ENGNG 2024 Electrical Engineering
E Levi, 2002
5
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
ENGNG 2024 Electrical Engineering
E Levi, 2002
6
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
ENGNG 2024 Electrical Engineering
E Levi, 2002
7
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):
ENGNG 2024 Electrical Engineering
E Levi, 2002
8
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
ENGNG 2024 Electrical Engineering
E Levi, 2002
9
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.
ENGNG 2024 Electrical Engineering
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 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
ENGNG 2024 Electrical Engineering
E Levi, 2002
11
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
ENGNG 2024 Electrical Engineering
E Levi, 2002
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
ENGNG 2024 Electrical Engineering
E Levi, 2002
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
Không có nhận xét nào:
Đăng nhận xét