This document provides information on three-phase induction motors: - It discusses the construction, operation, and advantages/disadvantages of three-phase induction motors. The main components are the stationary stator and revolving rotor, which can be either a squirrel cage or wound type. - A balanced three-phase supply to the stator produces a rotating magnetic field that induces voltage in the rotor windings, generating torque. The motor runs slightly slower than the synchronous speed due to slip. - Equivalent circuits are presented for analyzing induction motors, accounting for variables like induced voltage and reactance that change with slip frequency. Power losses and relationships are also examined.

1. UNIT 4: Three Phase
Induction Motors

2. Introduction
 Three-phase induction motors are the most common and
frequently encountered machines in industry.
Advantages
Simple and rugged in construction.
Generally cheaper due to absence of
brushes, commutators and slip rings.
Lower maintenance.
Can safely be operated in hazardous
environments.
RPM can be changed without gear
box.
Disadvantages
Draws a lot of current
at starting.
Low starting torque
Harder to control

3. Construction of 3-Phase IM
 An induction motor has two main
parts
- (a) stationary stator
• Consisting of a steel frame
that supports a hollow,
cylindrical core
• core, constructed from
stacked laminations, having
a number of evenly spaced
slots, providing the space for
the stator winding Stator of IM

4. Construction
(b) Revolving rotor
• composed of punched laminations, stacked to
create a series of rotor slots, providing space
for the rotor winding.
Two types of rotors:
Squirrel-Cage rotor: conducting bars laid into slots and shorted
at both ends by shorting rings (aluminum rings), forming a
squirrel-cage shaped circuit.
wound-rotor: Complete set of three-phase windings exactly as
the stator. Usually Y-connected, the ends of the three rotor wires
are connected to 3 slip rings on the rotor shaft. In this way, the
rotor circuit is accessible.

5. Construction
Squirrel cage rotor
Wound rotor
Notice the
slip rings

8. SCIM Vs SRIM
S.N
o
Squirrel Cage Induction
Motor
Slip Ring / Wound Rotor Induction
Motor
1 The rotor is simplest and most
rugged in construction.
The rotor is wound type and
construction is not simple.
2 Cylindrical laminated core rotor
with heavy bars or copper are
used for conductors.
Cylindrical laminated core rotor is
wound like winding on the stator.
3 Rotor conductors or rotor bars
are short circuited with end
rings.
At starting the 3 phase windings are
connected to a star connected rheostat
and during running, the windings are
short circuited at the slip rings.
4 Rotor bars are permanently
short circuited and hence it is
not possible to connect external
resistance in the circuit in series
with the rotor conductors.
It is possible to insert additional
resistance in the rotor circuit. Therefore
it is possible to increase the torque; the
additional series resistance is used for
starting purposes.

9. 5 Cheaper cost. Cost is slightly higher.
6 No moving contacts in the rotor. Carbon brushes, slip rings etc
are provided in the rotor
circuit.
7 Higher efficiency. Comparatively less efficiency.
8 Speed control by rotor resistance
is not possible.
Speed control by rotor
resistance is possible.
9 Starting current is 5 to 7 times
the full load current.
Less starting current
compared to squirrel cage
Induction Motor.

10. Rotating Magnetic Field in 3-Phase IM
 Balanced three phase windings, i.e.
mechanically displaced 120 degrees
from each other, fed by balanced three
phase source.
 A rotating magnetic field with constant
magnitude is produced, rotating with a
speed
Where fe is the supply frequency and
P is the no. of poles and nsync is called the
synchronous speed in rpm (revolutions
per minute)
120 e
sync
f
n rpm
P


11. Mathematical Proof:
 Applied Currents in all the phases
 Then flux will be produced in all the phases with
120 degree phase difference. Then the three mmfs are:

12. Rotating Magnetic Field
 The resultant mmf is
 Then the total mmf F
 This shows that the resultant magnetic field is
rotating with synchronous speed.

13. Synchronous speed
P 50 Hz 60 Hz
2 3000 3600
4 1500 1800
6 1000 1200
8 750 900
10 600 720
12 500 600
120 e
sync
f
n rpm
P


14. Principle of operation
 A 3-phase balanced AC supply is given to the 3-phase stator
winding, a rotating magnetic field will be developed;
 This rotating magnetic field cuts the rotor windings and
produces an induced voltage in the rotor windings
 Due to the fact that the rotor windings are short circuited, for
both squirrel cage and wound-rotor, and induced current
flows in the rotor windings
 The rotor current produces another magnetic field
 A torque is produced as a result of the interaction of those
two magnetic fields
Where ind is the induced torque and BR and BS are the magnetic
flux densities of the rotor and the stator respectively
ind R s
kB B
  

15. Induction motor speed
 At what speed will the IM run?
- Can the IM run at the synchronous speed, why?
- If rotor runs at the synchronous speed, which is the
same speed of the rotating magnetic field, then the rotor
will appear stationary to the rotating magnetic field and
the rotating magnetic field will not cut the rotor. So, no
induced current will flow in the rotor and no rotor
magnetic flux will be produced so no torque is
generated and the rotor speed will fall below the
synchronous speed
- When the speed falls, the rotating magnetic field will
cut the rotor windings and a torque is produced

16. Induction motor speed
 So, the IM will always run at a speed lower than
the synchronous speed
 The difference between the motor speed and the
synchronous speed is called the Slip
Where nslip= slip speed
nsync= speed of the magnetic field
nm = mechanical shaft speed of the motor
slip sync m
n n n
 

17. The Slip
sync m
sync
n n
s
n


Where s is the slip
Notice that : if the rotor runs at synchronous speed
s = 0
if the rotor is stationary
s = 1
Slip may be expressed as a percentage by multiplying the above
eq. by 100, notice that the slip is a ratio and doesn’t have units

18. Induction Motors and Transformers
 Both IM and transformer works on the principle of
Mutual Induction (induced voltage).
- Transformer: voltage applied to the primary windings
produce an induced voltage in the secondary windings
- Induction motor: voltage applied to the stator windings
produce an induced voltage in the rotor windings
- The difference is that, in the case of the induction motor,
the secondary windings can move.
- Due to the rotation of the rotor (the secondary winding
of the IM), the induced voltage in it does not have the
same frequency of the stator (the primary) voltage

19. Frequency
 The frequency of the voltage induced in the rotor is
given by
Where fr = the rotor frequency (Hz)
P = number of stator poles
n = slip speed (rpm)
120
r
P n
f


( )
120
120
s m
r
s
e
P n n
f
P sn
sf
 


 

20. Frequency
 What would be the frequency of the rotor’s induced
voltage at any speed nm?
 When the rotor is blocked (s=1) , the frequency of
the induced voltage is equal to the supply frequency
 On the other hand, if the rotor runs at synchronous
speed (s = 0), the frequency will be zero
r e
f s f


21. Torque
 While the input to the induction motor is electrical
power, its output is mechanical power and for that we
should know some terms and quantities related to
mechanical power
 Any mechanical load applied to the motor shaft will
introduce a Torque on the motor shaft. This torque is
related to the motor output power and the rotor speed
and
.
out
load
m
P
N m


 2
/
60
m
m
n
rad s

 

22. Horse power
 Another unit used to measure mechanical power is
the horse power
 It is used to refer to the mechanical output power
of the motor
 Since we, as an electrical engineers, deal with
watts as a unit to measure electrical power, there is
a relation between horse power and watts
746
hp watts


23. Example
A 208-V, 10hp, four pole, 60 Hz, Y-connected
induction motor has a full-load slip of 5 percent
1. What is the synchronous speed of this motor?
2. What is the rotor speed of this motor at rated load?
3. What is the rotor frequency of this motor at rated load?
4. What is the shaft torque of this motor at rated load?

24. Solution
1.
2.
3.
4.
120 120(60)
1800
4
e
sync
f
n rpm
P
  
(1 )
(1 0.05) 1800 1710
m s
n s n
rpm
 
   
0.05 60 3
r e
f sf Hz
   
2
60
10 746 /
41.7 .
1710 2 (1/ 60)
out out
load
m
m
P P
n
hp watt hp
N m

 

 

 
 

25. Equivalent Circuit
 The induction motor is similar to the transformer with
the exception that its secondary windings are free to
rotate.
As we noticed in the transformer, it is easier if we can combine
these two circuits in one circuit but there are some difficulties

26.  EMF induced in the Rotor
Where ER0 is the largest value of the rotor’s induced voltage
obtained at s = 1(locked rotor)
0
R R
E sE


27.  The same is true for the frequency, i.e.
 It is known that
 So, as the frequency of the induced voltage in the
rotor changes, the reactance of the rotor circuit also
changes
Where Xr0 is the rotor reactance
at the supply frequency
(at blocked rotor)
r e
f s f

2
X L f L
 
 
0
2
2
r r r r r
e r
r
X L f L
sf L
sX
 

 



28.  Then, we can draw the rotor equivalent circuit as
follows:
Where ER is the induced voltage in the rotor and RR is the
rotor resistance

29.  Now we can calculate the rotor current as
 Dividing both the numerator and denominator by s
so nothing changes, we get
Where ER0 is the induced voltage and XR0 is the rotor
reactance at blocked rotor condition (s = 1)
0
0
( )
( )
R
R
R R
R
R R
E
I
R jX
sE
R jsX




0
0
( )
R
R
R
R
E
I
R
jX
s



30.  Now we can have the rotor equivalent circuit

31. Equivalent Circuit
 Now as we managed to solve the induced voltage
and different frequency problems, we can combine
the stator and rotor circuits in one equivalent
circuit
Where
2
2 0
2
2
2
1 0
eff R
eff R
R
eff
eff R
S
eff
R
X a X
R a R
I
I
a
E a E
N
a
N






32. Power losses in Induction machines
 Copper losses
- Copper loss in the stator (PSCL) = I1
2R1
- Copper loss in the rotor (PRCL) = I2
2R2
 Core loss (Pcore)
 Mechanical power loss due to friction and windage
 How this power flow in the motor?

33. Power flow in induction motor

34. Power relations
3 cos 3 cos
in L L ph ph
P V I V I
 
 
2
1 1
3
SCL
P I R

( )
AG in SCL core
P P P P
  
2
2 2
3
RCL
P I R

conv AG RCL
P P P
 
( )
out conv f w stray
P P P P

   conv
ind
m
P




35. Power relations
3 cos 3 cos
in L L ph ph
P V I V I
 
 
2
1 1
3
SCL
P I R

( )
AG in SCL core
P P P P
  
2
2 2
3
RCL
P I R

conv AG RCL
P P P
 
( )
out conv f w stray
P P P P

  
conv RCL
P P
  2 2
2
3
R
I
s

2 2
2
(1 )
3
R s
I
s


RCL
P
s

(1 )
RCL
P s
s


(1 )
conv AG
P s P
 
conv
ind
m
P



(1 )
(1 )
AG
s
s P
s 




36. Power relations
AG
P
RCL
P
conv
P
1
s
1-s
: :
1 : : 1-
AG RCL conv
P P P
s s

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

38. Solution
1.
2.
3.
3 cos
3 480 60 0.85 42.4 kW
in L L
P V I 

    
42.4 2 1.8 38.6 kW
AG in SCL core
P P P P
  
   
700
38.6 37.9 kW
1000
conv AG RCL
P P P
 
  
&
600
37.9 37.3 kW
1000
out conv F W
P P P
 
  

39. Solution
4.
37.3
50 hp
0.746
out
P  
100%
37.3
100 88%
42.4
out
in
P
P
  
  

40. Example
A 460-V, 25-hp, 60 Hz, four-pole, Y-connected induction motor
has the following impedances in ohms per phase referred to
the stator circuit:
R1= 0.641 R2= 0.332
X1= 1.106  X2= 0.464  XM= 26.3 
The total rotational losses are 1100 W and are assumed to be
constant. The core loss is lumped in with the rotational losses.
For a rotor slip of 2.2 percent at the rated voltage and rated
frequency, find the motor’s
1. Speed
2. Stator current
3. Power factor
4. Pconv and Pout
5. ind and load
6. Efficiency

41. Solution
1.
2.
120 120 60
1800 rpm
4
e
sync
f
n
P

  
(1 ) (1 0.022) 1800 1760 rpm
m sync
n s n
     
2
2 2
0.332
0.464
0.022
15.09 0.464 15.1 1.76
R
Z jX j
s
j
   
     
2
1 1
1/ 1/ 0.038 0.0662 1.76
1
12.94 31.1
0.0773 31.1
f
M
Z
jX Z j
 
     
    
  

42. Solution
3.
4.
0.641 1.106 12.94 31.1
11.72 7.79 14.07 33.6
tot stat f
Z Z Z
j
j
 
     
     
1
460 0
3 18.88 33.6 A
14.07 33.6
tot
V
I
Z

 
     
 
cos33.6 0.833 lagging
PF   
3 cos 3 460 18.88 0.833 12530 W
in L L
P V I 
     
2 2
1 1
3 3(18.88) 0.641 685 W
SCL
P I R
   
12530 685 11845 W
AG in SCL
P P P
    

43. Solution
5.
6.
(1 ) (1 0.022)(11845) 11585 W
conv AG
P s P
    
& 11585 1100 10485 W
10485
= 14.1hp
746
out conv F W
P P P
    

11845
62.8 N.m
1800
2
60
AG
ind
sync
P

 
  

10485
56.9 N.m
1760
2
60
out
load
m
P

 
  

10485
100% 100 83.7%
12530
out
in
P
P
     

44. Torque-Slip (or) Torque- speed
Characteristics of 3-ph IM

45.  The torque slip curve for an induction motor gives us
the information about the variation of torque with the
slip.
 When the supply is given to the stator sides and the
motor always rotates below the synchronous speed.
 The induction motor torque varies from zero to full load
torque as the slip varies. The slip varies from zero to one.
It is zero at no load and one at standstill. From the curve
it is seen that the torque is directly proportional to the
slip.
 That is, more is the slip, more will be the torque
produced and vice-versa.

46. Starting Methods of an Induction Motor
 A three phase Induction Motor is Self Starting.
 When the supply is connected to the stator of a three-
phase induction motor, a rotating magnetic field is
produced, hence the rotor starts rotating and then the
induction motor starts.
 At the time of starting, the motor slip is unity, and the
starting current is very large.
 The purpose of a starter is not to just start the motor, but
it performs the two main functions. They are as follows.



47. There are three main methods of Starting of
Squirrel Cage Induction Motor.
They are as follows:

48. Direct Online Starter
 The direct on line starter method, of an induction motor is
simple and economical.
 In this method, the starter is connected directly to supply
voltage.
 By this method small motors up to 5 kW rating is started
to avoid the supply voltage fluctuation.

49. Star-Delta Starter
The Star-Delta Starter is a very
common type of starter and is used
extensively as compared to the
other type of starting methods of
the induction motor.
When the switch S is in the
START position, the stator
windings are connected in the star
as shown in Fig.

50.  When the motor picks up the speed, about 80 percent of
its rated speed, the switch S is immediately put into the
RUN position.
 As a result, a stator winding which was in star connection
is changed into DELTA connection now as shown in the
fig.
 At the time of starting ( Star connection)
 Under Running condition(Delta)

51. 1/3
1/3

52. Auto transformer Starter
 An Auto transformer
Starter is suitable for
both star and delta
connected motors.
 In this method, the
starting current is
limited by using a
three-phase auto
transformer to reduce
the initial stator
applied voltage.

53.  The primary of the auto transformer is connected to the
supply line, and the motor is connected to the
secondary of the auto transformer.
 When the motor picks up the speed of about 80 percent
of its rated value, the handle H is quickly moved to the
RUN position.
 Thus, the auto transformer is dis-connected from the
circuit, and the motor is directly connected to the line
and achieve its full rated voltage.
 where K is the transformation ratio.
 The star delta starter is equivalent to an auto
transformer starter of the ratio K =1/3= 0.58.

54. Starting method of Slip Ring Induction Motor
 In the Slip Ring Induction Motor starter, the full
supply voltage is connected across the starter. The
connection diagram of the slip ring induction motor
starter is shown below.

55.  Full starting resistance is connected and thus the supply
current to the stator is reduced.
 The rotor begins to rotate, and the rotor resistances are
gradually cut out as the speed of the motor increases.
 When the motor is running at its rated full load speed,
the starting resistances are cut out completely, and the
slip rings are short-circuited.
 In this way we can start the 3-Phase IM safely.

56. BRAKE TEST ON A THREE PHASE
SLIP RING INDUCTION MOTOR
Aim : To conduct the load test on a three phase slip ring
Induction motor and to draw the performance and
mechanical characteristics, i.e.(i) Efficiency Vs. Output
power, (ii) Torque Vs.Output power, (iii) Line current Vs.
Output power, (iv) Power factor Vs. Output, (v) Slip Vs.
Output power, and (vi) Torque Vs. Speed

57. Circuit Diagram
Circuit Diagram for Brake Test on 3-ph Induction Motor

58. Procedure:
1- Connect the circuit as per fig.
2- Ensure that the motor is unloaded and the variac of
autotransformer is set at zero output
voltage.
3- Switch-ON 3 phase AC mains and start the motor at
reduced applied voltage.
4- Increase the applied voltage, till its rated value.
5- Take-down the readings of all the meters and the speed
under no load running in table.

59. Table of Readings

60. 6- Increase the load on the motor gradually by turning
of the hand wheels, thus tighten the belt.
7- Record the readings of all the meters and the speed at
every setting of the load in above table.
8- Observation may be continued upto the full load
current rating of the motor.
9- Reduce the load on the motor and finally unload it
completely.
10- Switch-OFF the supply to stop the motor.
11- Measure the radius of the pulley (R) in (meter).

61. Calculations:
Input power is measured by the two wattmeters (W1 and
W2), properly connected in the circuit:

62. Results
The brake test on a three phase slip ring Induction motor
is performed and graphs have been plotted for various
characteristics.