Ch7 Induction Motor

CHAPTER 7 – INDUCTION MOTORS
 Induction machine – rotor voltage (that produces the rotor current

and the rotor magnetic field ) is induced in rotor windings. No
need for physical wires or dc field current (like in
synchronous machine). It can be motor or generator. But rarely
used as generator due to many disadvantages (eg: it always
consumes reactive power, low PF, not stand alone).
 The basic difference between an induction motor and a
synchronous AC motor is that in the latter a current is supplied onto
the rotor. This then creates a magnetic field which, through magnetic
interaction, links to the rotating magnetic field in the stator which in
turn causes the rotor to turn. It is called synchronous because at
steady state the speed of the rotor is the same as the speed of the
rotating magnetic field in the stator.
Electric Machines & Power Electronics 1

ENEE430
 The induction motor does not have any direct supply onto the rotor;
instead, a secondary current is induced in the rotor. To achieve this,
stator windings are arranged around the rotor so that when energised
with a polyphase supply they create a rotating magnetic field pattern
which sweeps past the rotor. This changing magnetic field pattern
induces current in the rotor conductors. These currents interact with
the rotating magnetic field created by the stator and in effect causes a
rotational motion on the rotor.
 However, for these currents to be induced, the speed of the physical
rotor and the speed of the rotating magnetic field in the stator must be
different, or else the magnetic field will not be moving relative to the
rotor conductors and no currents will be induced. If by some chance
this happens, the rotor typically slows slightly until a current is re-
induced and then the rotor continues as before.
 This difference between the speed of the rotor and speed of the
rotating magnetic field in the stator is called slip. It is unitless and is
the ratio between the relative speed of the magnetic field as seen by
the rotor (the slip speed) to the speed of the rotating stator field. Due
to this an induction motor is sometimes referred to as an
asynchronous machine.

ENEE430
CHAPTER 7 – INDUCTION MOTORS
 1. Induction Motor Construction:
 Induction motors have both a stator and rotor.
 There are basically 2 types of rotor construction:
 1) Squirrel Cage - no windings and no slip rings
conducting bars shorted at the end by shorting (or end) rings.
Stator winding casing

Cage rotor
IM

ENEE430
Wound Rotor Induction Motor (IM)
 Wound rotor - It has 3 phase 
windings, usually Y connected, and the
winding ends are connected via slip
rings on the rotor shaft. Slip rings
 The rotor windings are shorted

through brushes riding on the slip
rings
 Wound rotor motors have their rotor
currents accessible at the stator
brushes, where they can be
examined and extra resistance can Typical wound rotor for induction motors.
Rotor
be inserted into the rotor circuit to windings Stator
windings
modify the motor torque-speed
characteristic Shaft
 Wound rotor are known to be more

expensive due to its maintenance
cost to upkeep the slip rings,
carbon brushes and also rotor
windings. Brushes
Stator winding
connections to
 They are used to start high inertia power source
loads Cutaway diagram of a wound rotor induction motor.

ENEE430
BASIC CONCEPTS
 The Development of Induced Torque in an IM:
 The rotating stator magnetic field BS  induces voltage
eind in the rotor bars (windings)  this eind produces
rotor current flow IR (lagging)  IR produces BR lagging
by 90o  BR interacts with Bnet to produce torque
 When current flows in the stator, it will produce a
magnetic field in stator as such that Bs (stator magnetic
field) will rotate at a speed:
120 f e
nsync 
where fe = system frequency in Hz P
P = number of poles in machine

ENEE430
 This rotating magnetic field Bs passes over the rotor bars
and induces a voltage in them. The voltage induced in the
rotor is given by:
eind = (v x B) l
 Hence there will be rotor current flow which would be
lagging due to the fact that the rotor has an inductive
element. And this rotor current will produce a magnetic
field at the rotor, BR.
 Hence the interaction between both magnetic field
would give torque:
 ind  kBR  BS
 The torque induced would generate acceleration to the
rotor, hence the rotor will spin.
 However, there is a finite upper limit to the motor’s
speed.
ENEE430
E R
Net Voltage
Max induced voltage Max induced voltage Max induced current
BS IR
BS BS IR
Bnet
qR
w
w BR
w
X
X X
X
X
X
X X
X X X
X X X
X
IR produces BR (lagging 90 deg), BR

Rotating Magnetic field Bs produces a
rotor voltage produces a lagging IR due interacts with Bnet to produce a CCW
voltage in the rotor bars
to inductance of the rotor torque

ENEE430
Upper limit to the motor’s speed ?

ENEE430
Conclusion: Induction motor can speed up to near
synchronous speed but never actually reach it.
 
Note: Both BS and BR rotates at synchronous speed nsync
while rotor rotates at slower speed.

ENEE430
Concept of rotor Slip
 The induced voltage at the rotor bar is
dependent upon the relative speed between
the stator magnetic field and the rotor. This can
be easily termed as slip speed:
nslip  nsync  nm
Where nslip = slip speed of the machine
nsync = speed of the magnetic field.
nm = mechanical shaft speed
From this we can define slip (relative speed
expressed on a percentage basis):
 nslip nsync  nm
Slip, s   100%   100%
nsync nsync
ENEE430
 Slip may also be described in terms of angular velocity, .
 sync  m
s x100%
 sync
 Using the ratio of slip, we may also determine the rotor speed:
nm   1  s  nsync or m   1  s   sync
 Notice:
 rotor rotates at synchronous speed, s = 0
 rotor is stationary, s = 1
 Note: All normal motor speeds fall between s = 0 and s = 1
ENEE430
The Electrical Frequency on the Rotor
An induction motor is like a rotating transformer, i.e.
Stator (primary) induces voltage in the rotor

(secondary)
However, in induction motor:
secondary frequency is not the same as primary frequency
When rotor is locked, nm = 0 r/min,

s= 1 fr = sfe=fe
If rotor rotates synchronous to field, nm = nsync,

s= 0 fr = 0
Hence, at other rotor speeds, i.e. 0 < nm < nsync,

ENEE430
Hence, at other rotor speeds, i.e. 0 < nm < nsync,
f r  sf e
By substituting for s,
nsync  nm
fr  fe
nsync
Alternatively, since nsync = 120fe/P,
P
fr  (nsync  nm )
120
This shows that the relative difference between synchronous
speed and rotor speed will determine the rotor frequency.
ENEE430
Example 7.1
 A 208V, 10-hp, 4-pole, 60Hz, Y-connected
induction motor has a full load slip of 5%.
(a) What is the synchronous speed of the motor?
(b) What is the rotor speed of the motor at rated
load?
(c) What is the rotor frequency of the motor at rated
load?
(d) What is the shaft torque of this motor at rated
load?

ENEE430
The Equivalent Circuit of IM
 The operation of induction motor relies on
induction of rotor voltages and currents due to
stator circuit, i.e. transformer action.
  Hence, induction motor equivalent circuit
similar to that of a transformer. So to achieve the
final equivalent circuit, let’s start with…

ENEE430
Transformer model of an IM
 As in any transformer, there is certain resistance and
self-inductance in the primary (stator) windings, which
must be represented in the equivalent circuit of the
machine  R1 - stator resistance and X1 – stator
leakage reactance

ENEE430
IM mmf-flux curve
 Also, like any transformer with an iron core, the flux in the machine is
related to the integral of the applied voltage E1. The curve of mmf vs
flux (magnetization curve) for this machine is compared to a similar
curve for a transformer, as shown below:
 The slope of the induction motor’s mmf-flux curve is much shallower
than the transformer  This is because there must be an air gap in
an induction motor, which greatly increases the reluctance of the flux
path and thus reduces the coupling between primary and secondary
windings.
 The higher reluctance means that
a higher magnetizing current is
required to obtain a given flux level 
Therefore, the magnetizing reactance
Xm in the equivalent circuit will have
a much smaller value than it would in
a transformer.

ENEE430
 The primary internal stator voltage is E1 is coupled to the secondary
ER by an ideal transformer with an effective turns ratio aeff.
 The turns ratio for a wound rotor is basically the ratio of the conductors
per phase on the stator to the conductors per phase on the rotor.
 It is rather difficult to see aeff clearly in the cage rotor because there
are no distinct windings on the cage rotor.
 ER in the rotor produces current flow in the shorted rotor (or
secondary) circuit of the machine.
 The primary impedances and the magnetization current of the
induction motor are very similar to the corresponding components in a
transformer equivalent circuit.
Core loss

ENEE430
The Rotor Circuit Model
 When the voltage is applied to the stator windings, a
voltage is induced in the rotor windings.
 In general, the greater the relative motion between the
rotor and the stator magnetic fields, the greater the
resulting rotor voltage and rotor frequency.
 The largest relative motion occurs when the rotor is
stationary, called the locked-rotor or blocked-rotor
condition, so the largest voltage and rotor frequency are
induced in the rotor at that condition.
 The smallest voltage and frequency occur when the rotor
moves at the same speed as the stator magnetic field,
resulting in no relative motion.
 The magnitude and frequency of the voltage induced in the
rotor at any speed between these extremes is directly
proportional to the slip of the rotor.

ENEE430
 Therefore, if the magnitude of the induced rotor
voltage at locked-rotor conditions is called ER0, the
magnitude of the induced voltage at any slip will
be given by:
ER = sER0
 And the frequency of the induced voltage at any
slip is: fr = sfe
 The rotor resistance RR is a constant, independent
of slip, while the rotor reactance is affected in a
more complicated way by slip.

ENEE430
 The rotor contains both resistance and reactance.
However, only the reactance will be affected by the
frequency of the rotor voltage and current. Hence,

X R  r LR  2f R LR
 2  sf e  LR IR jXR=jsXR0
 s 2f e LR 
 X R  sX R 0
ER=sER0 RR
 where XR0 = blocked-rotor rotor reactance.

 Hence, the resulting rotor equivalent circuit is

ENEE430
Rotor Equivalent Circuit
ER ER ER0
 The rotor current flow is: IR   
RR  jX R RR  jsX R 0 RR  jX
s R0
 Therefore, the overall rotor impedance talking into account rotor slip would
be:
RR

Z R ,eq   jX R 0
s IR jXR0
 And the rotor equivalent

circuit using this convention is: ER0 RR/s
 The rotor circuit model with all the frequency (slip) effects concentrated in
resistor RR.
 In this equivalent circuit, the rotor voltage is a constant ER0 V and the rotor
impedance ZR,eq contains all the effects of varying rotor slip.

ENEE430
 Notice:
 at very low slips, RR /s >> XR0
rotor current varies linearly with slip
 at high slips, XR0 >> RR /s
rotor current approaches steady state value

ENEE430
The Final equivalent Circuit/refer rotor to stator
 To produce the final per-phase equivalent circuit for an
induction motor, it is necessary to refer the rotor part of the
model over to the stator side.
 In an ordinary transformer, the voltages, currents and
impedances on the secondary side can be referred to the
primary by means of the turns ratio of the transformer.
 Exactly the same sort of transformation can be done for
the induction motor’s rotor circuit. If the effective turns ratio
of an induction motor is aeff , then the transformed rotor
voltage becomes
E1  ER'  aeff ER 0
The rotor current: IR


I2 
aeff
 And the rotor impedance:  RR 
Z2  a 2
eff  jX R 0 
 s 
ENEE430
 If we make the following definitions:
R2 = a2eff RR (rotor impedance referred to stator)
X2 = a2eff XR0 (reactance referred to stator)
 The final per-phase equivalent circuit is as shown below:
IR jXR0
ER0 RR/s

ENEE430
Power and Torque in Induction Motor
 Losses and Power-Flow diagram
 An induction motor can be basically described as a rotating
transformer.
 Its input is a 3 phase system of voltages and currents.
 For an ordinary transformer, the output is electric power
from the secondary windings.
 The secondary windings in an induction motor (the rotor)
are shorted out, so no electrical output exists from normal
induction motors. Instead, the output is mechanical.
 The relationship between the input electric power and the
output mechanical power of this motor is shown below:

ENEE430
Power Flow Diagram

ENEE430
 The input power (electrical) of an induction motor:
 Losses encountered on stator side:

- Stator copper loss PSCL, i.e. I2R loss in stator windings.
- Hysteresis and eddy current losses Pcore
 Air gap power PAG  power transferred to the rotor across the air
gap
 Losses encountered on rotor side:
- Rotor copper loss PRCL, i.e. I2R loss in rotor windings.
 What is left?
 Power converted from electrical to mechanical form, Pconv.
 Final losses:
- Friction and windage losses, PF&W
- Stray losses, Pmisc

ENEE430
 The output power (mechanical) of the induction motor:
Pout =load m
 Special note on Pcore:

 The core losses do not always appear after PSCL.
 Pcore comes partially from the stator circuit and partially from the rotor
circuit. Usually the rotor core losses are very small compared to the
stator core losses.
 Pcore are represented in the induction motor equivalent circuit by the
resistor RC (or the conductance GC).
 If RC is not given but Pcore = X watts is given, then often add it
together with PF&W at the end of the power flow diagram.
 Note: Pcore , PF&W and Pmisc are sometimes lumped together and called
rotational losses Prot.

ENEE430
Example 7.2
 A 480V, 60Hz, 50hp, 3 phase induction motor is
drawing 60A at 0.85 PF lagging. The stator copper
losses are 2kW, and the rotor copper losses are
700W. The friction and windage losses are 600W,
the core losses are 1800W, and the stray losses
are negligible. Find:
 The air gap power PAG
 The power converted Pconv
 The output power Pout
 The efficiency of the motor

ENEE430
Power and Torque in an Induction Motor
 By examining the per-phase equivalent circuit, the power
and torque equations governing the operation of the
motor can be derived.

ENEE430
Power and Torque in an Induction Motor
 The input current to a phase of the motor is:
 V
I1 
Z eq
 Where 1
Z eq  R1  jX 1 
1
GC  jBM 
R2
 jX 2
s
 Thus, the stator copper losses, the core losses, and the
rotor copper losses can be found.
 The stator copper are: PSCL = 3 I12 R1
 The core losses : Pcore = 3 E12 GC
 So, the air gap power: PAG = Pin – PSCL - Pcore

ENEE430
 Also, the only element in the equivalent circuit where the
air-gap power can be consumed is in the resistor R2/s.
Thus, the air-gap power: 2 R
 PAG  312 2
s
 The actual resistive losses in the rotor circuit are given by:
PRCL = 3 IR2 RR
 Since power is unchanged when referred across an ideal
transformer, the rotor copper losses can also be expressed
as:
PRCL = 3 I22 R2

ENEE430
 After stator copper losses, core losses and rotor copper losses are
subtracted from the input power to the motor, the remaining power is
converted from electrical to mechanical form. The power converted,
which is called developed mechanical power is given as:
Pconv  PAG  PRCL

R2
2 2
 3I 2  3I 2 R2
s
2 1 
 3I 2 R2   1
s 
2 1 s 
Pconv  3I 2 R2  
 s 
 And the rotor copper losses are noticed to be equal to the air gap
power times the slip  PRCL = s PAG
ENEE430
Hence, the lower the slip of the motor the lower the rotor losses.
Also, if the rotor is not turning, the slip is s=1 and the air gap power is
entirely consumed in the rotor. This is logical, since if the rotor is not
turning, the output power Pout ( = τload ωm) must be zero.
Since Pconv = PAG – PRCL , this also gives another relationship between
the air-gap power and the power converted from electrical and
mechanical form:
Pconv = PAG – PRCL
= PAG – sPAG
Pconv = (1-s) PAG
Finally, if the friction and windage losses and the stray losses are
known, the output power:
Pout = Pconv – PF&W - Pmisc
ENEE430
The induced torque in a machine was defined as the torque generated by
the internal electric to mechanical power conversion. This torque
differs from the torque actually available at the terminals of the motor
by an amount equal to the friction and windage torques in the machine.
Hence, the developed torque is:
Pconv
 ind 
m
Other ways to express torque:
 ind 
 1  s  PAG
1  s sync
PAG
 ind 
 sync
ENEE430
Separating the Rotor Copper Losses and the Power Converted
in an Induction Motor’s Equivalent Circuit
A portion of power transferred via the air gap will be consumed by the
rotor copper loss and also converted into mechanical power. Hence it
may be useful to separate the rotor copper loss element since rotor
resistance are both used for calculating rotor copper loss and also the
output power.
Since Air Gap power would require R2/s and rotor copper loss require
R2 element. The difference between the air gap power and the rotor
copper loss would give the converted power, hence;
R2 1 s 
Rconv   R2  R2  
s  s 

ENEE430
For Info only
Therefore the equivalent circuit would be modified to be as follows:

ENEE430
Example 7.3
 A 460V, 25hp, 60Hz, 4 pole, Y-connected induction motor has the
following impedances in ohms per phase referred to the stator circuit:
 R1 = 0.641 Ω R2 = 0.332 Ω
 X1 = 1.106 Ω X2 = 0.464 Ω Xm = 26.3 Ω
 The total rotational losses are 1100W and are assumed to be
constant. The core loss is lumped in with the rotational losses. For a
rotor slip of 2.2% at the rated voltage and rated frequency, find the
motor’s
(a) speed
(b) stator current
(c) power factor
(d) Pconv and Pout
(e) τind and τload
(d) efficiency

ENEE430
7.5 Induction Motor Torque-Speed Characteristics
 The torque-speed relationship will be examined first from
the physical viewpoint of the motor’s magnetic field
behaviour and then, a general equation for torque as a
function of slip will be derived from the induction motor
equivalent circuit.

ENEE430
 Assume IM is rotating at no load conditions  its rotating
speed is near to synchronous speed.
 The net magnetic field Bnet is produced by the
magnetization current IM .
 The magnitude of IM and Bnet is directly proportional to
voltage E1.
 If E1 is constant, then Bnet is constant.
 In an actual machine, E1 varies as the load changes due
to the stator impedances R1 and X1 which cause varying
voltage drops with varying loads.
 However, the volt drop at R1 and X1 is so small, that E1 is
assumed to remain constant throughout.
Electric Machines & Power Electronics

ENEE430
No Load
 At no-load, slip is very small, and so the relative motion
between rotor and magnetic field is very small, and fr is also
very small.
 Since the relative motion is small, ER induced in the bars of
the rotor is very small, and IR is also very small.
 Since fr is small, the reactance of the rotor is nearly zero,
and the max rotor current IR is almost in phase with the rotor
voltage ER .
 The rotor current produces a small magnetic field BR at an
angle slightly greater than 90 degrees behind Bnet.
 c
ENEE430
 The stator current must be quite large even at no-
load since it must supply most of Bnet .
The induced torque which is keeping the rotor
running,
 ind  kBR  Bnet
 and its magnitude is

 ind  kBR Bnet sin 
 In terms of magnitude, the induced torque will be
small due to small rotor magnetic field.

ENEE430
At Heavy Load
 As the motor’s load increases  its slip increases, and the rotor
speed falls.
 Since the rotor speed is slower, there is now more relative motion
between rotor and stator magnetic fields.
 Greater relative motion means a stronger rotor voltage ER which in
turn produces a larger rotor current IR .
 With large rotor current, the rotor magnetic field BR also increases.
However, the angle between rotor current and BR changes as well.
 Since the rotor slip is larger, the rotor frequency rises (fr =sfe) and the
rotor reactance increases (ωLR).
 Therefore, the rotor current now lags further behind the rotor voltage,
and the rotor magnetic field shifts with the current.
 The rotor current now has increased compared to no-load and the
angle δ has increased.
 The increase in BR tends to increase the torque, while the increase in
angle δ tends to decrease the torque (τind is proportional to sin δ, and
δ>90º).
 Since the first effect is larger than the second one, the overall induced
torque increases to supply the motor’s increased load.
ENEE430
 As the load on the shaft is increased, the sin δ
term decreases more than the BR term increases
(the value is going towards the 0 cross over point
for a sine wave).
 At that point, a further increase in load decreases
τind and the motor stops. This effect is known as
pullout torque.

ENEE430
Modelling the torque-speed characteristics
of an induction motor
 Note:
 ind  kBR Bnet sin 
 Each Term can be considered separately to derive the overall torque

behaviour:

a) BR  I R (provided the rotor core is unsaturated). Hence,

BR increases with I R which in turn increases with slip
(decrease in speed).

b) Bnet  E1 and will remain approximately constant.
c) The angle  increases with slip. Hence, ‘sin ’ term

decreases. From the on load condition,

ENEE430
Modelling the torque-speed characteristics
of an induction motor
 Note:
   R  90 o
sin   sin( R  90 o )  cos  R  power factor of rotor
where,
sX R 0
 R  tan1
RR

ENEE430
 The torque-speed characteristic can be constructed from
the graphical manipulation of the three properties (a)-(c)
which is shown on the next page.
 The characteristic curve can be divided into three regions:
1. Low-slip region ( s  linearly, nm  linearly) :
 XR negligible  PFR  1
 I R increases linearly with s
Contains the entire normal operating range of an induction

motor.
2. Moderate-slip region:
 XR same order of magnitude as RR  PFR droops
 I R doesn’t increase as rapidly as in low-slip region
Peak torque (pullout torque) occurs in this region.

ENEE430
3. High-slip region:
 Increase in I R completely overshadowed by decrease
in PFR.
 Tind decreases with increase in load
Note:
 Typical pullout torque  200% to 250% of Trated.
 The starting torque 150% of the Trated.
Hence induction motor may be started at full load.

ENEE430
ENEE430
Deriving Induction Torque Equation
 Only the final result is shown without going
through the derivation
Pconv PAG
 ind  or  ind 
m  sync
 Second equation is more useful since it is referred
to sync
 Using the equivalent circuit of the motor and using
thevenin’s equivalent circuit techniques, an
expression for Torque can be found by first finding
an expression for the PAG

ENEE430
R2
PAG per phase  I 2
2
s
hence, total air gap power :
R2
PAG  3I 22
s

ENEE430
1) Derive the thevenin voltage (potential divider rule):
jX m
VTH  V
R1  jX 1  jX m
Hence the magnitude of thevenin voltage:
Xm
VTH  V
R12   X 1  X m 
2
Since Xm >> X1 , Xm >> R1, therefore the magnitude may be

approximated to:
Xm
VTH  V
X1  X m
ENEE430
1) Find the thevenin impedance
Take out the source and replace it with a short circuit, and derive the
equivalent impedances.
jX m  R1  jX 1 
Z TH 
R1  jX 1  jX m
Since Xm >> X1, Xm >> R1,
2
 Xm 
RTH  R1  
 X1  X m 
X TH  X 1
Representing the stator circuit by the thevenin equivalent, and adding
back the rotor circuit, we can derive I2,
VTH
I2 
R2
RTH   j ( X TH  X 2 )
s
Hence the magnitude will be,
VTH
I2 
 
2
R2
  X TH  X 2 
2
RTH 
s
ENEE430
Hence air gap power,
2
 
 
VTH  R2
PAG  3

R s
 s
2
R2
  X TH  X 2 
2
 TH  
 
Hence, induced torque,

2
 
 
VTH  R2
3

   s
2
R2
  X TH  X 2 
2
 RTH  
s
 ind   
 sync
If a graph of Torque and speed were plotted based upon changes in slip,
we would get a similar graph as we had derived earlier.

ENEE430
2
R 3VTH R2 s
PAG  3I 22 2 
s  RTH  R2 s  2   X TH  X 2  2
PAG
 ind 
 sync
2
3VTH R2 s

 sync [ RTH  R2 s    X TH  X 2  ]
2 2

ENEE430
ENEE430
Comments on the induction motor torque-
speed curve
 At synchronous speed, ind = 0.
 The curve is linear between no load and full
load.
 The maximum torque is known as pullout torque
or breakdown torque. It is approximately 2 to 3
times the rated full-load torque of the motor.
 The starting torque is slightly larger than its full-
load torque. So, IM will start carrying any load it
can supply at full power
 Torque for a given slip varies as square of the
applied voltage. This is useful as one form of IM
speed control.
ENEE430
 If rotor is driven faster than synchronous speed,
 ind direction reverses and machine becomes
a generator .
 If motor is turning backward relative to the direction of
magnetic fields (achieved by reversing the magnetic field
rotation direction),
 ind will stop the machine very rapidly (braking) and try
to rotate in the other direction.
 This can be achieved by switching two stator phases,
which causes the machine to change direction of rotation
which is called plugging (plug reversal).

ENEE430
Converted Power
 The power converted to mechanical form in an induction
motor is: Pconv   ind m
 Hence, a characteristic to show the variation of converted
power with speed (i.e. load) can be obtained.
Note that:
Peak power supplied by
the induction motor occurs
at different speed to
maximum torque.
No power is converted
when rotor speed = 0.

ENEE430
Comments on the Induction Motor Torque
Speed Curve
a) Induced Torque is zero at synchronous speed.
b) The graph is nearly linear between no load and full load (at near
synchronous speeds).
c) Max torque is known as pull out torque or breakdown torque
d) Starting torque is very large.
e) Torque for a given slip value would change to the square of the
applied voltage.
f) If the rotor were driven faster than synchronous speed, the motor
would then become a generator.
g) If we reverse the direction of the stator magnetic field, it would act as
a braking action to the rotor – plugging.
Since Pconv may be derived as follows:

Pconv   ind m
Hence we may plot a similar characteristic to show the amount of power
converted throughout the variation of load.
ENEE430
Maximum (Pullout) Torque in an Induction
Motor
Since induced torque is equal to PAG / ωsync , the maximum pullout

torque may be found by finding the maximum air gap power. And
maximum air gap power is during which the power consumed by the
R2/s resistor is the highest.
Based upon the maximum power transfer theorem, maximum power

transfer will be achieved when the magnitude of source impedance
matches the load impedance. Since the source impedance is as follows:
Z source  RTH  jX TH  jX 2

ENEE430
Maximum (Pullout) Torque in an Induction
Motor
Hence maximum power transfer occurs during:
R2
  X TH  X 2 
2
 RTH
2
s
Hence max power transfer is possible when slip is as follows:
R2
smax 
  X TH  X 2 
2 2
RTH
Put in the value of Smax into the torque equation,
2
3VTH
 max 
2 sync  RTH  RTH   X TH  X 2  
2 2
 

ENEE430
2
3VTH
 max 
2 sync  RTH  RTH   X TH  X 2  
2 2
 
From here we can say:
a) Torque is related to the square of the applied voltage

b) Torque is also inversely proportional to the machine impedances
c) Slip during maximum torque is dependent upon rotor resistance
d) Torque is also independent to rotor resistance as shown in the
maximum torque equation.
By adding more resistance to the machine impedances, we can vary:
a) Starting torque
b) Max pull out speed
ENEE430
 Torque is related to the square of supplied voltage.
 Torque is inversely proportional to stator impedances and rotor
reactance.
 smax is directly proportional to R2.
 max is independent of R2.
 As increase R2 (i.e. increase smax):
 pullout speed of motor decreases
 maximum torque remains constant
 starting torque increases

ENEE430
Example 7.4
 A 2 pole, 50 Hz induction motor supplies 15kW to
a load at a speed of 2950 r/min.
 What is the motor’s slip?
 What is the induced torque in the motor in Nm
under these conditions?
 What will the operating speed of the motor be if its
torque is doubled?
 How much power will be supplied by the motor
when the torque is doubled?

ENEE430
Example 7.5
 A 460V, 25hp, 60Hz, 4-pole, Y-connected wound rotor
induction motor has the following impedances in ohms per-
phase referred to the stator circuit:
 R1 = 0.641 Ω R2 = 0.332 Ω
 X1 = 1.106 Ω X2 = 0.464 Ω Xm = 26.3 Ω
 What is the max torque of this motor? At what speed and slip does
it occur?
 What is the starting torque?
 When the rotor resistance is doubled, what is the speed at which
the max torque now occurs? What is the new starting torque?

ENEE430
7.9. Speed Control of IM
 Skip 7.6, 7.7 and 7.8
 Before advances in Power electronics IM were not
good machines for applications requiring
considerable speed control .
 The normal operating range of a typical induction
motor was confined to less than 5% slip, and the
speed variation is more or less proportional to the
load on the motor shaft.
 If slip is made higher, motor efficiency becomes
very poor since rotor copper losses will be high as
well (PRCL = sPAG).
ENEE430
IM Speed Control
 There are basically 2 techniques to control the
speed of an induction motor:
 Varying speed of the stator and rotor magnetic
fields, i.e. synchronous speed by:
1) varying the frequency or
2) changing the number of poles

 Varying the slip by:
1) varying the rotor resistance or
2) varying the terminal voltage

ENEE430
IM speed control by pole changing
 1) Method of consequent
poles
 (requires motor with special
stator windings)
 General Idea:
 Consider one phase winding in
a stator. By changing the
current flow in one portion of
the stator winding such that it is
similar to the current flow in A two-pole stator winding for pole
the opposite portion of the changing. Notice the very short pitch
(60 to 90) of these windings.
stator, an extra pair of poles is
generated.

ENEE430
Close up view of one phase of a pole
changing winding.
(A) In the 2-pole configuration,

one coil is a north pole and the
other is a south pole.
(B) When the connection on one

of the two coils is reversed, they
are both north poles, and the
magnetic flux returns to the
stator halfway between the two
coils. The south poles are called
consequent poles. Hence the
winding is now 4-pole.

ENEE430
 Rotor must be of cage type.
 In terms of torque, the maximum torque
magnitude would generally be maintained.
 Disadvantage:
 speed changes must be in a ratio of 2:1.

ENEE430
2) Multiple stator windings
 To obtain variations in speed, multiple stator windings
have to be applied 
 extra sets of windings with different number of poles
that may be energized one at a time as required.
 Example:
 IM wound with 4-pole and 6-pole set of stator windings to
give switch in nsync from 1800 rpm to 1200 rpm on a 60-
Hz system)
 Unfortunately this is an expensive alternative.
 Combination of consequent poles and multiple stator
windings creates a four-speed induction motor.

ENEE430
 For some applications continuous speed control
may be an unnecessary luxury, and it may be
sufficient to be able to run at two discrete speeds.
 Among many instances where this can be
acceptable and economic are pumps, lifts and
hoists, fans and some machine tool drives.

ENEE430
Speed control by changing the line
frequency
 This is also known as variable frequency control.
 Changing the electrical frequency will change the
synchronous speed of the machine since

 Base speed = synchronous speed of motor at rated
conditions
 Hence, it is possible to adjust speed of motor either above
or below base speed.

 BUT it also requires terminal voltage limitation in order
to maintain the same amount of flux level in the machine

core.
 If not the machine will experience:
1) Core saturation (non linearity effects)

2) Excessive magnetization current will flow

ENEE430
 Varying Below base speed Above base speed
frequency with decreased linearly
Terminal
with decrease in stator held constant at rated
or without (stator)
frequency, i.e. derating value
voltage
adjustment to
the terminal Motor
varying varying
speed
voltage may
give 2 Flux level
Constant Decreasing
different Max
Decreasing
Operating constant
effects:
Torque
maximum power
rating must also be
Notes decreased to protect -
from overheating

ENEE430
Note:
Flux in the core of an induction motor is given by Faraday’s Law
d
v t    N .
dt
If v t   VM sin t is applied, the resulting flux  is:
1 1 VM
   v t dt   VM sin tdt   cost
N N N

ENEE430
Variable-frequency speed control in IM
 torque-speed curves for speed below base speed.
 V/F=constant to keep flux constant
for V/F (F decreases

and V decreases also)
If f is decreased while V

kept constant

ENEE430
Above Base Speed
 Variable-frequency speed control in an IM
 Above Base Speed  Frequency is increased ,
while V=Vrated  Flux is decreased  maximum
Torque is also decreased

ENEE430
 There may also be instances where both characteristics
are needed in the motor operation, hence it is possible to
combine both effects.
 With the arrival of solid state devices/power electronics,
variable frequency control has become the method of
choice for induction motor speed control.
 Advantage: can be used with any induction motor
V=const, f increase
V/f =const
ENEE430
Speed control by changing the line voltage
 Variable-line-voltage speed control in an induction motor
But it also causes variation of operating torque since
Tstart  VT2
However, it only allows for motor speed variations over limited

range.
Hence, this method is only suitable

for small motors driving fans

ENEE430
Speed control by changing the rotor
resistance
 This is only possible for wound rotor induction motors,
i.e we can add extra rotor resistance to vary the torque-
speed curve.
 BUT this causes
reduction in motor
efficiency. Hence,
it is used only for
short periods.

ENEE430

Ch7 Induction Motor

Uploaded by

Copyright:

Available Formats

Ch7 Induction Motor

Uploaded by

Document Information

Original Description:

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Ch7 Induction Motor

Uploaded by

Copyright:

Available Formats

CHAPTER 7 – INDUCTION MOTORS

 Induction machine – rotor voltage (that produces the rotor current

Electric Machines & Power Electronics 1

Electric Machines & Power Electronics 2

Stator winding casing

Electric Machines & Power Electronics 3

 The rotor windings are shorted

 Wound rotor are known to be more

loads Cutaway diagram of a wound rotor induction motor.

Electric Machines & Power Electronics 4

Electric Machines & Power Electronics 5

IR produces BR (lagging 90 deg), BR

Electric Machines & Power Electronics 7

Electric Machines & Power Electronics 8

Electric Machines & Power Electronics 9

Stator (primary) induces voltage in the rotor

When rotor is locked, nm = 0 r/min,

If rotor rotates synchronous to field, nm = nsync,

Hence, at other rotor speeds, i.e. 0 < nm < nsync,

Alternatively, since nsync = 120fe/P,

Electric Machines & Power Electronics 14

Electric Machines & Power Electronics 15

Electric Machines & Power Electronics 16

Electric Machines & Power Electronics 17

Electric Machines & Power Electronics 18

Electric Machines & Power Electronics 19

Electric Machines & Power Electronics 20

 where XR0 = blocked-rotor rotor reactance.

Electric Machines & Power Electronics 21

 And the rotor equivalent

Electric Machines & Power Electronics 22

Electric Machines & Power Electronics 23

The rotor current: IR

Electric Machines & Power Electronics 25

Electric Machines & Power Electronics 26

Electric Machines & Power Electronics 27

 Losses encountered on stator side:

Electric Machines & Power Electronics 28

 Special note on Pcore:

Electric Machines & Power Electronics 29

Electric Machines & Power Electronics 30

Electric Machines & Power Electronics 31

Electric Machines & Power Electronics 32

Electric Machines & Power Electronics 33

Pconv  PAG  PRCL

Pconv = (1-s) PAG

Electric Machines & Power Electronics 37

Therefore the equivalent circuit would be modified to be as follows:

Electric Machines & Power Electronics 38

Electric Machines & Power Electronics 39

Electric Machines & Power Electronics 40

Electric Machines & Power Electronics

 and its magnitude is

Electric Machines & Power Electronics

Electric Machines & Power Electronics

 Each Term can be considered separately to derive the overall torque

c) The angle  increases with slip. Hence, ‘sin ’ term

Electric Machines & Power Electronics

Electric Machines & Power Electronics