1. EC010504(EE) Electric Drives & Control
Dr. Unnikrishnan P.C.
Professor, EEE

2. Module II Transformer √ Three Phase Induction Motor
Single Phase Induction Motor
Alternator
Synchronous Motor

3. 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

4. Induction Motor Disadvantages Essentially a “fixed-speed” machine
Speed is determined by the supply frequency
To vary its speed need a variable frequency supply

5. Parts of an Induction Motor

6. 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

7. Construction- Stator

8. 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 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. (Slip-Ring Induction Motor)

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

10. Construction- Rotor

11. Construction-Wound Rotor Slip Ring Induction Motor
Slip rings
Cutaway in a typical wound-rotor IM. Notice the brushes and the slip rings
Brushes

12. 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
Where fe is the supply frequency and
P is the no. of poles and nsync is called the synchronous speed in rpm

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

14. Rotating Magnetic Field

15. Rotating Magnetic Field

16. 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
𝑇 𝑖𝑛𝑑 =𝑘 𝐵 𝑟 𝐵 𝑠
Where 𝑇 𝑖𝑛𝑑 is the induced torque and BR and BS are the magnetic flux densities of the rotor and the stator respectively.

17. 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

18. 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 𝑁 𝑠 = speed of the magnetic field
𝑁 𝑚 = mechanical shaft speed of the motor
S = The slip
S = 𝑁 𝑠 − 𝑁 𝑚 𝑁 𝑠

19. 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

20. 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)

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
𝑇 𝑙𝑜𝑎𝑑 = 𝑃 𝑜𝑢𝑡 𝜔 𝑚 and 𝜔 𝑚 = 2𝜋 𝑛 𝑚 60 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.
hp =746 𝑊𝑎𝑡𝑡𝑠

23. Torque-Slip Equation 𝑇 𝛼 𝐸 2 𝐼 2 𝑐𝑜𝑠 ∅ 2
𝑇 𝛼 𝐸 2 𝐼 2 𝑐𝑜𝑠 ∅ 2
𝑇=𝑘 𝐸 2 𝑆 𝐸 𝑅 (𝑆 𝑋 2 ) 𝑅 𝑅 (𝑆 𝑋 2 ) 2
𝑇= 𝐾𝑆 𝐸 𝑅 2 𝑅 (𝑆 𝑋 2 ) 2

24. Torque-Slip Characteristics
𝑻= 𝑲𝑺 𝑬 𝟐 𝟐 𝑹 𝟐 𝑹 𝟐 𝟐 + (𝑺 𝑿 𝟐 ) 𝟐
Low Slip Region: 𝑇𝛼 𝑆
High Slip Region: 𝑇𝛼 1 𝑆

25. Starters- Squirrel Cage IM
Direct-On-line Starting
Stator Rheostat Starting
Autotransformer Starting
Star-Delta Starter

26. Direct On-Line(DOL) Starter
Used for motors upto 2 kW
Starting torque about twice the full load torque

27. Stator Rheostat Starter
Advantages: High PF during start, Less Expensive, Smooth acceleration
Disadvantages: Losses in Resistors, Expensive, Low torque efficiency

28. 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

29. 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

30. Starters- Slip Ring IM Rotor Rheostat Autotransformer Starting
Star-Delta Starter

31. 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

32. 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?

33. Power Flow in Induction Motor

34. Power Relations 𝑇 𝑙𝑜𝑎𝑑 = 𝑃 𝑐𝑜𝑛𝑣 𝜔 𝑚 𝑃 𝑆𝐶𝐿 −𝑆𝑡𝑎𝑡𝑜𝑟 𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠
𝑃 𝐴𝐺 −𝐴𝑖𝑟 𝐺𝑎𝑝 𝑃𝑜𝑤𝑒𝑟
𝑃 𝑅𝐶𝐿 −𝑅𝑜𝑡𝑜𝑟 𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠
,𝑃 𝑐𝑜𝑛𝑣 −𝑇ℎ𝑒 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 𝑐𝑜𝑛𝑣𝑒𝑟𝑡𝑒𝑑 𝑡𝑜 𝑚𝑒𝑐ℎ𝑎𝑐𝑖𝑐𝑎𝑙
𝑇 𝑙𝑜𝑎𝑑 = 𝑃 𝑐𝑜𝑛𝑣 𝜔 𝑚

35. Torque-speed characteristics
Typical torque-speed characteristics of induction motor

36. Maximum torque
Effect of rotor resistance on torque-speed characteristic

37. Speed Control Methods of IM
Induction Motor Speed Control From Stator Side
By Changing The Applied Voltage
𝑇= 𝐾𝑆 𝐸 𝑅 2 𝑅 (𝑆 𝑋 2 ) 2 , T 𝑆 𝐸 2 2
By Changing The Applied Frequency
Constant V/F Control Of Induction Motor- Most Popular
Changing The Number Of Stator Poles

38. 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.

39. 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

40. Cascade Operation

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

42. Solution

43. 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:
The air-gap power PAG.
The power converted Pconv.
The output power Pout.
The efficiency of the motor.

44. Solution

45. Solution

46. Example
A two-pole, 50-Hz induction motor supplies 15kW to a load at a speed of 2950 rpm.
What is the motor’s slip?
What is the induced torque in the motor in N.m under these conditions?
What will be the operating speed of the motor if its torque is doubled?
How much power will be supplied by the motor when the torque is doubled?

47. Solution

48. Solution
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

49. Question? Q: How to change the direction of rotation?
A: Change the phase sequence of the power supply.

50. 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.