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5
speed is synchronous and no electromagnetic force can be generated. Therefore at synchronous
speed the developed torque in an induction machine equals zero. As certain amount of torque
is always necessary in a machine that operates as a motor in order to cover mechanical losses,
induction motor has to operate with certain amount of developed torque even when it is not
loaded at the shaft. Thus the rotor in motoring regime can never attain synchronous speed, i.e.
it can never catch with the revolving field. When the motor is unloaded, it runs under no-load
conditions and the amount of torque that is needed is determined with mechanical losses
(windage losses and friction losses in bearings). The torque that describes mechanical losses is
small, thus indicating that the induction motor will have the highest possible speed when it runs
unloaded; this is so-called no-load speed and it is only slightly smaller from synchronous speed.
Let us summarise the above given explanations: connection of three-phase stator
winding of an induction machine at standstill to a voltage source causes current flow in stator
windings; these currents give rise to production of revolving field; revolving filed cuts
conductors of both stator and rotor windings; emf is induced in stator and it provides voltage
balance to supply voltage; emf is induced in rotor as well and it causes current flow through
short-circuited rotor winding; an electromagnetic force is created which acts on every rotor
conductor, leading to the creation of the electromagnetic torque which pulls rotor into
rotation; the direction of rotation is the same as the direction of rotation of stator revolving
field; when a steady-state is established, rotor rotates with angular velocity equal to
rs
ωωω
−= ; rotor currents create another revolving field whose absolute speed equals
synchronous speed. Therefore, the torque is consequence of mutual interaction between stator
and rotor revolving fields. At synchronous speed rotor currents become zero and
electromagnetic torque disappears.
Windings are by the virtue of their construction of resistive-inductive nature. Reactive
power has to be provided for magnetisation of iron cores and air gap between stator and rotor.
The question is how this reactive power is provided in induction machines. The machine does
not contain any capacitances that could produce reactive energy. The only electrical
connection with outside world is the connection of the stator winding to the supply, as the
rotor winding is short-circuited. This means that there is no source of reactive power available
inside an induction machine. Therefore induction machine has always to absorb reactive energy
from the supply. Under all the possible operating conditions induction machine will act as a
reactive energy consumer. As there is no rotor winding connected to another electric source,
as is the case in synchronous machines, there is no way of exciting the induction machine in a
manner similar to synchronous machines. This is one of the main reasons why induction
machine is mainly utilised in motoring regime, while synchronous machine is used for
generation purposes. When an induction machine is applied as a generator, reactive power has
to be either taken from the power system or to be provided by a static VAr compensator (e.g.,
capacitor bank).
As already emphasised, during motoring induction machine has to rotate slower than
the revolving field, even under no-load conditions. The angular velocity of the rotor is given
with
rs
ωωω
−= . Revolving fields of stator and rotor rotate with angular velocity
ω
s
.The
difference between rotor speed and synchronous speed is characterised with the so-called slip.
The slip is expressed either as a percentage value of the synchronous speed or as a per unit
non-dimensional quantity. It is usually calculated out of the speeds given in [rpm] in the
following way:
[%]100or[p.u]
s
s
s
s
n
nn
s
n
nn
s
−
=
−
= (1)
where:
ENGNG2024 Electrical Engineering
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n
s
- synchronous mechanical speed, which is a function of the number of magnetic pole
pairs P and which is correlated with synchronous electrical speed 60f
s
as
P
f
n
s
s
60
=
(2)
For 50 Hz supply synchronous speeds are
P1234
n
s
[rpm] 3000 1500 1000 750
n - asynchronous mechanical speed of rotation of induction machine shaft.
Note that definition of the slip and the values are the same regardless of whether speeds in
[rpm] or angular speeds in [rad/s] are used.
Slip during normal operation of induction machines is in the range 10% to 2% for
induction machines with power ratings in the range 1 kW 100 kW. The value of the slip that
corresponds to the rated operating conditions, when speed is n
n
, will be denoted as s
n
.Indexn
will in general always define the rated (nominal) operating condition of the machine.
Let us now investigate correlation between stator and rotor frequencies with respect to
newly introduced notion of slip. From slip definition of (1) it follows that
()()
rs
ss
ss
ss
s
Psnnn
nnsn
ωωω
ωωω
π
+=
+=
+=
−=
timesameat theSince
1260bydivided
(3)
it follows that the rotor frequency is determined with
srsr
sffs ==
ωω
(4)
Example 1:
A 4-pole, 3-phase induction machine is fed from 50 Hz supply and operates in steady
state with slip equal to 0.03. Determine the rotor speed and frequency of rotor
currents.
Solution:
[Hz]5.15003.0
[rpm]14551500)03.01(1
[rpm]15002/5060
60
===
=−==
===
xsff
x-s)n(n
x
P
f
n
sr
s
s
s
Example 2:
A 60 Hz induction motor has one pole pair and runs at 3150 rpm. Calculate the
synchronous speed and slip in per unit and in percents.
Solution:
Note the this is an American machine, since the frequency is 60 Hz. Hence
%5.2100]p.u.[[%]
025.03600/)31503600(/)(]p.u.[
[rpm]36001/6060
60
==
=−=−=
===
xss
nnns
x
P
f
n
ss
s
s
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7
According to (4), frequency of the current in rotor is slip times frequency of stator
currents. For 50 Hz stator frequency and operating slips of 10% to 2% in an induction machine
rotor frequency is only 5….1 Hz. Consequently, as the losses in the iron core are proportional
to frequency and frequency squared, it follows that rotor iron losses are going to be negligibly
small and that the major part of the iron loss will take place in stator. The total iron loss in an
induction machine is for this reason always assumed to take place in stator only.
According to the slip definition, equation (1), slip is a variable determined with the
speed of rotation. This implies that frequency of rotor currents is, according to (4), a variable
as well, proportional to the slip. Characteristic slip values in motoring operation are:
n = 0 rotor at standstill s = 1
0<n<n
0
rotor rotates, machine is loaded 1 > s > s
0
n=n
0
<n
s
no-load, machine is unloaded s = s
0
n=n
n
<n
0
rotor rotates, rated load s = s
n
Normal operating range of induction machines in steady-states is in the speed range between
rated speed and no-load speed, the actual operating speed being dependent on the load torque
that the motor is driving.
Suppose now that a source of mechanical energy is connected to the induction machine
shaft and that the mechanical power provided by mechanical source is exactly equal to the
power which describes mechanical losses (i.e. mechanical source provides torque to overcome
mechanical loss torque). Then the speed of rotation will become equal to synchronous, as the
mechanical loss torque is equated by torque of the prime mover. Simultaneously the induction
machine torque will become equal to zero. Therefore at synchronous speed
00 ===
es
Tsnn
Suppose now that the power provided by the prime mover increases. The induction machine
then enters generating regime. Note that only real power will be generated, while the reactive
power is still absorbed. For generation
00 <<>
es
Tsnn
This means that in generation speed is above synchronous speed, slip is negative and the
machine’s electromagnetic torque is negative as well. In contrast to this, in motoring slip and
torque are positive since the reason for rotation is the machine’s electromagnetic torque. A
representation of induction machine operating modes is given in Fig. 5.
Motoring Generating
10-0.5s
0n
s
1.5n
s
n
Fig. 5 – Schematic representation of induction machine operating modes, in terms of slip and
speed of rotation.
Given the slip s in a steady state, speed can be determined from equation (1) as
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[]
()
[]
()
[rpm]100/%1
[rpm]p.u.1
s
s
nsn
nsn
−=
−=
(5)
Taking into account that during motoring slip is within the range from 1 to 0, it is obvious that
frequency of rotor currents varies as a function of slip in the range between stator frequency
and zero. At standstill rotor frequency equals stator frequency, while at synchronous speed
rotor frequency becomes equal to zero. Therefore it follows that frequency of rotor currents
and voltages is determined with slip.
3. ANALYTICAL THEORY OF INDUCTION MACHINE OPERATION
It can be shown that a balanced three-phase induction machine fed from symmetrical
sinusoidal three-phase supply can be treated in terms of per-phase equivalent circuit in steady
state. However, such a derivation is pretty involved and time consuming. As only steady states
under symmetrical supply conditions are of interest here, complex representatives of AC
sinusoidal quantities may be used (phasors). Furthermore, for the purpose of steady-state
analysis of a balanced induction machine fed from symmetrical source, the whole analysis can
be performed by utilising per phase representation with complex phasors. Such an approach is
utilised in what follows.
Let stator winding be connected to mains, which provide symmetrical three-phase
voltages and let rotor be at standstill, so that rotor speed is zero. Frequency of stator voltages
and currents is f
s
. An electromotive force will be induced in rotor winding that will cause
current to circulate around the rotor winding. As the rotor is at standstill, slip equals one and
the frequency in rotor winding equals stator frequency. When the rotor is at standstill,
difference between the speed of the rotating field and the rotor speed is of maximum value and
equals synchronous speed. This speed difference determines the induced emf, since the emf is
directly proportional to the speed at which the conductors are cut by the field (i.e. to the
difference between the synchronous speed and the rotor speed). Once when rotor rotates at
certain speed, the speed at which conductors are cut by the rotating field will be determined
with
ssr
sf
πωωω
2=−= and will be smaller than at standstill.
When the rotor is at standstill let the induced electromotive force in one rotor phase E
is identified with index rl. Its existence will cause current flow and rotor currents produce
corresponding revolving field and flux. One part of the flux dissipates around the rotor winding
(leakage flux), while major part links with stator windings contributing to the mutual flux.
Current flow through rotor winding, caused by induced emf, is opposed by the resistance of
the rotor winding and leakage reactance (which describes leakage flux). The value of the rotor
leakage reactance is again frequency dependent. At standstill rotor frequency, rotor leakage
reactance and modulus of rotor current are
22
22
rlr
rl
rl
rsrrrl
ssr
XR
E
I
LfLfX
fsff
+
=
==
==
γγγ
ππ
(6)
Suppose now that the rotor starts rotating, so that slip becomes smaller than 1 since
speed is greater than one. According to the fundamental expression for induced electromotive
force due to the relative movement of a conductor with respect to flux density, induced emf is
proportional to the relative speed of conductor with respect to flux density. As the rotor
rotates with certain speed, while revolving fields rotate with synchronous speed, relative speed
ENGNG2024 Electrical Engineering
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9
of the revolving field with respect to the rotor conductors is equal to the rotor angular
frequency. This means that induced electromotive force at slip s is proportional to frequency of
rotor currents. Consequently, at any other speed different from zero,
()
2
2
22222
1
1
rr
rl
r
rlr
rl
rr
r
r
rlr
rlr
rlr
XsR
E
I
XsR
sE
XR
E
I
sXX
sEE
ssEE
γ
γγ
γγ
+
=
+
=
+
=
=
==
≠=
(7)
The expression for modulus of the rotor current in (7) enables the following phasor equation
(phasors are identified with a bar over the symbol) to be written:
r
rl
r
r
rl
rlr
rl
r
r
r
IjXI
s
R
E
EsIjsXIRE
γ
γ
+=−
−=+=−
(8)
Equation (8) enables construction of the rotor per-phase equivalent circuit, shown in Fig. 6.
Let us consider now voltage balance for one stator phase winding. Stator phase
winding is characterised with resistance and stator leakage reactance. Note that for stator
rotating field always cuts the conductors at the same, synchronous speed. Hence the frequency
of the stator is constant (50 Hz) and the induced emf is proportional to this fixed frequency
regardless of the speed of rotation of the rotor. The induced emf exists in each stator phase and
it holds balance to the applied stator voltage. Following the same approach as for the rotor
phase, one can immediately write the phasor voltage equation for one stator phase as:
ss
s
s
s
EIjXIRV ++=
γ
(9)
Corresponding equivalent circuit for one stator phase is shown in Fig. 7.
jX
γ
rl
I
r
E
rl
R
r
/s
Fig. 6 - Rotor per-phase equivalent circuit
R
s
jX
γ
s
I
s
VE
s
Fig. 7 – Stator per-phase equivalent circuit.
By combining Figures 6 and 7, resulting complete equivalent circuit can be constructed.
It is shown in Fig. 8 and is described with the following two voltage phasor equations:
ss
s
s
s
EIjXIRV ++=
γ
(10)
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r
rl
r
r
rl
IjXI
s
R
E
γ
+=− (11)
R
s
jX
γ
s
I
s
VE
s
jX
γ
rl
I
r
E
rl
R
r
/s
Fig. 8 – Induction machine per-phase equivalent circuit.
Two circuits shown in Fig. 8 apply to two different voltage levels, since the induced
emf in stator is in general different from the induced emf in rotor. It is therefore not possible to
directly connect them. In order to be able to put the two circuits together, it is necessary to
apply transformer theory, which means that rotor voltage needs to be referred to stator voltage
level (as the secondary is referred to primary using the transformation ratio in transformers).
Correlation between the stator and rotor induced emf is established at standstill through the
transformation ratio
m. Transformation ratio is defined as
rls
rl
EEEEm
s
// == (12)
and in an induction machine it is dependent on design features. Rotor voltage equation (11) is
multiplied next with the transformation ratio
r
rl
r
r
rl
r
rl
r
r
rl
IjmXI
s
mR
Em
IjXI
s
R
E
γ
γ
+=−
•+=− m/
(13)
and new fictitious rotor variables, referred to stator, are then introduced respecting the
condition that power in terms of original variables and in terms of new (primed) variables has
to be the same:
mII
IEIE
EmE
rr
rrrr
rlrl
=
=
=
'
''
'
(14)
Rotor current and induced emf in (13) are replaced next with the new rotor current and new
induced emf
'''
2
2
r
rl
r
r
rl
IXjmI
s
Rm
E
γ
+=− (15)
Finally, new rotor parameters (primed symbols) are introduced and the equation is brought into
final form of
rlrlrr
r
rl
r
r
rl
XmXRmR
IjXI
s
R
E
γγ
γ
22
''
'''
'
'
==
+=−
(16)
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The left-hand side of (16) is now, by definition in (12), equal to the stator induced emf. It is
therefore possible to connect the two circuits into a single circuit, shown in Fig. 9, which is
described with the following two voltage equations:
()
'''
r
rlr
s
ss
s
s
s
IjXsRE
EIjXIRV
γ
γ
+=−
++=
(17)
R
s
jX
γ
s
jX
γ
rl
’
I
s
I
r
’
VE
s
= E
rl
’R
r
’/s
Fig. 9 – Connection of the stator and rotor per-phase equivalent circuit into a single electric
circuit after referring rotor to stator.
Note that the phasors are identified in all the equations with a bar over the symbol, while in
Figs. 6-9 (and subsequent figures as well) phasors are denoted with a bold symbol.
What remains to be done in Fig. 9 is to express the induced emf in terms of an
appropriate impedance. This can be done in the same manner as in transformer theory, by
utilising the notions of the magnetising current and magnetising reactance. Phasor sum of the
stator and referred rotor current is defined as the magnetising current, and induced emf is
expressed as a voltage across the magnetising reactance caused by the flow of the magnetising
current:
'
rsm
m
m
s
III
IjXE
+=
=
(18)
The final equivalent circuit, which suffices for most of the calculations, results in this way. It is
illustrated in Fig. 10.
R
s
jX
γ
s
jX
γ
rl
’
I
s
I
r
’
V
jX
m
R
r
’/s
I
m
Fig. 10 – Per-phase equivalent circuit of an induction machine, with rotor winding referred to
stator.
In all the considerations so far iron loss was neglected. As already noted, it occurs
predominantly in stator and its existence can be accounted for by adding so-called equivalent
iron loss resistance into the circuit, in parallel to the magnetising reactance. If iron loss needs
to be taken into consideration, then the equivalent circuit of Fig. 10 becomes as in Fig. 11.
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R
s
jX
γ
s
jX
γ
rl
’
I
s
I
r
’
VI
Fe
jX
m
R
r
’/s
R
m
I
m
Fig. 11 – Equivalent per-phase circuit of an induction machine with included iron loss
representation.
Note that the addition of the equivalent iron loss resistance in Fig. 11 changes the node current
balance equation. From Fig. 11
'
rsmFe
IIII +=+
(19)
Note as well that the fact that rotor parameters in the circuit of Fig. 11 are referred to stator is
irrelevant, since in a cage induction machine it is anyway not possible to determine actual rotor
parameter values. Since tests used to calculate parameters are performed by feeding the
machine from stator and by doing measurements at the stator side, parameter values obtained
by means of experiments (described in the next section) are anyway rotor parameters referred
to stator (i.e. as seen from the stator side).
Example 3:
A squirrel cage three-phase induction motor has the following parameters:
R0.39 R XX0.35 X=16
sr
'
sr
'
m
====ΩΩ ΩΩ014.
γγ
The motor is four-pole, three-phase, with star connected stator winding, rated
frequency is 60 Hz and rated voltage is 220 V. Rated speed is 1746 rpm. Mechanical
and iron loss may be neglected.
Calculate slip, power factor, stator current and input power of the motor when it
operates under rated conditions.
Solution:
Note that since the rated speed is 1746 rpm and the motor is four-pole, the operating frequency must
be 60 Hz, since rated slip has to be positive and of the order of a couple of percents.
03.01800/)17461800(/)(
[rpm]1746
[rpm]18002/6060/60
=−=−=
=
===
snsn
n
ss
nnns
n
xPfn
The motor is star connected. Hence the phase voltage is 220/√3=127V.
In order to find the stator current, it is necessary to solve the equivalent circuit of Fig. 10, in which the
magnetising reactance is included. From the circuit of Fig. 10 one has:
Ω+=+++=
+
++=
Ω=
Ω+=+=
)87.1525.4()52.1135.4()35.039.0(
16
)35.067.4(''
jjj
ZZ
ZZ
jXRZ
jZ
jjXsRZ
rm
rm
ss
in
m
rlnr
r
γ
γ
The modulus and the phase of the input impedance are
48.22)897.4/525.4(cos}/]{Re[cos
897.487.1525.4
11
22
===
Ω=+=
−−
in
in
n
in
ZZ
Z
φ
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Stator rated current and power factor are therefore
924.0cos
A269.4/127/
)(
=
===
n
inphnsn
ZVI
φ
The input power is
W91310924261273cos3
)(
=== xxxIVP
nsnphnin
φ
Example 4:
A squirrel cage three-phase induction motor has the following parameters:
0.35XX14.0R0.39R
'
rs
'
rs
Ω==Ω=Ω=
γγ
The motor is four-pole, three-phase, with star connected stator winding, rated
frequency is 60 Hz and rated voltage is 220 V. Rated speed is 1746 rpm. Mechanical
and iron loss may be neglected. Magnetising reactance may be regarded as of infinite
value.
Calculate slip, power factor, stator current and input power of the motor when it
operates under rated conditions.
Solution:
Note that all the data in this example are the same as in the previous one. The only difference is that
the magnetising reactance is now omitted from the equivalent circuit. Such an approximation enables
much simpler calculations. However, it is unrealistic for normal operating slip values, since it yields
an unrealistically high value of the power factor. Rated slip value is calculated as in the previous
example,
03.01800/)17461800(/)(
[rpm]1746
[rpm]18002/6060/60
=−=−=
=
===
snsn
n
ss
nnns
n
xPfn
and the rated phase voltage is again 127 V. Impedance calculation is however now much simpler,
W939899.09.241273
99.0108.5/06.5cos
A9.24108.5/127
108.57.006.5
)7.006.5()35.067.4()35.039.0(
)35.067.4(''
22
==
==
==
Ω=+=
Ω+=+++=++=
Ω+=+=
xxxP
I
Z
jjjZjXRZ
jjXsRZ
in
sn
in
r
ss
in
rlnr
r
φ
γ
γ
Comparing the results obtained with and without the magnetising branch, one can see that the error in
the stator rms current is rather small (26 A against 24.9). Similar conclusion applies to the real input
power (9131 W against 9398 W). However, the error in the power factor is significant (0.924 against
0.99) since when the magnetising branch is neglected reactive power taken by the machine is almost
zero (only reactive power for the leakage reactances exists now, in contrast to the real case when most
of the reactive power appears across the magnetising reactance).
4. NO-LOAD AND LOCKED-ROTOR REGIMES OF AN INDUCTION MACHINE
No-load and mechanical short-circuit (locked-rotor, rotor at standstill) are two
important regimes of an induction machine which occur during normal operation and which are
performed as standard tests on induction machines as well. When performed as tests, these two
tests enable calculation of the parameters of the equivalent circuit, determination of mechanical
losses and calculation of iron core losses. Here no-load state and rotor at standstill condition
are discussed only as regimes that appear in normal operation of the motor.
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Under no-load conditions the induction motor is connected during normal operation to
rated voltage supply and there is no mechanical load connected to the shaft. Therefore
000 ===
η
outL
PT (20)
and motor torque is just sufficient to overcome the mechanical losses. In this operating state
the motor draws from the mains reactive power for magnetisation of the magnetic circuit
(which is basically independent of the operating regime of the machine and can be regarded as
being the same, for the same voltage, for any load) and real power to cover its internal no-load
losses. The speed of rotation is very close to the synchronous speed, i.e. slip differs only
slightly from zero. No-load losses consist of mechanical losses, iron core losses and stator
winding copper no-load losses:
mechlossFeCusin
PPPPP
−
++==
00
(21)
Stator copper losses are determined with the no-load current that flows through stator
windings. Iron core losses exist practically only in the stator as rotor frequency is almost zero.
Iron losses, being dependent on frequency and flux density, are independent of mechanical load
and depend only on applied voltage and stator frequency. In other words, for rated voltage and
frequency, iron losses are the same for any load as under no-load conditions.
Since slip in no-load operation is practically zero, the resistance
sR
r
'intherotor
branch of the circuit in Fig. 11 tends to infinity. This means that there are hardly any rotor
currents flowing (rotor current is only of such a small value that is required to produce motor
electromagnetic torque to overcome mechanical losses). Hence the rotor circuit can be open-
circuited and the equivalent circuit for no-load operation becomes as shown in Fig. 12.
R
s
jX
γ
s
I
s0
I
r0
’=0
V
n
I
Fe
jX
m
R
r
’/s = infinity
R
m
I
m0
Fig. 12 – Equivalent circuit for no-load operation.
The equivalent circuit of Fig 12 neglects mechanical losses. Therefore, as far as the
equivalent circuit is concerned, no-load is characterised with slip equal to zero.
Power factor for no-load operation is
0
0
0
3
cos
sn
IV
P
=
φ
(22)
and its value is very small (typically 0.1 to 0.2), as the real power drawn from the supply is
small and just equal to losses which exist under no-load conditions (there is no output power
as there is no mechanical load connected to the machine). At the same time, the reactive power
drawn from the mains is basically the same as it is when machine delivers its mechanical
output (real) power equal to rated. No-load current in induction machines is in the range from
30% to 80% of the rated current. This is significantly higher than for transformers, the reason
being the existence of the air gap that requires large reactive power for magnetisation. Typical
value of mechanical losses is one third of rated iron core losses.
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