2024-ec

ENGNG 2024 Electrical Engineering

E Levi, 2002 1

FUNDAMENTALS OF ELECTROMECHANICAL ENERGYCONVERSION

1. PRELIMINARY CONSIDERATIONS

Electromechanical energy conversion is achievable in a number of ways. Thesepossibilities rely on different fundamental laws of electrical engineering. As the only methodthat has importance on the large scale is electromechanical conversion achieved by means ofelectromagnetic converters, this section is fully devoted to the analysis of basic principlesinvolved in electromagnetic electromechanical conversion.

Electromechanical energy conversion is achieved by devices that are usually calledelectric machines. In principle, laws of electromagnetics can be used to design converters withthe linear and with the rotary motion. Converters with linear motion are called linear electricmachines, while those that rely on rotating motion are called rotating electric machines. Vastmajority of existing machinery belong to the category of rotating electric machines. Theseinclude all the machines used to generate the electricity, as well as the most of the machinesused in industry to perform some useful work while converting electric into mechanicalenergy. Linear machines are used relatively rare for somewhat specialised applications. It isfor this reason that only rotating electric machines will be dealt with here. Prefix ‘rotating’will be omitted and the converters will be called simply electric machines, implying thatdevices under consideration are characterised with rotational movement.

Operating principles of electric machines involve two basic laws of electromagnetism,namely the law of the electromagnetic induction (Faraday’s law) and the law of force creationin an electromagnetic field (Bio-Savart’s law).

Consider the situation shown in Fig. 1. A conductor is connected to an electric sourceand it carries current I. It is placed in the magnetic field of certain flux density B (which is ofcourse a vector; hence the arrow above the symbol in Fig. 1). Interaction of the flux densityand the conductor current leads to the creation of an electromagnetic force

BlIF e ×= (1)where l is the conductor length. This electromagnetic force will cause the movement of theconductor, which will start travelling at certain linear speed v to the left. The electromagneticforce will be balanced by another, mechanical force that acts in the opposite direction (to theright). The equilibrium will be established when the two forces are mutually equal and theconductor will then travel at a constant speed. Note that the magnitude of force in (1) willsimply be Fe = IlB, since the angle between the conductor and the flux density is 90 degrees.

Once when the conductor moves, according to the law of electromagnetic induction anelectromotive force will be induced in the electric circuit,

( ) lBve •×= (2)The magnitude of this emf is simply e =vBl, since the angle between the speed vector and fluxdensity vector is 90 degrees, while the angle between the conductor length vector and thevector product is zero degrees.

A process of electromechanical conversion is established in this way. The energy willbe converted from electrical to mechanical and the process is called motoring. In order for themotoring to happen it is necessary to: i) establish flux density, using permanent magnets forexample; ii) create an electric circuit, that is connected to a voltage source and is placed in theflux density. This will lead to establishment of the electromagnetic force, which causes linearmovement of the conductor. This movement is counterbalanced by the applied mechanical


E Levi, 2002 2

force (not shown in Fig. 1) and the equilibrium is established when the conductor travels at aconstant speed. Under this condition the electromagnetic force and the mechanical force aremutually equal, but act in the opposite directions.

Consider next Fig. 2, where the same conductor is placed in the same flux density.However, the conductor is now not connected to the electric source; instead, the electriccircuit is closed by using, say, an external resistance. The conductor is now dragged throughthe flux density using mechanical force at certain speed and this is the origin of the movementin this case. The sequence of events now reverses. An electromotive force, given with (2), isat first induced in the conductor. Since the circuit is closed, a current starts flowing.Interaction of the current and the flux density causes creation of the electromagnetic force.This force again acts in the opposite direction to the mechanical force and the equilibrium isestablished when the two forces are equal but act in the opposite direction. Note that in thiscase the source of motion is the supplied mechanical energy. The mechanical energy is nowconverted into electrical energy and the process is called generation.

B

I

Fe

v

Fig. 1 – Illustration of motoring.

B

I

Fm

v

Fe

Fig. 2 – Illustration of generation.

It is important to note here that the process of electromechanical energy conversion isreversible. This means that either electric energy can be converted into mechanical energy, ormechanical energy can be converted into electric energy, by means of the same physicalassembly. Note as well that both the expression for electromagnetic force acting on aconductor and the expression for induced electromotive force due to relative movement ofconductor with respect to flux density, which are vectorial, reduce to very simple expressionsdue to the relative position of flux density vector, conductor and speed of motion. This isexactly the situation that is encountered in electric machines. Therefore equations (1) and (2),which contain scalar and vectorial multiplications, reduce to a very simple form of Fe = IlBand e = vBl.

Nothing changes in principle when rotational movement is under consideration insteadof the linear movement. Table I gives the analogy between the linear and the rotationalmovement. Creation of torque in the case of rotational movement is illustrated in Fig. 3.

Table I – Analogy between linear and rotational movement.

Speed Source of motion Road travelled PowerLinear motion Linear, v [m/s] Force, [N] Linear, s [m] F v

Rotational motion Angular, ω [rad/s] Torque, [Nm] Angle, θ [rad] T ω

Suppose once more that there is a certain flux density, in which a structure is placed.This structure can rotate and is of radius r. Assume that there are two conductors placed onthe structure, 180 degrees apart, as shown in Fig. 3, and let these two conductors carry currentin designated (mutually opposite) directions. An electromagnetic force, Fe = IlB, is created oneach of the two conductors. However, one of these two forces acts to the left, while the other


E Levi, 2002 3

one acts to the right (due to opposite directions of the current flow in the two conductors).Now, a torque is created on each of the two conductors, that equals the product of the forceand the radius. However, since forces act in opposite directions at opposite sides of thestructure, the torques will both act in anticlockwise direction, initiating the rotation of thestructure in anticlockwise direction. The total electromagnetic torque will in general be thesum of all the individual torques acting on individual conductors.

Current in

Current out

B

Fe

Fe

Fig. 3 – Torque creation in the rotating structure.

Every electric machine consists of ferromagnetic iron cores and windings mounted onthe iron cores, these elements being of essential importance for electromechanical conversion.An electric machine consists of a stationary element, called stator, and a rotating element(such as the one in Fig. 3) called rotor. The winding is placed in slots of the stationary statorand/or in slots of rotational rotor. The winding consists of an appropriate number of turns. Aturn is composed of two conductors which are placed in such a way that the inducedelectromotive forces in them sum up. The current therefore flows in the opposite direction, asillustrated in Fig. 3.

As already noted and explained, the operation of electric machines relies on Faraday'slaw of electromagnetic induction and on Bio-Savar's law of electromagnetic force (torque).One important point to note is that the induced emf will be described with (2) only if thecurrent in the system is pure constant DC current. A more general expression for the inducedemf says that, if the total flux through the electric circuit is changed, an electromotive force isinduced,

( )

θω

θθψ

ψ

d

dLi

dt

diLe

dtdddLidtdiLdtdLidtdiLe

Li

dtde

+=−

−−=−−==

−=

(3)

The first term in this expression will exist only in circuits with AC currents and it is calledtransformer emf. The second term is what corresponds to (2) and it is the induced emf due tothe movement of a conductor in certain flux density. It is called rotational emf. Note that,according to (3), a rotational emf will be induced only if the inductance of an electromagneticstructure is a function of the rotor position θ. This may sound awkward but will be clarifiedlater on. In deriving (3) the use was made of the correlation between the angle travelled by therotor and its speed of rotation,


E Levi, 2002 4

�= dtωθ (4)

that reduces for a constant speed of rotation to θ = ωt. Chain differentiation rule was appliedas well. The total flux of the winding is called flux linkage and is denoted with ψ in (3). Itdepends on the flux seen by each conductor Φ and on the number of turns N. Flux linkage isψ = NΦ.

Electromotive force in an electric machine is induced either due to rotation of awinding in the flux density, or due to rotation of the flux density with respect to a stationarywinding. Change of flux linkage can be caused either by mechanical motion or by change ofcurrent in time. This is reflected in (3) and will be elaborated in detail later on.

Let us further clarify the two operating regimes of electric machines, generating andmotoring. Generating is discussed first. Due to the action of the prime mover (which deliversmechanical energy to the machine’s shaft) rotational part of the machine is forced to rotate(Fig. 4). Consequently, the speed of rotation is constant (n = const.) and Te = TPM. Voltage atmachine terminals and induced emf differ because of the voltage drop in the winding; forgenerating induced emf is greater than terminal voltage (in the sense of rms values in ACmachines, i.e. in the sense of average values in DC machines). Note that in generationdirection of the speed of rotation coincides with the direction of the mechanical (primemover’s) torque, while the electromagnetic torque of the machine opposes motion.

During motoring (Fig. 4) created electromagnetic torque, which is a consequence ofelectric energy delivered to the machine, acts as the source of motion, i.e. it causes the rotorrotation. In this case the direction of speed and the electromagnetic torque coincide, while themechanical torque (that is now load torque) acts against the direction of rotation. Once morethe speed of rotation is constant (n = const.) and Te = TL. During motoring induced emf hasthe opposite polarity since it balances the applied voltage. It is therefore usually calledcounter-electromotive force. The terminal voltage is greater than the counter-emf in motoring.

n n

TPM

TLTe

Te

Fig. 4 – Torque and speed directions in generation (left) and motoring (right).

In what follows a generalised electromechanical converter is discussed at first. Theanalysis is valid for any type of electric machine; the only constraint is that there is only onedegree of freedom for mechanical motion (i.e. rotor can rotate along one axis only).

2. GENERAL MODEL OF AN ELECTRIC MACHINE

2.1 Losses and efficiency

Efficiency of an electric machine is defined in the same way as for any other device,as ratio of the output to input power


E Levi, 2002 5

11 <+

−=+

==lossout

loss

lossout

outinout PP

P

PP

PPPη (5)

where the difference between the input and the output power is the loss in the machine, thatconsists of three components: winding loss (or copper loss) that is caused by the current flowin windings of the machine (in general, it appears at both stator and rotor), iron (or core) lossthat appears in the ferromagnetic structure of the machine that is exposed to AC flux, andmechanical loss that takes place due to friction in bearings and rotation of the rotor in the air:

mechlossFeCuoutinloss PPPPPP −++=−= (6)

Copper losses are of standard RI2 form and total winding loss is given with the summation oflosses for all individual windings. Iron or core loss depends on the flux density and frequencyand comprises hysteresis and eddy-current losses,

( ) 22mFeFe BffmP ξς += (7)

It is, according to (7), proportional to the mass of the ferromagnetic material. Mechanical losscan be taken as proportional to the speed of rotation squared,

2ωkP mechloss =− (8)

Since mechanical power is a product of torque and speed, this means that the mechanical losstorque is taken as proportional to the speed of rotation.

One important point to note is that the nature of the input and output power dependson the role of the machine. In motoring the input power is electrical, while the output power ismechanical. In generation it is the other way round, the input power is mechanical while theoutput power is electrical (in generation, there may be some windings that take electricalpower as well, while some other windings generate electrical power). It has to be rememberedthat the rated power of the machine (power for which the machine has been designed), whichis always given on the nameplate of the machine, is the output power. Hence, in generationthe known rated power (always identified further on with an index n) is the output electricalpower, while in motoring it is the output mechanical power.

2.2 Power flow in an electric machine

Since the role of the input and the output power is dependent on the function that themachine performs, the two cases are treated separately. In what follows lower case symbolsare used for all the quantities, meaning that instantaneous time-domain variables are underconsideration. The idea behind the subsequent development is to develop a generalmathematical model that is valid for any rotating electric machine. It is for this reason that thenumber of windings is not specified. Instead, it is taken as being equal to n, where this is anarbitrary number. The electric machine is for the time being a black box. There are two doorsthat enable access to the machine, electrical door and mechanical door. The power can beeither delivered to the machine, or taken away from the machine, through these two doors.Electromechanical conversion takes place inside the box and the converted power is pc. Fig. 5illustrates power flows inside an electric machine for motoring and generating. Apart fromthese two doors there are two windows that are unwanted outputs only. These windows areoutputs for winding losses and mechanical losses, which are inevitably created within amachine, and which represent lost power. Note that the iron (or core) loss is omitted from thisrepresentation. The reason is that it is of electromagnetic nature and it does not take place inthe windings. The existence of the iron loss can be accounted for at a later stage, in anapproximate manner, as it is done in transformer theory. Both doors can be either inputs oroutputs (depending on whether the machine operates as a motor or as a generator), whilewindows are outputs only. Normally, one door will be the input while the other door will bethe output, although in generation some windings make consume electrical energy, while


E Levi, 2002 6

other windings are generating it (as shown in Fig. 5). The role of doors is thus reversible, asthe machine can operate both as a generator and as a motor.

Electrical inputpower Mechanical

output power

Copperlosses inwindings

Mechanicalloss

Convertedpower

Electromagneticenergy storage

Mechanical energystorage

Electrical outputpower Mechanical input

power

Copperlosses inwindings

Mechanicalloss

Convertedpower

Electromagneticenergy storage

Mechanical energystorage

Small electricalinput power

Fig. 5 – Power flow in an electric machine for motoring and generation, respectively.

As can be seen from Fig. 5, apart from input and output power and losses, there aretwo internal storages of energy inside the machine. The first one is the stored electromagneticenergy, while the second one is the stored mechanical energy. Stored mechanical energy is theenergy stored in rotating masses (kinetic energy) and it is in every aspect analogous to the

energy stored under linear movement (which is 2

2

1mvWmech = , where m is the mass of the

body). Rotating bodies are characterised with so-called inertia (that is function of the massand dimensions) J [kgm2], while instead of the linear velocity one uses angular velocity.Hence

2

2

1 ωJWm = (9)

Stored energy in the electromagnetic system is function of the inductances and thecurrents of the windings (or flux linkages and currents). For example, in the case of a singlewinding


E Levi, 2002 7

iLiWe ψ2

1

2

1 2 == (10a)

If the machine has two windings, then the stored energy is

( )

112222

212111

22112112222

211 2

1

2

1

2

1

iLiL

iLiL

iiiiLiLiLWe

+=+=

+=++=

ψψ

ψψ

(10b)

Taking index e for electrical power and index m for mechanical power in Fig. 5, one can writethe following power balance equations:Motoring:

mm

mechlossc

ce

Cue

pdt

dWpp

pdt

dWpp

++=

++=

−

(11)

Generation:

21 eee

Cuc

cm

mechlossm

ppdt

dWpp

pdt

dWpp

−++=

++= −

(12)

Note that storages are energies, as defined in (9)-(10). Powers are time derivatives of energiesand this is taken into account in formulation of (11)-(12). In generation some windings maketake the power (pe2), while other winding actually generate the power (pe1).

Equations (11)-(12) enable formulation of the converted power that is defined as

mec tp ω= (13)

in terms of other known powers and derivation of the equation for motion of rotating massesin terms of known parameters and inputs of the machine. This is a tedious procedure for thegeneralised n-winding converter and most of the derivations will be therefore omitted. Onlythe starting equations and the final equations are presented in the next sub-section. It is to benoted that all the powers, as well as all the other variables (currents, flux linkages) weredenoted with lower-case letters in this section. These are instantaneous time domainquantities, and the same approach is used in the following sub-section. This enables creationof a general mathematical model, in terms of time-domain instantaneous quantities, that isvalid for all possible existing types of electric machines with rotational movement.

2.3 Mathematical model

Each of the n windings of the machine is a piece of wire. Hence each winding can becharacterised with its resistance and inductance. In addition, there are mutual inductancesbetween any two windings. An induced emf appears in general in each winding. Hence thevoltage equilibrium equation for one particular winding can be written as

niniiii

iiiiii

iLiLiLiL

dtdiReiRv

++++=+=−=

...........21211ψψ

(14)

There is one flux linkage equation and one voltage equation for each of the n windings. It isconvenient to use further on matrix notation to express these and other equations, since matrixnotation will enable substitution of n equations with a single matrix equation. Hence for allthe n windings one has (matrices are underlined):


E Levi, 2002 8

iLdt

diRv

=

+=

ψ

ψ(15)

where

��

�

�

��

�

�

=

−

n

n

R

R

R

R

R

1

2

1

..

..

��

�

�

��

�

�

=

nnnnn

n

n

n

LLLL

LLLL

LLLL

LLLL

L

....

............

............

....

....

....

321

3333231

223221

113121

(16a)

��

�

�

��

�

�

=

nv

v

v

v

v

..

..3

2

1

��

�

�

��

�

�

=

ni

i

i

i

i

..

..3

2

1

��

�

�

��

�

�

=

nψ

ψψψ

ψ

..

..3

2

1

(16b)

Note that in any electrical machine Lij = Lji.Input electrical power in motoring is

viivivivp Tnne =+++= ..........2211 (17)

Note that current matrix in (17) has to be transposed to satisfy the rules of matrixmultiplication. In generation the output power is

viivivivivivp Tnnkkkke =−−−+++= ++ .............. 112211 (18)

where winding 1…k generate electricity, while windings k+1…n consume electric energy.Current vector in (18) has positive currents for the windings that generate and negativecurrents for the windings that consume electric power.

Winding losses can be expressed as

iRiiRiRiRp TnnCu =+++= 22

22211 ....... (19)

Stored electromagnetic energy is

iLiW

iiLiiLiiLiiLiiLiLiLiLW

Te

nnnnnnnne

21

.........2

1....

2

1

2

11)1(32231131132112

2222

211

=

++++++++++= −−

(20)

Current sign in voltage equation (15) is such that the current is positive when it flowsinto the winding. Hence in generation all the windings that generate will have negativecurrents since the current flow will be in the opposite direction from assumed positive currentflow.

Mechanical power and mechanical loss are governed with

ωω

ω

kt

tp

tp

mechloss

mechlossmechloss

mm

==

=

−

−− (21)

and the correlation between speed of rotation and the angle travelled is


E Levi, 2002 9

dtddt θωωθ == � (22)

Angular speed of rotation is related to the speed n in [rpm] through n60

2πω = . Angle θ is the

mechanical angle measured with respect to certain stationary defined axis in the machine’scross-section. Mechanical torque in (21) can be the load torque (in motoring) or the primemover torque (in generation).

Stored mechanical energy remains to be given with2

2

1 ωJWm = (23)

What remains to be done is to substitute all the powers and derivatives of storedenergies into the power balance equations (11)-(12). This enables, first of all, calculation ofthe converted power and the electromagnetic torque. Regardless of which of the two regimesis considered, the converted power is found to be

idt

Ldip T

c 2

1= (24)

Since according to (13) converted power is ωec tp = and since one can write using chain

differentiation rule that ( )( ) ( )ωθθθ dLddtddLddtLd ==/ , one finds the electromagnetictorque in the form

id

Ldi

pt Tc

e θω 2

1== (25)

Electromagnetic torque is positive for the motoring, while it has negative value in generation(as it opposes motion). From the second of (11) or the first of (12) one finds the equation ofmotion of the rotor in the form

generation

motoring

ωω

ωω

kdt

dJtT

kdt

dJTt

ePM

Le

+=−

+=−(26a)

On the left-hand one has the difference between the driving torque (electromagnetic torque inmotoring, prime mover torque in generation) and the opposing torque (load torque inmotoring and electromagnetic torque in generation). On the right-hand side the first term isthe acceleration/deceleration torque (that exists only during transients and is zero in steady-state) and the second term is the torque that describes mechanical losses. This particulartorque can be always taken as part of the load (or prime mover) torque since it is mechanicalin nature. One then arrives at the equation of mechanical motion in the frequently used form

generation

motoring

dt

dJtT

dt

dJTt

ePM

Le

ω

ω

=−

=−(26b)

which shows that in any steady state (at constant speed)

generation0

motoring0

=−=−

ePM

Le

tT

Tt(27)

The meaning of (27) is simple. It is the basic law of action and reaction. In any steady statethe two torques are of the same absolute value but act in the opposite direction.

The equations presented in this sub-section fully describe any rotational electricalmachine, in terms of the instantaneous time-domain variables. The full mathematical model issummarised in the following sub-section.


E Levi, 2002 10

2.4 Summary of the mathematical model

Any rotating electric machine, regardless of the actual structure of the stator and rotor andregardless of the number of windings, is completely described with the following set ofequations:

iLdt

diRv

=

+=

ψ

ψ

��

�

<>−

=+=+generation0

motoring0,

ePM

eLmme tT

tTtk

dt

dJtt ωω

(28)

id

Ldit T

e θ2

1=

dtdθω =

where J and k are parameters of the machine and

��

�

�

��

�

�

=

−

n

n

R

R

R

R

R

1

2

1

..

..

��

�

�

��

�

�

=

nnnnn

n

n

n

LLLL

LLLL

LLLL

LLLL

L

....

............

............

....

....

....

321

3333231

223221

113121

��

�

�

��

�

�

=

nv

v

v

v

v

..

..3

2

1

��

�

�

��

�

�

=

ni

i

i

i

i

..

..3

2

1

��

�

�

��

�

�

=

nψ

ψψψ

ψ

..

..3

2

1

(29)

Equations (28)-(29) constitute the mathematical model of a generalised n-windingelectromechanical energy converter. Note once more that all the variables (voltages, currents,flux linkages, electromagnetic torque, speed of rotation) are instantaneous time-domainquantities. Note as well that voltage equation is valid for current flowing into the winding.Hence in generation some of the currents will be negative since they will be flowing out of themachine.

2.5 Existence of converted power and electromagnetic torque and average torque

Equation (25) shows that power will be converted if and only if the machine rotates. Thismeans that at zero speed converted power is always zero. Further, one can see thatelectromagnetic torque can exist at zero speed (a machine will always start from standstill andthe torque at zero speed is called starting torque). In order for an electromagnetic torque toexist it is necessary that at least some windings carry current and that at least someinductances of the machine are functions of the rotor angular position. Note that unless this


E Levi, 2002 11

condition is satisfied, torque will be zero. The issue of dependence of machine’s inductanceson angular position of the rotor will be discussed later.

Although an electromagnetic torque will exist if appropriate currents flow in themachine and there are inductances that depend on the rotor position, this is still not sufficientto realise useful electromechanical energy conversion. Assume that the torque of ahypothetical electric machine varies as a sine function of time, with the period equal to theperiod of rotation. The instantaneous torque does exist. But, it is positive in the first half-cycleand negative in the second half-cycle. The average torque is zero and hence the averageconverted power will be zero even if the machine runs at a constant speed. The machine willdo motoring in the first half-cycle and generating in the second half-cycle, with a net zeroconverted power over one cycle. Thus it follows that, if useful electromechanical conversionis to take place, average torque of the machine must differ from zero. Averageelectromagnetic torque Te can only exist if the certain correlation between stator current(voltage) frequency, rotor current (voltage) frequency and the frequency of rotation issatisfied. It can be shown that Te will be of nonzero value if and only if

rs ωωω −= (30)

where indices s and r identify stator and rotor angular frequency. Note that DC case isencompassed by (30). Note as well that, according to (30), it is not possible to realise usefulelectromechanical energy conversion if both stator and rotor windings are supplied with DCcurrents. In such a case an average torque can only exist at zero speed. But converted powerequals zero at zero speed.

On the basis of (30) is it is now possible to classify the most commonly used electricmachines into three categories:

1. Synchronous machines: rotor frequency is zero. Hence frequency of rotationequals stator frequency.

2. Induction machines: both stator and rotor windings carry AC currents. Rotor speedis related with the two angular frequencies as rs ωωω −= .

3. DC machines: stator frequency is zero. Hence frequency of rotation must equalfrequency of rotor winding currents.

Note that even when (30) is satisfied, this still does not mean that the electromagnetic torqueis constant. However, if the machine’s torque varies in time, then it follows from the equationof the mechanical motion that the speed will constantly vary although mechanical torquemight be perfectly constant. It is therefore necessary to provide such arrangements inelectrical machines that not only (30) is satisfied but in addition

ee Tt = (31)

This implies that the instantaneous torque value equals the average torque. In other words,torque is just a constant time-independent quantity. The means of achieving (31) in ACmachines (induction and synchronous) will be discussed shortly. Chapter on DC machineswill explain how (31) is achieved in that particular case.

2.6 Fundamental and reluctance torque component

Consider as an example a rotating electric machine with one winding on stator and onewinding on rotor. General mathematical model (28)

iLdt

diRv

=

+=

ψ

ψ

dtdθω = (28)


E Levi, 2002 12

id

Ldit T

e θ2

1=

��

�

<>−

=+=+generation0

motoring0,

ePM

eLmme tT

tTtk

dt

dJtt ωω

(28)

remains to be valid in the same form. However, (29) reduces to

��

��

�=�

�

��

�=�

�

��

�=�

�

��

�=�

�

��

�=

rrs

srs

r

s

r

s

r

s

r

s

LL

LLL

R

RR

i

ii

v

vv

ψψ

ψ (32)

and the electromagnetic torque expression reduces to

θθθ d

dLii

d

dLi

d

dLit sr

rsr

rs

se ++= 22

2

1

2

1(33)

The first two components of the torque expression will have non-zero values only if thewinding self-inductance is a function of the rotor position. The third torque component is theone due to the interaction of the stator and rotor winding and this component will exist in allmachines that do have windings on stator and rotor. The component of the torque due tointeraction of the stator and rotor winding is called fundamental torque component. Thetorque component that exists only if self-inductances of the windings are functions of therotor position is called reluctance torque component. In general both torque components willcontribute to the average torque so that

lfundamentae

cerelucee TTT += tan (34)

Mutual inductance between a stator and a rotor winding is always a function of the rotorposition. However, self-inductances are in many cases constants, so that the reluctance torquecomponent will rarely exist. Fig. 6 illustrates two commonly met cross-sections of ACmachines. The air-gap is white, while the rotor and stator body are grey. The first cross-section is characterised with cylindrical stator and rotor. The air-gap is therefore uniform.Recall that an inductance is inversely proportional to the magnetic reluctance. Magneticreluctance comprises mostly of the air-gap reluctance. Since air-gap is uniform, the magneticreluctance seen by both stator and rotor winding is constant. Self-inductances of the stator androtor winding are in this configuration constant and they do not depend of the rotor position.The reluctance torque component is therefore zero and torque consists only of fundamentaltorque component. This cross-section corresponds to all induction machines, numeroussynchronous machines (so-called turbo-machines and permanent magnet synchronousmachines with surface mounted magnets) and modern DC machines.

Fig. 6 – Two frequently met cross-sections of AC machines.

The other cross section in Fig. 6 is characterised with cylindrical stator and non-cylindrical rotor. The rotor is of so-called salient pole structure. Assume that there is awinding on both stator and rotor. From the point of view of rotor the air-gap appears as


E Levi, 2002 13

constant so that rotor self-inductance is constant. However, stator self-inductance is a functionof the rotor position, since the magnetic reluctance seen by the stator winding depends onwhere exactly rotor is. This particular cross-section is met in certain types of synchronousmachinery (hydro-generators and synchronous reluctance motors). In this case the torquedeveloped by the machine contains both the fundamental and the reluctance torquecomponent.

Functional dependence of a self-inductance on rotor position is beyond the scope ofinterest here. However, it is necessary to explain how the mutual inductance between thestator and the rotor winding depends on rotor position. Consider Fig. 7, where magnetic axesof the two windings are identified with symbols s and r. Magnetic axis of a winding is theaxis along which a particular winding produces flux. Stator magnetic axis is obviouslystationary, while rotor magnetic axis rotates with rotor. Let the maximum value of the mutualinductance between the two winding be M. Flux linkage of the stator and the rotor windingcan be expressed as

ssrrrr

rsrsss

iLiL

iLiL

+=+=

ψψ

(35)

For the sake of explanation, let us assume that rotor current is constant DC and let usinvestigate the contribution of this rotor current to the flux linkage in stator winding. Themaximum value of the contribution of the rotor current to the flux linkage in stator winding,MIr, is shown along the magnetic axis of the rotor winding. Its projection on stator windingmagnetic axis is LsrIr. Table II lists values of the contribution for various rotor positions.When the two axes are aligned (zero angle) the contribution is of maximum value. When thetwo axes are perpendicular, the contribution is zero. When the two axes are aligned but 180degrees apart, contribution is of negative maximum value. The function that describes suchbehaviour is a cosine function. Hence the mutual inductance between the stator and the rotorwinding can be written as

s

r

θ

MIr

LsrIr

Fig. 7 – Magnetic axes of stator and rotor windings.

Table II – Contribution of rotor current to the stator flux linkage

Angle θ [°] 0 90 180 270 360Lsr Ir MIr 0 −MIr 0 MIr

θcosMLsr = (36)

The mutual inductance between a stationary and a rotating winding is therefore alwaysa function of the rotor position. Simple sine or cosine functional dependence suffices for


E Levi, 2002 14

machines with uniform air-gap. If the machine is with salient poles on rotor, the expressionfor the mutual inductance becomes more complicated.

As the next step, let us consider the induced emfs in the two windings of the machinein Fig. 7. By definition

dtde

dtde

rr

ss

ψψ

=−=−

(37)

Substitution of (35) into (37) yields to

ωθθ

ωθθ

θθθ

��

��

� ++��

��

� +=−

��

��

� ++��

��

� +=−

��

��

� ++��

��

� +=−

��

��

� ++��

��

� +=−

d

dLi

d

dLi

dt

diL

dt

diLe

d

dLi

d

dLi

dt

diL

dt

diLe

dt

d

d

dLi

d

dLi

dt

diL

dt

diLe

dt

dLi

dt

dLi

dt

diL

dt

diLe

srs

rr

ssr

rrr

srr

ss

rsr

sss

srr

ss

rsr

sss

srr

ss

rsr

sss

rotorforwaysamein theand

(38)

The first term is the transformer induced emf, while the second term is the rotational emf.Note that in the case of Fig. 7, when both stator and rotor self-inductance are constant, (38)reduces to

ωθ

ωθ

��

��

�+��

��

� +=−

��

��

�+��

��

� +=−

d

dLi

dt

diL

dt

diLe

d

dLi

dt

diL

dt

diLe

srs

ssr

rrr

srr

rsr

sss

(39a)

Assuming further that rotor current is constant (this corresponds to a synchronous machine)further simplifies this expression to

ωθ

ωθ �

�

��

�+��

��

�=−��

��

�+��

��

�=−d

dLi

dt

diLe

d

dLi

dt

diLe sr

ss

srrsr

rs

ss (39b)

Example 1:A two winding system has the following inductances: stator winding self inductance =0.8 H, rotor winding self-inductance = 0.2 H and mutual inductance between the statorwinding and the rotor winding = 0.4 cos θ [H]. The rotor revolves at constant angularvelocity of 40 rad/s and the initial value of the mechanical co-ordinate at zero timeinstant equals zero. Determine the instantaneous value of induced electromotive forcein open-circuited rotor winding if the current that flows through the stator winding isequal to 10cos(100t) A.

Solution:In this example rotor current and its derivative are zero since the rotor winding is open circuited.Moreover, stator and rotor self-inductances are constant. Hence the induced emf follows from (39b) as

ωθ

��

��

�+��

��

�=−d

dLi

dt

diLe sr

ss

srr

Derivatives of the stator current and the mutual inductance are


E Levi, 2002 15

ttdt

ddL

tdtdi

sr

s

ωδωωθ

θθ

=−==

−=−=

�

sin4.0

100sin1000

Substitution into the expression for the induced emf yields

( ) ( )

tte

BAMBAN

MNe

BAe

tttte

r

r

r

r

60sin120140sin280

)(5.0)(5.0

sinsin

sincoscossin

100cos40sin16040cos100sin400

+=−=+=

−++=+=

+=

βαβαβαβα

Example 2:A two-winding system has stator inductance of 0.1 H, rotor inductance of 0.04 H andmutual inductance of 0.05 cosθ [H].a) Rotor rotates at 200 rad/s and stator current is known to be 10sin200t. Find theinduced emf in rotor winding if it is open-circuited. The initial value of mechanical co-ordinate at zero time is zero.b) Current 10sin200t flows through both windings, which are connected in series. Findthe speeds at which average torque will exist; find the average torque values anddetermine the values of the load angle which yield maximum values of the averagetorque.

Solution:a) Rotor current and its derivative are again zero. Hence once more

ωθ

��

��

�+��

��

�=−d

dLi

dt

diLe sr

ss

srr

ttdt

ddL

tdtdi

sr

s

ωδωωθ

θθ

=−==

−==

�

sin05.0

200cos2000

V7.702/100

[rad/s]400

400cos100

200sin200sin100200cos200cos100

==

==−

−=−

r

r

r

r

E

te

tttte

ω

b) Condition of average torque existence yields speeds at which the average torque will have non-zerovalues:

( )

��

�=

==±=

]rad/s[400

][rad/s0

]/[200

ω

ωωωωω

sradrs

rs

Note that both self-inductances are constant. Hence the torque contains only the fundamental torquecomponent. Instantaneous torque is

( ) ( )[ ]( ) )sin()400cos1(5.2sin200sin5

sin05.0200sin102

2

2

δωδω

δω

θθ

−−−=−−=

−−=

==

ttttt

ttt

ddLiddLiit

e

e

srsrrse

For the speed of zero [rad/s]:

2/for[Nm]5.2

[Nm]sin5.2)sin(5.21

max

0

πδ

δδ

==

=−−== �

e

T

ee

T

dttT

T


E Levi, 2002 16

For the speed of 400 [rad/s]:

2/for[Nm]25.1

[Nm]-1.25sin

1.25sin-)-0t1.25sin(80)-0t-2.5sin(40

)-sin(400t2.5cos400t)-0t-2.5sin(40

)400sin()400cos1(5.2

1

max

0

πδδ

δδδδδ

δ

−===

+=+=

−−−=

= �

e

e

e

e

e

T

ee

T

T

t

t

ttt

dttT

T

In this example both stator and rotor winding are supplied with single-phase currents. As the resultshows, the average torque will exist at 400 [rad/s] speed of rotation. However, apart from the averagetorque, there will exist two more time varying components. This is true for any single-phase ACmachine. It is not possible to realize conversion with constant time independent torque (i.e. it is notpossible to achieve equality of the average and instantaneous torque values). It is for this reason thatvast majority of AC machines in use nowadays are three-phase. It will be shown in the next section howutilization of a three-phase machine instead of a single-phase machine enables achievement of aconstant time independent torque.Note as well that the angle δ, taken initially as the initial value of the mechanical co-ordinate, is actuallymuch more than that. It is called load angle and its value depends on the loading of the machine. In thisexample machine’s average torque can be anything between zero and 1.25 [Nm] at 400 [rad/s]. Howmuch it will be depends on the load torque. The load angle value will adjust itself in such a way that themotor torque and the load torque are mutually equal at the given speed of rotation.

Example 3:An electromechanical converter has a three winding structure, with two windings onstator and one winding on rotor. The two stator windings are displaced in space by 90degrees. The winding self-inductances and mutual inductances are equal to:

δωθθθ

−==

===

tL

LL

LLL

ss

rsrs

rss

where0

sin9.0cos9.0=

H0.95=H1H1

21

21

21

Rotor winding current is constant DC, of 10 A. Stator windings are fed with two-phasesystem of currents, such that

titi ssss ωω sin10cos10 21 ==a) Sketch a cross-sectional view of the machine and identify the type of the machine.b) Develop the expression for the instantaneous and average torque produced by themachine under the assumption that the condition of average torque existence issatisfied. Calculate the average torque for load angle equal to 30 degrees and explainits nature.c) Sketch the dependence of average torque on load angle δ, identify motoring andgenerating part and define the region of stable operation.

Solution:In this example a so-called two-phase machine is considered. The example will show that with thespecific two-phase winding structure on stator it is possible to realize electromechanical energyconversion with the time independent constant torque developed by the machine. Two-phase structureis realized by displacing in space two stator windings by 90 degrees and by feeding the two-phasewinding with currents having phase displacement of 90 degrees. This example serves as an introductioninto the explanation of why three-phase machines are used nowadays and how time independent torqueis obtained, elaborated in the next section.a) Rotor frequency is zero. According to the condition of average torque existence, stator angularfrequency must in this case equal angular frequency of rotation. The machine is therefore a synchronousmachine with cylindrical stator and rotor structure. The cross section is shown in figure.


E Levi, 2002 17

S1 axis

r

θ

S1

r

S2

S2 axis

b) Self-inductances of all the three windings are constants. Mutual inductance between two statorwindings is zero. The instantaneous torque therefore contains only fundamental torque components andis given with

( )[ ][ ]

δωω

δωωδωωδωω

θωθωθθθθ

θθ

sin90

)sin(90

)sin(cos)cos(sin90

cos9.0sin10sin9.0cos10

cos9.0

sin9.0

2

1

2211

=≡=

+−=−−−=

+−==

−=+=

ee

s

se

sse

ssre

rs

rs

rsrsrsrse

Tt

ttt

ttttt

ttIt

ddL

ddL

ddLiiddLiit

Instantaneous torque is therefore time independent and equal to its average value. For the load angle of30 degrees the average torque is 45 [Nm].c) Average torque is a sine function of the load angle. Positive values correspond to motoring, whilenegative values describe generation. The machine can operate stably in the load angle region from -π/2to π/2. The region from -π/2 to zero is the generating region. The region from zero to π/2 is themotoring region. Stable operation is not possible outside these regions.

3. ROTATING FIELD IN THREE-PHASE MACHINES

Consider the stator winding of a three-phase AC machine. The winding is placed inslots with 120 degrees spatial displacement between any two respective phases. The windingis supplied with a system of three-phase currents, such that there is a phase displacement of120 degrees between any two respective stator phases. Let the three phase currents be givenwith:

( ) ( )3/4cos3/2coscos πωπωω −=−== tIitIitIi mcmbma (40)Current flow in each of the three phases causes an appropriate magneto-motive force (m.m.f.),that acts along the magnetic axis of the given phase. The situation is illustrated in Fig. 8,where individual phase m.m.f.’s are given with (N is the number of turns per phase):

( )( )

F NI t

F NI t

F NI t

a m

b m

c m

=

= −

= −

cos

cos /

cos /

ωω πω π

2 3

4 3

(41)

One observes that in terms of spatial dependence, all the three individual phase magneto-motive forces are stationary and they act along the defined magnetic axis of the winding.From (41) one notices that each of the three m.m.f.’s is varying in time. The values of the


E Levi, 2002 18

three phase m.m.f.’s in the given instant of time correspond to those met in any three phasesystem.

ac

b

Fa Fb Fc

bc

a

Fig. 8 - Individual phase m.m.f.’s of a three-phase winding.

The resultant magneto-motive force that stems from the three phase system of currentsflowing through spatially displaced windings is the sum of the individual contributions of thethree phases. The summation is done in the cross-section of the machine, and it is necessary toobserve the spatial displacement between the three m.m.f.’s. One may regard the cross-sectionof the machine as a Cartesian co-ordinate system in which phase a magnetic axis correspondsto x-axis, while y-axis is perpendicular to it. For the purposes of calculation this co-ordinatesystem may be treated as a complex plane, with x-axis corresponding to the real axis, and y-axis corresponding to the imaginary axis. In terms of the complex plane, spatial displacementof the m.m.f. by 120 degrees corresponds to the so-called ‘vector rotator’, ( )a j= exp /2 3π .Hence the resultant magneto-motive force can be written as

( )( ) ( )( )

F F F F e e

F NI t t t

res a b c

j j

res m

= + + = =

= + − + −

a a a a

a a

2

2

3 2

4

3

22 3 4 3

π π

ω ω π ω π

,

cos cos / cos /

(42)

The expression for the resultant magneto-motive force is most easily found if one recalls thewell-known correlation ( )cos . exp( ) exp( )δ δ δ= + −0 5 j j . Hence

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

( )

F NI e e ae ae a e a e

F NI e e aa e aae a a e a a e

a a

a a a a a a a

F NI e a a aa e

resj t j t j t j t j t j t

res mj t j t j t j t j t j t

res mj t j t

= + + + + + =

+ + + + +

+ + =

= = = =

= + + + +

− − − − − − −

− − −

−

1

21

2

1 0

1

1

21 1

2 3 2 3 2 4 3 2 4 3

2 2 2 2

2

2 2 3 4

2 2

ω ω ω π ω π ω π ω π

ω ω ω ω ω ω

ω ω

/ / / /

* *

* *

=

( )( )( ) ( )( )

a a

F NI e e

F NI e

res mj t j t

res mj t

2

1

23 0

3

2

+ =

= +

=

−ω ω

ω

(43)

Symbol * in (43) stands for complex conjugate. The result obtained in (5) is an important one.The equation


E Levi, 2002 19

F NI eres mj t=

3

2

ω (44)

describes a circular trajectory in the complex plane. This means that the resulting magneto-motive force, produced by three stationary, time-varying m.m.f.’s (often called pulsatingm.m.f.’s) is a time-independent and rotating magneto-motive force. Thus it follows that thethree-phase system creates a rotating field (called also Tesla’s field) in the machine. Figure 9illustrates the rotating field in different instants of time. The speed at which the rotating fieldtravels equals the angular frequency of the stator three-phase supply.

Imωt=90°

ωt=135° Fres ωt=45°

ωt=0°

Re (a)1.5NIm

Fig. 9 - Resultant field in the three-phase machine for sinusoidal supply conditions.

When the machine is synchronous, rotor winding carries DC excitation current and afield is produced by this current. This field is stationary with respect to rotor. Since the rotorrotates at synchronous speed, then, looking in from stationary stator, this rotor field rotates atsynchronous speed. This is always the case in any multi-phase AC machine: regardless ofwhether the rotor rotates synchronously or asynchronously, all the fields in the machine rotateat synchronous speed.

Since the resulting m.m.f. is responsible for the resulting flux density and ultimatelyresulting flux, this means that apart from the rotating m.m.f., there is a rotating flux densitywave and a rotating flux in the machine as well. The term rotating field in general denotes anyof the three.

Example 4:Consider a three-phase machine with cylindrical cross section of stator and rotor.Rotor carries a single winding, supplied with DC current. Mutual inductances betweenstator windings and rotor winding and three phase stator currents are given with:

θcosMLar = )3/2cos( πθ −= MLbr )3/4cos( πθ −= MLcr

tIi sma ωcos= )3/2cos( πω −= tIi smb )3/4cos( πω −= tIi smc

Derive the expression for the instantaneous and the average torque.

Solution:The instantaneous torque will once more be equal to the average torque, since the stator carries a three-phase winding. All the self-inductances are constant, as are the mutual inductances within the statorwinding. Hence the torque is:


E Levi, 2002 20

[ ]

δωω

δωθθω

θωωθθω

ωθθωωθ

θωθωθωθθθ

sin5.1

)sin(5.1

)]sin()120sin()sin(

)240sin()sin()[sin(5.0

)240sin()240cos()120sin()120cos(sincos

rmee

s

smre

sss

sssmre

sssrme

crrc

brrb

arrae

IMITt

t

tIMIt

ttt

tttIMIt

tttIMIt

d

dLIi

d

dLIi

d

dLIit

=≡=

−=−=

−−−++−−

−−++−−+−=

−−−−−−−=

++=

�

�

��

Note that the three terms with 0, 120 and 240 degrees mutual phase displacement in the instantaneoustorque expression sum to zero in any instant in time. This is a definition of a three-phase system. Thisleaves only the three terms that are mutually the same and each yield 1/3 of the total torque. Torque istime independent, this being the consequence of the three-phase winding existence on stator. Themachine is a synchronous machine with excitation winding on rotor.

Example 5:A three-phase synchronous machine is operated with constant DC current in the rotorwinding and it runs as a generator at constant speed of rotation ω. Stator windings areopen circuited. Determine induced emfs in the stator windings. Mutual inductancesbetween stator and rotor are θcosMLar = , )3/2cos( πθ −= MLbr , )3/4cos( πθ −= MLcr .

Solution:Since rotor current is constant and since the stator currents are zero, induced emfs in the stator windingsare:

)3/4sin()()3/4sin(

)3/2sin()()3/2sin(

sin)(sin

πωωπθωωθ

πωωπθωωθ

ωωθωωθ

−=−=−=

−=−=−=

==−=

tMIMId

dLIe

tMIMId

dLIe

tMIMId

dLIe

rrcr

rc

rrbr

rb

rrar

ra

It follows that a system of three-phase voltages is generated at open-circuited stator terminals. Thisexample shows the basic principle of three-phase electricity generation using synchronous three-phasegenerators.

4. TUTORIAL QUESTIONS

Q1. An electromechanical converter has a three winding structure. Winding resistances,self-inductances and mutual inductances are known to beR R R L L L L L L1 2 3 11 22 33 12 23 31, , ; , , ; , ,and , respectively. The windings are fed from voltagesources of known voltages v v v1 2 3, , and rotor inertia and friction coefficient are J andk. The machine runs at certain speed ω. Write the time domain matrix equations andequations in developed form for the following: i) voltage equilibrium and inducedelectromotive forces; ii) mechanical equilibrium; iii) power balance; iv) convertedpower and electromagnetic torque.

Q2. a) State the complete time-domain mathematical model of a generalised electro-mechanical converter with n windings and define all the matrices of the model.b) State the general condition of average torque existence in a two-winding structure(define all the symbols used).c) A two-winding system has stator inductance of 0.1 [H], rotor inductance of 0.04 [H]and mutual inductance of 0.05 cos ϑ [H]. If the rotor rotates at 300 rad/s and stator


E Levi, 2002 21

current is known to be 10 sin 300t, find the induced electromotive force in rotorwinding if it is open-circuited (the initial value of mechanical co-ordinate at zero timeis zero).

Q3. a) State the complete time-domain mathematical model of a generalisedelectromechanical converter with n windings and define all the matrices and vectors ofthe model.b) State the general condition of average torque existence in a two-winding structureand classify the electrical machines with respect to the way in which this condition issatisfied.c) An electromechanical converter has a three winding structure, with two windings onstator and one winding on rotor. The two stator windings are displaced in space by 90degrees. The winding self-inductances and mutual inductances are equal to:

δωθθθ

−==

===

tL

LL

LLL

SS

rSrS

rSS

where0

sin8.0cos.80=

H0.9=H1.1H1.1

21

21

21


i t i t1 210 10= =cos sinω ωSketch cross-sectional view of the machine and identify the type of the machine.Calculate the average torque for load angle equal to 45 degrees and explain its nature.Sketch the dependence of average torque on load angle δ, identify motoring andgenerating part and define the region of stable operation.

Q4. An electromechanical converter has two windings only, both mounted on stator anddisplaced in space by 90 degrees. The winding self-inductances and mutual inductancecan be given with

.where

2sin2

2cos22

2cos22

12

22

11

δωθ

θ

θ

θ

−=

−=

−−

+=

−+

+=

t

LLL

LLLLL

LLLLL

qd

qdqd

qdqd

Stator windings are fed with two-phase currents is1 = Im cosωst and is2= Im sinωst.a) Sketch the cross-sectional view of the machine and identify the type of the machine.b) Derive the expression for instantaneous torque developed in the converter and theexpression for average torque assuming that condition of average torque existence issatisfied.c) Calculate the average torque and explain the origin of the torque in this machine ifL L Id q m= = = =2 1 5 A and 30H, H, δ � .

Q5. a) State the complete time-domain mathematical model of a generalisedelectromechanical converter with n windings and define all the matrices and vectors ofthe model.b) Give graphical representation of the power flow in an electric machine for motoringand generation and define all the powers in this representation in terms of terminalquantities and parameters (use matrix form).


E Levi, 2002 22

c) State the general condition of average torque existence in a two-winding structureand classify the electrical machines with respect to the way in which this condition issatisfied.d) An electromechanical converter has a three winding structure, with two windingson stator and one winding on rotor. The two stator windings are displaced in space by90 degrees. The winding self-inductances and mutual inductances are equal to:

δωθθθ

−===

==

tL

LL

LLL

ss

rsrs

rss

where0

sin95.0cos95.0=

H1=H25.1H25.1

21

21

21


i t i t1 210 10= =cos sinω ωSketch cross-sectional view of the machine and identify the type of the machine.Calculate the average torque for load angle equal to 60 degrees and explain its nature.Sketch the dependence of average torque on load angle δ, identify motoring andgenerating part and define the region of stable operation.

Q6. a) State the complete time-domain mathematical model of a generalised electro-mechanical converter with n windings and define all the matrices and vectors of themodel.b) A two-winding system has stator inductance of 0.15 [H], rotor inductance of 0.05[H] and mutual inductance of 0.1 cos ϑ [H]. If the rotor rotates at 250 rad/s and statorcurrent is known to be 15 sin 250t, find the induced electromotive force in rotorwinding if it is open-circuited (the initial value of mechanical co-ordinate at zero timeis zero).c) The same system of part b) is again considered. However, current 15 sin 250tflows now through both windings that are connected in series. Find the speeds atwhich average torque will exist, determine corresponding average torque values andthe values of load angle, which yield maximum average torque values.