ANALYSIS OF INDUCTION MOTOR TORQUE

DISSERTATION

Presented in Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy in the Graduate School of the Ohio State University

By

Sakae Yamamura, B. E., M. S.

The Ohio State University

1953

Approved by;

Adviserv^ TABLE OF CONTENTS Page

Chapter I. Introduction. 1

Chapter II. The Principle of the Analyzer. 6

Chapter III. The Component Parts of the Analyzer. 12

1. Homopolar generator. 12

2. Amplifiers. 15

3* Power supplies. 20

4. Differentiating circuit. 23

5. Frequency response of the analyzer. 27

6. Cathod-ray scope. 31

7. Adjustments of the complete set. 33

Chapter IV. Results Obtained. 38

1. Three-phase induction motor; 220 V, 1/4 HP, 4 poles,

60-cycle. 38

2. Single-phase operation of the three-phase Induction

motor; 220 V, l/A HP, 4 poles, 60-cycle. 45

3. Three-phase induction motor; 220 V, 1/8 HP, 4 poles,

60—cycle. 48

4- Single-phase induction motor; 115 V, 1/6 HP, 4 poles,

60-cycle. 53

5. Shaded-pole motor; 115 V, 1/12 HP, 6 poles, 60-cycle. 58

Chapter V. Consideration of Steady States. 63

1. Three-phase induction motor. 63

2. Single-phase induction motor; 115 V, 1/6 HP, 4 poles,

60—cycle. 66

-i-

A 0 U 8 6 1 Page 3. Consideration of braking torque of the three-phase

induction motor; 220 V, 1/4 HP, 4 pOles, 60-cycle. 75

(a). Friction and windage. 79

(b). Iron loss try fundamental magnetic flux. 80

(c). Saturation of iron. 83

(d). Skin effect. 88

(e). Space harmonics of air gap magnetic flux. 92

(f). Iron loss due to zigzag leakage flux. 95

(g). Some additional considerations of the

extraordinary braking torque. 99

4- Single-phase operation of the three-phase induction

motor; 220 V, 1/4 HP, 4 poles, 60-cycle. 103

Chapter VI. Tr*isient Torque of Three-phase Induction Motors* 108

1. Differential equations of unbalanced two-phase induction

motors. 109

2. Transient phenomena of a balanced two-phase induction

motor. 112

3. Solution of the transient phenomena of the three-phase

induction motor (220 V, 1/4 HP, 4 poles, 60-cycle) by

the electronic analog computer. 126

Chapter VII. Conclusion and Acknowledgments. 133

Appendix I. Some experiments about the tooth-pulsation loss

due to zigzag leakage flux. 135

-11- Page

Appendix II. Derivation of the equivalent two-phase induction

motor for a three-phase induction motor. 140

Appendix III. Solution of transient phenomena of a balanced

three-phase induction motor by the ReeveB

Analog Computer. 14-9

Bibliography. 159

Autobiography. 160

-iii- CHAPTER I

IKTRODUCTIOli

Speed-torque characteristics sre among the most important characteristics of induction motors. They enable us to judge, when the

■peed-torque curve of the load is also known, whether a motor under consideration can start and carry expeoted loads successfully or not.

Here, we take the speed-torque characteristics to mean generally any relation between motor speed and motor torque. Usually they mean the relation between motor speed and time-average torque at steady state.

But there are many aspects of speed-torque characteristics, as we shall see in this dissertation; and not all of these have been thoroughly investigated, mainly because they are difficult to determine experi­ mentally or to compute. In order to measure torque at any given speed with a dynamometer, the induction motor under test must be kept at constant speed under a certain load. This is difficult for certain speed ranges, although not impossible. Labor and cost involved in such measurements are prohibitively high in some oases,

When attempts are made to oaloulate the torque, computations do not give accurate results for wide range of speed. This is probably due to variation of machine constants used in these computations. The machine constants are usually obtained from no load test and locked rotor test, or by calculations from formulas and design data. Because of magnetic saturation of the iron, these values are usually satisfactory for current up to or a little higher than rated current. Hence, for lower 2

speed, torque computed from these machine constants is different from

the measured torque due to magnetic saturation by larger current.

Several attemptshave been made to correct the machine con­

stants for magnetic saturation. But they are not accurate and are

often awkward to use for practical purposes.

Deep slot-oage motors or double-cate motors are of great practical

importance, but the speed-torque curves involving deep slot or double

cage effect are very difficult to oompute. It will be shown in Article

3 of Chapter V that stray load loss has a great influence on torque,

sometimes doubling or tripling the torque that would be developed

without it. But so far practioally no attempt has been made to com­

pute stray load loss.

From the above considerations we can see the need for devices which will enable us to obtain the speed-torque characteristics of in­ duction motors with more accuracy and ease. In order to meet the need, many attempts have been made.^®^^^®^ The principles underly­ ing these attempts are about the same, and the devices developed by these investigations may be called speed-torque analyzers. These analyzers trace out speed-torque curves on an oscilloscope while the motors under test are accelerated from standstill to no load speed.

Hence, speed-torque curves are determined with much less labor than by the other methods, such as the dynamometer method or computation. But in spite of their adequacy for certain purposes, these analyzers are not accurate enough and cannot indioate some of the fine details of the speed-torque curve. In the following section we shall briefly explain the main reasons why they are not satisfactory, if more details 5 of the speed-torque curve are desired. f 3 } Z. Set o'- J uses an ordinary d-c generator as a tachometer. In order to obtain instantaneous torque# the output voltage from the d-c generator is differentiated by a C-R series oirouit. But an ordinary d-c generator has commutation ripple or noise in its output voltage. Generally, when any voltage is differentiated# high frequency componente of the voltage become more pronounced, ^ence# in the torque signal the commutation noise becomes more predominant due to its high frequencies qnd masks the details of the speed-torque curves. YJhen the commutation noise is fil­ tered out# the higher harmonic components of the frequency spectrum of the speed-torque curves are also filtered out. Details of the curves are thus lost and the curves obtained do not give a complete or true picture.

H. Kondo^) uses a vacuum tube osoillator to measure instantaneous torque. The frequency of the oscillator is controlled by slight change of capacitance resulting from the twist of the motor shaft by the torque transmitted. Due to meohanic&l transient phenomena of the motor shaft# frequency change of the oscillator does not always represent the true instantaneous torque# and error is introduced thereby# especially for rapidly changing portions of the curve.

S. Chang^5^ uses a square wave generator consisting of a circular disc with evenly distributed slots on its periphery. Square wave voltage is generated photo-electrically and its frequency is proportional to instantaneous motor speed. By taking average value of the sequence of square waves by a filter# d-c voltage# which is proportional to motor speed# is obtained. But the filter# which takes the average value of the sequence of square wave voltages# suppresses the higher harmonics of the signal as well. Hence, curves obtained by Chang's analyzer are

distorted and not accurate.

In the present research a homopolar generator is used as a tachometer instead of an ordinary d-c generator. In a homopolar generator the electromotive force induced in a single oonductor is d-c while it is a-o in an ordinary d-c generator. So the commutator is not required for a homopolar generator. Thus, its output voltage does not contain any com­ mutation ripple or noise and is pure d-c voltage at any constant speed.

It mig^it seem strange that a homopolar generator has not been used as a tachometer in previous analysers, but this can be explained by the fact that it introduces a new difficulty. A homopolar generator is inherently a lew voltage device. Its output voltage is therefore too low for indicators of the type such as an electromagnetic oscillograph or a oathode-ray oscillograph, as long as the generator size remains within practical dimensions. Hence, a high-gain d-c amplifier is needed to produce enough deflection on an indicator. As is well known, a stable high-gain d-c amplifier is difficult to build. This is the reason afcy a homopolar generator has not been used successfully thus far. But proper design of the homopolar generator and the d-c ampli­ fiers, which will be explained in Chapter III, has resulted in the construction of a satisfactory speed-torque analyzer. This analyzer reveals on its cathode-ray scope such details of speed-torque curves as switching transient torque, pulsating torque of single-phase induction motors, eto.

In Chapters II and III of this dissertation the principle and the structure of the analyzer will be explained. The various interesting 5 results obtained "with it will be given in Chapter IV.

This analyzer revealed many features of speed-torque curves which had not been obtained by the dynamometer method or by the analyzer developed heretofore. Some of the curvos are quite different from the familiar one. In order to show that these new curves give a satis­ factory representation of the actual speed-torque curves, they will be compared with measured or calculated values in Chapters V and VI.

The analyzer revealed also that the braking torque of one of the three-phase induction motors tested was greatly influenced by stray load loss. The nature of the stray load loss will be clarified in

Article 3 of Chapter V. CHAPTER II

THE PRINCIPLE OF THE ANALYZER

In ordor to trace out a speed-torque curve during the acceleration

period of a motor under teat, a speed-torque analyzer must have some means of getting both the electrical quantity which is proportional to

instantaneous speed and the other electrical quantity which is propor­ tional to instantaneous torque. An electric tachometer rigidly coupled

to a motor gives voltage proportional to instantaneous motor speed.

If a motor under test carries no load except rotational inertia,

Instantaneous acceleration is proportional to instantaneous torque.

The time derivative of the output voltage of the electric tachometer is, then, proportional to instantaneous torque under the assumption of no load on the motor.

Thus, a speed-torque analyzer may consist of an eleotric tachometer to detect instantaneous motor speed, some means of differen­ tiating the tachometer output voltage for detecting torque, amplifiers to strengthen these weak signals, and a recorder to trace out the speed-torque curves. The underlying principle will now be explained.

When a motor under test carries no load except rotational inertia, the instantaneous acceleration OL is proportional to the instantaneous torque T. That is,

OC = t/J 2 .1 where J is the rotational inertia of the whole rotating part. If an electrio tachometer, which is essentially a d-c generator, with constant 7

field excitation, is rigidly coupled to the motor shaft, its induced voltage V is proportional to the rotational speed S of the motor. That

is

V = k S 2.2 lrfiere k is a oonstant.

From (2.1) and (2.2) we get

dv/dt = kds/dt ■ k tx, a (k/j)T 2.3 which shows that the time derivative dv/dt is proportional to instantan­ eous torque T.

From (2.2) and (2.3) w® know that a d-c generator coupled to the motor gives the electrical quantities necessary in obtaining the speed- torque curve of a motor under test. If a cathode-ray scope is given horizontal defleotion proportional to the generator voltage V and vertical defleotion proportional to the time derivative dv/dt, a spot on the scope screen will trace a curve relating instantaneous speed and instantaneous torque. VVhen a motor under test is started from

standstill, a spot on the cathode-ray scope will trace a speed-torque curve from zero to no load speed while the motor is being accelerated.

The fundamental principle explained above is shown sohematioally in Figure 1. This sohematio diagram is self-explanatory except for the method of obtaining dv/dt, which will now be explained. It should be understood at this point that the schematic diagram in Figure 1 is used only to explain the principle, since the C-R differentiating circuit will be replaced by a transformer differentiating circuit in the final analyzer (see Article 4 of Chapter III and Figure 8 )* 8

dV

o-J

d-c differentiating cathode-ray generator circuit oscillograph

Figure 1. Schematic diagram to show the principle of the analyzer.

The branch containing C and R in series is one form of differen­

tiating circuits. The current i in the branch satisfies the following

equation.

V ■ i J idt + Ri 2.4

If Ri idt, then (2.4) beoomoa

idt or i a cdV 2.5 dt

Thus, the -voltage drop across R is given by

V = Ri s CR dV 2.6 R dt which is a quantity proportional to torque from (2.5). Sinoe

Ri«-g-J\dt, then ■ Ri i# very small and high amplification is necessary for it or the torque signal.

If an ordinary d-c generator with a commutator is used as a tachometer, the output voltage V contains confutation ripple. Then

(2 .2 ) should be written as 9

V = kS + F(t) (2.2* )

where F(t) is a function of time t representing commutation noise. The

time derivative indicated in (2.3) now becomes

dv/dt » kOC + dF/dt « (k/j)T + dF/dt (2.3*)

As (2.21) and (2.3*) show, the output voltage V is no longer proportional

to the motor speed S, and dv/dt is no longer proportional to the

torque T, The noise term or error term dF/dt in (2.3’) may be quite

large because the differentiation magnifies the higher frequency

components in the commutation noise F(t)* If we attempt to filter out

dF/dt in (2.3*)* the higher frequency components of koc. is also

filtered out, and the actual form of the speed-torque ourve is distorted*

A homopolar generator does not have cosmutatlon noise in its out­

put voltage. If it is used as a tachometer, the problem of filtering

out the commutation noise is eliminated* But the use of a homopolar

generator presents a new problem, for Its electromotive force is too

small to produce enough defleotion on any type of indicator. As ss

shall see in Article 1 of the next chapter, a homopolar generator will have electroiaotive force of the order of 5QmV» and a cathode-ray tube may need 50V or higher voltage for deflection. Then, a d-c amplifier of at least 60 db (10* times) amplification will be needed*

Another d-o amplifier will be needed for the torque defleotion of a cathode-ray tube* This is because Vg ■ Ri of the differentiating oircuit, which is a measure of instantaneous torque, is very small*

For the C-R differentiating oircuit, the factor CR in (2*6) indicates the attenuation. For our purpose the factor CR should be about 10

10”®*, ■which means an attenuation of 60 db. Since an amplification of

60db is necessary merely to recover this attenuation, the total neces­

sary amplification for torque deflection may be of the order of 100 db.

The block diagram may now be represented as shown in Figure 2.

homopolar differentiating torque genera tor circuit j— o 100 _dk

speed deflection 60 db

Figure 2. Block diagram of an analyzer*

A stable d-c amplifier of 100 db amplification is not easy to build. V#e can reduce the necessary amplification of one of the amplifiers by modifying the block diagram in Figure 2 as that shown in Figure 3*

In Figure 3, Amplifier 1 of 60 db amplification gives the speed deflection. At the same time it gives output voltage to a differen­ tiating circuit, to which another amplification of 40 db is connected to give the torque deflection. That is. Amplifier 1 is used for both

♦The factor CR is the time constant of the differentiating circuit. The time constant should be much smaller than the period of harmonic components of the phenomenon under consideration. 11

deflections} accordingly, Amplifier 2 in Figure 3 has 60 db less amp 11-

fication than that in Figure 2.

torque Amp.l Amp • deflection — — C AO db

speed deflection

Figure 3. Block diagram of an analyzer*

The analyzer will be constructed aooording to the principle of

Figure 3, but the C-R differentiating circuit will bo replaced in the actual analyzer by a differentiating transformer because the former has more attenuation than the latter* It should be remembered that the C-R differentiating circuit was shown only to simplify the explan­ ation of the theory* CHAPTER III

THE COMPONENT PARTS OF THE ANALYZER

In this chapter the construction of the different parts of the analyzer will be disoussed*

1. Homopolar Generator.

The homopolar generator should have the following characteristics*

(i) Electromotive force should be as large as possible

(ii) Size of the generator should be small for convenience

(iii) Noise in output voltage should be as small as possible

(iv) iloment of inertia of the rotating part should be small

(v) Friction and windage loss should be as small as possible

Parts (i) and (ii) contradict each other, so a compromise is neces­ sary* Parts (iii) to (▼) depend on the design and mechanical construc­ tion of the generator*

The homopolar generator was constructed as shown in Figure 4*

The structure of the generator will now be explained*

In Figure 4, A is a cup-shaped inductor made of brass and fastened solidly on the shaft, and it rotates In an air gap of 0.074 inoh.

(The thickness of the inductor is 0.05 inch.) The exciting winding B excites the radial magnetic field in the air gap. Thus, the electro­ motive foroe is induced between the cup edge and -the shaft. The two oananonly-connected silver-carbon brushes Ci and Cg slide on the oup edge, and a carbon brush Cj slides on the shaft; these brushes form the

- 12 VjJ Sectionat X-X' Figure 4.Homopolargenerator.

~( ~( I = 1.9- — XV-I -.9 IP— Z 1 1 -- U) 14

output terminals of the generator* The two brushes forming a common

terminal on the cup edge seem to permit better contact than one brush

and thereby deorease contact noise*

The principal data for the homopolar generator are as follows* Exoiting winding

'Wire sire #27

Number of turns 8000

Resistance 350 ohms

Current 120 ma

Air gap 0.074 inoh

Output voltage 44 h i t at 1800 rpm

No load saturation eurre Figure 5

To obtain a satisfactory speed-torque curve, a rigid coupling and perfect alignment of the generator shaft and the test motor shaft are

at 1800 rpm JL 30

"O g 10

20 60 80 100 110 U O 160 180 100 Rxcitin^ current (ma)

Figure 5* No load saturation curve of the homopolar generator. 16

highly desirable* In Figure 6 the homopolar generator is on the left

and the n*o*cor under test is on the right* Note the coupling between

them* It is a kind of flexible coupling, but has negligible backlash*

Figure 6. The homopolar generator (left) and the motor under test (right)

2. Amplifiers

The amplifiers are built on the plan of Figure 3, although the C-R

differentiating circuit is finally replaced by a differentiating trans­

former, as explained in Article 4 of the present chapter*

As it was discussed with regard to Figure 3, two amplifiers are needed* The necessary amplification of oaoh amplifier may differ, but 16 they are constructed, almost identically* and the amplification of each is adjusted by potentiometers. High-gain d-c amplifiers tend to become unstable. There may be slow drift of the spot on a cathode-ray scope or there may be noise of 60 cps leaking from the a-c power supply used.

Every possible precaution should be taken to prevent these*

Each stage of the amplifiers is a bridge-balanced amplifier network^®^*^, as shown in Figure 7. The principle of the bridge- balanced amplifiers will now be briefly considered.

0 + 1 50V

input output

500

3K

O -150V

Figure 7. Bridge-balanced amplifier network. 17

In Figure 7 two triodes are in one twin-triode tube. The upper

triode amplifies the signal; the lower acts as a compensator for sup­

pressing noise and drift, ^et us first consider how a-c noise from a-e heating of the filament is being suppressed. This noise, entering both

triodes and amplified similarly* will emerge as identical voltages of

opposite signs at points A and B* and hence, theoretically, should not appear at the output terminal.

The d-o voltage of 150 V for the anodes may change according to the voltage changes in the a-c power supply to the power pack. Then, changes in magnitude of both d-c voltages will be the same, and thus no voltage change should appear at the output terminal.

By adjusting the 3 X ohm variable resistance, we can change the d-c potential of the output terminal and thus adjust the grid-bias of the next stage.

The completo connection diagram of both amplifiers is shown in

Figure 8. (it should be noted that in Figure 8 a transformer is used as a differentiating circuit in plaoe of the already-mentioned C-R differentiating circuit. The reason for this substitution will be given

In Artiolo 4 of the present chapter.) In this figure the upper half shows Amplifier 1 and the lowsr half shows Amplifier 2, both of which were mentioned with regard to Figure 3. The two amplifiers are about the same, containing three stages of the bridge-balanoed amplifier network shown in Figure 7, and are connected through a transformer dif­ ferentiating circuit.

In each amplifier, twin triodes 6SL7 are used for the first two stages, and twin triode 6SN7 is used for the third stage. These three \naJZL horizontal different defl. coil 6SN7 JW'i nrznrw— transf. +300V

10K 10K -O b +150V 100K 00K 50K

K 10K 100K 100K 1 m 10K 10QK 3K

10K vertical defl. coil 100K 100K 50K ttotp -----

100K 100 100 500 800

Fig. 8. Connection 100 diagram of the 3K

100K 50K 10K 100K.

6SL7 6SL7 6SN7 19

stages are essentially the same as the bridge-balanced amplifier network

of Figure 7. In the fourth stage of each amplifier, a 6V6 tube is

employed, which supplies current to & defleotion coil of the cathode-ray

tube used as an indicator*

The first two stages of Amplifier 1 and the first stage of Ampli­

fier 2 have a decoupling condensor of 30juf combined with a 10 K ohm

resistance, which will help to suppress parasitic oscillation, or noise

coming from the voltage regulator VR 150 in the d-c power supply.

Figure 11.

The frequency response of the amplifiers will be discussed next*

In order to obtain the frequency response of each of the amplifiers,

Amplifier 2 was disconnected from Amplifier 1 by the switch S shown on the left of Figure 3. Sinoe each of the scope defleotion coils aots as a load on the 6V6 tube, they are replaced by a 1300 ohm resistance, which is about the same as the resistance of the deflection coils. This

replacement eliminates the effect of the ooil inductance on the response*

Under these conditions the frequenoy response of each of the two ampli­ fiers was determined. The results are shown in Figure 9. The ordinate

is the amplification in db, and the amplification for d-c is set to zero db. The amplification of both amplifiers is relatively flat until

5000 cps, at which frequenoy the doorease in amplification is about 0.5 db* Aa will be seen in the next ohapter, this flatness of frequenoy response is adequate because the speed-torque curves can be expressed with sufficient accuracy without higher frequency components. But in the actual usage of the analyzer this desirable frequency response is 20 deteriorated by the presence of inductance in the defleotion coils, as will be explained in Article 5 of this chapter*

JO

Amp. 1

Amp. 2

« -2 1 3 i 7 g 9 10 Frequency (Kilocycles/eec.)

Figure 9. Frequenoy response of Amplifier 1 and Amplifier 2*

A photograph of the two amplifiers mounted on one chassis is shown in Figure 10*

3. Power Supplies*

All d-o voltages except the homopolar generator excitation are obtained by rectifier tubes 5Z3 and associated circuits, as shown in

Figure 11*

The noise frequencies of 60 and 120 ops arising from the a-e power supply are in the working frequency range of the amplifiers. 21

Figure 10. Amplifiers.

As will be seen in Chapter VI. these two frequencies are important frequency components of the speed-torque curves, since these are the main frequency components of the swltohing transient torque (60 ops) and the pulsating torque (120 ops) of single-phase induction motors.

Hence, every possible preoaution must be taken to filter out noise from the reotified voltages. For this purpose, filters with large oapaoitors and inductances are used to produce satisfactory reotifioation.

A photograph of the power supply pack is shown in Figure 12. 22

300V X 2 70m&& / — \ cO l0 li. 800 oTrcr ■ V s A r O * 115V r O 300V for 0V6 5M 16m VR 150 (hcriz. dt.fl.)

(a)

300V X 2 7Gma

115V 30OV for 6V6 VR 150 (vert, ue fl.)

25oVX 2 AOma 30 fl . 100 H . 1.3K + 150V | rogo^ - ■ — 1 — 115V fO -r O B yp 150 for Amp. 1 5S3 — |— 20m ~ p 20m “p /,0 m 4 Amp. 2 c to 150V

250VX 2 spot position adjustment 20 II. 1 K f-'TTC

^ V V \ X /~ ~ /Vv V “ 1 K 1 K 115V % I 100M U) Figure 11. Connection diagrams of the d-c power supplies, Figure 12* D-c power supply pack*

4. Differentiating Circuit

In Chapter II * differentiating circuit consisting of C-H in series was used to explain the principle of the analyzer. But there is another type of differentiating circuit* the transformer. For the

Transformer in Figure 13, the secondary voltage v is given by

v = U s.l at where i is the primary current and M is the mutual induotance between the primary and secondary colls• To show why the C-R type of differ­ entiating oircuit is replaced by the transformer type in the actual analyzer, the advantages and disadvantages of each type will now be oonp sidered. 24

Figure 13. Differentiating transformer.

One difference between the two types of differentiating circuits is noticed at once; that is, the input quantity to be differentiated by the C-R type is voltage, while in the transformer type it is ourrent.

In connection with (2.6), it was shown that the C-R type differ­ entiating circuit generally has large attenuation in itself. The total attenuation of such a differentiating cirouit placed in the analyzer will now be disoussed with numerical values.

One of the best ways to connect the C—R type is that shown in

Figure 14. The cirouit is connected to the cathode follower of 6 V 6 in

Amplifier 1. The cathode follower has the merit of giving low input

Impedance (in this case 300 ohms) to the C-R differentiating circuit and thus provides the least possibility of noise induction or feed back in the differentiating circuit.

As (2 .6 ) shows, the output voltage Vg from the differentiating circuit is CR times the derivative dv/dt. Hence, the factor CR is the attenuation factor. In figure 14, CR " 5x10"^, which corresponds 25

1 3 0 0 ohms ^ T T E r t T N -- +300V 6V6 horiz. defl. coil

input

*differenciateu ?00 voltage

Figure 14. Connection diagram of the C-R type differentiating circuit. to an attenuation of 66 db* This is the attenuation of the differen­ tiating circuit itself. ’-Then the attenuation between the output voltage from the 676 (s voltage drop across the horizontal deflection coil) and the output voltage from the differentiating cirouit is con­ sidered. the total is increased according to the attenuation in the cathode follower circuit. The additional attenuation is considered to be 300/1300, which is the ratio between the cathode follower resis­ tance (300 ohms) and the defleotion coil resistance (1300 ohms). This is an attenuation of 13 db. The overall attenuation then becomes

66 + 13 * 79 db, which is numerically more than the amplification of

Amplifier 1.

The quantity to be differentiated by a differentiating circuit of the transformer type is the current. Henoe, the connection should be that shown in Figure 15. The output voltage from the differentiating transformer is given by 26

V Q25/1

• *— *—► i • s ^ r tT ' '-TsfST n r-‘ +300V different. 1300 orirns trane• horiz. defl. coil m out

Figure 15. Connection diagram of the transformer typo differentiating circuit*

dl U dv dt 1500 dt 5.2 whore V i a voltage drop in the deflection coil of 1300 ohms approxi­ mately* The mutual inductance U of the transformer can be chosen quite large by increasing the number of secondary turns, *n the present case M - 1 henry. Then, the attenuation given by (3.2) is

1/1300 or 62 db.

As seen above, both types of differentiating circuits have large attenuation. Of the tyro types, the transformer type has less attenu­ ation (62 db compared with 79 db in the C-R type), and thus was adopted, as shown in Figure 8 (see the differentiating transformer in the anode circuit of 6V6 of Amplifier 1)* 27

5* Frequency He3ponse of the Analyzer*

In Article 2* the frequency response of each of the two amplifiers

was discussed* and the response curves are as shown in Figure 9* It

should be remembered that the response curves were obtained under the

oondition that the deflection coils in the anode circuit of 6V6 were

replaoed by 1300 ohm resistance* and the frequency response was deter*

mined by measuring the voltage drop In the 1300 ohm resistance*

How the frequency response of the amplifiers coupled with the

oathode-ray scope will be considered. Amplifier 2 was disconnected from

the differentiating transformer byssitch S in Figure 8 and the measure*

raent was performed separately for each amplifier* Constant voltage of various frequencies was applied to the input terminals of each amplifier*

and the deflection on the soope screen was read for various frequencies*

The results are shown in Figure 16(a) and (b). In the figure the

ordinate is the deflection in db* the d*c deflection being set to zero

db* and the abscissa is frequency*

Compared with Figure 9* the frequency response of each of the

amplifiers is deteriorated. This is because of the presence of induotance

in the deflection coils* which were replaoed by pure resistance of 1300 ohms for Figure 9. Amplifier 1 has a poorer response due to a larger inductance in the horizontal deflection coil* which is about one henry while the inductance in the vertical deflection coil is 0.7 henry.

It is desirable to discuss next the frequency response of the complete set* including the two amplifiers and the transformer differ­ entiating circuit between them. A constant voltage of various frequen- Relative aaplification (db) Relative amplification (db) -10 £ - -10 Figure 16 (b) 16 response of 2.FigureAmplifier Frequency . -6 -2 -6 iue1 a. Frequencyof1. response Amplifier (a). Fifure16 0 0 50 5070 60 60 7 60 (cycle/sec.) y c n e u q e r F rqec (cycles/aec.)Frequency 8 9 100 9 100 a. without condenserin withoutcoil. parallelwith a. deflection . ih0.25^f indeflectioncoil. parallel with ^ 5 2 . withb. 0 m. 2 Amp. 200 200 300 £00 5 £00 300 200 0 £00 50 7 9 1000 9 8 7 6 500 0 0 £ 300 6 7 8 ? 1000? 28 29

cies was applied to the input terminals of Amplifier 1, and the vertical

deflection on the acope screen was measured. In Figure 17 the relation

between frequency and vertical deflection Is plotted with vertical

defleotion at 100 cps as the base. A continuous straight line (a) would be the result, if the two amplifiers had a perfect flat frequency

response and the differentiating cirouit worked perfectly. But due to the frequency responses in Figure 16(a) and (b), which are not perfectly

flat but bent downward, the actual results are the curves (b) (c) of

Figure 17, which are also bent downward. Curve (b) is the case with a

6

5

4

3

a theoretical line b with .2 5 /if in parallel with the hoW^ontal deflection .5 coil. C without ,25yUf. >

1 * --- L- 100 120 150 200 300 4.00 500 600 700 800 1000 Frequency (cycle®/sec.)

Figure 17. Frequency response of the complete set* 50

0,Z5ytu f capacitor in parallel with the horizontal deflection coil, as indicated in Figure 15* and curve (c) is the case without the 0.25/t- f capacitor. This capacitor gives high frequency peaking to the vertical deflection, which will now be explained.

Let current i in the horizontal deflection coil be expressed in terms of a Fourier series, such as

i * I0 I sin(nwt+fl) 3.5 a

Then the secondary voltage v of the differentiating transformer is given by oO v ■ - M 21 cos (-n.u>t+fl) 3.4 dt =

Comparing (3.3) with (3.4), we notice at once that in (3.4) the higher harmonics becomes more conspicuous because of the factor "n" in each term. This means that the higher harmonies are of more relative importance in (3.4) than, in (5.3). Hence, it follows that the harmonics in (3.3), higher than a certain frequency, may be suppressed without causing appreciable distortion of the speed deflection. But these higher harmonics must be retained in (3.4) in order not to cause appreciable distortion of the torque deflection, which is produced by the voltage v in (3.4). This can be done by by-passing the horizontal deflection coil with a 0.25 yu. t capacitor as shown in Figure 15 by the broken line. Then the frequency response of Amplifier 1 is deteriorated slightly as shown in Figure 16(a), but the overall frequency response of the complete set now becomes curve (b) in Figure 17, which is 31 preferable to curve Co). This high frequency peaking is adopted in the actual analyzer, as shown in Figure 8 (300 0.25 ju f capacitor in parallel with the horizontal deflection coil).

The by-passing condensor has an additional function. The vertical and horizontal deflection coils are wound next to each other around the neck of the cathode-ray tube and have a mutual electrical coupling.

Through this mutual coupling feed-back occurs between the input and output terminals of Amplifier 2, which may lead to instability in the amplifier. The condenser of 0.25 f tends to suppress this feedback.

6 . Cathode-ray soope.

A cathode-ray scope is used as an indicator which traces out the speed-torque curves on its screen. In order to have a better visual observation, a cathode-ray tube of long persistence is desirable.

The cathode-ray tube 7BP7 used in the analyzer is of this type. With

4000 V for the anode voltage, the image persists from 0.5 to 2 minutes, depending on the electron beam intensity and the darkness of the room.

The tube i s focused and deflected magnetically. Conventional deflection coils for TV sets cannot be used because their current sensibility is not sufficient. Suitable coils were wound specially with large number of turns of fine wire. In order to adjust the spot position on the scope screen, position adjusting coils must also be provided.

Data for the coils are* 32

Horizontal deflection coils

.Tire size £35

Number of turns 2500 x 2

Resistance 575 x 2 ohms

Inductance 1 henry (total)

Current sensibility 4 mrry'mA

Vertical deflection coils

Wire size £35

Number of turns 2500 x 2

Resistance 575 x 2 ohms

Inductance 0.71 henry (total)

Current sensibility 5*5 mn/ma.

Position adjusting coils

For horizontal adjustment

Wire size £35

Number of turns 3000 x 2

Resistance 875 x 2 ohms

Current sensibility 15 mm/ma

For vertical adjustment

V/ire size £35

Number of turns 3000 x 2

Resistance 875 x 2 ohms

Current sensibility 11.3 mcy'ma

These are all oonoentrated coil typesand are placed around the neck of the tube in the following order, beginning with the one next to the tube; the horizontal deflection coils, the vertical deflection 33

coils, the horizontal position adjusting coils and the vertical position

adjusting coils*

The connection diagram of the cathode-ray tube 7BP7 is shown in

Figure 18* Brightness of the spot is adjusted by the grid-bias control

as shown in the figure. The connection diagram of the position adjust­

ing coils is shown in Figure 11 (d).

The focusing current is supplied b; the 6AG7 tube and adjusted by

grid-bias control as shown in Figure 19*

Data for the focusing coil arei

TO.re size ^35

Number of turns 17,000

Resistance 3 K ohms

Current 30 mA

Figure 20 shows the external view of the cathode-ray scope.

Figure 21 shows the entire view of the analyzer*

7* Adjustments of the Complete Set*

Having explained the individual components of the analyzer, we

shall now oonsider the adjustments of the complete sot*

The exciting current of the homopolar generator should be large enough so that the generator is slightly saturated. This will produce nearly the largest possible output voltage, and fluctuation of output voltage caused by fluctuation of the exciting current is suppressed by the saturation* The saturation curve in Figure 5 gives the suitable exciting current to be about 120 ma* 34

6KV d-c 7BP7

4500V O

10M 180K 115V a (6watts)

150K 100K

Figure 18. Connection diagram of the cathode-ray tube 7BP7.

500V d-c

60K

lOmH 100K 115V

9 5Z3 100K 2K

>S77/S77//S >*/>///

Figure 19* Connection diagram of the focusing coil circuit. Figure 20. External view of the cathode-ray scope.

Figure 21. Entire view of the analyzer. 36

The amplifiers should be adjusted so that; they have linear ampli­

fication and can display linear deflection over the entire screen of the

eathode-ray scope. This adjustment can be made in the following way.

In Figure 8 disconnect Amplifier 2 from Amplifier 1 with the

selector switch S located just after the differentiating transformer■

Apply a small a-c voltage of 60 cps to the input terminal of Amplifier

lj then the spot on the screen vri.ll display only horizontal deflection.

Adjust the a—c input voltage with the input potentiometer of 1 K ohms

indicated at the upper left co m e r of Figure 8 * so that the horizontal

defleotion is arbitrarily adjusted for a half inch in length. When the

3 K ohm variable resistance of the first 6SL7 is changed, the position

of the half inch oscillating deflection will be changed along a hori­

zontal line on the scope. Move the half-inch oscillating deflection

over the horizontal span of the screen and see whether its amplitude

of one-half inch remains the same. If the amplitude does not change,

the linearity of Amplifier 1 is good. If the amplitude changes from

the value of one-half inch, adjustment of the other two 3 K ohm variable

resistances of Amplifier 1 is necessary. Change the setting of the

last two 3 K ohm potentiometers arbitrarily. Repeat the above test

for the various settings until the amplitude of one-half inch of the

oscillating deflection remains the same over the horizontal span of the

screen when the position of the oscillating defleotion Is changed by the first 3 K ohm variable resistance*

The linearity of Amplifier 2 can be similarly established.

After the linearity of each of the two amplifiers has been adjusted, the zero position of the spot should be changed not by the 3 K ohm 37 variable resistance as above, but by the position adjusting potentio­ meters shown in Figure 11 (d).

The amplification of Amplifier 1 can be adjusted by the 250 K ohm potentiometer located between 6SN7 and 6V 6 ; and that of Amplifier 2 can be adjusted by the 100 K ohm potentiometer located between the second

6SL7 and 6SN7.

The input voltage to Amplifier 1 can be adjusted by the 1 K poten­ tiometer at the input terminals located at the extreme left of the upper half of Figure 8 . The input voltage to Amplifier 2 oan be adjusted by the 1O0 K ohm potentiometer at the input terminals located at the extreme left of the lower half of Figure 8# CHAPTER IV

RESULTS OBTAINED

The speed—torque curves of* various types of induction motors were

observed and recorded photographically with the analyzer. In this chapter the results obtained will be explained*

1 . Three-phase Induction ilotors 220 V, 1/4 HP, 4 poles, 60-cycle, stator slots/rotor slots ■ 3 6 / 4 8 ,

The motor was directly coupled to the homopolar generator and started. Then the spot of the cathode-ray scope traced out the curve for the acceleration period from standstill to no load speed (from left to right). This curve is shown in Figure 22. The acceleration period is about l/4 second*

In Figure 22 the horizontal deflection is speed, and the vertical deflection is torque• This convention is followed in all following figures* The curve is quite different from the familiar form of a peed- torque curves and has pronounced ripple on the first or left half of the curve. If we take the moan value of the two envelopes of the ripple and trace it, w o may get the familiar curve obtained by, say, a dynam­ ometer. Hence, we may say that- a damped oscillation is superimposed on the familiar speed-torque ourve, This damped oscillation is the transient torque due to electrical transient following switoh-in and has a frequency of 60 eps*

The form of the apeed-torque curve is the same as long as all three phases are closed simultaneously, indicating that the switching

- 38 - 39

I .

Figure 22. Unlocked case; three-p ha a© motor* 220 V, 1/4 HP, 4 poles, 50—cycle.

transient torque is independent of the switching instant. This fact will be demonstrated mathematically in Article 2 of Chapter VI, and will be demonstrated again by an electronic analog computer in Article 3 of Chapter VI.

It can be seen from Figure 22 that after 8 or 9 cycles, the damped oscillation dies out, and the curve takes on the familiar form, although it still oontains small ripple. The spot returns to the same level as at the starting point at no load speed and remains there. The horizontal distance of this final point from the starting point represents the no load speed* 40

Th© fact that th© damped oscillation is th© switching transient

torque is demonstrated by the following simple experiment. First lock

th© rotor and then close the three-phase supply switch. Release the

rotor to start. The time interval between the switch-in and the release

of the rotor must be long enough for the switching transient to die out.

The time interval needed is usually less than one second. The picture

of the released case is shown in Figure 23. This has the familiar form

of the a peed-torque curves similar to those taken by a dynamometer. The

first vertical rise at the left end represents the starting torque of the motor.

Here it should be pointed out that the vertical rise corresponding to the starting torque is quite sharp. Ihis shows that the frecpenoy

response of the set, as discussed in Chapter III, is satisfactory because the vertical rise can be only sharp when a sufficient number of higher harmonics is included in the curve.

To show the relation between the eurve of the unlocked case and that of the released oase, the two curves were photographed together as shown in Figure 24. To take this picture, the curve of the released case was t raoed out first, and the other was traced out iimediately afterward. Then the shutter of the camera was kept open and the com­ bined picture of Figure 24 was taken from the persisting images. This picture shows that the curve of the released ease traces the mean values

e This same experiment will be performed many times. This will be referred to as "the released case." The oase ahen the rotor is not looked will be referred to as "the unlocked case." 41

Figure 23. Released case. Three-phase motor; 220 V, 1/4 HP, 4 poles, 60-oycle.

of the two envelopes or the unlocked oase. This fact has practical value; that is, if the mean values of the two envelopes of the unlocked oase are plotted, then the speed-torque curve under the steady state is obtained.*

Next the measurement will be extended to the negative speed range.

The motor is started in the negative direction; and after reaching the no load speed, it is plugged by interchanging two of the three-phase

a In Article 1 of the next ohapter it will be shown that the curve of the released case is the same as the speed-torque curve under the steady state. 42

Figure 24. Unlocked case and released case* Three-phase motor* 220 V, l/4 HP, 4 poles* 60-cycle*

leads. The pioture taken in this ease is shown in Figure 2 5. The white spot in the middle of the lower half is the zero or initial point from which the spot traoes first the speed—torque curve as in Figure 2 2 in the third quadrant. This curve is upside down due to aooeleration in the negative direction. At point A* whioh corresponds to no lead speed in the negative direction* the switch is opened and then the spot rises slightly upward due to deceleration. At point B the switch is closed again with two phases Interchanged, and a transient torque similar to the 43

Figure 25. Plugged case; Three-phase motor, 220 V, l/4 HP, 4 poles, 60-cycle.

one at starting, which is shown in Figure 22, appears. But they are different in two respects. One difference is that the transient torque at plugging is damped more rapidly than that at starting. Another dif­ ference is that the plugging transient depends on the switching instant, as indicated in Figure 26 where a larger plugging transient torque is shown.

Refer to Figure 25 again. As the spot approaches the first quad­ rant, the torque decreases rapidly; and after entering the first quadrant, the spot traoes out exaotly the same c u m as Figure 23.

* W e , we have another way to obtain the steady-state speed-torqu e curve without switching transient torque for slip 1 to 0. The ordinate of the point of intersection between the curve and the ordinate axis 44

Figure 26* Plugged ease; Three-phase motor, 220 V, l/4 HP, 4 poles, 60—cycle*

gives the starting torque*

■tfhen we compare portions of the curve in the first and the second quadrants, we notice at once that the torque in the second quadrant is much higher than the maximum torque in the first quadrant* The highest average torque (mean value of the two envelopes) is about 1*7 times the ffls-rimurn in the first quadrant* in the second quadrant the direction of rotation and that of the torque are opposite. h«nce, the torque in this negative speed range is braking torque. The explanation of this unusually high braking torque will be given in Article 5 of Chapter V* 45

2. Single-phase operation of the Three-phase --lotor, 220 V, 1/4 HP, 4 poles, 60-cycle*

Single-phase operation of the same three-phase motor shows peculiar

phenomena, which will now be explained, ^f oourse the motor cannot be

started as a single-phase motor. In Figure 27 the motor was brought up

to no load speed (point A) as a three-phase motor by releasing the

locked rotor (released case). The upper curve is the speed-torque

Figure 27. Single-phase operation of the three-phase motor, 220 V, 1/4 HP, 4 poles, 60-oycle*

curve obtained in this oase* When the three-phase supply switch was opened, the motor was decelerated from A to B along the thick horizontal line in the figure* When the motor speed reached point B, single-phase of the three-phase voltage was applied* Then the motor was accelerated again to point A. The speed-torque curve obtained in this case has 46

pronounced ripple all along the curve. This is the oscillating torque

of 120 cps inherent to any unbalanced induction motor, including the

plain single-phase induction motor. The existence of the osoill&ting torque was known but it seems that it has not been previously indicated on the speed-torque curve.

The same procedure was repeated to obtain Figure 28, except that before single-phase voltage was applied, the motor speed reached point C,

Figure 28. Single-phase operation of the three— phase motor, 220 V l/4 HP, 4 poles, 60-cyole. . which wasslightly lower than speed B in Figure 27, Then an unexpected phenomenon happened; the motor was decelerated from C to 0 and c ame to standstill quickly. The speed-torque curve CO in Figure 28 was obtained in this experiment, It has an oscillating torque of 120 ops, and 47

its average value ia negative. This ia peculiar because it seems con­ trary to the ordinary theory that single-phase induction motors develop positive average torque as long as their speed is not zero.

The above two curves of single-phase operation in Figures 2 7 and 28 were combined into one picture, as shown in Figure 29. From this figure the average value for single-phase operation was taken and plotted in Figure 30. The average value is negative for slip * 1 to 0.5. This peculiar phenomenon will be explained in Article 3 of Chapter v.

Figure 29. Single-phase operation of the three-phase motor, 220 V, 1/4 HP, 4 poles, 60—oycle. 48

n 600

Three-phase running

200 •b ingle— phase cr running

-200 .8 slip

Figure 30. Speed-torque curres of the three-phase motor, 220 V, 1/4 HP, 4 poles, 60-cycle.

3, Three-phase induction motor; 220 V, l/8 HP, 4 poles, 60-oycle.

Two rotors are available and they can be interchanged for the same

stator. It appears that both of these rotors have deep slots. The number of scator slots is 27; that of one rotor is 17 and that of the

other rotor is 37.

First the rotor with 17 slots will be used. Figure 31 Is the speed-torque curve of the unlocked case. The large damped oscillation on the curve is the switching transient of 60 ops. After about 3 cycles, the transient torque disappears. Acceleration is quite rapid; the acceleration period being about 5 cycles (l/l2 sec.).

Figure 32 shows the speed-torque ourve of the released case.

There is no switching transient torque in this curve, which has the familiar form of the steady-state speed-torque curve* 4 9

Figure 31* Unlooked case; Three-phase motor* 220 V, 1/ 8 HP, 4 poles* 60-oyole* stator slots/rotor slots • 27/17,

In Figure 33 the motor was plugged and the speed-torque ourve was obtained for slip 2 to 0, The white spot in the middle of the lower half Is the initial point (slip » l). At the left end the motor was plugged. The left half of the ourve is the negative speed range

(slip ■ 2 to 1 )* and the right half is the positive speed range (slip ■»

1 to 0), The pronounced ripple at the left end is the switching transient torque due to plugging. The right half of the ourve is the same as that given in Figure 32, 50

Figura 52, Beleasad ease; Three-phase motor# 220 V, l/8 HP, 4 poles, GO-oyole# stator slots/rotor slots = 27/17,

The same experiments were repeated with the same 27-slot stator and the 17-slot rotor replaced by the 37-slot rotor. The following three pictures were taken. Figure 34 shows the unlocked case, Figure 35 the released oase, and Figure 36 the plugged oase. That is, they correspond to Figure 31, 32, and 33, respectively, where the 17-slot rotor was used*

The general feature of the corresponding figures are about the same. They show the high starting torque inherent to the deep-slot- oage-rotor. But there are some differences. Figure 35 has a flatter curve near the starting point than Figure 32, indicating that Figure 35 51

* !

Figure 53. Plugged oasej Three-phase motor, 220 V, l/8 HP, 4 polos, 60-eyole, stator slots/rotor slots * 27/l7.

has a larger relative starting torque than Figure 32. Comparing

Figure 36 with Figure 33, we notice that the braking torque is higher in Figure 36 than in Figure 33. This seems to indicate that a larger number of rotor slots produces a larger braking torque for the same stator, and this tendency can be recognized also in the experiment performed by Moller^8\ 52

\

Figure 3-4• Unlocked case; Three-phase motor, 220 V, 1/8 HP, U poles, 60-cycle, stator slots/rotor slots = 27/37

i f t ! ‘s \V \

< ' X

Figure 35* Locked case; Three-phase motor, 220 V, 1/8 HP, A. poles, 60-cycle, Bt&tor slots/rotor slots = 27/37. 55

Figure 36. Plugged case; Three-phase motor, 220 V, l/8 HP, 4 poles, 60-oyole, stator slots/rotor slots ■ 27/37*

4. Single-phase Induction Motor* 115 V, l/6 HP, 4 poles, 60-oycle,

This is a split-phase, resistance starting, single-phase induction motor•

When it is started without first locking the rotor, the speed- torque c u m obtained is as shown in Figure 37. This curve is peculiar in having pronounced ripple all along the curve.

There are two discontinuities in the ripple, one at slip = 0.75 and the other at slip = 0.24. The curve is divided into three portions by the two discontinuities. The ripple in the left portion of the curve dies out at the first discontinuity. Then the middle ripple 54 begin* to build up and continues to increase as the motor speed Increases towards the right. At the discontinuity of slip =. 0,24, the amplitude of the ripple is suddenly reduced to about one-half. In the right portion of the curve the amplitude of the ripple increases for a while and then deorease* to the low value at the no load speed.

Figure 37, Unlocked oaset Single-phase motor* 115 V, 1/6 HP* 4 poles, 60-oyole (Switoh was closed at the instant of sero volt.)

The ripple in the left portion is the switching transient torque and has the frequency of 60 ops. The ripple in the middle and right portion* of the ourve is the oscillating torque of the single-phase induotion 55 motor and has the frequency of 120 cps.* The discontinuity at slip

= 0.24 is caused by the opening of starting winding by the centrifugal switch.

The ordinary theory of single-phase induction motors recognizes the existence of such torque ripple with double frequency, which is superimposed on the average torque. But this may be the first time that it has been measured and visualized as shown in Figure 37.

The magnitude of the switching transient ripple in the left portion of the curve depends on the switching instant relative to the supply voltage wave. This phenomenon was observed, using a synchron­ izing switch whiah closes the motor circuit at any chosen instant on the voltage wave. For Figure 37 the switch was closed at the instant when the voltage wave was going through zero. This case gives the maximum switching transient torque ripple. For Figure 38 the switch was closed at the instant of about 0.7 maximum voltage. The switohing transient ripple in this figure is smaller than that in Figure 37.

For Figure 39 the switch was closed at the instant of maximum voltage.

In this figure there is no switohing transient ripple of 60 cycles/ sec., but the ripple of 120 cycles/sec. builds up gradually, starting from almost zero amplitude.

The above relation between the switching instant and the switohing transient torque is exactly the same as that between switching instant

*The difference between the two frequencies is not so easy to recognize because the abscissa of the ourve is not time but motor speed. However, when the left and the right portions are compared, we can easily see the difference between the two frequencies. Figure 38. Unlocked casej Single-phase motor, 115 V, 1/6 HP, 4 poles, 60-cycle, (Switch was closed at the instant of about 0.7 maximum voltage.)

and transient current for an inductive circuit with small reslatanoe.

It should be noticed that the maximum switching torque occurs several cycles after switch—in for the single-phase motor, while it occurs in the first half cycle for the three-phase motors. (Compare

Figure 37 with Figure 22.)

When the motor was started by releasing from the locked condition, the same curve as Figure 39 was obtained.

By taking the mean value of the two envelopes of the curve in

Figure 39, the speed-torque curve in Figure 40 is obtained. This ia the speed-average torque curve of the motor* The points in the figure are the values determined with a dynamometer* Figure 39* Unlocked case when theswitchcase when Unlocked closed atthe was 39*Figure Figure 4.0. Speed—average torquecurvethe4.0. single—phase Speed—average of Figure Torque(synch, watts) 200 1 cycle. 115Single-phase,volts,1/6 60- 4poles, Hp, instant of maximum voltage*or case*released induction motor* 1/660-cycle.115V, 4induction poles* Hp, .8 slip 0

58

5. Shaded-pole l^otor; 115 V# l/l2 HP# 6 poles# 60-oyole*

A shaded-pole motor has oharaoteristies similar to those of a split-

phase single-phase induction motor* This is because a shaded-pole motor

is a kind of split-phase single-phase induction motor. One difference between the two motors is that the auxiliary winding (shading coil) of the shaded-pole motor is inductively connected to the power supply# while that of the split-phase motor is connected conductively. The speed- torque curve of a shaded-pole motor has, therefore# the oscillating torque of 120 cps inherent in any kind of single-phase induction motor# as shown in Figure 41*

*

•/ v* j »V

i _ A

Figure 41* Unlocked case; Shade-pole motor# 115 V# l/l2 HP# 6 poles# 60-oyole# C9

Figure 41 is the curve for the unlocked case. This is similar to

Figures 37 and 38 of the split-phase single-phase motor, but there are several differences*

First, the switohing transient torque of 60 cps in Figure 41 is small and is difficult to recognise.

Second, there is no discontinuity on the curve, which results from from the opening of the auxiliary winding, and the oscillating torque remains quite large at no load speed. This is because the auxiliary winding (shading coil) remains connected.

Third, there is another smaller ripple near the upper envelope of the oscillating torque in Figure 41. Bach high peak of the oscillating torque is sharply cut down by the higher harmonic. As seen from the figure, frequency of the harmonic is twice that of the oscillating torque, that is, 240 ops*

When the magnetic bridges between the poles are taken out, there is no recognisable harmonic of 240 cps near the upper envelope, as shown in Figure 42* It seems that this harmonic of torque comes from saturation of the magnetic bridges*

When several harmonics exist in the speed-torque curves as in

Figure 41, we must be oareful in taking the average torque from the curve. The reason will now be explained*

e If the auxiliary winding of the split-phase motor remains con­ nected, the oscillating torque at no load speed is larger, and the no load speed 1s reduced* Figure 42* Unlocked case* without; magnetic bridges between poles* Shaded pole motor, 115 V, l/l2 HP, 6 poles, 60—cycle#

Figure 43 was taken with 0.1 f capacitor connected to the control-

grid of 6SN7 of Amplifier 1, Figure 8 , to suppress the higher harmonics*

It oan be seen from Figure 43 that there is no appreciable harmonic of

240 cps, while the ripple of 120 cps still remains* The mean values

of the two envelopes of Figures 41 and 43 will now be compared* To make comparison easier, both curves are duplicated together in Figure 44.

In this figure the full line indicates the mean value of the two

envelopes belonging to Figure 41, shils the broken line indicates the

mAftn value for Figure 43* The two mean value lines do not ooinoide* 61

Figure 43. Unlooked oase. the osoillating torque Is suppressed •with O . l ^ f in the amplifierj Shaded-pole motor* 115 V, 1/12 IIP, 6 poles, 60-cycle*

The mean-value broken line for Figure 43 ia higher than the mean-value full line for Figure 41. As it will be explained in Article 2 of

Chapter V, existenoe of even hamonics causes discrepancy between the average (with respect to time) torque line and the mean value line of the two envelopes. In Figure 44 the mean-value broken line belonging to Figure 43 gives the true average torque line*

^ h e lowest frequency of the torque ripples is considered fundamental frequency* Figure 44. Combination of Figure# 41 and 43 CHAPTER V

CONSIDERATION OF STEADY STATES

The speed-torque curves taken with the analyzer were presented In

Chapter IV. Attempt will now be made to show that these curves are satisfactory representations of the actual conditions. This attempt will be divided into two parts * consideration of steady states, and consid­ eration of transient states. The former will be treated In this chapter, and the latter will be treated in the next chapter.

It might be helpful at this point to explain briefly what are meant by steady states and transient states. Two kinds of transients are involved* one is electrical transient and the other mechanical transient. *n this dissertation electrical transient is meant by trans­ ient state, electrioal steady state is meant by steady state, and mechanical transient will be designated as mechanical transient.

The speed-torque curves are taken with the analyzer under the mechanical transient state, because motor speed is varying continually while the curves are being traced by the analyzer. But after the switohing electrical transient is over, the motor produces even under the mechanical transient state the same torque as under steady speed.

This is apparently a kind of electrical steady state, which is also meant by steady state in this dissertation.

1. Comparison of the Analyzer Curves with Dynamometer Values for Three-phase Induction Motors.

First, the speed-torque curves for the three-phase Induction motor,

220 V, l/4 HP, 4 poles, 60-cyole, will be compared with the values - 65 - 64

obtained with a dynamometer.

The torque determined with a dynamometer is given in Table 1,

Table 1

Slip .044 .1 -193 .355 .722 .889 1.0

Torque .75 1.5 2.075 2.2 (lb-ft) T oroue 191 385 530 561 (syn. watt)

Slip 1.194 1.361 1*5 1.945

Torque 2.9 3.6 3.7 3-75 (lb-ft) Torque 740 918 944 956 (syn. watt)

These values are shown as * points in Figures 23 and 25. In order

to plot these values in the figures, the speed soale and the torque

scale are necessary* The speed scale is immediately given by the

figures because the horizontal distance between the initial point and

the final point corresponds to the no load speed. The torque scale

is obtained by placing one arbitrary torque value in Table 1 on the

curve* In Figures 23 and 25 torque * 385 synchronous watts at slip =

0.1 was chosen, to determine the torque soale on the analyzer curves*

Other values are plotted, using the scales obtained above. The agree­

ment between the curves and the measured values is good. This agreement

shows two thingsi one is that we are using the eorreot scales and the

other is that the speed-torque curves are good representations of the 65

actual conditions.

In Figure 2 3 the measured starting torque is shown by two * points

at the left end. The starting torque changes between these two limits

due^ *to cogging (9) ,

In Figure 2 4 the curve of the unlocked case and that of released

case are superinposed. From this figure we see that the curve of the

released case is the mean ourve of the two envelopes of the unlocked case. lienee, we may conclude that the mean curve of the unlocked case gives the steady-state speed-torque ourve.

In Figure 2 5 the ripple on the left portion of the ourve is the transient ripple caused by plugging. A line drawn through the measured values by the dynamometer method for this portion shows the mean values of the two envelopes of the ripple.

N o w the analyzer curves of another three-phase induction motor,

220 V, l/8 HP, 4 poles, 60-cycle, stator slots/rotor slots = 27/37, will be compared with the values obtained with a dynamometer. The torques determined with a dynamometer are given in Table 2.

Table 2

Slip .028 .07A .1 .31 .597 1.0

Torque .IA .A17 .587 1.18 1.A4 1.31-1.63 (lb— ft) Torque 35.7 106 1A9 300 365 33A-A15 (syn. watt) 66

These values are shown as *points in Figure 35. The "calibrating point," which determines the torque scale, is given by torque = 300 synch, watts at slip * 0.31 as indioated in the figure. The agreement is good*

The acceleration of this motor is quite rapid. As can be seen from

Figure 34, the motor reaches nearly no load speed in l/l2 sec. (5 cycles).

The agreement in Figure 35 between the curve and the dynamometer values indicate that even such rapid acceleration does not affect the steady- state torque at each speed.

2. Single—phase Induction J.lotor, 115 V, 1/6 HP, 4 poles, 60—cycle*

As was explained in Article 3 of Chapter IV, Figure 39 shows the instantaneous speed-torque curve where the motor was started by releasing the rotor from locked condition or the switch was closed at the instant of maximum voltage. The average torque* and the oscillating torque will be checked separately. First the average torque will be compared with the values measured by a dynamometer*

The average torque determined with a dynamometer is given in Table 3*

Table 3

Main winding only Both winding

Slip .044 .056 .13 .18 .29 .35 .52 1.0

Torque .393 .59 .88 .95 1.25 1.3 1.12 .87 (lb-ft)

Torque 100 150 220 240 320 333 295 223 (syn. watt)

*"The average torque" means the time-average torque under steady state. A dynamometer gives the time-average torque under steady state. "The mean value" will be used for the mean value of the two envelopes of the speed-torque curve* 67

Figure 40 is the mean value line of the two envelopes of Figure 39, as explained in Article 3 of Chapter IV* In Figure 40 the * points indicate the measured torque in Table 3. The "calibrating point," which determines the torque scale, is given by torque 295 synch, watts at slip - 0.52. The measured torque in Table 3 is also indicated as * points in Figure 39. The agreement is fairly good. From the above comparison

.vo can conclude that the mean line of the two envelopes of the released case. Figure 3 9, gives the average torque or dynamometer torque values.

Next, the osoiHating torque in Figure 39 will be considered.

There seems to be no other simple way to measure the oscillating torque of 120 cps. So it will be computed for comparison with the oscillating torque in Figure 39.

In the double-revolving field theory of the capacitor motor by

-V. J. Morill^®), the average torque and the oscillating torque are given by the expressions!

2 2 2 Average torque ■ ® (^f ~ ^-b^ + 2 & +■ 5.1 Oscillating torque »lJj(Rf - Rb)cos2«> t - ljj(Xf - Xb)sin2<*>t

+ Ig a (Rf - Rb) cob2 (us t + ^ ) - Ig a2(Xf - Xb) sin2(a> t ) 5-2

These equations can be applied without alteration to the present

3ingle-phase motor, if the impedance of the condenser Is set equal to zero.

The machine constants of the l/6 HP single-phase motor were obtained from tests, following the procedure described by Puohstein and

Lloyd^*^ as shown below. 68

(a) Locked rotor test Main winding

VL = 39.2 V 1^ = 2.6 A Wl = 52 watts

Starting winding

VL = 23 V IL = 1.8 A WL = 34 watts

(b) No load test Main winding

Vn = 115 V In = 1.73 A Wn = LH watte

(c) D-c resistance

Main winding = 3.1 ohms

Starting winding 8 .1 ohms

From the above data we obtaine the following.

For the main winding

ZeM= VL/IL = 39.2/2.6 =15.2 ohms

ReM = Wl A l = 52/(2.6)2 = 7.7 ohms

2M — Rgjj — R-]m — 7.7 — 3*1 — 4-»6 ohms

xeM eM -~ R2neMu = 13-1 ohms

Xtu — X©m/ 2 - 6.55 ohms X0- 2Vn/ln= 133 ohms

xo - xeM 133 - 13.1 = 0.901 133

2 - * r 2 - 0.901 amp. mag. main = 1.73 = 0.9!

Xo= Tn^«ag. main = ^5/0.95 = 121 ohms jr _ *o - XeM — 121 ~ 13*1 = 0.89 r *o 121

R = _Rag - R1H = 7 -7 - 3-1 = 5 -2 ^ Kr 0.89

Ijjl ■* X2j|“ xeM^2 ~ 6.55 ohms 69

Xm = XQ - Xm = 121 - 6.55 = 1U.5 ohms (magnetizing reactance)

For the starting winding

ZeS = VL/lL= 23/1.8 =12.8 ohms

ReS “ Wl/ tL “ 34/(1.8)2 -10.5 ohms

*eS =/z2s - Ris = 7.35 ohms

a2 = X e s A e M = 7.35/13-1 = O .56

a = 0.748

Summarizing the above, we obtain the machine constants, which will be used for computing equations (5.1) and (5*2) as below.

R1M = 3.1 ohms Xd i = 6.55 ohms

a^ R-jp. — 8.1 ohms (actualresit tanoe of the staring winding)

a 2 x^s ~ Xeg/2 = 3-68 ohms (actual reactance of the starting winding)

The rotor machine constants and the magnetizing reactance for the double revolving field theory are half the corresponding values for

the cross field theory and are given as below.

Xjjj = Xjjj/2 = 114.5/2 = 57.2 ohms

r2 “ r 2m /2 = 5.2/2 = 2.6 ohms

x2 = X2m /2 = 6.55/2 = 3.28 ohms

We shall now compute for slip =1.

(rz/e) R f = Rb~ 2 / To ~ 2 “^2 ohms (t 2/ b)^ + (x2 + Xfli)^

% (r2/s)2 + Xgtxo + Xjg) xf — Rb “ *" ~ j = 3-21 ohms (r2 / ® ) 2 + (x 2 + 70

Denominator of the current eqpressions is

D =[a2 (R!s + R f + Rb) + j a2

[Rlli + Rf + Rb + ^(X1M + Xf + xb)]

- a ^ R f - Rb) + j(Xf- Xb)]*

= - 8.7 + j 197.9 = 198 /92.5°

The currents are

E ^ 2 (Hi£ + Rf + Rb) + j ®a(Xis + Xf+ X b )]

I m = d

JEg a [(Rf- Rb) + j (Xf - Xb)]

D

= 7.6 /- 58.5°

_ ^sjj|RlM + R frf~ Rb^ + ^ X1M+ X f4~ xbj ZS D

i % a [(Rf- Rb) + j(Xf- X b )]

D

= 8.82 /- 34°

The average torque and the oscillating torque are

Average torque — (if + a2I2)(Rf-R) + 2 Iylg a (Rf+Rb)sln^

= 202 synchronous watts

Oscillating torque = l£(Rf - Rb) 0062^ t - IjjjCXf - Xb)sin2u/ t

+ if a2 (Rf - Rb)cos 2(u» t + 0) - if a2 (Xf - Xb) sin2(^t^)

= 0 (R^ = R t> X, = Xt,)

Here is the angle by which Ig leads I|| (See Table A). 71

Proceeding in the same manner for other slip, we obtain the result shown in Table 4*

Table 4

Slip Rf Rb Xf x b III is * .Tav Tqsc (syn.

1 2.42 • 42 3.21 3.21 7 . 6 /-58.5° 8.82/-34° 24.5° 202 0 .8 2.9 1.92 3.25 3.16 7 . 8 /-56° 8.5 /-35.8° 20.2° 265 94 .6 3.86 I .65 3.38 3.15 7.82/-510 7*94 /-37.8° 13.2° 330 208 7 . 1 /-41° 1.2° 383 374 • 4 5.74 1.45 3.7 3.13 7.63 /-42.2° .2 11.1 1.29 5.48 3.12 7 . 0 1 /-30.4° 6.36/-56.5° -2 6.1° 300 565 1 9 .6 1.23 11.54 3-12 4. 1 5 /-10.75° 4.97/-60.250 -49.5° 80 467 .05 42.4 1.19 2 6 .1 3.12 2.69/4.4 3.26/-62.7° -67.1° 18 157 264 266 • ^n 19.6 1.23 11.54 3.12 3 . 5 9 / - ^ ° 0 .05 42 *4 1.19 2 6 .1 3.12 1 . 6 /-37.5° 0 105 120

In the last two rows the starting winding is disconnected and the motor runs as a plain single-phase induction motor (see 0 ).

Vihen we compare these two rows with the two rows above for the same slips, we notice at once the significances of disconnecting the starting winding at higher speed. The plain single-phase running without the starting winding has less current in the main winding, much larger average torque, and much smaller oscillating torque, which are all desirable. But this is true only for the higher speeds.

The above results of computation is shown as the speed-torque curve in Figure 45. In the figure the continuous lino shows the average torque Tav, and the broken lines show the envelopes of the instantaneous speed-torque curve. Vertical distance between the continuous line and the broken lines indicates the amplitude T of the oscillating torque 040 in Table 4. Compare Figure 45 with Figure 39 obtained with the 72

900 w i — 1 800 ✓ ✓ s / 700 7 ^ ✓ / a tOO — r. ■ ✓ 03 ✓ fc ✓ - 1JZ-3 500 ✓ \ ✓ V .c 1r £ 4.00 7" T ✓ i s t s 1 S> 300 4t► cr s main wiriuinr only ° 200 "* — ‘ - y \ /\ ' 100 -v. \ t S. N ■s. . 2 0 / X / \ -100 \ \ \ -200 V -

-300 1.0 .8 0.6 0.4. 0.2 blip

Figure 45. Calculated speed-torque curve of the single-phase motor, 116 V, 1/6 HP, 4 poles, 60-cycle. Continuous line indicates average torque and broken line indicates the envelopes of Instantaneous torque.

analyzer, and notice the good agreement. So we can conclude the f o l l o w ­ ing two facts. First, acceleration of the rotor does not affect either the average torque or the oscillating torque. Second, the analyzer shows the correct torque ripple of 120 cps as well as the correct average torque* 73

•T/hon wo observe Figure 39 carefully, we notice that the ripple is

not sinusoidal, but unsymmetrical above and below the mean line of the

envelopes (see Figure 46b alsc). Let us consider which kind of curve

it is*

At steady speed the total instantaneous torque is the sum of (5.1) and (5*2). That is

T s average torque ~h osoiHating torque

» T + T cos (2 iu t t 4>) 5*3 *v 080 7 where the four terms of (5.2) are combined into one term of a single cosine function. Then the acceleration is ^ Tav Togc - — -- + --- — cos( 2 w t t

(* Tav Tosc u S ■ I of d t =------+ ------s i n ( 2 ^ t ) + ^ 5 . 5 J° J J where A is an integration oonstant.

Equations (5.3) or (5.4) and (5.5) are the equations of the apeed- torque curve. These are the parametric equation of a trochoid.

In the actual speed-torquo curve as in Figure 39, T and T in O 80 (5 .3) are not oonstant but vary in value, as the motor speed changes.

But in Figure 39 we oan still recognise the trochoidal nature of the curve.

This might be an appropriate place to consider how to take the average torque from the speed-torque curve. Equation (5.3) gives the 74: curve in Figure 46a, which is the time-torque curve. In this case there is no difficulty in finding the (time) average torque T , because the ev average torque T of equation (5*3) and the mean line of the two envelopes in Figure 46(a) are the same* f^¥¥¥Vf: av. T I av,______time speed

Figure 46. (a) Time-torque curve. (b) Speed-torque curve*

Equations (5.3) and (5.4) combined give the speed-torque curve in

Figure 46(b). In this case the (time) average torque T is still given by the mean line of the two envelopes* as shown in Figure 46(b). But the d-c mean T* av in the figure is high®r than T av .

The conclusion that the mean line of the two envelopes of the speed* torque curve gives the (time) average torque is generally true when the time-torque equation contains only higher harmonics of odd orders,* but 4r this is not always true when higher harmonics of even orders are con­ tained in the time-torque equation. This is because the time-torque curve is symmetric above and below the (time) average torque when higher

The lowest frequency of the torque ripple isoonsidered tobe fundamental • 75

harmonics of odd orders only exist, but it is not necessarily symmetrio when higher harmonics of oven orders are contained. The latter case was

shown in Figure 41 and 44*

3. Consideration of Braking Torque of the Three-phase Induction l..otor, 220 V, 1/4 HP, 4 poles, 60-cycle*

In Article 2 of this chapter we confirmed the speed-torque curves of the three-phase induction motor by the values obtained with a dynam­ ometer. So we know that the curves in ^igure 2 3 and 25 are correct*

Figure 2 5 is peculiar in having large braking torque.* The curve in the

left half of this figure corresponds to the negative speed, and hence the torque in this region is braking torque. As Figure 2 5 shows, the maximum braking torque is much higher than the maxi mum torque in the positive speed range (the right half). In the following we shall try to find out the cause for this peculiar phenomenon of large braking torque* As the first step the motor torque will be calculated from the machine constants*

Measurement of the machine constants*

No load test

V a 193 V l a 0.495 A W « 03 watts n n n Locked rotor test

VL= 50 V IL= 1.29 A WL= 07 watts

D-c stator resistance ■ ■ 10.5 ohms

From -these data we get

*When direction of torque and direction of rotor rotation is opposite, the torque is called "braking torque." whore X^ is magnetizing reactance and X^ is stator leakage recatanoe. VL Z = -----= 22. I* ohme v 3 IL H « R-j + R-= --- = 1 7 . ohms 0 3 If **2 a Re - = 6.9 ohms Xe=/z^ - - 14.1 ohms

Xx = Xg = Xe/2 = 7.05 ohms (leakage reactance)

~ XQ - X s 225 - 7.05 - 218 ohms (magnetizing reactance )

Resistances are corrected 'for 75° C as below.

12.8 ohms

R = 8.42 ohms 2 The approximate equivalent circuit with the magnetizing reactance ( To ) moved to the input terminals gives some error. Hiehter * gives the moire accurate equation for the motor torque, which assumes the equivalent circuit shown in Figure 47, as below.

R1

Ic

% 1 o

O-

Figure 47. Equivalent cirouit of the three-phase induction motor 77

T = ------r------~ — syn. wattI. S-6 R2 r1 + xlo 8 2 R1 +0&xlo 2Rl+------2--- + — *20------i---- where 8 ^tb R2 x

T ~ motor torque in synchronous watte

number of stator phases

V = applied voltage per phase x_ = x ■+ x x_ = xo+- X lo 1 m zo 2 m xlox2 o ^ = --- 3 — - 1 AS s = slip

The result of computation of equation (5.6) is given in Table 5.

Table 5

Slip 2 1.8 1.6 1.4- 1.2 1 . 0 .8 .6 .4 .2 .1 .05

Torque 1 . 4 4 1.55 1.68 1.83 2.02 2.19 2.41 2.6 2.69 2.26 1.5 .833 (lb-ft) Torque 367 396 428 467 510 560 614 665 687 577 384 222 syn. watt)

This result of oaloulation is plotted as the continuous line in

Figure 48. The broken line in the figure is the curve obtained with a

dynamometer (Table l). Except along the small portion near the synchronous

speed, the two curves differ materially. in the positive speed range

the calculated curve is higher than the measured curve, and their

positions are reversed in the negative speed range. The difference between the two curves is plotted in Figure 49. it should be noted that the maximum difference in the negative speed range is 580 cynchronous watts, which is about the same as the measured maximum torque in the 78

1000

800

600 p 4 0 0 ntea 3ured

O’ 200

2.0 1,8 1.6 1,4 1.2 1,0 0.8 0.6 0.4 0.2 0 slip . , . (■ • :

Figure 48. : Speed-torque curves of the three-phase induction motor, - - - i 220 V T 1/4 Hp, 4 polefr,- 60-eyele.

\ r ’

600

200

200

; - 4 0 0 ______J 2.0 lj.8 1.6 1.4 1 ! : - r «iip ■ j * Figure 4 9 - » difference pf the two-tirque curves j in Fi$. 48. 79

positive speed range of Figure 48.

Roller's experiment# (^-S) show the same tendency for the same slot

combination of 36/48 as the present motor.

T. *4. IJorgan (19) attributes the extraordinary braking torque to stray

load loss at higher slip. But in the discussion of his paperthe opposing opinion is raised, which attributes the extraordinary braking torque to the effect of the space harmonics of magnetic flux distribu­ tion. V/e shall now try to determine what is the chief cause for the high braking torque of the present motor. As we shall see later in this article, stray load loss is the chief cause. But stray load loss is a broad term and includes many kinds of losses. Therefore v/e shall also try to find out which kind of stray load loss is the chief cause.

In computing the speed-torque curve in Figure 48, we resorted to the more accurate equivalent oirouit. Figure 48, with the magnetizing reactance at its original position. But even this equivalent circuit neglects some factors, which will be investigated here, in order to find out the causes of the discrepancy in Figure 49. The factors, which were neglected in the equivalent circuit of Figure 4 7, are listed in

Table 6 . In this table + sign means that the factor under consideration increases the calculated torque curve in Figure 48, and - sign means the opposite. The factors in Table 6 will now be considered individually.

(a) Friction and windage. Friction and windage loss causes braking torque. ^ence, it decreases the motor torque in the positive speed range and increases the motor braking torque in the negative speed range.

Thus, the sig?i combination - in Table 6 is understood. The friction and windage loss was determined by driving the motor with a dynamometer Table 6

Positive speed range Neg. speed range

Friction and windage - +

Iron loss by fundamental + + magnetic flux

Saturation of iron Change of magnetieing reactance + + Change of leakage reactance + +

Skin effect + +

Space harmonics of air gap magnetic flux + + (Motor or generator action)

Iron loss due to zigzag leakage flux Surface loss — + Tooth-pulaation loss - +

at synchronous speed. The loss is 22 watts at synchronous speed. This

value alone is too small to account for the discrepancy in Figure 49.

(b) Iron loss by fundamental magnetic flux. Iron loss on the stator

side increases the input to the motor, but does not have any apprecia­

ble effect on torque.

However, iron loss on the rotor side produces torque of sign com­ bination ♦ ♦ , as shown in Table 6 . ^t is divided into two parts; one

is hysteresis loss, which is proportional to slip s, and the other is

eddy current loss, which is proportional to s ^ ^ ^ *

Let hysteresis loss and eddy current loss of the rotor at line

frequency be denoted by and respectively, then the total rotor

iron loss Q at slip s is given by 81

Then torque T\ in synchronous ■watts due to this loss is

- q/b - ^ ♦ •% 5-e which states that hysteresis loss produces a oonstant torque and

eddy current loss produces & torque sQ^ proportional to the slip s.

■Wow the torque produced by the rotor iron loss will be considered

for speeds above synchronous speed. For speeds above synchronous

speeds, s<0 and equation (5.7) should be written as

Q "-eQjj + s2 Qft 5 . 9 because each term of iron loss should be positive. Then torque is

?i » Q/b * t 5 . 1 0

Comparing equations (5.8) and (5.10) we see that there should be a discontinuity of 2Q^ in the speed-torque curve at synchronous speed'(13 ) .

This fact enables us to determine the hysteresis loss This will now be done*

Driving the motor above and below synchronous speed by a dynamometer, we obtain the relation between the motor speed and the dynamometer torque reading as shown in Table 7.

T a b l e 8

Speed (rpm) 1781 1790 1794 1803 1805 1810

Torque (syn. watt) 37 18 0 -35 -52 -68

The result in Table 7 is shown as continuous line in Figuer 50*

To obtain the developed torque, the braking torque due to friction and windage must be added to the value in Table 7. In part (a) of this article the friction and windage loss at synchronous speed was given to 62

60

20 J L torque *due tq hysteresis

tts Tor friction! -60 and ^windage measured

1780 1790 1800 1810 1^20 ' motor speed (rpm) i , ...»■ *

Figure 50* Speed-torque curve in the neighborhood of* synchronous speed* be 22 watts* For the small range of speed in Figure 50, the friction and windage is considered to be oonstant and the braking torque caused by them is then 22 synchronous watts. Adding 22 synchronous watts to the continuous line in Figure 50, the developed torque is obtained and is shown as broken line* The developed torque shows the discontinuity of

20*4 synchronous watts at synchronous speed* As explained above, one- half of the discontinuity gives Hence ws get

Qh « 2 0 .-4/2 - 10.2 watts (5.11) which is the rotor hysteresis loss at power frequency* 83

Subtracting the stator copper loss (9 watts) from the input (39

watts) to the motor at synchronous speed, we obtain iron loss in the stator

to bd 30 wattsj we approximately assume the iron loss in the rotor at

slip ■ 1 to be 30 watts, because at the same power frequency the iron

loss does not differ materially on both stator and rotor. Then the

eddy current loss is given by

Q - 30 -Qh« 30-10.2 - IS.8 watts

Then from equation(5.6) the torque due to iron loss in the rotor

is computed as shown in Figure 51. The torque in Figure 51 is quite

small compared with the torque difference in Figure 49*

w** *>O ' * 20

2.0 islip

Figure 51. Torque due to rotor iron loss

(c) Saturation of iron. The machine constants in Figure 47 were ob­ tained from the no load test and the locked rotor test. But actually they are not oonstant. Due to saturation of the iron the reoatances change, as the running oondltbn of the motor changes, ■^ence, the torque computed under the assumption of oonstant reactances has error for 84

certain speed range. The effect of change of reactances due to the

saturation of iron on the torque may be divided into twoj that is, the

effect of the change of magnetising reactanoe and that of the leakage reactances. These will now be explained separately*

(1) Change of magnetising reactance X . In the equivalent circuit m of Figure 47, voltage on the magnetizing reactance X^ changes, as the stator current changes. Hence, if the stator current increases, X^ in­ creases due to decrease of saturation* Since the magnetizing current is small compared with the load current, the effect of the change of X^ on the torque is small.

(2) Change of leakage reactance. As the load current increases, iron paths for leakage flux are more saturated and hence leakage reac­ tances decrease. Usually leakage reactances are obtained from the locked rotor test for reduced current. Hence, if the leakage reactanoes are assumed constant, it causes error in the computation of torque for larger current. At first we shall find the relation between the stator current and the leakage reactances.

Figure 52 is the result of locked rotor test. The abscissa is the terminal voltage and the ordinate is the stator current. *f there were no saturation, the curve would be -the straight line indicated by the broken line. Actually the curve is bent upward slightly. From this curve leakage reactance is derived for current of various magnitude, using the following equations* VT Leakage impedance Z » — 5.IS

I 2 ~2 Leakage reactance Xe» X-^ + Xg*/Ze - (R3+R2) 5.14 05

Stator and rotor leakage reactance

~ ^2 s 5.15 The result of the computation is shown in Figure 53, which shows

decrease of leakage reactance with increase of current.

Figure 54 shows the input current measured as a function of slip.

Figures 53 and 54 provide the necessary data to compute torque, when

the effect of saturation is considered. The procedure will now be

explained*

Choose slip and read corresponding current in Figure 54. Read the

leakage reactances X^ and X^ corresponding to the above ourrent from

Figure 5 3. Substitute these X^ and Xg into equation (5.6) ( x X 2Q

and change), other constants being the same as before. The result

of computation is shown in Table 8#

Tahle 8

Slip 2.0 1.8 1.6 1,4 1.2 1.0 .8 .6 .4 .2 .1

Torque 1.46 1.58 1.71 1.85 2.o2 2.22 2.43 2.63 2.71 2.26 1.5 (lb-ft) Torque 373 403 435 473 517 566 619 671 690 578 384 (Byn. watt)

This table is plotted as the continuous line in Figure 55. In the figure the broken line indicates the values of Table 5, which was computed without taking saturation into account. The difference be­ tween the two curves is within 10 synchronous watts. So we can conclude that the effect of the change of leakage reaotanoes due to saturation is very small* 2 0 60 8p 100 120 UP 180 20p 220 terminal voltage (volts) Figure 52. Rocked rotorj test of the three^-phasej motor, 220 V, 1/4 HP, ! 7 . 0 4 poles, 60-cyCle. j

Figure 53. Variation of leakage1;e 1 reactances of the tbree-phase mbtor, 220 V], 1/4 HP es 60-cycle, f

rd tot-que {eyaoiv.; *afcte) :gr $. Coil] $9. F:.gure . 18 . 14 . 10 l 06 . 0.2 0.1 0.6 0l8 1.0 1.2 1.4 1.6 1.8 2.0 si. I . . slip. ■ i i F ig u r e 54. M e a s u r e d stator curriatjof th$ three-phase moto/, moto/, three-phase th$ curriatjof stator d e r u s a e M 54. e r u ig F C 220 ;V,. 1/4 HP* ppjkein, llue i o 20V l4$P, 4 ple[0cce I polee,[60-cycle. 4 , P $ l/4 V* |220 \ ■ ■ i , ■ i ■I itdnboilui 1 i■ iputed iputed [ speed

(atui-atlbp nc (atui-atlbp pls 6- , 60- poles, 4 torque cury< torque ■ i t ^ - a s ii]raJbl< a , c e r e d i s n o c t cycle.1 of te thifc the f o a -phase aoic 83

(d) Skin effect. Due to skin effect the rotor resistance chanres with

the motor speed. When we measured the rotor resistance Rg by the locked

rotor test, we obtained the rotor resistance corresponding to slip 1.

So in the positive speed range, the actual rotor resistance is smaller

than the locked rotor value due to lower frequency of rotor current.

In the nogative speed range, the actual rotor resistance is larger due

to higher frequency of rotor current. The change of rotor resistance R 2 causes error in the computation of torque, when the rotor resistance is

assumed constant. At first we shall compute the rotor resistance R^ for

various motor speed.

The cross-section of a rotor bar is shown in Figure 56(a). There

is no simple formula to take account of skin effect, which can be applied

immediately to this cross-section. Most formulas are given for a reotang-

ular cross-section. 3o the original cross-seotion is replaced by the TT — I- {7". 56mm,

H . ! 3 • zdmaL i Figure 56. (a) Actual rotor bar cross-section. (b) Equivalent cross-seotion rectangular cross-section in Figure 56(b), which has the same height and the same area as Figure 56(a). This replacement is on the safer side; that is, the skin effect is exaggerated. /16) Richter explained in his book how to calculate the skin effect for the reotangular conductor. Aocording to equation 556 in his book the dimensionless reduoed height of conductor \ is given by

X ~0Ch 5.16 89

OL = who re 5 5.17

f- ■ rotor frequency

f ~ specific resistivity ■ .0197 ohm mm^/m for Cu at 25°C

a = slot width

b = conductor width

h * conductor height

Then the ratio K between the actual resistance and the d-c resistance

is given by sinh2f + sln2 § 5.18 cosh23£ - cos21|

Slot leakage inductance is reduced by the skin effect. The ratio

K between the actual leakage inductance and the inductance without skin

effect is given by

3 sinh2^ - ein2^ K ------5.19 2 5 cosh23£ - cos23£

Now we shall compute K and for the locked rotor test.

Since fg« 60 cps, a « b, and h * 7.56 mm from Figure 56(b), we get

0t = 1.09 and '§-1.09*0.756

Then equation (5*13) gives

K ■ 1.04 for slip a 1.

Equation (5.19) gives

0.988 for slip =1 .

he shall here explain why the effect of can be neglected.

Ki has effect only on slot leakage inductance, which is one part of the total rotor leakage inductance. As the above value of Ki shows, the 90 deviation of from 1 is only 1.2/5 for slip * 1. So the change of rotor leakage inductance due to skin effect is very small. As we saw in part

(c) "Saturation of iron," change of leakage inductance of this order has negligibly small effect on torque. So in the following computation vve neglect the effect of change of leakage reactance due to skin effect.

Now we shall explain how to compute the rotor resistance for various slips. The ratio X oan be computed for different slips in the sane way as for slip = 1 above. From the locked rotor test, we got the rotor resistance to be 8.42 ohms for slip * 1. he have already

IE ■ 1.04 for slip ■ 1 . ^cnce, the actual rotor resistance R'g for other slips is given by , . 8 • 4k R* ■ (d-c resistances ■-----X z 1.0-4 Substituting the actual rotor resistance R* into equation (5.6), £+ we get torque oorrected for skin effect. The result of computation is given in Table 9. Table 9 Slip 2.0 1.8 1.6 1.4 1.2 1.0 .8 .6 .4 .2 .1 0

<*- 1.54 1.46 1.38 1.29 1.2 1.09 .975 .845 .69 .488 . 316 0

^ 1.17 1.11 1.04 .975 .904 . 824 .737 . 64 . 522 . 368 .261 0

K 1.16 1.13 1.11 1.08 1.06 1.04 1.03 1.02 1.01 1.0 1.0 1.0

Ki .955 .962 .968 .975 . 98 .988 .99 . 992 .995 1.0 1.0 1.0

R2 9.35 9.14 8.98 8.74 8.57 8.42 8.33 8.25 8.17 8.09 8.09 8.09

T 1.55 1.64 1.75 1.87 2.02 2.19 2.4 2.6 2.7 2.3 1.55 0 (lb-ft) T 396 418 447 477 516 560 612 663 688 587 395 0 (syn. watt) 91

The torque in Table 9 ia plotted as the continuous line in

Figure 57, whore the broken line indicates the uncorreoted torque in

Table 5*

■Ye see from the figure that due to skin effect, the torque is in­ creased slightly in the negative speed range, and is decreased very slightly to the left of the maximum torque and is increased very slightly to the right of the maximum torque in the positive speed range. It may be said that the change is small at any speed. At slip = 2 occurs maximum change, which is 29 synchronous watts. This value is too small to explain the torque difference in Figure 49 by itself.

*000

s m

600

200

Figure 57. Calculated speed-torque ourve* Continuous line - skin effeet considered. Broken line - skin effeot not considered. 92

(o) Space harmonics of air gap magnetic flux.

it is a well-known fact that space harmonics of the air gap magnetic

flux produce cusps on speed-torque ^et us con­

sider the magnitude of the cusps*

There are two kinds of space harmonics in the air gap flux. One is

space harmonics arising from phase belts* and the other is space

harmonics arising from slots. They will now be considered separately. (27 ) Richter gave the following equation for harmonic torque due to

phase belts* Tpm 1 Xm + J-2 f X&Lii K8k2# Kp2o \2 — a ------[------j 5.20 t „ 2* R2 lKdii K.kzi Here

^ = order of harmonic

T^m = maximum torque due to the th harmonic

T ■ starting torque due to the fundamental

X^- magnetizing reactance

Xg* rotor leakage reaotance

Rg* rotor reristance

= distribution factor of stator winding for >>th

harmonic

KtVih>- skew faotor of rotor for th harmonic

Kp2#" factor of rotor for Vth harmonic

The lowest harmonics due to phase belt are the 5th and 7th harmonics* which may have most effeot on torque. The result of computation of equation (5.20) for these two harmonics is shown in Table 10. As seen from the row 'L, /F, in the table* maximum torque Tg^ and T7m are about

10?S and 3.4,; of the starting torque respectively. These values are 93 small compared with the torque discrepancy in Figure 49.

Fundamental 5th 7th 17th 19th

Kdlv> .9601 .2176 -0.1774 .057 .14

.99 .84-5 .715

Kp2v> 1 1 1

W t s .0975 .0336

T»/T' .0032 .02

Next we shall investigate the effect of space harmonics due to stator slots. The orders of the spaoe harmonics caused by stator slots are given by 2N± P = k + 1 5.21 P where k is integer. In the present ease ■ 36 and p * 4. So the two lowest harmonios given by equation (5.2l) are 17th (k — l) and 19th

(k * 1). which may have most effeot on torque among the slot harmonics. (25 ) Qmoto Ishizuki gave a torque equation for the slot har­ monics. But they assume no skew either on stator or on rotor. The motor* which we are investigating* has skew on the rotor side, ^enoe* their formula eannot be applied to the present motor. In the following we shall derive a simple formula to estimate the torque due to t he slot harraoni cs •

Any harmonic flux due to stator slots is proportional to the stator current 1^. Hence, the voltage induced in the rotor due to the t? th har­ monic flux is given as 94

s, Ii 5 *22 vrhere ajj ia the slip of the rotor for the ^ th harmonic and Xmk, is the

magnetising reaotance for the v> th harmonic flux. Then the rotor current

Ig induced by the voltage given by (5.22) ia

6v XMk» Ii I2v|= i— 2 2 2 5,23 V R2y + x2*o where R^ ■ rotor resistance for th hamonio

Xg = rotor total reaotance for th harmonic

Then the torque due to v> th harmonic flux is given by

T* = (Vs*) “lR2w t 1*

2 = B» “1 2 -- 1-----— “7---!l H|v + «? 5 -24 where ia the number of stator phases*

Equation (5.24) holds both for unskewed motors and skewed motors*

•Ye shall now derive the ratio between the torque of the skewed motor

and that of the unakewed motor; both motors being identical to each

other except for rotor skew* The motor under consideration has rotor

skew. Then the magnetizing reactance X is given by

X » K . « X* 5.25 mi> sk2^ my where ^sk2»>= rotor skew factor for ^ th harmonio

x*m ^ ■ magnetizing reactance for v> th harmonic when

there is no skew.

The rotor bars become slightly longer due to rotor skew, and hence

is slightly increased. But this increase of Rg^can be negleoted as an

approximation. The total rotor reaotance Xgy0 is increased slightly 95

due to skew, because slot leakage reaotanoe Is slightly increased due to

the longer rotor bars. But this increase of oan be also neglected

as an approximation.

Now equations (5.24) and (s.25^ give the torque ratio between

the skewed motor and the unskewed motor to be

T, A i * K2<1c2i( 5.26 where Ty' is the torque of the unskewed motor. Equation (5.26) gives the row /ij in Table 10. As this row shows, if the rotor is skewed properly, the torque due to the slot harmonics is reduced to small

fraction of that of the corresponding unskewed motor by rotor skew.

Generally the torque of the unskewed motor is not large compared with the torque due to fundamental flux. So the effeot of slot harmonies oan be neglected for the present motor, which has rotor skew. f. Iron loss due to zigzag leakage flux.

Zigzag leakage flux is produced by concentrated ampere-oonductors in both stator slots and rotor slots. As the load current increases, zigzag leakage flux becomes more pronounced and is superimposed on the no load air gap flux.

Zigzag leakage flux arising from the stator slot ampere-oonduotors goes into the rotor teeth and causes surface loss and tooth-pulsation loss in the rotor. Also zigzag leakage flux arising from the rotor slot ampere-eonduotors goes into the stator teeth and causes surface loss and tooth-pulsation loss in the stator. Sinoe the frequencies of pulsa­ tion of both zigzag leakage fluxes are determined by motor speed and number of slots of stator and rotor respectively and are different from 96 the power frequency, these losses cannot be supplied directly from the power aouroe, but they are actually supplied by the mechanical motor (ia ) output Hence, their effect on torque is similar as that of mechanical loss, and their sign combination in Table 6 should be - * as that of friction and windage* &enoe, the torque caused by these losses is braking torque. *V© shall now compute these losses and the change in the motor torque caused by them* (2l) Richter treated these losses theoretically* According to his theoretical derivation, tooth-pulsation loss in the rotor is given by

Vt2 watts 5*27 Qt2 = “ 2— Pt2 where . _ / nH/ B1Yt lO-Wc,)^— _J(— ij(l - =oB«,) 5.28 and F a total weight of rotor teeth ■ 0.735 Kg t2 t 3 slot pitch of the stator ■ 0.87 ont

3 tooth width of the rotor * 0.321 cm

n ■ rpm

* number of slots of the stator ■ 36

(0^ * ampere-conductors per stator slot

£ « air gap length * 0.03 cm

Sj 3 slot opening of the stator a 0.23 ora

r \o&\a (J * given by Figure 58 * 33

B* a maximum flux density of the zigzag leakage

3 maximum value of the fundamental flux of the zigzag

leakage 97

#.z m 360°o2/t1

Equations (5.27) and (5.28) can also be applied to tooth-pulsation loss on the stator, if we interchange subscripts 1 and 2. Then

P ■ total weight of stator teeth « 1.1 Kg vi tg * slot pitch of the rotor « 0.65 cm

Cj 3 tooth width of the stator ■ 0.35 cm

- ntimber of slots of the rotor • 48

% “ ampere-conduotors per rotor slot

m fj-on Figure 58

Oj s tooth width of the stator * 0.35 cm

» 360° cj/t2

Surface loss in the stator is given by

v0^ l D ^ ~ ■ watts 5.29 tl where

v8l = 0.79 K 2 - t W t f

1 « stack length ■ 5.08 cm

0 = inner diameter of the stator * 9.84 cm

constant ■ 4 k. a k a given by Figure 59 a 0.4 2 u slot opening of the rotor ■ 0.1 cm *2 Equations (5.29) and (5.30) can also be applied to surface loss for the rotor, when subscripts 1 and 2 are interchanged* 0.4 40 98

30 0.3

10.2 cq 20 H 16 CQ 0.1 *—IO 10

0 0 10 20 30 40 50 10 20 30 40 50 t/S t / S t slot pitch Fig. 59 S air gap length Fig. 58 s slot opening Next ■wo shall consider how to compute the braking torque caused by these iron losses. Generally when any iron part of the motor and mag­ netic flux have relative motion, then iron loss and torque occur at the

same time. The iron loss and the torque are related by

Qi synchronous watts 5.31 8 ' where s' is equal to relative speed/synchronous speed*

For the iron losses under consideration, s' * 1 - s and

Q#2 • ^nee, the braking torque is given by 1 T. - 5.32 t - •CQtl + «t2 + Csl + Q s 2^ 1-8 Now we shall compute the iron losses and the braking torque caused by them. All necessary data were already given, except the slot ampere­ conductors and • They are the product of the current and the number of conductors per spot, that is

■ 861^ (No. of conductors per stator slot ■ 8 6 ) 5.33

\ - 86Ir 62 l x 5.34 lx (1/2) (2H2/P) where m number of stator phases

n-^ ■ number of slots per phase per pole on the stator = 3

winding faotor of the stator s 0.96 99

(Equation 5.34 "was obtained by equating the stator mmf and the rotor mmf equal to each other.)

The result of computation of these losses and the torque caused by them is given in Table 11* Table 11

Slip • .4 .6 .8 1.2 1.4 1.6 1.8 2.0

$tl 41.6 58 35.8 10 12.9 55-5 132 244 388 (watt) Qt2 39.4 55 33.6 9.49 12.2 52.2 126 230 368 (watt) Qal 1.61 2.65 1.98 .8 1.02 3-1 6.1 9-7 13.7 (watt) ?s2 6.7 10.7 8.2 3-3 4.22 12.8 25 40 56.8 (watt) Total loss 90.31 126.4 79.6 23.6 30.3 123.9 289.1 523.7 826.5 (watt) Torque -123 -210 -199 -115 152 309 480 655 826.5 (syn. watt)

The result is plotted in Figure 60 together with the curve in

Figure 49, which is shown as broken line. The agreement between the two curves is not bad in view of the many assumptions made in deriving and applying equations (s.27) to (5.52). The degree of the agreement is sufficient to oonfirm that these iron losses due to zigzag leakage flux are the main cause for the torque discrepancy in Figure 49, and that the effects of parts (b) to (6 ) are small.

(g) Some additional considerations of the extraordinary braking torque

After having found the main cause for the extraordinary braking torque in the proceeding parts, we shall now disouss the significances of the phenomenon* 100

1000

800

600

§. 200

-200

2.0 1.8 1.6 1 *4. 1.2 1.0 0.8 0*6 0.<4 0*2 slip

Figure 60. Calculated braking torque due to tooth-pulaation loaa and surface loss (continuous line). The broken line indicates the torque discrepancy in Figure 49.

Iron loss due to zigzag leakage flux increases the braking torque

in the negative speed range considerably* This high braking torque Is

of practical use. vVhen the motor is plugged* it comes to standstill very quickly due to the high braking torque* In the positive speed range the iron loss decreases the motor shaft torque and causes addi­ tional heat loss*

In order to see how much the input current and power are affected due to this phenomenon* their calculated values and measured values will be compared* 101

Resorting to the equivalent cirouit in Figure 47, the input current (23) is given by

R + j a X2o V — — ------— ------— 5*35 (Ri+ j X )R2+ s[j X2o(R!+ J Xx) + Xj^3 Input power is then given by

3 VI ^ (power factor) 5.36

The current and the input power computed from equations (5.35) and

(5.36) are given in Table 12*

Table 12

Slip 1 .8 1.4 1 .0 . 6 .4 .2

Input current 6.18 5.95 5.56 4 .8 4.04 2.82 (amp.) Power factor .715 •74 .78 •875 .93 .9 6

Input power 1645 I64O 1580 1563 1400 1056 (watt)

The result of computation and the measured values are plotted in

Figure 61. The agreements are fairly good. For instance, for

slip » 1.4 the difference in current is 0.1 amp. and the difference in

input power is 60 watts, while the difference in torque is 460 synchron­

ous watts for the same slip, as Figures 48 and 49 show. It is remark­

able that the increase in torque is quite large compared with the

small increase in current and in input power. This fact shows that

the equivalent circuit in Figure 47 is not practically affeoted by

this phenomenon, and that the additional iron loss is supplied by the meohanical power developed. This may be considered as some evidence that tooth-pulsation loss and surface loss due to zigzag leakage flux Xo 2

are the main cause of the phenomenon*

2800

2000

T=i.

Figure 61* Stator current and input power. Continuous line-measured value Broken line-computed value

The effeot of terminal voltage on this phenomenon was Investigated experimentally* Figure 62 shows speed-torque curves for three differ­ ent terminal voltages. For lower voltage the whole curve is lower* but still we recognise the high braking torque. For 100 V, the maximum braking torque is 2.4 times the maximum torque in the positive speed- range, while this ration is 1.8 for 215 V* 103

lOOOi

800

400 1507 200 10QV

1.6 ■1.2 0.8 Blip

Figure 62. Speed-torque curves for throe different voltages.

4. Single-phase Operation of the Three-phase Induotion l^otor, 220 V, l/4 HP* 4 poles, 60-cyole•

In Art!ole 2 of Chapter IV, the apeed-torque curves of single­ phase running of the three-phase motor ■sere taken. These a peed-torque curves are shown in Figures 27, 28 and 29. Pulsating torque of 120 ops in these figures will not be oonsidered here, because it was already discussed with the split-phase single phase motor in Article 2 of this chapter* * The average torque of these figures is shown in Figure 30. This speed-average torque curve has a peculiarity; that la, the average torque in the range of slip 1 to 0.5 is negative. This seems to be contrary to the ordinary theory of single-phase induction motor, aooord- ing to which the average torque of plain single-phase induction motors is positive, as long as the motor speed is not zero* We shall try to find out the cause of the negative average torque* 104

When the Y-cormected three-phase induotion motor is operated as a plain aingle-phase motor by connecting two of the three terminals to the power supply, two phases of the stator are connected in aeries*

So the number of turns in series on the stator is now twice as many as that of the three-phase operation. Accordingly the machine constants for the single-phase operation are different from those of the three- phase operation. In Table 13 the machine constants for both oases are tabulated (see Richter's boolc^24)). The machine constants for single­ phase operation are for "Double-Revolving field theory."

Table 13

Three-phase running Single-phase running

Stator resistance R^ 1 2 .8 2 5 .6

Stator leakage reactance X^ 7 .0 5 1 4 .1

Rotor resistance R£ 8 .4 2 8 .4 2

Rotor leakage reactance X^ 7 .0 5 7 .0 5

Magnetizing reactance Xa 218 218

Aocording to the double-revolving-field theory of single-phase induction motors input current and average torque are given by

V h s~------Rl+ Rft Rb* j (Il+ Xft Xb)

- R^) Synch* watts 5.38 where X.

(H2/*)2+ ( X ^ X.)2 f (R2/b)2+ (X2+ I.)2

5-39 105

-.2 X* CH /( 2-s)J Xm [(R2/2-s)* + X2(X2+ Xmj] «b = Xb= (R2/2-sf+(X2+ X j ; (R2/2-e) + (X2+ X m)‘ 5.4-0 The result calculated from these equations is given in Table 14.

Table 1 4

Slip .8 .6 .4- .2 .1 .05

Current (amp.) 4.-23 -4.07 3-7 2.81 1.91 1.29

Average torque (syn. watt) 58.5 123.-4 200 266 237 160

The average torque in Table 14 is plotted in Figure 63. This curve is positive for the whole positive speed range and is quite dif­ ferent from the single-phase running curve in Figure 30.

aoo

g 600

calculated I torq’ i#

braking jerque

Figure 63. Calculated torque and current in Table 14 and braking torque in Table 15 for the single-phase operation of the three-phase induotion motor. 106

..'hen this motor was run as a three-phase motor, the iron loss due to

zigzag leakage flux affected the motor torque considerably. Even when

the motor runs as a plain single-phase motor the same iron loss may

affect the motor torque considerably. We shall now estimate the iron

loss due to zigzag leakage flux and the braking torque caused by it

from the data obtained at the three-phase running.

The torque discrepancy of three-phase running in Figure 49 was mostly

caused by the iron loss due to zigzag leakage flux. This iron loss la

proportional to (stator current) • In an approximate manner we assume

that the iron loss for the single-phase running is given by multiplying

the iron loss for the three-phase running with square of the stator

current ratio. Then from the braking torque curve for the three-phase

running in Figure 49, the corresponding curve for the single-phase run-

ning is obtained by multiplying with \statort ourrent ratio) • The pro­

cedure is as follows.

Take the braking torque from Figure 49. Multiply it with Il)3] where (ij)^ ®^id (lp)j ®re corresponding stator current for the single-

phase running and three-phase running respectively, and they are taken

from Figures 63 and 61. The result of computation is given in Table 15.

Table 15

Slip Torque from Fig. 49 [(I^) 3/( 1 1 ) 3I Braking torque for single (syn. watt) phase running (syn. watt)

1.0 0 0 0.9 188 .658 123 0.8 230 .69 158.5 0.7 280 .735 206 0.6 260 .735 191 0.5 200 .75 150 0 . 4 130 .8 2 107 0.3 80 .92 73.5 0.2 4 0 1.0 40 107

The above result for braking torque under sin; 1e-phase operation

is plotted in Figure 63. Subtracting this braking torque from the

calculated torque in the same figure, we obtain the resultant torque of the single-phase running in Figure 64. The agreement between this curve and the average torque curve in Figure 30 is good in view of the assumptions made. (in Figure 64 the measured torque of the three-phase

running is also shown for comparison.)

o

Figure 64, Calculated speed-torque curve of single-phase running of the three-phase motor, 220 V, 1/4 HP, 4 poles, 60-cyole* Chapter VI

Transient Torque of Three-phase Induction Motors

In Chapter V the steady-state speed-torque curves obtained with

the analyzer were compared with the values determined by a dynamo­

meter and also with computed values. In this chapter the switching

transient ripples in the speed-torque curves will be considered.

The transient ripples under consideration are caused by switch­

ing. This fact was demonstrated in Chapter IV by locking the rotor

before starting, with the result that the switching transient ripples

were absent from the speed-torque curves.

Now numerical confirmation of the switching transient ripple

in the speed-torque curve will be performed. Since this may be the first time that such switching transient has been indicated in the speed-torque curve, an attempt will be made to compute the switching transient and compare it with that obtained with the analyzer.

The solution of switching transient phenomena of induction motors is not easy; in case of an unbalanced induction motors, which include single—phase split-phase induction motors, it is even more difficult.

Hence, the solution will be limited to that of a balanced three-phase induction motor.

Whenever the solution of switching transient phenomena of three- phase induction motors was attempted, it was found convenient to replace the three-phase motor by an equivalent two-phase m o t o r ^ ^ ^ ^ ^ ^ ^

(The derivation of the equivalent two-phase motor will be given in

Appendix II.) We shall take a similar approach, replacing the balanced

- 108 - 109

three-phase induction motor with an equivalent balanced two—phase

motor- Hence, in this chapter, the three-phase motor and its equi­

valent two-phase motor are the same machine and the terms will be

used interchangeably.

In Article 1, differential equations will be set up for an

unbalanced two-phase induction motor. This is done primarily for

the purpose that perhaps, in the future, solution of this case can

be obtained. Since the single—phase split-phase motor is an unbalanced

two—phase motor, the solution of the unbalanced two-phase motor is the

solution of the single—phase split—phase motor.

In Article 2, the differential equations will be solved for the

equivalent balanced two—phase motor at standstill, and the torque

calculated will be compared with the curves obtained with the analyzer.

In this article the effect of motor speed on the switching transient will also be considered. In Article 3, the switching transient of

the three-phase induction motor will be solved for varying motor

speed by an electronic analog computer.

1. Differential Equations of Unbalanced Two—phase Induction Motors.

The differential equations will now be set up for an unbalanced two—phase induction motor. The theory, according to which we set up the differential equations, is the "CrosB Field Theory". The cross field theory was originally developed for plaine single-phase induc­ tion motor, but was extended to explain capacitor motors(^), which are unbalanced two—phase induction motors.

The cross field theory gives the equivalent circuit of an un- 110

balanced two-phaee Induction motor as shown in Figure 6 5. In this

figure the subscripts x and y indicate the x-phase and y-phase res­

pectively and the symbols are as below.

Rlx* Ply = s-taltor resistance

^lx» ^ly “ stetor leakage inductance

^2x> ^2y = ro^or resistance (referred to tuestator)

^kxr L2y = ro'tor leakage inductance (referred to the stator)

ImXf Lmy = magnetizing Inductance

esx» esy ~ speed voltage in the rotor (referred to the stator)

esy

Figure 6 5. Equivalent circuit for an unbalanced two-phase induc- tion motor

The mesh equations of Figure 65 are

Here e6X and eBy are speed voltages and are given by Ill

©sx= ~ ® (jjjuyiiy + (^2y + Lmy)i2y] 6.5

eSy= S [jLmyiix + (^2x + Lmx)i2xJ 6.6 where S 1 b angular speed of the rotor. The + signs of equations

(6.5) and (6.6) assume that voltages and currents in the x-phase are leading those in the y-phase and the rotor rotates from the x-phase

to the y-phase.

Since the developed mechanical output of the motor is

W = — (i2x esx "*■ ^2y esy) 6.7

, the developed torque is given by

T ='s'= “ (a2x ®sx + *2y esy)/£

= *2x iiy ^my ~ *lx *2y + i2x ^-2y(^2y ~ I*2x +Ln^T ~ ^mx)

The equation of motion is given by

T = J -if +■ TL 6.9 where J — moment of inertia of the rotor

load torque on the motor

Equations (6.1) to (6.9) are the simultaneous differential equations to be solved, in order to explain the transient phenomena of the two-phase induction motor. The general solution of the simul­ taneous differential equations for the unbalanced two-phase induction motor is difficult. Hence, all solutions derived by other investi— g a t o r s ^ M ^ O O ) so far were obtained under certain assumptions, which made the simultaneous equations easier to solve. They were all derived for the balanced two-phase motor, where the correspond­ ing machine constants in the x-phase and in the y-phase were equal to each other. Stanley^^) simply gave the solution in terms of the 112 operator np", which does not help us bo understand the present problem.

Vi'ahl and Kilgore(^) gave the solution for the case, where the motor speed S in the above equations was kept to zero. Hence, their solu­ tion gives the switching transient of the motor at standstill.

Gilfillan and K a p l a n ^ O ) gave the solution for the case, where tne motor speed was kept at some arbitrary constant value.

These solutions cannot explain accurately our present cases, where the motor speed is not constant but changes continually during the switching transient phenomena. Only when the motor acceleration is not so rapid and the motor speed does not change materially during the switching transient under consideration , these solutions for the motor speed,either at standstill or at some constant value, may give fairly good approximation.

When the motor under consideration starts from standstill, as is the case with our tests to obtain the speed-torque curves for slip 1 to 0, the solution for switching transient with motor speed

S = 0 will give an approximation of the switching transient for lower motor speed. This will be done in the next article for the balanced two—phse induction motor, which is equivalent to the three- phase induction motor under consideration.

2. Transient Phenomena of a Balanced Two-phase Induction Motor.

As shown in Appendix II, a balanced three-phase induction motor can be replaced by a balanced two-phase Induction motor with respect to its behavior under both steady state and transient state.

From the transient solution for the latter, the transient solution 113 for the former can be derived immediately.

At first the switching transient of the balanced two-phase

induction motor with the rotor locked will be considered.

For the balanced two-phase induction motor at standstill, the speed voltages are zero and the equivalent circuit in Figure 65 is simplified as that shown in Figure 6 6.

Figure 66. Equivalent circuit for the balanced two-phase motor at standstill.

In this figure

R^_= stator resistance

stator leakage inductance

R.2= rotor resistance(referred to the stator)

L2- rotor leakage inductance(referred to the stator)

1*,,=: magnetizing inductance

For this circuit, equations (6.1) to (6.4) become

elx= LR1 + (L1 + Lm)p] i-ix + lb P *2x 6.10

0 = p ij* + QR2 + (1*2 + Lm)pJ A2x 6 .1 1

* As shown in later part of thiw article, Wahl's solution (^) is an approximation. We shall derive a more rigorous solution. 1 u

ely~“ [j*l + (Li + Ljn) pj 1 ly + Lm p ±2y 6.12

0 = i'm P 3-ly + [^^2 + ^2 + i*m)p} ^-2y 6.13

In these equations, (6.10) and (6.11) are independent of* (6.12) and

(6.13), and vice versa. This makes the solution much easier,

although still much work is involved, as seen below. In order to

make the final solution more convenient for numerical calculation,

equations will be expressed In terms of iix > ^ly» imx anci imy* From

Figure 66,

ijnx = iix + ^2x 6.L4

imy = ily = a 2y 6.15

where and i^y are the magnetizing currents at standstill. Then

equations (6.10) to (6.13) become

elx = (Ri + 1 P)iix + Lm p I** 6.16

e = - (R2 + L2 p)iix + [r 2 + (L2 + Lm)p3 inix 6.17

ely ~ t^l + i*l P)*-ly + i'm P imy 6.18

0 = - (R2 + L2 p)ily + [j*2 + (l2 + Lm)pJ i-ny 6*1^

The general solution of equations (6.16) to (6.19) have the

following form:

K e-<*± e-Rt Alx + Bx + ilx 6.20 e-At e-pt ijux = <*x + Dx + IfflX 6.21 -oit e-/3t ily = Ay e + By + iiy 6.22 e~*t e-pt H o + + iigy D7 6.23 Here A*s, B's, C's and D'S are integration constants and I*s are the instantaneous currents under steady state and are given by the following expressions. 115

Xlx " I1 sin(^ t + 6 ) 6.24

Jiy =■ - cos(“> t + 8 ) 6.25

Imx = Im siri(w t +6 -S ) 6.26

Imy = — Im cos(u> t + 6 — S ) 6.27

Here & Is the arbitrary initial phase angle, which indicates the

switching instant, and the phase difference S between Ilx and 1 ^

( or between I-^y and Iiny) is given from equations (6.17), (6.24) and

(6.26) to be

S = tan-1 “

The exponential coefficients *tand (5 in (6.20) to (6.23) are

the roots of the characteristic equation of the simultaneous equations

(6.16) - (6.19). The characteristic eauation is

Rl + Li p Lm p = 0 6.29 -(R2 + t>2 P) H2 + (l>2 + Lm)P

Hence oL and £ are given by

“[Rl(L2+ ^m) + r 2(^*2+ ^m)! oc or p = 2[Lm{Li+ L2) + L1L2]

V [Rl(L2+ La) + R2(^2 + 4 R i R ^ m(Li+ L2) + 6.30 2 [lb(Li + L2) +■ liL^I

The A'3, B*s, C's and D*s in (6.20) - (6.23) are not all inde­ pendent. By sustituting (6.20) and (6.21) into (6.17), we get

R2 — <*• L2 cx = 6 . 3 1 r 2 - ^(i^ + **») 116

R 2 ~ P^*2 Dx = ----- 1 B* 6.32 R 2 ” P (l2 + Lm) Similarly from (6.22), (6.23) and (6.18) we get

Ro — ^-Lq Cy = Ay 6.33 R2 — <*-(1*2 + 6m )

r2 ~ PhZ . D = By 6.34 Rg — piJ* 2 +

The factors Ax , Ay, Bx and By are independent integration cons­ tants and are determined by initial conditions. Let us assume that

all currents are zero at time — 0, when the switch is closed. Then

ilx = 0 and = 0 give

Ax + Bx + ^lxo ~ ® 6.35

Cx + Dx + Imxo = 0 6 .3 6

, where Ilxo ^lx ^=0) - *1 sin 6 from (6.24)

Inoco (= Imx at = Im ein(0 - S) from (6.26)

From (6.31), (6.32), (6.35) and (6.36) we get

I«xo C * 2 - [%- - Il*o [ « 2 - 'X (L 2+ L »>]

R2Lm( P -<*-) 6.37

- Xlxo(Rg-«.L2 )[R2 - p ( t a.f^)j Bx= ~ t & J f-*-) 6.38 Similarly we get _ Imyo(R2“

M w O -*) 6.39 117

Imjro[?2- {X'(I'2+L in)3 [h^-/3 (L^Lm)] - Ilyo(R2“ «■ L^) [^2* P(L 2+ L m )] B y ~ ------Z3 ~ a ) 6.4-0

, where Ilyo ( = *ly at — -Ii cos# from (6.25)

Imyo (=Imy at t=0) = -In, cos( 6 - S) from (6.27)

We have found all the integration constants in the current expression (6.20) - (6.23)* We shall now derive the standstill torque expression, using the above current expressions. The torque expression (6.8) becomes for the he lanced two—phase motor ( L ^ L ^ —Lj,, l 2x = L2 y = l 2)

T A m ~ *2x Ily - *lx *2y

— (Ijni “ ^Ix^ly “ ^Ix^^my ” *ly)

= ^mx *ly ~ *-iny ^lx 6 .4 1

Substituting (6.20) — (6.23) into (6.41) t WQ gat

— ^yAx)e + (DxBy — DyB^) e ^

+(CxBy + DxAy — CyB^ — DyAx)e ^ @^

+ (®xlly + Aylmx - Cyl^* - AxI,qy)e

+ (DxIly + ByImx “ Dy^lx ” Bx-tmy^°

+ (lm2lly - Inyll y) 6.42

Substituting the integration constants given by (6.31) — (6.34) and

(6.37) - (6.40) into the coefficients of (6.42), we get

^xAy “ CyAx — DxBy - DyBx — 0 6.43

^x®y + Bx^y ~ Cy®x - ByAx =

^mxo^iyo ” ^myo^lxo 6.44

Since the steady-state torque of the balanced two-phase motor is 118

constant, we get from (6.41)

Imx1 ly ~ ^ny^lx = ^mxo^lyo ~ ^ myo^lxo =

Illm sin£ = Tq / Lm 6.45

, where T0 is the starting torque under steady state. Then from

(6,LL) and (6.4 5) we get

CjBy t DxAy CyBj[ DyAx T=q /L jh 6 . 4 6

From equations (6.31) - (6.34.) and (6.37) - (6.40) , the coef­

ficient of e-(5L^ of (6.42) 1b given by

CxIly + Ayljux - CyljLx ~ AXIlJ5y =

Ao (R2 -«-L2) {R2- 0(^2 +Lm)f ein( w t +

Lm R2Lm( (J - <*) sin £

(R2 -/3L2){R2 - «:(L2 +Lm)} sin(u» t - JT )

RgL^p-*) Sin4

(R2 -

KR2Lm( ft -<* ) sin £

K [R2 - «(l2 +Lm)j { R2 - P(L2 +Lm)l Blnwt Sin 4* R2Lm( 0 - * ) 6.47

,where R2 + ( u> L2) K = 6.48 r | +U)2(L2 +Lm )2

Similarly, the coefficient of e~^^ in (6.42) is given by

Dx*iy + Bylmx - ^y^lx “ ^x^n^y =

*^o \ ^ 2 ~ P^2) j.^2 ~ 0t(IJ2 sin(*« t + 0

sin 4* 119

(1*2 — ^k 2) ^ ^2 ~ @ ^2 +km)J sin( u> t — £ ) + ^2^m( ** ~ P ) sin

K{R2 ~ cl(L2 +Lm)J{R2 ” ^(^2 +^*in)} sinu»t

R2Lm ( ^ - ^ )

Substituting (6.A3) - (6.A9) into (6.A2) w© get the standstill torque expression

(R2- 0£ 1 2 ) ^ 2- p (L2+Lm) sin( «> t + *" ) +(R2- /5L2){r2- * (L2+L,,,)} sin(^t-i ) + B2Lm(|3 - a£-) sinS -(lA)(R2-aL2 )(R2-PL2)ein«t - K{R2-t _ cC ------e

{H^L2+L,)j ,m(. t + t) „ ( L ^ S , ^ )

R^L^J cc - (3 ) sin &

-(l/K)(Ra-«L2)(Ra.-pLa)6in«.t - K{R*- (L*+L J}{R2-^(La+L,)} sina>t _

6.50

In (6.50) is given by (6.28). Usually **’(L2 + Then

(6.2.8) gives

- tan- 1 6.?1 2 R2 2

, where

6.52 120

In order to compare the standstill torque expression (6 .50) with

Wahl's^*^ standstill torque expression, £ is replaced by In

(6.51). Then (6.50) becomes

T = T0 £l + +

(R2-aL2 ){R2-p(L2+Lm)} c°s(^ t-

-(lA) (R2 - 1*2 ) (R2-0 L2) sin«*>t -K{R2-ot(L2+Lm)]{R2-^(L2+Lm)} sinu-t _ e

(R2- L2 ){R2-rt(Ii2+Lm)} cos (aJt-^p) -(R2-^L2){r2- K l24*I0 } c°s(»t+y)

RzL»v( ot ~ (3 ) cos?

-(l/K)(R2 -aL2 )(R2 -pL2) sina>t VSn)}{R2“^ L2+Lm)J sinwt ------— --- e P t

6.53

Comparing the torque expression (6.53) with the Wahl's

expression, we notice that some terms of (6 .53) are not present in

his torque expression. This is because his solution is an approxi­

mation. Notice that his current expressions are also approximations

and do not satisfy his original differential equation.

One important observation concerned with the torque expressions

(6 .50) and (6.53) is that these expressions are completely indepen­ dent of the initial conditions; that is, the initial steady—state currents Ilxo, ^Xyo* Imxo» -^myo aXid ^ phase angle 6 of the steady-state state current are not Included in the torque expression.

These torque expressions were derived for the balanced two—phase motor under the assumption that all initial currents are zero. Hence, we can conclude that the switching transient toraue is independent 121

of the switching instent for the balanced motor starting from stand­

still*

Let us calculate the torque expression (6.50) for the switching

transient torque of the three-phase motor, 220 V, l/A HP, 4. poles,

60-cycle, of Article 1 in Chapter IV. We had already the following

data.

R^ = 10.5 ohms = 6.9 ohms for cold state 7.05 L-, = “ ---- = 0.0187 henry 1 ^ 377 218 L = ---- = 0.578 henry 377

To change these machine constants into those of the equivalent two-

phase motor, they must be multiplied by 3/2. But actually this is

not necessery, because 3/2 is cancelled out from both denominators

and numerators of equations (6.30), (6.50) and (6.53). Then (6.30)

gives

o l -A. 66 ft — 6.15 6.54

And (6.28) gives

c o s ^ = 0.735 sini> = 0.678

Susstituting all the numerical values into (6.50), we get

T = To [l + e“^72*15t +(-cosu, t + 0.828 sin uj t)e“^66t

+ (-eosiut - 0.828 sin u> t) ^ - T© [l + e-472.15t + x 3

-1.3Bin(u> 14-50.3°) e~6 '151 J 6.55

* It is assume here also that all initial current are zero and all phases are closed simultaneously. Current in each phase depends on the switching instant. 1 ??

After half a cycle (1/120 sec.), e~4-7*:.15t e-466t are reciuced

to about 0.02. Hence, in equation (6.55) the terms with these factors

can be neglected for most proactical purposes. Then (6 .55) is simpli­

fied into

T = T0 [ 1 “ 1 . 3 s i n ( t + 50.3°) e-6*1^ ] 6.56

Equation (6 .56) gives the time-toraue curve in Figure 67. This

time

Figure 67. Time-torque curve given by (6 .56).

is a damped sine wave superimposed on the steady—state starting

torque T0 . The maximum torque occurs at the first peak of the curve

and its value is 2.24 TQ . The minimum values go into negative for

several cycles. Although Figure 67 is somewhat similar to Figure22

taken ty the analyzer for the same motor, they should not be the

same. In Figure 22 th« abscissa is motor speed, while in Figure 67 it is time. 123

Next, let us derive the parametric equation of the speed—torque

curve. From (6.56) the motor speed is given by

S =j— ^-dt =Ao J _... T e-6.15t =_LQ ^ + 1*3 ~o ~ " 2 f6.15sin( u> t + 50.3°) + cos( u) t+50.3°)i Cv -h (6.15) J

1.3 “i - (6.15ein50.3° + cos50.3°) +(6.15)*

T• o t + 0.003^5e~6"'t cos(K* t + 50.3°) - 0.0022 6.57 J

,where J is the moment of inertia of the motor. When (6.56) and

(6.57) are combined to get the speed-torque curve, the damped sine wave in Figure 67 is changed into the wave form somewhat like a damped trochoid, as it is in Figure 22. Aside from this difference in form, the calculated time-torque curve in Figure 67 explains well the first portion of the speed-torque curve in Figure 22.

Another difference between these two curve is the larger damping in Figure 22 than in Figure 67. This is due to the time constant of Figure 22 being smaller than that of Figure 67. In

Figure 67 the time constant of the curve is given by (6.56) as

'7' — 2" — — —— = 0.1626 sec. — 9*6 cycles * 6.15 For Figure 22 we should have a time constsnt of about A or $ cycles.

This difference in time constant is caused by the motor speed, which was assumed zero in our analysis, and this will be explained next.

When the motor speed, assumed to be held constant at any par­ ticular speed, is not zero, the simultaneous equations (6.1) - (6.6) 12A

become for the balanced two-phase motor

elx ” LR1 + (Ll + Lm) P]1^ Lm P *2x 6.58

esx = Lm p *lx + [R2 + p] *2x 6.59 ely = [R1 + (Ll + Lm) p] *iy+ LmP *-2y 6.60

esy = Lm p ^ y + LR2 +

esy = S CLm ilx + (L2 + L u ) ^ x ] 6.63 Substituting (6.62) and (6.63) into (6.59) and (6.61) respectively, we get

0 - -SI*m ilx + Lm P *ly ”^(L2 + Lm) *2x + CRr + (1*2 +I*m)p|12y 6.67

The simultaneous equations (6.64) - (6.67) are not so easy to solve as the simultaneous equations (6.10) - (6.13), because the equations

(6.64.) - (6.67) cannot be divided into two—independent sets of simul­ taneous equations, as was the case with the equations (6.10) - (6.1 3 ).

In the following we shall not derive the complete solution, but shall derive the characteristic roots of the equations (6.6-4) - (6.67), to see how the time constants of the transient phenomenon are affected

6y the motor speed.

The characteristic equation of the simultaneous equations (6.64.)

- (6.67) is 125

R1+ (LX + Lm)P O 0 0 0

S [LXL2 + Lm (Ll 4- L 2 )] +

2 CI*m^Ll + l 2) + l 1l 2] VTML2+Lm) R2fLl+I*m) — J s{LlL2+Lm(Ll+L2))J + ^lR2Lm 6.69 2 [Lm(Ll + L2 ) + Lll*2j

, where + signs in front of both jS is taken the same one at a time.

When S = 0, the p's in (6 .69) are reduced to ot- or /3 in (6.30).

Now the solution for current has the form given by

i = Aieplt + Ago^2^ + A3ep3t + A^eP4.t + I 6.70 where A 1s are integration constants and I is steady-state instan­ taneous current. Since the p’s are complex numbers, the transient components are now oscillatory, while the transient components given by (6.20) — (6.23) for standstill are nonosdilatory.

We shall now calculate the time constants of the three-phase induction motor, 220 V, 1/4. poles, 60—cycle, for synchronous speed and for standstill. For synchronous speed (S = 377), (6 .69) gives

p = - 381.8 + j 126.1 or - 90.2 + j 250.9 6.71 The time constants corresponding to ’these p* s are

— 1/381.8 — 0.00262 sec. = 0.16 cycle 6.72 126

^ - 1/90.2 = 0.0111 sec. = 0.67 cycle 6.72

The time constants for standstill are given from (6.54) as below.

n:fo~ 1 /«_ = 0.13 cycle 6.73 7-2i.-,= 1//3 = 9*8 cycle

Comparing (6.72) with (6.73) , we see that the time constant

at synchronous speed is slightly large than the corresponding time

constant 7r/a at standstill and the time constant at synchronous

speed is much smaller than the sorresponding time constant at

standstill. The same tendency holds for the motor speed other than

synchronous speed. The larger time constant *7^and determine

the duration tine of the switching transient and 7^ < This is

the reason why the switching transient torque ripple in Figure 22, where the motor speed iB changing continually, attenuates faster

than in Figure 67, where the motor is at standstill.

3* Solution of the Transient Phenomena of the Three-phase Induction

Motor (220 V, 1/4 HP, 4 poles, 60-cycle) by the Electronic Analog

Computer.

AB we saw in Article of this chapter, the solution of the switch ing transient of the balanced induction motor involves considerable amount of work even for the motor at standstill. If the motor speed changes continually, as Is the case with the actual tests, not only the number of equations to be solved Is increased to ei^ht (6.8),

(6.9) * (6.58) - (6.63) t but the new simultaneous equations become nonlinear. The only possible way of solving the equations might be 127

the use of an analog computer.

The Reeves Electronic Analog Computer was used to solve the

problem. After the analogy of the problem is set up in the analog

computer, it traces currents, voltages, motor speed, and torque of

the phenomena under consideration on the recording paper. The details

of the setting up of the analog computer are given in Appendix III.

Here the results recorded will be discussed.

In Figure 68 six curves are given. They are torque, motor

speed, stator current, magnetizing cureent, Impressed voltage per phase and speed voltage, as indicated in the figure. As explained in Appendix III, the equivalent two-phase motor is used to set up the analogy In the computer. But the scales given in Figure 68 are for the actual currents, voltages, torque and speed of the original three-phase motor.'

When we compare the time-torque curve in Figure 68 with the speed-torque curve on the same motor in Figure 22, we see that the switching transient torque ripple in Figure 68 is a damped sine wave, while that in Figure 22 is troehoidal. There is another small difference between the two curves. That is, the instantaneous torque in Figure 22 goes into negative values for the first several cycles, but it doeB not In Figure 68. This can be explained by the fact that in the actual motor steady-state torque is much lower that the theoretical value in the lower speed range, as Figure AS shows.

Except these two differences Figure 68 explains Figure 22 well.

Figure 68 was taken by switching in at the instant, when one 128

phase has maximum voltage and the other phaee has zero voltage in

the equivalent balanced two-pheee motor. Even when the voltage

phase at the switching instant was changed , the same torque curve

was obtained, although trie transient currents were changed* This

fact confirms the conclusion derived in Article 2 of this chapter

that the switching torque is independent of the switching instant,

if all initial currents are zero and the motor is balanced.

Figure 69 showB similar curves as Figure 68. But es the time- speed curve shows, the motor speed was kept at zero for a while after switch-in, and then it was allowed to increase continually after the point marked "release". This test simulates the released case as shown in Figure 23. The time—torque curve after the point marked "release" corresponds to the speed-torque curve in Figure23«

The time-torque curve before the point marked "release" represents the transient torque under the locked condition. The damping of the switching transient torque under the locked condition in Figure

69 is much slower than in Figure 68. From the torque curve in

Figure 69 we get the time constant to be about 10 cycles. In Article

2 of this chapter we got the time constant of 9-8 cycles as eqution

(6.73) shows. The agreement is good.

At the point of release in Figure 69 the switching transient is practically over and the time-torque curve remains smooth after that.

AXter the point of release, the motor speed increases as indicated in the time—speed curve. These two curves combined give the speed- torque curve shown in Figure 70, where the continuous line is the 70ty*€. / \ / y

i / \ / \ ■/ \ /' V./ ’ 32.Q^6yw. w

j i L___ W / o

"\ _/— — j./ \ / \ / \ _ ^ _ x ■ • / ■v ^ w -A- V , \ / ~V_

'X. ■ ' \ "\.. V / '

voltaic p&t. j*&*te.

"ffe. .4wC&&~ » «4 c.& + * a 1 = 0

SpW. wttaaP- /Tfc^-r U^tstcdnJ- CaML, ~TKw*l- 1 U . O V~7 '/+H P, &/>*{*•;

/' / \ r v y ./ J J

A / \ / \ ,A / J \ J \ J ' / \ / \ / \ I \ / \ / ' / V 3^0 4yi. wcto / ^ /___ V /___ V/ ^

J, SfW-

Z7f°|T - — ..

SW

8777 «*H'' \ 4 — \— /— a — t— v A

A X „ X V

JmpTtwi. p»v- I

17S.S L.&c&ad- iajL £LAJ>A- . ULm.-p£U*&— AZOV/ %-ttP, 4- />0£l6, .

VmPntKn _

'VWWWWWWV/

131

curve obtained from Figure 69 and the broken line is the calculated

curve given by Table 5 in Article V. The agreement is good.

1000 j4> oi * 800 o LOR 600 a> 3 uv* 4-00 o E-> 200

o .8 .6 •4 .2 0 Slip

figure 70. Speed-torque curve of the three-phase motor, 220 V, 1/4 HP, 4 poles, 60-cycle. Continuous line - the analog computer. Broken line - calculated value in Talbe 5.

The effect of acceleration on speed-torque curve was investi­

gated. For this purpose, the same test as in Figure 69 , which simu­

lates the released case of the actual motor, were performed with

quicker accelerztion after the release of the rotor. To get quicker

acceleration, the equivalent moment of inertia in the computer was decreased to one tenth of the original value. (Acceleration i* ten

times quicker.) But the speed-torque curves obtained from the

computer recordings were the same as Figure 70. From this fact we can conclude that if the rotor is released after the switching transient is over, the motor acceleration does not introduce an 132 electrical transient for practical cases, but the torque at each speed is the same as the steady-statetorque under corresponding constant speed. This is an important fact, which makes it possible for the analyzer to obtaine steady—state speed—torqu curves by the tests under mechanical transient. 133 Chapter VII

Conclusion and Acknowledgment

First, a brief summary will be given. The principle and the

construction of the speed-torque analyzer developed in the course of

the present researoh have been presented and explained in Chapters II

and III.

The results obtained with the analyzer have been presented in

Chapter IV and compared with values obtained by a dynamometer and with

computed values in Chapters V and VI, to show that the analyzer gives

true speed-torque curves.

Some of the speed-torque curves taken with the analyzer in Chapter

IV revealed features of the curves which had neither been measured nor

indicated previously. Theoretical explanations have been attempted to

these features, which include switching transent torque, pulsating

torque of single-phase induction motors , the extraordinarily high

braking torque of a three-phase induction motor in the negative speed range, and the negative average torque of a plain single-phase Induc­ tion motor.

It may be concluded that the usefulness and the accuracy of the analyzer have been demonstrated.

In the present research the analyzer was applied only to tests of induction motors, but it can also be used to test any kind of electric motors. It may be used to test a newly designed motor to check the adequacy of its design. Or it may be used in the production 134

line of electric motors to detect their defects.

The analyzer is essentially a speed-acceleration analyzer.

Therefore, it can be used to investigate the speed-acceleration rela­

tion of any rotating machines.

It should be pointed out that several improvements are still necessary in order to make the practical applications of the analyzer

easier. These consist in improvements in the stability of the ampli­

fiers, adjustments of the amplifiers, and inprovement ofthe brush

contact of the horaopolar generator.

It has been shown that the braking torque of the three-phase

induction motor is increased considerably due to tooth-pulsation loss

caused by zigzag leakage flux. In some cases the braking torque

caused by tooth-pulsation loss is much larger than that caused by

the rotor cage winding itself. More attention should be paid

to this phenomenon in order to take advantage of it.

The author expresses his sincere thanks to Professor K. Y. Tang

for his guidance throughout the research. He also thanks Mr L. V.

Lewis and Mr. J. R. Wolf for their advice and excellent work in building the homopolar generator, and Mr D. Ossing for his help in building the amplifiers. 135 Appendix I

Some Experiments about the Tooth-pulsation Loss due to Zigzag

Leakage Flux

In Article 3 of Chapter V we showed that tooth-pulsation loss due to zigzag leakage flux is quite large and is the main factor causing the extraordinary braking torque of the three-phase induction motor

(220 V, 1/4 HP, 4 poles, 60-cycle). There equations (5.Z7) - (5-30) were used to calculate the iron losses. But so far there is no ex­ periment supporting these formulas, because these losses are very difficult to measure directly. Here we shall present some experi­ ments, to show that these formulas give reasonable results.

The stator has open slots. A search coil of three turns was wound around one tooth-tip of the stator. When motor runs, voltage is induced in the search coil by magnetic flux passing through the tooth. The induced voltage was observed, using a cathode-ray oscil­ lograph. According to the observation, the voltage under no load is practically a sine wave of 60 cps, which is induced by the main air gap flux. As load current increases, a higher harmonic voltage began to appear superimposed on the fundamental sine wave. The wave form was as drawn in Figure 71. It was found from the observa-

Figure 71. Wave form of the voltage induced in a search coil wound around a tooth-tip of the stator. 136

tion that the frequency of the harmonic voltage was linearly pro­

portional to the rotor speed; that is, the frequency is given by

11 - s | 24 X 60 1* where s is slip. The number of rotor slots N£ is 48, and the

number of poles p is 4* Then number of dbts per pair of poles

2N2/P = 24* Hence equation (1) shows that this harmonic is caused

by the rotor slots.

But as explained above, no harmonic appears in the voltage

induced in the search coil under no load condition. This means that

the change of magnetic reluctance due to rotor slot openings is not the

cause for the marmonic voltage. As the load current increases, harmonic voltage appears in the induced wave form and it is caused by the zig­

zag leakage flux due to slot ampere-conductore of the rotor.

The relation between the harmonic voltage in the search coil and

the stator current was measured. The harmonic voltage was measured

from, the deflection length on the cathode-ray scope. The stator current was adjusted by changing the terminal voltage of the motor.

The motor speed was kept at slit = 2 and at slip = 1.5. The result is shown in Figure 72. The ordinate is the harmonic voltage (effective value) and the abscissa is the stator current. The curve shows satura­ tion.

Figure 73 shows the relation between the harmonic voltage (effec­ tive value) and the motor speed from slip=l to 2, while the stator

* At slip = 2 the number of the harmonic waves in Figure 71 is 24 for one wave length of 60 cps. 137

1300 rpm

0.3

0.2 900

B 0.1

1 2 3 5 6 7 Stator current (amp} Figure 72. Harmonic voltage induced in the search coil wound around a stator tooth. Motor speed was kept constant at slip 2 and 1.5.

be

20.1

200 4.00 600 800 1000 1200 U 0 0 1600 1800 Motor speed ( rpm ) Figure 73. Harmonic voltage induced in the search coil. Stator current was kept constant at 2 amp.. current was kept at 2 amp. by adjusting the terminal voltage of the motor. This figure shows that the harmonic voltage is proportional to the motor speed for constant stator current.

These two figures give the necessary data to compute the stator tooth-pulsation loss caused by zigzag leakage flux arising from the rotor. The zigzag leakage flux density will be computed first.

The wave form on the cathode-ray scope like that in Figure 71 shows that the harmonic voltage is practically sinusoidal. Hence, effective value V*> of the harmonic voltage is given by

/ 2 V„ = 3 “V ^ KBk2ls 10-S 2 where - effective value of the harmonic voltage

3 - number of turns of the search coil

4 ^ — maximum total zigzag leakage flux in one stator

tooth ariBing from the rotor

^sk2>> “ ro^or skew factor for the v'th harmonic voltage

- angular velocity of the v>th harmonic voltage

Now equation (2) will be applied to slip = 2 . For slip = 2 we have

stator current = 6.3 amp. from Figure 61.

— 0.^02 volt for 6.3 amp. from Figure 71.

W0 = 7 fr t\ 24 * 6 0

\) - 0.212

Substituting these numerical values into (2), we get

S VITa 0.402 x IQ8______— 9000 Maxwell ^ 3 x 2«X24 A 60X0.212 139

Then flux density = 9000/1.6 = 5630 Gauss, where 1.6 em^ is the cross section of one stator tooth.

Now the iron loss due to a flux density = 5630 Gauss will be computed. Since the frequency of the harmonic voltage is quite high, hysteresis loss is negligible compared with eddy current loss. Eddy current loss is given by (34)

(t£ watts/ Kg 3 where f = cps = 24 60 for slip 2

B - flux density = 5630 Gauss

^ = constant = 9 for 0.025 inch sheet

Then (3) gives _ ~ , 5630 24 * 60,2_____ ve“ 9 a 1(j3 ) = 590 wattB/ *3

Toatl eddy current loss is given by

590 a l.l * — = 325 watts 2 The corresponding figure in Table 7 in Article 3 of Chapter V is

383 watts for stator tooth-pulsation loss at slip 2. The agreement of the two values is not bad in view of the assumptions made in deriving equations (5*27) to (5*30), from which Table 7 was calculated.

This is an experimental proof that Table 7 gives reasonable values.

* Here 1.1 Kg is total weight of stator teeth. The factor 1/2 comes from the fact that the current is changing with the frequency of 60 cps, which is quite low compared with 24 60 cps of the harmonic frequency. The factor 1/2 gives the mean value of the maximum loss and zero loss around the inner surface of the stator. 140

Appendix II.

Derivation of the Equivalent Two—phase Induction Motor for

a Three-phase Induction Motor*

Whenever the solution of switching transient phenomena of a balanced three-phase induction motor was attempted, it was found convenient to replace the three-phase motor by an equivalent bal­ anced two— phase motor. But it seems that in the previous attempts derivation of the equivalent two—phase motor was not presented in detail(28)(29)(30)b Here its derivation will be presented in such a way that it is hoped a clearer picture of the equivalent balanced two-phase motor can be obtained*

N

2x VJ/2 n

L—o eb

i N/2 b

Fig. 74. A original three-phase Fig. 75# ^ equivalent three- squirrel—cage induction motor, phase motor.

Figure 74 shows a balanced three-phase squirrel-cage induction motor, which has effective number of turns N per phase on the stator< U 1

If the magnetic flux distribution in the air gap is assumed to be

sinusoidal , then the original motor of Figure 7 A can be replaced by

the motor in Figure 75 for the same phase currents ia , i^ and ic

and the same phase voltages ea , e^ and eo* In this figure, two of

the three windings are divided into component windings along the

two perpendicular axes, and the rotor cage winding is replaced by two

circuits along the two perpendicaular axes. The rotor currents ir2x

and i'2y are '^le values referred to the stator side.

For the five circuits in Figure 75, we get the following circuit

equations. d A*. ea - Rl 3*+ 1 dt d X b 2 ®b 3 R1 *b + dt d Ac ec = R1 ic + --- 3 dt d X 2%- II T so

e,sx k A

e * sy = R* 2 i-2, + d ^ 5 J dt where = stator resistance

R*2 = rotor resistance referred to the stator

X a , Ac, = total flux linkage of each stator circuit;

the subscripts indicate the circuit concerned.

* The number of turns 3/2 N and "(l/2)N of the component windings of the stator in Figure 75 are given by Ncos30° and Ncos 60° respectively, which are the projections of the original number of turns N on the two perpen­ dicular axes. The two components of each winding are connected in series, and the mmf1s of the two components of each winding produce the same air gap flux distribution as the original winding of N turns. U 2

®fsx* e,sy = speed voltage (voltage rise referred to the

stator); the second subscripts indicate the

circuit concerned.

t total flux linkage of each rotor circuit

(referred to the stator); the second subscripts

indicate the circuit concerned.

In order to express the flux linkages in terms of currents, self- and mutual inductances will now be determined. Let the magnetizing inductance of one stator circuit alone be denoted by L*m . Then the magnetizing inductance of one rotor circuit alone is also k' m» because all the rotor quantities, including the magnetizing inductance, are referred to the stator. With the aid of Figure 75, mutual inductances are given in, terms of L'm as below.

Mutual Inductance among stator circuits — L'm/2

w between rotor circuits x and y ~ 0

n between a-circuit and y-circuit = L'm

w between b— or c-circult and y-circult = L1 m/2

n between a-circuit and x-circuit = 0

" between b- or c-circuit and x-circuit = {/3/2)L*m

Summing up all self- and mutual flux linkages, we get

V a = (Li + Ii ) I* - (L*m/2)(ib + iC) + I* i^y 6

(Li + L*m) ib- (L'a/2)(ic + ±a) + (V3'/2)L'mi»23t-(L»n/2) i ’2y 7

\ . = (I*]. ^ k 1 m) ^ “(^dh/2) (^c *b) (^3/2)I*fin ^-2x “ ^f2y U 3

\ ' « . = (L-2 + L'm) 1'^+ (/3/2)L'm(ib- ic) 9

*z,= (L-2 + L'b) 1*^+ L>„ ia - (L-,/2)(ib + ic) 10

where L'm = magnetizing Inductance of one stator circuit alone

= stator leakage Inductance

L *2 - rotor leakage Inductance referred to the stator

The speed voltages (considered to be voltage rises) in (4) and

(5) are given by-

0,sx ~ ~ S 11

e 'sy = ® 12 Here S is angular velocity of the rotor and its positive direction

is indicated by an arrow in Figure 75 •

Here it should be pointed out that the primed quantities, such

as L'm , R'2» *'2x » ©be., are different from the corresponding

quantities used in the ordinary theory of three-phase induction motors.

This is because the theory used here is "Cross Field Theory", for

which number of stator phases should be assumed 2 instead of 3 in

referring rotor quantities to the stator side. If the corresponding

quantities for the ordinary theory of three-phase induction motor are

denoted by the same symbols without prime, then the relations between

the primed quantities and the unprimed quantities are as below.

R*2 = (2/3)R2 L»2 = (2/3) L 2 L'm = (2/3) Lm

i12x = (3/2) i2x 2y ~ (3/2) i2y 13 ®'sx = (2/3) ©sx 0,sy = (2/3) ®sy

* When the rotor quantities are referred to the stator, the reduction ratio to be multiplied has the factor n±/ m2= number of stator phase^/ 144 number of rotor phases (for resistances, inductances and voltages) or its inverse n^/®^ (for currents). For the ordinary theory of three- phase induction motors, m^ — 3, while for the cross field theory = 2. (This comes from the fact that the rotating magnetic field is 3/2 times the alternating magnetic field from one phase.) The factor 2/3 or 3/2 in (13) comes from the different number of the stator phases. The magnetizing inductance Lg of the ordinary theory of three-phase induc­ tion motors is 3/2 times the magnetizing inductance L ’m of one phase winding alone. i2y is one of the three rotor currents referred to the stator in the ordinary theory, and i2x * 1 /V3" (diffemce between two other currents). The same relation holds for esx and eBy. (^*ee equation 1^)

In (13) no stator quantities appear, because they are not referred quantities but the actual quantities and are the same for both theories.

Since this is a three-phase 3-wire induction motor, we have

ia l-b ^c = ® We have just set up all the necessary equations. We shall now transform these equations, in order to derive cirdfult equations for the equivalent balanced two-phase motor.

Subsituting (6) into (1) and replacing i^+ ic by-ia from (14), we get

ea = Rl i. + (Ll — + 14 'S dt dt

Subtracting , (3) from (2), and substituting (7) and (8) into the result of subtraction, we get

®b - ®c = R i U b " ic) + (Ll — — (ib - ic) + V J L ' m ^ ^ 16 2 dt dt

In order to obtain expressions for voltages and currents of the two-phase motor, voltages and currents in (15) and (16) are replaced by the following new voltages and currents on the left. Then we get

©ly=^R + — (L + ^L» ) d 1ly + 3 ______L1 d 1*2y18 y 2 l ly 2 1 o m ~ „ a 2 dt < dt

® l x = 4 Rl ilX +^

Having transformed equations (1), (2) and (3) into (18) and (19), we now transform equations (4) and (5)- Substituting (9),(10) into

(A) and (5), we get — d i ^ , V 3 e 'ax = R '2 ^ a , + (l,2 + * T L ; " ^ (ib “ ±o) 20

di'oy dia L* d

e;y =Ri + { L 4 + Lt-?— « - - i r t (ib + lo) 2 1

Eliminate ia , ib and ic from (20) and (21) , by substituting (14) and (17). Then we get (The result is multiplied by 3/2.)

3 3 _. . 3 ...... /3 \®. diix 2 •i* =—R4 2 i’ax*— 2 (t'2+ «••«>—- dt + tH-TT- 2 dt 22

3 3 3 , , dll2y / 3 \* dllv — e' = — R*2 i ' 2 y + — a * 2 + L'm) ------) Li ^ 2 y 2 ^ 2 4(1 dt 1 2 / dt

The speed voltages e,gx of (H) ©*6y of (12) are expressed in terms of currents, by substituting (9) and (10), as below. Eliminate ifl, and ic from (24-) and (25), by substituting (14) and

(17). Then we get (The result is multiplied by 3/2.)

3 3 (3 & e' = - S — (L» +• L ’ ) i* + /— IL* ii 26 2 £X 2 2 m I ? \ 2} m iy

+ L'b) 1J* + i< lx 27

Now in equations (18), (19), (22), (23), (26) and (27), the

primed quantities (quantities for the cross field theory) will be

replaced by the unprimed quantities (quantities for the ordinary

theory of three-phase motor) given by (13) • Then we get

dl2y 28 dt dt di. dl2x 29 dt 1 dt >dl2: 30 dt ldt % di. IBdh z 31 dt dt

esx s 32

e By = S (L2 + Lm) ±2 x 33

If we draw the mesh network for (28) - (33), we get the network in Figure 76. This is an equivalent circuit of a balanced two-phase induction motor. U 7

0 ■» 1< A A / -'THr---- 1 ^ cn---'s/s/'--- 1 (3/2)11! (3/2)Lx ^ (3/2)L£ (3/2) R. e ia1 y---- L 1(3/2) < L > sy

lx ,/qq. ■ jr V Oi—■/■pin- *yvv- -v \a - (3/2) R-l (3/2) Lx I (3/2) L£ (3/2) R,

"1X 3(3/2) l b 1. «;X

L_ ------L ------

Figure 76. Equivalent circuits of the balanced two—phase motor, which is equivalent to the three-phase induction motor.

In this equivalent circuit of a balanced two—phase motor all

machine constants are 3/2 times the corresponding machine constants

of the actual balanced three-phase motor. This two-phase motor can

be realized If the number of turns of the original stator windings is increased to J 3/2 times the original number of turns and the cross-section of the stator wire is reduced to-Jz/3 times the ori­ ginal cross-section, and on the stator only two such windings are installed. Thus, Figure 77 is the equivalent balanced two—phase motor, whose equivalent circuit is given in Figure 76. U 8

Figure 77. The equivalent balanced two—phase motor.

i lx 2y lx

The ratio between the quantities of the original three-phase motor ana those of the equivalent two-phase motor are given by (17) and Figures 76 and 77, and are tabulated in the table below.

Three-phase motor Two—phase motor

Number of effective turns 1 JV /2 in series per phase

Voltage 1 3/2

Current 1 1

Power 1 1

Impedance 1 3/2

Flux 1 3/2 U 9

Appendix III.

Solution of Transient Phenomena of a Balanced Three-phase

Induction Motor by the Reeves Analog Computer.

In order to solve the switching transient phenomena of the

balanced three-phase induction motor, the Reeves Electronic Analog

Computer was used, and the results were presented and discussed in

Article 3 of Chapter VI. Here the method of setting up the analog

computer will be explained.

As the name "Analog Computer" indicates, an electrical analogy of the equivalent balanced two-phase motor is set up in the computer.

The electrical analogy means that the circuit equations of the electric networks set up In the computer are analogous to the original differ­ ential equations of the problem. Hence variations of the electrical quantities in the computer are similar to the variation of the corre­ sponding quantities of the problem. After the analogy Is set up in the computer, it gives automatically recordings of time—variation of the quantities needed in solving the problem.

In setting up the analogy in the computer, the balanced three- phase induction motor is replaced by Its equivalent balanced two- phase motor derived in Appendix II. The equivalent circuit o£ the balanced two—phase motor is given in Figure 78. In this figure the factor 3/2 attached to all the machine constants in Figure 76 is included in the symbols (see Figures 76 and 78). When the numerical values are substituted, the factor 3/2 must be included. 150

Figure 78. Equivalent circuit of the two—phase motor, which is equivalent to the three-phase motor.

In Figure 78 the stator currents i^x and i^y and the magnetizing

currents i^* and imy are chosen as the independent currents for the mesh

equations. Thus

®lx = tRl + L1 P] i-lx + Lm P imx 1

esx = “ [R2 + L2 p] 1!^ + [r2 + (l2 + Lm) ^ 1b* 2

®ly = [R1 + L1 p3 Aly + ^ P iny 3

®sy = ~ |^2 + l2 p } 1 !* + {r2 + (l2 + L m) pl^my 4

The speed voltage rises eax and e Sy in (2) and (4) are given by

(See eq. 6.5 and eq. 6.6 in Chapter VI.)

esx = - S L2 1-ly + (L2 + Lm) imyj 5

°sy = S L2 ilx + (1*2 + I*n,) imxj 6

And the developed torque is given by (See eq. 6.8 in Chapter VI.)

T = LB [*1x *my - ^ly ^-mxj 7 Assuming no load on the motor, the equation of motion is

T = J ~ = J p S 8 dt where J is the moment of inertia of the rotating part and S is angular 151

velocity of the motor.

Now we shall transform the equations (1) to (4-) into such forms

so that the setting up of the analogy in the computer will be simplified.

Solving equations (1) and (2) for p i^x and p 1,^, we get

(L2 + Lm)®lx“ Lmesx “ ( ^ l + R1L2 + RlLm) ilx + ^n^-mx P *1* = Lm(Ll + L2^ + L1L2

^*2 elx + ^1 esx + (r 2l1 “ r 1l 2) ilx ~ R2L1 Imx p — 10 Lm(Ll + l 2^ + L1L2

Solving equations (3) ana (.4 ) for p i^y and p i ^ , we get

(L2 + Lia)®iy ~ esy “ (R2^1+ R1^2 + Rlkm)ily + R2^m Imy H p i]_y = — — - — I*m(I*l + 1*2 ) + ^1^2

l2 ®ly + L1 esy + (r2l1 “ R1L2) ily “ ^ L ^ m y p 1 = ------_ ------1 2 my , . Lm(Ll + L2^ + L1L2 For the three-phase motor (220 V, 1/4 HP, A poles, 60-cycle) ,

the machine constants are

= 10.5 X (3/2) Rj = 6.9 X (3/2)

Li = L2 = 0.0187 x (3/2) L„, = 0.578 x (3/2 )

Notice that all the constants are multiplied by 3/2, to change the

original values into those for the equivalent two—phase motor.

Substituting these numerical values into (9) - (12), we get

p *1* = £7-2 ®lx - 26-3 ®sx - A68 + 181.6 1 ^ 13 p V = 0.852 Blx + 0.852 «„ - 3.062 - 5.88 U

P V = 27-2 °ly " 26,3 V - 468 lly + 181.6 1,^. 15

p igy = 0.852 «iy + 0.852 e s y - 3.062 ily - 5.88 16 152

Substituting the numerical values for the machine constants, equations (5), (6 ) and (7) become

eBX = - S [- 0.0187 ily + 0.5967 i ^ ] 17

eSy = S [- 0.0187 iix +■ 0.5967 imx] 18

T = 0.867 [i-i* ira - ±xy i « ] 19

In the analog computer, all the original quantities are repre­ sented by voltages. The limits of the voltages are + 100 V. The ratio between the original numerical values and the corresponding voltages in the computer are called scale factors. In order to get adequate magnitudes of the voltages In the computer, the scale factors must be chosen properly. They are chosen as indicated in the talbe below.

Original quantities Scale factor Corresponding voltages in the computer elx» ®2x* esx» esy 1 Elx » E2x f Esx * Esy ilx t *-ly 1/20 I^x , I^y ^ - imy V200 imx • *my s a o o s

T 0.867A 0 0 T

t(tlme) 1/200 2 (time in the computer)

In this table the following numerical relations hold.

Original quantity = (scale factor)X(corresponding voltage in the computer)

For the time-derivative p, the following relation holds,

ft = P = 200-5 5 .= 20° P where P = ——d_ (derivative with respect to the computer time-zr) d'td' Replacing the original quantities with the corresponding voltages 153

in the computer by (20) and the time t wi-tfi t = eJ^d p with

P by (21), equations (13) - (19) become

PIlx = 2.72Elx - 2.63Esx - 2.34Iix + 0.09081^ 22

Pljnx = 0.852E1x + 0.852Eex - 0.153Hix - 0 . 0 2 9 ^ ^ 23

PIly = 2.72Ely - 2.63Esy - 2.3-41;^ + 0.09031w 2A

P I ^ = 0.852Ely + 0.852Eey - 0.1531Ily - 0.02941^ 25

EgX = — ^0.374liy + 1 • 19341 inyj 26

ESy = S [- o.37iIly + 1.193am ] 27

I = -^lx Iiny - Ily Ijnx 28

Now the original power supply voltages of the balanced two—phases

In Figure 78 are given by

elx = Ej, sin(377t +

®ly = - Em cos(377t + 9 ) 30 where is the maximum value of the voltage and

Elx = Em sin(1.8857r + SP) 31

Ely = ~ ^ cos( 1.885*2: + SP) 32 From (31) and (32) we get by differentiating with respect to °t

PZ E ± x = - (1.885)2 Elx 33

P Ely = 1.885 Ejj,. 34

Equations (22) - (28), (33) and (34-) are set up in the analog computer as the block diagrams in Figure 79 show. In this figure indicates an integrator, ^ an amplifier, - d > — a potentiometer, 15 Ar

M a servo. Numbers in these symbols in the diagram indicate amplification ( number > 1) or attenuation (number

Torque | , motor speed S » stator current , magnetizing current

Ijnx, impressed voltage E^x and speed voltage Egx were recorded on the running paper of the recorder. The results were presented and discussed in Article 3 of Chapter VI.

* A servo performs multiplication as indicated in the figure. Eix O (T2tT >

155

-crllx - I l x O —

O — { ^ 0 8 > PIlx“ 2.72Elx -2.63ESX- 2.34Ii;y+ .o908l•rax

E I jP ---

ES3p (TTl3> -O -I mx -ii9— d U i ) -O Imx -I o- mx PImx’ * 852E1x + * 852Egx- -15 3 H i x - .02941■mx

10

.263 10 o -I -I 58

my P^ly 2.72E2y-2.63ESy -2.341 j_y +.09081my

O-I my I l y O & 5 g ) ■my .0294 PIllly=.852E1y+.852ESy -.153H iy -.02941^

Figure 79* Block diagrams of the analog computer. *56 -O -E il y °----- C - w . ) ---- sx >----j------L A -O Eex I»y O----- C-1193*)-- H {> E6X = - S (-37411^+1.19341,

s O-

-I lx o---- ( 3 tZ > ---- O ■mx O---- (. 11934 sy

■O -E sy E6y= S(-.374Iix+ 1.1934IIDX)

ImxIly —Imx O- H

I mx—« o-

-Iiy O O T ■ 0

S -I H ^P>—° my

xmy ImyIl» T =IlxXnly-IlyImx T=(1/.928)p 5 -Ilx

3 >— rE>— -O E Figure 79- Block diagrams of lx the analog computer. -O E iy p 2e 1x ' PEiy^1.885Elx 157

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Springer, 1936, S. 49 - 57.

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Capacitor Motor. " Trans. A. I. E. E. ^8 (1929), pp. 614 — 629.

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Springer, 1936, S. 49-

14- Ditto, S. 53*

15* Ditto, S. 55 - 57.

16. Ditto, S. 219 - 220.

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19. Morgan, T. M. "Induction Motor Characteristics at High

Slip." Trans. A. I. E. E. j>2 (1940), pp. 464 - 468.

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E. E. 42, pp. 1209 - 1210.

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23. Ditto, S. 41.

24. Ditto, S. 66.

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Motor." Trans, of Institute of Electrical Engineers of Japan, pp. 127 - 134.

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Springer, 1936, S. 180. 159

• 27- Puchstein, A. F. and Lloyd, T. C. Alternating Current

Machines. New York: Wiley, 19-42, pp. 381 - 396.

28. Stanley, H. C. "An Analysis of the Induction Machines."

Trans. A. I. E. E. j>7 (1938), pp. 751 - 757.

29* Wahl, A. M. and Kilgore, L. A. "Transient Starting

Torques of Induction Motors." Trans. A. I. E. E. 60 (1941), pp. 603 - 607.

30. Gilfillan, E. S. Jr. and Kaplan, E. L. "Transient

Torques in Squirrel-Cage Induction Motors." Trans. A. I. E. E.

60 (1941), pp. 1200 - 1 2 0 9 .

31. Korn, G. A. and Kom, T. M. Electronic Analog Computer,

New York: McGraw-Hill, 1952.

32. Ditto, pp. 23-25.

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Springer, 1936, S. 49-

34* Richter, R. Elektrische Maschinen Bd. I. Berlin:

Springer, 1924, S. 157. Autobio graphy

I, Sakae Yamamura, was b o m In the city of Himeji, Japan,

February 26, 1918. I received ray secondary school education in the

Third High School in the city of Kyoto, Japan. My undergraduate training was obtained at the University of Tokyo, Japan, where I received the degree of Bachelor of Enineering in 19-41. In 19-47 I was appointed assistant professor in the Electrical Engineering Depart­ ment at the University of Tokyo, where I am now on leave of absence.

In 1950 I came to the United States, participating In the exchange of persons program sponsored by the United State Any. From Michigan

State College, I received the degree Master of Science in Electrical

Engineering in 1951. Since 1951, I have been a research assistant in the Electrical Engineering Department at the Ohio State University.

My research activities have been in the field of small electric motors.

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