Chapter
10
Induction machines and drives
Table of Contents 10.1 Induction machine basics ...................................................................................................... 2 10.2 Machine model and analysis ................................................................................................. 6 10.3 No‐load and blocked‐rotor tests ......................................................................................... 15 10.4 Induction machine motor drives ......................................................................................... 16 10.5 Proposed exercises .............................................................................................................. 21 10.6 Historical notes .................................................................................................................... 20 10.6.1 Ferraris’ short biography ............................................................................................................. 20 For the teacher
10.2
Chapter 10: Induction machines and drives
10.1 Induction machine basics
The induction machine is also called asynchronous machine. While the latter name in the
authors’ opinion is more correct (since induction is a phenomenon that plays a fundamental role
both in synchronous and induction machines) it has the drawback to be very similar to the one of
synchronous machines, thus creating some communication difficulties.
As the synchronous machine, the induction machine is a bidirectional electromechanical
converter, i.e. a system able to convert electrical energy into mechanical form (when it operates
as a motor) and vice versa (when it operates as a generator).
It is however much more frequently used as a motor than as a generator, for the reasons that
will be discussed later.
As regards its construction the machine stator is substantially identical to the one used for the
synchronous machine, and discussed in chapter 7 (cf. fig. 10.1 a). For the induction machine also,
the windings are normally distributed along the air-gap, although for simplicity’s sake in the
figures of this chapter they are represented concentrated in a single turn. Moreover, for this
machine as well, the windings can have a number of pole pairs p, different from one.
The rotor is constituted by a three-phase system of windings, such as the one reported in
figure 10.1b, but with conductors bars indeed distributed along the air-gap and not concentrated
in single turns. These three windings are connected to each other, or, as it is commonly said are
short-circuited, as per fig. 10.1 b), referred to the case in case of windings’ star connection.
a)
b)
a
b’
c
c’
b

stator
b’
c
a
a’
a
b
c’
b
short
circuit
c
a
b
c
star center
Fig. 10.1. The basic structure of an induction machine stator (left) and rotor (right).
To be more accurate, it must be said that the induction machine rotors can have two possible
constructions: wound and squirrel-cage. The wound rotor is constituted by three coils star- or
delta-connected, whose free ends are short circuited. A wound rotor with a single turn per phase
has the aspect reported in fig. 10.1 b). Normally, in a similar way to what discussed for the
synchronous machine, both stator and rotor coils are distributed along the air-gap, spanned
through several slots; and the different turns of the same coil are connected in series using
connections at the two machine ends.
Induction machines with wound rotors have the advantage of giving access, when needed, to
the rotor windings, mainly for starting the machine, as discussed in the more in depth text of sect.
10.2..
The more frequent rotor construction, however, is the so called squirrel-cage rotor, that is
depicted in fig. 10.2.
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
10.3
A-A view
A-A
Fig. 10.2. The structure of a squirrel-cage rotor:
a) cross section A-A showing iron and bars b) perspective view of only the squirrel cage.
In the right part of the figure a view orthogonal to the rotor axis is reported, while in the left
one a perspective view of just the conductors is shown. The longitudinal conductor bars are
connected to the two front rings. This construction makes the rotor equivalent to a multi-phase
system of windings: each phase is constituted by a single bar, one of the two rings is the centre of
the star connection of the bars, the other is the short circuiting connection.
The squirrel-cage construction gives no access to the rotor circuits, but is much simpler,
more rugged, and cheaper than the wound-rotor one.
Indeed the wound construction tends to be abandoned, as discussed later.
In cases in which frequent starting operations or even continuous variation of speed are
needed (e.g- in the operation of a lift1, or an electric car) a much more flexible operation of an
induction machine can be obtained by varying the frequency of feeding the stator, as discussed, in
sect. 10.3, although this will require the addition of a complex power electronic converter
between the machine and the power supply.
To understand the operation principle of an induction machine, imagine to connect a
machine, whose rotor is initially standstill, to an external three-phase source of power, that, for
the purpose of this reasoning, can be considered as being constituted by an ideal three-phase set
of ideal sinusoidal sources (fig. 10.3).
Ua
Ia
+

Ub
+
Ib
Uc
+
Ic
Asynchr.
machine
3
T, 
Uc
Ic
Ib
Ub
Ua
Ia
Fig. 10.3. An induction machine, neglecting transients, absorbs a three-phase set of currents, because of
the symmetry of the feeding voltages and of its inner construction.
We analyse the machine under the hypothesis that the electric parts of the machine are in
steady-state, i.e. all electrical transients can be neglected. Under these hypotheses we understand,
1
All lifts are subject to very frequent start-ups. Lifts of very high buildings, in addition, require gradual
acceleration/deceleration phases, which require variable-speed operation of the machine.
10.4
Chapter 10: Induction machines and drives
from what we know from the knowledge acquired in chapter 5, that all voltages and currents in
the machine are sinusoidally-varying quantities. In particular we infer that the three currents
drawn by the machine from the source are sinusoids. We can also consider that the machine has a
full three-phase symmetry, in the sense that there is no reason for thinking that what happens in
phase a should be different from what happens in phase b or c except for the time displacement
induced by the time displacement of the three sources.
Therefore, after this simple qualitative reasoning, we can conclude that the three currents are
a three-phase set , as shown in fig. 10.3 as well.
This set of currents, because of the rotating field theorem (chapter 7), does create a rotating
field that produces in any of the rotor windings sinusoidally varying flux linkages that, in turn,
induce a three-phase set of EMFs, by Faraday’s law.
The EMF creation of any of the rotor conductor can be evaluated using eq. (7.2) :
e PQ  v rel  B l  v rel B l
(7.2)
Where the speed vrel is the speed of the considered rotor conductor in a reference integral
with B. Eq,. (7.2) is therefore naturally applied to the machine in a frame of reference integral
with B that, indeed, we know being a rotating field. The reference polarity of ePQ is to be set in
such a way that v, B, and a vector going from Q to P have a right-hand orientation; therefore the
“+” marking on e must be set using the right-hand rule, as shown in fig 10.4, for which B is
considered rotating (in a stationary frame of reference) counterclockwise.
F
Br
Q -
vrel
ePQ(t)
B
+P
i(t)
abs
P’ +
eP’Q’(t)
- Q’
Fig. 10.4. Induction and force generation on a rotor of an induction machine
(B rotor rotating counterclockwise, vrel rotor speed in a frame rotating along with B).
The stator windings can be realised using just two poles, or several pole-pairs. The
construction of multi-pole-pair windings is the same of that of synchronous machines and some
details can be thus be found in chapter 7. Moreover, the stator coils are distributed along the airgap, so the voltage produced by N turns is kN times the voltage of a single turn, where k is the
adimensional factor discussed in sect. 7.2.3.
Therefore, the total voltage amplitude of a coil containing N distributed turns and p pole pairs can
be expressed using eq. (7.7), i.e.:
Eˆ
 kNSB  kNpΩSB
coil
The EMFs induced in the rotor, because the rotor windings are short-circuited, and since we
are neglecting all the electrical transients, create a three-phase set of currents, that create a new
rotating field, that combines with the one produced by the stator currents.
Any of the rotor conductors is thus subject to a field, due to the combination of stator and
rotor rotating fields, that belongs to the plane orthogonal to the conductors itself, and are
traversed by currents. Therefore they are subject to Lorentz’s force (7.5) :
F  I  B  l F  I  B  l
(7.5)
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
10.5
in which the direction of F is determined using the right-hand rule with vectors I (having the
same direction of the conductor, flowing towards the conductor end towards which positive
charges flow) and B.
The force has the direction shown in figure 10.4. It tends to move the rotor in a direction
which tends to oppose the flux variation, i.e. in the example given, counterclockwise as B, in a
stationary frame.
Now that the rotor rotates what happens into the machine? Indeed the reasoning just followed
can be repeated: there still exist three stator currents, that create a stator rotating field that
generates three EMFs on the rotor, which, in turn, create three currents, and forces on the rotor
conductors.
The field rotational speed is the same as earlier, since it depends only on the angular
frequency  of the supply voltages U and on the number of the machine’s pole pairs p:
0=/p,
The speed of the rotor conductors is now, if r is the rotor radius2:
vrel  (Ω0  Ω)r  (p -Ω)r
By effect of this speed EMFs are induced in the rotor that constitute a three-phase set. The
angular frequency of these voltages is related to the relative movement of rotor and stator field:
rot  p (Ω 0  Ω )
and therefore each of the three induced voltages will have as amplitude:
Eˆ
 kN SB .
coil
rot
These voltages will induce in the rotor coils currents that constitute a three-phase balanced set,
which in turn generate a rotating field which, in a frame reference rotating with the rotor, will
rotate at the speed:
rot / p  Ω0 -Ω .
The field produced by the rotor currents, evaluated in a stationary frame, will then rotate at a
speed equal to 0, i.e. the same speed of the stator field.
This is an important conclusion:
RESULT: Rotational speed of the magnetic field produced by the rotor
In an induction machine the rotating field generated by the rotor’s currents rotates at the
same speed as the field produced by the stator’s currents. This speed is called synchronous
speed, and is equal to the angular frequency of the source, divided by the number of
machine pole pairs.
As long as the rotor speed  remains lower than the synchronous speed 0, force will be
generated on the rotor conductors, that will globally produce a positive torque that will tend to
accelerate the rotor. But when the actual speed approaches 0, the reason for the torque to be
generated,. i.e. a relative motion between rotor and rotating fields will tend to vanish, and at
exactly the synchronous speed, the torque will be zero, because in a frame of reference integral
with the rotor the machine field will be seen to be stationary, and therefore no EMF is generated.
In the next section a very simple and effective mathematical model will be presented that
allows to evaluate quantitatively the electrical and mechanical quantities of the machine at the
various speeds.
2
more precisely, r is distance of the conductor’s cross section center from the rotor’s cross section center.
10.6
Chapter 10: Induction machines and drives
10.2 Machine model and analysis
In the previous section a description of the physical phenomena occurring within the
induction machine was reported under the hypothesis that all transients are neglected, there exists
physical symmetry of the machine, and the machine itself is connected to a three-phase balanced
system of voltages.
Under these hypotheses also the currents absorbed by the machine is a balanced three-phase
system of currents, and the behaviour of the machine can be analysed using the single-phase
equivalent concept (if the reader does not recalls about it they are suggested to return to the
relevant section of chapter 5).
Here this equivalent circuit is presented and its usage is discussed; it is not derived from the
electromagnetic equations only to reduce the mathematical burden to the reader.
Before introducing the single-phase equivalent of an asynchronous machine a very important
quantity is to be defined, the so-called slip s of the machine:
Ω Ω
s 0
Ω0
When a machine is standstill its speed  is zero, and the slip is unity. When the rotor rotates
at the same speed of the rotating fields 0, i.e. when no current is induced in the rotor and
therefore no torque is produced, slip is zero. In all the other cases some power is converted
between the electrical circuit and the mechanical shaft of the machine.
Example 1
A six-pole 60-Hz induction motor runs with a slip of 4%. Determine the synchronous
speed n0, the rotor speed n, the frequency fr of rotor currents, the speed of the rotor
rotating field with respect to the rotor (nrr) and with respect to the ground (nr).
Expressing speeds in Revolutions Per Minute (RPM):
60 f 60  60
n0 

 1200 RPM
p
3
n  (1  s )n0  (1  0.04) 1200=1152 RPM
fr= sf = 0.0460 = 2.4 Hz
60 f r 60  2.4
nrr 

 48 RPM
p
3
nr  nrr  n  48  1152  1200 RPM  n0
Hence stator and rotor fields are synchronous.
2f 2n0
Ω0 

 125.7 rad/s
60
p
2n
Ω  (1  s )Ω 0 
 120.6 rad/s
60
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
10.7
Using the slip, a single-phase equivalent of a synchronous machine can be created as
reported in fig. 10.5.
Is
Rs
Xs
R’ r
Pmg/3
Ii
+
Us
X’r
I’r
Xi
R’ r
Ri
1-s
s
stator
air-gap
rotor
Mechanical
shaft
Fig. 10.5. Single-phase equivalent circuit of an induction machine.
The reader has surely noted a strong similarity to the transformer circuit. Indeed the very mechanism of
power transfer from the stator to the rotor has a lot of similarity with the power transfer from primary to
secondary windings of a transformer. The reader is invited to find himself similarities between the two
machines, as soon as he reads and studies this chapter.
In the figure two curved-dashed lines are reported, to indicate important physical
transformations: the leftmost part of the circuit model the stator quantities, then the dashed curve
named air-gap is crossed, and the circuit section containing a description of what happens in the
rotor is shown. The right-most section of the circuit represents the mechanical shaft.
Let first analyse the circuit in terms of power. The powers shown in the circuit are,
obviously, always one third of the ones circulating in the actual machine. Therefore the power
entering at the left-most terminals:
P=Us Is cos
is one third of the power actually entering the machine through its terminals.
Part of the power entering the machine is dissipated in the resistance of the stator’s coils, and
that power is
Pls  3Rs I s2
Another dissipation occurs in the stator and rotor iron, because of parasitic currents and
hysteresis. this is simulated by the resistance Ri, and the power correspondingly dissipated
constitutes, the so-called iron losses 3:
Pli  3Ri I i2
then the power traverses the air-gap and the power entering the rotor, the so-called air-gap power
is
Pag=P-Pls-Pi
Part of the air-gap power is dissipated in the rotors coils, and the power lost there is:
Plr  3R' r I r2
It must be noted that the resistance R’r, and the current I’r are not exactly resistance and
current of the rotor’s coils, but they are is connected to them. It can be shown to be:
k N
1
(10.1)
  s s R ' r   2 Rr I ' r  I r
kr N r

in which Ns and Nr are the turns of stator and rotor coils, respectively and ks and kr are the
corresponding coil distribution factors, and  is the equivalent turns ratio.
3
the subscript “i” stands for iron.
10.8
Chapter 10: Induction machines and drives
These expressions imply that the power dissipated in the three resistors R’r by effect of the
three currents I’r is exactly the power dissipated in the rotor’s coils:
Plr  3R' r I ' 2r  3Rr I r2
The right-most part of the circuit models the power behaviour of the shaft. Indeed the
component indicates ad a resistor having as resistance:
1 s
(10.2)
Rm,eq  R2'
s
is a fictitious resistor, since the electrical power “dissipated” in it is actually converted into
mechanical form. The right-most dashed curve in the circuit is the borderline between the electric
domain of the machine and its mechanical domain. If the mechanical power generated in the
machine is called Pmg, it can be thus written:
Pmg  3Rm,eq I ' 2r
This mechanical power is also called “developed power”. With the obvious deduction of
mechanical losses in the bearings and in the resistance encountered by the rotor by effect of the
air surrounding it, we obtain the useful (net) mechanical power, available at the machine flange.
The power flows in the machine are summarised in figure 10.6.
flange
(useful)
mechanical
power
Input
electrical
power
rotor
copper
losses
mechanical
losses
iron
stator
losses
copper
losses
Fig. 10.6. A qualitative chart illustrating different types of machine losses.
Example 2
A 50-Hz three-phase induction motor has a rated voltage of 400 V (phase-to-phase).
When operating at full load, it develops its rated mechanical power of 18 kW at 705
RPM, absorbing a line-current of 35 A and an electrical power of 21 kW from the
electric grid. Calculate:
a) the synchronous speed n0
b) the slip s
c) the power factor cos
d) the torque T
e) the efficiency 
With p=1, the synchronous speed n0 would be 60f=3000 RPM.
Since n=705 RPM, 3000/n=4.25, then p=4.
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
10.9
60 f 60  50

 750 RPM
p
4
n  n 750  705
s 0
=
=0.06
n0
750
P
21000
 0.866
cos  el 
3UI
3  400  35
2n
 73.82 rad/s
Ω
60
P
18000
T = mech 
 243.8 Nm
Ω
73.82
P
18000
  mech 
 85.71%
Pel
21000
n0 
Example 3
In a four-pole 50-Hz induction motor, the power crossing the air gap and the developed
mechanical power are respectively 22 kW and 20.8 kW. Calculate the copper rotor
losses and the slip. If the rotational losses are 450 W, determine the net output torque.
The rotor losses are the difference between the power crossing the air gap and the gross
delivered power. Since in running conditions iron rotor losses are negligible (due to the
low frequency of rotor voltages), such a difference corresponds to the copper rotor
losses:
Pcu-r=Pag-Pd=22-20.8=1.2 kW
Since Pcu-r=sPag, s=1.2/22=5.45%
The net mechanical power can be calculated subtracting rotational losses from Pd:
Pmech=20.8-0.45=20.35 kW
2f
(1  s )  148.5 rad/s

p
P
20350
T = mech 
 137.0 Nm
148.5

Example 4
A 380-V three-phase wye-connected induction motor has a 0.7+j1.4  per phase stator
impedance. The rotor impedance referred to stator is 0.6+j1.5  per phase. The
magnetising reactance Xi is 40 , the transverse resistance Ri is 150 .
At 4% slip calculate the input current, the power factor, the power crossing the air-gap,
the mechanical power and the efficiency.
10.10
Chapter 10: Induction machines and drives
Is
+
Us
Xs
Rs
X’r
I’r
Ii
Zs
Xi
R’ r
Z’r
Rl
Ri
R’ r
1-s
s
stator
rotor
Zs=0.7+j1.4 
Z’r=0.6+j1.5 
Rl =R’r(1-s)/s=0.6(1-0.4)/0.04=14.4 
Z’r-tot= Z’r+Rl = 15+j1.5  (also R’r/s+jX’r)
The total impedance seen by the stator is:
Ztot=Zs+jXi||Ri||Z’r-tot= 12.27+j6.432  = 13.8727.66°
Hence:
Power Factor=cos(27.66°)=0.886
Us=380/3=219.4 V
Is=Us/Ztot=14.02-j7.351
I’r=(Us-IsZs)/Z’r-tot = 13.06-j2.272 =13.25-9.87°
Pmg=3Rl2I’r2=314.413.252=7590 W
Pag= Pmg/(1-s)=7906 W
Pel=3Re(UsIs*)=9230 W
= Pmg/Pel=7590/9230=82.22%
Alternative method:
Pag=3Re((Us-IsZs )Is*)=7906 W
Pmg=(1-s)Pag=7590 W
Let us now rapidly consider the meaning of the reactances present in the machine equivalent
circuit:
 reactance Xs=Ls, where Ls is the proportionality coefficient between the stator current and
the stator leakage flux i.e. the quote of the flux linkage created by the stator currents that does
not cross the air gap, and therefore does not interact with the rotor
 reactance X’r=2Lr , where Lr is the proportionality coefficient between the rotor current and
the rotor leakage flux i.e. the quote of the flux linkage created by the rotor currents that does
not cross the air gap, and therefore does not interact with the stator
 reactance Xi=Li where Li is the proportionality coefficient between the stator currents and the
flux linkage that crosses the air-gap.
The equivalent circuit has longitudinal components Rs, Xs, R’r, X’r, and transversal
components Xi, Ri. In the normal operation of the machine, all the longitudinal components are
much smaller than the transversal one. Only when s is near to one the situation is different, since
the load resistor Rm,eq is null. In this situation, that occurs at the start-up of the machine, very
large currents are drawn from the mains and flow through the machine, a situation that is similar
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
10.11
to short-circuit conditions in power systems. For this reason, the longitudinal impedance
(Rs+R’r)+j(Xs+X’r) is also called “short circuit impedance” (Zsc). In fact, dividing by Zsc the
supply voltage, we obtain the “short-circuit current”, which corresponds both to the starting
current and to the current absorbed by the machine when it is kept blocked and is fed from the
mains. As discussed below, this test is used by engineers to evaluate the numerical values of the
equivalent circuit parameters, obviously at reduced voltage to avoid damages due to the high
currents that would otherwise circulate.
This gives us a first indication that the start-up currents of this kind of machines are much
larger than the currents occurring in steady-state, and therefore either frequent start-ups must be
avoided, or the machine must be dimensioned to withstand these larger currents. A typical
application in which the machine operates with frequent start-ups is in lifts.
Since the single phase equivalent circuit contains only reactors and resistors, it is obvious
that in all its operating conditions it absorbs active and reactive power, and therefore its power
factor is always lagging.
Indeed the statement that the machine “always absorbs active power” is to be corrected.
Since the load resistance is a fictitious one and, for negative slips, it has a negative value. Under
these conditions, i.e. when the machine rotor rotates at a speed larger than 0, the resistor Rm,eq
actually delivers power, and the machine operates as a generator.
Even when the machine generates active power, however, is still absorbs reactive power,
since the reactive components of the equivalent circuit remain positive even at speeds larger than
0.
The single-phase equivalent enables to determine important parameters as a function of the
rotational speed  of the machine rotor. Here it is used to determine the current absorbed from
the supply network and the mechanical torque generated.
Here, for simplicity’s sake, this is done in an approximated way, i.e. neglecting the effects of
the transversal components Ri and Xi. This will introduce some error, but the results that will be
obtained are qualitatively correct. The simplified circuit can be expressed in one of the two forms
reported in figure 10.7, in which is, obviously, X=Xs+Xr’. Note that if the two resistances Rr’ and
Rm,eq are summed up, the resulting resistor will absorb a power that is the sum of the generated
mechanical power and the rotor’s copper losses, that can be imagined as the power crossing the
air-gap and entering the rotor. This power is therefore called Pag.
Is
+
X
Rs
R’ r
Is
+
Us
U1
-
-
1-s
R’ r
s
X
Rs
+
+
Us
U1
-
-
Pmg/3
R’ r
s
Pag/3
Fig. 10.7. Simplified circuits of an induction machine.
We want now to derive the current absorbed and torque produced under the condition
Us=const. Indeed, as a further approximation, we consider here constant the voltage U1, that can
be measured downstream Rs that does not differ significantly from Us.
Using the right circuit in figure 10.7, the following can be written:
I2 
U 12
X 2  R ' r2 / s 2
Pag  3
R' r 2
R'
U2
U 2 R' s
I  3 r 2 1 2 2  3 2 12 r 2
s
s X  R' r / s
s X  R' r
10.12
Chapter 10: Induction machines and drives
and:
Tmg  Pag /  0  Pmg /  
3 U 12 R' r s
 0 s 2 X 2  R' 2r
(10.3)
in which the symbol Tmg indicates the “mechanical, generated” torque. The useful torque Tmu will
be lower than this by effect of mechanical losses.
If we imagine the rotor speed  to go from zero to 0, the corresponding slip will go from 1
to 0; the denominator of the fraction describing I2 will steadily grow, and therefore the current I
will decrease monotonically.
To analyse the shape of Tmg() consider first that two important points of the curve are those
corresponding to s=0 and s=1. It is obviously:
Tmg ( s  0)  0
Tmg 0
3 U 12 R' r
3 U 12 R ' r
 Tmg (  0) 

 0 X 2  R ' r2  0 Z 2
It is also of interest to evaluate whether between these two points there is a peak. This is
done equating to zero the first derivative of Tmg(s):
Tmg
3U 12
(10.4)
 0  sˆ  R' r / X Tˆmg  Tmg ( sˆ) 
s
2 0 X
It can be easily verified that it is always Tˆmg  Tmg 0 and therefore the curve has a maximum
for s  sˆ . The corresponding angular speed will be called ̂ . Note that while ŝ depends on the
rotor’s resistance, Tˆmg does not.
Comparing the expression of I2 and equation (10.3), it is possible to observe that
sT
3R'r
= constant
(10.5)
=
2
I
0
for any working condition. This observation is very useful to correlate T(s=0), Tmax and Tfull-load
(Trated) to their correspondent currents.
Example 5
An induction motor has a slip of 4% at full load. The starting current is 6 times the fullload current. Calculate the ratio of the starting torque to the full-load torque.
From Eq (10.5):
2
Tstarting  I starting  s full-load
0.04

 62 
 1.44

1
Tfull-load  I full-load  sstarting
Example 6
For the motor of Example 4, calculate the maximum mechanical power and the
 , the frequency being 50 Hz and the pairs of poles
correspondent slip ŝ and torque T
in number of 2.
X= Xs+X’r=1.4+1,5=2.9 
s =R’r/X=0.6/2.9=0.207
0=2f/p=157.1 rad/s
3U 2
3  219.42

 158,5 Nm
T 
2Ω 0 X 2 157.1  2.9
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
10.13
Pmg=0(1-s) T =19750 W
Please note that the equation used to calculate T does not consider Rs.
More in general, it is possible to demonstrate that:
U12 R'r s
3
T
Ω 0 s 2 X 2  ( sRs  R'r ) 2
ŝ 
R'r
Rs2  X 2
, Tˆ 
3U12
2Ω 0 (Rs  Rs2  X 2 )
Typical shapes of the absorbed current (RMS value) and mechanical torque generated, as a
function of the rotational speed, are reported in fig. 10.8. These curves were derived using
realistic numerical parameters of a 50 kW squirrel-cage machine.
This machine has a peak torque that is 2.35 times the starting torque The starting torque can
be higher than this value, but this is to be obtained using a larger rotor resistance, and this reduces
the machine efficiency. To have high efficiency the copper losses, and the rotor resistance, must
be small, but this reduces the starting torque at equal peak torque.
The torque, as expected, becomes zero when the synchronism speed 0 is reached, then it
reverses, and the machine operates as a generator. Obviously, for the generator region to be
actually reached it is necessary that at the machine shaft a mechanical motor is attached, i.e. a
device able to supply mechanical power.
Normally, however, as already noted, induction machines are operated as motors.
current
T1
torque
P
Q
0
generator
operation
motor operation
̂
0
Fig. 10.8. A typical torque vs. angular speed induction machine shape
of an induction machine.
The curves reported in figure 10.8 clearly show that the normal operation zone of the
machine is between the peak torque (either positive or negative) and the synchronous speed. For
example the torque T1 can be delivered at the speed corresponding to point P and Q; but point P,
while corresponding to a much lower mechanical power, implies the absorption of a much higher
current and losses! In the next section it will be shown that using a motor drive it is possible to
deliver a given torque at different speeds, while absorbing roughly the same current from the
supply.
10.14
Chapter 10: Induction machines and drives
ˆ is normally referred to as an unstable region, while the
The zone between =0 and   
region between ̂ and 0 is considered stable. This is not totally correct since stability depends
on the dynamic behaviour of a system, while the torque curves we are discussing are stationary,
but tells some truth: in the more common cases, indeed, equilibrium points between the torque
characteristic of the machine and of the load are stable only (in the machine operation as a motor)
between ̂ and 0.
The operation of an induction machine as a motor with its mechanical load is depicted in
figure 10.9, where both the motive torque Tm produced by the machine and the resistive one
absorbed by the load Tld (just as an example here is a fan) are reported. The difference between
the two is the accelerating torque.
unst
st
P
m
ld
Q
M
T, 
3
0
Fig. 10.9. Traction and load torques of an induction machine with a mechanical load.
Just to give a qualitative explanation of “stable and unstable regions” of the torque curve of
the machine consider the system to operate at the equilibrium point P between the machine
torque and a possible load torque Tunst. In case a light increase in  occurs, the machine torque
becomes larger than the load, and the speed tends to further increase. Also in case of a speed
decrease the system tends to move far from the equilibrium point. This does not happen in case of
point P and curve Tst, nor in case of point Q, of intersection with curve Tld, that show a stable
behaviour.
From the figure one can even have an idea of a starting-up transient of the machine, using the
mechanical equation:

T  T  J
mg
ld
In which obviously J is the combined moment of inertia of machine and mechanical load.
It must however be clarified that this is valid only as a first approximation, since the curves
reported in figures 10.8 and 10.9 are computed using the equivalent circuit that was drawn
assuming steady-state operation of the machine, while during the transient, especially the first
part of it, the actual behaviour is rather different.
The shape of the torque of an induction machine has the inconvenience of a start-up value is
smaller, often much smaller, than the peak torque and that during starting-up the currents
absorbed by the machine are very high. This can be mitigated by the usage of wound-rotor
machines and solved, but a larger cost, adopting electric drives instead of simple machines.
As has been seen earlier, wound–rotor machines allow to access the rotor windings, to
connect them electrically to a stationary circuit, typically for limited durations. Consider now that
as shown in (10.4), the slip at which the machine torque has its maximum is s=Rr’/X=2Rr/X, and
is therefore proportional to the rotor resistance, while the corresponding maximum does not
depend on this resistance.
Wound-rotor machines can thus be started inserting in series with the rotor windings at slow
speed additional resistors R1, R2, ... that are progressively by-passed, as far as the speed increases.
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
10.15
This is done collecting the rotor coils currents through slip-ring/brush coupling, of the same type
of those present in synchronous machines, using three rings to get the three-phase set of currents.
When the system is near it steady-state speed, all the external resistors are by-passed, and, by
means of a special mechanical arrangement, the rotor windings are short.-circuited, and the
brushes lifted form the rings, to avoid friction and brush consumption (fig. below).
1 closes
2 closes
1
Supply
2
M
R1
R2
T, 
3
0
This system is now becoming obsolete, since has several disadvantages over the more
modern solution of the usage of induction motor drives, that are discussed in the next section: the
external resistors produce excess losses, the ring-brush connection is mechanically complicated
and costly, and requires maintenance; the change between the different torque curves is discrete
and not continuous.
When an induction machine is required to operate at continuously-variable speed, such as in
electric trains or cars, the mechanical characteristic of the machine fed by constant
voltage/constant frequency source, such as that shown in fig. 10.8 and 10.9, is totally
inadequate.
In these cases induction machines can be used, and are today indeed frequently used, using
special devices to feed them, that are able to generate voltages having desired frequency and
amplitude. The machine and the feeding system constitute a subsystem that is called “motor
drive”. Some basic information about induction machine-based motor drives is reported in the
following section.
10.3 No-load and blocked-rotor tests
A method for determining the parameters of the equivalent circuit of an induction machine
consists of two tests: the no-load test and the blocked-rotor test.
In the former, the rated voltage is applied to the machine, while living the rotor free of
rotating (no mechanical load is applied). The current I0 absorbed by the motor is measured, as
well as the correspondent active power P0 (absorbed by Rs and Ri). Since the current is small, P0
is a good estimation of stator iron losses4 (at low slip, rotor losses are always negligible in an
induction machine, since the rotor frequency is low). Such losses, which strictly depend on the
voltage applied to Ri, remain practically constant for any loading condition, provided that the
machine is supplied at its rated voltage.
Ri 
P0
U2
U2
U2
, Xi 

, where cos0 
Q0 P0  tan0
P0
3UI 0
The second test is performed increasing the supply voltage until rated current is absorbed by
the motor, the rotor being mechanically blocked. The voltage Usc is then measured, as well as the
active power Pcc. Since Usc is usually only some percents of the rated voltage, during this test Ii
can be neglected respect to Is (the core is poorly magnetised); thus the power absorbed by Ri is
4
If Rs is known, for example thanks to a previous DC measure, to estimate the iron losses P0 can be corrected by
subtracting 3RsI2.
10.16
Chapter 10: Induction machines and drives
much lower than the one consumed by Rs and R’r. For this reason, Psc is a good estimation of
copper losses at rated current; in any other loading condition:
Pcu  Psc  ( I/I rated ) 2
The test is unable to separate Rs+jXs from R’r+jX’r, but it allows calculating Zsc:
U sc
Psc
Z sc 
e jφsc , where cossc 
3I rated
3U sc I rated
Finally, electrical and mechanical powers are strictly related:
Pel  Pmech  PFe  PCu
The reader should note the strict analogy with open-circuit and short-circuit tests of a
transformer.
Example 7
A 380-V 8-kW three-phase 4-poles induction motor has issued the following results to
the no-load and blocked-rotor test:
I0=5%, P0=1.2%
Usc=12%, Psc=7%
The stator resistance is 0.12  (phase)
The rated line current is 15 A and the stator is delta-connected.
Calculate Ri, Xi and Zsc
I0=0.0515=0.75 A
P0=0.0128000=96 W
P0
cos0=
=0.194, tan0=5.044
3UI 0
Usc=0.12*380=45.6 V (phase-to-phase)
Psc=0.07*8000=560 W
Psc
cossc=
=0.473, sc=61,8°
3U sc I rated
The wye-connected equivalent stator resistance is 0.12/3=0.04 .
The iron losses can be calculated subtracting the no-load stator copper losses from P0:
PFe=96-30.040.752=95.93 W (no-load stator copper losses are negligible)
Ri=U2/PFe=1505 
Neglecting Xs:
Xi=U2/(P0tan0)=298.2 
U sc
=1.755 
3I rated
Zsc= Zscsc = 1.75561.8° = 0.829+j1.546  = (Rs+R’r)+j(Xs+X’r)
Zsc=
10.4 Induction machine motor drives
Consider the above derived expressions of the machine current and torque:
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
I2 
10.17
U 12
3 U 12 R' r s
T

P


P


/
/
mg
ag
mg
0
 0 s 2 X 2  R' 2r
X 2  R ' r2 / s 2
(10.3)
Let us evaluate what happens of the expressions when the machine is fed in such a way that
is:
U1  K f f  K   0
(10.6)
i.e., the machine is fed in such a way that the frequency of supply and voltage U1 remain always
proportional to each other (remember that 0=/p=2f/p).
Because of its importance a supply compliant with (10.6) has a name of its own: it is normally
called constant voltage/frequency supply.
Using constant voltage/frequency supply and considering that it is:
   
: s 0

0
0
the squared current becomes
I2 
s 2U 12
2   K 2
.

s 2 X 2  R' r2 2  L2 p 2  R ' 2r
(10.7)
In the latter equality it has been considered that X is proportional to the angular frequency :
X  X s   2 X r   (Ls   2 Lr )  p0 L .
The torque is:
Tmg  3
K 2 R' r 
K 2 R' r 

3
s 2 X 2  R ' 2r
2 L2 p 2  R ' 2r
(10.8)
Equations (10.7) and (10.8) indicate that when the machine is fed according to (10.6) the
current absorbed and torque produced depend only on    0   and not by 0 alone.
Since the shape of Tmg as a function of  does not depend on 0, it can be derived from
what seen in the previous section. Evidently if the function T() is the one of previous figures,
reported again in the left part of figure 10.10, changing 0 leaves the shape intact, just the curve
is translated in such a way that the torque becomes zero for the given value of 0.
01
02
03
04
05
Fig. 10.10. Translation of torques feeding the machine with the rule U1=Kf f.
This technique can be used to make the machine operate so that it operating point is always
in the more efficient region of the characteristic, i.e. between ̂ and 0. In this region, as was
seen earlier, the torque is delivered at the lowest current and highest efficiency.
As an example, fig. 10.11 shows the starting of the same machine under two conditions:
using a constant terminal voltage and the technique (10.6) During the start-up 0 is continuously
changed in such a way that =0-=constant, and therefore the torque generated according to
10.18
Chapter 10: Induction machines and drives
the theory developed is constant and larger than the load torque. When the target speed base is
reached, however, the supply system works at that speed and the system reaches its equilibrium
point. The figure is obtained using a simulation program that models the machine with a higher
level of detail.
To understand the shown plots it must first be noted that the time span is eight seconds,
therefore sinusoids operating at 50 Hz are seen as a thick band, since they evolve too rapidly to
be seen well in the shown scale. This is particularly true for the top-left plot, in which the uniform
band is the visual representation of a constant-amplitude, constant-frequency 50 Hz sinusoid.
Voltages
Currents
Torques
Constant voltage and frequency supply
Supply U1=Kf f; =const
Fig. 10.11. Simulation of the starting of an induction machine
either fed directly from the mains (left) or with variable voltage/frequency (right).
The following observations can be done:
 at the beginning of the transient, in the case of constant-frequency supply, the torque shows a
high frequency fluctuation around a steadily increasing value. This fluctuation is actual in real
machines, and are not reproduced by usage of the equivalent circuit-based model proposed in
this chapter, in which all electrical transients are neglected. In the variable-frequency supply
the torque fluctuation is reduced to a minimum.
 the variable frequency starting allows the machine to be started-up in the same time, but at a
much lower current and reduced torque fluctuations;
The supply voltages to obtain the curves reported in the right part of figure 10.11, are
obtained using a control logic of the type described in fig. 10.12
The reference torque during the startup T* is first converted in a  signal (taking into
ˆ to
account the inner parameters of the induction machine) but saturated at the value of 
impose operation of the machine in the more efficient zone of its torque. Adding the measured
value meas of  the synchronous speed 0 is determined, that is converted into frequency using
the pole pairs number p. Before this, the 0 signal is saturated to the reference value 0ss, since
the machine is intended to be started up to this value of synchronous speed. Once the frequency is
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
10.19
known, the voltage U1 =kf* is known and from it, using the estimate I* of the current computed
from T*, the voltage U* desired at the machine terminals is determined.
The values of the voltage U* and frequency f* to be applied to the machine are then
transferred to the supply system that creates a three-phase voltage system having that voltage and
frequency.
f*
ˆ

T*
 0ss

Saturation
+
0
p
2
+
 meas
f*
 0ss
U=kf*+RsI*
U*
I*
 meas
f*
U*
Supply
system
M
T, 
mechanical
load
3
tachometer
Fig. 10.12. A possible torque control logic
(used to obtain the plots reported in fig. 10.11).
It is apparent that the variable-frequency control allows the machine to stay stably at any
speed, because the currents are kept under control. The value of  determines the torque and the
current, the latter being the value corresponding to the more efficient region of the torque
characteristic of the machine, i.e. the one between * and 0.
The operation of the induction machine under (10.6) is called also constant flux operation.
This is because the voltage U1 corresponds in the machine to the EMF produced by the flux
linked with the stator . For each phase:
U1  j
And therefore U1/f= const implies = const.
As speed increases, the voltage to be applied to the machine terminals increases. When the
maximum continuous voltage of the machine is reached (normally called nominal voltage), this
quantity cannot be increased anymore, while it may happen that a further increase in speed is
required. The speed that corresponds to the nominal voltage is normally called the base speed of
the machine. Beyond the base speed the voltage is kept constant, while 0 is raised, increasing
the frequency of the supply voltages. In these conditions, the curve of the torque is progressively
reduced, being proportional to 1/02:
Tmg 
U12 R'r 
3 U12 R'r s
3

0 s 2 X 2  R'r2 02  2 L2 p 2  R'r2
The full picture of the machine, showing the constant flux and flux weakening parts, is shown
in figure 10.13. It shows the zone below base, in which the torque characteristics translate, and
the zone above this value where they reduce in proportion to 1/02. A possible shape of the load
 is also reported.
torque Tld, and a possible locus of the machine operating points Tm  Tld  J 
10.20
Chapter 10: Induction machines and drives
T, P
Tm
P
Tld
base

Fig. 10.13. Induction machine drive composed by the constant flux (<base)
and flux weakening (>base) regions.
Using this Tm, the mechanical power is linearly increasing in the constant flux region; when
base is overcome, the Tm locus shown in figure is such that the power P is maintained constant
and therefore the actual torque reduces in proportion to 0. Doing so, since the machine torque
shape reduces in proportion to 1/2, the machine operating point becomes more and more near to
the peak torque.
Rather obvious, the zone with the flux reduction cannot be very large, because of the
corresponding torque reduction. A typical ratio of max/base in a normal induction drive does
not overcome 2.
Some final words about induction motor drives.
The constant voltage/frequency control allows smooth operation of the machine, and usage
of it at all speeds with good efficiency. However, in fast changing conditions it is not optimal
because its theory comes from a steady-state model of the machine.
More advanced control techniques, normally called vector control, that keep under control
the inner machine magnetic field are today used in advanced drives, and show better behaviour in
dynamic conditions.
These techniques, however, are well beyond the scope of this book and will not be dealt
with.
10.5 Historical notes
10.5.1 Ferraris’ short biography
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
10.21
10.6 Proposed exercises
Where not specifically indicated, electromagnetic quantities are expressed in RMS values.
10.1. An eight-pole 50-Hz induction motor runs
with a slip s=5%. Determine the synchronous
speed n0, the rotor speed n, the frequency fr of
rotor currents, the speed of the rotor rotating field
with respect to the rotor (nrr) and with respect to
the ground (nr). Express all speeds in RPM
(Revolutions Per Minute).
10.2. A four-pole 60-Hz induction motor runs at
1750 RPM. Determine the synchronous speed n0,
the slip s and the frequency fr of rotor currents.
10.3 A six-pole 50-Hz induction motor has a fullload slip of 3%. Determine:
a) the synchronous speed n0
b) the rotor speed n at full load
c) the frequency of rotor currents at the instant of
starting
d) the frequency of rotor currents at full load
10.4 The nameplate speed of a 25-Hz is 720 RPM.
If the no-load speed is 745 RPM, estimate the
number of poles and calculate the synchronous
speed and the slip at full load.
10.5. A 50-Hz, three-phase induction motor has a
rated phase-to-phase voltage of 400 V. When
operating at full load, delivering its rated mechanical power of 30 HP, the line-current is 42 A, the
rotor speed is 710 RPM and the motor absorbs
25.7 kW from the electric grid. Calculate:
a) the synchronous speed n0
b) the slip s
c) the power factor pf
d) the torque T
e) the efficiency 
10.6. An eight-pole, 50-Hz, three-phase induction
motor has an efficiency of 86% and absorbs 48.8
kW from the electric grid. If the slip is 5%,
calculate the shaft torque.
10.7. Induction motors are often braked reversing
the phase sequence of the voltage supplying the
motor (plugging). A motor with 6 poles is
operating at 1150 RPM while supplied at 60 Hz.
Two of the stator supply leads are suddenly
interchanged. Calculate the new slip and the new
rotor current frequency.
10.8. In a four-pole 50-Hz wound-rotor induction
motor, stator and rotor have the same effective
numbers of turns per phase. If the motor has 220
V per phase across its stator and the voltage
induced in the rotor is 6.6 V per phase, calculate
the slip and the motor speed.
10.9. In a two-pole 60-Hz induction motor, the
power crossing the air gap and the developed
power are respectively 21.6 kW and 20.1 kW.
Calculate the slip s. If the rotational losses are 400
W, determine the net output torque.
10.10. A three-phase six-pole 50-Hz induction
motor develops its maximum torque of 612 Nm at
600 RPM. The rotor resistance reported to stator,
R’r, is 0.5 . Calculate the supply voltage (phase
to phase) and the torque developed by the motor at
750 RPM.
10.11. An induction motor develops a maximum
torque twice the starting torque. Calculate X/R’r. If
the full-load torque equals the starting torque,
determine the full-load slip.
10.12. An induction motor has a wounded rotor,
having an impedance of 0.2+0.6j per phase.
Which is the resistance to be added to each rotor
phase, in order to maximise the starting torque? If
no resistance is added, at which slip the motor
develops the maximum torque?
10.13. An induction motor has a slip of 4% at full
load. The starting current is 8 times the full-load
current. Calculate the ratio of the starting torque to
the full-load torque. If the motor is started at
reduced voltage, so that the starting current is only
4 times the full-load currents, which is the new
ratio of the starting torque to the full-load torque?
Comment this result.
10.14. An induction motor has a slip of 5% at full
load. At rated voltage, the starting current is 6
times the full-load current. Calculate the ratio of
the starting torque to the full-load torque.
If the motor employs a wye-delta starter, which
connects the motor phases in wye for starting and
in delta when already running, which is the new
ratio of the starting torque to the full-load torque?
Comment this result.
10.15. The resistance measured between two
stator terminals of an induction machine is 0.14 ;
the stator id delta-connected. The motor delivers a
10.22
mechanical power of 14.7 kW, while supplied at
220 V (phase-to phase), absorbing a line current of
54 A at 0.82 power factor (lagging). The no-load
test has issued the following results: I0=20 A,
cos0=0.12.
Calculate:
a) the resistance of each stator phase
b) the efficiency
c) the no-load power
d) the sum of iron and mechanical losses
e) the copper losses at the current load condition
10.16. A three-phase induction motor has a fullload slip of 4%. The wounded rotor is star
connected, with a phase resistance is 0.12 .
Calculate the additional rotor resistance required
to obtain the full-load torque at starting
conditions.
10.17. An eight-pole 220-V 42-Hz induction
motor absorbs a line current of 60 A with a 0.82
power factor (lagging) and a slip of 3.5%. The
resistance measured between two stator terminals
is 0.162 , the stator being delta-connected. The
no-load test has issued the following results: I0=18
A, cos0=0.12.
Calculate:
a) the resistance of each stator phase
b) the sum of iron and mechanical losses
c) the stator copper losses
d) the power crossing the air gap
e) the power and torque delivered
f) the efficiency
10.18. A six-pole induction motor delivers 20 kW
with an efficiency of 91%, a power factor of 0.9
and a slip of 3%. The motor is supplied by a 380
V 50 Hz phase to phase voltage. Calculate the line
current and the delivered torque.
10.19. A six-pole 220-V 50-Hz induction motor
absorbs 22 kW with a power factor of 0.68 and a
slip of 3.4%. The resistance measured between
two stator terminals is 0.023 , the stator being
delta-connected. The iron losses are 820 W, the
mechanical losses 440 W. Calculate:
a) the stator copper losses
b) the net mechanical power and torque
c) the efficiency
d) the capacity of delta-connected capacitors,
required to compensate the motor to a unit
power factor
10.20. A four-pole induction motor with a wyeconnected stator, having a phase resistance of
0.4 , is supplied at 380 V, 50 Hz. Iron and
Chapter 10: Induction machines and drives
mechanical losses equal respectively 350 W and
300 W. The motor moves a DC generator, which
delivers 70 A at 125 V. The motor absorbs 12.25
kW with a 0.9 power factor, rotating at 1446
RPM. Calculate:
a) the overall efficiency
b) the motor efficiency
c) the DC generator efficiency
10.21. A two-pole three-phase induction motor is
supplied by a 380 V 50 Hz voltage. The rotor
resistance reported to stator is 0.15 ; the overall
leakage reactance reported to stator is 1 .
Calculate:
a) the maximum torque and the correspondent
slip and line current
b) the full-load torque and line current, if the fullload slip is 3%
c) the starting torque and line current
10.22. The motor of the previous exercise
applied to a mechanical load requiring a torque
30+0.005 Ω 2mecc Nm. Using eq. (10.3) and
numerical technique, calculate:
a) the slip
b) the line current
c) the rotor speed in RPM
d) the mechanical power and torque delivered
the load
is
of
a
to
10.23. A 50-Hz 15-kV (phase-to-phase) network
supplies, by means of a MV/LV transformer, a
380 V three-phase 4-pole induction motor, with a
delta-connected stator. The transformer has the
following parameters:
15000/380 V (phase-to-phase)
delta/wye connection
18 kVA, 50 Hz
I0=2%, P0=0.4%
Usc=7%, Psc=3%
The motor has the following parameters:
Pmech rated=8 kW
Irated=16 A (line)
I0=6%, P0=1.2%
Usc=14%, Psc=8%
Rs=0.12  (phase)
Between the transformer and the motor there is an
80-m LV cable, with r=0.9 /km and x=0.15
/km. Thévenin’s impedance of the MV network
can be neglected.
Draw the single-phase wye-equivalent of this
system. Assuming that the motor is running with a
3% slip, calculate:
a) the motor speed in RPM
b) the mechanical power and torque delivered to
the load
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
c) the phase-to-phase voltage at motor mains
d) the stator phase current of the motor
e) the primary phase current of the transformer
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