Single Phase induction Motor [1/Ch. 36]

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1 Single Phase induction Motor [1/h. 6] Equivalent ircuit of a Single-Phase nduction Motor without ore Loss [1/6.5/p.17] A single-phase motor may be looked upon as consisting of two motors, having a common stator winding, but with their respective rotors revolving in opposite directions. The equivalent circuit of such a motor based on double-field revolving theory is shown in Fig Fig Here the single-phase motor has been imagined to be made up of (i one stator winding and (ii two imaginary rotors. The stator impedance is =R 1 +jx 1. The impedance of each rotor is (r 2 +jx 2 where r 2 and x 2 represent half the actual rotor values in stator terms (i.e. x 2 stands for half the standstill reactance of the rotor, as referred to stator. Since iron loss has been neglected, the exciting branch is shown consisting of exciting reactance only. Each rotor has been assigned half magnetizing reactance (i.e. x m represents half the actual reactance. The impedance of forward running rotor is jxm ( + jx2 s f = + j( xm + x2 s And it runs with a slip of s. The impedance of backward running motor is jxm ( + jx2 2 s b = + j( xm + x2 2 s And it runs with a slip of (2-s. Under standstill condition, V f =V b, but under running conditions Vf is almost 90 to 95% of the applied voltage. 2 The forward torque in synchronous watts ist f =. Similarly, backward torque s 2 5 ist b =. The total torque is 2 s (2 s 5 s 2 s( + 5 T = T f Tb = = =. s 2 s s(2 s s(2 s Equivalent ircuit of a Single-Phase nduction Motor with ore Loss [1/6.6/p.174] Prepared by Dr. M. A. Mannan Page 1 of 8

2 The core loss can be represented by an equivalent resistance which may be connected either in parallel or in series with the magnetizing reactance as shown in Fig Since under running condition V f is very high (and V b is correspondingly, low most of the iron loss takes place in the forward motor consisting of the common stator and forward-running rotor. ore-loss current w =core-loss/v f. Hence, half value of core-loss equivalent resistance is r c =V f / w. As shown in Fig 6.15, r c has been connected in parallel with x m in each rotor. (a Fig (b Example 6.. A 250 W, 20 V, 50 Hz capacitor-start motor has the following constants for the main and auxiliary winding: Main winding, m =(4.5+j.7 ohm, Auxiliary winding, a =(9.5+j.5 ohm. Determine the value of the starting capacitor that will place the main and auxiliary winding currents in quadrature at starting. Solution: Let X be the reactance of the capacitor connected in the auxiliary winding. Then a =9.5+j(.5-X =(9.5-jX ohm where, X is the net reactance Now, m =(4.5+j.7= o ohm Obviously, m lags behind V by 9.4 o. Since time phase angle between m and a has to be 90 o, a must lead V by (9.4 o - 90 o =-50.6 o. For auxiliary winding, tanφ a =(.5-X /R or tan(-50.6 o = (.5-X /9.5= Or (.5-X = = ohm X = =15.06 ohm Or 1/(2 π 50 =15.06 or =1/(2 π = F =211 µf. Prepared by Dr. M. A. Mannan Page 2 of 8

3 apacitor Start-and-Run Motor [1/6.8/181] This motor is similar to the capacitor-start motor except that the starting winding and capacitor are connected in the circuit at all time. The advantage of leaving the capacitor permanently in circuit are: (i improvement of over-load capacity of the motor, (ii a higher power factor, (iii higher efficiency, and (iv quieter running of the motor which is so much desirable for small power drivers in offices and laboratories. Some of these motors which start and run with one value of capacitance in the circuit are called single-value capacitor-run motor. Other which start with high value of capacitance but run with a low value of capacitance are known as two-value capacitor-run motor. Single-value apacitor Run Motor t has one running winding and one starting winding in series with a capacitor as shown in Fig Since the capacitor remains in the circuit permanently, this motor is often referred to as permanent-split capacitor-run motor and behaves practically like an unbalanced two-phase motor. Fig Fig Fig. 6.0 Obviously, there is no need to use centrifugal switch which has necessary in the case of capacitor-start motors. Since the same capacitor is used for starting and running, it is obvious that neither optimum starting nor optimum running performance can be obtained because value of capacitance used must be a compromise between the best value for starting and that for running. Generally, capacitors of 2 to 20 µf capacitance are employed and are more expensive oil or pyranol-insulated foil-paper capacitors because of continuous-duty rating. The low value of the capacitor results in small starting torque which is about 50 to 100% of the rated torque as shown in Fig onsequently, these motors are used where the required starting torque is low such as airmoving equipment i.e. fans, blowers and voltage regulator and also oil burners where quiet operation is particularly desirable. One unique feature of this type of motor is that it can be reversed by an external switch provided its running and staring winding are identical. One serves as the running winding and the other as a starting winding for one direction of rotation. For reverse rotation, the one that previously served as a running winding becomes the starting winding while the former starting winding serves as a running winding. As seen from Fig. 6.0 when the switch in the forward position, winding B Prepared by Dr. M. A. Mannan Page of 8

4 serves as running winding and A as starting winding. When switch is in reverse position, winding A becomes the running winding and B the starting winding. Such reversible motors are often used for operating device that must be moved back and forth very frequently such rheostats, induction regulations, furnace controls, valves and arc-welding controls. Two-Value apacitor-run Motor This motor starts with a high capacitor in series with the starting winding so that the starting torque is high. For running, a lower capacitor is substitute by the centrifugal switch. Both the running and starting windings remain in circuit. The two values of capacitance can be obtained as follows: 1. by using two capacitors in parallel at the starting and then switching out one for low-value run as shown in Fig. 6.1, or 2. by using a step-up auto-transformer in conjunction with one capacitor so that effective capacitance value is increased for starting purpose. n Fig. 6.1, B is an electrolytic capacitor of high capacity (short duty and A is an oil capacitor of low value (continuous duty. Generally, starting capacitor B is 10 to 15 times the running capacitor A. At the start, when the centrifugal switch is closed, the two capacitors are put in parallel, so that their combined capacitance is the sum of their individual capacitances. After the motor has reached 75% full-load speed, the switch opens and only capacitor A remains in the starting winding circuit. n this way, both optimum starting and running performance is achieved in such motors. f properly designed, such motors have operating characteristics very closely resembling those displayed by two-phase motors. Their performance is characterized by 1. ability to start heavy loads, 2. extremely quiet operation,. higher efficiency and power factor, and 4. ability to develop 25% overload capacity Hence, such motors are ideally suited where load requirements are severe as in the case of compressors and fire strokers etc. As motor speeds up, the centrifugal switch shifts the capacitor from one voltage tap to another so that the voltage transformation ratio changes from higher value at starting to a lower value for running. The capacitor which is actually of the paper-tinfoil construction is immersed in a high grade insulation like wax or mineral oil. The use of an auto-transformer and single oil-type capacitor is illustrated in Fig The transformer and capacitor are sealed in a rectangular iron box and mounted on top of the motor. The idea behind using this combination is that a capacitor of value connected to the secondary of a step-up transformer appears to the primary as though it had a value of K 2 where K is voltage transformation ratio. Prepared by Dr. M. A. Mannan Page 4 of 8

5 Fig. 6.1 Fig. 6.2 For example, if actual value of =4 µf and K=6, then low-voltage primary acts as if it had a 144 µf (=6 2 4 capacitor connected across its terminals. Obviously, effective value of capacitance has increased 6 times. n the start position of the switch, the connection is made to the mid-tap of the auto-transformer so that K=2. Hence, effective value of capacitance at start is 4 times the running value and is sufficient to give a high starting torque. Problem A 250 W, 20 V, 50 Hz capacitor-start motor has the following constants for the main and auxiliary winding: Main winding, m =(4.5+j.7 ohm, Auxiliary winding, a =(9.5+j.5 ohm. alculate (a the value of the starting capacitor that will place the main and auxiliary winding currents in quadrature at starting, and (b the magnitudes and phase angles of the currents in the main and auxiliary windings when rated voltage is applied to the motor under starting conditions. Solution: Let X be the reactance of the capacitor connected in the auxiliary winding. Then a =9.5+j(.5-X =(9.5-jX ohm where, X is the net reactance Now, m =(4.5+j.7= o ohm Obviously, m lags behind V by 9.4 o. Since time phase angle between m and a has to be 90 o, a must lead V by (9.4 o - 90 o =-50.6 o. For auxiliary winding, tanφ a =(.5-X /R or tan(-50.6 o = (.5-X /9.5= Or (.5-X = = ohm X = =15.06 ohm Or 1/(2 π 50 =15.06 or =1/(2 π = F =211 µf. Problem: A 220 V, 1.5 hp, 59 Hz, two poles, capacitor start induction motor has the following main winding impedances: R 1 =4 ohm, X 1 =2.0 ohm, X m =200 ohm, R 2 = ohm and X 2 = 4.04 ohm. At slip of 0.05, calculate (i stator current, (ii stator power factor, (iii input power, (iv forward torque, and (v backward torque. Prepared by Dr. M. A. Mannan Page 5 of 8

6 Example 6.. A 250 W, 20 V, 50 Hz capacitor-start motor has the following constants for the main and auxiliary winding: Main winding, m =(4.5+j.7 ohm, Auxiliary winding, a =(9.5+j.5 ohm. Determine the value of the starting capacitor that will place the main and auxiliary winding currents in quadrature at starting. Problem W, 120-V, 50-Hz capacitor start motor has main winding impedance m = + j4 Ω and auxiliary winding impedance a = 6 + j8 Ω at starting. Find the value of starting capacitance that will place the main and auxiliary winding currents in quadrature at starting. Problem [6]: The impedance of the main and auxiliary windings of a 50 Hz single-phase induction motor are +j Ω and 6+jΩ respectively. What will be the value of the capacitor to be connected in series with auxiliary winding to achieve a phase difference of 90 o between the currents of the two windings? V 0 V 0 V 45 V 0 V 0 V 26.5 m = = = a = = = + j j The current flowing through the auxiliary winding after connecting a capacitor in series should make an angle 90 o with m or make an angle 90 o -45 o = 45 o with the applied voltage V. Since the new current of auxiliary winding should be leading the voltage V by an angle of 45 o, the capacitive reactance of the auxiliary circuit is greater than the inductive reactance. Thus X X L (1/ ω tan 45 = 1 = (1/ ω = 6 (1/ ω = 9 R 6 ω = 1 / 9; = 1/ 9ω = 5. 6 µ F Example 6.2 [SKB]: A 50 Hz Split-phase induction motor has a resistance 5 Ω and an inductive reactance 20 Ω in both main and auxiliary winding. Determine the value of resistance and capacitance to be added in series with auxiliary winding to send the same current in each winding with a phase difference of 90 o. = 5 + j20 = V 0 V 0 V 76 m = a = = = 5 + j The current flowing through the auxiliary winding after connecting resistor R and a capacitor in series should make an angle 90 o with m or make an angle 90 o -76 o =14 o with the applied voltage V. Since the new current of auxiliary winding should be leading the voltage V by an angle of 14 o, the capacitive reactance of the auxiliary circuit is greater than the inductive reactance. Thus 5 + R cos14 = ; 5 + R = cos14 ; R = 20.6 cos14 5 = = Ω X X L Again sin 14 = ; X X = sin14 ; X = 20.6sin14 + X = 20.6sin = Ω ( 1/ ω = = 127 µ F X L Problem: A 50 Hz split-phase induction motor has a resistance 10 Ω and an inductive reactance 80 Ω in both main and auxiliary winding. Determine the value of resistance and capacitance to be added in series with auxiliary winding to send the same current in each winding with a phase difference of 90 o. Solution: = 10 + j80 = m = a V 0 V 0 V = = = 10 + j L Prepared by Dr. M. A. Mannan Page 6 of 8

7 The current flowing through the auxiliary winding after connecting resistor R and a capacitor in series should make an angle 90 o with m or make an angle 90 o o =7.1 o with the applied voltage V. Since the new current of auxiliary winding should be leading the voltage V by an angle of 14 o, the capacitive reactance of the auxiliary circuit is greater than the inductive reactance. Thus 10 + R cos7.1 = ; 10 + R = cos 7.1 ; R = = Ω X Again sin 7.1 = X L ; X 80 = sin 7.1 ; X = sin X = = Ω 1 ( 1/ ω = = = F Possible Questions 1. Write different classifications of single phase induction motors. 2. lassify the single-phase motors depending on their construction and method of starting.. What are the differences between single-phase and three-phase induction motor constructionally? 4. State and explain double field revolving theory. 5. Draw the forward, backward, resultant, torques versus slip curve for a single phase induction motor according to the double-field revolving theory. 6. Write the equations for forward, backward, resultant, torques at standstill conditions. 7. Using double field revolving theory, show that the forward torque and the backward torque developed by a single phase induction motor is the same in magnitude at unity slip. 8. Using double field revolving theory, show that the starting torque of a single-phase induction motor is zero. What is the remedy of this problem? 9. Prove that for a single phase induction motor running at a speed of N rpm, slip corresponding to the backward field (s b = 2 s, here s is the slip corresponding to the forward field. 10. State and explain cross-field theory. 11. Using the cross-field theory, briefly explain why the single phase induction motor is not selfstarting. 12. Using the cross-field theory, briefly explain why the single phase induction motor will continue to rotate; producing induction motor torque in a rotating magnetic field, once a rotational emf has been initiated. 1. Define a split phase induction motor. 14. Why is phase splitting necessary in a single phase induction motor? 15. Why two windings (one main and other auxiliary in a single-phase induction motor are necessary, briefly explain. 16. Draw the equivalent circuit of a single-phase induction motor without core loss. 17. Draw the equivalent circuit of a single-phase induction motor with core loss. 18. With necessary circuit and vector diagram, briefly explain the operation of the split-phase (or resistance-start split phase induction motor. 19. Draw the circuit diagram and vector diagram for a resistance-start split phase induction motor. 20. Draw the torque vs speed curve for a resistance-start split phase induction motor. 21. With necessary circuit and vector diagram, briefly explain the operation of the capacitor-start induction run motor. 22. Draw the circuit diagram and vector diagram for a capacitor-start induction run motor. 2. Draw the torque vs speed curve for a capacitor-start induction run motor. Prepared by Dr. M. A. Mannan Page 7 of 8

8 24. What are the differences between resistance-start and capacitor-start single-phase induction motors? 25. What are the advantages of leaving the capacitor permanently in the auxiliary circuit of a singlephase induction motor? 26. With necessary circuit diagram, briefly explain the operation of the single-value capacitor-run induction motor. 27. Explain why in the case of single-value capacitor-run induction motor, neither optimum starting nor optimum running performance can be obtained. 28. With necessary circuit diagram, briefly explain the operation of the two-value capacitor-run induction motor. 29. What are the advantages of two-value capacitor-run induction motor? References [1] B. L. Theraja, A. K. Theraja, A Textbook of ELETRAL TEHNOLOGY in S Units Volume, A & D Machines, S. hand & ompany Ltd., (Multicolour illustrative Edition. [2] A. F. Puchstein, T.. Lloyd, A.G. onrad, Alternating urrent Machines, 1942, Asia Publishing House, Third Edition (Fully revised and corrected Edition [] Jack Rosenblatt, M. Harold Friedman, Direct and Alternating urrent Machinery, ndian Edition (2 nd Edition, BS Publishers & Distributors. [4] A. E. Fitzgerald, harles Kingsley, Jr. Stephen D. Umans, Electric Machinery, 5 th Edition in S units, 1992 Metric Edition, McGraw Hill Book ompany. [5] rving L. Kosow, Electrical Machinery and Transformers, Second Edition, Prentice Hall ndia Pvt. Limited. [6] S. K. Bhattacharya, Electrical Machines, Second Edition, Tata McGraw-Hill Publishing ompany, New Delhi. Prepared by Dr. M. A. Mannan Page 8 of 8

EE 350: Electric Machinery Fundamentals

EE 350: Electric Machinery Fundamentals Lecture Schedule See Time Table Course Type, Semester Fundamental Engineering, Fifth Credit Hours Three + One Pre-requisite Physics Instructor Dr. Muhammad Asghar