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Chapter Three
Three Phase Induction Motors
 Introduction
 Induction Motor Construction
 Basic Induction Motor Concepts
 The Equivalent Circuit of an Induction Motor
 Power and Torque in Induction Motors
 Induction Motor Torque-Speed Characteristics and their Variations
 Speed Control of Induction Motors
 Determining Circuit Model Parameters
3-phase Induction Motor
2
Learning Objectives
 Understand the construction of an induction motor.
 Understand the concept of rotor slip and its relationship to rotor frequency.
 Understand and know how to use the equivalent circuit of an induction motor.
 Understand power flows and the power flow diagram of an induction motor.
 Be able to use the equation for the torque-speed characteristic curve.
 Understand how the torque- speed characteristic curve varies with different rotor designs.
 Understand the techniques used for induction motor starting.
 Understand how the speed of induction motors can be controlled.
 Understand how to measure induction motor circuit model parameters.
 Understand the induction machine used as a generator and induction motor ratings. 3
Introduction
 Induction motors are used worldwide in many residential, commercial,
industrial, and utility applications.
 Induction Motors transform electrical energy into mechanical energy.
 It can be part of a pump or fan, or connected to some other form of
mechanical equipment such as a winder, conveyor, or mixer.
 A induction machine can be used as either an induction generator or an induction
motor.
 Induction motors are popularly used in the industry
 Focus on three-phase induction motor
 Main features: cheap and low maintenance
 Main disadvantages: speed control is not easy 4
Cont.…
 Three-phase induction motors are the most common and frequently
encountered machines in industry.
 Simple design, rugged, low-price, easy maintenance
 Wide range of power ratings: fractional horsepower to 10 MW
 Run essentially as constant speed from no-load to full load
 Its speed depends on the frequency of the power source
 Not easy to have variable speed control
 Requires a variable-frequency power-electronic drive for optimal
speed control
5
Advantages
 3-phase induction motors are simple & rugged.
 Its cost is low & it is reliable
 Its has high efficiency.
 Its maintenance cost is low
 It is self starting motor
 Suitable for all environments like coal mines & chemical factories
6
Induction Motor Construction
 A 3-phase induction motor has
two main parts
1. Stator
2. Rotor
1. Stator
 It is stationary part of induction
motor.
 The stator of an induction motor
consists of Stator Frame, Stator
core, 3 phase winding, two end
covers, bearings etc.
 The stator and the rotor are electrical
circuits that perform as electromagnets.
7
Cont.…
 Stator core is a stack of cylindrical steel laminations.
 Slots are provided on the inner periphery.
 Three windings are space displaced by 120° electrical.
 Large size motors use open slots.
 Small size motors use semi closed slots.
 The air gap length should be as small as possible
 Stator frame provides only mechanical support to the stator core.
 The stator winding for a definite number of poles.
 Greater the poles, lesser the speed and vice versa
8
Cont.…
 The stator core of a National Electrical Manufacturers Association
(NEMA) motor is made up of several hundred thin laminations.
 Stator laminations are stacked together forming a hollow cylinder. Coils of
insulated wire are inserted into slots of the stator core.
 Electromagnetism is the principle behind motor operation.
 Each grouping of coils, together with the steel core it surrounds, form an
electromagnet.
 The stator windings are connected directly to the power source.
9
Cont.…
 consisting of a steel frame that supports
a hollow, cylindrical core.
 core, constructed from stacked
laminations (why?), having a number of
evenly spaced slots, providing the space
for the stator winding.
Stator of IM
10
2. Rotor
 The rotor is the rotating part of the electromagnetic circuit.
 There are two different types of induction motor rotors which can be
placed inside the stator.
Squirrel-cage rotor
Wound(Slip ring) rotor
 However, the most common type of rotor is the “squirrel cage” rotor.
Squirrel-cage rotor
The rotor winding is in form of copper bars.
The copper bars are revited, welded with end rings.
No external resistance can be inserted in the rotor circuit 11
Cont.…
 Conducting bars laid into slots and
shorted at both ends by shorting
rings.
 Rotor winding is composed of
copper bars embedded in the rotor
slots and shorted at both end by
end rings.
 Simple, low cost, robust, low
maintenance
 A cage induction motor rotor
consists of a series of conducting
bars laid into slots carved in the
face of the rotor and shorted at
either end by large shorting rings.
12
Wound (Slip ring) rotor
 A wound rotor has a complete set of three-phase windings that are similar
to the windings on the stator.
 The three phases of the rotor windings are usually Y-connected, and the
ends of the three rotor wires are tied to slip rings on the rotor's shaft.
 The rotor windings are shorted through brushes riding on the slip rings.
 Wound-rotor induction motors therefore have their rotor currents
accessible at the stator brushes, where they can be examined and where
extra resistance can be inserted into the rotor circuit.
 It is possible to take advantage of this feature to modify the torque-speed
characteristic of the motor.
13
Cont.…
 Wound-rotor induction motors are more expensive than cage induction
motors.
 They require much more maintenance because of the wear associated
with their brushes and slip rings.
 Wound-rotor induction motors are rarely used.
14
Cont.…
15
Cont.…
Squirrel cage rotor
Wound rotor
Notice the
slip rings
16
Cont.…
Cutaway in a typical wound-
rotor IM.
Brushes
Slip rings
17
Notice the brushes and the slip
rings
Comparison of wound and squirrel cage rotor
18
Sr.No. Wound or slip ring Squirrel cage rotor
1 Rotor consists of a three phase winding
similar to the stator winding
Rotor consists of bars which are shorted at the
ends with the help of end rings
2 Construction is complicated Construction is very simple
3 Resistance can be added externally As permanently shorted, external resistance cannot
be added
4 Slip rings and brushes are present to add
external resistance
Slip rings and brushes are absent
5 The construction is delicate and due to
brushes, frequent maintenance is necessary
The construction is robust and maintenance free
6 The rotors are very costly Due to simple construction, the rotors are cheap
7 Only 5% of induction motors in industry
use slip ring rotor
Very common and almost 95% induction motors
use this type of motor
Cont.…
Sr.No. Wound or slip ring Squirrel cage rotor
8 High starting torque can be obtained Moderate starting torque which can not be
controlled
9 Rotor resistance starter can be obtained Rotor resistance starter can not be used
10 Rotor must be wound for the same number
of poles as that of stator
The rotor automatically adjusts itself for the same
number of poles as that of stator
11 speed control by rotor resistance is possible speed control by rotor resistance is not possible
12 Rotor copper losses are high hence
efficiency is less
Rotor copper losses are less hence have higher
efficiency.
13 Used for lifts, hoists, cranes, elevators,
compressors etc.
Used for lathes, drilling machines, fans, blowers,
water pumps, grinders, printing machines etc.
19
The Development of Induced Torque in an Induction Motor
 A three-phase set of voltages has been applied to the stator, and a three-
phase set of stator currents is flowing.
 These currents produce a magnetic field 𝐵𝑠 , which is rotating in a
clockwise/counterclockwise direction. The speed of the magnetic field's
rotation is given by
𝑛 𝑠𝑦𝑛𝑐 =
120𝑓𝑒
𝑃
Where 𝑓𝑒 is the system frequency applied to the stator in hertz
P is the number of poles in the machine.
𝑛 𝑠𝑦𝑛𝑐 is called the synchronous speed in rpm (revolutions per minute)
20
Cont.…
 This rotating magnetic field 𝐵𝑠 passes over the rotor bars and induces a
voltage in them.
 The voltage induced in a given rotor bar is given by the equation
𝑒𝑖𝑛𝑑 = 𝑣 × 𝐵 . 𝑙
where 𝑣 = Velocity of the bar relative to the magnetic field
B = Magnetic flux density vector
𝑙= length of conductor in the magnetic field
 It is the relative motion of the rotor compared to the stator magnetic field
that produces induced voltage in a rotor bar.
21
Cont.…
 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 flow produces a rotor magnetic field 𝐵 𝑅.
 A torque is produced as a result of the interaction of those two magnetic
fields
Where 𝜏𝑖𝑛𝑑 is the induced torque and
k is a constant representing the construction of the machine.
BR and BS are the magnetic flux densities of the rotor and the stator respectively
ind R skB B  
22
Working Principle of 3 Phase Induction Motor
 When three phase supply is given to the three phase stator winding of the
induction motor, a rotating magnetic field is developed around the stator
which rotates at synchronous speed.
 This rotating magnetic field passes through the air gap and cuts the rotor
conductors which were stationary. Due to the relative speed between the
stationary rotor conductors and the rotating magnetic field, an emf is
induced in the rotor conductors. As the rotor conductors are short
circuited, current starts flowing through it.
 As these current carrying rotor conductors are placed in the magnetic field
produced by the stator, they experiences a mechanical force i.e. torque
which moves the rotor in the same direction as that of the rotating
magnetic field.
23
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.
 So, the IM will always run at a speed lower than the synchronous speed. 24
Rotor Slip
 The voltage induced in a rotor bar of an induction motor depends on the
speed of the rotor relative to the magnetic fields.
 The difference between the motor speed and the synchronous speed is
called the Slip speed.
Where nslip= slip speed of the machine
nsync= speed of the magnetic field
nm = mechanical shaft speed of the motor
 Note that in normal operation both the rotor and stator magnetic fields 𝐵 𝑅 and 𝐵𝑠 rotate
together at synchronous speed 𝑛 𝑠𝑦𝑛𝑐,While the rotor itself turns at a slower speed.
slip sync mn n n 
25
The Slip
 Slip is the relative speed expressed on a per-unit or a percentage basis.
Where s is the slip
This equation can also be expressed in terms of angular velocity 𝜔(radians
per second) as
 Notice that : if the rotor runs at synchronous speed, s = 0
if the rotor is stationary, s = 1
26
Cont.…
 The mechanical speed of the rotor shaft can expressed in terms of
synchronous speed and slip.
 From the above expression for slip, solving equations for mechanical speed
yields
Or
 These equations are useful in the derivation of induction motor torque and
power relationships.
27
Induction Motors and Transformers
 Both IM and transformer works on the principle of 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.
28
Frequency
 The frequency of the voltage induced in the rotor is given by
 Since the slip of the rotor is defined as
Where 𝑓𝑟= rotor frequency (Hz)
𝑓𝑒 = supply frequency(Hz) s= slip 29
Cont.…
 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.
30
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

 
31
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.
 There is a relation between horse power and watts.
Example: A 208-V, 10hp, four pole, 60 Hz, Y-connected induction motor has
a full-load slip of 5 percent
a) What is the synchronous speed of this motor?
b) What is the rotor speed of this motor at rated load?
c) What is the rotor frequency of this motor at rated load?
d) What is the shaft torque of this motor at rated load?
746hp watts
32
Solution
a)
b)
c)
d)
120 120(60)
1800
4
e
sync
f
n rpm
P
  
(1 )
(1 0.05) 1800 1710
m sn s n
rpm
 
   
0.05 60 3r ef sf Hz   
2
60
10 746 /
41.7 .
1710 2 (1/ 60)
out out
load
mm
P P
n
hp watt hp
N m

 

 

 
 
33
The Equivalent Circuit of an Induction Motor
 The induction motor is similar to the transformer with the exception that
its secondary windings are free to rotate.
Fig.The transformer model of an induction motor, with rotor and stator
connected by an ideal transformer of tums ratio 𝑎 𝑒𝑓𝑓 34
The Rotor Circuit Model
 In general, the greater the relative motion between the rotor and the stator
magnetic fields, the greater the resulting rotor voltage and rotor frequency.
 When the rotor is locked (or blocked), i.e. s =1, the largest voltage and rotor
frequency are induced in the rotor, Why?
 On the other side, if the rotor rotates at synchronous speed, i.e. s = 0, the
induced voltage and frequency in the rotor will be equal to zero, Why?
Where ER0 is the largest value of the rotor’s induced voltage obtained at
s = 1(locked rotor)
 The same is true for the frequency, i.e.
35
Cont.…
 It is known that
𝑋 = 𝜔𝐿 = 2𝜋𝑓𝐿
 So, as the frequency of the induced
voltage in the rotor changes, the
reactance of the rotor circuit also
changes.
 Where 𝑋 𝑅𝑂 is the rotor reactance
at the supply frequency(at blocked
rotor)
 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
36
Cont.…
 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


37
Cont.…
 Now we can have the rotor equivalent circuit
38
Cont.…
 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





39
Losses and the Power Flow Diagram
Power Losses in Induction Machines
 Copper losses
 Copper loss in the stator (PSCL) = I1
2
R1
 Copper loss in the rotor (PRCL) = I2
2
R2
 Core loss (Pcore)
 Mechanical power loss due to friction and windage
 How this power flow in the motor?
40
Power Flow in Induction Motor
41
Fig.The power-flow diagram of an induction motor.
Power Relations
 The input current to a phase of the motor can be found by
3 cos 3 cosin L L ph phP V I V I  
2
1 13SCLP I R
( )AG in SCL coreP P P P  
2
2 23RCLP I R conv AG RCLP P P 
( )out conv f w strayP P P P  
conv
ind
m
P



42
I1 =
V∅
Zeq
𝑍 𝑒𝑞 = 𝑅1 + 𝑗𝑋1 +
1
𝐺 𝐶 − 𝑗𝐵 𝑀 +
1
𝑅2
𝑆
+ 𝑗𝑋2
Pcore = 3E1
2
GC
Power Relations
3 cos 3 cosin L L ph phP V I V I  
2
1 13SCLP I R
( )AG in SCL coreP P P P  
2
2 23RCLP I R
conv AG RCLP P P 
( )out conv f w strayP P P P  
conv RCLP P  2 2
23
R
I
s

2 2
2
(1 )
3
R s
I
s


RCLP
s

(1 )RCLP s
s


(1 )conv AGP s P 
conv
ind
m
P



(1 )
(1 )
AG
s
s P
s 



43
Cont.…
 The input power to an induction motor 𝑃𝑖𝑛 in the form of three-phase
electric voltages and currents.
 The first losses encountered in the machine are 𝐼2 𝑅 losses in the stator
windings (the stator copper loss 𝑃𝑆𝐶𝐿) ' Then some amount of power is lost
as hysteresis and eddy currents in the stator (𝑃𝐶𝑜𝑟𝑒).
 The power remaining at this point is transferred to the rotor of the
machine across the air gap between the stator and rotor. This power is
called the air-gap power 𝑃𝐴𝐺 of the machine.
 After the power is transferred to the rotor, some of it is lost as 𝐼2 𝑅
losses(the rotor copper loss 𝑃𝑅𝐶𝐿), and
44
Cont.…
 The rest is converted from electrical
to mechanical form (𝑃𝐶𝑜𝑛𝑣).
 Finally, friction and windage losses
𝑃𝐹&𝑊 and stray losses 𝑃 𝑚𝑖𝑠𝑐 are
subtracted.
 The remaining power is the output
of the motor 𝑃𝑜𝑢𝑡. : :
1 : : 1-
AG RCL convP P P
s s
45
Separating the Rotor Copper Losses and the Power Converted in an
Induction Motor's Equivalent Circuit
 Since air gap power would require
𝑹 𝟐
𝑺
and rotor copper loss require 𝑹 𝟐 element.
Actual rotor
resistance
Resistance
equivalent to
mechanical load 46
𝑹 𝒄𝒐𝒏𝒗 =
𝑹 𝟐
𝑺
− 𝑹 𝟐 = 𝑹 𝟐(
𝟏−𝑺
𝑺
)
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.
47
Solution
1.
2.
3.
3 cos
3 480 60 0.85 42.4 kW
in L LP V I 
    
42.4 2 1.8 38.6 kW
AG in SCL coreP P P P  
   
700
38.6 37.9 kW
1000
conv AG RCLP P P 
  
&
600
37.9 37.3 kW
1000
out conv F WP P P 
  
48
Cont.…
4.
37.3
50 hp
0.746
outP  
100%
37.3
100 88%
42.4
out
in
P
P
  
  
49
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
50
Solution
1.
2.
120 120 60
1800 rpm
4
e
sync
f
n
P

  
(1 ) (1 0.022) 1800 1760 rpmm syncn 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
 
    
    
 
51
Cont.…
3.
4.
0.641 1.106 12.94 31.1
11.72 7.79 14.07 33.6
tot stat fZ Z Z
j
j
 
     
     
1
460 0
3 18.88 33.6 A
14.07 33.6tot
V
I
Z

 
    
 
cos33.6 0.833 laggingPF   
3 cos 3 460 18.88 0.833 12530 Win L LP V I      
2 2
1 13 3(18.88) 0.641 685 WSCLP I R   
12530 685 11845 WAG in SCLP P P    
52
Cont.…
5.
6.
(1 ) (1 0.022)(11845) 11585 Wconv AGP s P    
& 11585 1100 10485 W
10485
= 14.1hp
746
out conv F WP P P    

11845
62.8 N.m
18002
60
AG
ind
sync
P

 
  

10485
56.9 N.m
17602
60
out
load
m
P

 
  

10485
100% 100 83.7%
12530
out
in
P
P
     
53
Induction Motor Torque-Speed Characteristics
 Thevenin’s theorem can be used to transform the network to the left of
points ‘a’ and ‘b’ into an equivalent voltage source VTH in series with
equivalent impedance RTH+jXTH
54
Cont.…
1 1( )
M
TH
M
jX
V V
R j X X

 
1 1( )//TH TH MR jX R jX jX  
2 2
1 1
| | | |
( )
M
TH
M
X
V V
R X X

 
55
Cont.…
 Since XM>>X1 and XM>>R1
 Because XM>>X1 and XM+X1>>R1, the equivalent impedance
1
M
TH
M
X
V V
X X


2
1
1
1
M
TH
M
TH
X
R R
X X
X X
 
  
 
 56
ZTH =
jXM(R1 + jX1)
R1 + jX1 + jXM
Cont.…
Then the power converted to mechanical (Pconv)
2 2
22
2( )
TH TH
T
TH TH
V V
I
Z R
R X X
s
 
 
   
 
2 2
2
(1 )
3conv
R s
P I
s


and the internal mechanical torque (𝜏 𝑐𝑜𝑛𝑣)(Induced Torque)
conv
ind
m
P



(1 )
conv
s
P
s 


2 2
23
AG
s s
R
I
Ps
 
 
57
Cont.…
2
2
2
22
2
3
( )
TH
ind
s
TH TH
V R
sR
R X X
s


 
 
   
   
         
2 2
2
22
2
3
1
( )
TH
ind
s
TH TH
R
V
s
R
R X X
s


 
 
 
 
   
 
58
Torque -Speed Characteristics curve
Fig.Typical torque-speed characteristics of induction motor
59
Comments
1. The induced torque is zero at synchronous speed.
2. The curve is nearly linear between no-load and full load. In this range, the
rotor resistance is much greater than the reactance, so the rotor current,
torque increase linearly with the slip.
3. There is a maximum possible torque that can’t be exceeded. This torque is
called pullout torque and is 2 to 3 times the rated full-load torque.
4. The starting torque of the motor is slightly higher than its full-load
torque, so the motor will start carrying any load it can supply at full load.
5. The torque of the motor for a given slip varies as the square of the applied voltage.
6. If the rotor is driven faster than synchronous speed it will run as a generator,
converting mechanical power to electric power. 60
Complete Torque-Speed characteristics curve
61
Fig: Induction motor torque-speed characteristic curve. showing the
extended operating ranges (braking region and generator region).
Pushover
torque
Cont.…
7. If the motor is turning backward relative to the direction of the magnetic
fields, the induced torque in the machine will stop the machine very
rapidly and will try to rotate it in the other direction. Since reversing the
direction of magnetic field rotation is simply a matter of switching any
two stator phases, this fact can be used as a way to very rapidly stop an
induction motor.
The act of switching two phases in order to stop the motor very rapidly is
called plugging
62
Operating Regions
The torque-speed curve brakes down into three operating regions:
1)Braking, 𝑛 𝑚 < 0, 𝑠 > 0
 Torque is positive whilst speed is negative. Considering the power
conversion equation
 It can be seen that if the power converted is negative (from P = τ ω) then
the airgap power is positive. i.e. the power is flowing from the stator to the
rotor and also into the rotor from the mechanical system. This operation is
also called plugging
63
(1 )conv AGP s P 
Cont.…
 This mode of operation can be used to quickly stop a machine. If a motor
is travelling forwards it can be stopped by interchanging the connections to
two of the three phases. Switching two phases has the result of changing
the direction of motion of the stator magnetic field, effectively putting the
machine into braking mode in the opposite direction.
2) Motoring,0 < 𝑛 𝑚 < 𝑛 𝑠𝑦𝑛𝑐, 1 > 𝑠 > 0
 Torque and motion are in the same direction. This is the most common
mode of operation.
64
Cont.…
3)Generating 𝑛 𝑚 > 𝑛 𝑠𝑦𝑛𝑐, 𝑠 < 0
 In this mode, again torque is positive whilst speed is negative. However,
unlike plugging,
 indicates that if the power converted is negative, so is the air gap power. In
this case, power flows from the mechanical system, to the rotor circuit, then
across the air gap to the stator circuit and external electrical system.
65
(1 )conv AGP s P 
Cont.…
 If an induction motor is driven at a speed greater than 𝑛 𝑠𝑦𝑛𝑐 by an external
prime mover, the direction of its induced torque will reverse and it will act
as a generator.
 As the torque applied to its shaft by the prime mover increases, the amount
of power produced by the induction generator increases.
 There is a maximum possible induced torque in the generator mode of
operation. This torque is known as the pushover torque of the generator.
 If a prime mover applies a torque greater than the pushover torque to the
shaft of an induction generator, the generator will over speed.
66
Cont.…
67
Maximum(Pullout) Torque in an Induction Motor
 Maximum power will be achieved when the magnitude of source
impedance matches the load impedance.
 Maximum torque occurs when the power transferred to R2/s is maximum.
 This condition occurs when R2/s equals the magnitude of the impedance
RTH + j (XTH + X2)
max
2 22
2( )TH TH
T
R
R X X
s
  
max
2
2 2
2( )
T
TH TH
R
s
R X X

 
68
Cont.…
 The corresponding maximum torque of an induction motor equals
 The slip at maximum torque is directly proportional to the rotor resistance 𝑅2.
 The maximum torque is independent of 𝑅2 and related to the square of applied
voltage.
 Rotor resistance can be increased by inserting external resistance in the rotor of a
wound-rotor induction motor.
 The value of the maximum torque remains unaffected but the speed at which it
occurs can be controlled.
2
max 2 2
2
31
2 ( )
TH
s TH TH TH
V
R R X X


 
 
    
69
Cont.…
 It is possible to take advantage of
this characteristic of wound-rotor
induction motors to start very
heavy loads.
 If a resistance is inserted into the
rotor circuit, the maximum torque
can be adjusted to occur at starting
conditions.
 Therefore, the maximum possible
torque would be available to start
heavy loads. 70
Effect of rotor resistance on torque-speed characteristic
Example
A two-pole, 50-Hz induction motor supplies 15kW to a load at a speed of
2950 rpm.
1. What is the motor’s slip?
2. What is the induced torque in the motor in N.m under these
conditions?
3. What will be the operating speed of the motor if its torque is doubled?
4. How much power will be supplied by the motor when the torque is
doubled?
71
Solution
1.
2.
120 120 50
3000 rpm
2
3000 2950
0.0167 or 1.67%
3000
e
sync
sync m
sync
f
n
P
n n
s
n

  
 
  
3
no given
assume and
15 10
48.6 N.m
2
2950
60
f W
conv load ind load
conv
ind
m
P
P P
P
 



  

  

72
Cont.…
3. In the low-slip region, the torque-speed curve is linear and the induced
torque is direct proportional to slip. So, if the torque is doubled the
new slip will be 3.33% and the motor speed will be
4.
(1 ) (1 0.0333) 3000 2900 rpmm syncn s n     
2
(2 48.6) (2900 ) 29.5 kW
60
conv ind mP  


    
73
Example
A 460-V, 25-hp, 60-Hz, four-pole, Y-connected wound-rotor 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 
1. What is the maximum torque of this motor? At what speed and slip does it occur?
2. What is the starting torque of this motor?
3. If the rotor resistance is doubled, what is the speed at which the maximum torque
now occur? What is the new starting torque of the motor?
4. Calculate and plot the T-s c/c for both cases.
74
Solution
2 2
1 1
2 2
( )
460
26.3
3 255.2 V
(0.641) (1.106 26.3)
M
TH
M
X
V V
R X X

 

 
 
2
1
1
2
26.3
(0.641) 0.590
1.106 26.3
M
TH
M
X
R R
X X
 
  
 
 
   
 
1 1.106THX X  
75
Cont.…
1.
The corresponding speed is
max
2
2 2
2
2 2
( )
0.332
0.198
(0.590) (1.106 0.464)
T
TH TH
R
s
R X X

 
 
 
(1 ) (1 0.198) 1800 1444 rpmm syncn s n     
76
Cont.…
The torque at this speed is
2
max 2 2
2
2
2 2
31
2 ( )
3 (255.2)
2
2 (1800 )[0.590 (0.590) (1.106 0.464) ]
60
229 N.m
TH
s TH TH TH
V
R R X X



 
 
    


    

77
Cont.…
2. The starting torque can be found from the torque eqn. by substituting s
= 1
 
2 2
21
22
2
1
2
2
2 2
2 2
2
2 2
3
1
( )
3
[ ( ) ]
3 (255.2) (0.332)
2
1800 [(0.590 0.332) (1.106 0.464) ]
60
104 N.m
TH
start ind s
s
TH TH
s
TH
s TH TH
R
V
s
R
R X X
s
V R
R R X X
 





 
 
  
 
   
 

  
 

    

78
Cont.…
3. If the rotor resistance is doubled, then the slip at maximum torque doubles too
The corresponding speed is
The maximum torque is still
𝜏 𝑚𝑎𝑥= 229 N.m
max
2
2 2
2
0.396
( )
T
TH TH
R
s
R X X
 
 
(1 ) (1 0.396) 1800 1087 rpmm syncn s n     
79
Cont.…
The starting torque is now
2
2 2
3 (255.2) (0.664)
2
1800 [(0.590 0.664) (1.106 0.464) ]
60
170 N.m
start

 

    

80
Determination of Motor Parameters
 Due to the similarity between the induction motor equivalent circuit and
the transformer equivalent circuit, same tests are used to determine the
values of the motor parameters.
 DC test: determine the stator resistance R1
 No-load test: determine the rotational losses and magnetization current
(similar to no-load test in Transformers).
 Locked-rotor test: determine the rotor and stator impedances (similar to
short-circuit test in Transformers).
81
DC Test
 The purpose of the DC test is to determine R1. A variable DC voltage
source is connected between two stator terminals.
 The DC source is adjusted to provide approximately rated stator current,
and the resistance between the two stator leads is determined from the
voltmeter and ammeter readings.
82
Cont.…
then
 The current flows through two of the windings, so the total resistance in
the current path is 2R1 . Therefore,
 If the stator is Y-connected, the per phase stator resistance is
 If the stator is delta-connected, the per phase stator resistance is
DC
DC
DC
V
R
I

1
2
DCR
R 
1
3
2
DCR R
83
No-load Test
1. The motor is allowed to spin freely
2. The only load on the motor is the friction and windage
losses, so all Pconv is consumed by mechanical losses.
3. The slip is very small 84
 The no-load test of an induction motor measures the rotational losses of the
motor and provides information about its magnetization current.
Cont.…
4. At this small slip
The equivalent circuit reduces to…
85
Cont.…
5. Combining Rc and RF+W we get……
86
Cont.…
6. At the no-load conditions, the input power measured by meters must
equal the losses in the motor.
7. The PRCL is negligible because I2 is extremely small because R2(1-s)/s is
very large.
8. The input power equals
Where
9. The equivalent input impedance is thus approximately
If X1 can be found, in some other fashion,
the magnetizing impedance X will be known
&
2
1 13
in SCL core F W
rot
P P P P
I R P
  
  &rot core F WP P P 
1
1,
eq M
nl
V
Z X X
I

  
87
Locked-Rotor(Blocked-Rotor) Test
 In this test, the rotor is locked or blocked so that it cannot move, a voltage
is applied to the motor, and the resulting voltage, current and power are
measured.
88Fig: The locked-rotor test for an induction motor test circuit:
Cont.…
 Since the rotor is not moving, the slip s = 1, and so the rotor resistance Τ𝑅2 𝑠 is just
equal to 𝑅2(quite a small value). Since R2 and X2 are so small, almost all the input
current will flow through them, instead of through the much larger magnetizing
reactance XM .
 Therefore, the circuit under these conditions looks like a series combination of
X1, R1, X2 and R2 .
89
Fig. The locked-rotor test for an induction motor equivalent circuit
Cont.…
 The AC voltage applied to the stator is adjusted so that the current flow is
approximately full-load value.
 The input power to the motor is given by
 The locked-rotor power factor can be found as
 The magnitude of the total impedance
cos
3
in
l l
P
PF
V I
 
LR
V
Z
I


90
Cont.…
Where X’1 and X’2 are the stator and rotor reactances at the test
frequency respectively.
'
cos sin
LR LR LR
LR LR
Z R jX
Z j Z 
 
 
1 2
' ' '
1 2
LR
LR
R R R
X X X
 
 
2 1LRR R R 
'
1 2
rated
LR LR
test
f
X X X X
f
  
91
Rules of thumb for dividing rotor and stator circuit reactance
X1 and X2 as function of XLR
Rotor Design X1 X2
Wound rotor 0.5 XLR 0.5 XLR
Design A 0.5 XLR 0.5 XLR
Design B 0.4 XLR 0.6 XLR
Design C 0.3 XLR 0.7 XLR
Design D 0.5 XLR 0.5 XLR
92
Example
The following test data were taken on a 7.5-hp, four-pole, 208-V, 60-Hz, design A,
Y-connected IM having a rated current of 28 A.
DC Test:
VDC = 13.6 V IDC = 28.0 A
No-load Test:
Vl = 208 V f = 60 Hz Pin = 420 W
IA= 8.12 A IB = 8.2 A IC = 8.18 A
Locked-rotor Test:
Vl = 25 V f = 15 Hz Pin = 920 W
IA = 28.1 A IB = 28.0 A I 𝐶 = 27.6 A
a) Sketch the per-phase equivalent circuit of this motor.
b) Find the slip at pull-out torque, and find the value of the pull-out torque.
93
Speed Control of Induction Motors
 There are really only two techniques by which the speed of an induction
motor can be controlled.
a) Vary the synchronous speed, which is the speed of the stator and rotor
magnetic fields, since the rotor speed always remains near 𝑛 𝑠𝑦𝑛𝑐.
b) Vary the slip of the motor for a given load.
 The synchronous speed of an induction motor is given by
𝑛 𝑠𝑦𝑛𝑐 =
120𝑓𝑒
𝑃
94
Cont.…
 So the only ways in which the synchronous speed of the machine can be
varied are
a) by changing the electrical frequency and
b) by changing the number of poles on the machine.
 Slip control may be accomplished by varying either the rotor resistance or
the terminal voltage of the motor.
95
Induction Motor Speed Control by Pole Changing
 There are two major approaches to change the number of poles in an
induction motor:
1. The method of consequent poles
2. Multiple stator windings
Method of Consequent Poles
 Consider one phase winding in a stator. By changing the current flow in
one portion of the stator windings as such that it is similar to the current
flow in the opposite portion of the stator will automatically generate an
extra pair of poles.
96
Cont.…
97
A close-up view of one phase of a pole-changing
winding. In the two-pole configuration. one coil is
a north pole and the other one is a south pole.
Cont.…
98
When the connection on one of the two
coils is reversed. they are both north
poles. and the magnetic flux returns to
the stator at points halfway between the
two coils. The south poles are called
consequent poles. and the winding is now a
four pole winding
Cont.…
 By applying this method, the number of poles may be maintained (no
changes), doubled or halfed, hence would vary its operating speed.
In terms of torque, the maximum torque magnitude would generally be
maintained.
Disadvantage:
 This method will enable speed changes in terms of 2:1 ratio steps, hence to
obtained variations in speed, multiple stator windings has to be applied.
Multiple stator windings have extra sets of windings that may be switched
in or out to obtain the required number of poles. Unfortunately this would
an expensive alternative.
99
Speed Control by Changing the Line Frequency
 If the electrical frequency applied to the stator of an induction motor is
changed, the rate of rotation of its magnetic fields 𝑛 𝑠𝑦𝑛𝑐 will change in
direct proportion to the change in electrical frequency, and the no- load
point on the torque-speed characteristic curve will change with it.
 The synchronous speed of the motor at rated conditions is known as the
base speed.
 By using variable frequency control, it is possible to adjust the speed of the
motor either above or below base speed.
 A properly designed variable-frequency induction motor drive can be very
flexible. It can control the speed of an induction motor over a range from
as little as 5 percent of base speed up to about twice base speed. 100
Cont.…
 However, it is important to maintain certain voltage and torque limits on
the motor as the frequency is varied, to ensure safe operation.
 When running at speeds below the base speed of the motor, it is necessary
to reduce the terminal voltage applied to the stator for proper operation.
 The terminal voltage applied to the stator should be decreased linearly with
decreasing stator frequency. This process is called derating. If it is not
done, the steel in the core of the induction motor will saturate and
excessive magnetization currents will flow in the machine.
101
Cont.…
 Changing the electrical frequency would also require an adjustment to the
terminal voltage in order to maintain the same amount of flux level in the
machine core. If not the machine will experience:
a) Core saturation (non linearity effects)
b) Excessive magnetization current.
 Varying frequency with or without adjustment to the terminal voltage may
give 2 different effects:
a) Vary frequency, stator voltage adjusted – generally vary speed and
maintain operating torque.
b) Vary Frequency, stator voltage maintained – able to achieve higher
speeds but a reduction of torque as speed is increased. 102
Cont.…
 There may also be instances where both characteristics are needed in the
motor operation; hence it may be combined to give both effects.
 With the arrival of solid-state devices/power electronics, line frequency
change is easy to achieved and it is more versatile to a variety of machines
and application.
103
Variable-frequency speed control in an induction motor
104
a) The family of torque-speed characteristic curves for speeds below base
speed. assuming that the line voltage is derated linearly with frequency.
Cont.…
105
b)The family of torque-speed characteristic curves for speeds above
base speed. assuming that the line voltage is held constant.
Cont.…
 As with any transformer, the flux
in the core of an induction motor
can be found from Faraday's law:
 If a voltage v(t) = 𝑉 𝑀 sin 𝜔𝑡 is
applied to the core, the resulting
flux ∅ is
 If the electrical frequency applied to the
stator decreases by 10 percent while the
magnitude of the voltage applied to the
stator remains constant, the flux in the core
of the motor will increase by about 10 % and
the magnetization current of the motor will
increase.
 Induction motors are normally designed to
operate near the saturation point on their
magnetization curves, so the increase in flux
due to a decrease in frequency will cause
excessive magnetization currents to flow in
the motor. 106
Speed Control by Changing the Line Voltage
 The torque developed by an induction motor is proportional to the square
of the applied voltage.
 Varying the terminal voltage will vary the operating speed but with also a
variation of operating torque.
 In terms of the range of speed variations, it is not significant hence this
method is only suitable for small motors only.
 The speed of the motor may be controlled over a limited range by varying
the line voltage. This method of speed control is sometimes used on
small motors driving fans.
107
Cont.…
108Fig. Variable-line-voltage speed control in an induction motor
Speed Control by Changing the Rotor Resistance
 In wound-rotor induction motors,
it is possible to change the shape
of the torque- speed curve by
inserting extra resistances into the
rotor circuit of the machine.
 Changing the rotor resistance will
change the operating speed of the
motor. However, inserting extra
resistances into the rotor circuit of
an induction motor seriously
reduces the efficiency of the
machine.
109
110

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Chapter 3

  • 1. Chapter Three Three Phase Induction Motors  Introduction  Induction Motor Construction  Basic Induction Motor Concepts  The Equivalent Circuit of an Induction Motor  Power and Torque in Induction Motors  Induction Motor Torque-Speed Characteristics and their Variations  Speed Control of Induction Motors  Determining Circuit Model Parameters
  • 3. Learning Objectives  Understand the construction of an induction motor.  Understand the concept of rotor slip and its relationship to rotor frequency.  Understand and know how to use the equivalent circuit of an induction motor.  Understand power flows and the power flow diagram of an induction motor.  Be able to use the equation for the torque-speed characteristic curve.  Understand how the torque- speed characteristic curve varies with different rotor designs.  Understand the techniques used for induction motor starting.  Understand how the speed of induction motors can be controlled.  Understand how to measure induction motor circuit model parameters.  Understand the induction machine used as a generator and induction motor ratings. 3
  • 4. Introduction  Induction motors are used worldwide in many residential, commercial, industrial, and utility applications.  Induction Motors transform electrical energy into mechanical energy.  It can be part of a pump or fan, or connected to some other form of mechanical equipment such as a winder, conveyor, or mixer.  A induction machine can be used as either an induction generator or an induction motor.  Induction motors are popularly used in the industry  Focus on three-phase induction motor  Main features: cheap and low maintenance  Main disadvantages: speed control is not easy 4
  • 5. Cont.…  Three-phase induction motors are the most common and frequently encountered machines in industry.  Simple design, rugged, low-price, easy maintenance  Wide range of power ratings: fractional horsepower to 10 MW  Run essentially as constant speed from no-load to full load  Its speed depends on the frequency of the power source  Not easy to have variable speed control  Requires a variable-frequency power-electronic drive for optimal speed control 5
  • 6. Advantages  3-phase induction motors are simple & rugged.  Its cost is low & it is reliable  Its has high efficiency.  Its maintenance cost is low  It is self starting motor  Suitable for all environments like coal mines & chemical factories 6
  • 7. Induction Motor Construction  A 3-phase induction motor has two main parts 1. Stator 2. Rotor 1. Stator  It is stationary part of induction motor.  The stator of an induction motor consists of Stator Frame, Stator core, 3 phase winding, two end covers, bearings etc.  The stator and the rotor are electrical circuits that perform as electromagnets. 7
  • 8. Cont.…  Stator core is a stack of cylindrical steel laminations.  Slots are provided on the inner periphery.  Three windings are space displaced by 120° electrical.  Large size motors use open slots.  Small size motors use semi closed slots.  The air gap length should be as small as possible  Stator frame provides only mechanical support to the stator core.  The stator winding for a definite number of poles.  Greater the poles, lesser the speed and vice versa 8
  • 9. Cont.…  The stator core of a National Electrical Manufacturers Association (NEMA) motor is made up of several hundred thin laminations.  Stator laminations are stacked together forming a hollow cylinder. Coils of insulated wire are inserted into slots of the stator core.  Electromagnetism is the principle behind motor operation.  Each grouping of coils, together with the steel core it surrounds, form an electromagnet.  The stator windings are connected directly to the power source. 9
  • 10. Cont.…  consisting of a steel frame that supports a hollow, cylindrical core.  core, constructed from stacked laminations (why?), having a number of evenly spaced slots, providing the space for the stator winding. Stator of IM 10
  • 11. 2. Rotor  The rotor is the rotating part of the electromagnetic circuit.  There are two different types of induction motor rotors which can be placed inside the stator. Squirrel-cage rotor Wound(Slip ring) rotor  However, the most common type of rotor is the “squirrel cage” rotor. Squirrel-cage rotor The rotor winding is in form of copper bars. The copper bars are revited, welded with end rings. No external resistance can be inserted in the rotor circuit 11
  • 12. Cont.…  Conducting bars laid into slots and shorted at both ends by shorting rings.  Rotor winding is composed of copper bars embedded in the rotor slots and shorted at both end by end rings.  Simple, low cost, robust, low maintenance  A cage induction motor rotor consists of a series of conducting bars laid into slots carved in the face of the rotor and shorted at either end by large shorting rings. 12
  • 13. Wound (Slip ring) rotor  A wound rotor has a complete set of three-phase windings that are similar to the windings on the stator.  The three phases of the rotor windings are usually Y-connected, and the ends of the three rotor wires are tied to slip rings on the rotor's shaft.  The rotor windings are shorted through brushes riding on the slip rings.  Wound-rotor induction motors therefore have their rotor currents accessible at the stator brushes, where they can be examined and where extra resistance can be inserted into the rotor circuit.  It is possible to take advantage of this feature to modify the torque-speed characteristic of the motor. 13
  • 14. Cont.…  Wound-rotor induction motors are more expensive than cage induction motors.  They require much more maintenance because of the wear associated with their brushes and slip rings.  Wound-rotor induction motors are rarely used. 14
  • 16. Cont.… Squirrel cage rotor Wound rotor Notice the slip rings 16
  • 17. Cont.… Cutaway in a typical wound- rotor IM. Brushes Slip rings 17 Notice the brushes and the slip rings
  • 18. Comparison of wound and squirrel cage rotor 18 Sr.No. Wound or slip ring Squirrel cage rotor 1 Rotor consists of a three phase winding similar to the stator winding Rotor consists of bars which are shorted at the ends with the help of end rings 2 Construction is complicated Construction is very simple 3 Resistance can be added externally As permanently shorted, external resistance cannot be added 4 Slip rings and brushes are present to add external resistance Slip rings and brushes are absent 5 The construction is delicate and due to brushes, frequent maintenance is necessary The construction is robust and maintenance free 6 The rotors are very costly Due to simple construction, the rotors are cheap 7 Only 5% of induction motors in industry use slip ring rotor Very common and almost 95% induction motors use this type of motor
  • 19. Cont.… Sr.No. Wound or slip ring Squirrel cage rotor 8 High starting torque can be obtained Moderate starting torque which can not be controlled 9 Rotor resistance starter can be obtained Rotor resistance starter can not be used 10 Rotor must be wound for the same number of poles as that of stator The rotor automatically adjusts itself for the same number of poles as that of stator 11 speed control by rotor resistance is possible speed control by rotor resistance is not possible 12 Rotor copper losses are high hence efficiency is less Rotor copper losses are less hence have higher efficiency. 13 Used for lifts, hoists, cranes, elevators, compressors etc. Used for lathes, drilling machines, fans, blowers, water pumps, grinders, printing machines etc. 19
  • 20. The Development of Induced Torque in an Induction Motor  A three-phase set of voltages has been applied to the stator, and a three- phase set of stator currents is flowing.  These currents produce a magnetic field 𝐵𝑠 , which is rotating in a clockwise/counterclockwise direction. The speed of the magnetic field's rotation is given by 𝑛 𝑠𝑦𝑛𝑐 = 120𝑓𝑒 𝑃 Where 𝑓𝑒 is the system frequency applied to the stator in hertz P is the number of poles in the machine. 𝑛 𝑠𝑦𝑛𝑐 is called the synchronous speed in rpm (revolutions per minute) 20
  • 21. Cont.…  This rotating magnetic field 𝐵𝑠 passes over the rotor bars and induces a voltage in them.  The voltage induced in a given rotor bar is given by the equation 𝑒𝑖𝑛𝑑 = 𝑣 × 𝐵 . 𝑙 where 𝑣 = Velocity of the bar relative to the magnetic field B = Magnetic flux density vector 𝑙= length of conductor in the magnetic field  It is the relative motion of the rotor compared to the stator magnetic field that produces induced voltage in a rotor bar. 21
  • 22. Cont.…  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 flow produces a rotor magnetic field 𝐵 𝑅.  A torque is produced as a result of the interaction of those two magnetic fields Where 𝜏𝑖𝑛𝑑 is the induced torque and k is a constant representing the construction of the machine. BR and BS are the magnetic flux densities of the rotor and the stator respectively ind R skB B   22
  • 23. Working Principle of 3 Phase Induction Motor  When three phase supply is given to the three phase stator winding of the induction motor, a rotating magnetic field is developed around the stator which rotates at synchronous speed.  This rotating magnetic field passes through the air gap and cuts the rotor conductors which were stationary. Due to the relative speed between the stationary rotor conductors and the rotating magnetic field, an emf is induced in the rotor conductors. As the rotor conductors are short circuited, current starts flowing through it.  As these current carrying rotor conductors are placed in the magnetic field produced by the stator, they experiences a mechanical force i.e. torque which moves the rotor in the same direction as that of the rotating magnetic field. 23
  • 24. 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.  So, the IM will always run at a speed lower than the synchronous speed. 24
  • 25. Rotor Slip  The voltage induced in a rotor bar of an induction motor depends on the speed of the rotor relative to the magnetic fields.  The difference between the motor speed and the synchronous speed is called the Slip speed. Where nslip= slip speed of the machine nsync= speed of the magnetic field nm = mechanical shaft speed of the motor  Note that in normal operation both the rotor and stator magnetic fields 𝐵 𝑅 and 𝐵𝑠 rotate together at synchronous speed 𝑛 𝑠𝑦𝑛𝑐,While the rotor itself turns at a slower speed. slip sync mn n n  25
  • 26. The Slip  Slip is the relative speed expressed on a per-unit or a percentage basis. Where s is the slip This equation can also be expressed in terms of angular velocity 𝜔(radians per second) as  Notice that : if the rotor runs at synchronous speed, s = 0 if the rotor is stationary, s = 1 26
  • 27. Cont.…  The mechanical speed of the rotor shaft can expressed in terms of synchronous speed and slip.  From the above expression for slip, solving equations for mechanical speed yields Or  These equations are useful in the derivation of induction motor torque and power relationships. 27
  • 28. Induction Motors and Transformers  Both IM and transformer works on the principle of 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. 28
  • 29. Frequency  The frequency of the voltage induced in the rotor is given by  Since the slip of the rotor is defined as Where 𝑓𝑟= rotor frequency (Hz) 𝑓𝑒 = supply frequency(Hz) s= slip 29
  • 30. Cont.…  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. 30
  • 31. 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    31
  • 32. 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.  There is a relation between horse power and watts. Example: A 208-V, 10hp, four pole, 60 Hz, Y-connected induction motor has a full-load slip of 5 percent a) What is the synchronous speed of this motor? b) What is the rotor speed of this motor at rated load? c) What is the rotor frequency of this motor at rated load? d) What is the shaft torque of this motor at rated load? 746hp watts 32
  • 33. Solution a) b) c) d) 120 120(60) 1800 4 e sync f n rpm P    (1 ) (1 0.05) 1800 1710 m sn s n rpm       0.05 60 3r ef sf Hz    2 60 10 746 / 41.7 . 1710 2 (1/ 60) out out load mm P P n hp watt hp N m            33
  • 34. The Equivalent Circuit of an Induction Motor  The induction motor is similar to the transformer with the exception that its secondary windings are free to rotate. Fig.The transformer model of an induction motor, with rotor and stator connected by an ideal transformer of tums ratio 𝑎 𝑒𝑓𝑓 34
  • 35. The Rotor Circuit Model  In general, the greater the relative motion between the rotor and the stator magnetic fields, the greater the resulting rotor voltage and rotor frequency.  When the rotor is locked (or blocked), i.e. s =1, the largest voltage and rotor frequency are induced in the rotor, Why?  On the other side, if the rotor rotates at synchronous speed, i.e. s = 0, the induced voltage and frequency in the rotor will be equal to zero, Why? Where ER0 is the largest value of the rotor’s induced voltage obtained at s = 1(locked rotor)  The same is true for the frequency, i.e. 35
  • 36. Cont.…  It is known that 𝑋 = 𝜔𝐿 = 2𝜋𝑓𝐿  So, as the frequency of the induced voltage in the rotor changes, the reactance of the rotor circuit also changes.  Where 𝑋 𝑅𝑂 is the rotor reactance at the supply frequency(at blocked rotor)  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 36
  • 37. Cont.…  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   37
  • 38. Cont.…  Now we can have the rotor equivalent circuit 38
  • 39. Cont.…  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      39
  • 40. Losses and the Power Flow Diagram Power Losses in Induction Machines  Copper losses  Copper loss in the stator (PSCL) = I1 2 R1  Copper loss in the rotor (PRCL) = I2 2 R2  Core loss (Pcore)  Mechanical power loss due to friction and windage  How this power flow in the motor? 40
  • 41. Power Flow in Induction Motor 41 Fig.The power-flow diagram of an induction motor.
  • 42. Power Relations  The input current to a phase of the motor can be found by 3 cos 3 cosin L L ph phP V I V I   2 1 13SCLP I R ( )AG in SCL coreP P P P   2 2 23RCLP I R conv AG RCLP P P  ( )out conv f w strayP P P P   conv ind m P    42 I1 = V∅ Zeq 𝑍 𝑒𝑞 = 𝑅1 + 𝑗𝑋1 + 1 𝐺 𝐶 − 𝑗𝐵 𝑀 + 1 𝑅2 𝑆 + 𝑗𝑋2 Pcore = 3E1 2 GC
  • 43. Power Relations 3 cos 3 cosin L L ph phP V I V I   2 1 13SCLP I R ( )AG in SCL coreP P P P   2 2 23RCLP I R conv AG RCLP P P  ( )out conv f w strayP P P P   conv RCLP P  2 2 23 R I s  2 2 2 (1 ) 3 R s I s   RCLP s  (1 )RCLP s s   (1 )conv AGP s P  conv ind m P    (1 ) (1 ) AG s s P s     43
  • 44. Cont.…  The input power to an induction motor 𝑃𝑖𝑛 in the form of three-phase electric voltages and currents.  The first losses encountered in the machine are 𝐼2 𝑅 losses in the stator windings (the stator copper loss 𝑃𝑆𝐶𝐿) ' Then some amount of power is lost as hysteresis and eddy currents in the stator (𝑃𝐶𝑜𝑟𝑒).  The power remaining at this point is transferred to the rotor of the machine across the air gap between the stator and rotor. This power is called the air-gap power 𝑃𝐴𝐺 of the machine.  After the power is transferred to the rotor, some of it is lost as 𝐼2 𝑅 losses(the rotor copper loss 𝑃𝑅𝐶𝐿), and 44
  • 45. Cont.…  The rest is converted from electrical to mechanical form (𝑃𝐶𝑜𝑛𝑣).  Finally, friction and windage losses 𝑃𝐹&𝑊 and stray losses 𝑃 𝑚𝑖𝑠𝑐 are subtracted.  The remaining power is the output of the motor 𝑃𝑜𝑢𝑡. : : 1 : : 1- AG RCL convP P P s s 45
  • 46. Separating the Rotor Copper Losses and the Power Converted in an Induction Motor's Equivalent Circuit  Since air gap power would require 𝑹 𝟐 𝑺 and rotor copper loss require 𝑹 𝟐 element. Actual rotor resistance Resistance equivalent to mechanical load 46 𝑹 𝒄𝒐𝒏𝒗 = 𝑹 𝟐 𝑺 − 𝑹 𝟐 = 𝑹 𝟐( 𝟏−𝑺 𝑺 )
  • 47. 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. 47
  • 48. Solution 1. 2. 3. 3 cos 3 480 60 0.85 42.4 kW in L LP V I       42.4 2 1.8 38.6 kW AG in SCL coreP P P P       700 38.6 37.9 kW 1000 conv AG RCLP P P     & 600 37.9 37.3 kW 1000 out conv F WP P P     48
  • 49. Cont.… 4. 37.3 50 hp 0.746 outP   100% 37.3 100 88% 42.4 out in P P       49
  • 50. 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 50
  • 51. Solution 1. 2. 120 120 60 1800 rpm 4 e sync f n P     (1 ) (1 0.022) 1800 1760 rpmm syncn 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               51
  • 52. Cont.… 3. 4. 0.641 1.106 12.94 31.1 11.72 7.79 14.07 33.6 tot stat fZ Z Z j j               1 460 0 3 18.88 33.6 A 14.07 33.6tot V I Z           cos33.6 0.833 laggingPF    3 cos 3 460 18.88 0.833 12530 Win L LP V I       2 2 1 13 3(18.88) 0.641 685 WSCLP I R    12530 685 11845 WAG in SCLP P P     52
  • 53. Cont.… 5. 6. (1 ) (1 0.022)(11845) 11585 Wconv AGP s P     & 11585 1100 10485 W 10485 = 14.1hp 746 out conv F WP P P      11845 62.8 N.m 18002 60 AG ind sync P        10485 56.9 N.m 17602 60 out load m P        10485 100% 100 83.7% 12530 out in P P       53
  • 54. Induction Motor Torque-Speed Characteristics  Thevenin’s theorem can be used to transform the network to the left of points ‘a’ and ‘b’ into an equivalent voltage source VTH in series with equivalent impedance RTH+jXTH 54
  • 55. Cont.… 1 1( ) M TH M jX V V R j X X    1 1( )//TH TH MR jX R jX jX   2 2 1 1 | | | | ( ) M TH M X V V R X X    55
  • 56. Cont.…  Since XM>>X1 and XM>>R1  Because XM>>X1 and XM+X1>>R1, the equivalent impedance 1 M TH M X V V X X   2 1 1 1 M TH M TH X R R X X X X         56 ZTH = jXM(R1 + jX1) R1 + jX1 + jXM
  • 57. Cont.… Then the power converted to mechanical (Pconv) 2 2 22 2( ) TH TH T TH TH V V I Z R R X X s           2 2 2 (1 ) 3conv R s P I s   and the internal mechanical torque (𝜏 𝑐𝑜𝑛𝑣)(Induced Torque) conv ind m P    (1 ) conv s P s    2 2 23 AG s s R I Ps     57
  • 58. Cont.… 2 2 2 22 2 3 ( ) TH ind s TH TH V R sR R X X s                         2 2 2 22 2 3 1 ( ) TH ind s TH TH R V s R R X X s                 58
  • 59. Torque -Speed Characteristics curve Fig.Typical torque-speed characteristics of induction motor 59
  • 60. Comments 1. The induced torque is zero at synchronous speed. 2. The curve is nearly linear between no-load and full load. In this range, the rotor resistance is much greater than the reactance, so the rotor current, torque increase linearly with the slip. 3. There is a maximum possible torque that can’t be exceeded. This torque is called pullout torque and is 2 to 3 times the rated full-load torque. 4. The starting torque of the motor is slightly higher than its full-load torque, so the motor will start carrying any load it can supply at full load. 5. The torque of the motor for a given slip varies as the square of the applied voltage. 6. If the rotor is driven faster than synchronous speed it will run as a generator, converting mechanical power to electric power. 60
  • 61. Complete Torque-Speed characteristics curve 61 Fig: Induction motor torque-speed characteristic curve. showing the extended operating ranges (braking region and generator region). Pushover torque
  • 62. Cont.… 7. If the motor is turning backward relative to the direction of the magnetic fields, the induced torque in the machine will stop the machine very rapidly and will try to rotate it in the other direction. Since reversing the direction of magnetic field rotation is simply a matter of switching any two stator phases, this fact can be used as a way to very rapidly stop an induction motor. The act of switching two phases in order to stop the motor very rapidly is called plugging 62
  • 63. Operating Regions The torque-speed curve brakes down into three operating regions: 1)Braking, 𝑛 𝑚 < 0, 𝑠 > 0  Torque is positive whilst speed is negative. Considering the power conversion equation  It can be seen that if the power converted is negative (from P = τ ω) then the airgap power is positive. i.e. the power is flowing from the stator to the rotor and also into the rotor from the mechanical system. This operation is also called plugging 63 (1 )conv AGP s P 
  • 64. Cont.…  This mode of operation can be used to quickly stop a machine. If a motor is travelling forwards it can be stopped by interchanging the connections to two of the three phases. Switching two phases has the result of changing the direction of motion of the stator magnetic field, effectively putting the machine into braking mode in the opposite direction. 2) Motoring,0 < 𝑛 𝑚 < 𝑛 𝑠𝑦𝑛𝑐, 1 > 𝑠 > 0  Torque and motion are in the same direction. This is the most common mode of operation. 64
  • 65. Cont.… 3)Generating 𝑛 𝑚 > 𝑛 𝑠𝑦𝑛𝑐, 𝑠 < 0  In this mode, again torque is positive whilst speed is negative. However, unlike plugging,  indicates that if the power converted is negative, so is the air gap power. In this case, power flows from the mechanical system, to the rotor circuit, then across the air gap to the stator circuit and external electrical system. 65 (1 )conv AGP s P 
  • 66. Cont.…  If an induction motor is driven at a speed greater than 𝑛 𝑠𝑦𝑛𝑐 by an external prime mover, the direction of its induced torque will reverse and it will act as a generator.  As the torque applied to its shaft by the prime mover increases, the amount of power produced by the induction generator increases.  There is a maximum possible induced torque in the generator mode of operation. This torque is known as the pushover torque of the generator.  If a prime mover applies a torque greater than the pushover torque to the shaft of an induction generator, the generator will over speed. 66
  • 68. Maximum(Pullout) Torque in an Induction Motor  Maximum power will be achieved when the magnitude of source impedance matches the load impedance.  Maximum torque occurs when the power transferred to R2/s is maximum.  This condition occurs when R2/s equals the magnitude of the impedance RTH + j (XTH + X2) max 2 22 2( )TH TH T R R X X s    max 2 2 2 2( ) T TH TH R s R X X    68
  • 69. Cont.…  The corresponding maximum torque of an induction motor equals  The slip at maximum torque is directly proportional to the rotor resistance 𝑅2.  The maximum torque is independent of 𝑅2 and related to the square of applied voltage.  Rotor resistance can be increased by inserting external resistance in the rotor of a wound-rotor induction motor.  The value of the maximum torque remains unaffected but the speed at which it occurs can be controlled. 2 max 2 2 2 31 2 ( ) TH s TH TH TH V R R X X            69
  • 70. Cont.…  It is possible to take advantage of this characteristic of wound-rotor induction motors to start very heavy loads.  If a resistance is inserted into the rotor circuit, the maximum torque can be adjusted to occur at starting conditions.  Therefore, the maximum possible torque would be available to start heavy loads. 70 Effect of rotor resistance on torque-speed characteristic
  • 71. Example A two-pole, 50-Hz induction motor supplies 15kW to a load at a speed of 2950 rpm. 1. What is the motor’s slip? 2. What is the induced torque in the motor in N.m under these conditions? 3. What will be the operating speed of the motor if its torque is doubled? 4. How much power will be supplied by the motor when the torque is doubled? 71
  • 72. Solution 1. 2. 120 120 50 3000 rpm 2 3000 2950 0.0167 or 1.67% 3000 e sync sync m sync f n P n n s n          3 no given assume and 15 10 48.6 N.m 2 2950 60 f W conv load ind load conv ind m P P P P              72
  • 73. Cont.… 3. In the low-slip region, the torque-speed curve is linear and the induced torque is direct proportional to slip. So, if the torque is doubled the new slip will be 3.33% and the motor speed will be 4. (1 ) (1 0.0333) 3000 2900 rpmm syncn s n      2 (2 48.6) (2900 ) 29.5 kW 60 conv ind mP          73
  • 74. Example A 460-V, 25-hp, 60-Hz, four-pole, Y-connected wound-rotor 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  1. What is the maximum torque of this motor? At what speed and slip does it occur? 2. What is the starting torque of this motor? 3. If the rotor resistance is doubled, what is the speed at which the maximum torque now occur? What is the new starting torque of the motor? 4. Calculate and plot the T-s c/c for both cases. 74
  • 75. Solution 2 2 1 1 2 2 ( ) 460 26.3 3 255.2 V (0.641) (1.106 26.3) M TH M X V V R X X         2 1 1 2 26.3 (0.641) 0.590 1.106 26.3 M TH M X R R X X                1 1.106THX X   75
  • 76. Cont.… 1. The corresponding speed is max 2 2 2 2 2 2 ( ) 0.332 0.198 (0.590) (1.106 0.464) T TH TH R s R X X        (1 ) (1 0.198) 1800 1444 rpmm syncn s n      76
  • 77. Cont.… The torque at this speed is 2 max 2 2 2 2 2 2 31 2 ( ) 3 (255.2) 2 2 (1800 )[0.590 (0.590) (1.106 0.464) ] 60 229 N.m TH s TH TH TH V R R X X                     77
  • 78. Cont.… 2. The starting torque can be found from the torque eqn. by substituting s = 1   2 2 21 22 2 1 2 2 2 2 2 2 2 2 2 3 1 ( ) 3 [ ( ) ] 3 (255.2) (0.332) 2 1800 [(0.590 0.332) (1.106 0.464) ] 60 104 N.m TH start ind s s TH TH s TH s TH TH R V s R R X X s V R R R X X                                    78
  • 79. Cont.… 3. If the rotor resistance is doubled, then the slip at maximum torque doubles too The corresponding speed is The maximum torque is still 𝜏 𝑚𝑎𝑥= 229 N.m max 2 2 2 2 0.396 ( ) T TH TH R s R X X     (1 ) (1 0.396) 1800 1087 rpmm syncn s n      79
  • 80. Cont.… The starting torque is now 2 2 2 3 (255.2) (0.664) 2 1800 [(0.590 0.664) (1.106 0.464) ] 60 170 N.m start           80
  • 81. Determination of Motor Parameters  Due to the similarity between the induction motor equivalent circuit and the transformer equivalent circuit, same tests are used to determine the values of the motor parameters.  DC test: determine the stator resistance R1  No-load test: determine the rotational losses and magnetization current (similar to no-load test in Transformers).  Locked-rotor test: determine the rotor and stator impedances (similar to short-circuit test in Transformers). 81
  • 82. DC Test  The purpose of the DC test is to determine R1. A variable DC voltage source is connected between two stator terminals.  The DC source is adjusted to provide approximately rated stator current, and the resistance between the two stator leads is determined from the voltmeter and ammeter readings. 82
  • 83. Cont.… then  The current flows through two of the windings, so the total resistance in the current path is 2R1 . Therefore,  If the stator is Y-connected, the per phase stator resistance is  If the stator is delta-connected, the per phase stator resistance is DC DC DC V R I  1 2 DCR R  1 3 2 DCR R 83
  • 84. No-load Test 1. The motor is allowed to spin freely 2. The only load on the motor is the friction and windage losses, so all Pconv is consumed by mechanical losses. 3. The slip is very small 84  The no-load test of an induction motor measures the rotational losses of the motor and provides information about its magnetization current.
  • 85. Cont.… 4. At this small slip The equivalent circuit reduces to… 85
  • 86. Cont.… 5. Combining Rc and RF+W we get…… 86
  • 87. Cont.… 6. At the no-load conditions, the input power measured by meters must equal the losses in the motor. 7. The PRCL is negligible because I2 is extremely small because R2(1-s)/s is very large. 8. The input power equals Where 9. The equivalent input impedance is thus approximately If X1 can be found, in some other fashion, the magnetizing impedance X will be known & 2 1 13 in SCL core F W rot P P P P I R P      &rot core F WP P P  1 1, eq M nl V Z X X I     87
  • 88. Locked-Rotor(Blocked-Rotor) Test  In this test, the rotor is locked or blocked so that it cannot move, a voltage is applied to the motor, and the resulting voltage, current and power are measured. 88Fig: The locked-rotor test for an induction motor test circuit:
  • 89. Cont.…  Since the rotor is not moving, the slip s = 1, and so the rotor resistance Τ𝑅2 𝑠 is just equal to 𝑅2(quite a small value). Since R2 and X2 are so small, almost all the input current will flow through them, instead of through the much larger magnetizing reactance XM .  Therefore, the circuit under these conditions looks like a series combination of X1, R1, X2 and R2 . 89 Fig. The locked-rotor test for an induction motor equivalent circuit
  • 90. Cont.…  The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value.  The input power to the motor is given by  The locked-rotor power factor can be found as  The magnitude of the total impedance cos 3 in l l P PF V I   LR V Z I   90
  • 91. Cont.… Where X’1 and X’2 are the stator and rotor reactances at the test frequency respectively. ' cos sin LR LR LR LR LR Z R jX Z j Z      1 2 ' ' ' 1 2 LR LR R R R X X X     2 1LRR R R  ' 1 2 rated LR LR test f X X X X f    91
  • 92. Rules of thumb for dividing rotor and stator circuit reactance X1 and X2 as function of XLR Rotor Design X1 X2 Wound rotor 0.5 XLR 0.5 XLR Design A 0.5 XLR 0.5 XLR Design B 0.4 XLR 0.6 XLR Design C 0.3 XLR 0.7 XLR Design D 0.5 XLR 0.5 XLR 92
  • 93. Example The following test data were taken on a 7.5-hp, four-pole, 208-V, 60-Hz, design A, Y-connected IM having a rated current of 28 A. DC Test: VDC = 13.6 V IDC = 28.0 A No-load Test: Vl = 208 V f = 60 Hz Pin = 420 W IA= 8.12 A IB = 8.2 A IC = 8.18 A Locked-rotor Test: Vl = 25 V f = 15 Hz Pin = 920 W IA = 28.1 A IB = 28.0 A I 𝐶 = 27.6 A a) Sketch the per-phase equivalent circuit of this motor. b) Find the slip at pull-out torque, and find the value of the pull-out torque. 93
  • 94. Speed Control of Induction Motors  There are really only two techniques by which the speed of an induction motor can be controlled. a) Vary the synchronous speed, which is the speed of the stator and rotor magnetic fields, since the rotor speed always remains near 𝑛 𝑠𝑦𝑛𝑐. b) Vary the slip of the motor for a given load.  The synchronous speed of an induction motor is given by 𝑛 𝑠𝑦𝑛𝑐 = 120𝑓𝑒 𝑃 94
  • 95. Cont.…  So the only ways in which the synchronous speed of the machine can be varied are a) by changing the electrical frequency and b) by changing the number of poles on the machine.  Slip control may be accomplished by varying either the rotor resistance or the terminal voltage of the motor. 95
  • 96. Induction Motor Speed Control by Pole Changing  There are two major approaches to change the number of poles in an induction motor: 1. The method of consequent poles 2. Multiple stator windings Method of Consequent Poles  Consider one phase winding in a stator. By changing the current flow in one portion of the stator windings as such that it is similar to the current flow in the opposite portion of the stator will automatically generate an extra pair of poles. 96
  • 97. Cont.… 97 A close-up view of one phase of a pole-changing winding. In the two-pole configuration. one coil is a north pole and the other one is a south pole.
  • 98. Cont.… 98 When the connection on one of the two coils is reversed. they are both north poles. and the magnetic flux returns to the stator at points halfway between the two coils. The south poles are called consequent poles. and the winding is now a four pole winding
  • 99. Cont.…  By applying this method, the number of poles may be maintained (no changes), doubled or halfed, hence would vary its operating speed. In terms of torque, the maximum torque magnitude would generally be maintained. Disadvantage:  This method will enable speed changes in terms of 2:1 ratio steps, hence to obtained variations in speed, multiple stator windings has to be applied. Multiple stator windings have extra sets of windings that may be switched in or out to obtain the required number of poles. Unfortunately this would an expensive alternative. 99
  • 100. Speed Control by Changing the Line Frequency  If the electrical frequency applied to the stator of an induction motor is changed, the rate of rotation of its magnetic fields 𝑛 𝑠𝑦𝑛𝑐 will change in direct proportion to the change in electrical frequency, and the no- load point on the torque-speed characteristic curve will change with it.  The synchronous speed of the motor at rated conditions is known as the base speed.  By using variable frequency control, it is possible to adjust the speed of the motor either above or below base speed.  A properly designed variable-frequency induction motor drive can be very flexible. It can control the speed of an induction motor over a range from as little as 5 percent of base speed up to about twice base speed. 100
  • 101. Cont.…  However, it is important to maintain certain voltage and torque limits on the motor as the frequency is varied, to ensure safe operation.  When running at speeds below the base speed of the motor, it is necessary to reduce the terminal voltage applied to the stator for proper operation.  The terminal voltage applied to the stator should be decreased linearly with decreasing stator frequency. This process is called derating. If it is not done, the steel in the core of the induction motor will saturate and excessive magnetization currents will flow in the machine. 101
  • 102. Cont.…  Changing the electrical frequency would also require an adjustment to the terminal voltage in order to maintain the same amount of flux level in the machine core. If not the machine will experience: a) Core saturation (non linearity effects) b) Excessive magnetization current.  Varying frequency with or without adjustment to the terminal voltage may give 2 different effects: a) Vary frequency, stator voltage adjusted – generally vary speed and maintain operating torque. b) Vary Frequency, stator voltage maintained – able to achieve higher speeds but a reduction of torque as speed is increased. 102
  • 103. Cont.…  There may also be instances where both characteristics are needed in the motor operation; hence it may be combined to give both effects.  With the arrival of solid-state devices/power electronics, line frequency change is easy to achieved and it is more versatile to a variety of machines and application. 103
  • 104. Variable-frequency speed control in an induction motor 104 a) The family of torque-speed characteristic curves for speeds below base speed. assuming that the line voltage is derated linearly with frequency.
  • 105. Cont.… 105 b)The family of torque-speed characteristic curves for speeds above base speed. assuming that the line voltage is held constant.
  • 106. Cont.…  As with any transformer, the flux in the core of an induction motor can be found from Faraday's law:  If a voltage v(t) = 𝑉 𝑀 sin 𝜔𝑡 is applied to the core, the resulting flux ∅ is  If the electrical frequency applied to the stator decreases by 10 percent while the magnitude of the voltage applied to the stator remains constant, the flux in the core of the motor will increase by about 10 % and the magnetization current of the motor will increase.  Induction motors are normally designed to operate near the saturation point on their magnetization curves, so the increase in flux due to a decrease in frequency will cause excessive magnetization currents to flow in the motor. 106
  • 107. Speed Control by Changing the Line Voltage  The torque developed by an induction motor is proportional to the square of the applied voltage.  Varying the terminal voltage will vary the operating speed but with also a variation of operating torque.  In terms of the range of speed variations, it is not significant hence this method is only suitable for small motors only.  The speed of the motor may be controlled over a limited range by varying the line voltage. This method of speed control is sometimes used on small motors driving fans. 107
  • 108. Cont.… 108Fig. Variable-line-voltage speed control in an induction motor
  • 109. Speed Control by Changing the Rotor Resistance  In wound-rotor induction motors, it is possible to change the shape of the torque- speed curve by inserting extra resistances into the rotor circuit of the machine.  Changing the rotor resistance will change the operating speed of the motor. However, inserting extra resistances into the rotor circuit of an induction motor seriously reduces the efficiency of the machine. 109
  • 110. 110