Download Chapter 5 Induction Machines

Chapter 2
Induction, Or Asynchronous
Machines
 Three-phase induction motors
 Unbalanced operation of three-phase
induction motors
1
History & Application Of Induction Machines
• The invention of induction machines in the l880s completed the ac system of electrical
power production, transmission, and utilization, which at the time was in competition
with the dc system for general acceptance. The whole concept of polyphase ac,
including the induction motor, was the idea of the great Yugoslavian engineer Nikola
Tesla. (Tesla became a U.S. citizen in 1889.) The system was patented in 1888. The first
large-scale application of the Tesla polyphase ac system was the Niagara Falls
hydroplant, completed in 1895.
• Today, most industrial motors of one horsepower or larger are three-phase induction
machines. Induction machines require no electrical connection to the rotor windings.
Instead, the rotor windings are short-circuited. Magnetic flux flowing across the air gap
links these closed rotor circuits. As the rotor moves relative to the air gap flux,
Faraday's-law voltages are induced in the shorted rotor wind-ings, causing currents to
flow in them. The fact that the rotor current arises from induction, rather than
conduction, is the basis for the name of this class of machines. They are also called
asynchronous (i.e., "not synchronous") machines because their operating speed is
slightly less than synchronous speed in the motor mode and slightly greater than
synchronous speed in the generator mode.
2
• Induction machines are usually operated in the motor mode, so they are usually called
induction motors. As generators, their peculiar characteristics make them suitable for
only a few special applications. As motors, they have many advantages. They are rugged,
relatively inexpensive, and require very little maintenance. They range in size from a few
watts to about 10,000 hp. The speed of an induction motor is nearly, but not quite,
constant, dropping only a few percent in going from no load to full load. The chief
disadvantages of induction motors are:
1. The speed is not easily controlled.
2. The starting current may be five to eight times full-load current.
3. The power factor is low and lagging when the machine is lightly loaded.
For most applications, their advantages far outweigh their disadvantages.
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4
5
6
Construction
1- STATOR
A three-phase windings is put in slots cut on the inner surface
of the stationary part. The ends of these windings can be
connected in star or delta to form a three phase connection.
These windings are fed from a three-phase ac supply.
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8
9
10
Construction (Cont.)
2- Rotor
it can be either:
a- Squirrel-cage (brushless)
 The squirrel-cage winding
consists of bars embedded in
the rotor slots and shorted at
both ends by end rings.
 The squirrel-cage rotor is the
most common type because it
is more rugged, more
economical, and simpler.
11
Construction (Cont.)
b- Slip ring (wound-rotor)
The wound-rotor winding has
the same form as the stator
winding. The windings are
connected in star. The terminals
of the rotor windings are
connected to three slip rings.
Using
stationary
brushes
pressing against the slip rings,
the rotor terminals can be
connected to an external circuit.
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13
14
15
16
17
How The Induction Motor Works


When the stator windings are connected to a three-phase
supply; a rotating field will be produced in the air-gap.
This field rotates at synchronous speed. This field is
exactly like that of the stator of a synchronous machine.
Its speed of rotation is given by:
ωs = 4π f1/p
rad/s
ns = 120 f1/p rpm
where ωs or ns is called the synchronous speed, f1 is the
frequency of the stator voltages and currents, and p is the
number of poles of the winding.
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19
The strength of this field is proportional to the rms stator current, and is expressed by
where
Where: NI1 is the effective stator turns per pole, I1 is the rms stator phase current, and
Nφ1 and kw1 are the series turns per phase and winding factor of the stator winding.
20
Suppose for the moment that the rotor is being driven by an adjustable-speed motor
so that it is rotating at the same speed as the stator field. The relative velocity between
the stator field and rotor conductors is zero. There is a mutual flux between the rotor
and stator, due to the stator MMF F1.This flux links the rotor; but the flux linking any
one conductor is constant, hence dλ /dt = 0, and no voltages are induced in the rotor
circuits. There are no rotor currents, and no rotor MMF field. The situation is
illustrated in Slide 22.
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22
If the drive motor is now turned off, the friction and windage of the two machines will
cause the rotor of the induction machine to slow down. The rotor conductors begin to
slip backward through the stator flux field. The conductors now experience changing
flux linkages, and voltages are produced in them according to Faraday's law. As a result,
current flows in the closed rotor circuits. These rotor currents produce a MMF field
having magnetic poles on the rotor surface.
The new situation, with the drive motor turned off, is shown in Slide 24. Magnetic
forces between the magnetic poles on the rotor surface and the flux field produce
torque in the direction of field rotation. The rotor is wound to produce the same
number of poles as the stator winding. As the machine slows down, the rotor
conductors slip faster through the flux. Greater voltages are induced in the rotor
circuits, heavier currents flow in the rotor circuits, stronger magnetic poles are
produced on the rotor surface, and more torque is developed. When this torque
balances friction and windage load on the machine, the rotor speed stabilizes at a value
somewhat less than the field speed.
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24
When the motor is lightly loaded as in Slide 24, the speed is only slightly less than
synchronous speed (say, ω = O.999ωs). The conductors are slipping very slowly
backward through the flux field, and the rotor voltages and frequency are both very
low. Because the frequency is so low, rotor reactance is negligible, and the currents in
the rotor conductors are practically in phase with the induced voltages. Consequently,
the rotor MMF vector F2 is at right angles to the flux vector ɸ. With current flowing in
the rotor, the flux is no longer due to the stator MMF F1 but rather to R, which is the
resultant of F1 and F2.
Note in Slide 24 that the position of the pattern of rotor currents is determined by the
positions of the flux poles of the air gap field. The reason for this is that the rotor
voltages are being induced by air gap fluxes. Since the air gap field is rotating at
synchronous speed, the rotor current pattern is also revolving at synchronous speed.
The rotor itself is rotating at less than synchronous speed, however.
25
The pattern of rotor conductor currents is thus moving forward, relative to the
rotor, at a speed of (ωs – ω) rad/s. The position of the rotor MMF vector F2 is
determined by the rotor current pattern and is directed by the right-hand rule
along the axis of that pattern. The following statements thus may be made about
the rotation of the rotor current pattern and its MMF vector F2:
1. They rotate at synchronous speed with respect to the stator.
2. They are stationary with respect to the air gap field vectors ɸ and R.
3. They rotate forward at (ωs – ω) rad/s, relative to the rotor.
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The Concept Of Slip
The magnitudes and frequency of the rotor voltages depend on the speed of the
relative motion between the rotor and the flux crossing the air gap, ɸ. Let the angular
speed of the rotor, in radians per second (rad/s) be assigned the symbol ω, and let n be
its speed in revolutions per minute (rev/min). The slip speed expresses the speed of
the rotor relative to the field, and is given by
Slip speed = (ωs – ω) rad/s
= ns – n rev/min (rpm)
The per-unit slip, usually called simply the slip, is a very useful quantity in studying
induction machines. It is given the symbol s and is defined as follows:
27






n = the rotor speed (the motor speed) w.r.t. stator
ns = the speed of stator field w.r.t. stator or the synch. speed
s = the slip = ns - n/ns
nr = the speed of rotor field w.r.t rotor
f1 = the frequency of the induced voltage in the stator (stator
or supply frequency)
f2 = the rotor circuit frequency or the slip frequency
nr  nS  n  snS
f2 
Slip rpm
p
p
p
(nr ) 
(nS  n ) 
( s nS )  s f1
120
120
120
28
The Frequency Of Rotor Voltages & Currents
Consider a typical pair of rotor bars, labeled x and y in Slide 24. As the rotor slips
backward through the flux field, the flux linking these bars will vary cyclically. In this
Slide, these conductors are shown at the instant the rate of change of flux linkages is a
maximum, and hence the voltage induced in the rotor circuit composed of these two
bars and the end rings is at its peak. When y has slipped to the position now occupied by
x, the polarity will have reversed. Conductor y will have moved relative to the field a
distance of one pole pitch, and one half cycle of rotor voltage will have been generated.
Thus one cycle of rotor voltage is generated as a given rotor conductor slips past two
poles of the air gap flux field. In other words, one cycle of rotor voltage corresponds to
360 electrical degrees of slip. Then the frequency of the rotor voltages and currents is
given by
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The Induction Motor Under Load
When a heavy mechanical load is placed on the induction motor, the rotor slows still
more; that is, the slip increases somewhat. For most motors, the full-load slip will be
about 0.03. The frequency of the rotor voltages and currents will still be quite small
(say, about 2Hz), but the effect of leakage reactance can be neglected no longer. There
will be a time phase lag between the rotor currents and voltages. If Ll2 is the effective
leakage inductance of a typical rotor circuit composed of a pair of rotor bars and their
end ring connection, the reactance of this circuit will be
XRs = 2πf2Ll2 = 2πsf1 Ll2
The phase angle of lag between maximum voltage and maximum current in a typical
rotor circuit will be the power-factor angle:
θ2 = tan-1(XRs /RR)
30
Where RR is the ac resistance of the same circuit. Peak current in a pair of bars occurs θ2o
in the cycle after peak induced voltage. During this interval, the rotor will have slipped θ2
electrical degrees relative to the rotating air gap field. If maximum voltage is being
induced in conductors x and y, maximum current will not flow in these conductors until
they have slipped an additional θ2 electrical degrees. Thus in Slide 32, maximum current
occurs in conductors x' and y', which lag x and y in rotation by θ2 electrical degrees.
Under load, then, the entire rotor current pattern is shifted from the lightly loaded
position, with the result that the rotor MMF vector F2 is now at an angle of 90° + θ2
behind the flux, and therefore behind R, the MMF responsible for the flux.
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32
Circuit Model Of The Induction Machine
The induction machine, in certain aspects, is like a synchronous machine. In others, it
is like a transformer. Charles Steinmetz developed the most widely used circuit model
of the polyphase induction machine. In developing his model, Steinmetz took a
transformer approach. This model has proved very useful in understanding the
characteristics of the induction machines.
Similarities between the induction motor and the transformer are readily apparent.
Power from a sinusoidal source is supplied to a stator winding or primary. This
winding establishes a flux that mutually couples a rotor winding or secondary. The
cyclic mutual flux traverses a ferromagnetic material that gives rise to eddy current
and hysteresis losses. Not all of the flux established by the primary winding
necessarily links the secondary winding, suggesting that leakage reactance exists. Any
current that flows in the secondary winding acts to oppose changes in the primary
generated mutual flux, requiring the existence of an mmf balance.
33
As a result of these similarities, the practical transformer equivalent circuit seems like
suitable candidate for modeling the induction motor. However, there are three notable
dissimilarities between the induction motor and the transformer that would be
expected to impact direct application of the practical transformer model.
• The mutual flux path must cross the high-reluctance air gap, resulting in reduced
magnetizing reactance and increased leakage reactance for the induction motor as
compared to a transformer.
• Also, there is relative motion between the induction motor and secondary coils.
• Additionally, the secondary winding has its terminals shorted.
The magnetic conditions in the motor at the instant the current in phase a is a
maximum are illustrated in Slide 35.
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35
The relative motion of the stator flux and the rotor conductors induces voltages of
frequency f2 = s f1 called the slip frequency, in the rotor. Thus, the electrical behavior of
an induction machine is similar to that of a transformer but with the additional feature
of frequency transformation produced by the relative motion of the stator and rotor
windings. In fact, a wound-rotor induction machine can be used as a frequency changer.
The rotor terminals of an induction motor are short circuited; by construction in the
case of a squirrel-cage motor and externally in the case of a wound-rotor motor. The
rotating air-gap flux induces slip-frequency voltages in the rotor windings. The rotor
currents are then determined by the magnitudes of the induced voltages and the rotor
impedance at slip frequency. At starting, the rotor is stationary (n = 0), the slip is unity
(s = 1), and the rotor frequency equals the stator frequency f1. The field produced by
the rotor currents therefore revolves at the same speed as the stator field, and a
starting torque results, tending to turn the rotor in the direction of rotation of the
stator-inducing field. If this torque is sufficient to overcome the opposition of
rotation created by the shaft load, the motor will come up to its operating speed. The
operating speed can never equal the synchronous speed however, since the rotor
conductors would then be stationary with respect to the stator field; no current would
be induced in them, and hence no torque would be produced.
36
With the rotor revolving in the same direction of rotation as the stator field, the
frequency of the rotor currents is sf1 and they will produce a rotating flux wave which
will rotate at sns rev/min with respect to the rotor in the forward direction. But
superimposed on this rotation is the mechanical rotation of the rotor at n rev/min. Thus,
with respect to the stator, the speed of the flux wave produced by the rotor currents is
the sum of these two speeds and equals
sns + n = sns + ns(1-s) = ns
Therefore, we see that the rotor currents produce an air-gap flux wave which rotates at
synchronous speed and hence in synchronism with that produced by the stator currents.
Because the stator and rotor fields each rotate synchronously, they are stationary with
respect to each other and produce a steady torque, thus maintaining rotation of the
rotor. Such torque, which exists for any mechanical rotor speed n other than synchronous
speed, is called an asynchronous torque.
37
Induced EMF
Where
38
Equivalent Circuit Per Phase

At standstill (n = 0 , s = 1)
The equivalent circuit of an induction motor at standstill is the same as that of a
transformer with secondary short circuited.
x1
r1
I1
V1
x2
I2
r2
Iex
Ih+e
Rc
Iφ
Xm
E1 = E 2
All values are
per phase
Where
E2 = per-phase induced voltage in the rotor at standstill referred to stator side.
x2 = per-phase rotor leakage reactance at standstill referred to stator side.
r2 = per-phase rotor resistance referred to stator side.
39
Equivalent Circuit Per Phase (cont.)

At any slip s
When the rotor rotates with speed n the rotor circuit
frequency will be: f 2  s f1
Therefore induced voltage in the rotor at any slip s will
be ERs = sERB , similarly XRs = s XRB
and the rotor equivalent circuit per-phase will be:
XRB
XRs = sXRB
ERs = sERB
IR
IR
IR
RR
ERB
XRB
RR
RR/s
ERB
RR (1-s)/s
Where
IR 
sERB
ERB

RR  jsX RB ( RR / s)  jX RB
40
Equivalent Circuit Per Phase (cont.)
1-The Complete Equivalent Circuit per phase
Airgap &
Magnetic Circuit
Stator Circuit
I1
V1
r1
Iex
x1
E1
Rc X m
Load +
Rotation losses
Rotor Circuit
x2
I2
r2
r2(1-s)/s
41
Equivalent Circuit Per Phase (cont.)
3-Thevenin Equivalent Circuit
In order to simplify the computations, IEEE recommended an
equivalent circuit that can be replaced by the following Thevenin’s
equivalent values:
Vth 
Rth
Xm
r  ( x1  X m )
2
1
2
x2
Xth
V1
I2
r2 /s
Vth
Zth 
jX m (r1  jx1 )
 Rth  jX th
r1  j ( x1  X m )
gap Air
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POWER FLOW IN INDUCTION MOTORS
dev
= 3V1I1cos
43
POWER FLOW IN INDUCTION MOTORS
dev
= 3V1I1cos
r2
s
PRCL  3( I 2 ) 2 r2
PAG  3( I 2 ) 2
Pdev  3( I 2 ) 2
r2 (1  s)
s
 PAG : PRCL : Pdev  1 : s : (1  s)
x1
r1
x2
I2
I
I1
V1
Ic
Rc
Im
r2
I2’
r2 (1-s)
s
Xm
PAG
44
PERFORMANCE CHARACTERSTICS
Rth
x2
Xth
Torque / speed Curve
Consider Thevenin’s equivalent circuit
The rotor current referred to stator side is:
I2 
I2
r 2 /s
Vth
Vth
2
r 

2
 Rth  2   ( X th  x2 )
s

gap Air
And the developed torque T is :
PAG  (1  s) PAG 3( I 2 ) 2 r2
T



m s  (1  s) s
s
Pdev
r
T  3 2
ss 
r
 Rth  2
s

Vth2
2

  ( X th  x2 ) 2

Speed p.u.
45
Torque/Speed Curve (cont.)
Starting Torque:
Maximum torque
at starting s =1, so the starting torque is:
Vth2
TSt  3
s Rth  r2 2  ( X th  x2 ) 2
r2
Unstable
Stable
For maximum torque
dT
0 ,
ds
sT max 
Tmax
from which
r2
[ Rth2  ( X th  x2 ) 2 ]1/ 2
Vth2

2S Rth  [( Rth ) 2  ( X th  x2 ) 2 ]1/ 2
Starting torque
STmax
3
46
Torque/Speed Curve (cont.)
If the stator resistance is small, so Rth may be neglected
 sT max
r2

( X th  x2 )
and
Tmax
Vth2

2S ( X th  x2 )
3
and the ratio between the maximum torque and the torque developed at
any speed is given by:
Tmax
(r2 / s) 2  ( X th  x2 ) 2
(r2 / s) 2  (r2 / sT max ) 2
s
s

(
)

(
)
2
2
2
2
T
(r2 / sT max )  ( X th  x2 ) sT max
(r2 / sT max )  (r2 / sT max ) sT max
Tmax s 2T max  s 2

T
2sT max s
47
Torque/Speed Curve for varying R2
Tmax
Vth2

2S ( X th  x2 )
sT max 


3
ST max 
r2
( X th  x2 )
R2'
( X th  X 2' )
The maximum torque is independent of
the rotor resistance. However, the value
of the rotor resistance determines the slip
at which the maximum torque will occur.
The torque-slip characteristics for various
values of are shown.
To get maximum torque at starting::
sT max  1
i.e. r2  ( X th  x2 )
48
Stator current and input power factor

From the IEEE recommended equivalent circuit, the input impedance Zin
jx 2
jx1
r1
is:
r

jX m  2  jx 2 
 s

Z in  r1  jx1 
r2
 j X m  x2
s
Z in  Z in  

I1

V1
r2/ s
jXm
Maximum or
pull-out torque
T
I1

The stator current I1 is :
V1
I1 
Z in

And the input power factor is :
p.f. = cos ()
p.f.
h
Tst
art
1
s
Starting
Typical Induction Motor
Characteristics
0
No-load
49
Example (1)
50
Solution
51
52
53
54
55
Control Of Performance By Rotor Design
A simple way of obtaining a rotor resistance which is automatically
vary with speed make use of the fact that at standstill the rotor
frequency equals the stator frequency; as the motor accelerates, the
rotor frequency decreases to a very low value, perhaps 2 or 3 Hz at full
load in a 60 Hz motor. With suitable shapes and arrangements for
rotor bars (deep-bar rotors [B] or double-cage rotors [C]), squirrelcage rotor can be designed so that their effective resistance at 60 Hz
(at starting) is several times their resistance at 2 or 3 Hz (running
condition).
56
57
Deep-Bar Rotors
• The leakage flux paths for a deep-bar design are illustrated in Slide 59(a). Since
inductance is the number of flux linkages produced per ampere (L = λ/I), it is obvious
that the parts of each bar extending deeply into the iron have higher leakage
inductances than those parts of the bar cross section near the air gap, because more of
the leakage flux links the deeper parts of the bar. One may think of each bar as being
composed of several layers of equal cross section, and thus of equal resistance, but
with inductances increasing with depth, as in Slide 59(b).
• Now under running conditions the slip is quite small, and therefore the frequency of
the rotor currents is only 1 or 2 Hz. As a result, all of the leakage reactances are
negligible, and the current distribution is essentially uniform throughout each rotor
bar. The effective resistance of the rotor is that of all layers in parallel. This low
resistance makes for low slip and high efficiency at full load. As the motor is
overloaded, however, slip and rotor frequency increase. The reactances of the deeper
levels of the bar come into play with two effects: (1) The value of x2 in the circuit model
is larger than it would be for shallow bars like those of Design A, and the breakdown
torque is reduced. (2) The rotor current begins to be crowded into those layers of the
bars near the air gap, increasing the effective resistance of rotor circuits. Thus
breakdown torque occurs at a somewhat greater slip than for Design A.
58
• When the motor is first connected to the line for starting, the slip is unity a frequency
of the rotor currents equals the line frequency. Rotor-bar currents are concentrated near
the air gap. Effective bar resistance and reactance are both high, with the result that
starting current is only about 75% of that for Design A.
• Thus a deep-bar rotor achieves good full-load efficiency at low slip, together with
reduced starting current, at the expense of somewhat lower pull-out torque. A typical
speed/torque curve for a deep-bar motor is shown in Slide 60 as the Design Class B curve,
although this same characteristic may also be obtained with a double-cage rotor.
59
60
Double-Cage Rotors
• A double-cage rotor has two squirrel cages. The two cages may connect to the same
end rings, or there may be separate end rings for each cage. The usual arrangement is
that of NEMA Design C, as illustrated in Slide 57. The outer cage for C is composed of
small, high-resistance bars, while the inner cage is made up of larger, low-resistance
bars. Since the inner bars are almost totally surrounded by the iron of the rotor core,
their leakage inductance is very great, indeed. For this reason they have little effect on
motor performance except at low slips; that is, they come into play when the motor has
reached running speed. At running speed the two cages are essentially in parallel, and r2
in the circuit model is quite low. Full-load slip is low, almost as low as for Design A, and
full-load efficiency is almost as high as for Design A, see Slide 60.
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• At starting, the inner cage may nearly be disregarded, and the outer cage designed
for optimum starting characteristics. The iron between the outer cage and the inner
cage, together with the proximity of the outer cage with the stator windings, causes
the leakage reactance to be low and permits a large maximum torque. If desired, the
resistance of the outer cage may be chosen to realize this high torque at starting. The
high outer-cage resistance keeps the starting current low.
• Double-cage construction is the most costly of the squirrel cage designs, but it
allows many of the advantages of the still more costly and less rugged wound rotor to
be included in a squirrel cage machine.
62
Classes of squirrel-cage motors

According to the National Electrical Manufacturing Association
(NEMA) criteria, squirrel-cage motors are classified into class A, B,
C or D. The torque-speed curves and the design characteristics for
these classes are :
Class
Starting Current
Starting Torque
Rated Load Slip
A
Normal
Normal
< 5% (less than that of design B)
B
Low
Normal
< 5% (usually 0.02 to 0.03)
C
Low
High
< 5% (slightly larger than design B)
D
Low
Very High
8-17 %
63
Tests To Determine Circuit Model Parameters
Since Steinmetz's induction machine model is similar to his transformer model, the
tests usually employed to evaluate its impedances are analogous to the open-circuit
and short-circuit tests on transformers. In addition, a dc measurement of r1 is required.
Many factors must be taken into account in making these tests. Winding resistances
vary with temperature, and rotor resistance is a function of frequency. Test procedures
are presented in great detail in IEEE Standard 112-1984. That standard should be
consulted when accurate measurements are necessary. The following paragraphs,
however, provide an understanding of the fundamental theory underlying these tests.
64
The No-Load Test
The no-load test corresponds to the open-circuit test on a transformer. However the
test measures friction and windage losses in addition to core losses. Slide 66 shows the
test circuit and the effect of zero mechanical load on the model. The mechanical load
per phase, represented by r2(1-s)/s, is simply the internal windage and friction of the
machine. Since this amounts to only a few percent of the rated mechanical load, the
slip is very small, say, 0.001, and r2(1 - s)/s is very large compared to x2 and r2 . It is
thus essentially in parallel with Rc and jXm, where Rc represents the core loss per
phase. In Slide 60, Rfwc represents the parallel combination of Rc and r2(1 - s)/s, and
absorbs a power equal to one third of the total friction, windage, and core loss. Since
under no-load conditions r2 is so much less than r2(1 - s)/s, the no-load rotor copper
loss is negligible. The no-load input power Pnl , then, may be taken to be the sum of the
stator copper loss and Pfwc, the friction, windage, and core loss.
Pnl = P1 + P2 = SCLnl + Pfwc = 3I2nl r1 + Pfwc
This permits the sum Pfwc to be evaluated: Pfwc = Pnl - 3I2nl r1
65
66
Where Pnl is the total three-phase power input to the machine at rated voltage and
frequency, and Inl is the average of the three line currents, with no mechanical load
connected to the shaft.
Under no-load conditions, the power factor is quite low, 0.2 or less, so the circuit is
essentially reactive. The current will be in the neighborhood of one third the rated
current. These facts imply that r1 is small compared to Xm and that Rfwc is much larger
than Xm. The input Impedance at no load is thus approximately jx1 and jXm in series:
The value of r1 and x1 have yet to be determined
67
Direct-Current Test
In the case of a transformer, it was only necessary to obtain Req1 or Req2 to solve
practical problems. Knowledge of the individual winding resistances was not
essential. In the induction machine, however, r2 plays a crucial role, and r1 must be
known to evaluate the stator copper loss.
With a steady direct current flowing in the stator windings, the stator magnetic field
is fixed in space. If no external torque turns the shaft, the flux linkages with the
circuits are constant in time. As a result, no voltages are induced in the rotor, no
currents flow, and no torque is developed. There is no rotor MMF (F2 = 0), and the
rotor circuits have no electrical effect on those of the stator. Thus the dc stator
winding resistance may be measured independently of the rotor impedance.
68
The circuit is shown in Slide 70. The rheostat must be capable of handling rated
current of the motor. The current is adjusted to approximately rated value, so that
the temperature of the winding approximates that of running conditions. (The
winding temperature is often measured so that the test resistance may be corrected
to actual operating temperature.) Applying Ohm's law to the circuit of Slide 70,
This value of r1 may now be substituted in Pfwc = Pnl - 3I2nl r1 to obtain Pfwc.
69
70
The Blocked-Rotor Test
This test, which corresponds to the short-circuit test of a transformer, is also called the
locked-rotor test. The shaft is clamped so that it cannot turn; thus ω = 0 and s = 1
during this test. If full voltage at rated frequency were applied, the current would be
five to eight or more times rated value. For this reason, locked-rotor tests are not done
at full voltage except for small motors. Instead, the applied voltage is usually adjusted
upward from zero until the stator current is at rated value. Blocked-rotor test
measurements are, then, taken at rated current rather than at rated voltage.
Under running conditions the frequency of the stator currents is line frequency f1,
while that of the rotor currents is sf1 only a few hertz. In the blocked-rotor test,
however, the slip is unity and the rotor and stator currents are at the same frequency.
Since both rotor and stator quantities are to be evaluated, the frequency of the
blocked-rotor test voltage ft, is usually a compromise between the normal stator and
rotor frequencies.
71
The test circuit and the effect of blocking the rotor on the model are shown in Slide 73.
The voltage appearing across one phase of the stator winding VBR/√ 3, and the
magnitude of the impedance looking into the terminals of one ph of the machine is
Where IBR is the average of the three line currents and should be roughly equal to
rated current.
With unity slip, (r2 /s) = r2. The applied voltage is quite low and the iron core the
machine is unsaturated. As a result, Pp is greater than at rated voltage and Xm is larger
than normal. Even at rated voltage, Xm is typically 25 times the magnitude r2 + jx2. The
effect of Xm may thus be neglected, and the assumption made that
72
73
Where X’BR = x’1 + x’2 is the blocked-rotor reactance at the test frequency ft .The real
part of ZBR is given by the power per phase divided by the phase current squared:
Now r1 was found by the dc test, then
The imaginary part of ZBR is given by:
If rated frequency is fB and the frequency of the blocked-rotor test is ft, then
74
The designer of the machine knows, on the basis of his or her calculations, how much
of XBR is x1 and how much is x2. There is no way to determine this ratio from tests on a
squirrel cage motor, however. Experience with these machines has been condensed
into the table given which shows approximately how XBR divides between x1 and x2 for
rotors of different designs.
75
Example (2)
The following test data were obtained on a 5-hp, four-pole, 220-V, three-phase, 60-Hz
Design B induction motor having a rated current of 12.9-A.
Calculate the performance of this motor at a slip of 0.03.
76
Solution
77
78
79
80
81
Starting Induction Motors
82
83
84
Starting Induction Motors (cont.)
1. Direct-On-Line Starting
Induction motors when started by connecting them directly across the
supply line take 5 to 8 times their full-load current. This initial excessive
current may cause large line voltage drop that affects the operation of other
electrical equipment connected to the same lines. This method is suitable
for small motors up to 10-hp rating.
2. Auto-transformer
A three-phase step-down autotransformer may be employed as a reduced
voltage starter. As the motor approaches full speed, the autotransformer is
switched out of the circuit.
85
Starting Induction Motors (cont.)
3. Star-delta method
The normal connection of the stator windings is delta
while running. If these windings are connected in star
at starting, the phase voltage is reduced, resulting in
less current at starting. As the motor approaches the
full speed, the windings will be connected in delta.
4. Solid-State Controller
The controller can provide smooth starting.
86
Starting Induction Motors (cont.)
5. Using External Rotor Resistance
For wound rotor induction motor only. The external resistance can be
chosen to get high torque and low current at starting and can be
decreased as motor starts up.
87
Speed control of induction motors

1. Line Voltage Control
Recall that the torque developed in an induction motor is proportional to the square of
the terminal voltage. A set of T – n characteristics with various terminal voltages is
shown. If the motor drives a fan load the speed can be varied over the range n1 to n2 by
changing the line voltage. Note that the class D motor will allow speed variation over a
wider speed range.
T Vs2
88
The terminal voltage V1 can be varied by using a three-phase autotrans-former or a
solid-state voltage controller as shown. The auto-transformer provides a sinusoidal
voltage for the induction motor, whereas the motor terminal voltage with a solid-state
controller is nonsinusoidal. Speed control with a solid-state controller is commonly
used with small squirrel-cage motors driving fan loads. In large power applications an
input filter is required; otherwise, large harmonic currents will flow in the supply line.
T1
ia
iL
A
T4
T3
vAN
vBN
vCN
Rotor
C
ib
B
N
T6
T5
Rotor
ic
T2
89
The thyristor voltage controller shown in Slide 91 is simple. During the operation of the
voltage controller the command signal for a particular set speed fires the thyristors at a
particular firing angle (α) to provide a particular terminal voltage for the motor. If the
speed command signal is changed, the firing angle (α) of the thyristors changes, which.
results in a new terminal voltage and thus a new operating speed.
Open-loop operation is not satisfactory if precise speed control is desired for a particular
application. In such a case, closed-loop operation is needed. If the motor speed falls
because of any disturbance, such as supply voltage fluctuation, the difference between
the set speed and the motor speed increases. This changes the firing angle of the
thyristors to increase the terminal voltage, which in turn develops more torque. The
increased torque tends to restore the speed to the value prior to the disturbance.
Note that for this method of speed control the slip increases at lower speeds, making the
operation inefficient. However, for fans, or similar centrifugal loads in which torque
varies approximately as the square of the speed, the power decreases significantly with
decrease in speed. There-fore, although the power lost in the rotor circuit (=sPg) may be
a significant portion of the air gap power, the air gap power itself is small and therefore
the rotor will not overheat. The voltage controller circuits are simple and, although
inefficient, are suitable for fan, pump, and similar centrifugal drives.
90
91
Speed control of induction motors

2. Line Frequency Control
n=(1-s)ns , ns= 120 f1/p
The synchronous speed and hence the motor speed can be varied by changing the
frequency of the supply. Application of this speed control method requires a frequency
changer. The Figure shows a block diagram of an open-loop speed control system in
which the supply frequency of the induction motor can be varied.
92
From the emf equation the motor flux is
If the voltage drop across r1 and x1 is small compared to the terminal voltage V1, that is,
, then
To avoid high saturation in the magnetic system, the terminal voltage of the motor must be
varied in proportion to the frequency. This type of control is known as constant volts per
hertz. At low frequencies, the voltage drop across r1 and x1 (Slide 41) is comparable to the
terminal voltage V1 , and therefore
is no longer valid. To maintain the same air gap flux density, the ratio V/f is increased for
lower frequencies. The required variation of the supply voltage with frequency is shown in
Slide 94. In Slide 92 the machine voltage will change if the input voltage to the inverter Vi is
changed; Vi can be changed by changing the firing angle of the controlled rectifier. If the
output voltage of the inverter can be changed in the inverter itself (as in pulse-widthmodulated inverters PWM), the controlled rectifier can be replaced by a simple diode
rectifier circuit, which will make Vi constant.
93
The Torque-Speed characteristics for variable-frequency operation are shown in Slide 95.
At the base frequency fbase the machine terminal voltage is the maximum that can be
obtained from the inverter. Below this frequency, the air gap flux is maintained constant
by changing V1 with f1 ; hence the same maximum torques are available. Beyond fbase,
since V1 cannot be further increased with frequency, the air gap flux decreases, and so
does the maximum available torque. This corresponds to the field-weakening control
scheme used with dc motors. Constant-horsepower operation is possible in the fieldweakening region.
94
95
In Slide 95 the Torque-Speed characteristic of a load is superimposed on the motor
Torque-Speed characteristic. Note that the operating speeds n1 …. n8 are close to the
corresponding synchronous speeds. In this method of speed control, therefore, the
operating slip is low and efficiency is high.
The inverter in Slide 92 is known as a voltage source inverter. The motor line-to-line
terminal voltage is a quasi-square wave of 120° width. However, because of motor
inductance the motor current is essentially sinusoidal.
96
Speed control of induction motors

3. Rotor Resistance Control
For wound-rotor induction motors only.
a- Classical technique:
It was pointed out that the speed of a wound-rotor induction machine can be
controlled by connecting external resistance in the rotor circuit through slip rings, as
shown in Slide 98. The Torque-Speed charac-teristics for four external resistances are
shown. The load T - n characteristic is also shown by the dashed line. By varying the
external resistance 0 < R-ex < Rex4, the speed of the load can be controlled in the range
n1 < n < n5. Note that by proper adjustment of the external resistance (Rex = Rex2),
maximum starting torque can be obtained for the load.
The scheme requires a three-phase resistance bank, and for balanced operation all
three resistances should be equal for any setting of the resistances. Manual
adjustment of the resistances may not be satisfactory for some applications,
particularly for a closed-loop feedback control system.
97
98
b- An alternative methods of varying rotor-circuit resistance
Using three-phase diode bridge and a single variable resistor.
Solid-state control of the external resistance may provide smoother operation. A
block diagram of a solid-state control scheme with open-loop operation is shown in
Slide 100. The three-phase rotor power is rectified by a diode bridge. The effective
value Rex* of the external resistance Rex can be changed by varying the on-time (also
called the duty ratio λ).
When λ = 0.0, that is, when the chopper is off all the time, Rex* = Rex. When λ = 1.0,
that is, the chopper is on all the time, Rex is short-circuited by the chopper and so
Rex* = 0. In this case, the rotor circuit resistance consists of the rotor winding
resistance only. Therefore, by varying λ in the range 1 > λ > 0, the effective resistance
is varied in the range 0 < Rex* < Rex, and Torque-Speed characteristics similar to those
shown Slide 98 are obtained.
99
100
Speed control of induction motors

4. Rotor Slip-Energy Recovery System Control
For wound-rotor induction motors only.
If the no-load slip is so, then
101
Speed control of induction motors

5. Pole-Changing Control
For squirrel-cage induction motors only.
Because the operating speed is close to the synchronous speed, the speed of an
induction motor can be changed by changing the number of poles of the machine.
This can be done by changing the coil connections of the stator winding. Normally,
poles are changed in the ratio 2 to 1. This method provides two synchronous speeds.
This method is not practical for wound-rotor motors, because the rotor windings have
to be reconnected to have the same number of poles as the stator. A squirrel-cage
rotor automatically develops a number of magnetic poles equal to those of the air gap
field. It is obvious, however, that the speed can be changed only in discrete steps and
that the elaborate stator winding makes the motor expensive. The basic scheme is
illustrated in Slide 103, which shows what happens in a typical phase of a two-speed
polyphase winding.
102
103
MODES OF OPERATION
The induction motor can be operated in three modes: motoring, generating, and
plugging as shown in Slide 105. To illustrate these three modes of operation,
consider an induction machine mechanically coupled to a dc machine, as shown in
Slide 105.
1. Motoring 0 < S < 1
In normal motor operation, the rotor revolves in the direction of rotation of the
magnetic field produced by the stator currents. This is the natural (motoring) mode of
operation of the induction machine. The steady-state speed n is less than the
synchronous speed ns .
2. Generating S < 0 (Negative slip)
The induction machine will operate as a generator if its stator terminals are connected
to a three-phase supply and its rotor is driven above the synchronous speed by a
prime mover (resulting in negative slip).
The generating mode of operation is utilized in some drive applications to provide
regenerative braking. For example, suppose an induction machine is fed from a
variable-frequency supply to control the speed of a drive system. To stop the drive
system, the frequency of the supply is gradually reduced.
104
105
In the process, the instantaneous speed of the drive system is higher than the
instantaneous synchronous speed because of the inertia of the drive system. As a
result, the generating action of the induction machine will cause the power flow to
reverse and the kinetic energy of the drive system will be fed back to the supply. The
process is known as regenerative braking.
3. Plugging (Braking) S > 1
Suppose an induction motor is running at a steady-state speed. If its terminal phase
sequence is changed suddenly, the stator rotating field will rotate opposite to the
rotation of the rotor, producing the plugging operation. The motor will come to zero
speed rapidly and will accelerate in the opposite direction.
This mode of operation is sometimes utilized in drive applications where the drive
system is required to stop very quickly. Suppose an induction motor is running at a
steady-state speed. If its terminal phase sequence is changed suddenly, the stator
rotating field will rotate opposite to the rotation of the rotor, producing the plugging
operation. The motor will come to zero speed rapidly and will accelerate in the
opposite direction, unless the supply is disconnected at zero speed.
106
107
Example (4)
108
Solution
109
110
111
112
The Circle Diagram Of The Induction Machine
The circle diagram is based on the approximate equivalent circuit of the induction
motor shown in Slide 114. The shunt branch has been moved from points a, b to the
input terminals and includes a resistor Rfwc to represent rotational losses.
Since all rotational losses are taken care of by Rfwc, the approximate circuit assumes
zero slip at no load on the motor shaft. Then at no load, r2/s is an open circuit, and
the no-load line current Inl is V1/Zs. By Kirchhoff's current law, the line current under
load is I1 = Inl + I2, where
The current I2 will lag V1 by the impedance angle
Therefore I2 may be rewritten as follows:
113
114
This is the polar equation of a circle of the form ρ = D sin θ, where D is the circle
diameter. Thus the locus of I2 is a circle of diameter
Where, (x1 + x2) is the blocked-rotor reactance when the blocked-rotor test is made at
rated frequency.
Therefore, the equation for the circle diagram may be rewritten as follows:
115
The resulting diagram is shown in Slides 117 or 118. Since I1 is the sum of Inl and I2, the
point of the I1- phasor will coincide with that of I2. Thus the circle serves as the locus
of both I1 and I2, though they radiate from different origins.
At starting, I1 and I2 will be at some point H. As the motor accelerates, the tips of the
current phasors travel around the circle in a CCW direction until the output torque
matches the load torque. If there is no shaft load, the motor will continue to accelerate
to very near synchronous speed (according to the approximate model, synchronous
speed actually will be reached). At this point, I2 = 0 and I1 = Inl.
116
117
118
For example (Slide 123), the input power to the motor is given by
Then, for constant applied phase voltage the input power is proportional to I1 cos θφm
i.e. proportional to length AE. Then for an appropriate chosen scale, the length AE is
calibrated in terms of power. Similarly the length AB represents the input no-load
power (Prot = core loss + mechanical loss). Consequently, the length BE represents:
1) Stator copper loss + 2) Rotor copper loss + 3) Output mechanical power. The phasor
OH represents blocked-rotor stator current I1BR, therefore FH represents (rotor +
stator) copper losses (mechanical power output = zero in this case), i.e.
Triangles O’FH & O’BD are similar, therefore
119
Therefore, BD represents the (stator + rotor) copper loss. Then DE represents output
mechanical power. Therefore, line O’H is then called power line since the ordinate
bounded by this line and the circumference O’EH represents output power by the
motor. If added to these ordinates the rotor copper loss, then the new ordinates will
be proportional to air gap power Pag which equals motor torque in synchronous
watts. This is achieved by dividing FH in the ratio of r2 : r1, then line O’G is termed
torque line. Therefore for the operating point E shown in Slide 123:
120
121
Measurements of slip in this way give poor accuracy. The method described in Slide 123
gives results with better accuracy. To evaluate the slip accurately point Q is joined by
point H by the straight line QHS. Then from point S a line SK is drawn perpendicular to
the torque line. This line is divided in equal divisions from 0 (point K) to 100 (point S).
To determine the slip for the point E, for example, the line EQ is drawn which intersects
SK at a point W graduated to the value of slip.
122
123
124
125
Example (5)
A 22.38-kW, 500-V, 4-pole, 50-Hz, star connected three-phase induction
motor has the following test data:
No-load
: 500-V 8.2-A 1.43-kW
Blocked-rotor: 140-V 45-A 3.2-kW
The equivalent rotor resistance is equal to the stator resistance.
To a proper scale draw the circle diagram and find:
(a) Stator current and power factor at full load,
(b) Maximum torque and,
(c) Starting torque in terms of full-load torque.
Answers:
(a) Stator full-load current =32-A, Full-load power factor = 0.91 lagging
(b) Maximum torque = 374.9-N.m.
(c) Starting Torque = 0.868 x Full-Load Torque
126
Unbalanced operation of three-phase
Induction Motors
Three-phase induction motors may operate under unbalanced
conditions such as:
 Unequal magnitude of the three-phase supply voltages
 Unequal phase-shift between the three-phase supply voltages
 Disconnected one of the three phases.
Evaluation of the three-phase induction motor performance under
such conditions can be found using Symmetrical Components
Technique. In such technique the unbalanced voltage supply can
be analyzed into its equivalent three-sets of symmetrical
components [ positive, negative, and zero sequence].
127
Unbalanced Supply
For unbalanced supply voltages
Vas  V11
Vbs  V2  2
Vcs  V3 3
Then the equivalent sequence voltages are given by:
1
( Vas  a.Vbs  a 2 .Vcs )
3
1

V  ( Vas  a 2 .Vbs  a.Vcs )
3
1
V 0  ( Vas  Vbs  Vcs )
3
where a  1 120 
, a 2  1 240 
V 
128
Symmetrical Components
Vas
Vas
Vas
Vcs Vas0 0
Vbs
Vcs0
+
Vbs
Vcs

cs
V
Vbs
+
Vbs
129
Unbalanced operation of threephase Induction Motors (cont.)
The motor will then operates under the influence of the three
balanced sets of voltages.
The zero sequence voltage will produce no resultant torque.
Regarding the resultant torque due to positive and negative
sequence sets of voltages, it will depend on the relative magnitude
of these sets. If V+ > V- then the motor will rotate with the positive
sequence set of rotation, while if V+ < V- then the motor will rotate
with the negative sequence set of rotation.
Denoting the set of voltages of higher magnitude as the forward set
the other will be denoted as the backward one.
130
Motor equivalent circuits
The equivalent circuit per phase as seen by both forward and
backward components will be as shown
R1
X’2
X1
X’2
X1
’
Ib
If
Xm
Vf
Forward eq. circuit
where
R1
’
Sf 
n s  nr
S
ns
R2'/Sf
Xm
Vb
R2'/Sb
Backward eq. circuit
Sb 
n s  (nr )
 2S
ns
131
Motor equivalent circuits
The equivalent circuit per phase as seen by both forward and
backward components may be redrawn as follows:
R1
X1
R1
X1
If
Vf
Rf
Ib
Vb
Xf
R2
where Z f  R f  jX f  (  jX 2 ) //( jX m )
S
R
Z b  Rb  jX b  ( 2  jX 2 ) //( jX m )
2S
Vf
Vb
If 
and I b 
Z1  Z f
Z1  Z b
Rb
Xb
132
Power and Torque
From the equivalent circuits the air gap powers due to the forward
and backward fields are:
Pgf  3 I .R f
2
f
and
Pgb  3 I .Rb
2
b
And the resultant developed torque is given by:
T  T f  Tb 
PgF  Pgb
s

3 ( I 2f .R f  I b2 .Rb )
s
133
Example (4)
A 220 V, 3-phase, Y- connected, 50 Hz, 4-pole, squirrel cage
induction motor develops its full load torque at slip of 0.04 when
operated at rated voltage and frequency. The equivalent rotor
resistance and reactance referred to stator side are 0.35 and 1.4 Ω
respectively and the magnetizing reactance Xm is 30 Ω . The stator
phases are aa', bb', and cc', star point is obtained by connecting a',
b', and c'. Neglecting stator impedance,
i. Find the full load torque
ii. If the phase bb' is reversed (the star point is obtained by connecting
a', b, and c') calculate the developed torque at 0.04 slip.
134
Solution
i- With balanced supply, the equivalent circuit/phase is:
Vcc'
I1
Vaa'
V
Vbb'
I'2
0.35/S
J30
j1.4
220 
0  1270
3
120 f1 120 * 50
ns 

 1500 rpm
p
4
S  0.04
V
1270
 I 2 
 14.34  9.1
0.35
(
 j1.4)
0.04
2
Pg 3I 2 R2
3(14.34) 2 (0.35)
T


 34.36 Nm
2 .1500
s
S s
(0.04)(
)
60
135
ii- When reversing phase bb' (unbalanced supply)
Vcs
Vas  1270 
,Vbs  12760 
,Vcs  127120 
1
V   (Vas  a.Vbs  a 2 .Vcs )
3
127

(1  160 . 1120  1120 . 1240)
3
127

(1  (1)  1)  42.340 
3
1
(Vas  a 2 .Vbs  a.Vcs )
3
127

(1  160 . 1240  1120 . 1120)
3
127

(1  1300  1240)
3
127

(1  j1.7321)  84.67  60 
3
Since V   V  , V f  84.67  60  and Vb  42.340 
Vbs
Vas
V 
136
If
I'2f
Vf
0.35/S
Ib
Vb
J30
I'2b
J30
j1.4
j1.4
Forward
0.35/(2-S)
Backward
0.35
 j1.4)( j 30)
Z f  0.04
 8.1524.67   7.41  j 3.4
0.35
(
 j 31.4)
0.04
(
84.67  60
If 

 10.39  84.67
Z f 8.1524.67
Vf
Pgf  3 . I 2f . R f  3 (10.39) 2 7.41  2399.7 W
137
Ib
Vb
I'2b
0.35/(2-S)
J30
j1.4
Backward
0.35
 j1.4)( j 30)
Z b  2  0.04
 1.3583.41  0.155  j1.34
0.35
(
 j 31.4)
2  0.04
V
43.340
Ib  b 
 31.36  83.41
Z b 1.3583.41
(
Pgb  3 . I b2 . Rb  3 (31.36) 2 0.155  457.3 W
T
Pgf  Pgb
s

2399.7  457.3
 12.36 Nm
157.1
138