1. induction motor1 AC Machine Stator ‘a’ phase axis ‘b’ phase axis ‘c’ phase axis 120 0

2. induction motor2 1 Cycle Amp time t0t0 t1t1 t2t2 t3t3 t4t4 t 01 t 12 Currents in different phases of AC Machine

3. induction motor3 Axis of phase a a’a’ a’a’ -90-401060110160210260 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 FaFa Space angle (theta) in degrees t0t0 t 01 t 12 t2t2 a MMF Due to ‘a’ phase current

4. induction motor4 aFcFc -9310113216 -1.5 -0.5 0 0.5 1 1.5 a’a’ c’c’ b’b’ b c a a’a’ c’c’ b’b’ b c a a’a’ c’c’ b’b’ b c a a’a’ c’c’ b’b’ b c FbFb FaFa F FbFb FcFc F FaFa F FbFb FcFc FcFc FbFb F Space angle (  ) in degrees F FaFa FcFc FbFb t = t 0 = t 4 t = t 1 t = t 2 t = t 3 t = t 0 = t 4 RMF(Rotating Magnetic Field)

5. induction motor5 Video of the unfolded rotating magnetic field

6. induction motor6 RMF(Rotating Magnetic Field) -Analogy with DC machines The salient field structure in DC machines is mimicked along with speed in an AC machines by a multiphase (2 or more) winding. The number of poles are determined by winding distribution and is independent of the number of phases. The rotational speed is determined by the supply frequency and the number of poles, such that an observer in air-gap counts same number of poles per second, meaning the more the number of poles the slower the machine will run and vice-versa.

7. induction motor7 Induction Motor Most popular motor today in the low and medium horsepower range Very robust in construction Speed easily controllable using V/f or Field Oriented Controllers Have replaced DC Motors in areas where traditional DC Motors cannot be used such as mining or explosive environments Of two types depending on motor construction: Squirrel Cage or Slip Ring Only Disadvantage: Most of them run with a lagging power factor

8. induction motor8 Squirrel Cage Rotor

9. induction motor9 Slip Ring Rotor The rotor contains windings similar to stator. The connections from rotor are brought out using slip rings that are rotating with the rotor and carbon brushes that are static.

10. induction motor10 Torque Production in an Induction Motor In a conventional DC machine field is stationary and the current carrying conductors rotate. We can obtain similar results if we make field structure rotating and current carrying conductor stationary. In an induction motor the conventional 3-phase winding sets up the rotating magnetic field(RMF) and the rotor carries the current carrying conductors. An EMF and hence current is induced in the rotor due to the speed difference between the RMF and the rotor, similar to that in a DC motor. This current produces a torque such that the speed difference between the RMF and rotor is reduced.

11. induction motor11 Slip in Induction Motor However, this speed difference cannot become zero because that would stop generation of the torque producing current itself. The parameter slip ‘s’ is a measure of this relative speed difference where n s,  s,f 1 are the speeds of the RMF in RPM,rad./sec and supply frequency respectively n,  are the speeds of the motor in RPM and rad./sec respectively The angular slip frequency and the slip frequency at which voltage is induced in the rotor is given by 

12. induction motor12 Induction Motor Example A 100 hp, 460V, 8 pole, 60 Hz, star connected 3 phase induction motor runs at 891 rpm under full load. Determine the synchronous speed in rpm, slip, slip frequency (frequency of the rotor circuit),slip rpm at full load. What is the speed of the rotor field relative to (i) rotor structure, (ii) stator structure, (iii) stator rotating field? Voltage induced in rotor under full load? N 2 /N 1 =0.5 Solution on Greenboard

13. induction motor13 Induction motor Equivalent Circuit

14. induction motor14 Relation between air-gap, gross mechanical power and rotor copper loss Internal efficiency = Implies lower the slip higher is the induction motor efficiency

15. induction motor15 Example problem related to the formula shown in the previous

16. induction motor16 Approximate Equivalent Circuit Assumes negligible magnetizing current Note Rc has been removed. The sum of core losses and the windage and friction loses are treated as constant. This is because as speed increases rotor core loss decreases (lower f 2 ) but windage and frictionloses increase.With decrease of speed the converse is true. Thus the sum is constant at any speed and is termed as rotational loss. j

17. induction motor17 IEEE Equivalent Circuit Assumes 30-50% magnetizing current and drop across R 1 +jX 1 not negligible As before, the sum of core losses and the windage and friction loses are treated as constant.

18. induction motor18 Thevnin’s equivalent of the IEEE Equivalent Circuit This is done by applying Thevenin’s theorem and treating the rotor side as load

19. induction motor19 Determining equivalent circuit parameters j Uses no-load test and blocked rotor tests to determine them

20. induction motor20 Example problem related to no-load and blocked rotor test

21. induction motor21 Performance Characteristics(1)

22. induction motor22 Performance Characteristics(2)

23. induction motor23 Performance Characteristics(3) Case 1:

24. induction motor24 Performance Characteristics(4) Case 2:

25. induction motor25 Performance Characteristics(5) Combining case 1 and 2 the approximate torque speed characteristics would look approximately like: T mech Speed (n) T max nmnm nsns

26. induction motor26 Performance Characteristics(6) How to obtain T max? By differentiating the following equation with respect to s and equating it to zero. One can obtain the following: Slip at maximum torque =

27. induction motor27 Performance Characteristics(7) (Speed Control) Speed control by varying rotor resistance (vary T max by varying s Tmax ) (inefficient) Speed control by varying supply voltage and frequency (V th /  1 ) (efficient)

28. induction motor28 Performance Characteristics(8) Also using and for small R 1 one can write the following:

29. induction motor29 Example problem based on the formula on previous to express maximum torque and starting torque in terms of rated torque

30. induction motor30 Performance Characteristics(9)

31. induction motor31 Example problem related to efficiency calculation of induction motor based on equivalent circuit parameters

32. induction motor32 Related to the problem in the previous slide

33. induction motor33 Different modes of IM operation

34. induction motor34 Different modes of IM operation

35. induction motor35 Example problem on variable frequency supply using a slip-ring induction motor

36. 36 Speed control of SRIM with ext. resistors

37. induction motor37 Applications of SRIM

38. induction motor38 Wind Power applications of SRIM

39. induction motor39