Induction motor (1).pdf ioe electrical engineeing
- 2. Induction motor or machines are also called as asynchronous machine because they always run at speed other than synchronous speed (the speed of the rotating magnetic field in the stator). Induction motors are mainly of two types. It can be single phase or three phase induction motors. Single phase induction motor is usually built in small size (up to 3 H.P). The three-phase induction motors are the most commonly used AC motors in the industry because of their simple and rugged construction, low cost, good operating characteristics, high efficiency, reasonably good power factor, absence of commutator, good power regulation, reliability and low maintenance. Almost more than 90% of the mechanical energy used in the industry is provided by three phase induction motors. Three phase induction motors are mainly used in the industry for power conversion- electrical to mechanical power conversion in bulk or large quantity. But for small power conversion single phase induction motors are used. The induction motors perform a variety of services in the home, office, business, factories, etc. In all the domestic appliances such as refrigerators, fans, washing machines, hair dryers, mixer grinder, etc., single phase induction motor are used. Induction Motor
- 3. A three phase induction motor mainly consists of two parts- stator and rotor. The construction of the induction motor is explained below in detail. 1) Stator The stator is the stationary part of the induction motor. It has three main parts, namely stator frame, stator core and stator winding. a) Stator frame It is the outermost part of the motor whose main function is to support the stator core and to provide the mechanical strength and protection to all the inner parts of the machine. For small machines, the outer frame is casted, but for the large machine, it is fabricated. The frame should be strong and rigid as the air gap length in an induction motor is very small. Otherwise, the rotor will not remain concentric with the stator, which will give rise to an unbalanced magnetic pull. Constructional details of three phase induction motor
- 4. b) Stator Core Its main function is to carry the alternating flux which produces hysteresis and eddy current losses. In order to reduce these losses, the stator core is built of high-grade silicon steel stampings The stampings are fixed to the stator frame and the stampings are insulated from the other with a thin varnish layer. The thickness of the stamping usually varies from 0.35 to 0.65 mm. Slots are punched on the inner side of the stampings. c) Stator winding The slots on the periphery of the stator core of the three-phase induction motor carry three phase windings which are supplied by three phase ac supply. The six terminals of the windings (two of each phase) are connected in the terminal box of the machine. The three phases of the winding are connected either in star or delta depending upon type of starting method used. The stator of the motor is wound for a definite number of poles, depending on the speed of the motor. If the number of poles is greater, the speed of the motor will be less and if the number of poles is less than the speed will be high.
- 5. As the relationship between the speed and the pole of the motor is given as: NS = 120f P Fig: Stator of three phase induction motor Fig: Three phase four pole stator winding Fig: Punching slots on stator core
- 6. 1) Rotor The rotor is the rotating part of three phase induction motor. The rotor is also built of thin laminations of the same material as the stator. The laminated cylindrical core is mounted directly on the shaft. These laminations are slotted on the outer side to receive the conductors. There are two types of rotor- squirrel cage rotor and phase wound rotor. a) Squirrel Cage Rotor It consists of a laminated cylindrical core with slots nearly parallel to the shaft axis or skewed. The circular slots at the outer periphery are semi-closed. Each slot contains uninsulated bar conductor of aluminium or copper. At each end of the rotor, the conductors the short-circuited by a heavy ring of the same material and is called end ring. The rotor conductors and end rings form a closed circuit resembling like a cage commonly used for keeping squirrels and hence, it is named as squirrel cage motor.
- 7. The skewing of the rotor conductors offers the following advantages: i) It reduces humming and provide smooth and noise free operation. ii) It results in a uniform torque curve for different positions of the rotor. iii)The locking tendency of the rotor is reduced. As the teeth of the rotor and the stator attract each other and lock due to magnetic action. Fig: Squirrel cage rotor
- 8. a) Phase wound rotor The Phase wound rotor is also called as slip ring rotor. It consists of a laminated cylindrical core. The outer periphery of the rotor has a semi-closed slot which carries a 3 phase insulated windings. The number of poles of rotor are kept same to the number of poles of the stator. The rotor is always wound three phase even if the stator is wound two phase. The rotor windings are connected in star. The open ends of the star circuit are brought outside the rotor and connected to the three insulated slip rings which are mounted on the shaft with brushes resting on them. These three brushes are connected to an external star connected rheostat. This arrangement is done to introduce an external resistance in rotor circuit. The resistor enables the variation of each rotor phase resistance to serve the following purposes: i) It increases the starting torque and decreases the starting current from the supply. ii) It is used to control the speed of the motor.
- 9. When motor is running at its rated speed, slip rings are automatically short circuited by means of a metal collar and brushes are lifted above the slip rings to minimize the frictional losses. The rotor is skewed in this case too. Fig: Phase wound rotor
- 10. Fig: Squirrel cage and phase wound rotor
- 11. When the three phase stator windings are energized from a three phase balanced ac supply, the three phase currents flow simultaneously through the stator windings and will magnetize the stator core. If we study the nature of magnetic field produced by these three phase currents, we will get a rotating magnetic field of constant magnitude and is equal to 1.5 Φmwhere, Φm is the maximum value of flux due to any of three phases. This field is such that its poles do no remain in a fixed position on the stator but go on shifting their positions around the stator. For this reason, it is called a rotating field. Let us study the nature of magnetic field produced in stator in detail: When the three phase currents flow simultaneously through the windings and are displaced from each other by 120° electrical. Each alternating phase current produces its own flux which is sinusoidal. So all three fluxes are sinusoidal and are separated from each other by 120°. If the phase sequence of the windings is R-Y-B, then mathematical equations for the instantaneous values of the three fluxes ΦR, ΦY, ΦB can be written as Production of rotating magnetic field
- 12. ΦR = Φm sin ω t ΦY = Φmsin ωt − 120 ΦB = Φm sin ωt + 120 Fig: Waveform of three fluxes Fig: Phasor diagram of fluxes with their assumed positive directions The net or resultant magnetic flux at any time at the central space of the machine will be equal to the vector sum of these fluxes. Case 1: When Ɵ = ωt = 0˚ ΦR = Φm sin 0 = 0 ΦY = Φm sin 0 − 120 = −0.866 Φm ΦB = Φm sin 0 + 120 = +0.866 Φm
- 13. Therefore, at this instant, current flowing through the coil R-R’ is zero, current flowing through the coil Y-Y’ is 0.866 Φm but the direction is negative and current flowing through the coil B-B’ is 0.866 Φm but the direction is positive. The direction of the net magnetic flux is determined by right hand screw rule and the magnitude of the net magnetic flux can be calculated as ΦT = ΦY 2 + ΦB 2 + 2ΦYΦBcosƟ Putting values on above formula, we get ΦT = 1.5 Φm Case 2: When Ɵ = ωt = 60˚ ΦR = Φmsin 60 = +0.866Φm ΦY = Φm sin 60 − 120 = −0.866 Φm ΦB = Φm sin 60 + 120 = 0
- 14. Therefore, at this instant, current flowing through the coil B-B’ is zero, current flowing through the coil Y-Y’ is 0.866 Φm but the direction is negative and current flowing through the coil R-R’ is 0.866 Φm but the direction is positive. The direction of the net magnetic flux is determined by right hand screw rule and the magnitude of the net magnetic flux can be calculated as ΦT = ΦY 2 + ΦR 2 + 2ΦYΦRcosƟ Putting values on above formula, we get ΦT = 1.5 Φm Case 3: When Ɵ = ωt = 120˚ ΦR = Φmsin 120 = −0.866Φm ΦY = Φmsin 120 − 120 = 0 ΦB = Φm sin 120 + 120 = 0.866 Φm Therefore, at this instant, current flowing through the coil Y-Y’ is zero, current flowing through the coil R-R’ is 0.866 Φm but the direction is negative and current flowing through the coil B-B’ is 0.866 Φm but the direction is positive.
- 15. The direction of the net magnetic flux is determined by right hand screw rule and the magnitude of the net magnetic flux can be calculated as ΦT = ΦR 2 + ΦB 2 + 2ΦRΦBcosƟ Putting values on above formula, we get ΦT = 1.5 Φm Similarly, at all other instant such as 180˚, 240 ˚, 300˚ and 360˚, the magnitude of net magnetic field is again 1.5 Φm. Therefore, from the above analysis, it is clear that the stator winding produces a rotating magnetic field with a constant magnitude at all positions. The rotational speed of the rotating magnetic field is called synchronous speed and it depends upon the frequency and number poles of the machine. It can be expressed mathematically as: Snchronous speed, Ns = 120f P where, f = frequency of the spply and P = number of poles
- 16. Fig: Net magnetic flux at different time instant Ɵ = ωt = 0˚, 60˚ and 120˚
- 17. For sake of simplicity, let us consider the single conductor on the stationary rotor (i.e. rotor at rest at starting). Let this conductor be subjected to the rotating magnetic field produced when three phase supply is connected to the three phase winding of the stator. Consider that the rotating magnetic field rotates in the clockwise direction. According to Faraday’s Law of electromagnetic induction, an emf will be induced in the conductor. Since, the rotor circuit is short circuited or made complete either through end ring or an external resistance, the induced voltage causes the current to flow in the rotor conductor. Since the magnetic field is rotating clockwise and the conductor is stationary, we can assume that the conductor is in motion in the anticlockwise direction with respect to the magnetic field. By right hand rule, the direction of the induced rotor current is outwards (shown by dot). The induced current produces its own magnetic field or flux in the rotor. Working/Operating principle of three phase induction motor
- 18. When a current carrying conductor is put in a magnetic field resulting from the interaction of rotor and stator fluxes, a force is developed which acts on the rotor conductor. The direction of this force can be found out by Fleming’s left hand rule. Since, the rotor conductor is in a slot in the circumference of the rotor, the force acts tangentially on the rotor and hence induces a torque. Similar torques are produced on all the rotor conductors. Now, the rotor is free to move so, it starts rotating in the same direction of the rotating magnetic field. The rotor starts moving without any additional excitation system and because of this reason the motor is called the self-starting motor. The operation of the motor depends on the voltage induced on the rotor conductor, and hence it is called the induction motor. The direction of rotation can be determined by Lenz’s law. The direction of force will be in such a way that it opposes the cause by which the emf was induced in the rotor conductor. The main cause of rotor emf is the relative speed between the rotating magnetic field and the rotor. Hence, to reduce this relative speed, the rotor will rotate in the same direction as that of stator magnetic field. The rotor will try to catch up the speed of the rotating magnetic field but it never success to do so and always runs at speed less than synchronous speed.
- 19. If the rotor succeeds in catching up the stator field, there would be no relative speed between the two, hence no rotor emf, no rotor current and so no torque to maintain rotation. However, this won't stop the motor, the rotor will slow down due to loss of torque, the torque will again be exerted due to relative speed. That is why the rotor rotates at speed which is always less the synchronous speed in induction motor.
- 20. The difference between the synchronous speed and the actual speed of the rotor is known as the slip speed. It is a factor indicating the fraction by which the speed of the rotor is slightly less than the synchronous speed. It is usually expressed as a fraction or percentage of synchronous speed and is given by % Slip, S = NS − N NS × 100 The slip of the induction motor changes with respect to the load on the motor. When load on the motor increases, speed of motor decreases, then the slip of the motor will increase. Slip
- 21. At standstill (i.e. when the rotor is stationary), the speed is zero but the relative speed is maximum and the maximum emf will induce in the rotor (just like secondary winding in a transformer) and the frequency of the induced rotor current is the same as the supply frequency (f). But when the rotor starts revolving, then the rate at which the rotor conductors cut by the rotating flux depends upon the relative speed or slip speed. If ‘fR’ is the frequency of the rotor current, then NS − N = 120fR P … … … … . 1 Also, NS = 120f P … … … … … … . . 2 Dividing (1) by (2), we get NS − N NS = fR f ∴ fR = sf Frequency of rotor current
- 22. As we have seen that the rotating field set up by stator currents rotates at synchronous speed NS rpm relative to stator surface. Similarly, the rotor current having frequency fR when flows through each phase rotor winding, it gives rise to rotor field which rotates at a speed of sNS rpm relative to the rotor surface. But the rotor itself is running at a speed of N rpm with respect to stator surface. So the speed of rotor magnetic field with respect to stator surface is given as N + sNS = NS(1-s) + sNS = NS. It means that no matter what the value of slip, rotor and stator currents each produce a sinusoidally distributed magnetic field of constant magnitude and constant space speed of NS. This concludes that both the rotor and stator fields rotate synchronously which means that they are stationary with respect to each other. These two synchronously rotating magnetic fields superimpose on each other to give rise to actually existing rotating field, which corresponds to the magnetizing current of the stator winding. Speed of rotor field
- 23. Rotor emf At standstill (i.e. when the rotor is stationary), the speed is zero but the relative speed is maximum and the maximum emf will induce in the rotor. Thus, the induction motor at standstill condition is equivalent to a three phase transformer with secondary short- circuited. So, the emf induced per phase in the rotor at the instant of starting is given by E2 E1 = N2 N1 ∴ E2= E1 N2 N1 Where, N1 and N2 are the number of turns per phase in stator and rotor winding respectively. When the rotor starts running, the relative speed of the rotor with respect to the stator flux drops in direct proportion with slip s and the magnitude of the emf induced in the rotor will decrease as compared to the emf induced at standstill and hence, it is given by sE2.
- 24. Fig: Equivalent circuit (per phase) of induction motor at standstill condition Fig: Equivalent circuit (per phase) of induction motor at running condition
- 25. Rotor current, power factor and rotor torque Let us consider resistance and inductance per phase of rotor be R2 and L2 respectively. At standstill condition: Induced emf per phase in rotor winding at standstill condition = E2 Rotor winding reactance per phase, X2 = 2πfL2 where, f is supply frequency Rotor winding impedance per phase, Z2 = R2 2 + X2 2 Rotor current per phase, I2 = E2 R2 2 + X2 2 Power factor of rotor current , Φ2 = R2 R2 2 + X2 2
- 26. At running condition: Induced emf per phase in rotor winding = sE2 Rotor winding reactance per phase, XR = 2πfRL2 = 2πsfL2 = sX2 Rotor winding impedance per phase, Z2 = R2 2 + s2X2 2 Rotor current per phase, I2 = sE2 R2 2 + s2X2 2 Power factor of rotor current , Φ2 = R2 R2 2 + s2X2 2 The torque developed by the rotor of an induction motor is proportional to the product of stator flux per pole and active component of rotor current i.e. T α ΦI2 cosΦ2 Like, in the transformer flux remains constant and is independent on I1 and I2 .
- 27. It depends only on E1. But we know that E2α E1. So, rotoremf per phase at standstill, E2α Φ. Hence, torque developed, T = KE2I2 cosΦ2. Substituting values of I2 and cosΦ2, we get TR = KE2 × sE2 R2 2 + s2X2 2 × R2 R2 2 + s2X2 2 = 𝐾𝑠𝑅2𝐸2 2 R2 2 + s2X2 2 From above equation, it can be concluded that the running torque is proportional to the square of supply voltage because the rotor induced emf is proportional to supply voltage. At the instant of starting, slip is maximum (i.e. s = 1), so the starting torque is given by: TS = 𝐾𝑅2𝐸2 2 R2 2+X2 2 If the supply voltage is constant, then the flux Φ and hence E2 both are constant. Hence, the starting torque developed by the motor is given by ∴ TS= K1R2 R2 2 + X2 2 where, K1is another constant
- 28. Condition for maximum running torque We have, Torque under running condition is given by TR = KsR2E2 2 R2 2 + s2X2 2 From the above torque equation, it is clear that the torque developed by the motor depends upon the slip s provided E2 and R2 are constant. As slip changes with the speed, torque will also change with the speed. If the speed decreases, the value of slip will increase then the motor will develop more torque. But there is a limit so that after particular value of speed, the motor will not be able to produce more torque. The speed at which the torque is maximum can be determined as follows: Let Y = 1 TR = R2 2 + s2 X2 2 KsR2E2 2 = R2 2 KsR2E2 2 + s2 X2 2 KsR2E2 2 = R2 KsE2 2 + sX2 2 KR2E2 2 Then dy ds = −R2 Ks2E2 2 + X2 2 KR2E2 2
- 29. TR will be maximum when Y will be minimum. For this, dY ds = 0 or, −R2 Ks2E2 2 + X2 2 KR2E2 2 = 0 or, R2 Ks2E2 2 = X2 2 KR2E2 2 or, S2 = R2 2 X2 2 ∴ s = R2 X2 Hence , maximum torque will be developed at a speed corresponding to s = R2 X2 . If the motor is overloaded so that the speed goes below this value, the motor will not be able to develop more torque to overcome this increased load.
- 30. Hence, the maximum torque under running condition is given by TRmax = KsR2E2 2 R2 2 + s2X2 2 = KR2 2 E2 2 X2(R2 2 + R2 2 X2 2 × X2 2 ) ∴ TRmax= KE2 2 2X2 This relation shows that the maximum torque is independent of rotor resistance.
- 31. Torque-Slip and Torque-Speed characteristics Torque developed by an induction motor rotor is given by TR = KsR2E2 2 R2 2 + s2X2 2 From the above equation, it can be revealed that i. When the speed is synchronous i.e. N=NS, slip is zero then the torque is zero so that torque-slip characteristics curve starts from point A. ii. When the speed is very close to synchronous speed, the slip is very small. So, the value of s2X2 2 is very small compared to R2 2 and is negligible in comparison with rotor resistance. Therefore, the equation of torque for this range of operation can be simplified as: TR = KsR2E2 2 R2 2 or, TRα sE2 2 R2 ∴ TR α s (if R2 and E2 both are constant) Therefore, torque increases proportionately with increase in slip (or with decrease in speed) which is represented by a straight line AB on the characteristics curve.
- 32. iii. As the slip increases (or speed decreases with increase in load), torque goes on increasing and reaches its maximum value corresponding to slip 𝑠 = 𝑅2 𝑋2 . This maximum torque is known as breakdown or pull-out torque and the slip corresponding to this torque is called breakdown slip. iv. With the further increase in slip (or decrease in speed due to overload) beyond the point of maximum torque, the torque begins to decrease. The result is that the motor slows down and eventually stops. The motor operates for the value of slip between zero and breakdown slip. With the higher value of slip s, R2 2 is very small compared to s2X2 2 and is negligible in comparison with rotor reactance. Therefore, the equation of torque for this range of operation can be simplified as: TR = KsR2E2 2 𝑠2X2 2 or, TRα R2E2 2 sX2 2 ∴ TR α 1 s (if R2, X2 and E2 both are constant)
- 33. Therefore, torque decreases which is represented by rectangular hyperbola CD in the characteristics curve. When the motor stops at slip, s=1 or speed, N=0, the torque developed by the motor is given by : TS = KR2E2 2 R2 2+X2 2 This torque is known as starting torque which is not enough to drive the load on the motor. This part is represented by OD in the characteristics curve. Fig: Torque-slip and Torque-speed characteristics
- 34. From torque-speed or torque-slip characteristics, we have at constant supply voltage At starting, TRα R2 s At normal running condition, TRα s R2 Therefore, if we insert some resistance in series with rotor resistance (generally in case of slip-ring induction motor), the starting torque will increase but running torque will decrease. For given value of torque, slip is proportional to rotor resistance, so addition of external resistance in the rotor circuit does not lower the torque curve but slightly stretches it so that same torque value occurs at lower speeds or higher slip as shown in T-s characteristics curve. Effect of rotor resistance upon torque-slip or torque-speed characteristics
- 35. Full load torque and maximum torque Let, the full load slip of the motor, full load torque and maximum torque be sf, TRf and TRmax respectively. We have, TRf = KsfR2E2 2 R2 2+sf 2X2 2 and, TRmax = KE2 2 2X2 So, TRf TRmax = KsfR2E2 2 R2 2+sf 2X2 2 KE2 2 2X2 = 2sfR2X2 R2 2+sf 2X2 2 = 2sfR2 X2 R2 2 X2 2+sf 2 ∴ TRf TRmax = 2sfsb sb 2 + sf 2 where, sb = R2 X2 is the value of slip at maximum torque Similarly, TS TRmax = 2sb sb 2 +1
- 36. Testing of an Induction Motor As the circuit model of the three phase induction motor is similar to that of transformer with the exception that its secondary windings are free to rotate. As we have already noticed in the transformer that it is easier to combine two circuits as single one but there are some difficulties in induction motors due to slip. The equivalent circuit parameters of three phase induction motor can be determined by no-load test (corresponding to open circuit test of a transformer) and blocked rotor test (corresponding to short circuit test of a transformer).
- 37. No load test of an Induction Motor No load test is an indirect method used for determining the efficiency and the circuit parameters of the equivalent circuit of the three-phase induction motors. In this method, the motor is run without load on its shaft at the rated voltage and frequency. With the help of the two wattmeter, the input power of the motor is measured. The circuit diagram of the no- load test is shown below:
- 38. Let, the input power to the motor at no load be sum of two wattmeter readings i.e. W0 = W1 + W2 V0 = Input line voltage (voltmeter reading) I0 = No load stator current (average of reading of three ammeters) In this case, the motor runs at a speed very close to synchronous speed, therefore, the value of slip is very small at no load. Hence, the emf induced in the rotor circuit is sE2 which is very less and is negligible, thus, very small amount of current will flow through the rotor circuit as well as stator circuit. As the series impedance in the equivalent circuit is very small with compared to the impedance of magnetizing shunt branch and voltage drop in the series impedance can be neglected for no load operation. Hence, the equivalent circuit at no-load condition can be simplified as shown below:
- 39. The no-load current Io will lag the applied voltage V1 by some angle ɸ0 less than 90° as shown in the phasor diagram. As seen from the phasor diagram, the no-load current Io has two components: Energy component of no-load current, IW = I0cosΦ0 Magnetizing component of no-load current, Iµ = I0sinΦ0 The input power to the motor at no-load (W0) represents the iron loss and friction loss of the motor and is given by
- 40. W0 = 3V0I0cosΦ0 Where, the no-load power factor is given by: cosΦ0 = W0 3V0I0 Once the no-load power factor cosΦ0 is calculated, then we calculate Iµ = I0 2 − Iw 2 Thus, equivalent resistance, R0 = V1 Iw = V0 3Iw and equivalent reactance, X0 = V1 Iµ = V0 3Iµ Friction and windage loss can be separated from the no-load loss W0. At no load, various readings of the no-load loss are taken at the different stator applied voltages. The readings are taken from rated to the breakdown value at rated frequency.
- 41. A curve plotted between W0 and V is shown below. The curve is almost parabolic at the normal voltage. Since, iron losses are almost proportional to the square of the flux density and therefore, the applied voltage. The curve is extended to the left to cut the vertical axis at the point A. At the vertical axis V = 0 and hence the intercept OA represents the independent voltage loss i.e. the loss due to friction and the windage Pfw.
- 42. Blocked rotor test of an Induction Motor The blocked rotor test of an induction motor is similar to the short circuit test of a transformer. In this test, the shaft of the motor is locked so that it cannot move and the rotor winding is short-circuited. In the slip ring motor, the rotor winding is short- circuited through the slip rings. In the case of the cage motors, the rotor bars are permanently short-circuited. A reduced voltage at the reduced frequency is applied to the stator through a three-phase auto-transformer so that motor draws rated full- load current in the stator. The power input on the locked rotor test is equal to the sum of copper losses of the stator and the rotor for all three phases. The core and the mechanical losses are negligible as the reduced voltage is applied to the stator and, as a result, the rotation of the rotor is not allowed.
- 43. The circuit diagram of the blocked rotor test is shown below:
- 44. Let, WSC = W1 + W2 (sum of two wattmeter readings) VSC = Line voltage applied during short circuit test (voltmeter reading) ISC = Line current during short circuit test (average of reading of three ammeters) Since the rotor is blocked, the speed is zero and the value of slip is unity so the voltage VSC required to circulate the full load current is very small. Therefore, the magnetizing current through the shunt branch parameters(R0 and X0)in the equivalent circuit model will be negligible with respect to ISC. Hence, the equivalent circuit can be simplified as shown below:
- 45. The blocked rotor test is performed under normal operating conditions when the rotor current and the frequency are on the same conditions. Generally, the slip of the induction motor varies between 2 to 4 percent, and the resulting rotor frequency is in the range of 1 to 2 Hz for the stator frequency of 50 Hz at the normal conditions. Equivalent resistance of the motor referred to stator is given by: R01 = Wsc 3Isc 2 Equivalent impedance of the motor referred to stator is given by: Z01 = Vsc 3Isc Equivalent reactance of the motor referred to stator is given by: X01 = Z01 2 − R01 2
- 46. Losses and efficiency An induction motor takes electrical power input through the stator windings and the shaft of the motor gives mechanical power output. The output power is always less than the power input due to the power losses taking place in different stages of an induction motor. The power stages in an induction motor is shown below.
- 47. Figure: Power stages in induction motor
- 48. Out of the total VA input to the stator, some power loss as copper loss will take place due to heating of stator winding resistance and some power loss as iron loss will take place due to heating of stator core because of eddy current loss and hysteresis loss which depends on the supply frequency and the flux density in the core and the iron loss is practically constant. Some of the reactive power(VAR) will also be lost to supply the reactive power requirement of the machine to maintain required air-gap flux. The remaining active power will be transferred to the rotor by means of electromagnetic induction like in a transformer. Some power loss as a copper loss will take place in rotor due to heating of rotor winding resistance. The iron loss of the rotor, however, is negligible because frequency of emf induced in the rotor circuit is very small with compared to supply frequency. Now, rest of the power will be converted into the mechanical power in terms of mechanical rotation of the rotor and give rise to gross torque.
- 49. Out of this gross torque developed, some is lost due to windage and friction losses in the rotor and the rest appears as the useful or shaft torque which is provided by the net power available in the shaft of the motor. Let, Tg be the torque developed by the rotor at a speed of N rpm Then, the mechanical power developed by the rotor or rotor gross output, Pm=2πN × T g If there were no copper losses in the rotor circuit, then all the power input to the rotor will be converted into mechanical power and in such a situation the rotor will rotate at synchronous speed, NS. The power input to the rotor, P2 =2πNS × Tg Thus, rotor copper loss, Pcr =P2 - Pm = 2πNS × Tg - 2πN × T g = 2π Tg (NS-N)
- 50. Now, rotor copper loss rotor power input = 2π Tg NS − N 2πNS × Tg = NS − N NS = s Therefore, rotor copper loss is equal to slip times rotor input power (or power across air-gap). Also, rotor gross output is the difference of rotor input and rotor copper loss and is given by Pm = 1 − s P2 Pm P2 = 1 − s = N NS Hence, rotor efficiency, ηrotor = rotor gross output rotor input = N NS And, overallefficiency, ηo = power developed by the shaft stator input power
- 51. Starting methods of an Induction Motor The equivalent circuit of the three phase induction motor at the time of starting is similar to that of transformer with short circuited secondary winding. At the time of starting, the rotor is stationary and the back emf due to the rotation is not developed yet, hence, the motor draws the high starting current if normal supply voltage is applied. Once the rotor begins rotating, the value of slip decreases and accordingly the emf induced in the rotor circuit will decrease, thereby, decreasing the values of rotor and stator current. The purpose of a starter is not to just start the motor, but it performs the two main functions. They are as follows: ● To reduce the heavy starting current, ● To provide overload and under-voltage protection. The three-phase induction motor may be started by connecting the motor directly
- 52. to the full voltage of the supply or applying a reduced voltage to the motor. The torque of the induction motor is proportional to the square of the applied voltage. Thus, greater torque is exerted by a motor when it is started on full voltage than when it is started on the reduced voltage. An induction motor, when directly switched on with full supply voltage, draws 5 to 7 times of their load current and develops 1.5 to 2.5 times of their full load torque, the actual values depending upon the size and design of the motor. This high starting current is objectionable because it will produce large line voltage drop, which in turn, will affect the operation of the electrical equipment connected in the same line. A Direct-On-Line starter is generally used for motors that are rated below 5kW. Hence, for large induction motors, various starting methods are available to reduce the starting current 1) Using primary resistors method 2) Auto transformer method 3) Star-delta starter
- 53. Using Primary Resistors method In this method, variable resistances are connected in series with the stator windings as shown in the figure below. The purpose of primary resistors is to drop some voltage and hence to reduce the voltage applied across the stator windings. Only a fraction of the voltage (x) of the supply voltage is applied at the time of starting of the induction motor. The value of x is always less than one. Due to the drop in the voltage, the starting torque also decreases. As the motor speeds up, the resistor is gradually cut out from the circuit and finally the resistors are short circuited when the motor reaches to its normal operating speed. Now let us derive the expression for starting torque in terms of full load torque for the stator resistor starting method.
- 54. We know, Current drawn by the motor ∝ voltage Torque developed by the motor ∝ voltage 2 Id ∝ V and Ist ∝ Xv Td ∝ V2 and Ist ∝ xV 2 ∴ Tst∝ x2 Td Let, V is the voltage per phase in stator at running condition xV is the voltage per phase in stator at starting condition Td is torque developed by the motor when directly started Tst is torque developed by the motor when primary resistors are used Id is current drawn by the motor per phase when directly started Ist is current drawn by the motor per phase when primary resistors are used
- 55. If the voltage is reduced by 50% at starting by using primary resistors (i.e. x = 0.5) Then, Ist = 0.5Id and Tst = 0.52 Td = 0.25Td Therefore, if the voltage applied is reduced by 50% at starting, the starting current will get reduced by 50% but the torque developed is reduced by 25%.
- 56. Auto transformer method The principle used in this method is similar to that of primary resistors method. The auto transformer is a step down transformer use to give reduced voltage at starting. When the switch in on 'Start' position, per phase supply voltage is reduced from V to xV in the stator windings. The reduction in voltage reduces current from Is to xIs. After the motor reaches to 80% of its normal operating speed, the auto transformer is disconnected and then full supply voltage is applied across the stator windings. The switch making these changes from 'Start' to 'Run' may be air break contactors (for small motors) and may be oil-immersed type (for large motors) to reduce sparking.
- 58. Star-Delta starter This method is used for the motors designed to run normally with delta connected stator windings. This method uses Triple Pole Double Throw (TPDT), a two way switch which connects the stator winding in 'Star' for starting period and then in 'Delta' for normal running period. When the stator winding is star connected, voltage over each phase in motor will be reduced by a factor 1/√3 of that would be for delta connected winding. The starting torque will 1/3 times that it will be for delta connected winding. If the motor is started with delta connection (i.e. direct starting) (Id)∆ = 3 V Z If the motor is started with star connection (Ist)Y = V/ 3 Z = V 3Z
- 59. (Ist)∆ (Ist)Y = 3 V Z × 3Z V = 3 ∴ (Ist)Y = 1 3 (Ist)∆ Therefore, the starting current will be reduced by 1/3th of direct Delta starting current. Hence a star-delta starter is equivalent to an auto-transformer of ratio1/√3 or 58% reduced voltage.
- 60. Slip Ring Induction Motor Starter Method
- 61. In this method, the full supply voltage is connected across the starter. The starting current is reduced by introducing a variable resistance in the rotor circuit externally through slip-rings. Full value of external rheostat is connected in the rotor circuit at starting and thus the supply current to the stator is reduced. The rotor begins to rotate, and the rotor resistances are gradually cut out as the speed of the motor increases. When the motor is running at its rated full load speed, the starting resistances are cut out completely, and the slip rings are short-circuited. The starting resistance also helps to develop higher starting torque which is clear from T-s characteristics. Rotor starting current per phase without external rheostat, IR(d) = E2 R2 2 + X2 2 Rotor starting current with external rheostat, IR st = E2 R2 + Rs 2 + X2 2
- 62. Speed Control Methods Of Induction Motor ● A three phase induction motor is practically a constant speed motor more or less like a dc shunt motor. This means, for the entire loading range, change in speed of the motor is quite small. ● There is one major difference between these two. DC shunt motor can be made to run at any speed within wide range of limits very easily with good efficiency simply by changing the field rheostat, but in case of induction motor, the speed control is done at the expense of decrease in efficiency and poor power factor. ● As induction motors are widely being used, their speed control may be required in many applications. Speed control means to change the drive speed as desired by the process to maintain different process parameter at different load.
- 63. ● Energy saving is the main requirement while performing speed control. ● Speed control is a different concept from speed regulation where there is natural change in speed due change in load on the shaft. ● Speed control is either done manually by the operator or by means of some automatic control device. ● The speed control of induction motor from stator side are as follows: a) By changing the applied voltage b) By changing the applied frequency c) Constant V/f Control d) By changing the number of stator poles ● The speed control of induction motor from rotor side are as follows: a) Rotor rheostat control b) Cascade operation
- 64. 1) By changing the applied voltage The torque produced by three phase induction motor is given by: TR = KsE2 2 R2 R2 2 + s2X2 2 = 3 2πNS × sE2 2 R2 R2 2 + s2X2 2 Where, E2 is the rotor emf, R2 is the rotor resistance and X2 is the rotor inductive reactance. From the above torque equation, for constant rotor resistance R2 if slip s is small then s2 X2 2 is so small that it can be neglected. Therefore, TRα sE2 2 and E2 ∝ V. Thus, TRα sV2 which means the developed torque in the rotor decreases if supplied voltage is decreased. Hence, for providing the same load torque, the slip increases with Speed Control of induction motor from stator side
- 65. decrease in voltage and consequently, the speed decreases. This method is cheaper and simple but rarely used because a large change in supply voltage is required for relatively small change in speed. A large change in supply voltage will result in a large change in flux density thereby seriously disturbing the magnetic conditions of the motor. 2) By changing the applied frequency The synchronous speed (speed of the rotating magnetic field in the stator) of an induction motor is given as: NS = 120f P where, f is frequency and P is the number of pole. Hence, the synchronous speed changes with change in supply frequency. Actual speed of an induction motor is given as N = NS 1 − s .
- 66. However, this method is not widely used. It may be used where, the induction motor is supplied by a dedicated generator so that frequency can be easily varied by changing the speed of prime mover. Also, at lower frequency, the motor current may become too high due to decreased reactance. And if the current is increased beyond the rated value, the maximum torque developed falls while the speed rises. 3) Constant V/f control of induction motor This is the most popular method for controlling the speed of an induction motor. In three phase induction motor, emf induced in the rotor is similar to that of emf induced in secondary of transformer at standstill condition which is given by E = 4.44f∅KT ⇒ ∅ = E 4.44fKT
- 67. If the supply frequency is reduced keeping the rated supply voltage, the air gap flux will increase and this change in value of flux causes saturation of rotor and stator cores which will further cause excessive stator current and distortion of the stator flux wave. Therefore, the stator voltage should also be reduced in proportional to the frequency so as to maintain the air-gap flux constant. The magnitude of the stator flux is proportional to the ratio of the stator voltage and the frequency. Hence, if the ratio of voltage to frequency is kept constant, the flux remains constant. Thus, by keeping V/f constant, the developed torque also remain nearly constant. This method gives higher run-time efficiency. Therefore, majority of AC speed drives employ constant V/f method (or variable voltage, variable frequency method) for the speed control. Along with wide range of speed control, this method also offers 'soft start' capability.
- 68. 4) Changing the number of stator poles The synchronous speed (speed of the rotating magnetic field in the stator) of an induction motor is given as: NS = 120f P where, f is frequency and P is the number of pole. Actual speed of an induction motor is given as: N = NS 1 − s It can be seen from above equations that synchronous speed (and hence, running speed) can be changed by changing the number of stator poles. This method is generally used for squirrel cage induction motors, as squirrel cage rotor adapts itself for any number of stator poles. Change in stator poles is achieved by two or more independent stator windings wound for different number of poles in same slots and with the help of switching arrangement, at a time, supply is given to one winding only and hence motor can operate at different speeds but smooth change in speed is not possible.
- 69. 1) Rotor Rheostat Control In this method of speed control of three phase induction motor, an external resistance is introduced in the rotor circuit. The equation of torque for three phase induction motor is given by TR ∝ sE2 2 R2 R2 2 + s2X2 2 The three phase induction motor operates in a low slip region. In low slip region, term s2 X2 2 becomes so small as compared to R2 2 that it can be neglected. And, E2 is also constant then the equation of torque after simplification becomes, TR ∝ s R2 Speed Control of induction motor from rotor side
- 70. Therefore, if the rotor resistance R2 is increased, torque decreases but to supply the same load, torque must remain constant. So, for a given constant load torque, increase in R2increases slip, which will further result in the decrease in rotor speed. Thus, by adding additional resistance in the rotor circuit, we can decrease the speed of the three-phase induction motor. The main advantage of this method is that with an addition of external resistance starting torque increases but this method of speed control have some disadvantages : ● The speed above the normal value is not possible. ● Large change in speed requires a large value of resistance, and if such large value of resistance is added in the circuit, it will cause large copper loss and hence reduction in efficiency. ● This method cannot be used for squirrel cage induction motor i.e. it is applicable to slip ring induction motor only.
- 71. 2) Cascade control method In this method, the two three-phase induction motors having different number of poles are mounted on a common shaft and hence called cascaded motor. One motor is the called the main motor, and another motor is called the auxiliary motor. The three-phase supply is given to the stator of the main motor while the auxiliary motor is supplied by the voltage induced in the rotor of main motor through slip rings (i.e. auxiliary motor is derived at a slip frequency from the slip ring of the main motor). Therefore, the two motors will try to run with two different speed corresponding to two different frequencies but as the two motors are coupled to a common shaft, the system will run at a new speed. Four different speeds can be obtained in this method of speed control of three phase induction motor.
- 72. 1) When only main induction motor works, then the complete set will run at a speed corresponding to N1 = 120f Pm 2) When only auxiliary induction motor work, then the complete set will run at a speed corresponding to N2 = 120f Pa 3) When cumulative cascading is done, then the complete set will run at a speed corresponding to N3 = 120f Pm + Pa 4) When differential cascading is done, then the complete set will run at a speed corresponding to N4 = 120f Pm − Pa
- 73. Note: If the torque produced by the main and auxiliary motor will act in same direction, resulting in number of poles Pm + Pa , then such type of cascading is called cumulative cascading. If the torque produced by the main and auxiliary motor will act in opposite direction, resulting in number of poles Pm − Pa , then such type of cascading is called differential cascading.
- 74. Double cage rotor of an Induction motor It is an induction motor with two rotor windings or cages used for obtaining high starting torque at low value of starting current. The stator of a double cage rotor of an induction motor is the same as that of a normal induction motor. In the double cage rotor of an induction motor, there are two layers of the bars as shown in the figure below:
- 75. Each layer is short-circuited by the end rings. The outer cage bars have a smaller cross-sectional area than the inner bars and are made of high resistive materials like brass, aluminium, bronze, etc. The inner cage bars are made of low resistance copper. Thus, the resistance of the outer cage is greater than the resistance of the inner cage. There is a slit between the top and the bottom slots. The slit increases permeance for leakage flux around the inner cage bars, thus, the leakage flux linking the inner cage winding is much larger than that of the outer cage winding. Therefore, the inner winding has a greater self-inductance. At starting, the voltage induced in the rotor is the same as the supply frequency that is (f2 = f1). Hence, the leakage reactance of the inner cage winding as compared to that of the outer cage winding is much larger.
- 76. Therefore, most of the starting current flows in the outer cage which offer low impedance to the flow of current. The high resistance outer cage winding, therefore, develops a high starting torque. As the rotor speed increases, the frequency of the rotor emf (fr =sf) decreases. At normal operating speed, the leakage reactance of both the windings become negligibly small. The current in the rotor divides between the two cages and is governed by their resistances. The resistance of the outer cage is about 5 to 6 times that of the inner cage. Hence, the torque of the motor developed mainly by the low resistance inner cage and is developed under normal operating speed. For the low starting torque requirements, an ordinary cage motor is used. For higher torque requirements a deep bar cage motor is used. A double cage motor is used for higher torques. The slip ring construction is used for large sized motors having very large starting torque and exceptionally long starting periods.
- 78. Just like a DC Machine, an induction machine can be used as an induction motor as well as an induction generator, without any internal modifications. Induction generators are also called as asynchronous generators. Before starting to explain how an induction (asynchronous) generator works, we must know the working principle of an induction motor. In an induction motor, the rotor rotates because of slip (i.e. relative speed between the rotating magnetic field and the rotor). Rotor tries to catch up the synchronously rotating field of the stator but never succeeds. If rotor catches up the synchronous speed, the relative velocity will be zero, and hence rotor will experience no torque. But what if the rotor is rotating at a speed more than synchronous speed? Working Principle of Induction Generator
- 79. If the induction machine is working as a motor, an AC supply is connected to the stator terminals of an induction machine so that the rotating magnetic field is produced in the stator that pulls the rotor to run behind it. Now, if the rotor is accelerated to the synchronous speed by means of a prime mover, the slip will be zero and hence the net torque will be zero. The rotor current will become zero when the rotor is running at synchronous speed. Now, if the rotor is made to rotate at a speed more than the synchronous speed by prime mover, the slip becomes negative. A rotor current is generated in the opposite direction, due to the rotor conductors cutting stator magnetic field. This generated rotor current produces a rotating magnetic field in the rotor which pushes (forces in opposite way) onto the stator field. This causes a stator voltage which pushes current How Induction motor works as an Induction generators ?
- 80. flowing out of the stator winding against the applied voltage. Thus, the induction machine is now working as an induction generator.
- 81. An induction machine with capacitors connected across its terminals builds up voltage in a similar manner like a dc shunt generator does. The build-up process in a dc machine depend upon the residual magnetism in the field poles and the final voltage is determined by the resistance of the field circuit. But, in case of induction machine, the residual flux present in the machine provides the initial excitation. In the absence of the residual flux, the machine is momentarily run as an induction motor to create the residual flux. The motor is running slightly above the synchronous speed at no load by a prime mover. Now, the residual magnetism in the magnetic circuit is sufficient to induce a small ac voltage in the stator at a frequency proportional to the rotor speed. Such a voltage appearing across the three phase capacitor bank connected to stator terminals causes a lagging magnetizing current to flow through the stator terminals or leading current through the capacitor. The flux set up by this current assists the initial residual flux, so, if a proper value of capacitor Voltage build up in the induction generator
- 82. is selected, the magnetizing current would be sufficient to increase the existing air- gap flux, thereby causing an increase in induced voltage. This increase in voltage causes further increase in exciting current, causing further increase in terminal voltage. This process of voltage build-up continues until induced voltage reaches a limit constrained by the saturation curve of the machine and the reactance of the capacitor. The equivalent circuit of an induction machine in such a condition is represented by figure below: Where, Im = magnetizing current Xm = magnetising reactance X1 = stator leakage reactance Xc = capacitive reactance E = emf induced in the stator Hence, self − excitation occurs when XC ≤ (Xm + X1)
- 83. Fig: No-load saturation curve and final build up voltage
- 84. The figure above shows the machine excitation characteristics and C1, C2, C3, C4 representing volt-ampere characteristics of the capacitors of different rating used for excitation. In the figure above, ‘Oa’ is generated small terminal voltage across the stator terminal due the residual magnetism when the rotor of the induction machine runs at the required speed (greater than synchronous speed). Due to this voltage ‘oa’, the capacitor current ‘ob’ is produced. This magnetizing current increases the flux thus increasing generated voltage to ‘bc’ with corresponding value of capacitor current ‘od’. The cumulative process of voltage generation continues till the saturation curve of the induction generator cuts the capacitor load line at some point. This point is marked as ‘f’ in the given curve. For any point on the excitation curve, E Im = Xm + X1 − XC
- 85. This equation is true when the reactive volt-amperes supplied by capacitor bank (leading VA rating of the capacitor) is equal to the reactive volt-amperes demanded by the generator (lagging or magnetizing VA of the machine) for reactive VA balance of the system. Thus, the slope of the curves of C1, C2, C3, C4 gives the reactance of the capacitor required to produce voltage corresponding to points of intersection with excitation curve. Obviously, as the value of capacitor decreases, its reactance increases i.e. the slope increases with resulting decrease in terminal voltage until the coincident point is obtained. The curve C3 gives infinite number of possible solutions. Below this value of capacitor, the machine will not excite and hence, will be inoperative. The capacitance corresponding to C3 represents the critical value of capacitance below which the machine cannot build up the voltage. E1 and E2 are the final build up voltage in the stator with corresponding magnetizing current Im1 and Im2 due to excitation capacitance C1and C2 respectively.
- 86. Power stages in an induction generator
- 87. Induction generator is not a self-excited machine. Therefore, when induction machine is running as a generator, the machine takes reactive power from the AC power line and supplies active power back into the line as shown in the figure above. Reactive power is needed for producing rotating magnetic field. Isolated and grid connected mode
- 88. The active power supplied back in the line is proportional to slip above the synchronous speed. It is clear that, an induction machine needs reactive power for excitation, regardless whether it is operating as a generator or a motor. When an induction generator is connected to a grid, it takes reactive power from the grid. But if we use an induction generator to supply a load without using an external source (e.g. grid) then a capacitor bank is used which is connected across the stator terminals to supply reactive power to the machine as well as to the load. When the rotor is rotated at an enough speed, a small voltage is generated across the stator terminals due to residual magnetism. Due to this small generated voltage, capacitor current is produced which provides further reactive power for magnetization.