulation on the rotary convertor.
This was explained in detail in the general discussion.

A.P.Bocg in his t.esis on A D.C.Neutral for a Three

Phase Convertor ,gives the following comparitive data

on this point. The graphical results are also included.

Converter R1 with Interconnected

Star Transformers.

to Neutral. P.P.Y.

0 1600

10 1600

15 1605

81 1610
** **

Convertor R1 without I.S.T.
Using Simple Star Connection.

to neutral. R.P.M.
O 1600
10 1680
15 16:0
50 1670

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Bibliography.

"Electrical Engineering"

EElements of Electrical Engineering"
"Transformer Practice"

"Alternating Currents"

"Experimental Electrical Engineering“

"Transformer Design"

Berg & Upeon.
Cook.
Reed.
Kagnusaon.
Kart‘ip et 0 ff.

Still.

 

31293 0232