Tài liệu Fundamentals of electromechanical energy conversion docx

ENGNG 2024 Electrical Engineering
 E Levi, 2002
1
FUNDAMENTALS OF ELECTROMECHANICAL ENERGY
CONVERSION
1. PRELIMINARY CONSIDERATIONS
Electromechanical energy conversion is achievable in a number of ways. These
possibilities rely on different fundamental laws of electrical engineering. As the only method
that has importance on the large scale is electromechanical conversion achieved by means of
electromagnetic converters, this section is fully devoted to the analysis of basic principles
involved in electromagnetic electromechanical conversion.
Electromechanical energy conversion is achieved by devices that are usually called
electric machines. In principle, laws of electromagnetics can be used to design converters with
the linear and with the rotary motion. Converters with linear motion are called linear electric
machines, while those that rely on rotating motion are called rotating electric machines. Vast
majority of existing machinery belong to the category of rotating electric machines. These
include all the machines used to generate the electricity, as well as the most of the machines
used in industry to perform some useful work while converting electric into mechanical
energy. Linear machines are used relatively rare for somewhat specialised applications. It is
for 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 that
devices 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 creation
in an electromagnetic field (Bio-Savart’s law).
Consider the situation shown in Fig. 1. A conductor is connected to an electric source
and it carries current I. It is placed in the magnetic field of certain flux density B (which is of
course a vector; hence the arrow above the symbol in Fig. 1). Interaction of the flux density
and 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 the
conductor, which will start travelling at certain linear speed
v to the left. The electromagnetic
force will be balanced by another, mechanical force that acts in the opposite direction (to the
right). The equilibrium will be established when the two forces are mutually equal and the
conductor will then travel at a constant speed. Note that the magnitude of force in (1) will
simply be
F
e
=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 an
electromotive 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 flux
density vector is 90 degrees, while the angle between the conductor length vector and the
vector product is zero degrees.
A process of electromechanical conversion is established in this way. The energy will
be converted from electrical to mechanical and the process is called motoring. In order for the
motoring to happen it is necessary to: i) establish flux density, using permanent magnets for
example; ii) create an electric circuit, that is connected to a voltage source and is placed in the
flux density. This will lead to establishment of the electromagnetic force, which causes linear
movement of the conductor. This movement is counterbalanced by the applied mechanical
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force (not shown in Fig. 1) and the equilibrium is established when the conductor travels at a

constant speed. Under this condition the electromagnetic force and the mechanical force are
mutually 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 electric
circuit is closed by using, say, an external resistance. The conductor is now dragged through
the flux density using mechanical force at certain speed and this is the origin of the movement
in this case. The sequence of events now reverses. An electromotive force, given with (2), is
at 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 is
established when the two forces are equal but act in the opposite direction. Note that in this
case the source of motion is the supplied mechanical energy. The mechanical energy is now
converted into electrical energy and the process is called generation.
B
I
F
e
v
Fig. 1 – Illustration of motoring.
B
I
F
m
v
F
e
Fig. 2 – Illustration of generation.
It is important to note here that the process of electromechanical energy conversion is
reversible. This means that either electric energy can be converted into mechanical energy, or
mechanical energy can be converted into electric energy, by means of the same physical

assembly. Note as well that both the expression for electromagnetic force acting on a
conductor and the expression for induced electromotive force due to relative movement of
conductor with respect to flux density, which are vectorial, reduce to very simple expressions
due to the relative position of flux density vector, conductor and speed of motion. This is
exactly 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 F
e
=IlB
and e=vBl.
Nothing changes in principle when rotational movement is under consideration instead
of the linear movement. Table I gives the analogy between the linear and the rotational
movement. 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 Power
Linear motion Linear, v [m/s] Force, [N] Linear, s [m] Fv
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 on
the structure, 180 degrees apart, as shown in Fig. 3, and let these two conductors carry current
in designated (mutually opposite) directions. An electromagnetic force,
F

e
=IlB,is created on
each of the two conductors. However, one of these two forces acts to the left, while the other
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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 force
and the radius. However, since forces act in opposite directions at opposite sides of the
structure, the torques will both act in anticlockwise direction, initiating the rotation of the
structure in anticlockwise direction. The total electromagnetic torque will in general be the
sum of all the individual torques acting on individual conductors.
Current in
Current out
B
F
e
F
e
Fig. 3 – Torque creation in the rotating structure.
Every electric machine consists of ferromagnetic iron cores and windings mounted on
the 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 stator
and/or in slots of rotational rotor. The winding consists of an appropriate number of turns. A
turn is composed of two conductors which are placed in such a way that the induced
electromotive forces in them sum up. The current therefore flows in the opposite direction, as
illustrated in Fig. 3.
As already noted and explained, the operation of electric machines relies on Faraday's
law 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 the
current in the system is pure constant DC current. A more general expression for the induced
emf says that, if the total flux through the electric circuit is changed, an electromotive force is
induced,
()
θ
ω
θθ
ψ
ψ
d
dL
i
dt
di
Le
dtdddLidtdiLdtdLidtdiLe
Li
dtde
+=−
−−=−−=
=
−=
(3)
The first term in this expression will exist only in circuits with AC currents and it is called
transformer emf. The second term is what corresponds to (2) and it is the induced emf due to
the 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 electromagnetic
structure is a function of the rotor position
θ

. This may sound awkward but will be clarified
later on. In deriving (3) the use was made of the correlation between the angle travelled by the
rotor and its speed of rotation,
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 E Levi, 2002
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= dt
ωθ
(4)
that reduces for a constant speed of rotation to
θ
=
ω
t. Chain differentiation rule was applied
as well. The total flux of the winding is called flux linkage and is denoted with
ψ
in (3). It
depends on the flux seen by each conductor
Φ
and on the number of turns N. Flux linkage is
ψ
=N
Φ
.
Electromotive force in an electric machine is induced either due to rotation of a
winding in the flux density, or due to rotation of the flux density with respect to a stationary
winding. Change of flux linkage can be caused either by mechanical motion or by change of
current 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 and
motoring. Generating is discussed first. Due to the action of the

prime mover (which delivers
mechanical 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.)andT
e
= T
PM
.Voltageat
machine terminals and induced emf differ because of the voltage drop in the winding; for
generating induced emf is greater than terminal voltage (in the sense of rms values in AC
machines, i.e. in the sense of average values in DC machines). Note that in generation
direction of the speed of rotation coincides with the direction of the mechanical (prime
mover’s) torque, while the electromagnetic torque of the machine opposes motion.
During motoring (Fig. 4) created electromagnetic torque, which is a consequence of
electric energy delivered to the machine, acts as the source of motion, i.e. it causes the rotor
rotation. In this case the direction of speed and the electromagnetic torque coincide, while the
mechanical torque (that is now load torque) acts against the direction of rotation. Once more
the speed of rotation is constant (n =const.)andT
e
= T
L
. During motoring induced emf has
the opposite polarity since it balances the applied voltage. It is therefore usually called
counter-electromotive force. The terminal voltage is greater than the counter-emf in motoring.
n n
T
PM
T
L
T
e

T
e
Fig. 4 – Torque and speed directions in generation (left) and motoring (right).
In what follows a generalised electromechanical converter is discussed at first. The
analysis is valid for any type of electric machine; the only constraint is that there is only one
degree 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
ENGNG 2024 Electrical Engineering
 E Levi, 2002
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11 <
+
−=
+
==
lossout
loss
lossout
out
inout
PP
P
PP
P
PP
η
(5)

where the difference between the input and the output power is the loss in the machine, that
consists of three components: winding loss (or copper loss) that is caused by the current flow
in windings of the machine (in general, it appears at both stator and rotor), iron (or core) loss
that appears in the ferromagnetic structure of the machine that is exposed to AC flux, and
mechanical 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
RI
2
form and total winding loss is given with the summation of
losses for all individual windings. Iron or core loss depends on the flux density and frequency
and comprises hysteresis and eddy-current losses,
()
22
mFeFe
BffmP
ξς
+= (7)
It is, according to (7), proportional to the mass of the ferromagnetic material. Mechanical loss
can be taken as proportional to the speed of rotation squared,
2
ω
kP
mechloss
=
−
(8)

Since mechanical power is a product of torque and speed, this means that the mechanical loss
torque is taken as proportional to the speed of rotation.
One important point to note is that the nature of the input and output power depends
on the role of the machine. In motoring the input power is electrical, while the output power is
mechanical. In generation it is the other way round, the input power is mechanical while the
output power is electrical (in generation, there may be some windings that take electrical
power as well, while some other windings generate electrical power). It has to be remembered
that the rated power of the machine (power for which the machine has been designed), which
is always given on the nameplate of the machine, is the
output power. Hence, in generation
the known rated power (always identified further on with an index
n) is the output electrical
power, 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 the
machine performs, the two cases are treated separately. In what follows lower case symbols
are used for all the quantities, meaning that
instantaneous time-domain variables are under
consideration. The idea behind the subsequent development is to develop a general
mathematical model that is valid for any rotating electric machine. It is for this reason that the
number of windings is not specified. Instead, it is taken as being equal to
n, where this is an
arbitrary number. The electric machine is for the time being a black box. There are two doors
that enable access to the machine, electrical door and mechanical door. The power can be
either 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
p
c
.Fig.5
illustrates power flows inside an electric machine for motoring and generating. Apart from

these two doors there are two windows that are unwanted outputs only. These windows are
outputs for winding losses and mechanical losses, which are inevitably created within a
machine, and which represent lost power. Note that the iron (or core) loss is omitted from this
representation. The reason is that it is of electromagnetic nature and it does not take place in
the windings. The existence of the iron loss can be accounted for at a later stage, in an
approximate manner, as it is done in transformer theory. Both doors can be either inputs or
outputs (depending on whether the machine operates as a motor or as a generator), while
windows are outputs only. Normally, one door will be the input while the other door will be
the output, although in generation some windings make consume electrical energy, while
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other windings are generating it (as shown in Fig. 5). The role of doors is thus reversible, as
the machine can operate both as a generator and as a motor.
Electrical input
power
Mechanical
output power
Copper
losses in
windings
M
echanical
loss
Converted
power
Electromagnetic
energy storage
M
echanical energy

storage
Electrical output
power
Mechanical input
power
Copper
losses in
windings
Mechanical
loss
Converted
power
Electromagnetic
energy storage
Mechanical energy
storage
Small electrical
input 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 are
two internal storages of energy inside the machine. The first one is the stored electromagnetic
energy, while the second one is the stored mechanical energy. Stored mechanical energy is the
energy stored in rotating masses (kinetic energy) and it is in every aspect analogous to the
energy stored under linear movement (which is
2
2
1
mvW
mech
= ,wherem is the mass of the

body). Rotating bodies are characterised with so-called
inertia (that is function of the mass
and dimensions)
J [kgm
2
], while instead of the linear velocity one uses angular velocity.
Hence
2
2
1
ω
JW
m
= (9)
Stored energy in the electromagnetic system is function of the inductances and the
currents of the windings (or flux linkages and currents). For example, in the case of a single
winding
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iLiW
e
ψ
2
1
2
1
2
== (10a)
If the machine has two windings, then the stored energy is

()
112222
212111
22112112
2
22
2
11
2
1
2
1
2
1
iLiL
iLiL
iiiiLiLiLW
e
+=
+=
+=++=
ψ
ψ
ψψ
(10b)
Taking index
e for electrical power and index m for mechanical power in Fig. 5, one can write
the following power balance equations:
Motoring:
m

m
mechlossc
c
e
Cue
p
dt
dW
pp
p
dt
dW
pp
++=
++=
−
(11)
Generation:
21 ee
e
Cuc
c
m
mechlossm
pp
dt
dW
pp
p
dt

dW
pp
−++=
++=
−
(12)
Note that storages are energies, as defined in (9)-(10). Powers are time derivatives of energies
and this is taken into account in formulation of (11)-(12). In generation some windings make
take the power (
p
e2
), while other winding actually generate the power (p
e1
).
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 masses
in terms of known parameters and inputs of the machine. This is a tedious procedure for the
generalised
n-winding converter and most of the derivations will be therefore omitted. Only
the starting equations and the final equations are presented in the next sub-section. It is to be
noted that all the powers, as well as all the other variables (currents, flux linkages) were
denoted with lower-case letters in this section. These are instantaneous time domain
quantities, and the same approach is used in the following sub-section. This enables creation
of a general mathematical model, in terms of time-domain instantaneous quantities, that is
valid 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 be
characterised with its resistance and inductance. In addition, there are mutual inductances
between any two windings. An induced emf appears in general in each winding. Hence the
voltage equilibrium equation for one particular winding can be written as
niniiii
iiiiii
iLiLiLiL
dtdiReiRv
++++=
+=−=

21211
ψ
ψ
(14)
There is one flux linkage equation and one voltage equation for each of the
n windings. It is
convenient to use further on matrix notation to express these and other equations, since matrix
notation will enable substitution of
n equations with a single matrix equation. Hence for all
the
n windings one has (matrices are underlined):
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iL
dt
d
iRv
=

+=
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)
=
n
v
v
v
v
v


3
2
1
=
n
i
i
i
i
i

3
2
1
=
n
ψ
ψ
ψ
ψ
ψ


3
2
1
(16b)
Note that in any electrical machine
L
ij
= L
ji
.
Input electrical power in motoring is
viivivivp
T
nne
=+++=
2211
(17)

Note that current matrix in (17) has to be transposed to satisfy the rules of matrix
multiplication. In generation the output power is
viivivivivivp
T
nnkkkke
=−−−+++=
++

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 negative
currents for the windings that consume electric power.
Winding losses can be expressed as
iRiiRiRiRp
T
nnCu
=+++=
22
22
2
11
(19)
Stored electromagnetic energy is
iLiW
iiLiiLiiLiiLiiLiLiLiLW
T
e
nnnnnnnne

2
1

2
1

2
1
2
1
1)1(32231131132112
22
22
2
11
=
++++++++++=
−−
(20)
Current sign in voltage equation (15) is such that the current is positive when it flows
into the winding. Hence in generation all the windings that generate will have negative
currents since the current flow will be in the opposite direction from assumed positive current
flow.
Mechanical power and mechanical loss are governed with
ω
ω
ω
kt
tp
tp

mechloss
mechlossmechloss
mm
=
=
=
−
−−
(21)
and the correlation between speed of rotation and the angle travelled is
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dtddt
θωωθ
== (22)
Angular speed of rotation is related to the speed
n in [rpm] through n
60
2
π
ω
= .Angle
θ
is the
mechanical angle measured with respect to certain stationary defined axis in the machine’s
cross-section. Mechanical torque in (21) can be the load torque (in motoring) or the prime
mover torque (in generation).
Stored mechanical energy remains to be given with
2

2
1
ω
JW
m
= (23)
Whatremainstobedoneistosubstituteallthepowersandderivativesofstored
energies into the power balance equations (11)-(12). This enables, first of all, calculation of
the converted power and the electromagnetic torque. Regardless of which of the two regimes
is considered, the converted power is found to be
i
dt
Ld
ip
T
c
2
1
=
(24)
Since according to (13) converted power is
ω
ec
tp = and since one can write using chain
differentiation rule that
()()()
ωθθθ
dLddtddLddtLd ==/ , one finds the electromagnetic
torque in the form
i

d
Ld
i
p
t
T
c
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 of
motion of the rotor in the form
generation
motoring
ω
ω
ω
ω
k
dt
d
JtT
k
dt
d
JTt

ePM
Le
+=−
+=−
(26a)
On the left-hand one has the difference between the driving torque (electromagnetic torque in
motoring, prime mover torque in generation) and the opposing torque (load torque in
motoring and electromagnetic torque in generation). On the right-hand side the first term is
the 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 particular
torque can be always taken as part of the load (or prime mover) torque since it is mechanical
in nature. One then arrives at the equation of mechanical motion in the frequently used form
generation
motoring
dt
d
JtT
dt
d
JTt
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 state
thetwotorquesareofthesameabsolutevaluebutactintheoppositedirection.
The equations presented in this sub-section fully describe any rotational electrical
machine, in terms of the instantaneous time-domain variables. The full mathematical model is
summarised in the following sub-section.
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2.4 Summary of the mathematical model
Any rotating electric machine, regardless of the actual structure of the stator and rotor and
regardless of the number of windings, is completely described with the following set of
equations:
iL
dt
d
iRv
=
+=
ψ
ψ
<
>−
=+=+

generation0
motoring0,
ePM
eL
mme
tT
tT
tk
dt
d
Jtt
ω
ω
(28)
i
d
Ld
it
T
e
θ
2
1
=
dtd
θω
=
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
=
n
v
v
v
v
v


3
2
1
=
n
i
i
i
i
i


3
2
1
=
n
ψ
ψ

ψ
ψ
ψ


3
2
1
(29)
Equations (28)-(29) constitute the mathematical model of a generalised n-winding
electromechanical energy converter. Note once more that all the variables (voltages, currents,
flux linkages, electromagnetic torque, speed of rotation) are instantaneous time-domain
quantities. 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 the
machine.
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. This
means that at zero speed converted power is always zero. Further, one can see that
electromagnetic torque can exist at zero speed (a machine will always start from standstill and
the torque at zero speed is called starting torque). In order for an electromagnetic torque to
exist it is necessary that at least some windings carry current and that at least some
inductances of the machine are functions of the rotor angular position. Note that unless this
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condition is satisfied, torque will be zero. The issue of dependence of machine’s inductances
on angular position of the rotor will be discussed later.
Although an electromagnetic torque will exist if appropriate currents flow in the
machine and there are inductances that depend on the rotor position, this is still not sufficient
to realise useful electromechanical energy conversion. Assume that the torque of a

hypothetical electric machine varies as a sine function of time, with the period equal to the
period of rotation. The instantaneous torque does exist. But, it is positive in the first half-cycle
and negative in the second half-cycle. The average torque is zero and hence the average
converted power will be zero even if the machine runs at a constant speed. The machine will
do motoring in the first half-cycle and generating in the second half-cycle, with a net zero
converted power over one cycle. Thus it follows that, if useful electromechanical conversion
is to take place, average torque of the machine must differ from zero. Average
electromagnetic torque T
e
can only exist if the certain correlation between stator current
(voltage) frequency, rotor current (voltage) frequency and the frequency of rotation is
satisfied. It can be shown that T
e
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 is
encompassed by (30). Note as well that, according to (30), it is not possible to realise useful
electromechanical energy conversion if both stator and rotor windings are supplied with DC
currents. In such a case an average torque can only exist at zero speed. But converted power
equals zero at zero speed.
On the basis of (30) is it is now possible to classify the most commonly used electric
machines into three categories:
1. Synchronous machines: rotor frequency is zero. Hence frequency of rotation
equals stator frequency.
2. Induction machines: both stator and rotor windings carry AC currents. Rotor speed
is related with the two angular frequencies as
rs
ωωω

−= .
3. DC machines: stator frequency is zero. Hence frequency of rotation must equal
frequency of rotor winding currents.
Note that even when (30) is satisfied, this still does not mean that the electromagnetic torque
is constant. However, if the machine’s torque varies in time, then it follows from the equation
of the mechanical motion that the speed will constantly vary although mechanical torque
might be perfectly constant. It is therefore necessary to provide such arrangements in
electrical 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 AC
machines (induction and synchronous) will be discussed shortly. Chapter on DC machines
will 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 one
winding on rotor. General mathematical model (28)
iL
dt
d
iRv
=
+=
ψ
ψ
dtd
θω
= (28)
ENGNG 2024 Electrical Engineering
 E Levi, 2002

12
i
d
Ld
it
T
e
θ
2
1
=
<
>−
=+=+
generation0
motoring0,
ePM
eL
mme
tT
tT
tk
dt
d
Jtt
ω
ω
(28)
remains to be valid in the same form. However, (29) reduces to
=====

rrs
srs
r
s
r
s
r
s
r
s
LL
LL
L
R
R
R
i
i
i
v
v
v
ψ
ψ
ψ
(32)
and the electromagnetic torque expression reduces to
θθθ
d
dL

ii
d
dL
i
d
dL
it
sr
rs
r
r
s
se
++=
22
2
1
2
1
(33)
The first two components of the torque expression will have non-zero values only if the
winding self-inductance is a function of the rotor position. The third torque component is the
one due to the interaction of the stator and rotor winding and this component will exist in all
machines that do have windings on stator and rotor. The component of the torque due to
interaction of the stator and rotor winding is called fundamental torque component. The
torque component that exists only if self-inductances of the windings are functions of the
rotor position is called reluctance torque component. In general both torque components will
contribute to the average torque so that
lfundamenta
e

cereluc
ee
TTT +=
tan
(34)
Mutual inductance between a stator and a rotor winding is always a function of the rotor
position. However, self-inductances are in many cases constants, so that the reluctance torque
component will rarely exist. Fig. 6 illustrates two commonly met cross-sections of AC
machines. 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. Magnetic
reluctance comprises mostly of the air-gap reluctance. Since air-gap is uniform, the magnetic
reluctance seen by both stator and rotor winding is constant. Self-inductances of the stator and
rotor 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 fundamental
torque component. This cross-section corresponds to all induction machines, numerous
synchronous machines (so-called turbo-machines and permanent magnet synchronous
machines 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 a
winding on both stator and rotor. From the point of view of rotor the air-gap appears as
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13
constant so that rotor self-inductance is constant. However, stator self-inductance is a function
of the rotor position, since the magnetic reluctance seen by the stator winding depends on
where exactly rotor is. This particular cross-section is met in certain types of synchronous
machinery (hydro-generators and synchronous reluctance motors). In this case the torque
developed by the machine contains both the fundamental and the reluctance torque

component.
Functional dependence of a self-inductance on rotor position is beyond the scope of
interest here. However, it is necessary to explain how the mutual inductance between the
stator and the rotor winding depends on rotor position. Consider Fig. 7, where magnetic axes
of the two windings are identified with symbols s and r. Magnetic axis of a winding is the
axis along which a particular winding produces flux. Stator magnetic axis is obviously
stationary, while rotor magnetic axis rotates with rotor. Let the maximum value of the mutual
inductance between the two winding be M. Flux linkage of the stator and the rotor winding
can be expressed as
ssrrrr
rsrsss
iLiL
iLiL
+=
+=
ψ
ψ
(35)
For the sake of explanation, let us assume that rotor current is constant DC and let us
investigate the contribution of this rotor current to the flux linkage in stator winding. The
maximum value of the contribution of the rotor current to the flux linkage in stator winding,
MI
r
, is shown along the magnetic axis of the rotor winding. Its projection on stator winding
magnetic axis is L
sr
I
r
. 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 the

two axes are perpendicular, the contribution is zero. When the two axes are aligned but 180
degrees apart, contribution is of negative maximum value. The function that describes such
behaviour is a cosine function. Hence the mutual inductance between the stator and the rotor
winding can be written as
s
r
θ
MI
r
L
sr
I
r
Fig. 7 – Magnetic axes of stator and rotor windings.
Table II – Contribution of rotor current to the stator flux linkage
Angle θ [°]
0 90 180 270 360
L
sr
I
r
MI
r
0
−MI
r
0MI
r
θ
cosML

sr
= (36)
The mutual inductance between a stationary and a rotating winding is therefore always
a function of the rotor position. Simple sine or cosine functional dependence suffices for
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14
machines with uniform air-gap. If the machine is with salient poles on rotor, the expression
for the mutual inductance becomes more complicated.
As the next step, let us consider the induced emfs in the two windings of the machine
in Fig. 7. By definition
dtde
dtde
rr
ss
ψ
ψ
=−
=−
(37)
Substitution of (35) into (37) yields to
ω
θθ
ω
θθ
θ
θθ
+++=−
+++=−
+++=−

+++=−
d
dL
i
d
dL
i
dt
di
L
dt
di
Le
d
dL
i
d
dL
i
dt
di
L
dt
di
Le
dt
d
d
dL
i

d
dL
i
dt
di
L
dt
di
Le
dt
dL
i
dt
dL
i
dt
di
L
dt
di
Le
sr
s
r
r
s
sr
r
rr
sr

r
s
s
r
sr
s
ss
sr
r
s
s
r
sr
s
ss
sr
r
s
s
r
sr
s
ss
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
ω
θ

ω
θ
++=−
++=−
d
dL
i
dt
di
L
dt
di
Le
d
dL
i
dt
di
L
dt
di
Le
sr
s
s
sr
r
rr
sr
r

r
sr
s
ss
(39a)
Assuming further that rotor current is constant (this corresponds to a synchronous machine)
further simplifies this expression to
ω
θ
ω
θ
+=−+=−
d
dL
i
dt
di
Le
d
dL
i
dt
di
Le
sr
s
s
srr
sr
r

s
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 stator
winding and the rotor winding = 0.4 cos
θ
[H]. The rotor revolves at constant angular
velocity of 40 rad/s and the initial value of the mechanical co-ordinate at zero time
instant equals zero. Determine the instantaneous value of induced electromotive force
in open-circuited rotor winding if the current that flows through the stator winding is
equal 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
ω
θ
+=−
d
dL
i
dt
di
Le
sr
s
s
srr
Derivatives of the stator current and the mutual inductance are

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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
+=
−=+=
−++=
+=
+=
βαβα
βαβα
Example 2:
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].
a) Rotor rotates at 200 rad/s and stator current is known to be 10sin200t. Find the
induced 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. Find
the speeds at which average torque will exist; find the average torque values and
determine the values of the load angle which yield maximum values of the average
torque.
Solution:
a) Rotor current and its derivative are again zero. Hence once more
ω
θ
+=−
d
dL
i
dt
di

Le
sr
s
s
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-zero
values:
()
=
==
±=
]rad/s[400
][rad/s0
]/[200
ω
ωω
ω
ω
ω
srad
rs
rs
Note that both self-inductances are constant. Hence the torque contains only the fundamental torque
component. Instantaneous torque is
() ()
[]
()
)sin()400cos1(5.2sin200sin5
sin05.0200sin10
2
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(5.2
1
max
0
πδ
δδ
==
=−−==
e
T
ee
T
dtt
T
T
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 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
πδ
δ
δδδ
δδ
δ
−==
=
+=
+=
−−−=
=
e
e
e
e
e
T
ee
T

T
t
t
ttt
dtt
T
T
In this example both stator and rotor winding are supplied with single-phase currents. As the result
shows, the average torque will exist at 400 [rad/s] speed of rotation. However, apart from the average
torque, there will exist two more time varying components. This is true for any single-phase AC
machine. It is not possible to realize conversion with constant time independent torque (i.e. it is not
possible to achieve equality of the average and instantaneous torque values). It is for this reason that
vast majority of AC machines in use nowadays are three-phase. It will be shown in the next section how
utilization of a three-phase machine instead of a single-phase machine enables achievement of a
constant time independent torque.
Note as well that the angle
δ
, taken initially as the initial value of the mechanical co-ordinate, is actually
much more than that. It is called load angle and its value depends on the loading of the machine. In this
example machine’s average torque can be anything between zero and 1.25 [Nm] at 400 [rad/s]. How
much it will be depends on the load torque. The load angle value will adjust itself in such a way that the
motor 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 on
stator and one winding on rotor. The two stator windings are displaced in space by 90
degrees. The winding self-inductances and mutual inductances are equal to:
δωθ
θθ
−==
=

==
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-phase
system 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 the
machine under the assumption that the condition of average torque existence is
satisfied. Calculate the average torque for load angle equal to 30 degrees and explain
its nature.
c) Sketch the dependence of average torque on load angle δ, identify motoring and
generating 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 the

specific two-phase winding structure on stator it is possible to realize electromechanical energy
conversion with the time independent constant torque developed by the machine. Two-phase structure
is realized by displacing in space two stator windings by 90 degrees and by feeding the two-phase
winding with currents having phase displacement of 90 degrees. This example serves as an introduction
into the explanation of why three-phase machines are used nowadays and how time independent torque
is obtained, elaborated in the next section.
a) Rotor frequency is zero. According to the condition of average torque existence, stator angular
frequency must in this case equal angular frequency of rotation. The machine is therefore a synchronous
machine with cylindrical stator and rotor structure. The cross section is shown in figure.
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17
S1 axis
r
θ
S1
r
S2
S2 axis
b) Self-inductances of all the three windings are constants. Mutual inductance between two stator
windings is zero. The instantaneous torque therefore contains only fundamental torque components and
is given with
()
[]
[]
δ
ωω
δωω
δωωδωω
θωθω

θθ
θθ
θ
θ
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 of
30 degrees the average torque is 45 [Nm].
c) Average torque is a sine function of the load angle. Positive values correspond to motoring, while
negative values describe generation. The machine can operate stably in the load angle region from -π/2
to π/2. The region from -π/2 to zero is the generating region. The region from zero to π/2 is the
motoring 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 in
slots with 120 degrees spatial displacement between any two respective phases. The winding
is supplied with a system of three-phase currents, such that there is a phase displacement of
120 degrees between any two respective stator phases. Let the three phase currents be given
with:
() ()
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):
()
()
FNI t

FNI t
FNI t
am
bm
cm
=
=−
=−
cos
cos /
cos /
ω
ωπ
ωπ
23
43
(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
ENGNG 2024 Electrical Engineering
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18
three phase m.m.f.’s in the given instant of time correspond to those met in any three phase
system.
a
c
b
F
a

F
b
F
c
b
c
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 currents
flowing through spatially displaced windings is the sum of the individual contributions of the
three phases. The summation is done in the cross-section of the machine, and it is necessary to
observe the spatial displacement between the three m.m.f.’s. One may regard the cross-section
of the machine as a Cartesian co-ordinate system in which phase a magnetic axis corresponds
to x-axis, while y-axis is perpendicular to it. For the purposes of calculation this co-ordinate
system 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 displacement
of the m.m.f. by 120 degrees corresponds to the so-called ‘vector rotator’,
()
aj= exp /23
π
.
Hence the resultant magneto-motive force can be written as
()
()()
()
FFFF e e
FNI t t t
res a b c
jj
res m

=++ = =
=+−+−
aa a a
aa
2
2
3
2
4
3
2
23 43
π
π
ωωπ ωπ
,
cos cos / cos /
(42)
The expression for the resultant magneto-motive force is most easily found if one recalls the
well-known correlation
()
cos . exp( ) exp( )
δδδ
=+−05 jj. Hence
()() () ()
( )
()
()
FNIee ae ae ae ae
FNIeeaaeaaeaaeaae

aa
aa a a a aa
FNIe aaaae
res
jt jt jt jt jt jt
res m
jt jt jt jt jt jt
res m
jt jt
=+++ + + =
++ + + +
++ =
====
=++++
−−−− −−−
−− −
−
1
2
1
2
10
1
1
2
11
23 23
2
43
2

43
22 22
2
22 3 4
22
ω ω ωπ ωπ ωπ ωπ
ωω ω ω ω ω
ωω
/// /
**
**
=
()
()
() ()
()
aa
FNIe e
FNIe
res m
jt jt
res m
jt
2
1
2
30
3
2
+=

=+
=
−
ωω
ω
(43)
Symbol * in (43) stands for complex conjugate. The result obtained in (5) is an important one.
The equation
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19
FNIe
res m
jt
=
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 pulsating
m.m.f.’s) is a time-independent and rotating magneto-motive force. Thus it follows that the
three-phase system creates a rotating field (called also Tesla’s field) in the machine. Figure 9
illustrates the rotating field in different instants of time. The speed at which the rotating field
travels equals the angular frequency of the stator three-phase supply.
Im
ω
t=90
°
ω

t=135° F
res
ωt=45°
ω
t=0
°
Re (a)
1.5NI
m
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 a
field is produced by this current. This field is stationary with respect to rotor. Since the rotor
rotates at synchronous speed, then, looking in from stationary stator, this rotor field rotates at
synchronous speed. This is always the case in any multi-phase AC machine: regardless of
whether the rotor rotates synchronously or asynchronously, all the fields in the machine rotate
at synchronous speed.
Since the resulting m.m.f. is responsible for the resulting flux density and ultimately
resulting flux, this means that apart from the rotating m.m.f., there is a rotating flux density
wave and a rotating flux in the machine as well. The term rotating field in general denotes any
of 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 between
stator windings and rotor winding and three phase stator currents are given with:
θ
cosML
ar
= )3/2cos(
πθ
−= ML

br
)3/4cos(
πθ
−= ML
cr
tIi
sma
ω
cos= )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 stator
winding. Hence the torque is:
ENGNG 2024 Electrical Engineering
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20
[]
δ
ωω
δωθ
θω
θωωθθω

ωθθωωθ
θωθωθω
θθθ
sin5.1
)sin(5.1
)]sin()120sin()sin(
)240sin()sin()[sin(5.0
)240sin()240cos()120sin()120cos(sincos
rmee
s
smre
sss
sssmre
sssrme
cr
rc
br
rb
ar
rae
IMITt
t
tIMIt
ttt
tttIMIt
tttIMIt
d
dL
Ii
d

dL
Ii
d
dL
Iit
=≡
=
−=
−=
−−−++−−
−−++−−+−=
−−−−−−−=
++=
Note that the three terms with 0, 120 and 240 degrees mutual phase displacement in the instantaneous
torque expression sum to zero in any instant in time. This is a definition of a three-phase system. This
leaves only the three terms that are mutually the same and each yield 1/3 of the total torque. Torque is
time independent, this being the consequence of the three-phase winding existence on stator. The
machine is a synchronous machine with excitation winding on rotor.
Example 5:
A three-phase synchronous machine is operated with constant DC current in the rotor
winding and it runs as a generator at constant speed of rotation
ω
.Statorwindingsare
open circuited. Determine induced emfs in the stator windings. Mutual inductances
between stator and rotor are
θ
cosML
ar
= , )3/2cos(
πθ

−= ML
br
, )3/4cos(
πθ
−= ML
cr
.
Solution:
Since rotor current is constant and since the stator currents are zero, induced emfs in the stator windings
are:
)3/4sin()()3/4sin(
)3/2sin()()3/2sin(
sin)(sin
πωωπθωω
θ
πωωπθωω
θ
ωωθωω
θ
−=−=−=
−=−=−=
==−=
tMIMI
d
dL
Ie
tMIMI
d
dL
Ie

tMIMI
d
dL
Ie
rr
cr
rc
rr
br
rb
rr
ar
ra
It follows that a system of three-phase voltages is generated at open-circuited stator terminals. This
example shows the basic principle of three-phase electricity generation using synchronous three-phase
generators.
4. TUTORIAL QUESTIONS
Q1. An electromechanical converter has a three winding structure. Winding resistances,
self-inductances and mutual inductances are known to be
RRR L L L L L L
123 11
2
2
3
31223
3
1
,,; ,
,
;,

,
and , respectively. The windings are fed from voltage
sources of known voltages vv
v
12
3
,
,
and rotor inertia and friction coefficient are J and
k. The machine runs at certain speed
ω
. Write the time domain matrix equations and
equations in developed form for the following: i) voltage equilibrium and induced
electromotive forces; ii) mechanical equilibrium; iii) power balance; iv) converted
power 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
(defineallthesymbolsused).
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
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21
current is known to be 10 sin 300t, find the induced electromotive force in rotor
winding if it is open-circuited (the initial value of mechanical co-ordinate at zero time
is zero).
Q3. a) State the complete time-domain mathematical model of a generalised

electromechanical converter with n windings and define all the matrices and vectors of
the model.
b) State the general condition of average torque existence in a two-winding structure
and classify the electrical machines with respect to the way in which this condition is
satisfied.
c) An electromechanical converter has a three winding structure, with two windings on
stator and one winding on rotor. The two stator windings are displaced in space by 90
degrees. The winding self-inductances and mutual inductances are equal to:
δωθ
θθ
−==
=
==
tL
LL
LLL
SS
rSrS
rSS
where0
sin8.0cos.80=
H0.9=H1.1H1.1
21
21
21
Rotor winding current is constant DC, of 20 A. Stator windings are fed with two-phase
system of currents, such that
i
t
i

t
12
10 10==cos
s
in
ω
ω
Sketch 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 and
generating part and define the region of stable operation.
Q4. An electromechanical converter has two windings only, both mounted on stator and
displaced in space by 90 degrees. The winding self-inductances and mutual inductance
can be given with
.where
2sin
2
2cos
22
2cos
22
12
22
11
δωθ
θ
θ
θ

−=
−
=
−
−
+
=
−
+
+
=
t
LL
L
LLLL
L
LLLL
L
qd
qdqd
qdqd
Stator windings are fed with two-phase currents i
s1
= I
m
cosω
s
tandi
s2
= I

m
sinω
s
t.
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 the
expression for average torque assuming that condition of average torque existence is
satisfied.
c) Calculate the average torque and explain the origin of the torque in this machine if
LLI
dqm
=== =215Aand30H, H,
δ
.
Q5. a) State the complete time-domain mathematical model of a generalised
electromechanical converter with n windings and define all the matrices and vectors of
the model.
b) Give graphical representation of the power flow in an electric machine for motoring
and generation and define all the powers in this representation in terms of terminal
quantities and parameters (use matrix form).
ENGNG 2024 Electrical Engineering
 E Levi, 2002
22
c) State the general condition of average torque existence in a two-winding structure
and classify the electrical machines with respect to the way in which this condition is
satisfied.
d) An electromechanical converter has a three winding structure, with two windings
on stator and one winding on rotor. The two stator windings are displaced in space by
90 degrees. The winding self-inductances and mutual inductances are equal to:
δωθ

θθ
−==
=
==
tL
LL
LLL
ss
rsrs
rss
where0
sin95.0cos95.0=
H1=H25.1H25.1
21
21
21
Rotor winding current is constant DC, of 15 A. Stator windings are fed with two-phase
system of currents, such that
i
t
i
t
12
10 10==cos
s
in
ω
ω
Sketch 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 and
generating 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 the
model.
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 stator
current is known to be 15 sin 250t, find the induced electromotive force in rotor
winding if it is open-circuited (the initial value of mechanical co-ordinate at zero time
is zero).
c) The same system of part b) is again considered. However, current 15 sin 250t
flows now through both windings that are connected in series. Find the speeds at
which average torque will exist, determine corresponding average torque values and
the values of load angle, which yield maximum average torque values.