electric motors and embedded electronics volume...

ElectricMotorsandEmbeddedElectronicsVol.3

Feb.19,2016/KL Motors_and_embedded_electronics_Vol03_v042.docx Page1

Electric Motors and Embedded Electronics

Volume 3

Kai LindgrenElectrical Engineering

Helsinki Metropolia UAS



ContentsForeword ................................................................................................................................................. 8

Acknowledgement ................................................................................................................................... 8

Version history ......................................................................................................................................... 9

1. Electric motors ................................................................................................................................12

1.1. Some criteria for classification .................................................................................................12

1.2. Synchronous motors ................................................................................................................17

1.2.1. Synchronous 1-phase motors ...........................................................................................17

Exercise 1. ...................................................................................................................................21

Problem 1. ...................................................................................................................................21

Exercise 2. ...................................................................................................................................22

Exercise 3. ...................................................................................................................................22

Problem 2. ...................................................................................................................................22

1.2.2. Synchronous 3-phase motors ...........................................................................................23

1.2.3. Synchronous 3-phase reluctance and salient pole motors ................................................29

1.2.4. The vector control of synchronous 3-phase motor ...........................................................32

Exercise 1. ...................................................................................................................................33

Exercise 2. ...................................................................................................................................35

Exercise 3. ...................................................................................................................................36

Exercise 4. ...................................................................................................................................37

Exercise 5. ...................................................................................................................................37

Exercise 6. ...................................................................................................................................38

Exercise 7. ...................................................................................................................................39

1.2.5. Power-angle characteristic of a round-rotor machine .......................................................39

Exercise 8. ...................................................................................................................................43

Exercise 9. ...................................................................................................................................44



Exercise 10. .................................................................................................................................46

Exercise 11. .................................................................................................................................48

Exercise 12. .................................................................................................................................49

Exercise 13. .................................................................................................................................50

Exercise 14. .................................................................................................................................51

Exercise 15. .................................................................................................................................51

Exercise 16. .................................................................................................................................51

Exercise 17. .................................................................................................................................52

1.2.6. Variable frequency drive ..................................................................................................56

1.2.7. Motor standards ..............................................................................................................59

Exercise 17. .................................................................................................................................62

Exercise 18. .................................................................................................................................62

Exercise 19. .................................................................................................................................64

1.3. Induction motors .....................................................................................................................66

1.3.1. A shaded-pole motor .......................................................................................................66

1.3.2. Principles of the induction motor .....................................................................................67

1.3.3. Stationary state of the induction motor ...........................................................................71

1.3.4. Characteristics of induction motor ...................................................................................74

1.3.5. Example induction motor 1 ..............................................................................................76

Exercise 1. ...................................................................................................................................77

Exercise 2. ...................................................................................................................................80

1.3.6. Casio Class Pad program for ex 2. .....................................................................................82


Exercise 3. ...................................................................................................................................83

Exercise 4a. .................................................................................................................................85

Exercise 4b. .................................................................................................................................86


1.3.9. Circle diagram of a polyphase induction motor ................................................................91



1.3.10. One phase equivalent circuit of the three phase induction motor ....................................95

Exercise 5. ...................................................................................................................................98

Exercise 6a. .................................................................................................................................99

Exercise 6b. ...............................................................................................................................100

Exercice 6c. ...............................................................................................................................101

Exercise 7. .................................................................................................................................102

Exercise 8. .................................................................................................................................103

Exercise 9. .................................................................................................................................104

Exercise 10. ...............................................................................................................................104

Exercise 11. ...............................................................................................................................105

Exercise 12. ...............................................................................................................................106

Exercise 13. ...............................................................................................................................106

Exercise 14. ...............................................................................................................................108

1.3.11. Analysis of the one phase induction motor/9/ ................................................................109

Exercise 15. ...............................................................................................................................112

Exercise 16. ...............................................................................................................................113

Exercise 17. ...............................................................................................................................114

Exercise 18. ...............................................................................................................................115

1.4. DC motors .............................................................................................................................116

1.4.1. Old structure ..................................................................................................................116

1.4.2. Cup shaped rotor with permanent magnet stator ..........................................................116

1.4.3. Disk shaped permanent magnet motor ..........................................................................118

1.4.4. Basic equations of permanent magnet motor ................................................................118

1.4.5. Characteristics of permanent magnet motor ..................................................................119

1.4.6. Brushless DC-motor .......................................................................................................119

1.4.7. Schematic models for DC-motor.....................................................................................121

Exercise 1. .................................................................................................................................130

1.4.8. EMF equation for DC-motor ...........................................................................................131



1.4.9. Torque equation for DC-motor .......................................................................................132

1.4.10. Speed equation for DC-motor ........................................................................................133

Exercise 2. .................................................................................................................................133

Exercise 3. .................................................................................................................................134

Exercise 4. .................................................................................................................................134

Exercise 5. .................................................................................................................................134

Exercise 6. .................................................................................................................................134

Exercise 7. .................................................................................................................................135

Exercise 8. .................................................................................................................................135

Exercise 9. .................................................................................................................................135

1.5. Stepper motors......................................................................................................................136

1.5.1. Half stepping ..................................................................................................................138

1.5.2. Micro stepping ..............................................................................................................138

1.5.3. Reluctance stepper motor ..............................................................................................139

1.5.4. Hybrid stepper motor ....................................................................................................139

1.5.5. Unipolar control .............................................................................................................141

1.5.6. Bipolar control ...............................................................................................................142

1.5.7. Serial resistance .............................................................................................................142

1.5.8. Chopper control .............................................................................................................143

1.5.9. Two voltage control .......................................................................................................143

1.5.10. Different control units ....................................................................................................145

1.5.11. Resonance phenomenon................................................................................................145

1.6. Servo motors .........................................................................................................................148

1.6.1. Servo amplifiers .............................................................................................................151

1.6.1.1. Linear amplifiers ........................................................................................................152

1.6.1.2. Chopper amplifiers .....................................................................................................153

1.6.1.3. Common control tasks ...............................................................................................153

1.6.1.4. Combined servo elements ..........................................................................................155



1.7. Exotic motors ........................................................................................................................156

1.7.1. Ball bearing motor .........................................................................................................156

1.7.2. Homo-polar motor .........................................................................................................156

1.7.3. Electrostatic motor ........................................................................................................157

1.7.4. Non-electric nanomotors ...............................................................................................158

1.7.5. Molecular motor ............................................................................................................159

1.7.6. Piezo motor ...................................................................................................................159

1.8. Model of a motor system .......................................................................................................162

1.8.1. Gears .............................................................................................................................164

1.8.2. Selecting the right gear for a servo motor ......................................................................169

Exercise 1. .................................................................................................................................171

Exercise 2. .................................................................................................................................172

1.9. Conclusions ...........................................................................................................................174

1.9.1 Synchronous Motors ......................................................................................................174

1.9.2. Direct Current Motors ....................................................................................................176

1.9.3. Stepper Motors ..............................................................................................................178

1.9.4. Servo Motors .................................................................................................................180

1.9.5. Gears .............................................................................................................................180

Appendix A: Most important equations.................................................................................................182

Appendix B: General rotating field ........................................................................................................184

B.1. The machine geometry ..........................................................................................................184

Exercise 1. .....................................................................................................................................186

B.2. Transformer Model................................................................................................................191

B.3. Phasor Model ........................................................................................................................195

Exercise 2. .....................................................................................................................................195

Exercise 3. .....................................................................................................................................196

Exercise 4. .....................................................................................................................................196

Exercise 5. .....................................................................................................................................197

Exercise 6. .....................................................................................................................................198



Exercise 7. .....................................................................................................................................198

Exercise 8. .....................................................................................................................................200

Exercise 9. .....................................................................................................................................200

Exercise 10. ...................................................................................................................................201

Exercise 11. ...................................................................................................................................201

Exercise 12. ...................................................................................................................................202

Exercise 13. ...................................................................................................................................203

B.2. Winding factors .....................................................................................................................204

B.3. Winding inductance calculation .............................................................................................206

B.3.1. Inductance expressions ..................................................................................................208

Problem 1. ....................................................................................................................................209

Problem 2. ....................................................................................................................................209

Problem 3. ....................................................................................................................................209

B.4. Rotating Model of a General Motor .......................................................................................209

B.4.1. General Non-rotating Model ..........................................................................................210

B.4.2. General Rotating Model .................................................................................................212

B.4.3. The phasor model of a multi-phase motor .....................................................................213

B.4.4. Simple dispersive model of a general machine ...............................................................214

B.4.5. Simple rotating voltage model .......................................................................................216

B.4.6. Realistic rotating voltage model .....................................................................................217

Exercise 14. ...................................................................................................................................219

B.5. Converting the general machine to the mechanical world ......................................................220

Appendix C: Machine Sizing ..................................................................................................................221

Appendix D: EV Design Example ............................................................................................................225

Exercise 1. .....................................................................................................................................233

Appendix E: Servomotor Specifications, Oriental Motor Catalogue 2012-2013. .....................................235

Literature ..............................................................................................................................................242



ForewordThe exercises and problems marked with grey background colour, if there are any, are not recommended,for they use phasor model of motor or otherwise are not considered to be useful.

AcknowledgementThe model for this writing has been the booklet “Guide to choose a small motor” written in Finnish by Mr.Tapio Verho. Now the text is here freely translated in English and pictures are almost directly copied fromthe original text. This material is included by the kind permission of the companies Stig Wahlström Oy andMovetec Oy. The new writing here, however, includes a wider perspective into the subject.

Many pages of the book:

”Matti Mård: Sähkökäyttö ja tehoelektroniikka, 354 sivua, Otatieto, teosnumero 889. Jyväskylä1993.Or in English:Matti Mård: Electric Drives and Power Electronics, 354 pages in Finnish edition, Otatieto, opusnumber 889. Jyväskylä 1993.”

are freely translated here to English in 2011-2013 by Kai Lindgren with the kind permission of the professorMatti Mård.

Wikipedia is used as a source to find new perspectives for the development of electric motors.



VersionhistoryVersion 01 was started on 1th of July in 2012.

Version 03 was written on 11th of July in 2012.

Version 04 was written on 12th of July in 2012. This is the first version, which includes all the originalmaterial in /1/.

Version 06 was written on 23th of August in 2012. In this version, there is included the text of permission ofthe copyright owners to use the original text /1/.

Version 07 was started on 4th of November in 2012 using the text of Vol. 3 version 06 as the starting point.

Version 07 was continued on 7th of Nov. 2012.

Version 07 was continued on 8th of Nov. 2012.



Version 08 was made on 9th of Nov. 2012.

Version 09 was made on 18th of Nov. 2012.

Version 10 was made on 23rd of Nov. 2012.

Version 11 was published on 2nd of Dec. 2012.

Version 12 was written on 10th of Dec. 2012; but unchecked and full of errors and contradictions.

Version 13 was completed on 15th of Dec. 2012; contradictions between main models largely corrected.

Version 14 was completed on 15th of Dec. 2012 with the intention to clarify the presentation.

Version 15 was completed on 15th of Dec. 2012. Pictures were added to the induction motor circle diagrampresentation.

Version 16: Some exercises added or corrected. Dec. 24, 2012.

Version 17: Some exercises and theory added and some exercises corrected. Dec. 27, 2012.

Version 18: Some errors are corrected. Dec. 27, 2012.

Version 19: Some errors are corrected. Dec. 28, 2012.

Versions 20-21. Texts in Figures have been translated to English. Dec. 29, 2012.

Version 22: Some new exercises are added. Essentially all kind of exercises needed are presented. Jan. 1,2013.

Version 23: Small corrections. Jan. 5, 2013.

Version 24-25: Corrections. Jan. 11, 2013.

Versions 26-28: New examples and some theory are added mainly from /9/. Jan. 25, 2013.

Version 29: Theory of gears added. Feb. 6, 2013.

Version 30-31: Some theory of dc and synchronous motors added. Some exercises of synchronous and dcmotors added or developed. Exercises of choosing motor-gear pair are added. Feb. 23, 2013.

Version 32-34: May 15-16, 2013. The theory of rotational coordinate systems of D. Y. Ohm in /8/ is added.Approximate methods are compared against more accurate method in some exercise solutions. Somemotor parameter sets are newly interpreted to represent wye-connected motors as opposed to the oldinterpretation of delta-connection.

Version 35: Some texts are refined. May 17, 2013.

Version 36: The purpose of this version has been to make the vocabulary more coherent. Aug. 7, 2013.

Version 37: Appendixes added according to Kwang Hee Nam /11/: “Machine Sizing” and “EV DesignExample”.



Version 38: Some model names are changed. Some theory of windings added, some exercises added, someminor error corrected. Aug. 25 - Oct. 19, 2013.

Version 39: Started on 20th of Oct. 2013.

Version 40: Started on 29th of Dec. 2013, ended on 3rd of Jan. 2014: Intention to calculate again theunsuccessful exercise solutions. In some there was confusion between mechanical and electrical frequency.Frequencies in some stationary equations checked. (More careful examination of stationary frequenciesmade in the section “B.4. Rotating Model of a General Motor.”)

Version 41: Started on 3rd of Jan. 2013. Same new exercises are intended to be added.



1. Electricmotors

1.1. Somecriteriaforclassification

Automatic control systems arenot as easily available for allkind of motors. It is probablethat most versatile systems aremade by many manufacturersto DC motors, both for modelswith brushes and brushless orelectrically commutatedmodels. You may take thefeedback signal to somecontrollers as a proportionalvoltage and to others as a pulsetrain. Galil Motion Control andPhoenix Contact are examplesof controller manufacturerswhich are gladlyrecommended. You can definewith the control parameters acomplicated movement and

the controller makes the motor to obey it. Also, you may easily define yourmovements as a cascaded e.g. speed-position control with lots of parameters andmany kind of options. In the Figure 1 you see as an example a piece of a controllerprogram.

The performance characteristics, as seen in the Figure 2, are of course an essentialcriterion for choosing the motor. However, it is maybe not so readily recognized thatall motors are not as free to maneuver: some can rotate only in one direction and

Figure 1. Program example. Stig Wahlström Oy.



some only with one or some specific speeds; also the power may be fed as DC-, ac-,or pulsed currents.

Figure 2. Example characteristics. Stig Wahlström Oy.

There may be combined control electronics or gear in the motor. The mechanicalendurance of the gear must be chosen to tolerate the torque generated in allsituations; see Figure 3 for different kind of gears. Also, the possible combined controlelectronics determines in some extent the control actions available.



Figure 3. Different gears in size and translation

In the Figure 4, there are seen different sensors to feel the state to be controlled bythe motor. These may be e.g. rotational speed or position or torque, or manyparameters of the motor may be controlled in a cascade.



Figure 4. Sensors used to feel the state of the motor

In the Figure 5, there is shown an example of electrical control used to control in thiscase a brushless DC-motor. Notice that current (torque) and speed are herecontrolled in a cascade.



Figure 5. An example of control electronics used to control an electric motor.



1.2. Synchronousmotors

Figure 6. Taxonomy of AC motors.

1.2.1. Synchronous1-phasemotors

Synchronous motors rotate in general in the pace of the alternating supply current.There are two main categories of synchronous motors. In the Figure 7, there is seen amotor with a small number of pole pairs: typically 5, 6, 10 or 12. This kind of motorrotates thus with a 50 Hz supply with corresponding speeds of 600, 500, 300 or 250rpm. If a synchronous motor is rotated with a one phase supply it starts to rotate inan undefined direction, if there is not used any means to define the direction. E.g. itis possible to use a hook mechanism so that if the motor starts to rotate in wrongdirection, it bounces from the hook back to the right direction. If the one phasesynchronous motor needs to rotate in two directions, a common structure is to usetwo similar pole series in such a phase difference that the poles of one series arebetween the poles of the other. This kind of synchronous motors are economical tomanufacture; for enhance the economy they usually use bushings. The motors



generate typically a week torque and are supplied with a gear which makes them tobe still more durable: life time from three to ten years is typical. Power range maybe from a fraction of a watt to some watts.

Figure 7. Synchronous motor with a small number of pole pairs1. Outer end of the first stator winding2. First stator winding with bushing to the rotor axis3. Permanent magnet rotor4. Inner ends of the stator windings5. Second stator winding with bushing to the rotor axis6. Outer end of the second stator winding

A multi-pole synchronous motor has typically 50 pole pairs; a typical structure ispresented in the Figure 8. The rotational speed is thus 60 rpm with 50 Hz supply. Thiskind of motor has two pole series as mentioned earlier; the structure is presented inmore details in the section of stepper motors. The torque of the multi-pole motor isoften big enough without a gear and the structure of the motor is robust using firmcasing materials and ball bearings to endure the higher torque. A fewer pole motorhas often a casing made of plate.



Figure 8. Structure of a multi-pole one-phase synchronous motor.1. Cover for the connection box2. Connectors3. End of the motor4. Ball bearing5. Permanent magnet rotor6. Wired stator7. Casing of the motor8. End of the motor.

The maximal torque, as a function of the rotational frequency, has a simple form asseen in the Figure 9. With constant supply frequency there is torque only atcorresponding rotational speed, if the maximal torque is not exceeded. If exceededthe motor drops from pace but starts again to work properly when the load hasbeen reduced sufficiently.

If you want your synchronous motor to develop the same torque with differentspeeds and constant stator current, you must keep the ratio VS/ω constant, for themotor obeys the following equations:

ΨS = VS/jω,

T = iR x ΨS.

Here ΨS is the rotating flux linkage or winding flux, VS the stator supply voltage, ω =pN · ωs the electrical angular velocity or the angular velocity of the supply (in acontinuous state the rotational speed and the synchronous one are the same, ωm =ωs), T the machine torque and iR the rotor current. - So, if you want more motorspeed, ωm, you must add voltage also, if you want the torque to remain in the oldvalue.



Figure 9

When changing the frequency of the power supplied to an AC-motor, the ratio ofthe applied voltage to the applied frequency (V/Hz) is generally maintained at aconstant value (constant torque operation) until the maximum of the supply voltageis reached. Operation at a constant voltage (maximum voltage), above this limitprovides a reduced torque capability (V/Hz) and a constant power. The speed withwhich the back voltage has the same magnitude as the supply voltage is called asthe base frequency or speed, see the Figure 10. Function with larger than base speedis possible with the so called weakening of field method if the field (magnetizingfield, not the armature field) is not made with permanent magnets.

Figure 10.

If your one phase synchronous motor is made to rotate in one direction only, youconnect the ac-power to it with two leads without any additional componentsexcept the hook mechanism explained earlier.

However, if your one phase synchronous motor is made to rotate in two directionswith two separate pole sets, you connect your one phase supply to one of the coilsets directly and to the other with a serial capacitor, as seen in the Figure 11. You canchange the direction of the motor with the switch shown in the figure.



Exercise 1. If the impedance of a coil set in one phase synchronous motor is ZL = 10 Ωand ZR = 10 Ω, what should be the impedance of the serial capacitor?

Solution. The current in the coil with the serial capacitor should be in 90˚ phase shift with the bare one. Ifthe impedance of the capacitor is ZC = -j20 Ω, the currents are as follows:

IL = Vac/(10 Ω + j10 Ω) and IC = Vac/(10 Ω - j10 Ω),

which means there is a phase shift of 90˚. For the voltages in the coils we get, also, the same 90˚ phaseshift:

Vbare = Vac and Vserial = (10 Ω + j10 Ω) · Vac/(10 Ω - j10 Ω) = jVac.

However, if ZC is kept constant and ZR -> 0, we get: Vbare = Vac and Vserial = -Vac; thus the phase shift goes to180˚.

Problem 1. Compute the right value of ZCn without guessing it beforehand.

Figure 11. One phase synchronous motor with two directions.

It is interesting to think the triangle connected three phase synchronous motor ofthe Figure 12. Obviously it is easiest to use with conventional three phase current.However, it can also be used with one phase current if an extra capacitor is added asin the Figure 13.



Figure 12. Three phase synchronous motor

Exercise 2. In the Figure 13 we need the following condition to apply, if the function isto be with one phase exactly as with three phases:

Vn = (eSj120˚Vn/ZL)[ZCnZL/(ZCn+ZL)]+[eSj120˚Vn]

ð 1 = (eSj120˚)[ZCn/(ZCn+ZL)]+[ej120˚]

ð 1 - eSj120˚= (eSj120˚)[ZCn/(ZCn+ZL)]

ð e-Sj120˚-1 = ZCn/(ZCn+ZL)

ð sqrt(3)e-Sj150˚= ZCn/(ZCn+ZL)

ð ZCn(1-sqrt(3)e-Sj150˚)= sqrt(3)e-Sj150˚ ZL

ð ZCn = 0.65465e-Sj169.11˚ZL.

We can also write:

ð ZL = 1.527525 eSj169.11˚ZCn.

This means that S = +1 and arg(ZL) = 79.89˚; so ZL may not be purely reactive.

Exercise 3. These conditions could also be found as follows here next.

ILej120˚ = ej120˚Vn/ZL = Vn/(ZCn||ZL + ZL) => ej120˚/ZL = 1/(ZCn||ZL + ZL)

ð ej120˚/ZL = 1/[ZCn·ZL/(ZCn+ZL) + ZL]

ð ej120˚/ZL = (ZCn+ZL)/[ZCn·ZL + ZL(ZCn+ZL)]

ð ej120˚[ZCn·ZL + ZL(ZCn+ZL)] = ZL (ZCn+ZL)…

Problem 2. Simplify this equation for the relationship between ZL and ZCn and showthat it is the same what we got with the first method.

This equation tells us that adding a vector B = ZCn·ZL to vector A = ZL(ZCn+ZL) does notchange the vector A at all, if we also rotate the sum vector 120˚ counter clockwise…We have now seen that a real three phase motor can be used as well with one



phase only if its impedance obeys the rule: With some capacitor Cn the impedance ZL

of the coil of one phase of the motor actualises the equation ZL = 1.527525eSj169.11˚ZCn. This poses firstly a condition to the frequency of the ac current and thenfor the capacitor Cn. However, if this condition is not met exactly or is only veryroughly obeyed, the one phase usage could still work satisfactory in many cases.

Figure 13. Conventional three phase motor used with one phase supply.

1.2.2. Synchronous3-phasemotorsThe structure of a general 3-phase electric motor is seen in the Figure 14. Thisprincipled motor has only one pole pair. The three stator coils, a0, a1 and a2, are fedwith 3-phase ac-current and generate the rotating stator field. The rotor coil F is fedwith DC-current and it generates the rotor field. The coils Q and D represent thedamping coils in the rotor. The damping coils are connected to form a shortcircuited cage-winding, similar to the one in the rotor of the induction motor. Ifrotor and stator field move, in respect to each other, a current is generated in thecage-winding, which counter acts that movement; that’s what gives the name to thedamping winding. In steady operation, the rotor and stator field rotate with thesame velocity and there is no current in the damping winding.

The damping winding may also be used in the starting up the synchronous motor asan induction motor. In this operation the rotor current is not connected on beforethe slip velocity has become sufficiently small. The current in damping windingproduces always power losses, so a smooth control-electronics can save in powerconsumption. If the motor has control electronics, which can control and start itsmoothly, the damping winding is not needed.



Figure 14. Coils in a general salient pole 3-phase motor.- a0, a1 and a2 are the stator coils,- F is the magnetising rotor coil,- Q and D are the damper coils perpendicular and in direction to the pole flux.

The damping winding and the stator winding can be seen to constitute atransformer as shown in the left side of the Figure 15. On the right side thetransformer is depicted as the magnetising inductance only.

Figure 15. The Stator and damping winding create the transformer in the left picture. In the right picture, the device is thought tobe a 1:1 transformer and represented only by the magnetising inductance. This model does not include the magnetising rotorcurrent source.

From the right side model, the function of the damping winding is described withthe following equations. This model, however, does not include the mutuallyrotating fields and describes therefore a non-rotating motor:

.

These can be solved for Δu as:



This has as the solution:

Im = (ΔV/RS)(1-e-t/τ).

Here the time constant τ is:

τ = Lp/R|| and R|| = RS||RD.

The τ gets the value ∞, when RD = 0, in which case the machine does not magnetiseat all, for pΨ = Lppim = 0. This implies also: p(iS + iD) = 0, and ΔiS = ΔV/RS = -ΔiD. Thusthe stator current can be controlled directly against RS with ΔV for a short time, andnothing prevents of continuing the control as needed.

If we add to our circuit the magnetising supply, VF, of the rotor we get the circuit ofFigure 16. This model includes the rotation of the rotor. In this particular model theco-ordinate system rotates in synchrony with the magnetic field of the stator. Seemore details in the appendix section B.4 “Rotating Model of General Motor”. Thecircuit applies for every time point, t, separately; this means that jωΨ and jωRΨneed to be considered as constant voltages for every time point. The angular slipfrequency ωr here, must be interpreted in electrical unit; i.e. ωr = pN(ωS-ωm) = ω -pNωm = ω - ω0. Here ω0 is the electrical angular frequency of the rotor. If there isonly one pole pair in the stator and rotor, the electrical unit of ωR is the same as themechanical. ω is the angular frequency of the supply voltage and the stator field.

Figure 16. Damped synchronous machine without magnetic dispersion. Coordinate system rotates with the stator field with theangular frequency ω, which is the electrical frequency of the supply. ωr is the electrical slip frequency.

If we have good control system, we do not need damping winding and our circuitmodel becomes as in the Figure 17. Note that we do not have any more the



damping resistor RD; and RR corresponds now to RF and VR to VF. This circuit alsoapplies for every time point, t, separately.

Figure 17. Non-damped full rotor machine in a circuit model with magnetic dispersion. Coordinate system rotates with the statorfield with the angular frequency ω = ω1. ωr is the electrical slip frequency. ωr = ω - ω0 = ω - pNωm.

The phasor diagram of this circuit in a stationary state is depicted in the Figure 18.Now pL = jωL and we denote:

uX = jωLσSiS and uR = RSiS.



Figure 18.

We can control the flux with stator voltage as this:

If RS = 0 we get for the flux linkage or flux linkage:

We use the notation:

.

We also denote: iS’ = (LS/Lm)iS. The flux linkage ΨS = VS/ω which is determined by theconstant ratio is represented in the current diagram with a constant flux linkagecircle as in the Figure 19.



Figure 19. Current drawing based on the magnetising of the flux linkage of the stator.

The Torque is:

In the permanent magnet machines the torque can be easily controlled with thevector control by controlling the current IS directly against the rotor magnet:

Of course, the rotor position must be estimated or measured.

In network use, synchronous motor works as single controlled (as opposed todouble controlled). If RS = 0, we get stationary values for stator flux linkage andmagnetising current:

The torque is:

The allowable working region is shown in the Figure 20 below.



Figure 20. Constrains.- Pole angle limit- Rotor current limit as continuous current (dashed line) and constrained by supply devises (bold arc).- Stator current limit as nominal current (fine arc) and over current constrain (bold semicircle)

1.2.3. Synchronous3-phasereluctanceandsalientpolemotorsThe field winding in a round rotor (left) and salient rotor (right) are compared in theFigure 21 according to Nasar /9/.

Figure 21

The torque of a reluctance motor is calculated with the cross field principle:



Here we have:

With some trigonometry the torque becomes:

The torque is represented in the Figure 22 as the function of the pole angle δ. Thediameter of the circle is D = |imd-imq| which is considered here to be a constant. Thetorque is ½ΨD·sin2δ.

Figure 22.The diameter of the circle is D = |imd-imq| which is considered here to be a constant. The torque is proportional to ½D·sin2δ.

If we have a salient pole motor, it generates the sum moment which is the sum ofthe electromagnetic and reluctance moment, as shown in the Figure 23.



Figure 23

The analysis is done now be noticing that the magnetising current is (if rotor currentiR = 0, we have the reluctance motor):

The generation of the torque is show in the Figure 24.

Figure 24.The diameter of the circle is D = |imd-imq| which is considered here to be a constant. The reluctance torque is proportional to½D·sin2δ,and the electromagnetic torque is, as usual, proportional to iR·sinδ.The total torque is proportional to the T* shown in the figure.



1.2.4. Thevectorcontrolofsynchronous3-phasemotorHow to control a 3-phase synchronous motor is shown in the Figure 25 in the co-ordinate system of the stator field.

Figure 25. The control of synchronous 3-pahase motor in the co-ordinate system of the stator field.

Torque control gives the instruction current i*, which is represented with the poleco-ordinates as:

This decomposition can be traced from the graph below.



Figure 26. Stator current in rotor co-ordinate system.

In practice, it is given the value 0 for iQ. We get from the Figure 26:

.

In practise, iSq must be constrained to the value corresponding to the highest poleangle possible, iSqmax

*.

In a salient pole machine the torque has the value:

The transverse stator current is constrained to the value determined by theallowable transverse flux linkage as here:

Here iQest is the estimated transverse current in the damping winding caused by theconstrained transverse control.

Exercise 1. Let’s think that in a round rotor example motor, the winding basedparameters (the impedances are for one phase in line frequency) are as follows:

· Rating 300 hpVoltage 440 Vrms 1-1

254 Vrms 1-nStator resistance R1 .0073 ΩRotor resistance R2 .0064 ΩStator reactance X1 .06 ΩRotor reactance X2 .06 Ω



Magnetising reactance Xm 2.5 ΩMagnetising stator reactance XS 2.5 ΩSynchronous speed Ns 1200 rpm

Convert these parameters to phasor model for use in electric utilities (so, use theline frequency).

Solution. The new set of parameters is:

· Rating[kW] = 0.7457kW/hp x 300hp = 224 kWVoltage-p = sqrt(2)*440 = 622 Vp 1-1

sqrt(2)*254 = 359 Vp 1-nStator resistance R1-r = .0073*(2/3) = .00487 ΩRotor resistance R2-r = .0064*(2/3) = .00427 ΩStator reactance X1-r = .06*(2/3) = .04 ΩRotor reactance X2-r = .06*(2/3) = .04 ΩMagnetising reactance Xm-r = 2.5*(2/3) = 1.67 ΩMagnetising stator reactance XS-r = 2.5*(2/3) = 1.67 ΩSynchronous speed Ns 1200 rpmSynchronous speed Ns-e = 3*1200 = 3600 rpm

Check:

- We think that the motor is wye connected.The original magnetising power = 3 x 2542/2.5 = 77.4 kW ( << 224kW).The new “phasor” modelled magnetising power = 3592/1.67 = 77.2 kW.



Exercise 2. In a round rotor example motor, the parameters for rotating flux modelare as follows:

· Magnetising reactance Xm-r 2.5 ΩSynchronous speed Ns 1200 rpmSynchronous speed Ns-e 3600 rpm

The electrical torque must be T = 1 Nm. The maximum values are: pole angle δmax <60°, IRmax < 30 A, ISmax < 30 A. Calculate the control isd

*, isq*, iR*. Start with iR

*= 10 ADC.Also, assume that dispersed fluxes do not make any work. Do not care of iQ value.

Solution. We use line frequency in our model to get compatibility to electric supply.

Xm = 2πfLm => Lm = Xm/(2πf) = 2.5Ω/(2π60Hz) = 6.63mΩs = 6.63mH.

Let’s start with iR and iS orthogonal, or iR · iS = 0 and the angle between them, φRS = 90°.

T = iR x Ψ = iR x Lm(iS + iR) = Lm·iR x iS = Lm·iR iS sin φRS

ð iS = T/(Lm·iR sin φRS) = 1Nm/(6.63mΩs · 10A · 1)

ð iS = 1Nm/(6.63mVs 10 · 1) = 1Nm/(6.63·10-3(Nm/C)s 10 · 1)

ð iS = 15.1A

This makes for the pole angle:

δ = arctan(iS/iR) = arctan(15.1/10) = 56.5°,

which is clearly smaller than the maximum, 60°. If it had been too big, we had have e.g. to increase therotor current, e.g. to the maximum 30 A. Now we calculate values for iP and iQ:

· iP = iR · sinδ = 10A · sin56.5° = 8.34A.

· iQ = IS · sinδ = 15.1A · sin56.5° = 12.59A

δ

iR = 10AIS

Ψ

T

IP

IQ



So clearly, iQ is not zero with this method.

· iSd* = -iPsinδ + iQcosδ = -8.34A · sin56.5° + 12.59A · cos 56.5° =

= -6.95A + 6.95A = 0

· iSq* = iPcosδ + iQsinδ = 8.34A · cos56.5° + 12.59A · sin56.5° =

4.60A + 10.5A = 15.1A

· iR* = 10A. □

Of course the final result for isd* = 0, isq

* = IS, iR* = iR(initial) was clear from the beginning because of the

construction of iS.

Exercise 3. In the round rotor motor of exercise 2, calculate the rotating current,voltage, and flux linkage in the stator and, also, the corresponding phase quantities.Calculate also the electrical power and torque for the model motor.

Solution. Here iS is clearly 90° ahead the iR. So, if iR = iR·sinωt, we get iS = iS·cosωt. The phase currents are:

iA = (2/3)·iS = (2/3) · 15.1A·cosωt = 10.1 A·cosωt.iB = 10.1 A·cos(ωt-120°).iC = 10.1 A·cos(ωt+120°).

Now, we need to calculate the stator voltage VS. If we estimate that the back voltage is the dominant one inthe stator winding, we get approximately, in the nominal speed.

VS = jωΨ = jωLm · im · sin(ωt + 56.6°) = jXm · (cosδ·iR +IQ) · sin(ωt + 56.6°)= j2.5Ω · (cos56.5° · 10A + 12.59A) · sin(ωt + 56.6°) = 45.27V · sin(ωt + 146.6°).

The phase voltages are:

VA = VS = 45.27V · sin(ωt + 146.6°).VB = 45.27V · sin(ωt + 146.6° - 120°) = 45.27V · sin(ωt + 26.6°).VC = 45.27V · sin(ωt + 146.6° + 120°) = 45.27V · sin(ωt - 93.4°).

The rotor voltage is VR = jωRΨ = 0, for the slip velocity ωR = 0.

The rotating flux linkage is:

Ψ = Lm · im · sin(ωt + 56.6°)= Lm · (cosδ·iR +IQ) · sin(ωt + 56.6°)= 6.63mVs/A · (cos56.5° · 10A + 12.59A) · sin(ωt + 56.6°)= 0.120 Wb · sin(ωt + 56.6°).

The flux linkages in the phases are:

ΨA = ΨS = 120mWb · sin(ωt + 56.6°).ΨB = 120mWb · sin(ωt + 56.6° - 120°) = 120mWb · sin(ωt - 63.4°).ΨC = 120mWb · sin(ωt + 56.6° + 120°) = 120mWb · sin(ωt + 176.6°).



The mechanical power for model motor is ω·T = 2π60(1/s) · 1Nm = 377W. The electrical power is VS·IS* =

45.27V · sin(ωt + 146.6°) · [15.1A·sin(ωt+90°)]* = 684W<56.6° = 376W + j571W.

Exercise 4. For the motor in exercise 3 and 4, calculate the flux linkage, power andtorque for real mechanical axis rotating 1200 rpm.

Solution. The real electrical flux linkage for a motor with pN pole pairs:

Ψ = pN · Ψp = pN · (VS/ω1) = VS/ωmS.

So in our case:

Ψ1200 = VS/jω1200 = 3VS/jω3600 = 3Ψ3600.

Mechanical torque is in our case:

T1200 = iR x Ψ1200 = iR x 3Ψ3600 = 3 · T3600 = 3 · 1Nm = 3Nm.

Mechanical power is our case:

P1200 = ω1200 · T1200 = ω1200 · 3T3600 = 3ω1200 · T3600 = ω3600 · T3600 = P3600.

So, when the mechanical frequency is one third of the electrical, the mechanical flux and torque are 3 timesgreater than the electrical quantities. The mechanical and electrical power is the same.

Exercise 5. In the round rotor motor of exercise 2, the torque must be T = 1 Nm. Themaximum values are: pole angle δmax <= 60°, IRmax < 30 A, ISmax < 30 A. Calculate thecontrol isd

*, isq*, iR*. Start with angle δ = 60° and φRS = 30 °. Also, assume that

dispersed fluxes do not make any work. What is the value of iQ you get?

Solution. Xm = 2πfLm => Lm = Xm/(2πf) = 2.5Ω/(2π60Hz) = 6.63mΩs = 6.63mH.

T = iR x Ψ = iR x Lm(iS + iR) = Lm·iR x iS = Lm·iR iS sin φRS = Lm·iR2·sinδ·cosδ

ð iR = sqrt[T/(Lm·sinδ·cosδ)] = sqrt[1Nm/(6.63mΩs · sin60° · cos60°)]

iR

IS

Ψ

T

δ = 60°



ð iR = sqrt[1Nm/(6.63·10-3(Nm/AC)s ·sqrt(3)/2·1/2)]

ð iR = sqrt[1/(6.63·10-3 ·sqrt(3)/4)]A = 18.7A.

Now we calculate values for iP and iQ:

· iP = iS = iR·sinδ = 18.7A · sin60° = 16.2A.

· iQ = 0.

So clearly, iQ is zero with this method. Now we get

· iSd* = -iPsinδ = -16.2A · sin60° = -14.0A,

· iSq* = iPcosδ = 16.2A · cos60° = 8.1A,

· iR* = 18.7A.

Exercise 6. In the round rotor motor of exercise 5, calculate the rotating current,voltage, and flux linkage in the stator and, also, the corresponding phase quantities.Calculate also the electrical power and torque for the model motor.

Solution. Here iS is clearly (δ+90°) = (60°+90°) = 150° ahead the iR. So, if iR = iR·sinωt, we get iS =iS·cos(ωt+60°). The phase currents are:

iA = (2/3)·iS = (2/3) · 16.2A ·cos(ωt+60°) = 10.8 A·cos(ωt+60°).iC = 10.8 A·cos(ωt-60°).iB = 10.8 A·cos(ωt+180°).

Let’s calculate the magnetising current, |im|, next:

|im| = cosδ·iR +IQ = cos60° · 18.7A + 0 = 9.35A.

The rotating flux linkage is:

Ψ = Lm · |im| · sin(ωt + 60°)= 6.63mVs/A · 9.35A · sin(ωt + 60°)= 62.0 mWb · sin(ωt + 60°).

The flux linkages in the phases are:

ΨA = Ψ = 62.0 mWb · sin(ωt + 60°) = 62.0 mWb · sin(ωt + 60°).ΨB = 62.0 mWb · sin(ωt + 60° - 120°) = 62.0 mWb · sin(ωt - 60°).ΨC = 62.0 mWb · sin(ωt + 60° + 120°) = 62.0 mWb · sin(ωt + 180°).

Now we need to calculate the voltage VS. If we estimate that the back voltage is the dominant one in thestator winding, we get approximately, in the nominal speed.

VS = jωΨ = jωLm · im = jXm · |im| · sin(ωt + 60°)= j2.5Ω · 9.35A · sin(ωt + 60°) = 23.38V · sin(ωt + 150°).

The phase voltages are:



VA = VS = 23.38V · sin(ωt + 150°).VB = 23.38V · sin(ωt + 150° - 120°) = 23.38V · sin(ωt + 30°).VC = 23.38V · sin(ωt + 150° + 120°) = 23.38V · sin(ωt - 90°).

The mechanical power for model motor is ω·T = 2π60(1/s) · 1Nm = 377W. The electrical power is VS·IS* =

23.38V · sin(ωt + 150°) · [16.2 A·sin(ωt+60°+90°)]* = 379W<0° = 379W + j0VAr.

Exercise 7. For the motor in exercise 5 and 6, calculate the flux linkage, power andtorque for real mechanical axis rotating 1200 rpm.

Solution. As in the exercise 4, the mechanical frequency is one third of the electrical and the mechanicalflux and torque are 3 times the electrical quantities. The mechanical and electrical power is the same.

1.2.5. Power-anglecharacteristicofaround-rotormachineHere the concept of power-angle is explained according to Nasar /9/. In the Figure 27

there is shown the difference between the power-factor angle, φ, and power-angle,δ.

· The power-angle is defined as the angle δ by which the internal voltage, Vo,leads the terminal (connector) voltage Vt.; δ > 0 for generator action.

· The power-factor angle is the angle φ by which the armature current, Ia, leadsthe terminal voltage Vt.

Figure 27.Phasor digrams for a synchronous generator.

We obtain from the Figure 27 the approximate equation (it is supposed that Ra = 0):

The power generated (per phase) by the generator, Pd, is the power supplied to theload, or:



This latter approximate equation equally applies for a motor for in this case δ, sinδand Pd change sign.

Figure 28

The Figure 28 explains the difference in the power direction between generator andmotor. If Vback -> 0, the power angle becomes the same angle δ, which gives birth forthe torque in the rotor. For, as you see in the following, if the terminal voltage isVtejωt, the internal voltage is Vo = Vback = aVtej(ωt+δ), where a and δ are someconstants, for:

ddww

d

d

www

w

www

www

jt

tjt

S

tjt

Sj

rotorbacko

Sj

rotor

S

tjt

SS

tjt

SS

backtj

tSSSS

tjtt

etaVeaVLj

eVLaejtjtVtV

teatLj

eVL

LjeV

LLj

VeVLtiLt

eVtV

)()()()(

)()(

0)()(

)(

)( ==×=Y==

Y×=Y

×=-

×=-

×=×=Y

=

+

After this exercise, we can calculate the more exact result: Angle(ΨR) = angle(IR) = δ -90◦, for current lags voltage in an inductance. Angle(ΨS) = angle(IS) = φ, from thedefinition of φ. Thus γ = angle(ΨR,ΨS) = φ - (δ - 90◦) = φ - δ + 90◦.

The developed power of a motor is, of course, negative and may also be calculatedas well known:



Also we have for the motor:

This is explained in more detail in the Figure 29 and Figure 30; the latter is the Théveninmodel of the first one.

Figure 29.

)'()()(

1

1

famaa

maat

IIjXIjXRVIjXRV

++×+=+×+=

Figure 30.

oaSa

oamaat

VIjXRVIjXIjXRV

+×+=++×+=

)()( 1

At constant power, also:



An under-excited motor operates at a lagging power factor (Ia lagging Vt), whereasover-excited motor operates at a leading power factor. Thus the operating powerfactor is controlled by varying the field excitation thereby altering Vo. The locus ofthe back voltage, Vo, at a constant load for varying field current, If, is shown on theleft side (a) in the Figure 31. From this we can obtain the armature current as the fieldcurrent changes; results for several different loads are plotted on the right side (b)of the Figure.

Figure 31

For round rotor generator we get the synchronous impedance from themeasurements of Figure 32, for terminal short circuited we have:



Figure 32

Thus we get from the figure:

The voltage regulation is defined for generator as for transformer:

Exercise 8.



Figure 33.

oaSa

oamaat

VIjXRVIjXIjXRV

+×+=++×+=

)()( 1

Exercise 9.



a) We get:

9989.0cos736.2)arg(

736.216.115325.

1513281328

=Þ°-==Þ

°-Ð=+

°-Ð-=

-=Þ

+=

ff a

S

ota

oSat

I

AjZ

VVI

VZIV

.

b) Now power factor is simply cosφ = 0.9989.

Alternative solution:

a) From the phasor diagram in the Figure 34, we get:

Figure 34

b) Neglecting Ra, we get:

Or



Exercise 10.

Figure 35

φ = cos-1(0.8) = 36.87°. => Ia = 22.4A<36.87°.

ZS = 0.15Ω + j3Ω.



VVV

VZIVVVZIV

o

o

Sato

oSat

36.274||73.11)arg(

73.1136.274

=Þ°-==Þ

°-Ð=-=Þ

+=

d.

b) From Figure 35, the field current corresponding to Vo = 274 V is If = 9.2 A.


For more special cases, we develop some new theory, which is known as the cosinelaw. As you have seen from the solution above, we do not need this new theorynow, but anyway, we shall need it later. So we start with a phasor equation: A = B +C, or equivalently: C = A - B. We think that after the solution for this equation, weknow all the sides of the triangle or A, B, and C and one angle between them, αAB.Because we do not know all the phasors entirely (that is length and phase), wedevelop our equation to include only the known information. The unnecessaryinformation is deleted by taking the absolute values of the phasors as follows:

ABABBACBABBAAC

BABAC

BABACC

BAC

acos2

)()(

)(

222

222

2

-+=Þ

+×-×-=Þ

-×-=Þ

-×-=×Þ

-=

Figure 36



b) From Figure 35, the field current corresponding to Vo = 274 V is If = 9.2 A.

Exercise 11. Calculate the minimum line current for the motor of Exercise 10. Also,determine the corresponding field current. Neglect Ra.

Solution. The minimum armature current is:

Arg(Ia) = 0°.

AIVVV

VjZIVVVZIV

fo

o

Sato

oSat

2.62.237||12.13)arg(

12.1319.237394.17231

=Þ=

°-==Þ

°-Ð=×-=-=Þ+=

d


corresponding to cosφ = 1. From Figure 36, with θ = 90° and φ = 0°, we have

which gives Vo = 237.2 V. Then from Figure 36, If = 6.2 A.



Exercise 12. The motor of Exercise 10 runs with an excitation of 10 A and takes anarmature current of 25 A. Neglecting the armature resistance, determine thedeveloped power and the power factor.

Solution a). From Figure 35, if If = 10A, Vo = 280 V.

From Figure 36, since θ = 90°,

We have |sinφ| = 0.5602886 and φ = sin-1(.5602886) = 34.076° and cosφ = 0.82830.The parasitic input power is:

QIN = Im{sqrt(3)Vt·conj(Ia)} = sqrt(3)(400)conj[25sin(34.076°)] = -9704.5 VAr.

Solution b). We can solve the power also from the output side. The sine law givesfor solving power angle δ:

)0(81858.1222186.0||sin

)076.3490sin(||sin

||sin||sin

<Þ°-=Þ=Þ

°+°=Þ

+=

ddd

d

fqd

Motor

VZIVZI

o

Sa

oSa

Now we get for the apparent output power, So, in Vo:

ooo

aoo

jQPVArjWSIVS

+=-=Þ°-°-Ð××=°Ð×°-Ð×=××=

1533214350)076.3481858.12(252803076.3425)81858.12280(33

So, the missing parasitic power (QZS) must be in the ZS:

VArAQZS 56253)25(3 2 =W××=



Thus, together the “output side” parasitic power is

QOUT = Qo + QZS = -15332VAr + 5625VAr = -9707VAr, as it should be.

Exercise 13. The motor of Exercise 10 has 4 poles and is rated at 60 Hz. If It isadjusted so that Vo = Vt, the motor takes 20 A in current. Neglecting Ra, determinethe torque developed.

Solution. Proceeding as in Exercise 12, we have:

More precise solution: We also take into account the angle θ now.

Θ = arctan(3/.15) = 87.14°

|ZS| = sqrt(.152 + 32) = 3.004 Ω

|Zs|Ia = 3.004 · 20 A = 60.08 A

ð (231)2 = (231)2 + (60.08)2 -2(231)(60.08)cos(θ+φ)

ð 2(231)cos(θ+φ) = 60.08

ð cos(θ+φ) = 60.08/462

ð θ+φ = cos-1(60.08/462)

ð φ = cos-1(60.08/462) - θ

ð φ = 82.53° - 87.14° = -4.612°

ð cosφ = 0.997, lagging.



Exercise 14. The motor of Exercise 10 has 4 poles and is rated at 60 Hz. Magnetisingis adjusted so that Vo = Vt at 60 Hz. When used at 50 Hz, the motor takes 20 A incurrent. Neglecting Ra, determine the torque developed with 50 Hz.

Exercise 15. The motor of Exercise 10 has 4 poles and is rated at 60 Hz. Magnetisingis adjusted so that Vo = Vt at 60 Hz. What is the motor current if the motor is used at50 Hz and the load torque remains in old value.

Exercise 16. A 3-phase, wye-connected synchronous motor is depicted in theequivalent circuit below. It has 4 poles and is rated at 60 Hz 400 V. X1=0.15Ω,Xm=2.85Ω and Ra=0.15Ω (Rm is considered to be ∞). Magnetising is done withcurrent If(DC,space)’ = (3/2)141.4 A. The motor is used at 40 Hz and it takes 30 A incurrent. Neglecting Ra, determine the torque developed.

Solution. A space vector value corresponds (3/2) times the phase value in a 3-phasesystem:

If(DC) = (2/3)If(DC,space) = (2/3)·(3/2)141.4 A = 141.4 A. Furthermore:

If’ = If(rms)’ = If(DC)’/sqrt(2) = 141.4 A/sqrt(2) = 100 A.



Exercise 17. This is the exercise 16 again, but now we solve it with magnetic model.A 3-phase, wye-connected synchronous motor is depicted in the equivalent circuitbelow. It has 4 poles and is rated at 60 Hz 400 V. X1=0.15Ω, Xm=2.85Ω andRa=0.15Ω (Rm is considered to be ∞). Magne sing is done with current If(DC,space)’ =(3/2)141.4 A. The motor is used at 40 Hz and it takes 30 A in current. Neglecting Ra,determine the torque developed.

Solution.



The the magnetic torque model is motivated in the next section. The electricsolution is presented here for comparison.



Some motivation for the magnetic torque model

We think that the structure of the coils and rotors are similar in the case of one polepair and in the case of pN pole pairs, except that the area of air gap belonging to onepole pair is of course only A/pN in the case of pN pole pairs; here A is the cross-sectional area of the air gap for only one pole pair. The Lm is indeed the same in bothcases as is easily checked:

For one pole pair: Lm = N2*μ*A/l. Here l is the width of the air gap.

For pN pole pairs: Lm,pN = pN*Lp= pN*(N2*μ*(A/pN)/l) = N2*μ*A/l = Lm. This

follows from the facts that now we have pN pole pairs in series and cross-sectional area for one is only A/pN. So the Lm is the same for bothconfigurations.

Also the magnetic flux density is the same for both configurations:

For one pole pair: B = I·Lm/(N*A) = I·(N2*μ*A/l)/(N*A) = I·N*μ/l.

For pN pole pairs: B= I·Lp/(N*A/pN) = I·(N2*μ*(A/pN)/l)/(N*A/pN)

= I·N*μ/l.

However, the torque is for pN pole pairs pN times so large as for one pole pair. Thisfollows from the fact that, on the receiving side, wire loop count is pN*N and thetorque is proportional to the receiving loop count.

Also a pair of words is needed to explain why in the magnetic torque expressionstator current is îa but Ψr’ = IfDCSpace’*Lm. Ψr’ is here considered as a space-vector withsize (3/2) times the amplitude of the phase value. However, îa is here considered tobe a space-vector with magnitude only one time the amplitude of the phase value.Easiest way to understand this is to think Ψr’ to be voltage-like quantity and îa is ofcourse amplitude of a current. So, to get 3-phase power of the amplitudes ofcurrent, î, and voltage, û, we have to use equation: P = (3/2)*û*î.

Kun roottorin ja staattorin suureet ovat omilla puolillaan, on momentti:



SRRSSSRRRRRRRR iilANNiN

lAiNiN

AiNABiNAT ´=´×=F´×=

F´××=´××=

mm

Kun roottorin ja staattorin suureet on siirretty samalle puolelle, momentille saadaanerilaiset lausekkeet:

SRRSS

RR

R

SmRS

SR

SmRS

SRRSRmRR

SSRmR

RS

SRRSSRR

S

S

RS

R

SS

S

RmSR

RS

RmSR

SRRSSRSS

RSmS

S

RRSmS

SR

iilANN

NNi

NNLiiLiT

iilANNiL

NNiiLiT

iilANNii

NN

NN

lAN

NNi

NNLiiLiT

iilANNii

lAN

NNiL

NNiiLiT

´=´-=´-=

´=´-=´-=

´=´=´=´=

´=´=´=´=

m

m

mm

mm

2

2

2

22

2

2

2

Cross product is defined with the expression

Here is the angle between the vectors a and b in their common plane (here theangle is always taken to be between 0° and 180°), and are the lengths ofthe vectors a and b and n the unit vector perpendicular to their plane. The directionof n is defined by the rule of right hand as shown in the picture below.



1.2.6. VariablefrequencydriveHere are seen some pictures of a variable frequency drive from Wikipedia. In theFigure 38, there is such a device and in the Figure 39 its chassis when the cover isremoved. However, simple drive models are probably not suitable for full vectorcontrol of synchronous motors; but intended for simpler control of synchronous orinduction motors.

Figure 38

Figure 37



Figure 39

In the Figure 40, there is seen how a variable-frequency drive is used to control an AC-motor.

Figure 40

The method to produce an arbitrary sinusoidal power voltage is shown in the Figure

41. The height of the wanted sinusoid is compared to a constant saw-tooth wave.The sinusoidal power is generated as a PWM-signal the duty cycle of which isdetermined by comparing the magnitude of the sinusoid wanted with the saw-toothwave.



Figure 41

In the Figure 42 the sinusoidal power is generated as a voltage signal and in the Figure

43 as a current.

Figure 42

Figure 43

In the Figure 44 the sinusoidal is generated with a simpler, so called six-pulse inverter.This does not give as high quality power as the triangular generation, if all the pulseshas equal length. However, if the pulse lengths are varied we get the same accuracy,see more in Chapter 10 in the book of Kwang Hee Nam /11/.



Figure 44

In the Figure 45 we see a so called cycloconverter or direct matrix converter, which donot use DC-power in the intermediate circuit conversely to the above mentionedsolutions, but converts ac-current directly to a different frequency ac-current.However, these devises are suitably only for very slow motor frequencies like 25 Hzand under; e.g. they can be used to drive large ship motors.

Figure 45

1.2.7. MotorstandardsInternational standard IEC 34-1 defines 8, IEC 60034-1 10, different stampings fordifferent ways to use a motor: S1, S2… S10. These are named as follows:

S1 - Continuous running duty. Stamp: S1.

S2 - Short-time duty. Stamp e.g.: S2 10 min. Thermal equilibrium not reached. De-energised rest to reach the coolant temperature within 2 K.

S3 - Intermittent periodic duty. Stamp e.g.: S3 25%. Identical cycles. The startingcurrent may not significantly affect the temperature rise.

S4 - Intermittent periodic duty with starting. Identical duty cycles, significant startingtime, constant load, de-energized at rest. FI = (JL+JM)/JM. Stamp e.g.:



S5 - Intermittent periodic duty with electric braking. This is like S4 but with electricbrake.

S6 - Continuous-operation periodic duty. Identical cycles, also the idling power isobserved, no time in rest.

S7 - Continuous-operation periodic duty with electric breaking. Like S6 but eachcycle includes also electrical breaking. Stamping like in S4.

S8 - Continuous-operation periodic duty with related load/speed changes. Stampingexample:

S9 - Duty with non-periodic load and speed variations. The duty includes frequentlyapplied overloads that may greatly exceed the reference load (Pref is appropriatelybased on duty type S1).

S10 - Duty with discrete constant loads and speeds. Duty consists of a specificnumber of discrete values of load (and if applicable, speeds), each for sufficient timeto allow the machine to reach thermal equilibrium. The parameter TL expresses thethermal life expectancy of the insulation system as compared to the value of dutytype S1. Stamping example:

S10: p/Δt = 1,1/0,4; 1/0,3; 0,9/0,2; r/0,1 TL = 0,6

In the Figure 46, there is shown an example of a value plate of a motor. Theconnection possibilities to the supply are shown in the Figure 47. Notice, that in bothconnections the voltage over the windings is 400 V and the current through is 3.8 A.The decimal comma is used in the plate and the pictures, instead of decimal point,according to the IEC standard.



Figure 46

Figure 47

The nominal electrical power can be calculated as shown here:

Pelectric = sqrt(3)VIcosφ = sqrt(3)·400V·6.6A·0.82 = 3750 W.

With the nominal load the efficiency is:

η = Pshaft/Pelectric = 3000W/3750W = 0.80 = 80 %.

In the USA, NEMA-standards are followed.

The heat energy balance of a motor is represented as follows:

Here PkN is the average cooling power during active use and P0N in idling. PK(t) is theinstantaneous heating power in active use and P0(t) in idling. tjE is the equivalentcooling time and tj the total heating time. This can be solved for allowable heatingenergy:

This can be written in the form:

PK·αtj = PKN·tjE[1+k0N(1-p0j)],



where PK is the average heating power during active operation and P0 during idling;k0N = P0N/PkN; P0j = P0·tj/(P0N·tjE). The proportion of active working time of the wholeperiod, tj, is α.

In practice k0N(1-p0j) << 1 and we get with PK=RI2 and PKN=RIN2:

.

If PA = ηMVMIP, and ηM and VM are constants (like in field weakening), we get fromthis:

.

Here PA is the axial power and ηM, VM, and Ip are the absolute values of efficiency,supply voltage and real current, respectively.

Exercise 17. Choose a motor for usage “S3 20% 1min”. The cooling power of themotor is during idling 20% of nominal value. What is your choice if an external fan isused?

Solution. Here α = 20% and tjE = (0.2+0.2·0.8)tj = 0.36tj. Thus working time is from equivalent cooling time0.2tj/tjE = 0.56. With the values PK = IS3

2Rk and PkN = IN2Rk, we get:

PK = (tjE/0.2tj)PKN·[1+k0N(1-p0j)]

ð IS32 = (tjE/0.2tj) IN

2·[1+k0N(1-p0j)]

We suppose that k0N = 0.3, which gives:

IN = IS3·sqrt[0.56/(1+0.44k0N)]=0.707·IS3.

With external fan tjE = tj and the nominal power is

IN = IS3·sqrt[0.2/(1+0.8k0N)]=0.40·IS3.

Exercise 18. Two trains, train 1 and train 2, travel a same distance in the same timewith different speed profiles, as shown in the Figure 48. Both drives use constant field.The top speed of train 2 is (2/3) of the corresponding value of the train 1. Theacceleration of train 2 is (4/3) of the acceleration of train 1. Train 1 only accelerates



and decelerates, but train 2 travels half of the time with constant speed. Which oneof the drives is more economic?

Figure 48

Solution. For train 1, for maximal power and torque, P1 and T1, respectively, we get clearly:

P1 = T1ω1 and T1 = T1.

For train 2 we get:

P2 = (4/3)T1(2/3)ω1 = (8/9)T1ω1 and T2 = (4/3)T1.

Now les’s take the equivalent cooling time to be the run time, T. For train 1 we get, αtj = T/2:

PAN1 = sqrt[(1/T)∫(0->T/2){P1·(t/0.5T)}2dt]

= sqrt[(1/T)(1/3)P12(1/0.5T)2(0.5T)3] - 0

= sqrt[1/6] P1

TAN1 = PAN1/ω1.

For train 2 we get, αtj = T/4:

PAN2 = sqrt[(1/T)∫(0->T/4){(8/9)P1·(t/0.25T)}2dt]

= sqrt[(1/T)(1/3)(8/9)2P12(1/0.25T)2(0.25T)3] - 0

= sqrt[1/12] (8/9)P1

= (8/9)PAN1/sqrt(2)

TAN2 = (8/9)[PAN1/sqrt(2)](3/2ω1) ≈ 0.94 PAN1/ω1· = 0.94·TAN1.

The motor of train 2 is therefore more economical (power is smaller and torque is smaller). Do you thinkthat it was justified to use the squared power in our latter analysis? - If you calculate again withoutsquaring, you’ll get a more radical result. We could also check what happens, if the motor speed of train 2



is increased to the same what the motor of train 1 has, keeping at the same time the velocity profile of thetrain invariable; here we would need help from a gear.

Exercise 19. A brushless round rotor synchronous motor has the following stampedproperties /5/:

Nominal power Pout 7500 kWSupply voltage VS 6000 VSupply current IS 823 ASupply frequency f 50 HzRotational speed n 1500 rpmPower factor cosφ 0.90 over magnetisedStator resistance RS 0 ΩStator dissipation reactance X1 = XσS 0.97 Ω

The idling curve is linear and Vo = VN and IF 4.0 A.The short circuit curve is linear and Ik = IN with IF = 7.6 A.

Compute in the nominal point:

a) The magnetising current

b) The largest parasitic power fed by the motor to the supply network

c) The largest real power taken by the motor from the supply network

Figure 49

Solution. Slope, ku, of the idling curve is

Ku = Vo/IFo = 6000/(sqrt(3)·4) = 866 V/A.

The voltage Vok, which corresponds to the short circuit magnetisation, is:

Vok =kuIFk = 866 · 7.6 A = 6582 V.

Synchronous reactance is:

Xd = Vok/Ik = 6582 V/823 A = 8.0 Ω.



Magnetising reactance is got with subtraction:

Xm = Xd - XσS = 8.0 - 0.97 = 7.03 Ω.

Stator current is reduced to rotor with the factor g,

IS’ = Xm IS /ku => g = IS’/IS = Xm/ku = 7.03/866 = 0.00812

Stator current is

IS = 823·(0.90 + j0.435) = 740.0A + j358.8A

IS’ = g·IS = 6.01A + j2.91A.

The total voltage, Vh, in the magnetising reactance Xm; and the total magnetising current, Im’, correspondingto the main flux (stator and rotor currents combined), are:

AkVI

VAjAjVIjXVV

u

hm

SSSvh

48.48663879||'

66.103879)8.3580.740(97.0)3/6000(

===

-<=+×-=×-= os

The magnetising current is 90° behind the voltage Vh, or:

Im’ = 4.48A<-100.66° = -0.83A - j4.4A

a) The nominal magnetising current is:

IF’ = Im’ - IS’ = -0.83A - j4.4A - 6.01A - j2.91A = 10.01A<-133.1°.

The corresponding internal magnetising voltage VF over Xm is:

o1.438674)8.3580.740(0.8)3/6000( -<=+×W-=×-= VAjAjVIjXVV SdSvF

b) The biggest possible parasitic power to the network with the nominal magnetisation is certainly the+ sign between the absolute values, which gives, in this case, the result ISQmax = 1517A. (Counterexample: if |VF|·= |VSv|, ISQMax would be zero with - sign which, of course, is not generally true):

AVVVVX

VVX

I SvFd

SvFd

SQ 1517)34648674(81|)||(|1)(1

max =+W

=+£-=

Qmax = sqrt(3)·6000V·1517A = 15.8 MVAr.

c) The biggest real power from the network with the nominal magnetisation is. - You must here think,according to the superposition theorem, separately the generation of the stator current andmagnetising voltage:

Pmax = 3|VF|·(|VSv|/Xd) = 3 · 8674 · (3464/8)=11.27 MW.



1.3. Inductionmotors

In an 3-phase induction motor or, as it also is called, non-synchronous or short-circuited motor, the structure is similar to a 3-phase synchronous motor with onlythe damping wiring left in the rotor. This wiring forms a short circuited cage windingin the rotor, which gives one of the names for the motor type. The more the rotorlags the speed of the rotating field, the more the field induces voltage in the cagewiring and the more there is short circuited current in the cage and the more therotating field generates torque in the cage.

1.3.1. Ashaded-polemotorOne small type of induction motors is the so called shaded-pole motor. This isbasically an ac single-phase induction motor with squirrel-cage rotor. In the stator,there is an auxiliary winding which is composed of copper rings surrounding aportion of each pole.[1] This auxiliary single-turn winding is called a shading coil.Currents induced in this coil by the magnetic field delay the phase of magnetic fluxlinkage for that pole (a shaded pole) enough to provide a rotating magnetic field.The direction of rotation is from the non-shaded side to the shaded (ringed) side ofthe pole.[2] Since the phase angle between the shaded and non-shaded sections issmall, 1-phase shaded poles produce only a low starting torque, so they are onlyused to start the rotor to spin in non-self-starting motors. Shaded-pole motors arenot normally reversible.

Figure 50. Small C-frame shaded-pole squirrel-cage motor. With the shaded poles shown, the rotor will rotate in the clockwisedirection.



Figure 51. Shading coils (copper bars).

These motors only have one winding, no capacitor nor starting switch, making themeconomical and reliable. Because their starting torque is low, they are best suited todriving fans or other loads that are easily started. Moreover, they are compatiblewith TRIAC-based variable-speed controls, which often are used with fans. They arebuilt in power sizes up to about 1/6 hp or 125 watts output. For larger motors, otherdesigns offer better characteristics.

The most common type of shaded-pole motors in fractional horsepower use is thesquirrel-cage induction motor. This has a rotor that consists of a laminated steelcylinder with conductive copper or aluminium bars embedded lengthwise in itssurface and connected at the ends. The rotating magnetic field of the stator inducescurrent in the bars that creates a magnetic field in the rotor that causes it to rotate.

1.3.2. PrinciplesoftheinductionmotorIn the Figure 52 we see a schematic model of an induction motor. In this model thecoordinate system is considered to rotate with the stator field or with the angularfrequency of the supply current, ω. This model describes the motor for any timepoint t; the sources jωΨS and jωRΨR are considered to be constant for the time pointin question; note that the flux linkages are not the same Ψ, as in in the Figure 16 andFigure 17 for the multipliers of the dispersed inductances have also changed and donot anymore include the parts jω and jωR. However, these two representations areof course equivalent. The angular slip frequency ωr here, must be interpreted inelectrical unit; i.e. ωr = pN(ωS-ωm) = ω - pNωm = ω - ω0. Here ω0 is the electricalangular frequency of the rotor. If there is only one pole pair in the stator and rotor,the electrical unit of ω0 is the same as the mechanical ωm. If you calculate withcorrect relative slip, but with pN = 1, you get correct output power but torque is only1/pN of the correct value.



Figure 52.Schematic diagram for an induction motor. Coordinate system rotates with the stator field with the angular frequency ω. ωr isthe electrical slip frequency.

The winding fields for the stator and rotor, ΨS and ΨR have the formulas:

In this model the short circuit inductance, Lk, of the machine is

Here the σ is so called total dispersion factor and LS the inductance of the statorcoil. σ is defined as follows:

,

Where LR is the inductance of the rotor cage and Lm is the magnetising inductance.

In the Figure 53, we see the circuit model of the induction motor, which is based to LS

inductance or stator field. With great rotational or slip frequency in the stationarystate, where p=0 and ωRLσ >> RR

S, LS and LσS are seen in parallel behind the stator

resistance RS. Thus we get:

Lk = σLS = LS||LσS = LSLσ

S/(LS + LσS)

ð σ = LσS/(LS + Lσ

S)

ð σLS + σLσS = Lσ

S



ð σLS = (1-σ)LσS

ð LσS= σLS/(1-σ)

Figure 53. The so called LS model of induction motor is based on the stator flux. Coordinate system rotates with the stator fieldwith the angular frequency ω. ωr is the electrical slip frequency.

When we measure with a no-load measurement the value for LS, and Lk = σLS =LS||Lσ

S from a high frequency (slip is high, s=1) “short-circuit” measurement, we cancompute σ and then Lσ

S= σLS/(1-σ); so the inductances of the LS-model are all gotquite accurately.

The per-phase equivalent circuit for with stator field rotating model is easilygenerated following D.Y. Ohm /8/ as described in the section “B.4. Rotating Modelof a General Motor”. We call the result as “Rotating model”. The model is easilysolved as follows. To get a positive value for the current iR

S, we take here thedirection of it in the clockwise direction, so oppositely to the one presented in theFigure 52 and Figure 53, /8/. Here the stator field and supply current are considered torotate with the angular frequency ω1 = pNωS. The electrical angular frequency of therotor is ω0 = pNωm, where ωm is the mechanical angular frequency of the rotor; pN isthe number of pole pairs. We also define the electrical angular slip frequency, ωr, asωr = pNωR where ωR = ωS - ωm is the mechanical slip frequency.

aRm

RaSm

aR

aRm

aSmS

aS

aRa

aSm

aRm

RR

aSa

aRm

aSmSS

as

iLLiLiLiLL

jpiLipLpLRjpiLipLpLRV

)(

)(

)()(0

)(

0

+-=Y

-+=Y

Y---++=

Y+-++=

s

s

s

s

ww

w



ωa is the, for the time being, arbitrary angular velocity of our co-ordinate system.We consider now the stationary case and the current phasors to be of the formIs

a=Isa·ej(ω1-ωa)t and IR

a=IRa·ej(ω1-ωa)t We denote the difference ω1-ωa with ω1a: ω1a = ω1-

ωa.

aS

aamR

R

aamaR

aR

aam

aamR

RaS

aSaam

aRaam

RaRR

aRaam

aSaamS

aSSS

ijjjLLR

jjjLi

ijjjL

jjjLLRi

ijjjLijjjLLiRijjLijjLLiRV

))(()()(

))((

)())((0)())((

01

01

01

01

0101

11

wwwwwwwww

www

wwwwww

wwww

s

s

s

s

-+++-+

=Þ

-+-+++

=Þ

-+--+++=

+-+++=

Þ

aS

aaaam

aamR

RaamSSSaam

RR

aR

aaaam

aamR

RaamSSSaam

ijjjjjL

jjjLLRjjLLRVjjjLLR

ijjjjjL

jjjLLRjjLLRVjjjL

úúû

ù

êêë

é

-++-

-++++++=-+++

úúû

ù

êêë

é

-++-

-++++++=-+

))((

)))(())()((()))(((

))((

)))(())()((()(

0112

01101

0112

01101

wwwww

wwwwwwww

wwwww

wwwwwwww

sss

ss

))(()))(())()((()))(((

))(()))(())()((()(

0112

011

01

0112

011

01

wwwwwwwwwwwww

wwwwwwwwwwwww

ss

s

ss

jjjjjLjjjLLRjjLLRVjjjLLRi

jjjjjLjjjLLRjjLLRVjjjL

i

aaaamaamR

RaamSS

aamR

RaS

aaaamaamR

RaamSS

aamaR

-++--++++++

-+++=

-++--++++++

-+=

For easy programming these expressions can be grouped as follows. Notice also, that the variable nameshave been slightly changed:

2121

2

2121

2

01222

1111

012

11

)()()()(

)()(

FFWWWVI

FFWWFVI

jLLRWjLLRW

jLFjLF

SS

SR

aam

aam

aam

aam

×-×=

×-×=

-+××++=+××++=

-+××=+××=

wwwww

wwwww

If ωa = 0, we call the model as “Stationary Model”, for the co-ordinate system isstationary. We get in this case (ω1a = ω1):



rmrmR

RmSS

rmR

RaS

rmrmR

RmSS

rmaR

jjLjLLRjLLRVjLLRi

jjLjLLRjLLRVjL

i

wwwww

wwwww

ss

s

ss

×-++++

++=

×-++++=

12

1

12

1

))()()(())((

))()()((

If ωa = ω1, i.e. co-ordinate system rotates synchronously with the stator field, themodel is called “Synchronous Model”. We get in this case (ω1a = 0):

rmrmR

RmSS

rmR

RaS

rmrmR

RmSS

rmaR

jjLjLLRjLLRVjLLRi

jjLjLLRjLLRVjLi

wwwww

wwwww

ss

s

ss

×-++++

++=

×-++++=

12

1

12

1

))()()(())((

))()()((

We think that the equivalent circuit is per-phase so that we need to include thephase count, 3, as a multiplier to the expressions of power and torque.

SRmRRR

RSRmaRR

m

out

RRR

mRRSmmRaRmout

RRRm

RRm

out

RRout

sSin

aRm

RaSm

aR

r

iiLiiLiiiLiSreT

iiLjiiiLjijS

Ris

sRiPT

ssRiP

iVS

iLLiL

s

´×=´×--´×=Y´×==

×--×=Y×=

=-

×==

-×=

×=

+-=Y

=-

=

33)(33)(3)(33

)313(

)13(

3

)(

***

22

2

*

11

01

s

s

s

w

www

www

ww

www

1.3.3. StationarystateoftheinductionmotorIn the stationary state, we can replace operator p with angular velocities and theschematic of Figure 53 gets the form shown in the Figure 54. Now we have number ofpole pairs pN, so that ω = ω1 = pN·ωS and ωr = pN·ωR.; in the following we denote,however, somehow confusingly ωr = ωR, as if there were only one pole pair. We callthis model with different names according to the method used toestimate/approximate/calculate the parameter values:

· “The rough model with inductances”. Here LS = Lm and LσS = LσR

S and LσS = 0.

· “LS-model”. Here LS = Lm + LσS and LσS = LσR

S and LσS = 0.



· There may be better/alternative ways to find values for LS and LσS.

Figure 54. For real numerical calculations, we interprete this followingly: Coordinate system rotates with the statorfield with the angular frequency ωmS=ω1. ωR = ωr is the electrical slip frequency. Also please notice, that you should notreplace jωRLσS with LσSp=0 (as is done in the case of inductance LS), for because the field on the rotor side is only thestator field, so not the total rotor field, exactly voltage jωRLσS·iRs were missing. However, jωrΨs also includes (ifLS=Lm+LσS) the term jωrLσSis, which should not be there.

Now we get the equations

You should notice that there is no voltage caused directly by currents iS and iRS in themagnetizing inductor LS in these voltage equations; this is of course because as atransformer coil, the LS causes the magnetising influences by iS and iRS, expect theidling effect, cancel each other away. The stator current iS = imS -iRS can be calculatedas follows:

,

.

The rotor frequency is controlled by the supply frequency according to the equation:



.

Here ω1 is the supply frequency, pN is the number of pole pairs, ωR is the slipfrequency, ωm the rotational speed of the rotor and ωmS is the synchronousfrequency or the frequency of the rotational field. The schematic in the Figure 54 canbe drawn in the form of Figure 55, where the slip frequency, synchronous frequencyand mechanical frequency may all be seen (ωR = ωmS - ωm).

Figure 55.Schematic for calculating the power division.The slip, mechanical and synchronous speed is seen here.

From the Figure 55 we get:

,

From the power induced from the stator flux to the rotor the portion correspondingto the mechanical speed transforms immediately to the mechanical power and the

proportion to the thermal dissipation of the rotor. The power mediated bythe stator may also be calculated:

Stator gives to the air gap the same power, which is taken by the rotor asmechanical power and thermal power. Thus we can infer that the power division isthe same as speed division or:

Air gap power: Mechanical power: rotor dissipation = ωmS : ωm : ωR.



The torque may be estimated with the formula:

In this, the mechanical power transferred from stator to rotor is calculated bysubtracting the thermal and no-load power of the rotor from the real power of thestator. iP is here the real stator current.

1.3.4. CharacteristicsofinductionmotorWe remember that:

The negative of rotor current is important. Let’s study it more carefully:

The rotor frequency is so called tip frequency, at which the rotorresistance and reactance are equal. At this frequency the induction motor workswith its biggest real power and reactive current:

There are two components in this current: The imaginary component, iQσ, which is inthe direction of the stator flux:

.

The other is the real current, iP, which is perpendicular to the stator flux:

Now we get “the Rough Results” which are together with the expression to the tipfrequency ωRh, presented above, as follows: The reactive current approaches thevalue



,

when the slip frequency approaches infinity. In the tip point the real and thereactive current are equal in magnitude and half of the maximal value ikσ of thereactive current.

Figure 56.Theoretical characteristics of an induction motor.- Rotor current- Power factor (……)

- Real current (……)- Parasitic current- Relative rotor frequency (ωR/ωRh); the slip frequency divided with the tip frequency.

The torque is got with the cross field principle from the real current:

If the equation LσS= σLS/(1-σ) is inserted here, we get:

, where

= ½ΨS2/Lσ

S.

Th is the tip torque at tip frequency ωRh. We expand with ω2 and get:

.



Here Xk is the short circuit reactance, ωLσS, seen from the stator. The expressions for

the tip torque reveal the importance of the supply voltage or flux and the shortcircuit reactance to the value of tip torque. Particularly, the value of the supplyvoltage has a great influence to the tip torque and the torque curve depicted in theFigure 57.

Figure 57. Theoretical torque of an induction motor on the motor region when the flux is constant.

1.3.5. Exampleinductionmotor1Impedances are for one phase winding.

· Rating 300 kWVoltage 440 Vrms 1-1

254 Vrms 1-nStator Resistance R1 .0073 ΩRotor resistance R2 .0064 ΩStator reactance X1 .06 ΩRotor reactance X2 .06 ΩMagnetising reactance Xm 2.5 ΩSynchronous speed Ns 1200 rpmSupply frequency fS 60 Hz



Exercise 1. Calculate the peak torque and power for the example motor 1. Use TheRough results of the flux linkage theory. Use the equivalent circuit in the picturebelow.

Solution 1-a. We use in the solution the electrical rotating field model. The dispersed inductance of rotor is:

LσS = (2/3)X2/(2πfS) = (2/3).06Ω/(2π60Hz) = 106μH.

The tip frequency of the rotor is

ωRh = (2/3)RRS/Lσ

S = (2/3)(.0064Ω/106μΩs) = 40.25rad/s.fRh = ωRh/2π = 6.41 Hz = 385 rpm.s = 385/3600 = 10.7 %ωm = 112.3 rad/s; fm = 17.87 Hz.

We get for Lm:

Lm = Xm/(2πfS) = 2.5Ω/(2π60Hz) = 6.63 mH.

We can estimate ΨS to be (It does not matter if we use value 6.63 mH or (2/3)6.63 mH for Lm):

ΨS = LmIm = Lm(VS/Xm) = 6.63mΩs(sqrt(2)254V/[j2.5Ω]) = -j0.9526Vs, orΨS = VS/ω = VS/(2π60Hz) = sqrt(2)·254V/(2π60Hz) = -j0.9528Vs.

The electrical torque is:

T = ½ΨS2/Lσ

S = ½(0.9528Vs)2/106 μΩs = 4282 Nm. (Good value)P = ω·T ≈ 2π60Hz·4282 Nm = 1614 kW. (Bad value, for the real rotational speed need to be used!)

Other method for electrical torque:

T = ½(1/ω)(VS2/Xk) = ½·[1/(2π60)][2·2542/((2/3).06)]=4278 Nm. (Good value)

(We can now compare the torque in the rough model, we have here, to the accurate one: The electricalfrequency in the coil Xk, in the real motor, is ω1 = pN·ωmS; so it is independent of the number of the polepairs, which keeps the current VS/Xk independent, too. The flux linkage Ψ = VS/ω, however, is in realitypN(uS/ω1) = uS/ωmS and proportional to the number of pole pairs).

P3600 = ω·T ≈ 2π60Hz·4282 Nm = 1614 kW. (Bad value, for the real rotational speed need to beused!))Pm = ωm·T ≈ 2π·17.87 Hz·4282 Nm = 480.8 kW. (Good value)

Check for perfect system:



Nms

kWPT

kWs

VXV

XVP

m

CheckCheckmech

SSS

Check

25680/202

3227

3227/06.0

4403

)(

2222

=×

==

=W

===

pw

ss => Far from perfect.

Solution 1-b. We calculate I2 for s=10.7% from the expression:

I2 = V1/(R1+R2/s+X1+X2) = 254/(.0073+.0064/.107+2*j.06)A = 853A-j1685A = 1889A<-63.16°.

|I2| = 1847A

Pshaft = 3|I2|2*R2(1-s)/s = 3·18472*.0064*(1-.107)/.107= 548.2 kW.

T = Pshaft/ωm = 548.2 kVA/[2π(1-s)ωS] = 4883 Nm

Solution 1-c. It is interesting to compare these approximate results with the precise Rotating Model. Weget for pN = 3:



When we try with pN = 1 we get:



So, with the help of more accurate calculations we can choose quite good starting values of the manypossibilities for the approximate method. - But does this apply in all cases?

Exercise 2. Calculate the torque and power for the example motor 1 at the tipfrequency. Assume here that the inductances are calculated with networkfrequency ω1 = 3600 rpm. Use the Rough Method with Inductances.

Solution 2-a. The parameters for rotating field are:

· Rating 300 kWPeak terminal voltage (line-line) sqrt(2)·440V = 622VPeak terminal voltage (line-n) sqrt(2)·440V/sqrt(3) = 359VFrequency fS 60 HzStator Resistance RS = R1 (2/3)·.0073Ω = 0.00487ΩRotor resistance RR

S = R2 (2/3)·.0064Ω = 0.00427ΩStator leakage reactance X1 (2/3)·.06Ω = 0.04Ω

o Lσ1 = 0.04Ω/(2π60/s) = 106μH

· Rotor leakage reactance X2 (2/3)·.06Ω = 0.04Ω

o Lσ2 = 0.04Ω/(2π60/s) = 106μH

o Total rotor reactance LσS = Lσ1 + Lσ2 = 106μH + 106μH = 212μH

· Magnetising reactance LS = Xm (2/3)·2.5Ω = 1.67Ω

o LS = 1.67Ω/(2π60/s) = 4.421mH

We use in the solution the electrical rotating field model. The dispersed inductance of rotor is:


ωRh = RRS/Lσ

S = .00427Ω/106μΩs = 40.21rad/s.fRh = ωRh/2π = 6.40 Hz = 384 rpm.s = 384/3600 = 10.67 %1-s=0.8933

We get the equation for iS as follows:

SSR

SR

R

SS LjR

jL

i Y+

+= )1(sw

w



mSSR

SR

SR

S

S

SS

SmSSRSR

SRSR

SmSSRSR

SRSR

SR

SRSR

SmSSR

SR

SR

S

SS

SmSSSS

jLjR

RjLR

u

jRjLjRLj

jRjLjRLjLjRLj

jLjR

RjLR

u

jiRu

ww

w

wwww

wwww

ww

ww

ww

s

s

s

s

s

++

+=YÞ

Y×++

=

Y×++

++=

Y++

+=Þ

Y+=

))(||)(

1(

))(

(

)(

VsjVs

sjAVssjAV

AVsjAmVsAV

V

S

S

6329.0203804.02

/602/106/21.40/00427.0

/00487.0/21.40/421.4/00487.0

359

×-×=YÞ

+×+

×+

=YÞp

m

Note: Without of those square roots of 2, the ΨS have the value corresponding to the usual model of onephase only.

Note: ωmS must be taken to be the electric synchronous value: ωmS = ω1. Also ωR is the electric value.

Now, we get the Ttip:

NmAVs

sVL

T SS

tip

3789/

3789

222

2

=

×=

Y=

s

kWNm

TfP tipmtip

4.425378987.172

2

=×××=

×××=

p

p

This method also gives good results if you know what parameter values you must choose. It is easy, if youknow the right solution beforehand, but otherwise very error prone.

Solution 2-b. We calculate I2 for s=16.2 % from the expression:

I2 = V1/(R1+R2/s+X1+X2) = 254/(.0073+.0064/.1067+j.06+j.06)A = 903.2A-j1610A.

|I2| = 1846 A

Pshaft = 3|I2|2*R2(1-s)/s = 3·18462*.0064*(1-.1067)/.1067= 548.2 kW.

T = Pshaft/ωm = 24.8 kVA/[2π(1-s)ωS] = 4883 Nm.



Comment: The results with this method are more precise. The difference with the solution 3a-a is causedby the approximations in that model; e.g. the value used for LS = Lm and Lσ

S could have better value. Alsothe LS model as such is approximate.

1.3.6. CasioClassPadprogramforex2.The results for solutions a and b of exercise 2 are calculated with the following Class Pad program; note,however, that the original program is copied here and a little edited by hand. This program is based, on thecontrary to the treatment in the exercises, on one phase equivalent circuit of the motor.

print “Induction motor 1 - real speed (not electrical speed)”print “One phase equivalent circuit”440/sqrt(3)=>V1 60=>f 3=>m 3=>pN pN=>pE.0073=>Re1 .0064=>Re2 .06=>Xe1 .06=>Xe2 2.5=>Xm2*π*f=>ω1 Xe1/ω1=>L1 Xe2/ω1=>L2 Xm/ω1=>Lm ω1/pN=>ωSRe2/L2/pN=>ωRh ωRh=>ωR ωS-ωR=>ωm ωm/(2*π)=>fm ωR/ωS=>s

print “Method with inductances”V1/(Re1/Lm+i*pE*ωR*Re1/(Re2+i*pE*ωR*L2)+i*pE*ωS)=>Ψsqrt(Ψ*conjg(Ψ))=>Ψabs3*Ψ*conjg(Ψ)/(2*L2)=>Tip2*Tip/(ωRh/ωR+ωR/ωRh)=>Tωm*T=>P

print “Method with reactances”V/(Re1+Re2/s+i*Xe1+i*Xe2)=>I2sqrt(I2*conjg(I2)=>I2abs3*I2*conjg(I2)Re2*(1-s)/s=>P2P2/ωm=>T2

1.3.7. Exampleinductionmotor2Impedances are for one phase winding.

· Number of phases m 3Number of pole pairs pN 3RMS terminal voltage (line-line) 230 VFrequency fS 60 HzStator Resistance R1 .06 ΩRotor resistance R2 .055 ΩStator leakage reactance X1 .34 ΩRotor leakage reactance X2 .33 ΩMagnetising reactance Xm 10.6 Ω



Exercise 3. Calculate the torque and power for the example motor 2 at the tipfrequency. Assume here that the inductances are calculated with networkfrequency ω1 = 3600 rpm. Use the Rough Method with Inductances.

Solution 3-a. The parameters for rotating field are:

· Peak terminal voltage (line-line) sqrt(2)·230V = 325VPeak terminal voltage (line-n) 325V/sqrt(3) = 188VFrequency fS 60 HzStator Resistance RS = R1 (2/3)·.06Ω = 0.04ΩRotor resistance RR

S = R2 (2/3)·.055Ω = 0.0367Ω

· Stator leakage reactance X1 (2/3)·.34Ω = 0.227Ω

o Lσ1 = 0.227Ω/(2π60/s) = 602μH

· Rotor leakage reactance X2 (2/3)·.33Ω = 0.22Ω

o Lσ2 = 0.22Ω/(2π60/s) = 583.6μH

o Total rotor reactance LσS = Lσ1 + Lσ2 = 602μH + 584μH = 1.186mH

· Magnetising reactance Xm (2/3)·10.6Ω = 7.07Ω

o LS = 7.07Ω/(2π60/s) = 18.8mH

We use in the solution the electrical rotating field model. The dispersed inductance of rotor is:

LσS = X2/(2πfS) = .22Ω/(2π60Hz) = 583.6μH.


ωRh = RRS/Lσ

S = .0367Ω/583.6μΩs = 62.8rad/s. We take ωRh to be electrical angular speed.fRh = ωRh/2π = 10 Hz = 600 rpm.s = 600/3600 = 16.2 %. So this is a ratio of electrical speeds.1-s=0.838

We get the equation for ΨS as follows (ωmS = ω1:)

SSR

SR

R

SS LjR

jL

i Y+

+= )1(sw

w



mSSR

SR

SR

S

S

SS

SmSSR

SR

SR

S

SS

SmSSSS

jLjR

RjLR

u

jLjR

RjLRu

jiRu

ww

w

ww

ww

s

s

++

+=YÞ

Y++

+=Þ

Y+=

)(

o94.843216.023204.0202836.02

/602/584.0/8.62/0367.0

/04.0/8.62/8.18

/04.0188

-<×=×-×=YÞ

+×+

×+

=YÞ

VsjVs

sjAmVssjAV

AVsjAmVs

AVV

S

S

p

Note: Without of those square roots of 2, the ΨS have the value corresponding to the usual model of onephase only.

Note: ωmS must be taken to be the electric synchronous value: ωmS = ω1. Also ωR is the electric value.

Now, we continue with the tip torque and power:

NmAVs

sVL

T SS

tip

3.177/105842

4554.02

22

6

2

2

=

×××

=

Y=

-

s

kWNms

TfP tipmtip

6.183.177/)162.01(202

2

=×-×××=

×××=

p

p

The results are again quite good. The right values are:

.

Solution 3-b. We calculate I2 for s=16.2 % from the expression:

I2 = V1/(R1+R2/s+X1+X2) = 133/(.06+.055/.162+j.34+j0.33)A = 87.23A-j146A = 170A<-59.2°.

|I2| = 170A

Pshaft = 3|I2|2*R2(1-s)/s = 3·1702*.055*(1-.162)/.162= 24.2 kW.

T = Pshaft/ωm = 24.8 kVA/[2π(1-s)ωS] = 231 Nm.



Comment: The results with this method (Solution 3-b) are quite certainly more precise. The difference withthe Solution 3-a is caused by the approximations in that model; e.g. the value used for LS = Lm and Lσ

S couldhave better value. Also the LS model as such is approximate.

Exercise 4a. Calculate the torque and power for the motor which have Xσ1 = Xσ2 =0.46Ω; R1 = R2 = 0.2Ω; Xm = 16.5Ω; V = 400 V, pN = 2; s = 3%. Use the Rough Methodwith Inductances.

Solution 4a-a. The parameters for rotating field are:

Xσ1 = Xσ2 = 0.307Ω;R1 = R2 = 0.133Ω;Xm = 11Ω;V = 566 V;L1 = L2 = 976μH;Lm = 35mH.You must take for ωmS to be ωmS = ω1.


ωRh = RRS/Lσ

S = .133Ω/976μΩs = 136.6rad/s.

fRh = ωRh/2π = 21.7 Hz = 1302 rpm.

ωR = 9.424778 rad/s.

s = ωR/(pN·2π·25) = 0.03 %

1-s=0.97

We get the equation for ΨS as follows. Note that we use for ωmS the value ωmS = ω1:

SSR

SR

R

SS LjR

jL

i Y+

+= )1(sw

w



mSSR

SR

SR

S

S

SS

SmSSR

SR

SR

S

SS

SmSSSS

jLjR

RjLR

u

jLjR

RjLRu

jiRu

ww

w

ww

ww

s

s

++

+=YÞ

Y++

+=Þ

Y+=

)(

VsjVsS 7137.02009827.02 ×-×=YÞ

Now, we proceed with the tip and real torque and power

NmL

T SS

tip

52182

2

=

Y=

s

Note that Ttip would be the maximum certainly only if ΨS were constant; but it depends on ωR! To get theabsolute maximum, you have to calculate Ttip for ωRh! The power is not necessarily in maximum with ωRh. -So be careful with these things.

NmT

TRhrrRh

tip 3.141//

2=

+=

wwww

kWNmpNpN

TP

r

m

54.213.141)//( 1

=×-=

×=

www

Solution 4a-b. We calculate I2 for s=3.0 % from the expression:

I2 = V1/(R1+R2/s+X1+X2) = 133/(.2+.2/.3+j0.46+j0.46)A = 33.04A-j4.427A.

|I2| = 33.33A

Pshaft = 3|I2|2*R2(1-s)/s = 3·33.332*.2*(1-.03)/.03= 21.56 kW.

T = Pshaft/ωm = 24.8 kVA/[2π(1-s)ωS] = 141.5 Nm.

Comment: The results with this method (Solution 4a-b) are quite certainly more precise. The differencewith the Solution 4a-a is, however, astonishingly small. Please remember still that parameters in 4a-a areapproximations; e.g. the value used for LS = Lm and Lσ

S could have been better optimised; and electricalfrequencies are used except for ωRh.

Exercise 4b. Calculate the torque and power for the motor which have Xσ1 = Xσ2 =0.46Ω; R1 = R2 = 0.2Ω; Xm = 16.5Ω; V = 400 V, pN = 2; s = 3% /5/. Use the preciseRotating method.

Solution. (Mathcad, HiePsiA2):



Figure 58

We have here used the equations:

2121

2

2121

2

01222

1111

012

11

)()()()(

)()(

FFWWWVI

FFWWFVI

jLLRWjLLRW

jLFjLF

SS

SR

aam

aam

aam

aam

×-×=

×-×=

-+××++=+××++=

-+××=+××=

wwwww

wwwww



Figure 59

Figure 60



Figure 61

Figure 62



Figure 63

This example gives almost the same result as given in the book of L. Hietalahti /4/, pp. 68-69. Exercise 7below repeats the solution of Hietalahti. Some remarks:

· The Electrical Model and the Stationary Method (coordinates not rotating, ωa=0) arevisually identical and shown below for input (Sin) and output power (Sout):

The same applies to output torques, which are here for both methods:

· The results of Hietalahti’s method are really near:I1 = VS/Zkok = V/(sqrt(3)Zkok) = (31.790887-j17.052138)AS = 3VV·I1

* = sqrt(3)V·I1* = 22kW+j11.8kVAr.

I2’ = I1 · Zm/(Zm+Z2) = 32.3 A <-6.7°.Pmech = 3[(1-s)/s]R2’·(I2’)2 = 20.5 kW.T = Pmech/ωm = 133.6 Nm.

· If the Rough Method and the Stationary Method (coordinates not rotating, ωa=0) arecalculated with 1 pole pair only (pN = 1), we get otherwise similar results, but the correctpower values are got with equal relative slip or double real slip:



Also, only half of the real torque is got with equal relative slip or double real slip:

· The s = 3 % moment and frequency are quite the same with the Rough Method withinductances in Exercise 4a. There we got:Pout(max) = 21.54 kW with ωR = 4.712 rad/s,Tout(max) = 141.3 Nm with ωR = 4.712 rad/s.

1.3.8. Exampleinductionmotor3Impedances are for one phase winding. This motor is a capacitor-start motor withthe parameters:

· Stator resistance R1 2 ΩRotor resistance R2 4 ΩStator leakage reactance X1 1.3 ΩRotor leakage reactance X2 1.1 ΩMagnetising reactance Xm 40 ΩCore parallel resistance Rc 1540 ΩNumber of pole pairs p 2Terminal voltage V 120 VElectrical frequency f 60 HzAuxiliary winding factor α 1.05Auxiliary winding resistance Rb 4.6 ΩAuxiliary winding leakage reactance Xb 1.2 ΩStarting capacitance C 150 μF

1.3.9. CirclediagramofapolyphaseinductionmotorAccording to the Figure 54, the model of the rotor circuit in the stationary state is aspresented in the Figure 64.

Figure 64



Thus the rotor circuit obeys the equation:

We divide the equation with jωRLσS, and denote the tip angular frequency with ωRh =

RRS/Lσ

S, and get:

With constant flux linkage the ratio ΨS/LσS is considered to be constant (as flux is for

all good transformers) and with infinite slip frequency, ωR = ∞, ΨS/LσS = -iRS and thus

parallel to the flux and magnetizing current imS. Because the left side of the equationis constant and the components on the right side are perpendicular, the equation isan equation of a circle as represented in the Figure 65.

Figure 65

According to the book of Beaty and Kirtley, /3/, the induction motor may berepresented as in the Figure 66.



Figure 66

The corresponding, so called, circle diagram is represented in the Figure 67.

Figure 67

The voltage-current relations of the poly-phase machine are roughly indicated bythe circuit of Figure 66. The magnetising current

of Figure 67 provides a picture of motor behaviour. The x-axis measures the parasiticand the y-axis the real current of the stator. The total current of rotor isapproximately equal to the parasitic part, as seen in the Figure 56. The stator currentis almost the same as rotor current, for a good transformer the total magnetisingcurrent im = iS + iR ≈ 0 is constant and equal to the small idling value. The voltage overthe input resistance in the power circuit is considered to be approximately constant,for the change of the input voltage is thought to link to the mechanical power of themotor. Thus the primary copper loss is considered to be proportional to the iS ≈ -iR ≈-iR(parasitic), as seen from the linear function of this value in the Figure 67. The statorcopper loss is e.g. to point A: m·ICD·VEO, where VEO is the line voltage and m = 3 isusually the number of phases.



The torque of the motor can be extracted with two different methods from the Figure

67. If we e.g. examine the motor state in point A, the torque is T = m·IAB·VEO/ωm,where ωm is the angular motor speed and VEO the line voltage. Here m·IAB·VEO is theshaft power, P, to the resistor (s-1)R2/s:

sRmIsP 22

2)1( -=

But also the torque is T = m·IAC·VEO/ωS. Here m·IAC·VEO is the air gap power, Pg, to theresistor RR/s (including the warming power) and ωS is the synchronous angularmotor speed:

sRmIPg

222=

Here m = 3 is the number of phases in winding. This latter method must be used ifidling torque is needed; and also if the maximum torque and its position are needed.



1.3.10. OnephaseequivalentcircuitofthethreephaseinductionmotorThe equivalent circuit in the Figure 66 is got by the following calculation. At standstill,the secondary current is equal to the air-gap voltage, E2, divided by the secondaryimpedance at line frequency, Z2 = R2 + jX2, where R2 is the effective secondaryresistance and X2 the secondary leakage reactance at primary frequency.

The so called slip is defined as follows:

Here N is the actual and NS is the synchronous rotor speed. As the rotor speeds up,with a given air-gap field, the secondary induced voltage and the frequency bothdecrease in proportion to s. Thus the secondary voltage becomes sE2 and thesecondary impedance R2 + jsX2, or

Thus we get the circuit in the Figure 66. We call this as the “Electrical Model”.

This can also be drawn for the new equivalent circuits in the Figure 68 which is muchused.

Figure 68



The total power, Pg1, transferred across the air-gap from stator is

.

m is here the number of phases. The Rotor copper loss is:

The mechanical power generated by the motor is therefore:

On the other hand, the power P is, as expressed with synchronous speed, ωS, andslip, s:

.

ωS is here the synchronous angular velocity of the rotor in mechanical radians persecond Therefore,

The following table lists the notations used in the book of Beaty and Kirtley, /3/.



In the book of Nasar the model for an induction motor is as shown in the Figure 69

/9/.

Figure 69

This is sometimes approximated by the circuit in the Figure 70.



Figure 70

Exercise 5. Calculate the peak torque and power for the example motor 1 with helpof the graphs in Figure 71 and Figure 72.

Solution. To get identical results with the original source of the motor data, we must suppose that ourmotor is wye connected. In fact in this exercise, we could use also delta connected model but in thefollowing exercises that choice would produce discrepancies.

Voltage 440 Vrms 1-1254 Vrms 1-n

In the Figure 71, we see that the shaft rotates with the speed 1130 rpm, which makes for the slip s = 6 % ascalculated below. With this slip value we calculate for the torque 5862 Nm, when we take the impedance ofI2 to be 0.1674Ω.

Figure 71

s = 70/1200 = 0.06

T = (1/ωS)mI22(R2/s).

ð T = [1/(2π·20)]3[254V/(0.1674Ω]2·.0064Ω · (1/.06)= 5862 Nm

Now we continue with computing the power:

P = (1-s)ωST.



ð P = [1130/1200]·2π·20·5862 = 693669 W = 694 kW.

The value we get corresponds nicely to the result in Figure 72.

Figure 72

For a comparison with the magnetising reactance: P = 3(254V)2/2.5Ω = 77.4 kW.

Exercise 6a.Estimate the torque and power in the example motor 1 with slip s =70/1200 = 5.83 %.

Solution. We suppose that our motor is wye connected. At first we compute new values for the resistancesand reactances as follows:


Stator Resistance R1 = .0073 ΩRotor resistance R2/s = .0064 Ω/(70/1200) = 0.1097 ΩStator reactance X1 = .06 ΩRotor reactance X2 = .06 ΩMagnetising reactance Xm = 2.5 Ω

Now, first we calculate the parallel impedance Z||:

Z|| = jXm||(R2/s + jX2) = [j2.5||(0.1097+j.06]Ω == 0.1044Ω + j0.0631Ω

Now we get for E2:

E2 = VIN (Z||/(R1+jX1+Z||)

= 254V·(0.1044Ω + j0.0631Ω)/(.0073+j.06+0.1044Ω + j0.0631Ω)

= 178.6V-j53.4V

Now we get for I2 the value:

I2 = E2/(R2/s + jX2)



= (178.6-j53.4)/(0.1097+j.06)

= 1048A-j1060A

|I2|=1491A

s = 70/1200 = 0.0583

Now, ωS is the mechanical angular velocity and m is the number of phases m=3 in the mechanical torqueexpression:

T = (1/ωS)mI22(R2/s).

ð T = [1/(2π·20)]·3·[2583]2·.1097= 5820 Nm


P = (1-s)ωST.

ð P = [1-0.0583]·2π·20·5820 = 689 kW.

ð Ptot = 2π·20·689 = 731 kW.

The results look good. We can compare them with exercise 1. If we had used line-to-line voltage, we hadgot results 3 times as big.

Exercise 6b.Estimate the torque and power in the example motor 1 with fR = 570rpm = corresponding to the slip 570/3600 = 15.8%.

Solution. We compute new values for the resistances and reactances as follows:


Stator Resistance R1 = .0073 ΩRotor resistance R2/s = .0064 Ω/.158 = 0.0405 ΩStator reactance X1 = .06 ΩRotor reactance X2 = .06 ΩMagnetising reactance Xm = 2.5 Ω

Now, first we calculate the parallel impedance Z||:

Z|| = jXm||(R2/s + jX2) = [j2.5||(0.0405+j.06]Ω == 0.0386Ω + j0.0592Ω

Now we get for E2:

E2 = VIN (Z||/(R1+jX1+Z||)

= 254V·(0.0386Ω + j0.0592Ω)/(.0073+j.06 + 0.0386Ω + j0.0592Ω)



= 137.5V-j29.3V

Now we get for I2 the value:

I2 = E2/(R2/s + jX2)

= (137.5V-j29.3V)/( 0.0405+j.06)

= 726.3A-j1040A

|I2|=1942A

s = 570/3600 = 15.8%

Now, ωS is the mechanical angular velocity and m is the number of phases m=3 in the mechanical torqueexpression:

T = (1/ωS)mI22(R2/s).

ð T = [1/(2π·20)]·3·[1942]2·.0405= 3645 Nm


P = (1-s)ωST.

ð P = [1-.158]·2π·20·3645 = 386 kW.

ð Ptot = 2π·20·10935 = 458 kW.

Exercice 6c. Estimate the torque and power in the example motor 1 graphically withslip values , s = 1, 3, 5.8, 9.2, 15.98 and 47.5 %.

Solution. The computed values are all collected in the table below and shown graphically in the two graphs.

Slips R2/s[Ω] Ptot[kW] P[kW] T[Nm]1 0.64 273 270 21723 0.213 630 611 5014

5.8 0.1097 731 689 58209.2 0.070 648 589 5158

15.8 0.0405 458 386 364547.5 0.0135 172 90.3 1368



0100200300400500600700800

-50 -40 -30 -20 -10 0 - Slip %

Ptot[kW]P[kW]

01000200030004000500060007000

-50 -40 -30 -20 -10 0- Slip %

T[Nm]

The data in the graphs shows that the values of power and torque, we have calculated, are about the samemagnitude as in the original graphs shown in the exercise 5. - So, these calculations are made correctlyaccording to the theory.

Exercise 7. Calculate the torque and power for the motor which have Xσ1 = Xσ2 =0.46Ω; R1 = R2 = 0.2Ω; Xm = 16.5Ω; V = 400 V, pN = 2; s = 3%.

Solution. Let’s calculate the combined inductances:

Z1 = R1 + jXσ1 = (0.2 + j0.46)Ω

Z2 = R2’ +[(1-s)/s]R2’ + jXσ2’ = (6.67 + j0.460)Ω

Z3 = Rfe||jXm = j16.5Ω

Zkok = Z1 + Z2||Zm = (5.641300+j3.025906)Ω

Now we get the current:

I1 = VS/Zkok = V/(sqrt(3)Zkok) = (31.790887-j17.052138)A

S = 3VV·I1* = sqrt(3)V·I1

* = 22kW+j11.8kVAr.

I2’ = I1 · Zm/(Zm+Z2) = 32.3 A <-6.7°.



Pmech = 3[(1-s)/s]R2’·(I2’)2 = 20.5 kW.

T = Pmech/ωm = 134.5 Nm.

Exercise 8. For an induction motor the results from the idling and short circuit testare /5/:

Idling (n = nN, s = 0) Short circuit (n = 0, s = 1)Vo = 400V Vk = 400VIo = 14A Ik = 230APo = 160W cosφk = 0.40

In addition the resistance R1 = 0.2 Ω is measured. Calculate the one phase modelcircuit.

Solution. We consider at first the idling results and get the components in the transverse branch:

W===Þ== 10001604003

2222

VAV

PV

RRV

RV

Po

ofe

fe

o

fe

vo

From the reactive power we get the magnetizing impedance:

W===

=-=

=×=

5.169698400

9698

96993

22

22

VAV

QV

X

VArPSQ

VAIVS

o

om

ooo

ooo

The values for the longitudinal branch are got from the short circuit measurement.

W=-=W==W==

W===

20.0'92.0sin||40.0cos||

0.13

||

12 RRRZXZR

IV

IVZ

k

kkk

kkk

k

k

k

kvk

jj

We think that the magnetic dispersion is divided evenly between stator and rotor:

W=== 46.021

12 kXXX ss

The so called component of the axis power is defined as:



S

Rh

NSN

RhRhS

R

r

r

rh

rrrmech

ppLR

XR

s

RRs

Rs

sR

ww

ww

ww

w s

×=====

-=-

=

1'''

''1'1

11

Exercise 9. The nominal power of an induction motor is 100 kW and the efficiency0.94. The torque is as depicted in the Figure 73. The motor starts in time 0.6 s. Thesynchronous speed is 1500 rpm and the frequency of the power network 50 Hz.What is the moment of inertia, JM?

Figure 73

Solution. Clearly

J = T/(dωm/dt) = T/a. The average torque is according to the Figure 73 2.6·TN. The nominal speed isfrom the figure obviously 95% from the synchronous speed which is .95·1500 ≈ 1430rpm = 150 rad/s. Nowthe nominal torque is:

TN = PN/ωmN = 667VAs.

The average acceleration a is

a = (0.95·150rad/s)/(0.6s) = 237 rad/s2.

Thus:

J = (2.6·667VAs/(237 rad/s2) = 7.35 VAs3 (= kgm2).

Exercise 10. Nominal power of an induction motor is 11 kW, power factor is 0.8 andefficiency 90 %. What is the real power and parasitic power the machine takes fromthe supply network? /5/



Solution. Nominal power means the mechanical shaft output (axis power). From the network, thedissipation power of the machine itself must also be taken.

kVArPSQ

kVAkWPSkWkWPP

in

inin

outin

15.92.1225.15

25.158.0/2.12cos/2.129.0/11/

2221

21 =-=-=Þ

===Þ===

fh

Exercise 11. /9/

The phasor addition of voltages (for q = 3) is shown in the Figure 74, from thegeometry of which we get

, = ξq = sin(τv/2)/[qsin(τv/2q)]

Here α = τv/q.



Figure 74

Exercise 12. /9/

In a sinusoidal distributed flux density we show a full-pitch and a fractional-pitch coilin the Figure 75. The coil span of the full-pitch coil is equal to the pole pitch, τ. Let thecoil span of the fractional coil be β < τ, as shown. The flux linking the fractional-pitchcoil will be proportional to the shaded area in Figure 75 whereas the flux linking thefull-pitch is proportional to the entire area under the curve. The pitch factor istherefore the ratio of the shaded area to the total area:

= ξj = sin(γ/2), where γ = πβ/τ.

Figure 75

Notice that in kp, β and τ may be measured in any convenient unit.

Exercise 13. /9/



Figure 76



.

Exercise 14. /9/

To obtain a high starting torque in a cage-type motor, a double-cage is used. Theforms of a slot and the bars of the two cages are shown in the Figure 77. The outercage has a higher resistance than the inner one. At starting, because of thedifferences in the reactance-resistance ratios and the skin effect, the influence ofthe outer cage dominates, thus producing a high starting torque. An approximateequivalent circuit for such a rotor is given in the Figure 78. Suppose that, for a certainmotor, we have the per-phase values:

Ri = 0.1Ω, Ro = 1.2 Ω, Xi = 2Ω, Xo = 1Ω.

Figure 77

Figure 78

Determine the ratio of the torques provided by the two cages at (a) starting and (b)2% slip.

a) From the Figure 78, at s=1,



b)

1.3.11. Analysisoftheonephaseinductionmotor/9/



The torque relationship of the poly-phase induction motor is applicable to each ofthe rotating fields of the single-phase motor, except that the amplitude of the eachrotating field is one-half that of alternating field. This results in an equal division ofmagnetising and leakage reactances, and the approximate equivalent circuitbecomes as shown in the Figure 79 (a). The torque speed characteristic is shown in the(b) part of the figure.



Figure 79

Different structures for starting a one-phase induction motor automatically arepresented in the Figure 80, Figure 81, Figure 82.

Figure 80



Figure 81

Figure 82

Exercise 15. /9/

From Figure 79 (a):



(c) Proceeding as with poly-phase induction motor, we have:

Exercise 16. /9/



Exercise 17. /9/



Exercise 18. /9/



1.4. DCmotors

1.4.1. OldstructureIn the Figure 83 we see the structure of a conventional DC motor.

Figure 83.

1. Coiled rotor2. Casing of motor3. Permanent magnet or core of the stator winding4. Field winding5. The brushes6. Commutator

Because of the existence of the field coil, we can control the motor also by changingthe strength of the stator field; which however is a slow control method for becauseof the usual construction, the time constant of the stator coil is usually big.

1.4.2. CupshapedrotorwithpermanentmagnetstatorThe most common structure of DC motors nowadays is the permanent magnetmotor: It is small in size and easy to control. Anyway, the available torque isdependent of the strength of the magnets and stronger magnets are moreexpensive.

If we want a permanent magnet rotor to accelerate fast, we can make it without anyiron core as shown in the Figure 84 looking from two opposite directions.



Figure 84.1. The end of the motor2. The holder of the brush3. The brushes4. Casing of the motor5. The ironless rotor6. Spacer7. Permanent magnet8. Bearing9. Lock ring10. Electrical connectors

The rotor winding is a cylindrical tube, which holds its shape because of an epoxyresin coating. The cup shaped rotor guarantees good acceleration and also goodefficiency for because of the lack of an iron core, there is no iron losses. The motor isalso light. In some models there may be a stationary iron core in side of the rotor.Also in this structure, there is little iron loss, for the field do not change in the ironcore. However, the motor does not tolerate over currents, for the heat capacity ofthe rotor is really small: This implies that the motor burns rapidly if stopped withforce.



1.4.3. DiskshapedpermanentmagnetmotorThis motor is particularly short in structure. Also, in the basic structure of this type,the winding is in the rotor, which is now made on a round circuit board; the magnetsare in the stator on both sides of the rotor. The properties of this motor type arequite similar to the cup shaped one. The structure of the motor is seen in the Figure

85.

Figure 85.1. Front end of the motor2. Ball bearing3. Rotor4. Casing of the motor5. Support plate for the bearing6. Ball bearing7. Rear end of the motor8. Support for the brushes9. Brushes10. The cover of the brush support

1.4.4. BasicequationsofpermanentmagnetmotorThe function of a permanent magnet DC-motor is explained with three basicprinciples:

1. The torque (T) is directly proportional to the current.

2. The rotational speed (n) is directly proportional to the voltage (V) whenthe load is constant.

3. The motor obeys the Ohm’s law: V = R · I.

These may be rewritten as equations:

1. T = kT · I



2. n = (1/kn) · V

3. V = R · I + kn · n.

1.4.5. CharacteristicsofpermanentmagnetmotorThe characteristics of a permanent magnet DC motor are seen in the Figure 86.

Figure 86

1.4.6. BrushlessDC-motorBrushless or electrically commutated DC-motor is one type of DC motors, where therotor is made of permanent magnets. The rotor may be outside of the winding, as inthe Figure 87, or inside of the winding, as in the Figure 88. The electronics switches thecurrent to separate windings according to the position of the rotor. In the motor ofFigure 87 the rotor position is sensed from the current generated in the winding andin the motor of the Figure 88 the position is sensed by three internal Hall sensors.With three Hall sensors it is usually achieved a speed control with speeds over 700-900 rpm. With slower speeds some other means, like more precise external pulsegenerator, are needed.



For the brushless motor differs from the brushed one only with the control methodof commutation – electrical or mechanical – the basic equations are the same. Thecharacteristics differ only in the fact that the brushless motor may rotate fasterwithout the danger of wearing of the brushes.

Figure 87.1. Rear plate2. Electrical leads3. Control electronics4. Casing5. Winding6. Ball bearing7. Spacer8. Spring9. Permanent magnet10. Outer rotor11. Axis12. Front plate



Figure 881. Rear plate2. Electrical board3. Hall sesor4. Support plate5. Ball bearing6. Spacer7. Axis8. Permanent magnet9. Electrical board10. Winding11. Plate spring12. Space ring13. Ball bearing14. Casing

1.4.7. SchematicmodelsforDC-motorIn the Figure 89, there is seen the wiring diagram and cross-section of a DC-Motor.The stator field is not rotating and is made here by magnetising wiring, F, but alsopermanent magnets may be used. If the magnetising current is taken from an ownsupply, the motor is called as out-awaken. If the wiring is connected in series withthe rotor wiring, the motor is called as a series motor; if the wiring is in parallel tothe rotor, the motor is called as a parallel machine. In the past the machine couldhave even all three magnetising wirings and was called as a compound motor.

In the demanding machines, like here, there is also so called compensating wiring,KO, in the stator, which is connected in series with the rotor; this wiring is placed inthe same direction with the rotor winding in the main stator field (see Figure 89),



which generates similar torque in it as in rotor, but of course the torque can’tproduce any rotation for the wiring is stationary; the role of the compensatingwiring is simply to prevent the dispersed field of rotor wiring from disturbing themain field, which is achieved because the current, and thus also the flux, incompensating wiring is opposite to the rotor current, and field.

Typically, there is still a third wiring in the stator which is called as polarity changeor commuting wiring, KN, which, also, is connected in series to the rotor wiring. Thisproduces a narrow flux perpendicularly to the main field in the position to the statorcircle, where current in the loops of rotor wiring changes direction; it producesthere a loop voltage which opposes the old current and stops it.

In the rotor, there is the armature winding, A. It produces a field which is constantlyperpendicular to the main field produced by the stator, even when the rotor isrotating, which is achieved by the to the rotor axis attached so called commutatorwhich changes the current direction in the loops of rotor wiring when the rotorrotates.

In electrically commutated or brushless DC-motors the permanent magnets areplaced in the rotor and rotors field orientation is constantly monitored; a rotatingmagnetic field perpendicular to the rotor field is produced electrically with a 3-phase wiring in the stator. So, no rotating contacts or brushes are needed.

Figure 89.Connection diagram (left) and cross-section for a DC-motor (right)A Armature wiringKN Polarity change wiringKO Compensating wiringF Magnetising wiring



The to-the rotor reduced magnetising current, IF, is defined with the real rotorcurrent, if, as here:

.

In theory, the coil factor ξA is the diameter of a circle divided by the half circle or2/π; this is when the covering factor is 1. Now the produced field is:

In the Figure 90 we see the flux density of a DC-motor when loaded and also withoutload.

Figure 90. Field density in a DC-machineField density verticallyRing angle horisontallyq-axis in the direction 0 degreesd-axis in the direction 90 degreesWithout load: horizontal lineWith load: tilted line.

Figure 91



The rotor field now produces the torque:

The loop wiring of the rotor is shown spread on a plane in the Figure 92; there are 7loops with current in the figure and the loops marked black are changing theirdirection. In the Figure 93 there is shown how current is oriented in the halves of thewiring loops. In the Figure 94 there is shown how the wire loops of rotor areconnected to the lamellae.

Figure 92. One branch of the rotor loop wiring in the brush distance.The loops marked with black are changing their current direction.

Figure 93. Position of the rotor wires and the direction of the currents.



Figure 94. Connections of the rotor wires (blue) to the lamellae.

The rotational voltage VAi = Viq develops to the armature (rotor) wiring and it isperpendicular to the main field (winding field in fact) Ψd:

.

In the stationary state the armature current can be calculated from the followingformula, which is here written to be scalar in type:

The commuting wiring produces the voltage needed for change in the armaturecurrent direction:

Vk = .

The “rotating” winding field, Ψ, in a DC-motor is build up by the magnetising wiring,compensating wiring and the perpendicular commuting wiring in the stator. In factthe field Ψ is stationary and the armature rotates in respect to it. The field has theformula:

The transverse air gap inductance gets the value Lq ≈ α2/3Ld where Ld is thelongitudinal inductance. fKO describes the amount of compensation and in practicegets the values 1 or 0. We see from the equation that Ψ depends only frommagnetising current in the stator if the motor is compensated. For theuncompensated machine we get:



The current iF has the formula:

,

where if is the magnetising current. Here ξf and ξA get the approximate values 1 and2/π. Now we add the dispersed fluxes and get the formulas which have beenreduced to the rotor:

The formula for compensating flux linkage seems to be complex, but the complexitycomes only for the requirement that the flux disappears when there is nocompensating or fKO = 0.

If rotor is not rotating, we get for the magnetising voltage VF:

In practice we suppose that Ld is not changing and write this equation in scalar form:

.

Here σF is the dispersion factor for magnetising wiring. The possible compensatingand commuting wirings have zero-valued fields and their voltage formulas are:

In practice these voltages are in series with the armature voltage.



The field Ψ = Ψd + Ψq induces in the closed rotor winding the following componentvoltages:

.

Here ωR = ωΨ -ωm, when the flux is not rotating. These voltages are presented in theFigure 95.

Figure 95. Movement voltages in the commuting winding.

We see from the Figure 95 that in the clockwise rotating machine the rotor rods feelthe rotation of the flux to be counter clockwise. In a DC-machine the brushes are onthe q-axis so that the longitudinal voltage VRid can’t be utilised. When only thetransverse motion voltage is utilised, the longitudinal field Ψd = Ψ is the working realfield. Thus the motion voltage in the armature is as a vector and scalar:

Here Ψ = pNΨp is the total flux of all the pole pairs together and ωm is the real speedof the machine. Ψq does not participate to the creation of the working armaturevoltage and thus has the position of dispersion flux in a DC-machine. The dispersedinductances and the dispersed inductance voltage are:



The covering factor gets values α = 0.6… 0.7, so that compensating diminishes thedispersed inductance to 30… 50 % making the time constant of armature circuitsmaller. The voltage equation of armature circuit in scalar form is:

.

Here LA is the total dispersed inductance of the armature circuit and RA thecorresponding resistance in the armature.

We get the torque for the machine as:

.

This is in scalar form:

The equations of a DC-machine for flux linkage, voltage, torque equations for a timepoint t are:

Here J is the moment of inertia and Tw the counter torque.

A commonly used equivalent circuit of a DC-motor is represented in the Figure 96; E isthe back voltage (arrow in the direction of emf), Ra the armature resistance, Va thesupply (or armature) voltage, and Ia the armature current.



The Kirchhoff’s Voltage Law gives for the voltages:

Va = IaRa + E.

There is a special expression for the torque of a foreign magnetised machine, whichwe get with the following calculations:

aa

mcacmca

a

c

a

mca

m

mc

a

a

m

mc

a

aa

m

mc

m

a

m

ee

IRKUKKU

RK

RKUK

REUK

RRIKEIPT

Y=-

×=-×=

-×=

-×=×===

fwffw

f

fwwfw

wfw

wfw

ww

)(

So, Ψ=Kcφ, φ=KfIf. The idling speed is

fw

c

amo K

U=

,which gives us the final result:

)()( 2

mmoa

ce R

KT wwf

-×=.

For a parallel magnetised motor we get the torque:

affcacae IIKKIKIT ==Y= f .

E

Ra

Va

Ia

Figure 96



Figure 97. Torque of a parallel magnetised motor.

For a serially magnetised motor we get the torque

22)( a

mfca

fce V

KKRKK

T ×+

=w

.

Figure 98. Torque and speed of a serially magnetised motor as functions of the armature current withdifferent degrees of magnetisation, β=If/IA.

Exercise 1. A foreign magnetised DC-motor, in Figure 99, is connected in the electricnetwork of Va = 330 V. The loading is such that the armature current is Ia = 20 A. Therotational speed is n = 1420 rpm. The armature resistance is 0.50 Ω. Compute theaxis power and torque. Compute, also, the rotational speed in idling when themagnetising current is kept constant. /5/

Solution. We compute at first the back emf, E, with the load mentioned.

E

Ra

Va

Ia

Figure 99



VAVIRVE aaa 3202050.0330 =×W-=-=

The angular speed is

sradfm /71.14860

142022 === ppw

The idling speed, no, is got with the following facts: The machine constant Kc does not change; andmagnetising φ is kept constant; and the armature current Ia = 0 in idling:

rpmn

sradVs

VKEKE

VsEKKKKE

mo

cmomoco

mcbmbmc

1464602

3.153602

/31.153152.2

330

152.27.148

320

0

=×=×=Þ

===Þ=Þ

====Þ==

ppw

fwfw

wfwfw

Here we have defined Kb = Kcφ. The power is easily got as here:

kWWWIKEIP acmae 4.62032020152.27.148 =×=××=== fw

The mechanical torque is:

NmIKIKEINmPT atacm

a

m

ee 04.4320152.2

71.1486400

=×====== fww

.

Here we have defined Kt = Kcφ. Thus E = Kbωm and T = KtIa. Note that surprisingly Kb = Kt = Kcφ.

1.4.8. EMFequationforDC-motorConsider a conductor rotating at n rpm in the field of p poles having a flux φ perpole (p = 2·pN, where pN is the number of pole pairs). The total flux cut by theconductor in n revolutions is pφn; hence, the absolute flux cut per second, giving theinduced voltage e, is

60|| np

dtde ff

== .

This is seen by thinking at first that initially the current loop is in the angle 0° and theflux (p=2) goes directly through it in the positive direction, so with the value +φ.Then after a rotation of 90°, there is no flux through the loop and the flux haschanged by absolute value Δ|φ| = φ. After all 90° turnings of the loop, the flux andthe sum, Δ|φ|, of the absolute changes in its magnitude are:

Angle Flux Sum of the absolute changes Δ|φ|0° φ 0



90° 0 φ180° -φ 2φ270° 0 3φ360° φ 4φ.

Thus the total average voltage around the loop is 2e = Δ|φ|/T = 4φf = 2pφf. Here eis the voltage over half of the conductor loop or over “one conductor cutting theflux”, or

e = pφf = pφn/60,as stated earlier. If there is a total of z conductors on the armature, connected in aparallel paths, then the effective number of conductors in series is z/a, whichproduce the total voltage E in armature winding. Hence, for the entire winding weget the emf equation:

mazp

aznpE fw

pf

260=×= ,

where ωm = 2πn/60 (rad/s). This may also be written as

nkkE gma ffw ==

where ka = zp/2πa (a dimensionless constant) and kg = zp/60a. If the magnetic circuitis linear (i.e. if there is no saturation), then

ff ik=f

where if is the field current and kf is a proportionality constant; and so we get

mbmf KkiE ww ==

where k ≡ kfka, a constant. For a nonlinear magnetic circuit, E versus if is a nonlinearcurve for a given speed. The other constant, Kb ≡ kif = kfkaif, is called the back EMFconstant.

1.4.9. TorqueequationforDC-motor

The mechanical power may be equated to the electrical one as:

ame EiT =w ,where the armature losses are ignored. This becomes:

aae ikT f= .



This is known as the torque equation. For a linear magnetic circuit this yields:

atafe iKikiT == ,

where k = kfka. Thus k may be termed the electromechanical energy-conversionconstant. The other constant, Kt ≡ kif = kfkaif, is called the torque constant. It issurprising that the constants of back EMF and torque are equal, or

Kb = Kt.

1.4.10. SpeedequationforDC-motor

If Vt is the terminal voltage of the DC motor, then Vt = IaRa + E or

E = Vt - IaRa

From this we get:

fw

a

aam k

RIV -=

which, for a linear magnetic circuit, becomes to the speed equation:

f

aam Ik

RIV×

-=w .

An alternative form of this is:

f×-

=×

-=

g

aa

fm

aa

kRIV

IkRIVn (rpm)

where km = 2πk/60 (Ω·min).

Exercise 2. Calculate the voltage induced in the armature winding of a 4-pole, lab-wound, dc machine having 728 active conductors and running at 1800 rpm. The fluxper pole is 30 mWb. /9/

Solution. Because the armature is lab wound, p = a, and

VapnzE 2.655

60)728)(1800)(1030(

60

3

=×

=×=-f

.



Exercise 3. Calculate the voltage induced in the armature winding of a 4-pole, wave-wound, dc machine having 728 active conductors and running at 1800 rpm. The fluxper pole is 30 mWb. /9/

Solution. For a wave wound armature, a = 2. Thus:

VapnzE 4.1310

24

60)728)(1800)(1030(

60

3

=××

=×=-f

Exercise 4. If the armature of the lab-wound machine above is designed to carry amaximum of 100 A, what is the maximum electromagnetic power developed by thearmature? /9/

Solution. Because there are 4 parallel paths (a = p = 4) in the lap-wound armature, each path can carry amaximum current of:

AII a 254max == .

Nevertheless, the power developed by the armature is (all the paths are parallel):

kWEIP ad 5.65)100)(2.655( === .

Exercise 5. If the armature of the wave-wound machine above is designed to carry amaximum of 25 A for a path, what is the maximum electromagnetic powerdeveloped by the armature? /9/

Solution. Because there are now 2 parallel paths (a = 2) in the armature, each path can carry a maximumcurrent of:

AAI 50252max =×= .

Hence,

kWEIP ad 5.65)50)(2.1310( === .

So the power is the same for lap and wave wound machines.

Exercise 6. Calculate the electromagnetic torque developed by the armature of thelab-wound machine above. /9/



Solution. The rotational speed is n = 602/(p/2) rpm = 1800 rpm. We get from the energy-conversionequation:

NmEITm

ae 6.347

60/)1800(2105.65 3

=×==pw

.

Of course we get the same result to the wave-wound machine, for the pole count is the same, p = 4, andthe developed power is the same.

Exercise 7. A dc-machine has a 4-pole, wave-wound armature with 46 slots and 16conductors per slot. If the induced voltage in the armature is 480 V at 1200 rpm,determine the flux per pole. /9/

Solution. Here z = 16 x 46 = 736, and so, from the emf equation:

mWbpa

nzE 3.16

42

73612004806060

=××

×=×=f .

Exercise 8. A 4-pole, lap-wound armature has 144 slots with two coil sides per slot,each coil having two turns. If the flux per pole is 20 mWb and the armature rotatesat 720 rpm, what is the induced voltage? /9/

Solution. Substitute p = a = 4, n = 720, φ = 0.020, and z = 144 · 2 · 2 = 576 in the emf equation to obtain

VapnzE 24.138

44

60)576)(720)(020.0(

60=×=×=

f.

Exercise 9. A 10-turn square coil of side 200 mm is mounted on a cylinder 200 mm indiameter. The cylinder rotates at 1800 rpm in a uniform 1.1 T field. Determine themaximum value of the voltage induced in the coil. /9/

Solution. From Faraday-Henry law we get:

VBNAE 94.8260/)18002()200.0)(10)(1.1( 2max === pw .



1.5. SteppermotorsThe mechanical structure of a stepper motor is the same as that of the synchronousmotor shown in the Figure 7, which is reproduced here as the Figure 100.

Figure 100. Synchronous motor with a small number of pole pairs1. Outer end of the first stator winding2. First stator winding with bushing to the rotor axis3. Permanent magnet rotor4. Inner ends of the stator windings5. Second stator winding with bushing to the rotor axis6. Outer end of the second stator winding

In the Figure 101, there is shown how the permanent magnet rotor follows therotation of electrical magnet. This simple motor takes four steps for a full rotationand the angle of one step is 90˚. In general motors of this type have 20, 24 or 48poles. The step angle is calculated by dividing 360˚ by the number of the poles.However, the motor is quite inefficient because only one winding (A or B) hascurrent in the same time.



Figure 101

In the Figure 102 the motor is got to work 40 % more effectively for both the windingsare active all the time.

Figure 102



1.5.1. Halfstepping

If we use the stepper motor windings alternatively one and two at a time we get socalled half stepping control, which is depicted in the Figure 103. The accuracy isdoubled but the torque is estimated to be 15-30 % smaller.

Figure 103

1.5.2. Microstepping

Often the half stepping is not accurate enough or we want to get less jerking in aslow movement. In the micro stepping the full step is divided e.g. to 5, 10, 100 oreven 250 equal sized steps. The principle is easy to understand if we for examplestart from the state 1 in the Figure 103 and make the current in the A phase smaller in



small steps, which we call as micro steps. When the current A has reached the value0, we have arrived to the state 2 in the figure.

1.5.3. Reluctancesteppermotor

In the Figure 104 we see a so called reluctance stepper motor. In this motor there arefour phases in the stator winding through the rotor, which is made of magneticallysmooth iron in a form of a cylinder which has oppositely placed spikes in equaldistances. One phase at a time gets current and attracts the nearest spike pair in thesame line. The principle is that the rotor rotates in the position where the air gapsbetween the rotor and stator are as small as possible. This way we can get a precisecontrol but the torque is not as great as in permanent magnet motors. In the motorof the Figure 104 we get the accuracy of 15˚ (180˚/3 – 180˚/4 = 60˚ – 45˚ = 15˚).

Figure 104

1.5.4. Hybridsteppermotor

The hybrid stepper motor is a hybrid of reluctance and permanent magnet steppermotors; an example is presented in the Figure 105 and Figure 106. In the rotor there istwo around the same axis sequentially placed equal rotor-stator packages. Both ofthe rotors are divided across the axis in two separate equal parts which are the



north and south pole of a permanent magnet. The poles are toothed axially withnarrow teeth as in a reluctance motor and the teeth groups of the poles are in halftooth phase difference to each other as seen in the Figure 105.

The function of the motor is explained in the Figure 106 where it is drawn the southpole of one of the rotors with the surrounding stator winding. The stator is windedso that when there is current in the phase A, the poles marked with A are north andthe poles marked with A’ are south. In this situation the teeth of the half rotorapproach the A poles and avoid the A’ poles and the rotor rotates in the position ofFigure 106; let’s call this position as sate 1. Next, current is connected to phase B, thenin A’, then in B’ and finally again in A so that we have got back to the state 1 and onestep sequence has become ready and the rotor has rotated the angle of one tooth.

The opposite half of the rotor is the north and in the state 1 it approaches the A’pole of the winding, which supports the movement of the south part of the rotorbecause the teeth of the parts are in half tooth phase shift, as you might remember.

There are typically 50 teeth in the rotor of a hybrid motor; to move the rotor withone tooth takes 4 steps so that a whole circle takes 200 steps. The step angle istherefore 1.8˚. In half stepping the step angle is 0.9˚.

Figure 105

The half stepping is performed as explained in the Figure 103: connecting two phaseson simultaneously rotates between the full steps. Also micro stepping is possible asexplained earlier in the context of the Figure 103, too.



Figure 106

1.5.5. UnipolarcontrolThe old method to control stepper motors is the so called unipolar method. Thereare in this case 6 leads connected in the motor; they make up two phases with amiddle contact in both of them. The connection is seen in the Figure 107. Comparethis with the topmost picture in the Figure 114.

Figure 107

In the application where two phases are kept ON, other of the transistors E and Fand also of the pair G and H is conducting. In this case half of the copper in thewinding is in use.



1.5.6. BipolarcontrolThe bipolar technology is nowadays the dominant one. In this case only four leadsare needed in a motor. Four transistors are used in the circuit as seen in the Figure

108. The motor takes a step every time the direction of one of the phases isreversed; e.g. in the figure the direction of the current in the AB phase.

Figure 108

1.5.7. SerialresistanceThe time constant τ for the connection of the current in the motor winding iscalculated with the expression:

τ = L/(R + Rs), where

L is the winding inductance,R is the resistance of the motor winding andRs is the serial resistance

So we see that the time constant shortens when the serial resistance is increased.The traditional method to increase the top speed of an stepper motor is to add aseries resistance which prevents the motor from burning even if the supply voltageis made in size many times the nominal voltage. In the Figure 109 we see how thecurrent develops in a stepper motor winding.



Figure 109

1.5.8. Choppercontrol

With the developing transistor technology the stepper motor controllers havebecome chopper controlled. In this technology the motor winding is supplied with arelatively high frequency current. Typical chopper frequency is from couple ofthousand Hz to 45 kHz. How the motor current is born is presented in the Figure 110.Usually it is possible to adjust the maximum current value.

Figure 110

1.5.9. TwovoltagecontrolIn the two voltage control supply voltage is connected to the wiring at first with highvoltage value for guaranteeing fast rising of the current. After some time or when a



suitable level of current is reached, the voltage is automatically adjusted to a smallervalue. In the higher coordinate system in the Figure 111 we see how the supplyvoltage changes in time and in the lower system the development of current at thesame time. In some systems the voltage may change according to differentsituations: e.g. the voltage may be low with small velocities and high with largerones. – higher voltages get the current grow faster and thus increase the motorspeed.

Figure 111

In the Figure 112 is seen the torque characteristics with different voltages in anexample motor.

Figure 112



1.5.10. DifferentcontrolunitsThe basic block of a control is the inverter which takes in pulses and controls themotor to take one step for each of them. Inverter might take in also other controlinformation as the direction, current level, step type (half or full) and so on.

A position control might in its simplest form drive a certain amount of steps in themotor. Also, it is possible in some cases determine the structure of acceleration anddeceleration ramps and the basic speed of the motor. The basic concepts in thespeed profile determination are defined in the Figure 113.

Figure 113

Ta1 Acceleration time with the basic speedTd1 Deceleration time with the basic speedTa2 Acceleration time without the basic speedTd2 Deceleration time without the basic speedVh Accelerated speedVb Basic speed

1.5.11. ResonancephenomenonA stepper motor might lose its torque totally in a resonance state: it might onlyshiver in place and give some unpleasant voice.

The first resonance point is in hybrid motors in the speed interval 20-200 steps/s.The next point is in the region 50-1500 steps/s and the last one in the region 2500-



5000 steps/s. The first resonance is the most difficult one and has typically the widthof 20-50 steps/s. All the resonance phenomena in stepper motors occur in evenspeed or slow acceleration; the first symptom is lower torque.

The simplest method to dampen resonances is to set the base speed over the firstresonance point. Using micro stepping is the best way to avoid the resonances, forthen the resonances disappear totally. Half stepping might also help. Earlier,resonances were damped with a viscous damper in which a wheel is placed to rotatein a suitable fluid. Often only the adding some friction to the system is enough todampen the resonances; using lower supply voltage may influence similarly.

The connector wiring in a stepper motor may vary. In a hybrid motor the number ofleads may be 4, 6 or 8. Different wirings are seen in the Figure 114. The eight-leadconnection makes it possible to connect the windings externally in series or inparallel.



Figure 114



1.6. ServomotorsSmall servomotors are electric motors which are combined internally or externallywith suitable sensors and control electronics to make them capable to controldifferent kind of behaviour. Nowadays, they are typically permanent magnetmotors, either DC motors with or without brushes, or synchronous ac motors. In theFigure 115 we see a schematic diagram of a control system, where a velocity iscontrolled, in the Figure 116, we see the mechanical structure of a real servomotor, inwhich a tachometer is integrated.

Figure 115

Figure 116



The feedback sensor in the servomotor may be a tachometer, as in the Figure 115 andFigure 116, or also a position sensor like the optical realisation of an incremental pulsesensor in the Figure 117. An optical pulse sensor might also give the absolute positionas a parallel set of pulses. The code disk of an absolute sensor is shown in the Figure

118.

Figure 117

Figure 118

A pulse sensor might be also magnetic as the Hall sensor depicted in the Figure 119.Incremental sensors are good for positional and speed measurements. The absolutesensors are suitable to determine a position. The maximal resolution of the firstones is 5000 pulses for a full rotation and for the last ones 13 bits.



Figure 119

With a resolver you get a sine shaped signal from a rotation. The principle ofoperation is seen in the Figure 120. Through a stator winding of the resolver which isconnected as a transformer to the rotor winding, it is input an ac-current offrequency typically from 400 Hz to 10 kHz. This signal is induced to the outputwindings of the resolver which are fixed on the stator. There are two outputwindings in the stator in 90˚ phase angle. From one of these is got the input signalmultiplied by the cosine of the resolver angle; from the other multiplied by the sine.

Resolver is good to be used in connection with an ac-servomotor because it is freeof maintenance and gives sine shaped output signal. It is also mechanically durableagainst hits and tremor. Additional combined electronics may change the signal to10, 12, 14 or 16 bit parallel positional signal with measuring range of one turn - butan index pulse can be added to the start of a new round.



Figure 120

1.6.1. ServoamplifiersThe most simple servo amplifiers work only in one fourth of torque-frequency plane,as seen in the Figure 121. In this situation the amplifier tries to follow the controlcommand, which is typically a voltage signal from 0 V to 10 VDC, only when themotor lags the control signal: the torque is always in the same direction. In thesecond fourth of the co-ordinate system the motor should brake, but it can’t reallydo it; the passive braking is, however, achieved by short circuiting the motorwinding with suitable load resistor: now the motor works as a generator and theenergy produced is dumbed in the resistor.



Figure 121

A real servo control works in all the quadrants of the torque-frequency plane asshown in the Figure 122.

Figure 122

1.6.1.1. Linear amplifiers

Linear amplifiers are usually not used in the power range over 500 W because of thestrong heat generation in the amplifier. They have, however, definite advantages.Firstly, they do not generate electrical or electro-magnetic disturbances. Secondly,there is no need for serial choke if used with low impedance motors.



1.6.1.2. Chopper amplifiers

As is commonly known, the waste power in an amplifier is Pwaste = V·I. In chopperamplifiers either the amplifier voltage V or current I is zero, and thus there is nowaste power at all. This makes it to be the dominant one in many applications andthe only sensible alternative in large power applications. In reality the efficiency of achopper amplifier is typically 80-90 %. One much used power control method is thePWM or Pulse Width Modulation.

1.6.1.3. Common control tasks

Here are only listed some common control tasks or parts of them.

· Speed control. The real speed may be monitored with a tachometer orresolver or some pulse sensor. The real speed and the speed instruction arecompared and made equal.

· Position is not always possible to be controlled directly but is implemented asa so called cascaded control, which has two nested control loops, see Figure

123. – The inner one controls the speed or torque to be linearly dependent ofthe position control signal (dependent of position error); or speed or torque isconstrained to some limits. The outer loop makes the position control signalto be linearly dependent of the position instruction; or position is constrainedto some limits.

A control amplifier obeys often so called PID algorithm. It does not have to bepowerful, for the control signal got out can be amplified with a separate poweramplifier of linear or chopper type, as described earlier. The output voltage of acontrol (PID) amplifier may be in the range ±5 V or ±10 V with maximal outputcurrent in 100 mA level. The voltage V needed out from the power amplifier can beroughly estimated from the equation:

V = R · I + kn · n,

where R and I are the armature resistant and current of the motor, respectively; kn isthe speed factor and n is the rotational speed.



Figure 123

Often it is important to make sure that the maximum allowable torque is notexceeded, or it is needed to keep torque in a constant value; one example of the lastsituation is waltzing. In simple situations (simple DC motor) torque is proportional tothe armature current and the control of torque is realised as a control of current, asshown for electrically commutated DC motor in the Figure 124. Here the motor hasthree built-in Hall sensors, which give the axis position to the control amplifier, asseen in the figure.

Figure 124



Very often or even usually, the servo amplifier is not a plain PID controller butincludes also so called positioning functions, which can automatically obeypredefined velocity profiles, like in the Figure 113. Or the servo amplifier may evenaccomplish so called trajectory control.

1.6.1.4. Combined servo elements

In principle, the motor, the servo amplifier, the power amplifier and the sensorsneeded for feedback are all separate. However, they are all needed in a servoapplication, which makes one think if they could be combined to a single servomotor-component. The servo applications are yet so different that in general it can’tbe done economically. The most common combination might be Hall sensor and ECDC-motor. There are still some servo motors with complete in-built electronics forcontrol; in the Figure 125 we see a motor with a speed or position control in its tail.

Figure 125



1.7. Exoticmotors

1.7.1. BallbearingmotorA ball bearing motor is an unusual electric motor that consists of two ball-bearing-type bearings, with the inner races mounted on a common conductive shaft, and theouter races connected to a high current, low voltage power supply. An alternativeconstruction fits the outer races inside a metal tube, while the inner races aremounted on a shaft with a non-conductive section (e.g. two sleeves on an insulatingrod).This method has the advantage that the tube will act as a flywheel. Thedirection of rotation is determined by the initial spin which is usually required to getit going.

There are two rivalling theories for explaining the principle of operation for thismotor type: one thermodynamic and the other electro-magnetic.

1.7.2. Homo-polarmotorCould the homo-polar motor be the simplest electric motor? One is shown in theFigure 126 below.

Figure 126. Homo-polar motor. Simple homo-polar motor made with drywall screw, an alkaline battery cell, a wire, and aneodymium disk magnet. The screw and magnet make contact with the bottom of the battery cell and are held together by themagnet's attraction. The screw and magnet spin.

Like any other electric motor, the homo-polar motor is driven by the Lorentz force:as they move through an external magnetic field, the current carriers in theconductor experience a push that is perpendicular to both their velocity and to theexternal magnetic field. This force induces a torque around the axis of rotation.Because the axis of rotation is parallel to the external magnetic field, nocommutation is required for the conductor to keep turning, see Figure 127 for theprinciple.



Like most electro-mechanical machines, a homo-polar motor is reversible: if theconductor is turned mechanically, then it will operate as a homo-polar generator,producing a direct current voltage between the two terminals of the conductor. Thedirect current produced is an effect of the homo-polar nature of the design.

Figure 127. Magnetic field lines and Lorentz force on Homo-polar motor

1.7.3. ElectrostaticmotorAn electrostatic motor or capacitor motor is a type of electric motor based on theattraction and repulsion of electric charge. Usually, electrostatic motors are the dualof conventional coil-based motors. They typically require a high voltage powersupply, although very small motors employ lower voltages. Conventional electricmotors instead employ magnetic attraction and repulsion, and require high currentat low voltages. In the 1750s, the first electrostatic motors were developed byBenjamin Franklin and Andrew Gordon. Today the electrostatic motor finds frequentuse in micro-mechanical (MEMS) systems where their drive voltages are below 100volts, and where moving, charged plates are far easier to fabricate than coils andiron cores. Also, the molecular machinery which runs living cells is often based onlinear and rotary electrostatic motors.

Nanotube nanomotor. Researchers at University of California, Berkeley, recentlydeveloped rotational bearings based upon multiwall carbon nanotubes. By attachinga gold plate (with dimensions of the order of 100 nm) to the outer shell of asuspended multiwall carbon nanotube (like nested carbon cylinders), they are able



to electrostatically rotate the outer shell relative to the inner core. These bearingsare very robust; devices have been oscillated thousands of times with no indicationof wear. These nano electro-mechanical systems (NEMS) are the next step inminiaturization and may find their way into commercial applications in the future.

Electrostatic ion drive. Electric motors, in general, produce motion when poweredby electric currents. The common type of spacecraft ion drive uses electrostaticforces to accelerate ions to generate forces to create motion, and thus can beconsidered as unconventional electric motors. Gridded electrostatic ion thrusterscommonly utilize xenon gas. This gas has no charge and is ionized by bombarding itwith energetic electrons. These electrons can be provided from a hot cathodefilament and accelerated in the electrical field of the cathode fall to the anode(Kaufman type ion thruster). Alternatively, the electrons can be accelerated by theoscillating electric field induced by an alternating magnetic field of a coil, whichresults in a self-sustaining discharge and omits any cathode (radiofrequency ionthruster).

1.7.4. Non-electricnanomotorsHere, a nanomotor is a molecular device capable of converting energy intomovement. It can typically generate forces on the order of piconewtons. A proposedbranch of research is the integration of molecular motor proteins found in livingcells into molecular motors implanted in artificial devices. Such a motor proteinwould be able to move a "cargo" within that device, via protein dynamics, similarlyto how kinesin moves various molecules along tracks of microtubules inside cells.Starting and stopping the movement of such motor proteins would involve cagingthe ATP in molecular structures sensitive to UV light. Pulses of UV illuminationwould thus provide pulses of movement. DNA nanomachines, based on changesbetween two molecular conformations of DNA in response to various externaltriggers, have also been described. Chemically powered or externally poweredartificial nanomotors have also been prepared using synthetic materials andchemical methods.

Nanotube and nanowire motors. The first Nanotube nanomotor has beendeveloped in 2003 by the group of Alex Zettl at UC Berkeley Catalytic nanowiremotors and micro-tube micro-engines exhibit autonomous self-propulsion in thepresence of a hydrogen peroxide fuel. The latter offers efficient bubble-inducedpropulsion in relevant biological fluids. These catalytic motors are autonomous in



that they do not require external magnetic, electric, or optical fields as energysources. Researchers led by Joseph Wang have made a breakthrough developmentin 2008 by making a new generation of fuel-driven catalytic nanomotors that are upto 10 times more powerful than existing nanomachines. It is a major step forward toa practical energy source for powering tomorrow's nanomachines. The large force ofrecently developed nanomotors holds major promise for important cargo-towingapplications ranging from cell sorting in microchip devices to directed drug delivery.

1.7.5. MolecularmotorMolecular motors are biological molecular machines that are the essential agents ofmovement in living organisms. In general terms, a motor may be defined as a devicethat consumes energy in one form and converts it into motion or mechanical work;for example, many protein-based molecular motors harness the chemical freeenergy released by the hydrolysis of ATP in order to perform mechanical work. Interms of energetic efficiency, this type of motor can be superior to currentlyavailable man-made motors. One important difference between molecular motorsand macroscopic motors is that molecular motors operate in the thermal bath, anenvironment in which the fluctuations due to thermal noise are significant.

1.7.6. PiezomotorIn the following is presented piezo motors of Faulhaber Minimotor SA.



1.8. Modelofamotorsystem

We see in the Figure 128 a typical system, which uses an electric motor.

Figure 128.ControlMotorTransmissionLoad

The mathematical model of this system is as follows:

TTotal = J·α+B·ω+TFriction+TWork, where

· TTotal is the total motor torque

· J is the moment of inertia of the system

· α is the angular acceleration

· B is the viscous factor

· TFriction is the total torque of friction in the system

· TWork is the torque needed by the load

The moment of inertia is defined in the following equation as the sum of masspoints m multiplied by their squared distance r from the axis:

For some specific objects J is gotten as simpler equations as seen here in the Figure

129 and Figure 130 and Figure 131. The symbols used in these figures are:



· l Length of the object

· m Mass of the object

· r radius of the object

Figure 129Under the right picture, there is the textJ = m(r1

2-r22)/2 for the mass is concentrated.

Figure 130



Figure 131

We continue with the same symbol definitions: A mass m hanging from a drum ofradius r causes the torque T:

T = m·g·r, where g is the acceleration of gravitation.

The centrifugal force F is calculated as:

F = m·r·ω2, where m is the mass, where the force is generated and ω is theangular velocity.

1.8.1. Gears

When we deal with gears, we are interested in the following things:

1. Translation N

2. Mechanical strength

3. Efficiency η

4. Life expectancy

5. The moment of inertia J



The translation (N) is defined as the quotient of the rotational speeds of the input(ninput) and output axis (noutput) as follows

N = ninput/noutput.

The output torque (Toutput) of a gear is got from the input torque (Tinput) as shownhere:

Toutput = N·Tinput.

The moment of inertia of the load (JLoadOut) is seen to the motor as (JLoadIn)

JLoadIn = JLoadOut/N2.

The efficiency varies from gear to gear as listed here:

· Planetary gear: 0.6-0.9

· Cog gear 0.4-0.85

· Screw gear 0.1-0.25-0.5

· Precision gear 0.9-0.95

Force F by a belt causes a torque T in a cylinder of radius r as explained in the Figure

132.

Figure 132

T = F·r/η.

Mass m on a belt causes the moment of inertia J in a cylinder of radius r as explainedin the Figure 133.



Figure 133

In Figure 132 and Figure 133 the belt cylinder system is seen as a gear with efficiency η.

For a screw transmission we get torque T from a longitudinal force F as shown in theFigure 134.

Figure 134

The inertial moment caused by a mass m to the axis of a screw transmission can becalculated as explained in the Figure 135. Here q is the rise of the screw.

Figure 135



In the Figure 136 we see an example of dividing a motor load in small parts.

Figure 136

In the Figure 137 we see characteristics for an example motor of allowed torque as afunction of motor speed for different conditions.



Figure 137

In the following pictures, there is presented the gear series for Faulhaber mini-motors.



1.8.2. Selectingtherightgearforaservomotor

According to a video of Yaskawa, you should choose a suitable servo motor with thefollowing 7 step procedure:

1. Check what is the available voltage to your application.

2. Determine the needed motion profile:

a) Repetitive operations, plot out motor speed throughout cycle, includingacceleration/deceleration times.

b) For non-repetitive (like milling) operations calculate peak speed andacceleration

3. Define the amount of torque “muscle” needed (difficult to calculate accuratelybut most forgiving):

a) Inertia J (influences to the acceleration)

b) Friction

c) External forces and torques.

4. Calculate inertial ratio, IR, (often overlooked but important for determiningservo performance).

2

1aJ

JIRrotor

load ×= , where a is the gear reduction.

· Basic servo drives may require inertia ratios of 3:1 or smaller



· Advanced features (auto-tuning, vibration suppression, resonance filters,disturbance compensation functions) allow up to 30:1

· A ratio of 1:1 gives excellent performance (but usually oversized motor).

· Ratios less than 1:1 waste power with no performance advantage.

5. Find a motor and drive that:

· Matches supply voltage

· Has rated speed, continuous torque, peak torque, that exceed calculatedvalues and

· Satisfies inertia ratio requirement in Step 4

6. If there is a motor which is a close match you are finished. Otherwise continue tothe step 7 to add gearing.

7. Matching the motor to load with a gear. Gear matches the servo to the load byreducing speed, increasing torque and lowering inertia ratio:

· Gear box manufactures list the inertia of the gear box to be added to the load.

· Divide the motor speeds by required speed to get a starting gear ratio.

· Divide the required torque by the gear ratio to find the new required torque.

· This will narrow choices to a couple of motors. Find a motor with anacceptable inertia ratio. If two motors look equal choose the one with smallerinertia ratio.

· Repeat this step a couple of times.

The different loads are reduced to the motor using the following equations, wherethe notations are obvious. The reduction is based to the equality of powers.



The inertial quantities are reduced to the motor with the equations. Here thereduction is based to the equality of the kinetic energies.

Exercise 1. Find a servomotor of NX type of Oriental Motor with the followingcaracteristics.

1. Available power supply: 3 phase wye connected 230 V per phase.

2. Rotational speed n = 300 rpm

3. Load torque TL = 10 Nm

4. Inertial moment JL = 4000 kg(cm)2

5. Non-backlash

Solution:

· Non-backlash means we must use PJ geared type.

· Power is 2π(300/60)·10 Nm = 314 W. NX1040(640)AS/MS gives power 400W, which is enough.

· The load torque is (reduced to motor TL/a):- TL = 10 Nm > 1.27 Nm (NX640AS/MS): Not good- TL = 10 Nm > 5.08 Nm (NX1040AS/MS-J5): Not good- TL = 10 Nm < 10.2 Nm (NX1040AS/MS-J10): Fine, Also rotational speed300 rpm is just enough- TL = 10 Nm < 25.4 (NX1040AS/MS-J25): Spins too slowly.



· The inertial moment is (reduced to the motor is JL/a2):- JL = 4000 kg(cm)2 = 0.4 kgm2 > 0.2680 (NX1040AS/MS-J10): The 400 W-series is not good enough

· We must now seek a motor-gear pair, which has large enough JL capability:The 750 W series is enough:- JL = 4000 kg(cm)2 = 0.4 kgm2 < 0.4710 (NX1075AS/MS-J10):

Thus NX1075AS/MS-J10 fulfils all the requirements; and it is the only motor-gear pair in the series which does this.

Exercise 2. Find a servomotor of NX type of Oriental Motor with the followingcaracteristics.

1. Available power supply: 3 phase wye connected 230 V per phase.

2. Rotational speed: Pulling wheel has radius rL = 20 cm.

3. Load force FL = 10 N, load speed is vL = 10 m/s.

4. Inertial mass mL = 1 kg,

5. Small or no backlash.

Solution:

· Small or no backlash means standard or PS/PJ-geared solutions are allacceptable.

· Rotational speed is got from the equation vL = rL · ωm. => ωm = vL/rL =(10m/s)/(0.2m) = 50 rad/s = 478 rpm. So only the standard or gear-ratio 5is suitable.

· Power is PL = vL · FL = 10 m/s · 10 N = 100 W.

· The load torque is (reduced to motor TL/a):- TL = rL · FL = 0.2 m · 10 N = 2 Nm.- So Standard (50…400W) and NX610A/M-PS5 (TL = 1.43 NM) or



NX810A/M-J5 (TL = 1.27 Nm) are not acceptable.- But Standard with power of 750 W (2.39 Nm) or NX920A/M-PS5 (200 Wand 2.87 Nm) and NX820A/M-J5 (200W and 2.54 Nm) are good.

· The inertial moment is (reduced to the motor JL/a2):- JL = mL · rL

2 = 1 kg · (.2m)2 = 0.04 kgm2.- Standard is not accessible, only NX1040AS/MS-J5 (400W and 5.08 NMand 0.0669 kgm2) and NX1075AS/MS-J5 (750W and 9.56 Nm and 0.1180kgm2) are possible solutions.

Thus the smaller one or NX1040AS/MS-J5 is the best choice.



1.9. Conclusions

1.9.1 SynchronousMotors

The speed maximal torque characteristic in Figure 138 is very simple: the maximaltorque is independent of the speed if flux linkage Ψ = VS/(jω) is kept constant. Ascommonly known, the speed is determined by the electrical frequency of the supplyvoltage. In the sizing of the maximal torque of the motor, a safety factor from 1.5 to2 could be good. The inertial moment which can be accelerated with the motorcould be increased to be 4 to 5 times larger if flexible joints are used between themotor and the load.

Figure 138

If the synchronous motor is simply switched on in an arbitrary point of time, theinduced stator field could be in whatever position in respect of the rotor magnet.Thus the rotor starts at first rotate in wrong direction in half of the switching times.In the Figure 139 we see as curve A how a rotor starts to rotate in the optimal case andin curve B how it rotates at first 5 ms in wrong direction in the worst case.



Figure 139

In the Figure 140 we see how the maximal torque is dependent of the applied supplyvoltage. However, using higher voltage generates more heat, which might destroythe motor immediately or reduce its life time even considerably. In the planning ofthe motor system, the basis is the knowledge of the greatest allowable motortemperature.



Figure 140

1.9.2. DirectCurrentMotorsIn the Figure 141 we see the characteristics of an DC motor with nominal voltage. Theallowable working region is determined by the maximal current (and maximaltorque) and is marked with grey colour. The motor works always on the so calledwork line (line n0-TS), which tells the allowable torque-speed combinations (T;n), orpoints in the T-n plane.



Figure 141

You can’t increase the supply voltage permanently from the nominal one used in theFigure 141, but you can use any lower voltage as you wish; this makes you thepossibility to change your work line lower from the nominal one to whatever level asshown in the Figure 142. And you really need to lower you supply voltage, if you wantto operate in a point (Tyours; nyours) which is under the nominal work line, for youmust change the work line to go through your operating point to get to this point.

Figure 142

You probably can get the needed supply voltage to your work line from thecharacteristics in the Figure 141 and Figure 142, but it is also easy to calculate from theequations below. At first you solve, using the first equation, the needed current Ifrom the needed torque T. Then you get the Supply voltage V from the secondequation.



T = kM·I

U = R·I + kn·n

In reality, your current I is not constant, but could vary as presented in the Figure 143.

Figure 143

Because of this you must represent your current with the so called effective valueIRMS, which is the root of the mean square of the real current and can be calculatedas shown here:

1.9.3. StepperMotors

You start the planning of your stepper motor system by calculating the load. It isrecommendable to use a safety factor 2. At first you subtract the frictional forcesfrom your motors torque characteristic, as shown in the Figure 144. The torque leftover is usable for accelerating your motor; if this is not enough, you might increasethe supply voltage. In the Figure 145, you see how a change in the voltage influencesto the performance of the motor.



Figure 144

Figure 145

The stepper motor manufacturers do not tell how it is possible to over load yourmotor for a while. The only method is to experiment carefully. Of course the firstrequirement is that the rms-value of your current may not exceed the nominalvalue.

You should also think about the resonance regions. You should either:



1. Choose your base velocity over the first order resonances,

2. Or, you should accelerate fast through the resonance region,

3. Or, you should use micro stepping. This method also helps to avoid thehigher order resonances.

1.9.4. ServoMotorsIn servo motors, also, the rms-value of the current must always be lower than thenominal value. – If the period of the motor is long, the rms-value needs to beconsiderably lower than the nominal value. The maximal torque of the system mustnot be exceeded in any time. In the Figure 146 we see how the motor temperature isaffected by the loading situations.

Figure 146- Lowest curve correspons to 65˚ C of rise in the motor temperature- Second curve from the bottom corresponds to 100 ̊C of rise in the motor temperature- Highest curve corresponds to a use time ratio of 20 % in cycles of 5 minutes.

1.9.5. GearsIt is most important that you do not exceed the torque allowed to the gear. Youshould remember, that gear reducer increases the torque with the reduction factor.



The gear efficiency might make the gear to be self-inhibitory. The torque can growenormously if an external device stops the load very rapidly.

The forces in axial and radial direction must be in the allowable region. If theallowable speed is exceeded the life time of the gear is reduced.



AppendixA:Mostimportantequations



AppendixB:Generalrotatingfield

B.1. ThemachinegeometryThe most illustrative description of machines which have rotating magnetic field isprobably got by the presentation in the rotating field a machine which in fact stopsthe field. - A little surprisingly the DC machines also benefit from this theory. In theFigure 147, there is shown how a 3-phase motor is connected to the power supplyaccording to the IEC standard.

Figure 147. Connection of the armature voltage to a 3-phase motor

In the Figure 148, there is seen the phase winding for one zone with its magnetic axis,a = a0. In the machine in the figure, there are also two other phase windings.



Figure 148

The winding in the figure magnetizes the simple flux, Φv:

.

Here:

· a0iv is the phasor for the phase current.

· Λp is the main permeance of the machine.

· Nv and ξ are the loop number and winding factor of the phase winding.

Nv and ξ are dependent of the geometry of the machine, as:

· The equivalent air gap, δ”.

· Zone and width angles of the winding, τv and γ, respectively.

· The number of tracks q.

· μ0, D, and L are the magnetic permeability of air and diameter and length ofthe air gap.

· ξq and ξj are the Distribution or Breadth factor and Pitch factor.

These are bound together with the following equations:



, see section Winding factors later in this appendix.

.

.

The simple physical flux, Φv pierces the Ne equivalent loops of the winding,

Ne = ξvNv.

Here ξv = ξ. The flux linkage Ψv of the winding is the product

Ψv = ξvNvΦv.

The voltage needed to change the Ψ is:

Here Lp is the so called main inductance of the winding:

.

Exercise1. A cylindrical coil has the following properties: radius r = 10 cm, length isl=5 cm, relative permeability is µr = 100, and the number of turns is N=100. Calculatethe:

· Reluctance Rm

· Permeance Λ

· Inductance L.

· Reactance X, when frequency is f = 50 Hz.

· Winding flux or flux linkage Ψ, when current is 10 A.

· Flux Φ.

· Density of the magnetic flux B.



Solution.

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m ppmmmm

sm

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-

m

ppmmmm

7905.0

)1.0(/102566.1100 26200

HssNL 79.079.07910022 =W=W×=L= m

W=W××=== - 24879.05022 1 ssfLLX ppw

WbVsAAVsIL 9.79.710)/(79.0 ==×=×=Y

WbWbN 079.0100/9.7/ ==Y=F

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mVs

rAB 51.251.2

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×=

×F

=F=pp

. This is really much!

Figure 149. Definition of the angles γ, which corresponds to the distance τ, and τv = q·α. /12/q is the number of tracks,γ is the pitch angleα is the track angle: α = 60°/q for a 3-phase motor.



Figure 150. a) is a diameter winding andb) chord winding, which is got by moving the lower tracks by one to the left. /12/a is the width of the coil andτ is the pitch width.

Figure 151. Example of 3-phase cross-winding. Q=24, pN=2, q=2 = 24/(3·2·2). /12/In a cross-winding the ends of a coil go in opposite directions, which makes the coil structure strong; the length of the coils areequal.In an overlapping winding the ends of the coils go in the same direction, which makes all the loops equal.



Figure 152. Example of a three phase two-level winding. /12/Number of phases m = 3Number of tracks is Q = 36Number of pole pairs is pN = 2Number q = Q/(2pNm) = 3.a) Spreading of the winding on a plane. b) Organizing of the endings of the windings.

Figure 153. Here a 3-phase one-layer winding has an integer track count (on top): Q=18, q=3, pN=1. /12/The winding is modified to two-layer fractional track count winding (on bottom) with Q=9, q=3/2, pN=1.

In overlapping windings, it is possible to make the coils chord-like by many differentmethods, see Figure 155. The benefit of chording is the elimination or weakening ofover-tones, see Figure 154. Also the connecting wires become shorter. With doublechording two over tones can be attenuated.

Figure 154



Figure 155. Modifying an integer track count overlapping winding to a fractional oneby changing diameter coils to chord ones. /12/a) Diameter winding.b) Step shortening.c) Changing pitch width.d) Pitch mixing.e) Step shortening and pitch mixing.

Figure 156. Example of a phase-winding of a 3-phase winding where the coil sets are in series. /12/There are N coil loops, A is the cross sectional area of the wire, pN = 2, q=2.



Figure 157. Example of a phase-winding of a 3-phase winding where the coil sets are in parallel. /12/There are 2N coil loops, A is the cross sectional area of the wires together, pN = 2, q=2. Let’s suppose that the voltage is now halfof the serial case => current per path is half with same number of turns per path => only half of the cross-sectional wire area, soA/2, is needed. Still field produced per pole is the same for the number of turns is doubled; and the voltage per pole is the sameas in the serial case. => However, to keep the back emf per pole in the old value, the magnetizing field needs to be halved for theturn count per pole is doubled.

B.2. TransformerModelThe equivalent circuit of on electric motor is based on the model of the transformer/9/.

Figure 158.

The symbols used in the Figure 158, Figure 159, and Figure 160 are listed here below.



The parameters are transferred to one side of the transformer using the usualtransformations, see Figure 159:

2

2

1

2

1

2

1

2

1

2

1

1

aZZ

RR

aII

aVV

NN

a

==

=

=

=

Figure 159. Equivalent circuits of a non-ideal transformer.



Figure 160. A phasor diagram for the circuit a in the Figure 159.

To calculate torque of magnetic field we need Lorentz force, see Figure 161: F=qv x B =li x B.

Figure 161. F = li x B.

See Figure 162 for definition of directions in torque calculation T= r x F:

Figure 162. τ=r x F.

Torque for one conductor loop in rotor is calculated in the Figure 163.



F´=´=´=Þ

´´=´=

iBiABirlTonconstructi

BilrFrT

2

)(22

Figure 163. Torque for one current loop of the rotor.

Note that angle(i -> Φ) = angle(r -> F). Thus we get for the torque, when we havecylindrical rotor as shown in the Figure 164:

T = NRiR x ΦS = iR x ΨSR, for

NRiR x ΦS = NRiR x NSΛiS = iR x NRNSΛiS = iR x MRSiS

= iR x (NR/NS)2NS2Λ(NS/NR)iS = iR x (NR/NS)2LS(NS/NR)iS = iR x LS

RiSR = iR x ΨSR

Figure 164.

You get the same torque, even if you have moved all the components and signals tothe same side of the motor, but you need to use the expression like iR x ΨS

R. Notethat the torque is got independently of how you transfer components and signalsfrom one side to the other. You get the torque with Lorentz force without transferfrom side to side with the expression like NRiR x ΦS as:

SRRSSSRRRRRRRR iilANNiN

lAiNiN

AiNABiNAT ´=´×=F´×=

F´××=´××=

mm



When everything is on the same side, you still get the same result independently ofwhat you have transferred.

S

RR

R

SmRS

SR

SmRS

RmRR

SSRmR

RS

SRRSSRR

S

S

RS

R

SS

S

RmSR

RS

RmSR

SRRSSRSS

RSmS

S

RRSmS

SR

NNi

NNLiiaLiT

iLNNiiLiT

iilANNii

NN

NN

lAN

NNi

NNLiiLiT

iilANNii

lAN

NNiL

NNiiLiT

2

22

2

22

2

2

2

´-=´-=

´-=´-=

´=´=´=´=

´=´=´=´=

mm

mm

If you have a stator field of m phases, the total flux, Φ, is (m/2) times the phase flux,Φv. E.g. if you have 3-phase stator, the total flux, Φ, is

Φ = (3/2)Φv.

B.3. PhasorModelIn a symmetrical 3-phase winding the magnetic axes of the phase-windings (1,2,3)are in the directions a0, a1, a2, where a = exp(j2π/3). The winding creates the mainflux linkage (or sum flux linkage) Ψp:

Here i is the geometric sum of the currents in the phase windings:

Here ωi is the angular velocity and φi the phase angle at the zero time, of the currenti. We denote the axes of a three-phase winding with the phasors a0 = ej0˚, a = ej120˚

and a2 = ej240˚ = e-j120˚, where the angles are measured from the magnetic axis of A-phase.

Exercise2. Show for a three phase system that the expression for the phasor current iis:

iS = iA + iB + iC = (3/2) ejωt·î.



Solution. We assume that

From this it follows:

ie

ijjeie

ijjeie

ieeeieiii

tj

tjtj

tjtj

tjtjCBA

ˆ23

ˆ)23

21

23

211(ˆ

23

ˆ))60sin()60cos()60sin()60cos(1(ˆ23

ˆ)1(ˆ23 120120

w

ww

ww

ww

=

+---+=

+--+-+=

+++=++

-

-

--

oooo

oo

□

Exercise3. Deduce expression for phase-to-phase voltages in a wye connected threephase system.

Solution. V10 = V - Vexp(-j120°) = V[1 - cos(-120°) - jsin(-120°)]

= V[1-(-cos(60°))-(-jsin(60°))]

= V[1-(-½)-j·(-sqrt(3)/2)] = V[+3/2+j·sqrt(3)/2] = sqrt(3)V·exp(j30°). □

Exercise4. Deduce expression for connecting currents in a delta (triangle) connectedthree phase system.

Solution. V10 == V·exp(30°).

V20 = V·exp(30°-60°) = V·exp(-30°)

I0 = [V·exp(30°) + V·exp(-30°)]/Z

= [cos(30°)+jsin(30°)+cos(-30°)+jsin(-30°)]·V/Z

= [cos(30°)+jsin(30°)+cos(30°)+(-jsin(30°))]·V/Z

= [cos(30°)+jsin(30°)+cos(30°)-jsin(30°)]·V/Z

= [2(sqrt(3)/2)]·V/Z



= sqrt(3)·V/Z

= sqrt(3)·I,

where I is the current in one coil. □

Figure 165. Magnetic induction in an electric machine.

Exercise5. a) In a synchronous motor of one pole pair, calculate torque, and powerwhen the total flux (from all stator phases together) is ΦS = 1.0 T·(.01 m2) and therotor coil has turns NR = 100 and current IR = 100 ADC; ΦS and IR are orthogonal (sothe maximum torque is calculated). The rotational speed of the motor is n = 3600rpm. b) Calculate back EMF for a phase, when maximum ΦR = 0.8 T·(.01 m2) and NS =100; speed is as in the a-case.

Solution.

The torque in the rotor is:

a) T = NRiR x ΦS = - NSiS x ΦR.

T = sin(90◦) · IR · NR · ΦS = 1 · 100A · 100 · 1 Vs/m2 · .01m2= 100 A·Vs = 100 Nm.

P = ω · T = n (2π min/60 s) · T = 2π60/s · 100 Nm = 37.7 kW.

b) ÊBack = jωmNSΦR (ÊBack = NSdΦR(t)/dt where ΦR(t) = φRexp(jωmt) , and φR is considered to be aconstant).

|ÊBack| = n (2π min/60 s) NS · ΦR = 3600 (2π/60 s) 100 ·0.8 Vs/m2 · (.01 m2)



= 2π60/s · 0.8 Vs = 2π60 · 0.8 V = 302 V. □

Exercise6. In a synchronous motor of one pole pair, calculate torque, back EMF, andpower when (in amplitude value) the phase current in stator is iS = 20 A and iRS = 20A, and for one phase LmS = LmR

S = 0.1 H. The power angle is δ = -30◦ and power factor

angle is φ =-60◦, see Figure 166. Note that mSSRR

SSmR LNN

NNL =L=L= 22

2

2

. The rotational

speed of the motor is n = 3600 rpm = 60/s.

Figure 166.

Solution.

Angle(ΨR) = angle(IR) = δ - 90◦, for current lags voltage in an inductance. Angle(ΨS) = angle(IS) = φ, from thedefinition of φ. Thus γ = angle(IR,IS) = φ - (δ - 90◦) = -60◦ - (-30◦) + 90◦ = 60◦.

The torque in the rotor is:

T = iR x ΨSR = - iS x ΨR

S. (iRS is DC-current.)

T = iRS x LmS · iS = sin(60◦) · 20 A · (3/2) 0.1 Ωs · 20 A = sqrt(3)/2 · 60 VAs = 51.96 Nm.

ÊBack = jωmΨRS (EBack = dΨR

S(t)/dt where ΨR(t) = ψRSexp(jωmt+δ-90◦) , and ψR

S is considered to be aconstant).

|ÊBack| = n (2π min/60 s) LmS · iRS

= 2π60/s · 0.1 Ωs · 20A = 754 V, EBack = 754V·exp(jωmt-30◦).

P = ω · T = n (2π min/60 s) · T = 2π60/s · 51.96 Nm = 19.59 kW. □

Check by simulation: P = (3/2)Ia·Vt cosφ = (3/2) 20A · 1306V · 0.5 = 19.59 kW. □

Exercise7. The structure of a synchronous motor is shown in the Figure 167. The linearsurface current densities in the stator and rotor, Kz

S and KzR respectively, are

.



Show that the radial magnetic flux, Br, is:

.

Figure 167

Solution. We assume that the azimuthal (or tangential) magnetic flux, Bθ, isconsidered separately and is therefore here thought to be zero.

The Ampère-Maxwell law in integral form is:

ò ò ×´+=´C S

encfree danDdtdIdlH ,

The rotation direction is in the line integral according to the right handed role, asshown in the Figure 168.

Figure 168

In our case, the A-M equation becomes:



RKKgBgBB Rz

Szrrr ×¶×+=×+×úû

ùêëé ¶×

¶¶

+- qmqqqq

q )()()()( 0

.)()(

)()(

0

0

Rg

KKB

RKKgB

Rz

Sz

r

Rz

Szr

×+

-=¶¶

Þ

×+=×¶¶

-Þ

mqq

mqq

The solution for this is, as is easy to check by derivation:

□

Exercise8. Show that in the motor of Figure 167 the azimuthal magnetic field has on thesurface of the stator the value:

SzKB 0)( mqq =

On the rotor surface the value is:

RzKB 0)( mqq -= .

Solution. We assume that the radial magnetic flux, Br, is considered separately andis therefore here thought to be zero.

The Ampère-Maxwell law in integral form is:

ò ò ×´+=´C S

encfree danDdtdIdlH ,

Because μr = ∞, there is no azimuthal magnetic field in the stator and A-M lawbecomes around a slice of the stator surface:

RKRB Sz ×¶×=×¶× qmqqq 0)(

SzKB 0)( mqq =Þ . □

Similarly for the rotor (azimuthal magnetic field in the rotor is zero):

- RKRB Rz ×¶×=×¶× qmqqq 0)(

RzKB 0)( mqq -=Þ . □

Exercise9. Show that the torque in the rotor is:



Solution. The areal shear force density, τθ, is:

τθ = Trθ = Br x KzR = -Br x Hθ

R.

=

The average of this, <τθ>, is simply:

,

for <sinα·cosα> = <½sin(2α)+ ½sin0> = 0 over all α; and <sin(pθ)·cos[p(θ-φ)]> =<½sin(2pθ+φ) +½cos(pφ)>=½cospφ.

Now we get for the total torque T:

□

Exercise10. Show that the turns distributions below for the stator and rotor winding,nS and nR indeed fulfil the condition that the total turn numbers are NS and NR,respectively :

Solution. By integrating for a half circle e.g. around the field axis (zero angle) andcounting the number of pole pairs to be p, we get immediately the result for thestator; the rotor is calculated similarly:

. □

Exercise11. Show that the flux Φi(θ) through an individual round of the stator coil is:



.

Here, Is is the current in the stator coil.

Solution. At first we notice that the stator current density can be presented with theturn distribution, nS, by supposing that KS = NSIS/2R:

KzS = KS·cos pθ = (NSIS/2R) cos pθ = nS·IS.

Now we start with integrating the radial magnetic field over half circle for one polepair to get the corresponding flux:

, here -π/p <= θ <= π/p and

.

When θ = 0, the sign of sin pθ is minus for the whole region of integration -π/p < θ’<=0; when θ = π/p, the sign is plus for whole region. The result of integration is:

,

For cos(θ)-cos(θ-π) = cos(θ)+cos(θ) = 2cos(θ).

Exercise12. Show that the self-inductance of the stator and rotor windings, LS and LR

respectively, are:

,

.

Solution. The total flux linkage of the stator, λS, is the whole stator flux integratedfor a symmetric half circle, -π/2…π/2 if p=1. If the number of pole pairs, p, is higherthan 1, we must multiply the integral of one pair with the number of pairs.



or

ò-

××=p

p

SSSS Rdp

RNp

gplRINp

2/

2/2

0 cos2

cosp

p

qqqm

l .

Because aaaa 2sin41

21cos2 +=ò d , we get:

=> ,

=> .

Similar calculation gives the result for LR. □

Exercise13. Show that the mutual inductance between stator and rotor windings,M(θ) has the expression:

.

Solution. To get the mutual inductance we e.g. excite the stator winding andcalculate the flux linkage through the rotor winding.

ò-

××F=p

pRiRS Rdnp

2/

2/

)()(p

p

qqql

[ ]ò-

×-×=p

p

RSSRS Rdp

RNp

gplRINp

2/

2/2

0 )(cos2

cosp

p

qfqqm

l .

Now we can continue with the fact:

)cos(21)cos(

21coscos bababa ++-=× .

We continue with the integral of cos pθ·cos[p(θ-φ)]:



[ ]

[ ]

)cos(4

)sin()sin(221)2/2/)(cos(

21

)2(cos21)cos(

21 2/

2/

2/

2/

fp

fpfpppf

fqqfp

p

p

p

p

ppp

pp

pdp

p

p

p

=

----×

++=

-+ òò--

This substituted in λRS gives the result.

B.2. WindingfactorsThe simple definition of winding factor, kw, is the ratio of flux linked by an actualwinding to flux that would have been linked by a full-pitch, concentrated windingwith the same number of turns. That is

Kw = λactual/λfull-pitch.

What is interesting to note, although we do not prove it here, is that the windingfactor of any given winding is the product of the pitch and breadth factors, kp and kb

respectively:

.

Pitch factor is found by considering the flux linked by a less-than-full pitchedwinding. Consider the situation in which radial magnetic flux density is

.

A winding with pitch α will link flux

Pitch α refers to the angular displacement between sides of the coil, expressed inelectrical radians; the coil doesn’t see the angle (π-α)/2 on both sides of themagnetic flux, if we have only one pole pair (p=1). The integral sums the sinesymmetrically around the maximum point at π/2 (p=1). For a full-pitch coil α = π. Inthe Figure 148, the pitch angle α is denoted with γ, so α = γ. We develop as follows:



[ ]

)2/sin()2/sin(2

)(21sin)2(

21sin2

)2/2/cos()2/2/cos(

)cos()cos( 2/2/2/2/

awp

awp

wapwap

wfwfl apap

ntnnp

RNlB

ntnnp

RNlB

tnntnnnp

RNlB

tnptnpnp

RNlB

n

n

n

ppppn

--==

--=

----+-=

----= -+

So,

2sin

2sin ga nnk pn == .

Now for breadth factor. This describes the fact that a winding may consist of anumber of coils, each linking flux slightly out of phase with the others. A regularwinding will have a number (say m) of coil elements, separated by electrical angle γ(not the same as in the Figure 148). A full pitch coil with one side at angle ξ will, in thepresence of sinusoidal magnetic flux density, link flux

fwflxp

x

RdtnpBNlpp

pn ×-= ò

+ //

/

)sin(

If we have one pole pair (p=1), ξ here counts how much the coil is off from theoptimal angle to the flux. If ζ = 0, the integral sums the whole positive region of thesine. We evaluate this as follows:

[ ]

[ ]

[ ]

[ ])(

)(

///

Re2

Re2

)cos(2

)cos()cos(

)cos()cos(

xw

wx

xxp

wxl

wxwxp

wfwfl

ntjn

tnjn

n

n

pppn

enp

RNlB

enp

RNlB

tnnp

RNlBoddn

tntnnnp

RNlB

tnptnpnp

RNlB

-

-

+

=

=

-=Þ=

---+-=

----=

□

Now if the winding is distributed into m sets of slots and the slots are evenly spaced,the angular position of each slot will be:



, i = 0…m-1.

If the breadth is τv, and the number of tracks is q (see Figure 148) then we use in theabove equation the values: m=q and γ = τv/(m-1)

The number of turns in each slot will be N/mp, so that actual flux linked will be

You do not need to divide with p, for there is no summation over p-slots here. Nowwe see that the breadth factor is:

)2/sin()2/sin(

)2/sin()2/sin(

111

11

)1(

1

2/2/

2/2/

2/)1(

0

1

0

2/)1(

1

0

2/)1(

1

0

)2/)1((

qnqn

nmnm

eeee

m

ee

mek

xx

eem

e

em

e

em

k

v

v

jnjnjn

jnmjnmjnm

jn

jnmmjn

bi

i

i

jnijnmmjn

m

i

jnimjn

m

i

mijnb

tt

gg

ggg

ggg

g

gg

ggg

gg

gg

==

--

=

--

=Þ-

=

-=

=

=

+-

+-

-

--¥

=

¥

=

---

-

=

--

-

=

---

å

å

å

å

.

B.3. WindinginductancecalculationNow we want to get expressions which give the inductances from the geometricproperties of the coils. We suppose here that the distribution of the acting magnetomotive force (MMF), F, in the rounds of the coils is rectangular as a function of angleφ. We use here the definition for rectangular MMF as shown in the Figure 169.

h

-hπ 2π-π 0

F(φ)

Angle φ



This rectangular MMF may be represented as a Fourier series as follows:

å¥

=

=

oddnn n

nhF:

1

sin4)( fp

f

If the current in the wire of the winding is I (which may be time-dependent) andthere are p pole pairs (and 2p poles) and the total count of turns for all the polestogether is N, we get the magnitude of the MMF, h, for each pole to be h = NI/2p.Because every pole pair takes the angle of 2π of current in the angle 2π/p on themotor circumference, we get for the MMF around the circumference theexpression:

Because we assume the permeability of the magnetic material in stator and rotor tobe infinite, we get for the reluctance, Rm, in the magnetic circuit of one pole thevalue Rm = g/(μ0·A), where A is the cross-sectional area of the magnetic circuit and gis the width of the air gap. The magnetic flux density is Br = F/(A·Rm), which gives:

We have two basic situations for the phase currents. For the so called balancedcurrent we have:

In this case the sum of the magnetic fluxes is:

Figure 169



,

where

and the sign rule is:

.

The other basic situation is the zero-sequence currents:

.

The sum of the phase fluxes is non-zero only for the triplen harmonics:

The third different case is the rotor winding. It will have the same flux form as asingle armature winding, but its direction will depend on rotor position:

.

Now we substitute for the rotor angle φ’ = φ - ωt/p:

.

B.3.1. InductanceexpressionsWe now get the winding inductances when we combine the winding factors to theflux expressions. For balanced currents, with Ia = cos(ωt) we get:



)cos(

)cos(423

)cos(2

4232

)cos(2

...7,5,122

220

...7,5,1

20

1

tIL

tIpn

RlkNg

knp

tp

NIng

RlN

tnpBRlNN

d

n

wn

nwn

n

rn

w

wm

p

wp

m

wfl

×=

=

=

==

å

å

å

¥

=

¥

=

¥

=

So we have for the inductance of one phase when the poly-phase winding is drivenby a balanced set of currents:

å¥

=

=...7,5,1

22

2204

23

n

wnd pn

RlkNg

L mp

For zero-sequence currents we get the inductance L0 to be:

.

For the mutual inductance between a field winding (f) and an armature winding (a)we get M(θ), which depends on the mutual angle:

Problem1. Show that the field φ used in the derivation of the expression of Ld has theform:

Problem2. Derive the expression for L0 above.

Problem3. Derive the expression for M(θ) above.

B.4. RotatingModelofaGeneralMotor



B.4.1. GeneralNon-rotatingModelThe conventional transformer based per-phase equivalent circuit of general electricmotor model is shown in the Figure 170.

Figure 170

The Rs and Rr are the stator and rotor resistance and Lm is called the magnetisinginductance of the motor. The stator and rotor inductances, Ls and Lr, are defined by:

where Lls and Llr are the stator and rotor leakage inductances. All the rotorparameters are referred to the stator here. We define the slip, s, needed withinduction motors as:

s = (ωe - ω0)/ωe = (ωe - pNωm)/ωe = pN(ωS - ωm)/(pNωS) = ωr/ωe = ωR/ωS.

Here ωe is the angular frequency of the supply voltage; ω0 = pNωm is the electricalangular frequency of the rotor circuit; pN the number of the pole pairs; ωm themechanical angular frequency of the rotor; ωS the synchronous angular frequency ofthe rotor; ωr = pNωR is the electrical angular rotor slip frequency; ωR the mechanicalangular rotor slip frequency.

This time we define in stator all the space vectors or phasors, Yss, similarly

independently of the meaning of the quantity as:

.

Here α = exp(j2π/3). The above transformation is reversible and each phase quantitycan be calculated from space vector by,

.



The voltage equations for a general machine are:

.

In these equations we have used the actual rotor quantities Vr’, Ir’, λr’. The stationaryvalues are: Vs

s = eφvejωe·tVss, Is

s = eφIsejωe·tIss, Ir’ = eφIrej(ωe-ωo)tIr’. Now we translate them

to stator frame with definitions of new quantities:

.

Here we have rotated the phasors with angle θ = pNωm·t + C, where C is a constant.Also, we define referred rotor impedances as:

Zrs = n2·Zr’,

and we get:

. (base equations)

This is verified by inserting the expressions for the new phasors and simplifying as:

''''

''1''

'')('1''

')('1''

)(

2

2

2

rrrr

rj

rj

rrj

rj

rj

mNmNrj

rrj

rj

mNrj

rrj

srmN

srr

sr

pIRV

pneIen

RnVne

pnenejpjpIen

RnVne

nejppIen

RnVne

jppIRV

l

l

llww

lw

lw

qqq

qqqq

qqq

+=Þ

+=Þ

+-+=Þ

-+=Þ

-+=

We will also use for the flux linkages the notations:

The stationary values are: Vss = eφvejωe·tVs

s, Iss = eφIsejωe·tIs

s, Irs = eφIrejωe·tIr

s. Ifexpressions of the flux linkages are inserted in the equations of the motor, we get:



.

This pair of equations can be presented with the equivalent circuit in the Figure 171.

Figure 171

If we have steady state operation with excitation (i.e. supply) frequency ωe, operatorp corresponds to jωe Thus we get:

.

This can be represented with Figure 170.

B.4.2. GeneralRotatingModelNow we put our coordinate system to rotate with speed ωa. Therefore we definenew space vectors as:

.

Our base equations now become:

,

where the new flux linkages are linked to the new currents as:

.

The stationary values are: Vsa = eφvej(ωe-ωa)tVs

s, Iss = eφIsej(ωe-ωa)tIs

s, Irs = eφIrej(ωe-ωa)tIr

s. Wesubstitute the flux linkages into the pair of equations to get:



This pair of equations may be presented with the equivalent circuit in the Figure 172.

Figure 172

If ωa = ωe, we get the stationary values: Vss = ej0Vs

s, Iss = ej0Is

s, Irs = ej0Ir

s, so that all thederivatives disappear; and this is equivalent to inductances L ls, Llr and Lm todisappear everywhere except in the expressions of the flux linkages.

B.4.3. Thephasormodelofamulti-phasemotorThis model represents the multi-phase motor as a phasor machine; so you do notneed to solve your one-phase model and then separately combine all the phases tothe whole machine, but your solution is right away for the whole machine. I reallydo not know if this model is advantageous and we will use in this book mainly thereal phase-based motor models.

Now, we forget, for simplicity, the rotation of the rotor and coordinate system. Wethink also that there is only one winding of one pole pare in the stator and in therotor. The corresponding current vectors for stator and rotor are denoted with iS

and iR. Thus the main flux linkage (for one pole pair) is:

The current sum, iS + iR, is the magnetising current, im, of the whole machine:

.

We use the following definitions for some flux linkages:

Ψ, Lpim = Ψp, Ψδ.

Here Ψ is simply “a rotating flux linkage”, Ψp the main flux linkage, and Ψδ is the airgap flux linkage. This situation is presented in the Figure 173.



Figure 173. Flux model without magnetic dispersion.

B.4.4. SimpledispersivemodelofageneralmachineFor simplicity, the rotation of the rotor and coordinate system is still considered tobe negligible. If we want a more precise magnetic model of the machine we need totake in to account the magnetic dispersion. This is done in the Figure 174.

Figure 174. Flux model with magnetic dispersion.

Now we get for the flux linkages of the stator and rotor:

The real flux linkages of the phase coils are the projections of the windings. Thesmall subscripts point to rotor winding:

In symmetrical situations, we can sum the projections back to the rotating vectors;the factor (2/3) is needed to get the right result, for the directions a0, a1, and a2 arenot orthogonal:



The vectors of the dispersed flux linkages are:

These may also be composed from the components, e.g.

So, the dispersed inductance of the rotating vector model and the dispersedinductances of the separate phases have the connection:

For the magnetizing inductance we have:

In real machine, the magnetising inductance Lm is got from the number of pole pairsand the main inductance Lp corresponding a pole pair, using a method which takesinto account the way the pole pairs are connected together: typically in series or inparallel.

It should be firmly emphasized here that for the purpose of calculating power andtorque, the amplitudes of the rotating model for current, voltage and flux linkage, I,V, and Ψ correspondingly in a three phase system, are related to the amplitudes(peak values) of the one phase quantities, Iv, Vv, Ψv, as (for magnitudes):

I = (3/2)Iv

V = Vv andΨ = Ψv.



This guarantees, that the apparent power |S| = IV = (3/2)IvVv = (3/2)IvωΨv = ωT andV = VV = dΨv/dt = ωΨv and T = I x Ψ = (3/2)Iv x Ψv. This way for the rotating model,the electric power could be calculated from the peak-valued phasors directly.However, it is probably more common to use the equivalent circuit of one phase assuch and change the space vector magnitudes to correspond directly the rms-valuedphase quantities: (for magnitudes) I = Iv, V = Vv, Ψ = Ψv so that |S| = 3IV, V = VV =dΨ/dt = ωΨ and T = 3I x Ψ.

B.4.5. SimplerotatingvoltagemodelNow finally, the coordinate system is considered to rotate with the angular velocityof the supply voltage, or ωa = ωe = ω, but resistances and the dispersed inductancesare neglected. The voltage model of the simple rotating machine is seen in the Figure

175.

Figure 175. Simple voltage model of rotating general machine

The angular velocities of the flux linkage, ωΨ, and the rotor ωm, and the slip velocityωR are defined in the Figure 176.

Figure 176- Stator axis- Rotor axis- Flux axis.



The stator voltage, Vδ, is:

The rotor voltage is similarly:

B.4.6. RealisticrotatingvoltagemodelIn the real machine the voltage loss in resistances need to be taken into account, asin the Figure 177. The coordinate system is considered to rotate with the angularvelocity of the supply voltage, or ωa = ωe = ω, where ωa = angular velocity of the co-ordinate system, ωe = ω1 = ω is the angular velocity of the electrical signal

Figure 177. Realistic rotating voltage model for a general machine.

ωm is the angular velocity of the rotor. The air gap voltages for stator and rotor,correspondingly Vδ and VδR, are:

So, the natural rotating vector for a voltage is composed from the phase voltages as:

We see that the multiplier for voltage is kv = (2/3). For the multiplier for current is ki

= 1 and for voltage kv = (2/3), we get the resistances for the model as:



For a m-phase machine the multipliers are:

ki = 1 and kv = (2/m).

For example the stator parameters are calculated as follows:

The magnetising inductance is for in series connected pole pairs:

Here pN is the number of pole pairs. And for parallel connected pole pairs:

Lm = Lp/pN.

The one axis model of a general machine is presented in the Figure 178.



Figure 178. One axis model of general machine.

The equations for the magnetising current, flux linkages, voltages and torque of themodel are as here:

For a salient pole rotor:

,

And for round rotor, also:

Exercise14. Proof the equation

.

Solution. Let’s denote the rotating current amplitude with I and the phase current amplitude with Iv.Thus:



RSI2 = RS[(3/2)iv]2 = RS(9/4)iv2 = (3/2)RSviv2.

The claim follows from the last equality. □

B.5. ConvertingthegeneralmachinetothemechanicalworldIf the electrical calculations are made with the general machine, the quantities inthe axis do not correspond to the mechanical reality. The speed of the generalmachine is the angular frequency of the electrical supply, ω1, and the speed of themechanical machine is the synchronous angular frequency of the mechanical motor,ωmS. The quantities must be converted with the following equations:

Here Te, Je are the electrical torque and moment of inertia; T and J are thecorresponding mechanical quantities. pN is the number of the pole pairs in themotor.



AppendixC:MachineSizing

Figure 179. A single-pole section of a PM synchronous motor.

Here we have assumed that the windings of all poles are in series; thus the totalnumber of coil turns for the phase is P·(Nc/P) = Nc; in fact we could even better callthe expression of λ as phase linkage (coil flux) and denote it by λphase; this is alsoessential to understand when we next calculate the induced voltage for the wholephase winding.



Figure 180. Magnetic and electric loadings.

Figure 181. Torque is proportional to the rotor volume.



Figure 182. Size comparision of high and low-speed motors.



AppendixD:EVDesignExample

See Figure 183.

Figure 183. A voltage source inverter with six switching devices.



966.0

2*260sin2

)2/60sin()2/sin(

)2/sin(=

×==

===

o

o

qnqn

kk

v

v

qdb

tt

x

.

Note that τv=m*q*360°/(P*m*q)=3*2*360°/(6*3*2)=3*2*10°=60°, and for baseharmonics n=1.

Figure 184. Table of specifications for a prototype EV motor.

This can be seen as follows:

strmmwSgap

str

mmwS

gap

r

gaprotormmwe

lDABknP

lDABkn

PPVolABkT

22

2

)(

2

42

p

ppw

=Þ

=×

===



The equation to the volume Dr2lst is derived as here:

mmwS

shaftstr

strmmwSgapshaft

ABkn

PlD

lDABknPP

2cos

2cos1cos1

22

22

pfh

e

pfhe

fhe

=Þ

==

It should be a multiple of 6; therefore Nc = 24 was chosen. Note that the number ofpoles for every phase is P = 2·pN = 2·3 = 6 and every pole must have the samenumber of turns. The full pitch winding is selected for convenience of coil windingand insertion. The motor specifications are summarized in the Table of Figure 184.



Figure 185. Experimental back EMF waveform at 350 rpm.

Figure 185 shows an experimental waveform of the back EMF at 350 rpm. It is a line-to-line voltage. From the plot, the back EMF constant is read ψm = 0.123 V/rad(peak). Figure 186 shows flux lines and the air gap field without stator current. Note



Figure 186. (a) Flux lines and (b) air gap field density without stator current.



Figure 187. Torque, power, and current versus speed.

Figure 188 shows voltage angle, current angles, and the power factor while the motorspeeds up with the maximum power. Note that the current and power angles



Figure 188. Current and voltage angles and power factor along the maximum power operation.



Figure 189. Torque components: magnetic torque plus reluctance torque.



Figure 190. Photos of a sample motor: (a) rotor after PM insertion, (b) rotor with shaft, (c) stator core, and (d) theirassemply.

Exercise1. Design a PMSM with the following properties: base speed 15 000 rpm,power 60 kW, battery voltage 900 V and no voltage booster.

Solution.

The maximum phase voltage (rms) is

Va = 900V/sqrt(6) = 367V.

We assume that cosφ = 0.87 and efficiency is η = 0.92. The rated current is:

Ia = Pe/(3Vaηcosφ) = 60000/(3·367·0.92·0.87) = 68.1 A.



For the great speed we choose pN = 1 or P = 2. Also we choose a 24-slot stator and full pitch winding, whichmakes kp = 1. We get for the number of slots per pole per phase, q = 24/(3·2) = 4. The distribution anglecould be:

τv = m·q·360°/(P·m·q) = 360°/2 = 180° and the distribution factor is

Kd = sin(τv/2)/[qsin(τv/2q)] = sin90°/[4·sin(180°/2·4)] = 1/[4·sin(22.5°)] = 0.653.

We get kw = kd · kp = 0.653.

We choose ε = EBack/Va = 1 and Am = 60000A/m and Bm = 0.8T. So we get:

32

22

1094.187.092.0600008.0

6015000653.0

2

60000

cos2

-×=××××××

=

=

p

fhpe

mmsw

shaftstr

ABnk

PlD

We choose Dr = 30 cm which gives lst = 1.94·10-3/(0.32) = 0.0215m = 21.5 mm.

τp = πDr/P = π0.3m/2 = 47.1 cm.

mWb

lB stpmf

16.50215.0471.08.02

2

=×××=

=F

p

tp

Now we get the number of turns Nc:

06.981016.5653.01

60150002

3672 3

=××××××

=F×

=-pp fwe

Backc kf

EN

96 was chosen, for it is multiple of P = 2, but also of 3 and many other integer number.



AppendixE:ServomotorSpecifications,OrientalMotorCatalogue2012-2013.



Literature

/1/ Tapio Verho: Pienmoottoreiden valintaopas (Guide to choose a small motor),Optimaatiosanomat, Stig Wahlström Oy, www.swoy.fi, ESPOO.

/2/ Matti Mård: Sähkökäyttö ja tehoelektroniikka, 354 sivua, Otatieto, teosnumero 889. Jyväskylä1993. English translation: Matti Mård: Electric Drives and Power Electronics, 354 pages in Finnishedition, Otatieto, opus number 889. Jyväskylä 1993.

/3/ H. Wayne Beaty and James L. Kirtley, Jr.: Electric Motor Handbook. McGraw-Hill, New York1998.

/4/ Many articles in Wikipedia.

/5/ Lauri Hietalahti: Muuntajat ja sähkökoneet, 1. painos. Direct, Vantaa 2011.

/6/ K. Järvinen and S. Varis: Tasavirta- ja vaihtovirtakoneet. Tekniikan käsikirja Vol. 3, pages 573-615. Gummerus, Jyväskylä (Finland) 1975.

/7/ K. Järvinen: Sähkökoneet. Sähkötekniikan käsikirja Vol. 1, pages 309-407. Tammi, Helsinki1978.

/8/ Dal Y. Ohm: Dynamic Model for Induction Motors for Vector Control. Drivetech, Inc.,Blacksburg, Virginia. In Internet on 11th of Jan, 2013.

/9/ Syed Nasar: Electric Machines and Electromechanics. Schaum’s Outline Series, McGraw-Hill,New York 1998.

/10/ Oriental Motor: General Catalogue 2012-2013. Oriental Motor (Europa), Düsseldorf 2012.

/11/ Kwang Hee Nam: AC Motor Control and Electric Vehicle Applications. CRC Press, Boca Raton(FL) 2010.

/12/ Lauri Aura and Antti J. Tonteri: Sähkökoneet ja tehoelektroniikan perusteet. WSOY, Porvoo,1996.

/13/ Juha Pyrhönen: Electrical Drives 2010-2011, Lecture Notes, L983: Lappeenranta University ofTechnology, Lappeenranta.

/14/ Juha Pyrhönen, Tapani Jokinen, Valéria Hrabovcová: Design of Rotating Electrical Machines,Wiley, Chichester UK 2008.

/15/ A. E. Fitzgerald, Charles Kingsley Jr., Stephen D. Umans: Electric Machinery, sixth edition.McGraw-Hill, New York, 2003.

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