1. Induction motors janaka

2. Principles of Operation The principle of operation for all AC motors relies on the interaction of a revolving magnetic field created in the stator by AC current, with an opposing magnetic field either induced on the rotor or provided by a separate DC current source

3. Rotating Field

4. Slip It is virtually impossible for the rotor of an AC induction motor to turn at the same speed as that of the rotating magnetic field. If the speed of the rotor were the same as that of the stator, no relative motion between them would exist, and there would be no induced EMF in the rotor. (Recall from earlier modules that relative motion between a conductor and a magnetic field is needed to induce a current.) Without this induced EMF, there would be no interaction of fields to produce motion. The rotor must, therefore, rotate at some speed less than that of the stator if relative motion is to exist between the two. The percentage difference between the speed of the rotor and the speed of the rotating magnetic field is called slip. The smaller the percentage, the closer the rotor speed is to the rotating magnetic field speed. Percent slip can be found by SLIP: NS -NR / NS x 100%

5. Stator design. The stator is the outer body of the motor which houses the driven windings on an iron core. In a single speed three phase motor design, the standard stator has three windings, while a single phase motor typically has two windings. The inner surface of the stator is made up of a number of deep slots or grooves right around the stator. It is into these slots that the windings are positioned.

6. The arrangement of the windings or coils within the stator determines the number of poles that the motor has. It has poles in multiples of two. i.e. 2 pole, 4 pole, 6 pole etc. The winding configuration, slot configuration and lamination steel all have an effect on the performance of the motor. Stator design contd….

7. The voltage rating of the motor is determined by the number of turns on the stator and the power rating of the motor is determined by the losses which comprise copper loss and iron loss, and the ability of the motor to dissipate the heat generated by these losses. The stator design determines the rated speed of the motor and most of the full load, full speed characteristics.

8. Rotor Design. Two types of designs are available: 1. Squirrel cage rotor 2. Wound rotor (also known as slip ring induction motor)

9. Rotor Design contd… The Rotor comprises a cylinder made up of round laminations pressed onto the motor shaft, and a number of short-circuited windings. The rotor windings are made up of rotor bars passed through the rotor, from one end to the other, around the surface of the rotor The bars protrude beyond the rotor and are connected together by a shorting ring at each end.

10. Rotor Design contd… The bars are usually made of aluminium or copper The position relative to the surface of the rotor, shape, cross sectional area and material of the bars determine the rotor characteristics. the rotor windings exhibit inductance and resistance, and these characteristics can effectively be dependant on the frequency of the current flowing in the rotor.

11. Rotor Design contd… A bar with a large cross sectional area will exhibit a low resistance, while a bar of a small cross sectional area will exhibit a high resistance Positioning the bar deeper into the rotor, increases the amount of iron around the bar, and consequently increases the inductance exhibited by the rotor

12. Rotor Design contd… Two bars of equal dimensions will exhibit a different A.C. impedance depending on their position relative to the surface of the rotor. The rotor design determines the starting characteristics

13. Slip Ring Motors. The slip ring motor has a set of windings on the rotor which are not short circuited, but are terminated to a set of slip rings for connection to external resistors and contactors They are wound for 3 phase, star connection.

14. The slip ring motor enables the starting characteristics of the motor to be totally controlled and modified to suit the load Resistances can be added externally and these resistances can be varied from zero to a maximum value. This results in a very high starting torque from zero speed to full speed at a relatively low starting current.

15. This type of starting is ideal for very high inertia loads allowing the machine to get to full speed in the minimum time with minimum current draw.

19. Equivalent Circuit. The induction motor can be treated essentially as a transformer for analysis.

20. The induction motor has stator leakage reactance, stator copper loss elements as series components, and iron loss and magnetising inductance as shunt elements. The rotor circuit likewise has rotor leakage reactance, rotor copper (aluminium) loss and shaft power as series elements.

21. The transformer in the centre of the equivalent circuit can be eliminated by adjusting the values of the rotor components in accordance with the effective turns ratio of the transformer.

22. Starting Characteristics. In order to perform useful work, the induction motor must be started from rest and both the motor and load accelerated up to full speed. The starting torque of an induction motor starting with a fixed voltage, will drop a little to the minimum torque known as the pull up torque as the motor accelerates, and then rise to a maximum torque known as the breakdown or pull out torque at almost full speed and then drop to zero at synchronous speed. The curve of start torque against rotor speed is dependant on the terminal voltage and the motor/rotor design

24. Starting characteristics contd… If the terminal voltage to the motor is reduced while it is starting, the current drawn by the motor will be reduced proportionally The torque developed by the motor is proportional to the voltage squared, and so a reduction in starting voltage will result in a reduction in starting current and a greater reduction in starting torque

25. Starting characteristics contd…. If the start voltage applied to a motor is halved, the start torque will be a quarter, likewise a start voltage of one third will result in a start torque of one ninth.

26. Frame Classification. Totally Enclosed Forced air Cooled (TEFC): The TEFC motor is totally enclosed in either an aluminium or cast iron frame with cooling fins running longitudinally on the frame. A fan is fitted externally with a cover to blow air along the fins and provide the cooling. These motors are often installed outside in the elements with no additional protection and so are typically designed to IP55 or better.

27. Drip proof motor: Drip proof motors use internal cooling with the cooling air drawn through the windings. They are normally vented at both ends with an internal fan. This can lead to more efficient cooling, but requires that the environment is clean and dry to prevent insulation degradation from dust, dirt and moisture. Drip proof motors are typically IP22 or IP23.

28. Types of enclosures o ODP – An open drip proof motor is the most common one used in industrial settings. This motor doesn't have a fan and has openings which can be penetrated by dirt and water, so it is ideal for indoor use. o TEFC – A totally enclosed fan cooled motor has a fan mounted to blow air across the motor frame. Because the fan is sealed, no external elements can get in. TEFC fans are often used with conveyors.

29. Types of enclosures contd…. TENV – A totally enclosed non-ventilated motor is also used in materials handling applications where another source of ventilation to cool the motor is available o TEBC – A totally enclosed blower-cooled motor includes a fan that is controlled apart from the motor. These are typical in applications calling for larger horsepower, like a crane and hoist, or in variable speed applications where the motor may operate near zero revolutions per minute. EPFC – An explosion proof fan cooled motor is used in environments where combustible elements like gasoline, oil, ammonia, coal or combustible dust are present

32. Auto transformer start

34. Star-delta starter

35. Shaded pole motor

36. Numericals

37. Numerical 1 A three phase induction motor is wound for 4 poles and is supplied from a 50 Hz system. Calculate (1) synchronous speed (2) speed of rotor when slip is 4 % (3) rotor frequency when speed of rotor is 600 rpm

38. Solution for numerical 1 Synchronous speed N S = 120 f / P = 1500 rpm Slip = 4%, hence speed N = N S (1-s) = 1440 rpm f’ = s.f ; s at 600 rpm = 1500 – 600/1500 = 0.6 f’ = 0.6 x 50 = 30 Hz = rotor frequency

39. Numerical 2 A 440 V, 3 ph, 4 pole, 50 Hz, 37.3 kW, star connected induction motor has the following parameters : R 1 = 0.1Ω, X 1 = 0.4 Ω, R 2 ’ = 0.15Ω, X 2 ’ = 0.44Ω. Motor has stator core loss of 1250 watts and rotational losses of 1000 watts. It draws a no load current of 20A at a p.f. of 0.09 (lag). When motor operates at a slip of 3%, calculate (a) input line current and its p.f. (b) torque developed in N-m, ( c ) Kw output and (d) efficiency of motor

40. Solution for numerical 2 Draw the equivalent circuit I 2 ’ = s E 2 / √R 2 ² + (sX 2 )² Phase voltage = 440 / √3 = 254 V. Hence I 2 ’ = V 1 / √(R 1 +R 2 /s)² + (X 1 +X 2 ’)² = 49.1 A, Φ = 9.3 (lag) No load current I 0 = 20 A, cos Φ = 0.09 Φ = 84.9°. Input current I 1 = vector addition of I 0 & I 2 ’ = 57.4A, Φ = 29° Hence p.f. = cos Φ = cos29° = 0.875 (lag)

41. Solution for numerical 2 contd… (ii) power in the rotor = 3 x I2’² (R2’/s) = 36,160 watts NS = 1500 rpm power developed = 2 NST/60 N-m = 230 N-m (iii)Maximum mechanical power output = (1-s) Protor = 0.97 x 36,160 = 35,075 w. Losses = 1000 watts Hence power output = 34,075 watts

42. Solution for numerical 2 contd… Core loss = 1250 watts Stator copper losses = 3 I 1 ² R 1 w = 988 w Rotor copper losses = 3 I 2 ’² R 2 ’ = s P rotor = 0.03 x 36,160 w Friction and windage losses = 1000 w Total losses = add all above = 4323 w Efficiency = output / output + losses = 88.7 %