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

Synchronous Generator 101 Number of turns per phase, T = Zp = 480 22 = 240 Number of slots per phase per pole, m = 144 =3 16 × 3 Number of slots per pole, η = 144 = 9 16 Angular displacement between slots, β = 1800= 1800 η 9 = 200 Distribution factor, sin m β sin 3 × 200 =2 2 Kd = m sin β 3 sin 200 22 = 0.96 Eph = 4.44 × 1 × 0.96 × 50 × 0.03 × 240 = 1534 V Ans. EL = √3 Eph = √3 × 1534 = 2658 V Ans. Prob.5. A 3-phase star connected alternator supplies a load of 10MW at 0.85 lagging pf at 11 kV terminal voltage. Its resistance is 0.1 ohm per phase and synchronous reactance 0.66 ohm per phase. Calculate the line value of emf generated. (2012, 14, S/15) Sol. Full load output current = 10 × 106 √3 × 11000 × 0.85 = 618A IRa drop = 618 × 0.1 = 61.8 V IXs drop = 618 × 0.66 = 408 V Terminal voltage per phase = 11000 = 6350 V √3 φ = cos–1 (0.85) = 31.80 Hence, sin φ = 0.527 E0 = √(V cos φ + IRa)2 + (V sin φ + IXs)2

102 Electrical Machines–II = √(6350 × 0.85 + 61.8)2 + (6350 × 0.527 + 408)2 = 6625 V Line emf = √3 × 6625 = 11474.8 V Ans. Prob.6. A three-phase, 16-pole alternator has star connected winding with 144 slots and 10 conductors per slot. The flux per pole is 0.03 Wb distributed and the speed is 375 rpm. The coil span is 150 electrical degrees. Find: (i) frequency, (ii) electromotive force per phase, (iii) line electromotive force. (1984) Sol. Given, Number of poles = 16 Number of turn phase = Z 2 = 144 × 10 2×3 = 240/phase Flux per pole = 0.03 Wb N = Speed in rpm = 375 (i) Frequency, f = PN 120 = 16 × 375 120 = 50 Hz Ans. Number of slots/pole/phase, m = 144 = 3 16 × 3 β = 180 n n = Number of slots per pole = 1800 9 = 200 n = 144 = 9 16 α = Angle by which coil has been short pitched = (1800 – 1500) = 300 (ii) Induced emf per phase E = 4.44 KcKdφf T V

Synchronous Generator 103 Kc = cos α 2 300 = cos 2 = cos 150 = 0.966 Kd = sin m β/2 m sin β/2 sin 3 × 200 2 = 3 sin 100 = sin 300 3 sin 100 = 0.96 ∴ Induced emf per phase E = 4.44 × 0.966 × 0.96 × 0.03 × 50 × 240 = 1482.3 V Ans. (iii) Line value of induced emf = √3 E √3 × 1482.3 = 2567.4 V Ans. Prob.7. Calculate the emf of a 4-pole, 3-phase star-connected alternator running at 1500 rpm from the following data: Flux per pole = 0.1 Wb, Total number of slots = 48, Conductors per slot (in two layers) = 4 coil, span = 1500. Sol. Flux per pole, φ = 0.1 Wb Frequency, f = P×N 120 = 4 × 1500 120 = 50 Hz Number of turns/phase, T = Zs 2 = 48 × 4 = 32 3×2 ( )Coil span factor, Kc = cos 1800 – 1500 = 0.966 2 Number of slots per pole,

104 Electrical Machines–II n = 48 = 12 4 Number of slots per pole per phase, m = n = 12 = 4 33 Angular displacement between the slots, β = 1800 n = 1800 12 = 150 Distribution factor, Kc = sin m β m sin 2 β 2 sin 4 × 150 2 = 150 4 sin × 2 = 0.9577 Erms/Phase = 4.44 × KdKcφf T = 4.44 × 0.9577 × 0.966 × 0.1 × 50 × 32 = 657.2 V EL = √3 × 657.2 Ans. = 1138 V Prob.8. A 3-φ, 50 Hz star connected 2000 kVa, 2300 V alternator gives a short-circuited current of 600 A for a certain field excitation. With the same excitation, the open circuit voltage was 900 V. The resistance between a pair of terminals was 0.12 Ω. Find full load regulation at (i) UPF (ii) 0.8 pf lagging. Sol. OC volts/phase Zs = SC current/phase = 900/√3 600 = 0.866 Ω Resistance between the terminals is 0.12 Ω. It is the resistance of two phases connected in series ∴ Resistance/phase = 0.12 2 = 0.06 Ω Effective resistance/phase = 0.06 × 1.5

Synchronous Generator 105 = 0.09 Ω Xs = √(0.866)2 – (0.09)2 = 0.86 Ω Full load current, I = 2000000 √3 × 2300 = 502 A IRa = 502 × 0.06 = 30.12 V IXs = 502 × 0.86 = 431.72 V Rated voltage/phase = 2300 √3 = 1328 V (i) At Unity pf, E0 = √[(V cos φ + IRa)2 + (IXs)2] = √[(1328 + 30.12)2 + (431.72)2] = 1425 V % regulation = (1425 – 1328) × 100 1328 = 0.073 or 7.3% Ans. (ii) At 0.8 pf lagging E0 = √[(V cos φ + IRa)2 + (V sin φ + IXs)2] = √[(1328 × 0.8 + 30.12)2 + (1328 × 0.6 + 431.72)2] = 1644 V % regulation = (1644 –1328/1328) × 100 = 0.237 or 23.7% Ans. Prob.9. A 500V, 50 kVA, 1 phase alternator has an effective armature resistance of 0.2 Ω. An excitation current of 10 A produces armature current of 200 A on short circuit and emf of 450 V on open circuit. Calculate the synchronous reactance. (S/2010) Sol. Given, Armature resistance, Ra = 0.2 Ω = 200 A Short circuit current, Isc = 450 V Open circuit emf, Voc Synchronous impedance, Zs = Open circuit voltage Short circuit current = 450 = 2.25 Ω 200

106 Electrical Machines–II So, Synchronous reactance, Xs = √(Zs)2 – (Ra)2 = √(2.25)2 – (0.2)2 = 2.24 Ω Ans. Prob.10. A 3 phase, 1200 kVA, 6600 volt, star connected alternator has resistance of 0.4 Ω and synchronous reactance 6 Ω per phase. This alternator supplies full load at 0.8 pf lagging and at normal voltage. Then at same excitation and 0.8 pf leading. Find full load current and full load voltage regulation. (2013) Sol. Full load = 1200 kVA = 1200000 VA Phase voltage, Vp = 6600 = 3810 V √3 Full load current, I = 1200000 √3 × 6600 = 105 A Ans. Effective armature resistance per phase, Ra = 0.4 Ω Synchronous reactance per phase, Xs = 6 Ω When pf is 0.8 lagging, cos φ = 0.8 and sin φ = 0.6 E0 = √(V cos φ + IRa)2 + (V sin φ + IXs)2 = √(3810 × 0.8 + 105 × 0.4)2 + (3810 × 0.6 + 105 × 6)2 = 4248 V Full load voltage regulation, ( )% regulation = 4248 – 3810 × 100 3810 = 11.5% Ans. Prob.11. A 50 kVA, 440 V, 50 Hz, 3 phase star connected alternator has effective armature resistance of 0.25 Ω/phase. The synchronous reactance is 3.2 Ω/phase and leakage reactance of 0.5 Ω/phase. Determine at rated load and unity pf: (i) internal emf, Ea (S/2013) (ii) no-load, E0 (iii) percentage regulation on full load. Sol. Rated load = 50 kVA = 50000 VA Line voltage, VL = 440 V Phase voltage, VP = 440 = 254 V √3

Synchronous Generator 107 Full load current, I = 50000 = 65.6 A √3 × 440 Effective armature resistance per phase, Ra = 0.25 Ω Leakage reactance per phase, XL = 0.5 Ω Synchronous reactance per phase Xs = 3.2 Ω Power factor, cos φ = 1 (i) Internal emf per phase, Eap = √(V + IRa)2 + (IXL)2 = √(254 + 65.6 × 0.25)2 + (65.6 × 0.5)2 = 272 V Line value of internal emf, Ea = √3 × 272 Ans. = 471 V (ii) No-load emf per phase, Eop = √(V + IRa)2 + (IXs)2 = √(254 + 65.6 × 0.25)2 + (65.6 × 3.2)2 = 342 V Line value of no-load emf, E0 = √3 × 342 Ans. = 593 V (iii) % regulation = 593 – 440 × 100 440 = 34.6% Ans. Prob.12. A 3-phase, 800 kVA, 11 kV, star connected alternator has resistance of 1.5 ohms and a synchronous reactance of 25 ohms per phase. Find the regulation in percent for a load of 600 kW at 0.8 pf leading. (S/2003) Sol. VL = 11 KV P = 800 KV Resistance, Ra = 1.5 Reactance, Xs = 25 Phase voltage at load = 11000 = 6350.85 V = 6351 (approx) √3 Load current/phase at 600 kW and 0.8 pf I = 600 × 1000 √3 × 11000 × 0.8 = 39.36 A cos φ = 0.8 leading ∴ sin φ = – 0.6 Open circuit voltage per phase, E0 = √(V cos φ + IRa)2 + (V sin φ + IXs)2

108 Electrical Machines–II = √(6351 × 0.8 + 39.36 × 1.5)2 + (6351 × (– 0.6) + 39.63 × 25)2 = 5865.80 V Percentage regulation = E0 – V V = 5865.80 – 6351 6351 = – 7.64% Ans. Prob.13. A 600 V, 60 kVA, 1 φ alternator has an effective resistance of 0.25 Ω. A field current of 8 A produces an armature current of 150 A on short-circuit and an emf of 330 V on open-circuit. Determine percentage regulation. Assume pf 0.8 lag. (1992) Sol. Effective armature resistance, Ra = 0.25 Ω Synchronous impedance, Zs = Open circuit voltage Short circuit current = 330 150 = 2.2 Ω Synchronous reactance, Xs = √Zs2 – Ra2 = √(2.2)2 – (0.25)2 = 2.186 Ω Load current, I = 60 × 1000 600 = 100 A Power factor, cos φ = 0.8 lag sin φ = 0.6 Open-circuit voltage, E0 = √(V cos φ + IRa)2 + (V sin φ + IXs)2 = √(600 × 0.8 + 100 × 0.25)2 + (600 × 0.6 + 100 × 2.186)2 = 768 volts ∴ % regulation = E0 – V × 100 V = 768 – 600 × 100 600 = 28% Ans. Prob.14. A 3-φ star connected alternator is rated at 1600 kVA, 13.5 kV. The armature resistance and synchronous reactance are 1.5 Ω and 30 Ω respectively per phase. Calculate the percentage regulation for a load of 1280 kW at 0.8 (lagging) power factor. (2000)

Synchronous Generator 109 Sol. Given, kVA = 1600 VL = 13.5 KV R = 1.5 Ω Xs = 30 Ω √3 VI cos φ = 1280000 √3 × 13500 × I × 0.8 = 1280000 I = 1280000 √3 × 13500 × 0.8 ∴ I = 68.43 A IRa = 68.43 × 1.5 = 102.6 V IXs = 68.43 × 30 = 2052 V Voltage/phase = 13500 √3 = 7794.23 V cos φ = 0.8 sin φ = 0.6 For lagging power factor, E0 = √(V cos φ + IRa)2 + (V sin φ + IXs)2 = √(7794.23 × 0.8 + 102.63)2 + (7794.23 × 0.6 + 2052)2 = √(40170421 + 405286681) = 9244.30 V % regulation = [(9244.30 – 7794.23/7794.23] × 100 = 18.60% Ans. Prob.15. An alternator on open-circuit generates 360 V at 60 H, when the field current is 3.6 A. Neglecting saturation, determine the open circuit emf when the frequency is 40 Hz and the field current is 2.4 A. Sol. As seen from the emf equation of an alternator, E α φf ∴ E1 = φ1f1 E2 φ2f2 Since saturation is neglected, φ α If is the field current ∴ E1 = If1f1 E2 If2f2 or 360 = 3.6 × 60 E2 2.4 × 40 ∴ E2 = 160 V Ans.

110 Electrical Machines–II Prob.16. A 4 pole, 50 Hz, star connected alternator has 15 slots per pole and each slot has 10 conductors. All the conductors of each phase are connected in series. The winding factor being 0.95. When running on no-load for a certain flux per pole, the terminal emf was 1825 volt. If the windings are lap connected as in a DC machine, what would be the emf between the brushes for the same speed and the same flux per pole. Assume sinusoidal distribution of flux. Sol. Here, Kf = 1.11 Kd = 0.95 Kc = 1 (assumed) f = 50 Hz emf/phase = 1825 V √3 Total number of slots = 15 × 4 = 60 ∴ Number of slots/phase = 60 = 20 3 Number of turns/phase = 20 × 10 = 100 2 ∴ 1825 = 4 × 1.11 × 1 × 0.95 × φ × 50 × 100 √3 φ = 1825/√3 × 4 × 1.11 × 1 × 0.95×50×100 φ = 49.97 mWb When connected as DC generator φZN P 60 A ( )Eg= Z = 60 × 10 = 600 N = 120 f P = 120 × 50 4 = 1500 rpm ∴ Eg = 49.97 × 600 × 10-3 × 1500 × 4 60 4 = 750 V Ans. Prob.17. From the following test results, determine the voltage of a 2000 V, 1 phase alternator delivering a current of 100 A at 0.8 leading pf. Test results - Full-load current of 100 A is produced on short circuit by a field excitation of 2.5 A. An emf of 500 V is produced on open-circuit by the same excitation. The armature resistance is 0.8 Ω (1993)

Synchronous Generator 111 Sol. Given, Open circuit voltage = 500 V Short circuit current, I = 100 A Armature resistance, Ra = 0.8 Power factor, cosφ = 0.8 (leading) Synchronous impedance, Zs = Open circuit volts Short circuit current For same excitation = 500 = 5 Ω 100 Xs = √Zs2 – Ra2 = √52 – (0.8)2 = 4.936 Ω Fig. 3.23 IRa = 100 × 0.8 = 80 volts IXs = 100 × 4.936 = 494 volts ∴ pf = 0.8 lead (given) Hence, Ea = √(V cos φ + IRa)2 + (V sin φ + IXs)2 √(2000 × 0.8 + 80)2 + (2000 × 0.6 + 494)2 = 2385.8 volts % regulation = E0 – V × 100 V = 2385.8 – 2000 × 100 2000 = 19.29% Ans. FF

112 Electrical Machines–II 4 SINGLE PHASE INDUCTION MOTORS Chapter SUBJECTIVE QUESTIONS AND ANSWERS Q.1. What is single phase induction motor? Write the constructional details of single phase induction motor. Ans. The motor which are working on single phase supply are known as AC single phase motor, this converts AC single phase into mechanical energy. This motor also having fractional horse power because of low power or fractional horse power, sometime called as fractional horse power motors. Since, the performance requirements of the various applications are differ so widely, the motor-manufacturing industry has developed many different types of such motors, each being designed to meet specific demands. Construction of Single Phase Induction Motor: The construction of single phase induction motor is more or less similar to that of a polyphase induction motor. It has the two main parts which are as follows: (i) Stator: As its name indicates stator is a stationary part of induction motor. A single phase AC supply is given to the stator of single phase induction motor. The stator of the single phase induction motor has laminated stamping to reduce eddy current losses on its periphery. The slots are provided on its stamping to carry stator or main winding. In order to reduce the hysteresis losses, stamping are made up of silicon steel. When the stator winding is given to a single phase AC supply, the magnetic field is produced and the motor rotates at a speed slightly less than the synchronous speed. The construction of stator of a synchronous motor is similar to that of three phase induction motor except there are two dissimilarity in the winding part of the single phase induction motor as follows: (a) Firstly the single phase induction motors are mostly provided with concentric coils. As the number of turns per coil can be easily adjusted with the help of concentric coils, the mmf distribution is almost sinusoidal. (b) Except for shaded pole motor, the a synchronous motor has two stator windings namely the main winding and the auxiliary winding. These two windings are placed in space quadrature with respect to each other. (112)

Single Phase Induction Motors 113 (ii) Rotor: The construction of the rotor of the single phase induction motor is similar to the squirrel cage three phase induction motor. The rotor is cylindrical in shape and has slots all over its periphery. The slots are not made parallel to each other but are bit skewed as the skewing prevents magnetic locking of stator and rotor teeth makes the working of induction motor more smooth and quieter. The squirrel cage rotor consists of aluminium, brass or copper bars. These aluminium or copper bars are called rotor conductors are permanently shorted by the copper or aluminium rings called end rings. In order to provide mechanical strength these rotor conductor are braced to the end ring and hence, form a complete closed circuit resembling like cage and hence got its name as squirrel cage induction motor. Q.2. Explain why the single phase induction motor is not self-starting? Also write the ways of making single phase motor self-starting? (2008, S/16, S/17) Why single phase motors are not self-starting? Name the various methods of making them self-starting. (2010, 16) Explain why a single phase induction motor is not self start? What are the methods to make itself start? (S/2012) Explain why single phase induction motor is not self- starting? (2017) What are different methods of self-starting single phase induction motor? (2017) Write different types of single phase induction motor. (2017) Ans. Working Principle and Reason for not Self-Starting of Single Phase Induction Motor: 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. Methods of Self-Starting and Types of Single Phase Induction Motor: There are several methods which have been developed for the starting and types of single phase induction motors may be classified as follows:

114 Electrical Machines–II (i) Split-phase starting, it has four types as follows: (a) resistor-split phase motors, (b) capacitor-split phase motors, (c) capacitor start and run motors, (d) capacitor run motors. (ii) Shaded-pole starting. (iii) Repulsion-motor starting. (iv) Reluctance starting. A single phase induction motor is commonly known by the method employed for its starting. Q.3. Write short note on double revolving field theory. (2008) Explain double revolving field theory. Draw torque-speed characteristics of a single phase induction motor and explain it. (2009) Explain the double revolving field theory of the single phase induction motor. (S/2010, 11) Explain double revolving field theory. (2012, S/15, 15, S/17, 17) Explain double revolving field theory. Draw torque-speed characteristics of 1-φ induction motor and write applications. (2013) Write application of 1-φ induction motor. (S/2017) Ans. The explanation of the single phase induction motor can be easily, and clearly understood by the double revolving field theory or the cross field theory. According to the double field revolving theory, the pulsating field is resolved into two opposite rotating field each of Fig. 4.1 Double Revolving Field Theory

Single Phase Induction Motors 115 the half magnitude. Each field induces a torque but being equal and opposite the net torque during the complete cycle is the sum of both torque is achieved and rotor is standstill. So the motor are not self- starting one. According to the cross field theory, the emf produced by state fed by AC single phase, but pulsates in magnitude and varies sinusoidally with time. Now the emf induced is because of the transformer action resulting no torque. The single pulsating field having its maximum value of φm is resolved into two magnetic field, each having modulus of φm/2 at different instances. At 1 step, the pulsating field φm is positive maximum, and it is resolved into two fields acting in the same positive direction of the pulsating field. After 600 i.e., at 2 step, the pulsating field attains a value of φm/2 and it is said to be equivalent to field moving clockwise by 600 contributing a horizontal component of φm/2 cos 600 = φm/4 and another field moving anticlockwise by 600 contributing another φm/4 along the horizontal direction. Their vertical components cancel each other. At 3 step, θ = 900, the values of pulsating field is zero and is equivalent fields in vertical direction, cancel each other. Next moved by 900 clockwise from 1 step and moved by the same angle but in anticlockwise direction as shown in fig. 4.1. On further rotation of 4 step, 5 step, 6 step, 7 step and 8 step makes one complete cycle which is shown in fig. 4.2. Fig. 4.2 Rotation of Different Steps Makes One Cycle During running condition, in addition to the transformer emf, there is one more emf known as special emf. The generated rotor emf varies in space with the stator current and flux. Now, both the fields are at right angle to each other resulting the rotating field revolving at synchronous speed. If an induced motor is switched on and given some starting torque, it will continue to run in the same direction in which the torque is given.

116 Electrical Machines–II Torque-Speed Characteristics of Single Phase Induction Motor: Torque developed by the two rotating fields are acting in opposite directions, each field develops a torque that tends to rotate the rotor in the direction in which the field rotates. The resultant torque developed in the rotor is the summation of the torque produced by the two rotating fields. It may be noted that torque-speed curves have been drawn for a speed range of – Ns to + Ns. The resultant torque-speed curve is shown in the fig. 4.3. Fig. 4.3 Torque-Speed Characteristics of Single Phase Induction Motor as Derived from Double Revolving Field Theory From the resultant torque-speed curve the following can be observed which are as follows: (i) Average torque at standstill is zero and therefore, the motor is not self-starting. (ii) When the rotor is given an initial rotation is any direction, the average torque developed causes the rotor which continue to rotate in the direction in which it is started. (iii) The average torque becomes zero at some value of speed below the synchronous speed. This indicates that a single phase motor operates with a generator percentage of slip at full load than a corresponding three phase induction motor. Application of Single Phase Induction Motor: The application of single phase induction motor are as follows: (i) These motors are used extensively in industrial, commercial and domestic applications. (ii) They are used in clocks, refrigerators, freezers, fans, blowers, pumps, washing machines, machine tools and range in size from a fraction of a horse power to 13 hp. Q.4. State and explain cross-field theory for single phase induction stator. (1998, 99) Ans. As a single phase induction motor having a single phase winding on the stator as shown in fig. 4.4. The rotor is of squirrel cage

Single Phase Induction Motors 117 type. When a single phase supply is applied across the stator winding an alternating field, φs is produced along the horizontal axis. With the rotor at standstill, this field will induced a transformer emf in the rotor winding. This emf will cause a current to flow in the rotor winding. As the axis of both the stator and rotor fields lies in the horizontal direction no torque will be developed. Assume that the rotor is Fig. 4.4 now given an initial rotation, means in the clockwise direction. An emf, called rotational emf, will be induced in the rotor winding by virtue of its rotation in the stationary stator field. The direction of emf induced in the rotor conductors are shown in fig. 4.4. Emf induced in the rotor conductors is in one direction on one side of the vertical axis and in the other direction on the other side of the vertical axis. As the rotor circuit is closed, the rotational voltage so induced will produce a component rotor current and a rotor emf wave whose axis is displaced 900 electrical from the stator axis. Unlike a polyphase induction motor, the frequency of the rotor induced emf is high and therefore, the rotor reactance is high (X ∝ f ). The rotor current will lag the rotor induced emf by about 900. The field produced by rotor current, φr known as crossfield will have a time phase difference of 900 with the stator field φs . Thus the stator flux, φs and rotor flux, φr are in space and time quadrature. These two fields will produce a revolving fields which will rotate in the direction in which the rotor was given an initial rotation. Thus, the torque produced will be in the same direction as that of rotation. Q. 5. Draw the equivalent circuit of a single phase induction motor. Ans. Let, R1 = stator resistance X1 = stator reactance r2, x2 = half of rotor resistance and standstill reactance in stator term xm = half of the total magnetising reactance assigned to each rotor

118 Electrical Machines–II Fig. 4.5 Equivalent Circuit of Single Phase Induction Motor s = slip of the first rotor 2 – s = slip of the second rotor A single phase motor as consists of two motors, having a common stator winding, but with their respective rotors revolving in opposite directions. The equivalent circuit is based on double field revolving theory is shown in fig. 4.5. Here, the single phase motor has been imagined to be made-up of one stator winding and two imaginary rotors. The stator impedance is Z = R1 + jX1. The impedance of each rotor is (r2 + jX2) where r2 and X2 represents half the actual rotor values in stator terms. Q.6. Explain the basic principle of starting a 1-φ induction motor by split-phase method. (1998) Ans. The principle of split-phase method is to create at starting a condition similar to a two-phase stator winding carrying two phase currents, so that a rotating magnetic field is produced. This is achieved by providing, in addition to main single phase winding, a starting winding which is displaced in space by 900 with the main winding on the stator slots. The resistance and reactance of the two winding circuits are made such that, when connected across a single phase supply the currents flowing through the windings will have a considerable time- phase difference. Thus, although a supply is a single phase one current flowing through the two windings are to some extent similar to two phase currents. A rotating magnetic field will therefore, be produced which will develops starting torque on the rotor. Once the motor has started, this extra starting winding may be cut out from the supply by some device.

Single Phase Induction Motors 119 Q.7. How the direction of rotation split-phase single phase induction motor can be reversed? Ans. The direction of rotation of such a motor can be reversed by interchanging the connections to the supply of either starting (auxiliary) or running (main) winding. The worth noting point is that reversal of a split-phase induction motor takes place only when it is at rest and the starting switch is in its normally closed de-energised position. Q.8. Explain the working principle of resistor split-phase single phase induction motor. Draw the phasor diagram and torque-speed characteristics. Ans. In this method a high resistance is connected in series with the starting winding, as shown in fig. 4.6 (a). Because of low resistance and high inductive reactance of main winding and high resistance (c) Torque-Speed Characteristics Fig. 4.6

120 Electrical Machines–II and low inductive reactance of starting (auxiliary) winding, the starting winding current Is does not lag so much as the main winding current Im. At starting the main winding current Im lags behind the applied voltage V by 65-75 degrees, whereas the starting winding current Is lags by 35-45 degrees. Thus, the phase displacement between main winding current and starting winding current is about 20-30 degrees as shown in phasor diagram 4.6 (b). Since, the currents in two windings are different in magnitude and constants of two windings are different therefore, the arrangement forms imperfect two phase motor that produces rotating magnetic field that is not uniform in either time or space, but it is sufficient for starting the motor. The best starting condition will be obtained if Is and Im differ in time about 900, but this condition is not possible in this type of motor. When the motor picks up the speed 75 or 80 percent of synchronous speed, the starting winding (or auxiliary winding) is taken out of circuit by a centrifugal switch. It is important that this should be done because otherwise, it will result in overheating and burning out of auxiliary winding owing to its low current carrying capacity and inefficient noisy performance. A typical torque-speed characteristics is shown in fig. 4.6 (c). The starting torque is 1.5 times to twice the full load starting torque and starting current is 6 to 8 times full load current. The speed regulation of resistance-start single phase induction motor is very good. The present slip in most split-phase motors is about 4-6%. Such a motors may operate with a pf of about 0.55-0.65 and efficiency of 60-65%. Q.9. What are the applications of resistor split-phase induction motor? Ans. These motors are made in fractional kilowatt (1/20 to 1/4 kW) ratings with speed ranging from 2,875 to 700 rpm. Such motors find its wide applications for low intertia loads, continuous operating loads and applications requiring moderate starting torque such as for driving washing machines, fans, blowers, centrifugal pumps, domestic refrigerators, duplicating machines, office appliances, food processing machines, wood working tools, grinders etc. Because of low starting torque these motors are seldom employed in sizes larger than 1/4 kW. Q.10. Describe the construction and working of a capacitor- start single phase induction motor with suitable diagram. (2010) Write short note on capacitor-start single phase induction motor. (2011) Ans. Construction: The schematic diagram of a capacitor-start single phase induction motor is given in fig. 4.7 (a). It is also known as capacitor-start, induction-run motor. This is an improved form of a resistance-start split-phase motor. Fundamentally this motor is very

Single Phase Induction Motors 121 similar to the resistance-start split-phase motor, except that the starting winding has a few more turns and consists of a heavier wire than the starting winding of a resistance-start motor, there is also a large electrolytic capacitor in place of resistor, connected in series with the starting winding. By using a capacitor of proper value, the starting (a) Schematic Diagram (b) Phasor Diagram (c) Torque-Speed Characteristics Fig. 4.7 winding current Is at standstill, can be made to lead the main winding current Im by 900 (i.e. θ = 900). The phasor diagram of a capacitor-start motor is given in fig. 4.7 (b) and torque-speed characteristic is shown in fig. 4.7 (c). Working: The capacitor-start motor, like resistance-start motor, has the starting winding disconnected by means of a centrifugal switch

122 Electrical Machines–II as the motor picks up speed. Because of short duty service rating of electrolytic capacitor used in capacitor-start motor, proper operation of the starting (or centrifugal) switch is doubly important. A faulty switch may keep the capacitor in the circuit for too long time during each starting period and thereby appreciably shorten its life. Of course, the starting winding itself is affected by the switch in the same manner as in the resistance-start motor. Q.11. Write application of capacitor for split-phase induction motor. Ans. Capacitor start motors are manufactured in rating ranging from 1 kW to 3 kW, but larges sizes are also available. They are 10 4 widely used for heavy duty general purpose applications like refrigerator units, air conditioners, compressors, jet pumps, sump pumps, centrifugal pumps, compressors, conveyors, fans, blowers, farm and home workshop tools, oil burners etc. Q.12. Write a short note on capacitor-start and run-induction motor. (2013) Ans. In order to have the advantage that result from the use of a capacitor during the normal running operation as well as starting period, the capacitor-start, capacitor-run motors have been developed. The capacitor-start, capacitor-run motors is similar to capacitor-start, induction-run motor, except that the auxiliary winding and a capacitor remain connected in the circuit of all times. In such a motor two different capacitors one small value oil impregenated paper continuous rating capacitor CR and another much larger value electrolytic short-duty capacitor CS are employed. It also provide a centrifugal switch to take out the electrolytic capacitor out of auxiliary winding circuit when the motor has attained a pre-determined speed. The schematic diagram of a capacitor-start, capacitor-run motor is given in fig. 4.8 (a). When the motor attains 75% of synchronous speed, the starting capacitor CS is taken out of the circuit by the operation of centrifugal or transfer switch, which is normally closed. It is to be noted that when the centrifugal switch is closed during the starting period, the two capacitors CS and CR are in parallel, so that total capacitance in the auxiliary winding circuit is sum of their values, after the switch opens, only continuous oil type capacitor CR remains in the auxiliary winding circuit. In practice starting capacitor CS is about 10-15 times as large as the running capacitor CR. Such a motor operates as a two phase motor from single phase supply, thereby, producing a constant torque and not a pulating torque,

Single Phase Induction Motors 123 as in other single phase motors. Besides their ability to start heavy loads they are extremely quiet in operation, have better efficiency (55-65%) and pf (0.80-0.95) when loaded and develop upto 25% greater over-load capacities. The disadvantage of such machines is only high cost. Typical torque-speed characteristic is an shown in fig. 4.8 (b). (a) Schematic Diagram (b) Torque-Speed Characteristics Fig. 4.8

124 Electrical Machines–II Q.13. Write various applications of capacitor-start, capacitor-run motors (two value capacitor motor). Ans. These machines are very useful where the load requirement are severe as in case of compressors, refrigerators, fire-strokers, pumps, conveyors etc. These motors are often employed requiring a quiet operating motors such as in hospitals, studies etc. In recent years, capacitor-start, capacitor-run motors have become very popular and are now manufactured in the larger sized single phase motors where previously only repulsion-start motors were available. For these motors both the starting and running currents are low. Q.14. Explain with the help of a neat connection diagram and working of permanent capacitor single phase induction motor. Draw its torque-speed characteristics. Write its advantages. Ans. These motors are also sometimes called the single value capacitor-run motor, has two stator windings placed mutually 90 electrical degree apart. The main or running winding is connected directly across the single phase AC supply. A capacitor in series with the auxiliary winding is also connected across the supply lines. These are no centrifugal switch, since, the auxiliary winding is energised at all times when the motor is in operation. The auxiliary winding are such motors usually have almost the same size of wire and almost as many turns as the main winding, infact in some motors the two windings are identical. The schematic diagram of a permanent capacitor motor is given in fig. 4.9 (a). The auxiliary winding is always in the motor circuit and therefore, the operation of the motor when loaded resembles more closely to that of a two phase motor. One advantage of such arrangement/operation is that the rotating field developed by the two stator windings is a more (a) Schematic Diagram (b) Torque-Speed Characteristics Fig. 4.9

Single Phase Induction Motors 125 nearly uniform and therefore, the motor is quitter in operation. The other advantages of such motors are as: (i) increased pull out torque, (ii) a higher pf (0.8-0.95), (iii) higher operation efficiency, (iv) smaller full load line current. Such motors have drawbacks also. Since, in such motors the same capacitor is used for starting and running therefore, neither optimum starting nor running performance can be obtained. A typical performance characteristic for such a motor is given in fig. 4.9 (b). Generally capacitors of 2-20 µF capacity are employed. The electrolytic capacitor can not be employed owing to the requirement of continuous, operation and therefore, more expensive oil or pyranol insulated foil paper capacitor is to be used. Under starting conditions, the capacitor is too small and so the starting torque tends to be low (50-80% of full load torque) but the manufactures can partly compensate for this by using a high resistance squirrel cage motor. Q.15. How the direction of rotation permanent capacitor single phase induction motor can be reversed? Write applications of permanent capacitor motor. Ans. A unique feature of such a motor is that it can be very easily reversed by an external switch, provided by the two winding are identical. Because of absence of starting switch it is possible to reverse the leads of either winding while the motor is running. The torque is thus reversed, the motor comes to a standstill and then starts in the opposite direction. Applications: Such motors are widely used in table, ceiling fans, blowers, oil burners, room coolers, portable tools, other domestic and commercial electrical appliances, where low starting torque is required. Because of the simplicity of reversal of motor, such motors are often used to operate devices that must be moved back and forth frequently such as induction regulators, rheostate, furnace controls, valves and arc welding controls. Because of the absence of starting switch, it is possible to run a permanent (or single value) capacitor-run motor over a wide range of speeds and such a motors has many applications where adjustable speed is required. Q.16. Write a short note on shaded pole motor. (2013, S/16, S/17) Ans. Shaded pole motor is a split-phase type single phase induction motor. The stator has salient poles, each provided with its own exciting coil. A part of (about 1/4 to 1/3) each pole is wrapped by a copper strap forming a closed loop, known as shading coil, as shown in fig. 4.10 (a). Stator usually has 2 or 4 poles giving synchronous speed of 3000, or

126 Electrical Machines–II 1500 rpm for 50 Hz power supply. The rotor is a simple squirrel cage type with a 600 skew for optimum starting torque and for limiting the torque dip during run-up. A low cage resistance impairs both starting torque and the dip, and usually the resistance is greater than for any other type of single phase and cage motor, giving a starting torque in the range of 40-80% of full load and pull out torque of about 110-140% of full load torque. The gap length is of the order of 0.25-0.50 mm. Too short air gap may result in starting-torque variations due to rotor slotting. It has no commutator, brushes, collector rings, contactors, capacitors or moving switch parts, so it is relative by cheaper, simper and extremely rugged in construction and reliable. Operation: When the exciting winding is connected to a single phase AC supply source an alternating current flows through it and sinusoidally varying flux is set up in the core. This change of flux induces short circuit currents in the shading coil in such a direction as to appose the core flux. The flux in the shaded part of the pole thus, lags behind the flux in the main part of the pole. At the same time, the main flux and shaded pole flux are obviously displaced in space, although by less than 900. Since, two fields have both space and time displacement, the condition for setting up a rotating field are present. Torque is thus, developed in the squirrel cage rotor. As soon as the rotor starts rotating under the influence of starting torque, an additional torque is developed by single phase induction motor action. The motor ultimately attains a speed slightly less than the synchronous speed and runs as a single phase induction motor. (a) Schematic Diagram of Shaded Pole Diagram (b) Torque-Speed Characteristics Fig. 4.10

Single Phase Induction Motors 127 Operating Characteristics: A typical torque-speed characteristics of a shaded pole induction motor is shown in fig. 4.10 (b). Starting torque is very small (about 50 to 80% of full load torque). The efficiency ranges from 20 to 50%. Power factor is also very poor ranging from 0.5 to 0.6. It has little over-load capacity. Full load torque are developed at comparatively high values of slip, about 10 to 25%. Also speed falls with the decrease in applied voltage. Q.17. Write applications of shaded pole induction motor. Can the direction of rotating be reversed of these motors? Ans. Because of their poor starting torque, low power factor and poor efficiency, such motors are only suitable for low power applications such as for toys, small fans, electric clocks, hair dryers, time photographs, slide projectors, small business machines (photocopying machines, advertising displays, etc.) and other similar applications. They are manufactured in small sizes ranging from below 5 W to about 100-150 W. They are available with built in gear reducer to give almost any speed (as low as below 1 rpm). The direction of rotation of such a machine can not be reversed. Q.18. Compare single phase and three phase induction motor. (2010, S/11, 11, S/15) Compare single phase induction motor with three phase induction motor. (2014, S/17) Difference between single phase and polyphase induction motor. (2017) Ans. The comparison between single phase and three phase/ polyphase induction motor are as follows: S. No. Single Phase Induction Motor Three Phase/Polyphase Induction Motor (i) Single phase motor has Three phase motor has pulsating magnetic field. rotating magnetic field. (ii) These motor have no starting These motor have self torque. starting torque. (iii) These motor are heavier. These motor are lighter. (iv) It has low efficiency. It has high efficiency. (v) It has low pf. It has high pf. (vi) Stator have two windings Stator has no winding. one is for starting and other (vii) is for running. No starting device are needed. Some starting device are (viii) always needed. It has three phases. (ix) It has single phase. It requires six wire supply. It requires a two wire supply.

128 Electrical Machines–II Q.19. What are the advantages of three phase induction motor over single phase induction motor? (S/2012) Ans. The advantages of three phase induction motor over single phase induction motor are as follows: (i) It is more economical as it requires less conductor material as compared to single phase system. (ii) A three phase machine give more output as compared to a single phase machine of the same size. (iii) Three phase motors have uniform torque, whereas most of the single phase motors have pulsating torque. (iv) Domestic power and industrial or commercial power can be supplied from the same source. (v) A three phase motor produces more torque as compared to a single phase motor. (vi) Three phase induction motor are self-starting, whereas single phase motors are not self-starting. (vii) Voltage regulation is better in a three phase system. NUMERICAL PROBLEMS AND SOLUTIONS Prob.1. The name plate of a single phase, 4 pole induction motor gives the following data - output 373 W, 230 V, frequency 50 Hz, input current 2.9 A, power factor 0.71, speed 1410 rpm. Calculates the efficiency the slip of the motor when delivering the rated output. Sol. Power output, P0 = 373 W Power input, Pin = VI cos φ = 230 × 2.9 × 0.71 = 473.6 W Motor efficiency, ηm = P0 × 100 Pin = 373 × 100 473.6 = 78.76% Ans. Synchronous speed, NS = 120 f P = 120 × 50 4 = 1500 rpm Slip, s = NS – N NS

Single Phase Induction Motors 129 = 1500 – 1410 1500 = 0.06 or 6% Ans. Prob.2. A 230 V, 50 Hz split phase induction motor has main winding resistance 5 Ω and inductive reactance 12 Ω and auxiliary winding of resistance 12 Ω and inductive reactance 5 Ω. Determine at start: (i) the current in the starting winding, (ii) the current in the main winding, (iii) line current, (iv) phase displacement between two winding currents, (v) power factor of the motor. Sol. At start, (i) Current in starting winding, Is = V Zs = 230 √122 + 52 = 17.7 Amps Ans. Phase angle, φs = tan–1 Xs Rs = tan–1 5 12 = 22.620 (lag) (ii) Current in main winding, V Im = Zm = 230 √52 + 122 = 17.7 Amps Ans. Phase angle, φm = tan–1 Xm Rm = tan–1 12 5 = 67.380 (iii) Phase displacement between the two winding currents θ = φm – φs = 67.380 – 22.620 = 44.760 Ans.

130 Electrical Machines–II (iv) Line Current, IL = √Is2 + Im2 + 2IsIm cos θ = √(17.7)2 + (17.7)2 + 2 × 17.7 × 17.7 × √cos 44.760 √313.29 + 313.29 + 444.91 = 32.73 A Ans. (v) Motor Power factor, cos φ = Is cos φs + Im cos φm I = 17.7 cos 22.620 + 17.7 cos 67.380 32.73 = 0.707 (lag) Ans. Prob.3. The resistance and inductive reactance of each winding of a 50 Hz split phase induction motor are 80 Ω and 237.5 Ω respectively. In series with one winding there are additional resistance R and condenser C. Calculate their values to give the same current in each winding with a phase difference of 900. Sol. Resistance of each = 80 Ω stator winding = 237.5 Ω Impedance of each stator winding, Z = √(80)2 + (237.50)2 = 250.6 Ω Phase angle, φ = tan–1 237.5 80 = 71.40 The impedance of starting, after connecting resistance R and capacitance e in series with any of the given winding should be equal to its old impedance i.e., 250.6 Ω and phase angle 18.60 so, that the value of current in each winding is same and has phase difference of 900. These are shown in fig. 4.11. Starting winding resistance, Zs = 250.6 Ω φs = 18.60 Starting winding resistance, Rs = Zs cos φs = 250.6 × 0.9478 = 237.5 Ω The value of additional resistance = 237.5 – 80 R = 157.5 Ω Ans. Fig. 4.11 Starting winding capacitive reactance, Xs = Zs sin φs

Single Phase Induction Motors 131 = 250.6 × sin 18.60 = 250.6 × 0.319 = 80 Ω ∴ Required additional capacitive reactance, Xc = 80 – (–237.5) = 317.5 Ω The value of additional capacitance, C =1 2 πf Xc =1 2 π × 50 × 317.5 = 10 µF Ans. Prob.4. A 250 W, 230 V, 50 Hz capacitor-start motor has the following constant for the main and auxiliary windings: Main winding, Zm = (4.5 + j 3.7) ohms Auxiliary winding, Za = (9.5 + j 3.5) ohms Determine the value of the starting capacitor that will place the main and auxiliary winding currents 900 at starting. Sol. Let Xc be the reactance of the capacitor connected in the auxiliary winding Then Za = 9.5 + j 3.5 – jXc = (9.5 + jX) where X is the net reactance. Now, Zm = 4.5 + j 3.7 = 5.82 ∠ 39.40 ohm Obviously, Im lags behind V by 39.40. Since, time phase angle between Im and Ia has to be 900, Ia must lead V by (900 – 39.40) = 50.60. For auxiliary winding, tan φa = X or tan 50.60 R = X 9.5 or X = 9.5 × 1.217 = 11.56 Ω (capacitive) Xc = 11.56 + 3.5 = 15.06 Ω ∴ 15.06 = 1 314 C C = 211 µF Ans. FF

132 Electrical Machines–II 5 AC COMMUTATOR MOTORS Chapter SUBJECTIVE QUESTIONS AND ANSWERS Q. 1. What is AC commutator and write its types? Write short note on AC commutator motor. (2000) Ans. The AC commutator motors are single phase AC motors where commutator are existing just like DC motors. Some modifications are adopted in the design for satisfactory performance. The AC commutator motors can be made to operate nearer unity pf so as to achieve an economy in their running cost. For speed control and power factor improvement of low speed large induction motors, AC commutator machines are widely used. These are divided into following types: (i) AC series motor, (ii) universal motor, (iii) repulsion motor, (iv) repulsion induction motor. Q.2. Explain the working principle of AC series motor with the help of diagram. Ans. If an alternating voltage is applied to an ordinary DC series motor, then for one half cycle, the field flux φd and armature flux φq may be shown in fig. 5.1 (a). Their interaction results in the development of torque in the clockwise direction. For the other half of the cycle, the direction of the current in both field and armature is reversed simultaneously, because they are in series. As a result both the field and armature fluxes are reversed as shown in fig. 5.1 (b) and torque is Fig. 5.1 (132)

AC Commutator Motors 133 developed again in the clockwise direction. Thus, an ordinary DC series motor if energised from an AC source, would develop unidirectional torque and the armature would rotate. The electromagnetic torque Te is proportional to the product of field flux φd and armature current i, is unidirectional but pulsating at double the supply frequency. However, an ordinary DC series motor when energised from AC source would operate inefficienctly because of inferior commutation poor pf and increased iron loss in poles and yoke which are usually unlaminated. Q.3. Draw and explain a phasor diagram of a single phase AC series motor. Ans. The vector diagram of single phase AC series motor is shown in fig. 5.2, the resistance drop IRse, IRp, IRc and IRa due to reactances of series field, interpole winding, compensating winding and of armature respectively are in phase with current. The reactance drops IXse, IXp, IXc and IXa due to reactances of series field, interpole winding, compensating and of armature respectively are in quardrature with the current and leading. Fig. 5.2 Vector Diagram of Single Phase Series Motor Counter emf is essentially 1800 out of phase with the current. The terminal voltage V is vector sum of reversed counter emf and voltage drops in armature, interpole, main field and compensating winding. Q.4. Discuss the modifications necessary to operate a DC series motor satisfactorily on single phase AC supply. Explain why the speed regulation of a series motor is greater when fed from AC mains than DC mains? (1991) Ans. The modifications necessary to operate a DC series motor satisfactorily on single phase AC supply are as follows: (i) The number of its series field turns should be as small as possible so that the field reactance may be reduced.

134 Electrical Machines–II (ii) The number of its armature conductors should be large to account for the reduction in torque due to fewer series field turns. (iii) The larger number of armature conductors would increase the armature reaction ampere-turns, hence, compensating winding has to be provided. (iv) The number of field poles is kept large, so that transformer emf in the short circuited armature winding elements may be brought down considerably to improve commutator sparking. (v) The armature and field pole cores are to be laminated for reducing the eddy-current losses. (vi) Lower voltage ratings (about 375 voltage at 50 Hz) are to be employed for bringing down the inductive reactance and improve the poor power factor. (vii) DC series motors work more efficiently at lower frequencies are used for AC traction system in Europe and America. (viii) Comparatively thinner splited brushes are employed. This is done to increase the resistance of the brush which should short circuit more than two coils at the same time during commutation. (ix) Single turn coils should be employed for its armature winding. As, n = E – ΣIR – ΣIX , it is clear that the speed will decrease a Kφ large amount as the load changes from no-load to full-load. So, regulation for the AC motor will be greater than for a DC motor. Q. 5. Draw torque-speed characteristics of a DC series motor and AC series motor together on the same axis. What problem arises when DC series motor is allowed to work on AC supply? Ans. The speed-torque characteristic curves of a DC series motor and AC series motor together on the same axis is shown in fig. 5.3. Fig. 5.3 Torque-Speed Curve

AC Commutator Motors 135 The problem arises when a DC series motor is allowed to work on AC supply are as follows: (i) Because of the increase in hysteresis and eddy-current losses due to the alternating flux created by AC supply, efficiency of the motor will be poorer. (ii) Power factor of the motor will be less. This is because of the inductive reactance of the field and armature winding. (iii) Considerable sparking at the brushes will occur. This is because in addition to the causes of sparking that occurs in a DC motor, transformer action on a coil undergoing commutation further intensifies comutation difficulties. The coil, short circuited by the brushes, links part of the constantly changing main field flux and hence, a voltage is induced in it. Q. 6. Show the difference in the vector diagram when a series motor is applied with DC and AC. Ans. Fig. 5.4 (a) and (b) shows the voltage relations of the series motor when it is supplied with DC and AC respectively. Of course in case of DC the voltage is nothing but a scalar quantity and represented by the equation, Fig. 5.4 V = Ea + I (ra + rf) Where, V = applied voltage, Ea = voltage induced or back emf, I = full-load current, ra, rf = resistance of armature and field respectively. The induced voltage in case of DC is proportional to the flux per pole and speed of the machine. On the other hand, in case of AC supply, the large reactance drops which will occur both in the armature and field. Hence, the value of back emf will be less than the value of back emf in the DC operation. Q.7. Draw circuit diagram of conductively and inductively compensated AC series motor. (2002)

136 Electrical Machines–II Ans. In small AC series motors, the commutating difficulties are not much importance since, these motor operate at lower currents. In addition, these motors run at high speed and therefore, their operating pf is relatively good. In the view of small AC series motors are usually uncompensated. But for larger series motors, a compensating winding is necessary to get better pf, better speed-torque characteristics and improved commutation. The function of compensating windings in both AC and DC series motors are similar. There are two methods of connecting the compensating windings. In fig. 5.5 (a), the compensating winding is connected in series with the armature and this arrangement is called conductivity compensated machine. An alternative arrangement called on inductively compensated machine, is given in fig. 5.5 (b), where compensating (a) Conductively Compensated (b) Inductively Compensated Fig. 5.5 winding is disconnected from the main circuit and short circuited on itself. The compensating winding now acts as the short circuited secondary of a transformer, the armature winding acting as its primary. Q.8. Write applications of universal motor. (2007, S/08) Explain the working of universal motor. (S/2009) Describe the construction, torque-speed characteristics and applications of a universal motor. (2009)

AC Commutator Motors 137 Write short note on universal motor. (S/2011, 11, 13, 14) Describe the construction, principle and applications of a universal motor. (S/2012) Explain working principle of universal motor. (2012, S/15, S/17) Explain working principle of universal motor and write its uses. (2016) Explain operation of universal motor. Draw the characteristics curve showing power output, current, torque and speed. Write its application. (2017) Ans. A universal motor is defined as a motor which may be operated either on direct current or single phase AC supply at approximately the same speed and output. In fact, it is a smaller version (5 to 150 W) of the AC series motor. Being a series wound motor, it has high starting torque and a variable speed characteristics. It runs at dangerously high speed on no-load. Thats why such motors are usually built into the device they drive. Construction: Generally universal motors are manufactured in two types which are as follows: (i) Non-Compensated Type: The non-compensated motor has two salient pole and is just like a 2-pole series DC motor except that whole of its magnetic path is laminated. The laminated stator is necessary because the flux is alternating when the motor is operated from AC supply. The armature is of wound type and similar to that of a small DC motor. It consists essentially of a laminated core having either straight or skewed slots and a commutator to which the leads of the armature winding are connected. Fig. 5.6

138 Electrical Machines–II (ii) Distributed Field Compensated Type: The distributed field compensated type motor has wound armature similar to that of a small DC motor. The compensating winding is used to reduce the reactance voltage present in the armature when motor runs on AC supply. This voltage is caused by alternating flux by transformer action. In a 2-pole compensated motor, the voltage induced by transformer action in a coil during its commutation period is not sufficient to cause any serious commutation problem. Moreover high resistance brushes are used to aid commutation. Operation/Working Principle: Such motors develop unidirectional torque, regardless of whether they operate on DC or AC supply. The connection of unidirectional torque, when the motor runs on AC supply can be easily understood from fig. 5.6. The motor works on the same principle as DC motor, i.e. force between the main pole flux and the current carrying armature conductors. This is true regardless of whether the current is alternating or direct. Fig. 5.7 Characteristic Curve of Universal Motor Torque-Speed Characteristics: The characteristics curve of universal motor is very much similar to those of DC series motors which is shown in fig. 5.7. The no-load speed of a universal motor are unlike than other machines, is very high (of the order of 2000 rpm). Therefore, the motor is smaller in size and used for lower rating generally from 1 HP to 1 HP. 200 3 Uses/Application: Universal motors are used in vacuum cleaners, hair dryers where actual motor speed is the load speed. Other applications where motor speed is reduced by a gear train are drink and food mixer, portable drills, domestic sewing machines, etc. Q. 9. Draw and explain the phasor diagram for a single phase series motor. (S/2010)

AC Commutator Motors 139 Ans. The schematic diagram and phasor diagram for the conductively coupled single phase AC series motor are shown in fig. 5.8 and fig. 5.9 respectively. Fig. 5.8 Schematic Circuit Diagram of a Single Phase Series Motor Fig. 5.9 Phasor Diagram The resistance drops IaRse, IaRi, IaRc and IaRa due to resistance of series field, interpole winding, compensating winding and of armature respectively are in phase with armature current Ia. The reactance drop IaXse, IaXi, IaXc and IaXa due to reactance of series field, interpole winding, compensating winding and of armature respectively lead current Ia by 900. The generated armature counter emf is Eg. The terminal phase voltage Vp is equal to the phasor sum of Eg and all the impedances drops in series, Vp = Eg + IaZse + IaZi + IaZc + IaZa The power factor angle between Vp and Ia is φ.

140 Electrical Machines–II Q.10. How will you change the direction of rotation of an universal motor? (S/2010) Ans. The concentrated pole (or salient pole) type universal motor may be reversed by reversing the flow of current through either the armature or field winding. The usual method is to interchange the leads on the brush holders. The distributed field compensated type universal motor may be reversed by interchanging either the armature or field leads and shifting the brushes against the direction in which the motor will rotate. The extent of brush shifts usually amounts to several commutator bars. Q.11. What are the use of commutating poles in DC series motor? Ans. All DC series motor are used commutating poles for improving commutation. But only commutating poles will not produce satisfactory commutation unless some action is done to neutralize the huge voltage induced in the short circuited armature, coil by the transformer. In AC series motor, the flux produced by the field winding is alternating and it induces voltage (by transformer action) in the short circuited armature coil during its commutation period. The field winding connected with the armature coil under going commutation acts as primary and the armature coil during its commutating period acts as a short circuited secondary as shown in fig. 5.10 (a). This transformer action produces heavy current in the armature coil as it passes through it commutating period and occurs vicious sparking, unless transformer voltage is neutralized. Where, CW is commutation winding, FW is field winding. Fig. 5.10 Fig. 5.10 (b) shows the phasor diagram of a shunted commutator pole. The current Ic through the commutating pole resolve into two components Id and Iq. Id produces a flux which is in the phase with total motor current I whereas flux produced by Iq lags the I by 900. By

AC Commutator Motors 141 proper adjustment by shunt resistance, the speed voltage produced in short circuited coil by the cutting of the 900 lagging component of the commutating pole, flux can be made to neutralize the voltage induced by transformer action. Q.12. How will you speed up the commutation process? Where are the commutating poles provided? Ans. To speed up the commutation process, the reactance voltage must be neutralized by injecting a suitable polarity dynamical (speed) voltage into the commutating coils. In order that this injection is restricted to commutating coils, narrow interpoles (also called commutating poles or compoles) are provided in the interpolar region. Q.13. Write short note on repulsion motor. (S/2000, S/01, 04, 06, 08, S/16) Describe the construction and working of a repulsion motor. (S/2010) Describe the construction and working principle of repulsion motor. Write the short comings of repulsion motor. (S/2013) Explain the working principle of repulsion motor. (2015, 17) Ans. Repulsion motor consists of one stator winding, one rotor which is wound like DC armature, commutator and a set of brushes, which are short circuited and remains in contact with the commutator all times. It operates continuously on the repulsion principle. No short circuited mechanism is required for this type of motor. Construction: In AC series motors, the rotor and stator windings are conductively coupled i.e., the rotor current is obtained by conduction from the stator. Repulsion motor are similar to series motor except that rotor and stator windings are inductively coupled. It means the current is obtained by transformer action from the stator. The stator Fig. 5.11 Schematic Diagram of a Repulsion Motor

142 Electrical Machines–II carries a distributed winding like the main winding of the ordinary single phase induction motor. The rotor or armature has a drum type winding connected to the commutator. The brushes are fixed directly opposite to each other short circuited such a motor as shown in fig. 5.11. The rotor mmf setup by current is along the brush axis. Working: In fig. 5.11 when angle α is 900, then the magnetic axis of stator and rotor are in quadrature as shown in fig. 5.12 (a) and there is therefore, no mutual induction between stator and rotor windings. Consequently, the voltage across the brushes is zero, rotor-induced currents are zero and hence, no electromagnetic torque is developed. Therefore, motor will not run when the angle α = 900. In view of this, the brush position of α = 900 is called open circuit, no-load, high- impedance or neutral position. Fig. 5.12 When angle α = 00 as shown in fig. 5.12 (b), the stator and rotor magnetic axes coincide, with the result that the mutual induction between the two windings is maximum. The large rotor currents produce rotor mmf opposite to the stator mmf. Since, two mmf’s are along the same axis, no torque is developed. Further, this condition of α = 00, is similar to transformer, on short circuit. Therefore, this position of α = 00, is called the short circuit, maximum current or low impedance position. Fig. 5.13

AC Commutator Motors 143 Actual, the brush axis is neither along α = 900 nor along α = 00, but occupies some intermediate position as shown in fig. 5.13. If the stator mmf at any instant is directed from A to B, then the rotor-induced mmf must have a component opposite to the stator mmf at the same instant, i.e. the rotor induced mmf must be directed from C to D in fig. 5.13. The stator polarity at A is S1 and at the same instant, rotor-induced polarity at C is S2. Repulsion between like S1, S2 and N1, N2 results in the clockwise direction of rotation. It is obvious that the direction of rotation can be reversed merely by shifting the brush axis to the other side of the field winding axis. The principal characteristics of repulsion motor are as follows: (i) high-starting torque, (ii) wide range of speed variation by brush shifting. Short Comings or Disadvantages: The short comings or disadvantages of repulsion motor are as follows: (i) Occurrence of sparks at brushes. (ii) Commutator and brushes wear out quickly. This is primarily due to arcing and heat generated at brushes assembly. (iii) The pf is poor at low speeds. (iv) No-load speed is very high and dangerous. Q.14. Write a short note on repulsion-start induction-run motor. (S/2012) Ans. The repulsion-starting induction-running motor, called a repulsion-start motor, is a single phase motor which uses the repulsion principle for starting, and the induction principle for running. These motors have the general construction of a repulsion motor. The only difference is that in addition to the basic repulsion motor construction they are equipped with a centrifugal device which operates somewhat below Fig. 5.14 Speed-Torque Characteristic normal speed and short circuits all of Repulsion-Start Induction-Motor of the commutator bars. In some motors the centrifugal device also lifts the brushes from the commutator in order to reduce wear of the brushes and commutator bars and to make the running operation more quiet. When the commutator bars are short circuited the machine becomes a single phase induction motor. These motors, therefore, start as repulsion motors and after the action of the centrifugal device run as single-phase induction motors.

144 Electrical Machines–II The starting torque is 2.5 to 4.5 percent of full-load torque, and the starting current is approximately 3.75 percent of full-load current. These motors are made in the large functional-horsepower sizes and upto 15 hp. The applications of these motors include air compressors and refrigerators. The speed-torque characteristic of repulsion-start induction motor is shown in fig. 5.14. Q.15. What is a repulsion induction motor? Explain its principle of operation and draw its speed-torque characteristic. (S/2011) Draw speed-torque characteristics of a repulsion induction motor. (S/2017) Ans. The repulsion induction motor produces high starting torque through repulsion motor action. When running, it functions through a combination of induction motor and repulsion motor action. These motors have a single distributed stator winding. The rotor is provided with two independent windings. Fig. 5.15 Speed-Torque Curve of Repulsion Induction Motor Principle of Operation: One is a repulsion armature winding with induction motor winding. There is no centrifugal device, and the repulsion winding is actively functioning at all times. Thus, both rotor winding are operative when the machine is running, and its performance is due to the combined effects of the two windings. Its characteristics will be a combination of those for a single phase induction motor and a repulsion motor. The torque-speed characteristic for a repulsion induction motor is given in fig. 5.15. The repulsion-start type of motor will run at a speed somewhat below synchronous, since it runs like an ordinary induction motor, except during the starting period. The repulsion induction motor on the other hand reaches, at no-load, a speed slightly above synchronous because of the effect of the repulsion winding. At full-load the speed is

AC Commutator Motors 145 somewhat below synchronous, as in an induction motor. The speed regulation is about 6 percent. The starting torque is 225 to 300 percent of the full-load torque, the lower value being for the larger motors. The starting current is from 3 to 4 times the full-load current. NUMERICAL PROBLEMS AND SOLUTIONS Prob.1 A single phase half hp series motor has a resistance of 30 Ω and total induction of 0.5 H. When connected to a 250 V DC supply and loaded to take 0.8 A it runs at 2000 rpm. Calculate the speed and pf when connected to 250 V, 50 Hz supply and loaded to same current. Sol. When connected to DC supply, back emf developed, Eb = V – IR = 250 – (0.8) × (30) = 226 V Speed, N = 2000 rpm When connected to AC supply, Voltage, V = 250 V Line current, IL = 0.8 A Resistance drop, IR = 0.8 × 30 = 24 V Reactance drop, IX = I × 2πfL = 0.8 × 2π × 50 × 0.5 = 125.6 V From phasor diagram, fig. 5.16, sin φ = IX V φ = sin–1 IX = sin–1 125.6 V 250 = 30.150 Power factor, cos φ = cos 30.150 = 0.86 Ans. Developed back emf, Eb' = V cos φ – IR = 250 × 0.86 – 24 Fig. 5.16

146 Electrical Machines–II = 191 V Speed, N' = N × Eb' Eb = 2000 × 191 226 = 1690 rpm Ans. Prob.2. A 250 W, single phase, 50 Hz, 220 V universal motor runs at 2000 rpm and takes 1A when supplied from a 220 V DC supply. If the motor is connected to 220 V AC supply and takes 1 A (rms), calculate the speed, torque and power factor. Assume Ra is 20 Ω and La = 0.4 H. Sol. For DC operation, Eb = V – IaRa = 220 – 20 × 1 = 200 V For AC operation, as shown in fig. 5.17. Fig. 5.17 From this figure, Xa = 2π f La = 2π × 50 × 0.4 = 125.7 Ω V 2 = (Eb.ac + IaRa)2 + (IaXa)2 Eb.ac = –IaRa + √V2 – (IaXa)2 = –1 × 20 + √(220)2 – (125.7 × 1)2 = 160.5 V Since, the armature current is same for both DC and AC excitations, Eb.dc = Ndc Eb.ac Nac ∴ Nac = Ndc × Eb.ac Eb.dc = 2000 × 160.5 200 = 1605 rmp Ans.

AC Commutator Motors 147 Power factor, cosφ = AB = Eb.ac + IaRa OB V = 160.5 + 20 220 = 0.82 lag Ans. Pmech = Eb.ac × Ia = 160.5 × 1 = 160.5 W Torque, T = 9.55 × Pmech Nac = 9.55 × 160.5 1605 = 0.955 N-m Ans. Prob.3. A universal series motor has resistance of 30 Ω and an inductive of 0.5 H. When connected to a 250 V DC supply and loaded to take 0.8 A, it runs at 2000 rpm. Calculate its pf and speed, when connected to a 250 V, 50 Hz, AC supply and loaded to take the same current. Sol. Given, Frequency, f = 50 Hz Inductance, La = 0.5 H Resistance, Ra = 30 Ω For AC operation on armature reactance, Xa = 2π f La = 2π × 50 × 0.5 = 157 Ω IaRa = 0.8 × 30 = 24 Ω IaXa = 0.8 × 157 = 125.6 V Fig. 5.18

148 Electrical Machines–II The vector diagram is shown in fig. 5.18. V2 = (Eb.ac + IaRa)2 + (IaXa)2 (250)2 = (Eb.ac + 24)2 + (125.6)2 Eb.ac = 192.16 V For DC operation, Eb.dc = V – IaRa = 250 – 0.8 × 30 = 226 V Eb.ac = Nac or Nac= Eb.ac × Ndc Eb.dc Ndc Eb.dc Nac = 192.16 × 2000 226 = 1700.5 rpm Ans. ( )Power factor, cos φ = Eb.ac + IaRa V = 216.16 250 = 0.864 lagging Ans. FF

149 6 SPECIAL PURPOSE MACHINES Chapter SUBJECTIVE QUESTIONS AND ANSWERS Q.1. Write working principle, construction of linear induction motor. Write short note on linear induction motor. (2000, S/04, 05, S/16) Write short note on working principle and application of linear induction motor. (S/2003) Describe the working principle of linear induction motor and write its applications. (2016) Ans. If the sector of induction is laid out flat squirrel cage winding is brought near to it, is known as linear induction motor. In place of flat squirrel cage winding, an aluminium or iron or copper plate is used as rotor. The flat stator produces a flux that moves in a straight line from its one end to the other at a linear synchronous speed. It is given by, Vl = 2wf Where, Vl = linear synchronous speed (m/s), w = width of one pole pitch (m), f = frequency (Hz). In this motor, speed does not depend on number of poles but only on pole pitch and stator supply frequency. Working Principle: The principle of operation of a linear induction motor is the same as that of a rotary induction motor. The relative movement between the magnetic field and the short circuited conductors induces emf in the conductors that causes current to flows in them. This induced current produces electromagnetic forces, which tend to rotate or move the conductors, if they are free to rotate. If the conductors are stationary then a force will produce which will tend to move the system comprising the magnetic field. This happens because the conductors tend to move in such a way so as to eliminate the induced currents in accordance with Lenz’s Law. Now, if the movement of the magnetic field is rotational then the movement of the conductors is also rotational. But, if the movement of the magnetic field is rectilinear then the movement of the conductors will also be rectilinear. Construction : The construction of a linear induction motor may be viewed as the development of the cylindrical stator-bore and the (149)

150 Electrical Machines–II Fig. 6.1 cylindrical rotor surfaces about their shaft axis, with their respective windings and cores, as shown in the fig. 6.1. Applications: The applications of linear induction motor are as follows: (i) Linear induction motor can be used in conveyors, travelling cranes, haulers, electromagnetic pumps, high speed rail traction etc. (ii) These motors are used in which the field systems moves and the conducting plate remains stationary such as travelling craves motors. (iii) These motors can also be used where the field system remains stationary and the conducting plate moves such as in automatic sliding doors in electric trains, metallic belt conveyors, etc. (iv) It can be used on trolley cars for internal transport in workshop, as booster accelerator for moving heavy trains from rest or up the inclines or on clomes or as a propulsion unit in marshalling yards in place of shunting locomotives. (v) Linear induction motor has superiority over conventional rotary motor for speed over 2000 kmph. It provide excellent source of motive power for magnetically suspended trains where conventional rotary motor fails. Q.2. Write down various properties of linear induction motor. Ans. These properties are almost identical to those of a standard rotating machine. The various properties of linear induction motor are as follows: