Electrical Machine - II -English-QB Compose By Lalit Chaudhary

1 THREE PHASE INDUCTION MOTOR Chapter SUBJECTIVE QUESTIONS AND ANSWERS Q.1. Explain the working principle of an 3-φ induction motor. (2000, 11) Write down the principle of working of induction motor. (2005) Explain the principle of working of a three phase induction motor. (S/2007, S/08, 09, S/17) Explain how a three phase induction motor rotates? (2012, S/15) Ans. An induction machine is a singly excited alternating current (AC) machine. Its stator winding is directly fed by AC source, whereas its rotor winding receives energy from stator by means of induction, i.e., transformer action. In fact an induction motor can be treated as a rotating transformer i.e., one in which primary winding is stationary but the secondary is free to rotate. When the three phase stator winding, are fed by a three phase supply then a magnetic flux of constant magnitude, but rotating at synchronous speed, is setup. The flux passes through the air gap, sweeps past the rotor surface and so cuts the rotor conductors which, as yet, are stationary. Due to the relative speed between the rotating flux and the stationary conductors, an emf is induced in the latter, according to Faraday’s Laws of electro magnetic induction. The frequency of the induced emf is the same as the supply frequency. It’s magnitude is proportional to the relative velocity between the flux and the conductors and its direction is given by Fleming’s Right Hand Rule. Since, the rotor bars or conductors form a closed circuit, rotor current is produced whose direction, as given by Lenz’s Law, is such as to appose the very cause producing it. In this case, the cause which produces the rotor current is the relative velocity between the rotating flux of the stator and the stationary rotor conductors. Hence, to reduce the relative speed, the rotor starts running in the same direction as that of the flux and tries to catch up with rotating flux. Q.2. How the revolving magnetic field is produced in three phase induction motor? (2001) Write short note on production of rotating magnetic field in induction motor. (S/2009) (1)

2 Electrical Machines–II Explain with the help of suitable diagrams, the production of rotating magnetic field in 3-φ induction motor. (2010) Explain how rotating magnetic field is produced by 3-φ supply to a 3-φ induction motor? (S/2011, 14) Explain production of rotating magnetic field in 3-phase induction motor. (S/2012, 16) Explain with the help of suitable diagram, the production of rotating magnetic field in a 3-phase induction motor. (S/2013) Ans. When three phase winding displaced in space by 1200, are fed by three phase currents displaced in time by 1200, they produce a resultant magnetic flux which rotates in space as if actual magnetic poles were being rotated mechanically. Fig. 1.1 Let, the maximum value of flux due to any one of the three phase be φm. The resultant flux φr, at any instant, is given by the vector sum of the individual fluxes, φ1, φ2 and φ3 due to three phase. The value of φr at 4 different positions are as follows:

Three Phase Induction Motor 3 (i) At θ = 00, φ1 = 0, φ2= –√3 φm , φ3 = √3 φm and φr = 1.5 φm 2 2 (ii) At θ = 600, φ1 = √3 φm, φ2 = – √3 φm , φ3 = 0 and φr = 1.5 φm 2 2 (iii) At θ = 1200, φ1 = √3 φm, φ2 = 0 , φ3 = –√3 φm and φr = 1.5 φm 2 2 (iv) At θ = 1800, φ1 = 0, φ2= √3 φm , φ3 = – √2 φm and φr = 1.5 φm 2 2 All these 4 different positions are shown in fig. 1.1. Hence, user conclude that, (i) The resultant flux is of constant value = 1.5 φm, i.e., 1.5 times the maximum value of the flux due to any phase. (ii) The resultant flux rotates around the stator at synchronous speed. Q.3. Write in brief construction of 3-φ induction motor with diagram. (S/2017) Ans. A three phase induction motor consists of two main types which are as follows: (i) Stator: It is the stationary part of the motor. It has three main parts, are as follows: (a) Outer Frame: It is the outer body of the motor. Its function is to support the stator core and to protect the inner parts of the machine. For small machines the frame is casted but for large machines it is fabricated. To place the motor on the foundation, feet are provided in the outer frame as shown in fig. 1.2. Fig. 1.2 Fig. 1.3

4 Electrical Machines–II (b) Stator Core: The stator core is to carry the alternating magnetic field which produces hysteresis and eddy-current losses, therefore, core is built up of high grade silicon steel stampings. The stampings are assembled under hydraulic pressure and are keyed to the frame. Each stampings is insulated from the other with a thin varnish layer. The thickness of the stampings varies usually from 0.3 to 0.5 mm. Slots are punched on the inner periphery of the stampings as shown in fig. 1.3, to accommodate stator winding. (c) Stator Winding: The stator core carries a three phase winding which is usually supplied from a three phase supply system. The six terminals of the winding (two of each phase) are connected in the terminal box of the machine. The stator of the motor is wound for definite number of poles, the exact number being determined by the requirement of speed. It will be seen that greater number of poles, lower the speed and vice-versa, since, Ns ∝ 1 [ ]...Ns=120f P P The three phase winding may be connected in star of delta externally through a starter. (ii) Rotor: It is the rotating part of the motor. There are three types of rotor, which are employed in three phase induction motor. (a) Squirrel Cage Rotor: The motors employing this type of rotor are known as squirrel cage induction motors. Most of the induction motors are of this type because of simple and rugged construction of rotor. A squirrel cage rotor consists of a laminated cylindrical core having semi-closed circular slots at the outer periphery. Copper or aluminium bar conductors are placed in these slots and short circuited at each end by copper or aluminium rings called short circuiting rings, as shown in fig. 1.4. Thus, the rotor Fig. 1.4 winding is permanently short circuited and it is not possible to add any external resistance in the rotor circuit. (b) Double Cage Rotor: The stator of double cage rotor induction motor is same as that of ordinary induction motor whereas its rotor consists of two cages or two layers of bars short circuited by end rings since, the upper cage is having smaller cross-sectional area than the lower cage, the upper cage is having higher resistance than that of

Three Phase Induction Motor 5 lower cage with equal cross-sectional areas of two cages the upper cage is made up of high resistance material like brass, aluminium, bronze, etc and the lower cage is made up of low resistance material like copper. The upper cage and lower cage are seperated by a narrow slit or construction. This is shown in fig. 1.5. Fig. 1.5 Double Cage Rotor (c) Phase Wound Rotor or Slip Ring Rotor: Phase wound rotor is also called slip ring rotor and the motors employing this type of rotor are known as phase wound or slip ring induction motors. Slip ring rotor consists of a laminated cylindrical core Fig. 1.6 having semi-closed slots at the outer periphery and carries a three phase insulated winding. The rotor is wound for the same number of poles as that of stator. The three finish terminals are connected together forming star point and the three start terminals are connected to three copper slip ring fixed on the shaft. In this case, depending upon the requirement any external resistance can be added in the rotor circuit.

6 Electrical Machines–II In this case also the rotor is skewed. A mild steel shaft is passed through the centre of the rotor and is fixed to it with key. The purpose of shaft is to transfer mechanical power. This is shown in fig. 1.6. Q.4. What are the advantages of slip ring motors over squirrel cage induction motors? (2013) Ans. Advantages of Slip Ring Motors: The main advantages of slip ring motors are as follows: (i) It is possible to get high starting torque by introduction of resistances in the rotor circuit. (ii) Starting current is comparatively less and it is 2 to 2.5 times the full load current or even less, if required. (iii) It is capable of starting with load, demanding a high torque as in fans, pumps, compressors, hoists, conveyors, cranes, lifts and so on. (iv) Speed control is possible by varying the resistances in the external circuit of the rotor. (v) The motor can be started direct on line (DOL) without the necessity of starters, because of the low starting current of the motor. Advantages of Squirrel Cage Induction Motors: The main advantages of squirrel cage induction motors are as follows: (i) These motors are cheaper in cost as compared to slip ring induction motors. (ii) These motors requires less maintenance and rugged construction. (iii) By using a double cage rotor, it is possible to obtain higher starting torque. (iv) Losses are relatively less. (v) Squirrel cage motor have slightly higher efficiency than wound motor. Q.5. What are the disadvantages of squirrel cage induction motors and slip ring motors? Ans. Disadvantages of Squirrel Cage Induction Motors: The main disadvantages of squirrel cage induction motors are as follows: (i) Speed control is not possible in squirrel cage induction motor. (ii) This motors have poor starting torque. Starting torque will be in the order of 1.5 to 2 times the full load torque. (iii) These motors have high starting current. Starting current is about six to eight times the full load current. (iv) Because of the high starting current, it requires a starting device such as an autotransformer or a star-delta starter. (v) The total energy loss during starting of squirrel cage motor is more compared to slip ring motors. Disadvantages of Slip Ring Motors: The main disadvantages of slip ring motors are as follows:

Three Phase Induction Motor 7 (i) Initial and maintenance cost is more as compared to squirrel cage motor because presence of slip rings, brushes, short circuiting devices, etc. (ii) Speed regulation is poor when operated with external resistances in rotor circuit. (iii) Efficiency and power factor of slip ring motor is lower as compared to squirrel cage induction motor. (iv) Lower power factor is at the light loads. (v) Sensitivity to fluctuations in supply voltage. Q.6. Write the advantages of ‘skewed rotor slots’ of a squirrel cage induction motor. (2013, 17) Ans. The rotor slots are usually not parallel to the shaft but are skewed. Skewing of rotor have the some advantages which are as follows: (i) It reduces humming thus ensuring quiet running of a motor. (ii) It results in a smoother torque curves for different positions of the rotor. (iii) It reduces the magnetic locking of the stator and rotor. (iv) It increases the rotor resistance due to the increased length of the rotor bar conductors. Q.7. Write the applications of squirrel cage induction motors and slip ring motors. Ans. Applications of Squirrel Cage Induction Motors: These are widely used in industrial applications than slip ring induction motors due to cheaper in cost, rugged in construction, low maintenance. Squirrel cage induction motors are suitable for applications where the drive requires constant speed, low starting torque and no speed control drives. Applications of Slip Ring Motors: Slip ring induction motors have big advantage of high starting torque compared to squirrel cage motor. Therefore, slip ring induction motors are generally employed where load requires high starting torque or good speed control. They are employed in hoists, elevators, compressors, printings presses, large ventilating fans, loads requiring speed control such as for driving lifts and pumps. Q.8. Write a short note on comparison between squirrel cage and slip ring induction motor. (2010) Differentiate between squirrel cage and slip ring motor. (S/2011) Write the difference between squirrel cage and slip ring induction motor. (S/2017) Ans. The difference/comparison between squirrel cage and slip ring induction motor are as follows:

8 Electrical Machines–II S.No. Squirrel Cage Induction Motor Slip Ring Induction Motor (i) Construction is very simple. Construction is complicated (ii) due to presence of slip ring The rotor consists of rotor and brushes. (iii) bars which are permanently The rotor winding is similar shorted with the help of end to the stator winding. (iv) rings. (v) Since, the rotor bars are This motor can easily add (vi) permanently shorted, its not rotor resistance by using slip (vii) possible to add external ring and brushes. (viii) resistance. (ix) Starting torque is low and Due to presence of external (x) cannot be improved. resistance high starting (xi) torque can be obtained. Slip ring and brushes are Slip ring and brushes are absent. present. Less maintenance is Frequent maintenance is required. required due to presence of brushes. The construction is robust The presence of brushes and it is cheap as compared and slip ring makes the to slip ring induction motor. motor more costly. The squirrel cage induction This motor is rarely used motor is widely used. only 10% industries uses slip ring induction motor. Less rotor copper losses and Rotor copper losses are high hence high efficiency. and hence less efficiency. Speed control by rotor Speed control by rotor resistance method is not resistance method is possible. possible. This motor is used in lathes, These motors are used drilling machine, fan, blower where high starting torque printing machines, etc. is required, i.e., in hoists, cranes, elevators, etc. Q.9. What are the advantages and disadvantages of three phase induction motor over other types of motors? Ans. Advantages: The advantages of three phase induction motor over other motors are as follows: (i) It has very simple and extremely sugged, almost unbreakable construction especially squirrel cage type motor. (ii) Its costs is low and it is very reliable.

Three Phase Induction Motor 9 (iii) It has sufficiently high efficiency. In normal running condition, brushes are needed, hence frictional losses are reduced. It has a reasonably good power factor. (iv) It requires minimum cost of maintenance. (v) Its startup from rest and needs no extra starting motor and has not to be synchronised. Its starting arrangement is simple especially for squirrel cage type motor. Disadvantages: The disadvantages of three phase induction motor over other motors are as follows: (i) Its speed cannot be varied without sacrificing some of its efficiency. (ii) As DC shunt motor its speed decreases with increase in load. (iii) Its starting torque is inferior to that of a DC shunt motor. Q.10. Define slip. (S/2017) Ans. In practice, the rotor never succeeds in catching up with the stator field. If it really happens then there would be no relative speed between the two, hence, no rotor emf no rotor current and also no torque will maintain rotation. That is why the rotor runs at a speed which is always less than the speed of the stator field. Hence, slip can be defined as the difference between the synchronous speed Ns and the actual rotor speed N. It is expressed as a fraction or percentage of a synchronous speed, Fraction slip, s = Synchronous speed – Rotor speed Synchronous speed = Ns – N Ns and % Slip, s = Ns – N – × 100 Ns Sometimes, Ns – N is called the slip speed. Obviously rotor (or motor) speed is N = Ns (1 – s). Q.11. Draw the equivalent circuit of 3-phase induction motor. (S/2012) Draw and explain in brief equivalent circuit of 3-phase induction motor. (S/2016, 17) Ans. A equivalent circuit of a three phase induction motor is like a transformer equivalent circuit when it is standstill. However, it is rotating the frequency dependent on quantities like voltage, currents, reactants changes as they should change. But the induction motor, the input power and output power, when rotating are not same (because some power is given to the shaft) unlike transformer.

10 Electrical Machines–II Fig. 1.7 Equivalent Circuit Diagram where Rotor Quantities are Referred to Stator Circuit In transformer it is used properly to create equivalent circuit referred to primary side. The equivalent circuit of three phase induction motor is shown in fig. 1.7. In this diagram rotor quantities are referred to stator circuit. Here, R'2 = R2/K 2 X'2 = X2/K 2 and I'2 = KI2 Q.12. Draw the phasor diagram of an 3-φ induction motor on load and explain it. (S/2000) Ans. The phasor diagram of 3-φ induction motor is shown in fig. 1.8 in which V1 is applied voltage/stator phase, R1 and X1 are stator resistance and leakage reactance/phase respectively. The applied voltage V1 produces a magnetic flux which links both primary and secondary thereby producing a counter emf of self induction E1 in primary (stator) and a mutually induced emf Er (=sE2) in secondary (rotor). There is no secondary terminal voltage V2 in secondary because whole of the induced emf Er is used up in circulating the rotor current as the rotor is closed upon itself. Fig. 1.8 Phasor Diagram of Three Phase Induction Motor on Load

Three Phase Induction Motor 11 Obviously, V1 = E1 + I1R1 + jI1X1 The magnitude of Er depends on voltage transformation ratio K between stator and rotor and slip. As it is wholly absorbed in the rotor impedance. ∴ Er = I2Z2 = I2 (R2 + jsX2) In the phasor diagram, I0 is the no-load primary current. It has two components are as follows: (i) The working or iron loss component Iw which supplies the no-load motor losses. (ii) The magnetising component Iµ which sets up magnetic flux in the core and the air gap. Obviously, I0 = √Iw2 + I2µ In the phasor diagram I2 is the equivalent load current in primary and is equal to KI2. Total primary current is the vector sum of I0 and I'2. Q.13. Derive the torque equation for 3-φ induction motor. (S/2011, S/15) Derive condition for maximum starting torque of a three phase induction motor. (2012, 14) Derive the torque equation for 3-phase induction motor. (2013, S/16, S/17, 17) Derive the condition for maximum starting torque for 3-φ induction motor. (2016) Ans. The torque of an induction motor, being due to the interaction of the rotor and stator fields, depends on the strength of those fields and phase relation between them. Mathematically, T ∝ φI2 cos φ2 Where, T = torque φ = rotating stator flux I2 = rotor current per phase cos φ2 = rotor power factor Since rotor emf/phase at standstill, E2 ∝ φ ...(i) ∴ T ∝ E2I2 cos φ2 or T = KE2I2 cos φ2 Where, K is constant. Let, Er be the rotor emf per phase under running conditions. So, Er = sE2 ∴ Rotor current per phase under running conditions,

12 Electrical Machines–II I2 = Er Zr = sE2 √R22 + (sX2)2 Also, cos φ2 = R2 √R22 + (sX2)2 By putting the value of I and cos in equation (i). ∴ Torque, T = KE2 × sE2 × R2 √R22 + (sX2)2 √R22 + (sX2)2 or T = sKR2E22 ...(ii) R22 + (sX2)2 At standstill when s = 1, obviously, Tstarting = KR2E22 R22 + (sX2)2 The condition for maximum torque may be obtained by differentiating the equation (ii) with respect to s and then putting it equal to zero. Let, Y = 1 T ∴ Y = R22 + (sX2)2 KsE22R2 = R2 + sX22 KsE22 KE22R2 ∴ dY R2 + X22 =0 = ds Ks2E 2 KE22R2 2 or R22 = sX22 or R2 = sX2 Therefore, maximum torque is given by the expression, Tmax = KE22 2X2 K= 3 2πNs

Three Phase Induction Motor 13 Tmax = 3 E22 2πNs 2X2 Newton - metre Hence, from the above expression of maximum torque it is clear that: (i) The maximum torque is independent of rotor circuit resistance. (ii) However, the speed or slip at which maximum torque occurs is determined by the rotor resistance. As seen above, torque becomes maximum when rotor reactance equals its resistance. Hence, by varying rotor resistance maximum torque can be made to occur at any desired slip. (iii) Maximum torque varies inversely at standstill reactance. Hence, it should be kept as small as possible. (iv) Maximum torque varies directly as the applied voltage. (v) For obtaining maximum torque at starting (s = 1) rotor resistance must be equal to rotor reactance. Q.14. Explain the principle of working of a three phase induction motor. Draw its torque slip characteristics. Define slip of induction motor. (2009) Draw and explain slip torque characteristic. (2010) Write torque equation and explain torque slip characteristics of an induction motor. (S/2013) Draw and explain torque slip characteristics of 3-φ induction motor. (2013) Draw and explain torque slip characteristics of an induction motor. (S/2016) Draw and explain slip torque characteristics curve. (2016) Ans. Working Principle of a Three Phase Induction Motor: Refer Q.1. Slip of an Induction Motor: Refer Q.10. Torque Slip Characteristics: The torque speed curve of an induction motor is the curve of the internal torque of induction motor drawn against the slip of the motor. A family of torque/slip curves is shown in fig. 1.9 for a range of s = 0 to s = 1. Torque, T = KsE22 R2 √R22 + sX22 It is clear that when s = 0, T = 0, hence, the curve starts from point 0. For low values of slip, there is little change in the denominator of the above expression, and the torque varies nearly in direct proportion to the slip, so that this portion of the torque slip characteristic is nearly a straight line. As the slip increases beyond this range, the value of the denominator will increase more and more rapidly with change of slip.

14 Electrical Machines–II Fig. 1.9 At a certain value of slip, the denominator will vary in direct proportion to the slip. For any increase in slip beyond this value, the denominator will increase at a greater rate than the numerator, and the torque will decrease. The result is that the motor slows down and eventually stops. The motor operates for the slip value between zero and that corresponding to maximum torque. s = R2 X2 (a) Torque Speed Curves (b) Torque Slip Curves Fig. 1.10 With higher slip, R2 becomes negligible as compared to sX2 and torque varies as, T ∝ 1 , i.e. s speed torque or slip torque curves are rectangular hyperbola with the speed or slip beyond that corresponding to maximum torque. Torque speed and torque slip curves are shown in fig. 1.10 respectively for various values of rotor resistance. Q.15. Explain why an induction motor at no load operates at a very low pf ? Ans. The current drawn by an induction motor running at no-load is largely a magnetizing current, so no-load current lags behind the applied voltage by a large angle. Therefore, the pf of a lightly loaded or

Three Phase Induction Motor 15 no-load of an induction motor at a very low. At no-load, the value of pf is approximately 0.1 and power input is equal to core losses, stator copper loss and friction and windage losses. Q.16. Draw speed torque characteristics and explain how the starting torque of an induction motor can be increased? (2008) Ans. Speed Torque Characteristics: Refer Q.14. Starting Torque of a Induction Motor: The torque developed by an induction motor depends on its speed but the relation between them cannot be shown by a simple equation. But these relationship shown in the form of a curve as shown in fig. 1.11. Fig. 1.11 In fig. 1.11, T shows the nominal full-load torque of the motor. At starting torque (at N=O) is 1.5T and the maximum torque is 2.5T. At full load, the motor runs at a speed of N. When load increases, motor speed decreases till the motor torque again becomes equal to the load torque. As long as, the two torques are in balance, the induction motor will run at constant speed. In this way starting torque of an induction can be increased. Q.17. Explain why a 3-φ induction motor is self-starting whereas 1-φ induction motor is not self-starting? (S/2010) Ans. When the three phase stator winding, are fed by a three phase supply, then a magnetic flux of constant magnitude, but rotating at synchronous speed is set up. An emf induced in the rotor winding by the rotating magnetic field so that the current flows. Since, the rotor windings forms a closed circuit, rotor current is produced whose direction as given by Lenz’s Law is such as to oppose the cause producing it. By this torque is produced in the rotor it rotates the rotor in the direction of rotating field. Now, the relative speed of the rotor and stator is same of both. So, the rotor does not need any supply or source. Due to electro magnetic induction current is induced in the rotor and the

16 Electrical Machines–II motor starts run. That is why the three phase induction motor is self- starting. When the stator winding of single phase induction motor is connected to single phase AC supply, a magnetic field is developed, whose axis is always along the axis of stator coils. With alternating current in the fixed stator coil the mmf wave is stationary in space but pulsates in magnitude and varies sinusoidally with time. Currents are induced in the rotor conductors by transformer action these currents being in such a direction as to oppose the stator mmf. Thus, the axis of the rotor mmf wave coincides with that of the stator field, the torque angle, is therefore zero and no torque is developed at starting. However, if the rotor of such a motor is given a push by hand or by another means in either direction, it will pickup the speed and continue to rotate in the same direction developing operating torque. Thus, a single phase induction motor is not self-starting and needs a special starting means. Q.18. Describe double squirrel cage induction motor. (2002) Write short note on double squirrel cage induction motor. (S/2003, S/05) Ans. Among the many devices developed for improving the starting performance, the simplest is the double squirrel cage motor. The double squirrel cage motor is designed to provide a high starting torque with a low starting current. The rotor is so designed that the motor operates with the advantages of a high resistance rotor circuit during starting and a low resistance rotor circuit under running conditions. The starting torque of the double squirrel cage motor ranges from 200 to 250% of full load torque with a starting current of 400 to 600% of full load value. It is classified as a high-torque low starting current motor. The rotor of double squirrel cage motor carries two squirrel cage winding embedded in two rows of slots. The outer slots contain a high resistance and low leakage reactance winding and the inner slots contain a low resistance and high leakage reactance winding as shown in fig. 1.12. At starting, the rotor current has the same frequency as the supply current, therefore, leakage Fig. 1.12 reactance of inner winding is high. As a result the phase difference

Three Phase Induction Motor 17 between emf and current in the inner winding will be great and torque will be small. Because of high resistance and low leakage reactance of the outer winding, the phase difference between the emf and current in it will be small and therefore, high starting torque will be produced by it. At normal speed the leakage reactance of the squirrel cage windings is of minor importance and the current divides between them almost inversely as their resistances. Because of low resistance inner winding Fig. 1.13 carries bulk of the current and provides greater proportion of torque. Thus, the outer winding produces the high starting and accelerating torque, while the inner winding provides the running torque at good efficiency. The torque speed characteristics of individual cages as well as resultant are shown in fig. 1.13. The cost of double squirrel cage motor is 20 to 30% higher than that of ordinary squirrel cage motor. The motor is particularly suited where both high starting torque and a small slip on full load are required. Q.19. Draw the equivalent circuit of double cage induction motor. Also write the value of total impedance as referred to primary? Ans. The equivalent circuit of double cage induction motor is shown in fig. 1.14. Here, R1 and X1 are the resistance and reactance per phase of stator windings. R0'/s and Ri'/s are resistance of outer and inner rotors as referred to stator respectively and X0 and X1 their reactances. Since, both cages are completely link the main flux so the impedances of two cages considered in parallel. Total impedance as referred to primary is given by,

18 Electrical Machines–II Fig. 1.14 1 Z01 = R1 + jX1 + 1/Zi' + 1/Z0' = R1 + jX1 + Zi' Z0' Zi' + Z0' Q.20. Why the no-load current of a three phase induction motor is so high? While in case of a transformer it is very low. Explain. Ans. In the transformer the no-load current I0 is very low (about 1% of the full load current). The reason is that the magnetic flux path lies almost completely in steel core of low reluctance, hence magnetising current component Iµ of I0 is small, with the result that I0 is small. But in an induction motor, the presence of an air gap of high reluctance to the magnetic flux necessitates a large Iµ and hence, I0 is very large (approximately 40 to 50%) of the full current. Q.21. In case of a three phase motor, if the fuse blows off in one of the lines while the motor is running, what will happen to the motor? Explain your answer. Ans. In case of a three phase motor, if the fuse blows off in one of the lines while the motor is running some possibility may occur which are as follows: (i) If motor is already running and carrying half load or less, the motor will continue running as a single phase motor on the remaining two phases without damage because half loads do not blow normal fuses. (ii) If motor is very heavily loaded, then it will stop under single phasing (one fuse blows) and since, it can neither restart nor blow out the remaining fuses, the burnout is very prompt. A stationary motor will not start with one line broken. In fact, due to heavy standstill current it is likely to burn out quickly unless immediately disconnected.

Three Phase Induction Motor 19 Q.22. What is the effect of air gap of an induction motor on breakdown torque and why? Ans. The effect of air gap of an induction motor on breakdown torque is that it will increase the air gap and decreases the breakdown torque of an induction motor. Since, the leakage reactance of an induction motor increases with the increase of the length of air gap and the breakdown torque of an induction motor decreases with the increase in leakage reactance, the breakdown torque of an induction motor decreases with the increase in air gap. Q.23. How does the breakdown torque of an induction motor depend on the voltage regulation of the circuit at which it is connected? Ans. Since, the breakdown torque of an induction motor varies as the square of the applied voltage, good voltage regulation of the circuit at which the induction motor is connected, gives better breakdown torque of the same. Q.24. What is the reason of better performance of the low frequency induction motor than that of high frequency induction motor? Ans. Since, the leakage reactance of both stator and rotor windings of induction motor are proportional to the primary frequency of the induction motor and the maximum torque or breakdown torque of the induction motor decreases with the increase of the above reactances, the breakdown torque of the induction motor will decrease with the increases of supply frequency. Hence, better performance will be obtained from low frequency induction motor than that of high frequency induction motor. Q.25. Develop the relation among f, P and N. Ans. For P poles in the machine, there will be P/2 pair of poles hence, the magnetic field will rotate through f or 2f/P rotation f P/2 cycles of the current. Thus, synchronous speed Ns is given by, f 2f Ns = P/2 = P rpm or Ns = 120 f rpm P Q.26. Why do we need starter to start an induction motor? Explain with neat sketch DOL starter. (2010) Draw and explain direct on line starter for 3-phase induction motor. (2012, 15) What is the need of starter in an induction motor? Explain with neat sketch DOL starter. (2013)

20 Electrical Machines–II Ans. Need of Starter: A starter is needed for a three phase induction motor because at the time of starting, if induction motor is started directly, it will draw large amount of current which causes damage to adjoining equipments. Thus, a starter is needed in order to limit the starting current. After the motor has started it reduced starting current and hence reduced voltage, the connections are diverted towards the main supply so that the motor can run at higher starting current and voltage. Direct On Line (DOL) Starter: The push button type direct on line starter shown in fig. 1.15. This method is simple, in expensive and easy to install and maintain. It has a set of start and stop push buttons, a contactor (an electromagnet) with its associated contacts and usually an overload and under-voltage protection devices. The start button is normally open by a spring. The stop button is closed by a spring. When the start button is pressed the operating coil is energized through the overload relay contacts (normally closed). This closes the three main contacts M that connects the motor to the supply. At the same time a set of auxiliary or maintaining contacts MC are closed. When the maintaining contacts MC closed a new circuit is established through stop button, maintaining contacts MC and operating coil. Since, the operating coil circuit is maintained by the auxiliary contacts MC, the Fig. 1.15 Wiring Diagram of a Direct On Line Starter

Three Phase Induction Motor 21 start button may be released. When the stop button is pressed, the coil is de-energized, thereby opening all main contacts and auxiliary contacts. If the supply fails or line voltage drops below a certain value, the main contacts and the maintaining contacts are both opened. Upon return of the supply, the contactor cannot close untill the start button is again closed. Because a contactor is controlled by a three wire control circuit maintains the interruption of the circuit even after the supply is restored, it is said to provide under voltage protection for the motor. This protection is employed when it is desired to prevent the unexpected starting of a motor. Q.27. Write a short note on star delta starter. (S/2010) Explain the working of star delta starter with diagram. (2011) Explain star delta starter with neat sketch. (S/2012) Ans. Star delta starter, used for squirrel cage motors, utilises the principle of reduced voltage starting. All the six terminals of three phase induction motor are brought to the starter as shown in fig. 1.16. Infact star delta starter is simply a two way switch. The method is based upon the principle that with three windings connected in star, the voltage across each winding 1/√3 i.e., 57.7% of the line-to-line voltage whereas the same winding connected in delta will have full line-to-line voltage across each. The start delta starter connects three stator windings in star across the rated supply voltage at the starting instant. After the motor attains speed, the same windings through a change over switch are connected in delta across the same supply voltage. The stator is also provided with a mechanical interlocking device to prevent the handle from being put in the run position first. Such starters are employed for starting three phase squirrel cage induction motors of rating between 4KW and 15KW. The basic diagram of connection is shown in fig. 1.16. Fig. 1.16 Star Delta Switch

22 Electrical Machines–II Since, at starting instant, the stator windings are connected in star, so voltage across each phase winding is reduced to 1/√3 of line voltages and therefore, starting current per phase becomes equal to Isc /√3. Starting line current by connecting the stator windings in star at the starting instant = Starting motor current per phase = Isc /√3. Starting line current by direct switching with stator windings connected in delta = √3 Isc ∴ Line current with star delta starting = Isc /√3 = 1 Line current with direct switching √3 Isc 3 Hence, by star delta starting line current is reduced to one third of line current with direct switching. By this method no power is lost in starting the motor. Q.28. Explain the working of autotransformer starter with diagram. (2011) Describe autotransformer starter with the help of a neat diagram. (S/2013) Draw neat diagram and explain working of a manual autotransformer starter for 3-phase induction motor. (S/2016) Ans. Though the method of obtaining reduced voltage by connecting resistors in series with the motor leads is simple and cheap but lots of power is wasted in the external series resistance. Autotransformer starters may be either manually or magnetically operated. The circuit diagram of an autotransformer starter is shown in fig. 1.17. The manual autotransformer starter is essentially a multi-pole double throw switch. It consists of three sets of contacts starting, running and the movable contacts. The operating handle is a spring loaded lever mounted on the outside of a steel cabinet to stand vertically in the off position. When the operating handle is moved to the start position, the movable contants are moved against the starting contacts. This energizes the star connected autotransformer and impresses reduced voltage across the motor. After the motor has accelerated to about full speed, the operating handle is moved to the run position. This instantly opens the starting contacts and connects the motor directly to the line through the running contacts. A latch, fixed to the mechanism, is made to drop into a notch so that the operator is prevented from throwing the handle accidently to the run position first, however, when the handle is quickly pushed from the start to the run position, the latch is kicked up to make the lever free for its forward motion. The operating handle is held in the run position by an electromagnet or under voltage coil until the stop button is pressed. If the supply fails or supply voltage drops to low value, the electromagnet

Three Phase Induction Motor 23 Fig. 1.17 Circuit Diagram of an Autotransformer Starter will release and trip the holding mechanism. This type of starter is suitable for the star connected as well as delta connected motors. Q.29. Draw circuit diagram for rotor resistance starting of slip ring induction motor. (S/2001) Explain in brief construction of 3-φ slip ring type induction motor with diagram. (2017) Ans. An extra starting resistance is connected across the slip ring terminals of the rotor while starting. The connection diagram is shown in fig. 1.18. High starting torque can be obtained by connecting extra resistance in series with the rotor circuit at starting. This extra resistance is gradually cut out of the circuit. Under running condition the speed of the motor can also be varied by varying the rotor circuit resistance in steps. Three phase supply is given to the stator windings with full extra

24 Electrical Machines–II Fig. 1.18 Starting of Slip Ring 3-φ Induction Motor or Rotor Resistance Starter resistance on the rotor using contactors. The rotor resistance gets cut out in steps as the motor picks up speed. This type of motor always started with full line voltages applied across the stator terminals. The value of starting current is adjusted by introducing a variable resistance in the rotor circuit. Q.30. Explain different types of starters used in induction motor starting. (S/2011, S/15) Why do we need starter to start an induction motor? Write various types of starters. (2016) What are the different types of starter used for 3-φ starting induction motor? (2017) Ans. Need Starter to Start an Induction Motor: Refer Q.26. Types of Starters: The different types of starters used in induction motor starting are as follows: (i) Direct On Line (DOL) Starter : Refer Q.26. (ii) Star Delta Starter: Refer Q.27. (iii) Autotransformer Starter: Refer Q.28. (iv) Rotor Resistance Starter: Refer Q.29. (v) Contactor/Automatic Starter: The automatic starter uses the automatic switches called contactors. The contactor is a switch whose operation depends on the solenoidal coil. Thus, its operation is

Three Phase Induction Motor 25 controlled electromagnetically. When the current passes through the coil, coil gets energized and operates the contacts of the contactor, due to magnetic field produced by it. Hence an automatic starter is also called contactor starter. The fig. 1.19 shows such type of contactor starter, with two steps of starting resistance R1 and R2. Fig. 1.19 Automatic Starter When the main supply is on, the field winding F1 and F2 gets supply as gets directly connected across the supply. It produces the required working flux. The C1 is the line contactor which is in series with the armature and connects the armature to the line with R1 and R2 when closed. The coil energized C1 as O1. The contactors A1 and A2 are accelerating contactors which provide short circuit across R1 and R2 in sequence, to cut out them from the armature circuit, when motor gets accelerated. The control of the entire circuit is achieved using start and stop push buttons named as S1 and S2 respectively, in fig. 1.19. The O2 is the coil to operate A1 while O3 is the coil to operate A2. When S1 is pushed which is normally open, the O1 gets energized and operates C1. Thus, current I2 starts flowing through R1 and R2 as A1 and A2 are open and is less due to R1 and R2. M1 is auxiliary contactor which operates when C1 operates. Thus, through S1 becomes open, O1 continues to be energized through M1 as S2 is normally off. As motor gathers speed, the back emf Eb increases. The coils O2 and O3 are directly connected across the armature so current through them is proportional to Eb. The coil design is such that when motor speed reaches upto 60% of rated speed then O2 gets energized sufficiently to operate A1. When A1 is closed, it provides short circuit

26 Electrical Machines–II across R1 and it gets cut out from armature. Similarly, at about 80% of rated speed O3 gets sufficiently energized, operating A2. This short circuits R2. Thus, motor safely attains a rated speed without any starting resistance in series with the armature. Q.31. Explain crawling in 3-φ induction motor. (2010) Ans. Sometimes, squirrel cage induction motors exhibits a tendency to run at very low speeds (as low as one seventh of their synchronous speed). This phenomenon is called as crawling of an induction motor. This action is due to the fact that, flux wave produced by a stator winding is not purely sine wave. Instead, it is a complex wave consisting a fundamental wave and add harmonics like 3rd, 5th, 7th, etc. The fundamental wave revolves synchronously at synchronous speed Ns whereas 3rd, 5th, 7th harmonics may rotate in forward or backward direction at Ns/3, Ns/5, Ns/7 speed respectively which is shown in fig. 1.20. Hence, harmonic torque are also developed in addition with fundamental torque. 3rd harmonics are absent in a balanced three phase system. Hence, 3rd harmonics do not produce rotating field and torque. The total motor torque consists of three components are as follows: (i) The fundamental torque with synchronous speed Ns. (ii) 5th harmonic torque with synchronous speed Ns/5. Fig. 1.20

Three Phase Induction Motor 27 (iii) 7th harmonic torque with synchronous speed Ns/7 (provided that higher harmonics are neglected). Now, 5th harmonic currents will have phase difference of 5 × 1200 = 6000 = 2 × 3600 – 1200 = –1200. Hence, the revolving speed setup will be in reverse direction with speed Ns/5. The small amount of 5th harmonics torque produces braking action and can be neglected. The 7th harmonic currents will have phase difference of 7 × 1200 = 8400 = 2 × 3600 + 120 = 1200. Hence, they will set up rotating field in forward direction with synchronous speed equal to Ns/7. If we neglect all the higher harmonics, the resultant torque will be equal to sum of fundamental torque and 7th harmonics torque. 7th harmonics torque reaches its maximum positive value just before 1/7th of Ns. If the mechanical load on the shaft involves constant load torque, the torque developed by the motor may fall below this load torque. In this case, motor will not accelerate upto its normal speed, but it will run at a speed which is nearly 1/7th of its normal speed. This phenomenon is called as crawling in induction motors. Q.32. Enlist speed control methods of 3-φ induction motor and explain variable frequency control method. (S/2010) Classify various methods of speed control of three phase induction motor. (2012, 14, S/15, 15) Write different methods of speed control of induction motor. (S/2013) Explain different methods of speed control of induction motor (3-φ). (S/2017) State the factors on which speed of 3-φ induction motor depends? Explain various methods of speed control. (2017) Ans. The different methods in which speed control of induction motor depends are as follows: (i) By Changing Applied Voltage: This method is the cheapest and the easiest, and rarely used because for the reason are as: (a) A large change in voltage is required for a relatively small change in speed. (b) This large change in voltage will result in a large change in the flux density thereby seriously disturbing the magnetic conditions of the motor. (ii) By Changing the Applied Frequency: This method is also used very rarely. However, this method could only be used in cases where the induction motor happens to be the only load on the generators, in which the case supply frequency could be controlled by controlling the speed of the prime movers of the generators. But, here again the

28 Electrical Machines–II range over which the motor speed may be varied is limited by the economical speeds of the prime movers. This method has been used to some extent on electrically-driven ships. (iii) By Changing the Number of Stator Poles: This method is easily applicable to squirrel cage motors because the squirrel cage rotors adopts itself to any reasonably number of stator poles. From the equation Ns = 120 f/P it is clear that the synchronous speed (and hence the running speed) of an induction motor could also be changed by changing the number of stator poles. This change of number of poles is achieved by having two or more entirely independent stator windings in the same slots. Each windings gives a different number of poles and hence, different synchronous speed. For example, a 36 slots stator may have two 3-φ windings, one with 4 poles and the other with 6 poles. With a supply frequency 50 Hz, 4 poles winding will give Ns = 120 × 50 = 4 1500 rpm and the 6 poles winding will give Ns = 120 × 50 = 1000 rpm. 6 Motors with four independent stator winding are also in use and they give four different synchronous speeds. Of course, one winding is used at a time, the others being entirely disconnected. This method has been used for elevator motors, traction motors and also for small motors driving machine tools. (iv) Rotor Rheostate Control: This method is applicable to slip ring motors alone, the motor speed is reduced by introducing an external resistance in the rotor circuit. For this purpose, the rotor starter may be used, provided it is continuously rated. We now that near synchronous speed, T ∝ s R2 . It is obvious that for a given torque, slip can be increased, i.e., speed can be decreased by increasing the rotor resistance R2. One serious disadvantage of this method is that with increase in rotor resistance, I2R losses also increase which decrease the operating efficiency of the motor. In fact, the loss is directly proportional to the reduction in speed. Because of this wastefullness of this method, it is used where speed changes are needed for short period only. (v) By Injection an EMF in the Rotor Circuit: In this method the speed of an induction motor is controlled by injecting a voltage of slip frequency in the secondary circuit. If injected voltage is in phase with the induced emf in the secondary circuit, it is equivalent to decreasing of resistance in the secondary circuit, therefore, slip decreases or speed increases. If injected voltage is in phase opposition

Three Phase Induction Motor 29 to the induced emf in the secondary circuit, it is equivalent to increasing of resistance in the secondary circuit, hence, resulting in increase in slip or reduction in speed. (vi) Cascade or Tandem Control: In this method two motors are required at least one of which must have a wound rotor. It is shown in fig. 1.21. The two motors may be mechanically coupled together to drive a common load. If the stator winding of one of them, a wound rotor motor is connected to three phase AC supply, and its rotor winding is connected to the stator winding of the second motor the speed of the combination will be determined by the sum or difference of the number of poles in the two machines. In practice, it is customary to connect the rotor output of the first machine to the stator of second machine in such a way that the revolving field of both are in the same direction, under the condition, the resulting synchronous speed will be determined by the following expression, Ns = 120 f rpm P1 + P2 Where, f = supply frequency P1 and P2 = number of poles on motor I and II respectively. Fig. 1.21 Cascade Connection of Induction Motors (vii) Speed Control by Kramer System: This method is used in large motors of 4000 kW or above. It is shown in fig. 1.22. This system has a rotary converter which converts the low-slip frequency AC power into DC power which is used to drive a DC shunt motor, mechanically coupled to the main rotor. The main motor is coupled to shaft of DC shunt motor. The slip ring of main motor are connected to rotary converter. The DC output of rotary converter is used to drive DC motor. Both these motors are excited from the DC bus bars or from an exciter. There is a field regulator which governs the back emf. Eb of DC motor and hence, the DC potential at the commutator of rotary converter. It further controls the slip ring voltage and speed of main motor.

30 Electrical Machines–II Fig. 1.22 Circuit Diagram of Kramer System The advantage of this method is that if the rotary converter is over excited, it will take a leading current which compensates for the lagging current drawn by main motor and hence, improves the power factor of the system. (viii) Speed Control by Scherbius Method: This system is also used for controlling the speed of large induction motors. The slip energy is not converted into DC and then fed to a DC motor. It is fed directly to a special three phase or six phase AC commutator motor called a scherbius machine. It is shown in fig. 1.23. The polyphase winding of motor B is supplied with the low frequency output of motor A through a regulating transformer. The commulator Fig. 1.23 Circuit Diagram of Scherbius Method

Three Phase Induction Motor 31 motor B is a variable speed motor and its speed is controlled by either varying the tappings on regulating transformer or by adjusting the position of brushes on motor B. (ix) Variable Frequency Control Method: For a three phase induction motor, stator voltage per phase is given by, V = √2π f1Nph1 φ1kw1 If the ratio of supply voltage V to supply frequency f1 is kept constant, the air gap flux φ1 remains constant. The starting torque is given by, Test = 3. V12 r2 ωs (r1 + r2)2 + (x1 + x2)2 Here, (r1 + r2) << (x1 + x2) and ωs = 2ω1 P So, Test = 3ρ V12 r2 . 2ω1 [ω12 (l1 + l2) 2] ( )=3ρ . V1 2 r2 2ω1 ω1 (l1 + l2) 2 The maximum torque is given by, Tem = 3 . V12 ωs 2 (x1 + x2) = 3ρ . V12 2ω1 2ω1 (l1 + l2) ( )= 3ρ . V1 2 1 4 ω1 (l1 + l2) If V1 of air gap flux φ1 is kept constant the maximum torque shown ω1 by above equation remains unaltered. The equation of Test shows that the starting torque increases even if air gap flux is kept constant. At low value of frequencies, the effect of resistances cannot be neglected as compared to reactances reduce the effect of magnitude of maximum torque. This method of speed control and its speed torque characteristics are shown in fig. 1.24.

32 Electrical Machines–II Fig. 1.24 For induction motor, the variable frequency control below the rated frequency is generally carried out at rated air gap flux by varying terminals voltage with frequency to maintain v/f ratio constant at the rated value. It operates in constant torque mode. Q.33. Explain blocked rotor test of an induction motor. (S/2010) Explain in details with circuit diagram: (i) no load test of induction motor, (ii) blocked rotor test of induction motor. (S/2011, S/15) Explain no load test of 3-φ induction motor. (2012, S/15) Explain no load and blocked rotor test in 3-φ induction motor. (S/2017, 17) Ans. No Load Test of Three Phase Induction Motor (Open Circuit Test): This test is performed to determine no load current I0, Fig. 1.25 No Load Test on Induction Motor

Three Phase Induction Motor 33 no load power factor, cosφ0, windage and friction losses, no load core losses, no load input, and no load resistance R0 and reactance X0. The no load test is performed with different values of applied voltage below and above rated voltage, while the motor is running light (without load). The power input W0, voltage V and current I0 are measured by watt meter W, volt meter V and ammeter A connected in the circuit as shown in fig. 1.25. Since motor is running without load, the power factor would be less than 0.5, hence, total power input will be equal to the difference of the two reading of the watt meters. At no load power input is equal to core losses stator copper losses and friction and windage losses. To determine friction and windage losses, the curve drawn between input watts and applied voltage is extended to intersect vertical axis at any point B as shown in fig. 1.26. Point B corresponds to power input when applied voltage is zero. Since, when applied voltage is zero, the stator copper losses and core losses are zero, therefore power input at no load and zero voltage given by OB represents the windage and friction losses. Fig. 1.26 If windage and friction losses determined that stator copper losses = 3I20R1 are subtracted from input at no load test, core losses corresponding to no load current I0. The total core losses Wc at current per phase I0 and applied voltage per phase is V. Im is the value of magnetising current, Ie is the energy current, R0 is the no load resistance, X0 is no load reactance will determined by following relation: No load power factor, cosφ0 = Wc 3VI0 Ie = I0 cosφ0 = I0 × Wc = Wc 3VI0 3V

34 Electrical Machines–II Im = √I02 – Ie2 V R0 = Ie and X0 = V Im Blocked Rotor Test (Short Circuit Test): This test is performed to determine the short circuit current Isc with normal applied voltage to stator, power factor on short circuit, total equivalent resistance and reactance of the motor as referred to stator. In this test the rotor is held firmly (rotor windings are short circuited at slip rings in case of wound rotor motor) and stator is connected across supply of variable voltage. This test is just equivalent to short circuit test on transformer. The connection diagram remains the same as in case of no load test. The graph between stator current and applied voltage if drawn, will be a straight line, which is shown in fig. 1.27. If Vs is the applied voltage per phase and causes current Is per phase in the stator winding and Ws is the total input voltage at short circuit then. Fig. 1.27 Short circuit current per phase Isc with normal voltage V applied across motor is given by, V Isc = Is Vs and power factor, cosφs = Ws √3 VscIsc Since, input on short circuit meets with stator and rotor copper losses and core losses. Core losses being very small can be neglected, therefore, Input on short circuit, Ws = 3Is2 R'1

Three Phase Induction Motor 35 or equivalent resistance per phase as referred to stator, R'1 = Vsc /√3 = r1 + r'2 I2sc Motor equivalent impedance per phase as referred to stator, Z'1 = Vsc /√3 Isc Motor equivalent reactance per phase as referred to stator X'1 = √(Z'1)2 – (R'1)2 = x1 + x'2 (2010) Q.34. Explain the braking of an induction motor. Explain regenerating types of braking in an induction motor. (S/2012) Write a short note on braking of 3-φ induction motor. (S/2013, 15) Ans. In many applications, it is very necessary to stop an induction motor and its coupled load quickly. During electrical braking conditions, a motor operates under severe duty cycle. Therefore, braking is used only when it is highly desirable to control the braking time, deceleration etc. There are three methods of braking which are as follows: (i) Regenerative Braking: This method is used when the load forces the motor to run above synchronous speed. When the rotor speed is more than rotating field speed, the slip becomes negative. Its means machine now operates as an induction generator in second quadrant as shown in fig. 1.28. It returns power to the supply line. This method Fig. 1.28 Regenerative Braking

36 Electrical Machines–II is useful only in squirrel cage motors because in this motors the stator and rotor poles must be changed simultaneously. (ii) Plugging: In this method any two supply leads are interchanged so that the direction of rotating field gets reversed. A forward rotating field is that in which it rotates in the direction of rotor and soon after the interchange of any two leads, this field becomes backward rotating field. In induction motor, electromagnetic torque is developed in the direction of rotating field. Therefore, backward rotating field torque T1 is opposite to the motor rotation and the braking torque brings the rotor to stop quickly. The terminals should be disconnected as soon as the motor comes to stop, unless motor will start running in backward direction. It is shown in fig. 1.29. Plugging involves heavy loss of power but the motor can be brought to rest quickly. Fig. 1.29 (iii) Dynamic or Rheostatic Braking: This method is achieved if its stator winding, after being disconnected from three phase supply, is switched on to a DC source. Under normal operating conditions, the stator field moves faster than the rotor conductors by slip speed sNs. As a consequence poles created on the rotor interact with stator poles to produce anticlock wise torque which is called dynamic braking torque. It is shown in fig. 1.30. (a) (b) Fig. 1.30 Speed Torque Characteristics during DC Dynamic Braking

Three Phase Induction Motor 37 Under normal running, the rotor frequency is sf1 and at the instant dynamic braking is applied, the rotor frequency becomes (1 – s) f1 = sf1 During dynamic braking, the entire power developed in the rotor is dissipated in the rotor circuit resistance. It gives smooth stop, results in less rotor circuit loss and has no tendency to reverse. Q.35. Write down the various losses occur in three phase induction motor. Ans. The various losses occur in three phase induction motor are as follows: (i) Iron or Core Losses: Iron or core losses are further divided into hysteresis and eddy current losses. Eddy-current losses are minimized by using laminations. Since, by laminating the core, area decreases and hence, resistance increases which results in decrease in eddy-currents. Hysteresis losses are minimized by using high grade silicon steel. The core losses depend upon frequency. The frequency of stator is always supply frequency, f and the frequency of the rotor is slip times the supply frequency, (sf) which is always less than the stator frequency. Hence, the rotor core loss is very small as compared to stator core loss and is usually neglected in running conditions. (ii) Mechanical and Brush Friction Losses: Mechanical losses occur at the bearing and brush friction loss occurs in wound rotor induction motor. These losses occurs with the change in speed. In three phase induction motor the speed usually remains constant. Hence, these losses almost remains constant. (iii) Variable or Copper Losses: These losses occur due to current flowing in stator and rotor winding. As the load changes, the current flowing in rotor and stator winding also changes and hence, these losses also changes. Therefore, these losses are called variable losses. The copper losses are obtained by performing blocked rotor test on three phase induction motor. Q.36. Write the efficiency of three phase induction motor. Ans. Efficiency is defined as the ratio of the output to that of input. Efficiency, η = Output Input Rotor efficiency of three phase induction motor, = Rotor output Rotor input = Gross mechanical power developed Rotor input = Pm P2

38 Electrical Machines–II Three phase induction motor efficiency, = Power developed at shaft Electrical input to the motor η = Pout Pin NUMERICAL PROBLEMS AND SOLUTIONS Prob.1. A 4-pole, 50 Hz induction motor runs at 1455 rpm. Calculate its slip and also the rotor frequency. (2017) Sol. Given, P =4 Supply frequency, f = 50 Hz Actual speed of rotor, N = 1455 rpm Slip, s = Ns – N Ns and Ns = 120 f P = 120 × 50 4 = 1500 rpm s = Ns – N Ns = 1500 – 1455 1500 = 45 = 0.03 1500 Hence, Slip, s = 0.03 or 3% Ans. Rotor frequency, f ' = sf f ' = 0.03 × 50 f ' = 1.5 Hz Ans. Prob.2. A 4 pole, 3 phase induction motor operates from a supply whose frequency is 50 Hz. Calculate: (i) rotor speed at 5% slip, (ii) rotor current frequency at standstill. (S/1990) Sol. Given, P =4 f = 50 Hz (i) Ns = 120 f P

Three Phase Induction Motor 39 or Ns = 120 × 50 4 = 1500 rpm Slip, s = Ns – Nr Ns 0.05 = 1500 – Nr 1500 Nr = 1500 – 75 Ans. = 1425 rpm (ii) Rotor current frequency at standstill f ' = sf Ans. = 1 × 50 = 50 Hz Prob.3. As 8 pole alternator runs at 750 rpm and supplies power to a 6 pole induction motor which has full-load slip of 3%. Find the full- load speed of the induction motor which has full-load slip of 3%. Find the full-load speed of the induction motor and the frequency of its rotor emf. (S/2009) Sol. Supply frequency, f = Pa × Na 120 = 8 × 750 120 = 50 Hz Full-load slip, s = 3% = 0.03 Ns = 120 f Pm 120 × 50 = 6 = 1000 rpm Full-load speed, Nr = Ns (1 – s) = 1000 (1 – 0.03) = 970 rpm Ans. Frequency of rotor emf f ' = sf = 0.03 × 50 = 1.5 Hz Ans. Prob.4. A 3-φ, 4 pole, 50 Hz induction motor has a slip of 4%. Calculate the rotor speed and the frequency of rotor current. (2011) Sol. Given, P =4 f = 50 Hz

40 Electrical Machines–II Slip, s = 4 = 0.04 100 Ns = 120 f P = 120 × 50 = 1500 rpm 4 Rotor speed, N = Ns (1 – s) = 1500 (1 – 0.04) = 1440 rpm Ans. Frequency of rotor f' = s×f = 0.04 × 50 = 2 Hz Ans. Prob.5. A 550 V, 50 Hz, 6 pole, 3-φ induction motor develops 30HP including mechanical losses of 2HP at a speed of 950 rpm. If the power factor is 0.88, then calculate: (i) the slip, (ii) the rotor Cu loss, (iii) the total input when stator losses are 200 Watts. (2010) Sol. Synchronous speed, Ns = 120 f = 120 × 50 P6 = 1000 rpm Motor speed, N = 950 rpm Gross mechanical power developed, Pmech = 30HP = 30 × 746 W Mechanical losses = 22380 W = 2HP = 2 × 746 W = 1492 W (i) Slip, s = Ns – N Ns = 1000 – 950 1000 = 0.05 or 5% Ans. (ii) Rotor copper loss Pm = Pmech + Mechanical losses = 22380 + 1492 = 23872 W ∴ Rotor copper loss = s × Pm 1–s = 0.05 × 23872 1 – 0.05 = 1256.42 W Ans.

Three Phase Induction Motor 41 (iii) Total power input = Input power to rotor + Stator losses = 23872 + 200 = 24072 W Ans. Prob.6. A 440 V, 50 Hz, 6 pole, 3 induction motor develops power of 30 HP including mechanical losses of 2HP at a speed of 950 rpm. If the power factor is 0.8, then calculate: (i) the rotor copper loss, (ii) the slip, (iii)the total input when stator losses are 200 watts. (2015) Sol. Given that, Mechanical losses, = 2 HP Power developed, = 2 × 746 W = 1492 W = 30 HP = 30 × 746 W = 22380 W Motor speed, N = 950 rpm Synchronous speed, Ns = 120f P = 120 × 50 6 = 1000 rpm (i) Slip, s = Ns – N Ns = 1000 – 950 1000 = 0.05% = 5% Ans. (ii) Total rotor developed = 22380 + 1492 power Pm = 23872 W ∴ Rotor copper loss = 1 s s × Pm – = 1 0.05 × 23872 – 0.05 = 1256.42 W Ans. (iii) Total power input = Rotor power + Stator losses = 23872 + 200 = 24072 W Ans. Prob.7. The rotor emf of a 3 phase, 6 pole, 400 V, 50 Hz induction motor has a frequency of 3 Hz. Compute the speed and percentage slip of the motor. Find the rotor Cu loss per phase if full input of the rotor is 111.9 kW. (2012) Sol. Stator supply frequency, f = 50 Hz Rotor emf frequency, fr = 3 Hz

42 Electrical Machines–II Percentage slip, s = fr × 100 = 3 × 100 f 50 = 6% Ans. Speed of motor, N = Ns (1 – s) = 120 f (1 – s) P = 120 × 50 (1 – 0.06) 6 = 940 rpm Ans. Rotor input = 111.9 kW Rotor copper loss = s × Rotor input = 0.06 × 111.9 = 6.714 kW Rotor copper loss per phase = 6.714 3 = 2.238 kW Ans. Prob.8. A 400 V, 50 Hz, 6 pole, 3-φ induction motor runs at 950 rpm and develops 14.92 kW. If windage and friction losses 1 HP and stator losses are 1 kW and if the power factor is 0.85 then, calculate: (i) the slip, (ii) the rotor Cu loss, (iii) input power, (iv) line current. (2013) Sol. Given, P =6 f = 50 Hz Stator losses = 1 kW (i) Ns = 120f P = 120 × 50 6 = 1000 rpm Slip, s = Ns – N Ns = 1000 – 950 1000 = 0.05 or 5% Ans. (ii) Rotor copper losses = s × Rotor gross output 1–s = 0.05 × 14.92 1 – 0.05 = 0.05 × 14.92 0.95 = 0.78 kW Ans.

Three Phase Induction Motor 43 (iii) Total input power to stator = Rotor input + Stator losses ∴ Rotor input = Rotor Output 1–s = 14.92 = 15.70 kW Ans. 1 – 0.05 ∴ Total input to stator = 15.70 + 1 = 16.70 kW (iv) Line current, IL = 16.70 × 103 400 = 41.75 A Ans. Prob.9. The power input to a 500 volt, 50 Hz, 6 pole, 3-φ induction motor running at 975 rpm is 40 kW. The stator losses are 1 kW and friction and windage losses are 2 kW. Calculate: (i) the slip, (ii) rotor copper loss, (iii) number of complete cycles of rotor emf per minute. (S/1990, 97, 2001) Sol. Ns = 120 f P = 120 × 50 6 = 1000 rpm (i) Slip, s = Ns – Nr Ns = 1000 – 975 1000 = 0.025 Ans. (ii) Motor input = 40 kW Stator losses = 1 kW ∴ Rotor input = 40 – 1 Ans. = 39 kW ∴ Rotor copper loss = s × Rotor input = 0.025 × 39 = 0.975 kW (iii) Frequency of rotor emf f ' = sf = 0.025 × 50 = 1.25 Hz Number of complete cycles of rotor emf per minute = 1.25 × 60 Ans. = 75

44 Electrical Machines–II Prob.10. A 415 volt, 6 poles, 50 Hz, 3 phase induction motor develops 20 kW include of mechanical losses when running at 995 rpm and 0.8 power factor lagging. Calculate: (i) the slip, (ii) the total input, if the stator losses are 500 watts, (iii) the line current, (iv) the rotor copper losses. (S/1992) Sol. (i) Ns = 120 f P = 120 × 50 6 = 1000 rpm Slip, s = Ns – Nr Ns = 1000 – 995 1000 = 0.005 Ans. (ii) Rotor input = Rotor output 1–s = 20 1 – 0.005 = 20.1 kW ∴ Total input to stator = Rotor input + Stator losses = 20.1 + 0.5 = 20.6 kW Ans. (iii) Line current, IL = 20.6 × 103 415 = 49.64 A Ans. (iv) Rotor copper losses = s × Rotor gross output 1–s = 1 0.005 × 20 – 0.005 = 0.1 kW Ans. Prob.11. The power input to a 3-phase induction motor is 60 kW. The stator loss is 1 kW. Find the total mechanical power developed and the rotor copper loss per phase, if the motor is running with a slip of 3%. (1989) Sol. Given, Slip, s = 0.03 Motor input = 60 kW Stator loss = 1 kW Power input to rotor = Motor input – Stator loss = 60 – 1 = 59 kW

Three Phase Induction Motor 45 Rotor copper loss = Slip × Power input to rotor = 0.03 × 59 = 1.77 kW Rotor copper loss/phase = 1.77 3 = 0.59 kW Ans. Total mechanical power developed, = Rotor input (1 – s) = 59 (1 – 0.03) = 57.23 kW Ans. Prob.12. A 3φ, 415 V, 50 Hz induction motor with 6 poles develops 14.92 kW at 950 rpm at a pf of 0.86 lagging. The total mechanical losses are 0.746 kW, stator losses 150 W. Calculate the line current. (1993) Sol. Synchronous speed, Ns = 120 f P = 120 × 50 6 = 100 rpm Slip, s = Ns – Nr Ns = 1000 – 950 1000 = 0.05 or 5% Mechanical power developed by rotor, Pm = 14.92 + 0.746 = 15.666 kW Rotor copper loss =s × Pm 1–s = 0.05 × 15.666 1 – 0.05 = 0.824 kW Rotor input = 15.666 + 0.824 = 16.49 kW Stator input = Rotor input + Stator losses = 16.49 + 1.5 = 17.99 kW Line current, IL = Stator losses √3 × V × cos φ ∴ IL = 17990 √3 × 415 × 0.86 Line current, IL = 29 A Ans.

46 Electrical Machines–II Prob.13. The power input to the rotor of a 400 V, 50 Hz, 6 pole, 3-φ induction motor is 20 kW. The slip is 3% Calculate: (i) the frequency of the rotor current, (ii) rotor speed, (iii) rotor Cu losses, (iv) rotor resistance / phase if rotor current is 60 A. Sol. (i) Frequency of rotor current = sf = 0.03 × 50 Ans. = 1.5 Hz (ii) Ns = 120 × 50 6 = 1000 rpm ∴ Rotor speed, Nr = Ns (1 – s) = 1000 (1 – 0.03) = 970 rpm Ans. (iii) Rotor Cu loss = s × rotor input = 0.03 × 20 = 0.6 kW = 600 W Ans. (iv) Rotor Cu loss/phase = 600 = 200 W 3 ∴ I22R2 = 200 602R2 = 200 ∴ R2 = 200 60 × 60 = 0.055Ω Ans. Prob.14. A, 3-φ, 6 pole, 50 Hz induction motor develops 5 HP at 950 rpm. What is the stator input and efficiency of the motor if the stator loss is 300 W? (S/2010) Sol. Given, Number of poles, P = 6 Frequency, f = 50 Hz Speed of motor, N = 950 rpm Developed mechanical power, Pm = 5 HP = 5 × 746 W = 3730 W Stator loss = 300 W Synchronous speed, Ns = 120 f P = 120 × 50 = 1000 rpm 6

Three Phase Induction Motor 47 Slip, s = Ns – N = 1000 – 950 Ns 1000 = 0.05 Rotor input = Pm = 3730 1–s 1 – 0.05 = 3926.3 W Stator input = Rotor input + Stator loss = 3926.3 + 300 = 4226.3 W Ans. Efficiency of motor, η = Pm × 100 Stator input = 3730 × 100 4226.3 = 88.2% Ans. Prob.15. A 440V, 6 pole, 50Hz induction motor run at 950 rpm. Its output power at 0.8 power factor is 50 kW. Its stator loss is 1 kW and friction and windage loss is 2 kW. Calculate: (i) slip, (ii) rotor copper loss, (iii) efficiency. (S/2011, S/17) Sol. Given, Number of poles, P = 6 Frequency, f = 50 Hz Speed of rotor, N = 950 rpm Pf, cosφ = 0.8 Input power = 50 kW Stator losses = 1 kW Friction and windage losses = 2 kW (i) Synchronous speed, Ns = 120 f P = 120 × 50 = 1000 rpm 6 Slip, s = Ns – N × 100 Ns = 1000 – 950 × 100 1000 = 5% or 0.05 Ans. (ii) Developed mechanical power, Pm = Input power + Friction and windage losses = 50 + 2

48 Electrical Machines–II = 52 kW Rotor Output = Pm = 1 52 1–s – 0.05 = 54.74 kW Motor input = Output + Stator loss = 54.74 + 1 = 55.74 kW Rotor copper losses = Rotor output – Mechanical power = 54.74 – 52 = 2.74 kW Ans. (iii) Efficiency, η = Output power of motor × 100 Input power of motor = 50 × 100 55.74 = 89.7% Ans. Prob.16. The power input to a 3-phase induction motor is 40 kW. The stator losses are 1 kW and windage and friction losses are 2 kW. Find the mechanical power developed by motor and motor efficiency if the motor is running at 3% slip. (S/2012) Sol. Given, Slip, s = 3% or 0.03 Stator losses = 1 kW Windage and friction losses = 2 kW Power input to rotor = Motor input – Stator loss = 40 – 1 = 39 kW Total mechanical power developed, = Rotor output = Rotor input – Rotor copper loss = Rotor input (1 – s) = 39 (1 – 0.03) = 37.83 kW Ans. Rotor input = Pm = 37.83 × 103 1 – s (1 – 0.03) = 39000 W Stator input = Rotor input + Stator loss = 39000 + 1000 = 40000 W Efficiency of motor, η = Pm × 100 Stator input = 39000 × 100 40000 = 97.5% Ans.

Three Phase Induction Motor 49 Prob.17. A 100 kW (output), 3300 V, 50 Hz, 3-φ star connected induction motor has a synchronous speed of 500 rpm. The full-load slip is 1.8% and full load pf 0.85, stator Cu loss = 2440 W, iron loss = 3500 W, rotational losses = 1200 W. Calculate: (i) the rotor Cu loss, (ii) the line current, (iii) the full load efficiency. Sol. Mechanical power developed, Pm = Output + Rotational losses = 100 + 1.2 = 101.2 kW (i) Rotor Cu loss = s × Pm 1–s = 0.018 × 101.2 1 – 0.018 = 1.855 kW Ans. (ii) Rotor input, P2 = Pm + Rotor Cu loss = 101.2 + 1.855 = 103.055 kW Stator input = P2 + Stator Cu and iron losses = 103.055 + 2.44 + 3.5 = 108.995 kW ∴ √3 VLIL cosφ = 108.995 Line current, IL = 108.995 √3 × 3300 × 0.85 IL = 22.4 A Ans. (iii) Full load efficiency = 100000 108995 = 0.917 or 91.7% Ans. FF

50 Electrical Machines–II 2 SYNCHRONOUS MOTOR Chapter SUBJECTIVE QUESTIONS AND ANSWERS Q.1. Write the principle of operation of a synchronous motor. (2008, S/16, S/17) What is synchronous motor? Explain the operation of synchronous motor with neat diagram. (2011) Explain the working principle of synchronous motor. (S/2013, 13) Why a synchronous motor is not self-starting? Ans. A synchronous electric motor is an AC motor in which, motor is at steady state, the rotation of the shaft is synchronized with the frequency of the supply current, the rotation period is exactly equal to an integral number of AC cycles. Synchronous motor contain electromagnets on the stator of the motor that create a magnetic field which rotates in time with the oscillation of the line current. The rotor turns in step with this field, at the same rate. Principle of Operation and Reason for Synchronous Motor is not Self-Starting: For the sake of simplicity let take a bipolar machine i.e., the stator winding is designed for two polar winding and the rotor has also two poles i.e., North and South. When three phase supply is given to the stator winding, the stator poles starts rotating at synchronous speed in clockwise direction. Let, the rotor position is such that North falls under North and South under South as shown in fig. 2.1 (a), (b) and (c). Being like poles they will repel each other and with the result the rotor tends to rotate in the anticlockwise direction as shown in fig. 2.1 (b). Fig. 2.1 (50)