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EC010504(EE) Electric
Drives & Control
Dr. Unnikrishnan P.C.
Professor, EEE
Module II





Transformer √
Three Phase Induction Motor
Single Phase Induction Motor
Alternator
Synchronous Motor
Induction Motor
Advantages
•
•
•
•
•
Robust; No brushes. No contacts on rotor shaft
High Power/Weight ratio compared to DC motor
Lower Cost/Power
Easy to manufacture
Almost maintenance-free, except for bearing
and other mechanical parts
• wide range of power ratings: fractional HP to 10
MW
Induction Motor
Disadvantages
• Essentially a “fixed-speed” machine
• Speed is determined by the supply frequency
• To vary its speed need a variable frequency
supply
Parts of an Induction Motor
Construction
• 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 (why?), having a
number of evenly spaced slots, providing the space for the
stator winding
Construction- Stator
Construction-Rotor
• Two basic designs
– Squirrel-Cage: conducting copper or aluminium
bars laid into slots and shorted at both ends by
shorting rings. (Squirrel Cage Induction Motor)
– Wound-Rotor: complete set of three-phase
windings exactly as the stator. Usually Yconnected, 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. (Slip-Ring
Induction Motor)
Construction
Squirrel cage rotor
Wound rotor
Notice the
slip rings
Construction- Rotor
Construction-Wound Rotor
Slip Ring Induction Motor
Slip rings
Cutaway in a
typical woundrotor IM.
Notice the
brushes and
the slip rings
Brushes
Rotating Magnetic Field
• Balanced three phase windings, i.e.
mechanically displaced 120 degrees
form each other, fed by balanced
three phase source
• A rotating magnetic field with
constant magnitude is produced,
rotating with a speed
120 f e
Ns 
P
Where fe is the supply frequency and
P is the no. of poles and nsync is called
the synchronous speed in rpm
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
Rotating Magnetic Field
Rotating Magnetic Field
Principle of operation
• 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, a 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
𝑇𝑖𝑛𝑑 = 𝑘𝐵𝑟 𝐵𝑠
Where 𝑇𝑖𝑛𝑑 is the induced torque and BR and BS are the
magnetic flux densities of the rotor and the stator respectively.
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
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
S=
𝑁𝑠 −𝑁𝑚
𝑁𝑠
Where 𝑁𝑠 = speed of the magnetic field
𝑁𝑚 = mechanical shaft speed of the motor
S = The slip
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
Frequency
• The frequency of the voltage induced in the
rotor is given by
Pn
fr 
120
Where fr = the rotor frequency (Hz)
P = number of stator poles
n = slip speed (rpm)
P  (ns  nm )
fr 
120
P  sns

 sf e
120
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
𝑇𝑙𝑜𝑎𝑑 =
𝑃𝑜𝑢𝑡
𝜔𝑚
and 𝜔𝑚 =
2𝜋𝑛𝑚
60
rad/s
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.
hp = 746 𝑊𝑎𝑡𝑡𝑠
Torque-Slip Equation
𝑇 𝛼 𝐸2 𝐼2 𝑐𝑜𝑠∅2
𝑆𝐸2
𝑇 = 𝑘 𝐸2
𝑅2
2
𝑅2
+ (𝑆𝑋2
)2
𝑅2
2
𝑇=
𝐾𝑆𝐸2 𝑅2
𝑅2
2
+ (𝑆𝑋2 )2
2
+ (𝑆𝑋2 )2
Torque-Slip Characteristics
𝟐
𝑻=
𝑲𝑺𝑬𝟐 𝑹𝟐
𝟐
𝑹𝟐 + (𝑺𝑿𝟐 )𝟐
Low Slip Region: 𝑇𝛼 𝑆
1
High Slip Region: 𝑇𝛼 𝑆
Starters- Squirrel Cage IM
•
•
•
•
Direct-On-line Starting
Stator Rheostat Starting
Autotransformer Starting
Star-Delta Starter
Direct On-Line(DOL) Starter
• Used for motors upto 2 kW
• Starting torque about twice the full load
torque
Stator Rheostat Starter
Advantages: High PF during start, Less
Expensive, Smooth acceleration
Disadvantages: Losses in Resistors,
Expensive, Low torque efficiency
Auto-Transformer Starter
Advantages: High torque/Ampere, Long
starting period, Can be used for star and
delta connected motors,
Disadvantages: Low PF (Lagging), Higher cost
Star-Delta Starter
• Reduced starting current but the starting torque is
also reduced by the same amount
• Limited to applications where high starting torque is
not necessary
Starters- Slip Ring IM
• Rotor Rheostat
• Autotransformer Starting
• Star-Delta Starter
Rotor Rheostat Starter for Slip Ring IM
• By increasing rotor resistance, the rotor current is
reduced at starting. The torque is increased due to
improvement in PF
• Such motors can be started under load
Power losses in Induction
Machines
• Copper losses
– Copper loss in the stator (PSCL) = I12R1
– Copper loss in the rotor (PRCL) = I22R2
• Core loss (Pcore)
• Mechanical power loss due to friction and
windage
• How this power flow in the motor?
Power Flow in Induction Motor
Power Relations
Pin  3 VL I L cos   3 Vph I ph cos 
PSCL  3 I12 R1
𝑃𝑆𝐶𝐿 − 𝑆𝑡𝑎𝑡𝑜𝑟 𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠
PAG  Pin  ( PSCL  Pcore )
𝑃𝐴𝐺 − 𝐴𝑖𝑟 𝐺𝑎𝑝 𝑃𝑜𝑤𝑒𝑟
PRCL  3I 22 R2
𝑃𝑅𝐶𝐿 − 𝑅𝑜𝑡𝑜𝑟 𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠
Pconv  PAG  PRCL
, 𝑃𝑐𝑜𝑛𝑣 −𝑇ℎ𝑒 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 𝑐𝑜𝑛𝑣𝑒𝑟𝑡𝑒𝑑 𝑡𝑜 𝑚𝑒𝑐ℎ𝑎𝑐𝑖𝑐𝑎𝑙
Pout  Pconv  ( Pf  w  Pstray )
𝑇𝑙𝑜𝑎𝑑 =
𝑃𝑐𝑜𝑛𝑣
𝜔𝑚
Torque-speed characteristics
Typical torque-speed characteristics of induction motor
Maximum torque
Effect of rotor resistance on torque-speed characteristic
Speed Control Methods of IM
Induction Motor Speed Control From Stator
Side
• By Changing The Applied Voltage
2
𝑇=
𝐾𝑆𝐸2 𝑅2
2
, T 𝑆𝐸2
2
𝑅2 + (𝑆𝑋2 )2
120 f e
Ns 
P
• By Changing The Applied Frequency
• Constant V/F Control Of Induction MotorMost Popular
• Changing The Number Of Stator Poles
Constant V/F Control Of Induction Motor
• If the supply frequency is reduced keeping the
rated supply voltage, the air gap flux will tend to
saturate causing excessive stator current and
distortion of the stator flux wave.
• If the ratio of voltage to frequency is kept constant,
the flux remains constant. Also, the developed
torque remains approximately constant.
Speed Control Methods Of IM …..
Induction Motor Speed Control From Rotor
Side
• Rotor Rheostat Control
• Cascade Operation
• By Injecting EMF In Rotor Circuit: By changing
the phase of injected emf, speed can be
controlled
Cascade Operation
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?
Solution
1.
2.
nsync
120 f e 120(60)


 1800 rpm
P
4
nm  (1  s)ns
 (1  0.05) 1800  1710 rpm
f r  sfe  0.05  60  3Hz
3.
 load
4.
Pout
Pout


m 2 nm
60
10 hp  746 watt / hp

 41.7 N .m
1710  2  (1/ 60)
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.
2.
3.
4.
The air-gap power PAG.
The power converted Pconv.
The output power Pout.
The efficiency of the motor.
Solution
1.
Pin  3VL I L cos 
 3  480  60  0.85  42.4 kW
PAG  Pin  PSCL  Pcore
 42.4  2  1.8  38.6 kW
Pconv  PAG  PRCL
2.
700
 38.6 
 37.9 kW
1000
Pout  Pconv  PF &W
3.
600
 37.9 
 37.3 kW
1000
Solution
37.3
Pout 
 50 hp
0.746
4.

Pout
100%
Pin
37.3

100  88%
42.4
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?
Solution
120 f e 120  50
nsync 

 3000 rpm
P
2
1.
nsync  nm 3000  2950
s

 0.0167 or 1.67%
nsync
3000
no Pf W given
2.  assume Pconv  Pload and  ind   load
 ind 
Pconv
m
15 103

 48.6 N.m
2
2950 
60
Solution
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
nm  (1  s )nsync  (1  0.0333)  3000  2900 rpm
Pconv   ind m
4.
2
 (2  48.6)  (2900  )  29.5 kW
60
Question?
• Q: How to change the direction of rotation?
• A: Change the phase sequence of the power
supply.
Question?
• Q: Why rotor core loss in a three phase induction motor
negligible?
• A: Usually we operate motor at slip in the range 0.04 ~
0.06, thus for a stator frequency of 50 Hz
, rotor frequency is around 2~3 Hz. As core losses are
proportional to square of frequency (Eddy
current losses) or proportional to frequency ( Hysteresis
losses) , therefore the value of the Core
loss is negligible for rotor.