Chapter (11) Synchronous Motors Introduction It may be recalled that a d. c. generator can be run as a d. c. motor. In like manner, an alternator may operate as a motor by connecting its armature winding to a 3-phase supply. It is then called a synchronous motor. As the name implies, a synchronous motor runs at synchronous speed (Ns = 120f/P) i. e. , in synchronism with the revolving field produced by the 3-phase supply. The speed of rotation is, therefore, tied to the frequency of the source.
Since the frequency is fixed, the motor speed stays constant irrespective of the load or voltage of 3phase supply. However, synchronous motors are not used so much because they run at constant speed (i. e. , synchronous speed) but because they possess other unique electrical properties. In this chapter, we shall discuss the working and characteristics of synchronous motors. 11. 1 Construction A synchronous motor is a machine that operates at synchronous speed and converts electrical energy into mechanical energy. It is fundamentally an alternator operated as a motor.
Like an alternator, a synchronous motor has the following two parts: (i) a stator which houses 3-phase armature winding in the slots of the stator core and receives power from a 3-phase supply [See (Fig. (11. 1)]. (ii) a rotor that has a set of salient poles excited by direct current to form alternate N and S poles. The exciting coils are connected in series to two slip rings and direct current is fed into the winding from an external exciter mounted on the rotor shaft. The stator is wound for the same number of poles as the rotor poles.
As in the case of an induction motor, the number of poles determines the synchronous speed of the motor: Fig. (11. 1) 293 Synchronous speed, N s = where 120f P f = frequency of supply in Hz P = number of poles An important drawback of a synchronous motor is that it is not self-starting and auxiliary means have to be used for starting it. 11. 2 Some Facts about Synchronous Motor Some salient features of a synchronous motor are: (i) A synchronous motor runs at synchronous speed or not at all. Its speed is constant (synchronous speed) at all loads. The only way to change its speed is to alter the supply frequency (Ns = 120 f/P). ii) The outstanding characteristic of a synchronous motor is that it can be made to operate over a wide range of power factors (lagging, unity or leading) by adjustment of its field excitation. Therefore, a synchronous motor can be made to carry the mechanical load at constant speed and at the same time improve the power factor of the system. (iii) Synchronous motors are generally of the salient pole type. (iv) A synchronous motor is not self-starting and an auxiliary means has to be used for starting it. We use either induction motor principle or a separate starting motor for this purpose.
If the latter method is used, the machine must be run up to synchronous speed and synchronized as an alternator. 11. 3 Operating Principle The fact that a synchronous motor has no starting torque can be easily explained. (i) Consider a 3-phase synchronous motor having two rotor poles NR and SR. Then the stator will also be wound for two poles NS and SS. The motor has direct voltage applied to the rotor winding and a 3-phase voltage applied to the stator winding. The stator winding produces a rotating field which revolves round the stator at synchronous speed Ns(= 120 f/P).
The direct (or zero frequency) current sets up a two-pole field which is stationary so long as the rotor is not turning. Thus, we have a situation in which there exists a pair of revolving armature poles (i. e. , NS ? SS) and a pair of stationary rotor poles (i. e. , NR ? SR). (ii) Suppose at any instant, the stator poles are at positions A and B as shown in Fig. (11. 2 (i)). It is clear that poles NS and NR repel each other and so do the poles SS and SR. Therefore, the rotor tends to move in the anticlockwise direction. After a period of half-cycle (or ? = 1/100 second), the polarities of the stator poles are reversed but the polarities of the rotor poles remain the same as shown in Fig. (11. 2 (ii)). Now SS and NR attract 294 each other and so do NS and SR. Therefore, the rotor tends to move in the clockwise direction. Since the stator poles change their polarities rapidly, they tend to pull the rotor first in one direction and then after a period of half-cycle in the other. Due to high inertia of the rotor, the motor fails to start. Fig. (10. 2) Hence, a synchronous motor has no self-starting torque i. e. , a synchronous motor cannot start by itself.
How to get continuous unidirectional torque? If the rotor poles are rotated by some external means at such a speed that they interchange their positions along with the stator poles, then the rotor will experience a continuous unidirectional torque. This can be understood from the following discussion: (i) Suppose the stator field is rotating in the clockwise direction and the rotor is also rotated clockwise by some external means at such a speed that the rotor poles interchange their positions along with the stator poles. (ii) Suppose at any instant the stator and rotor poles are in the position shown in Fig. 11. 3 (i)). It is clear that torque on the rotor will be clockwise. After a period of half-cycle, the stator poles reverse their polarities and at the same time rotor poles also interchange their positions as shown in Fig. (11. 3 (ii)). The result is that again the torque on the rotor is clockwise. Hence a continuous unidirectional torque acts on the rotor and moves it in the clockwise direction. Under this condition, poles on the rotor always face poles of opposite polarity on the stator and a strong magnetic attraction is set up between them.
This mutual attraction locks the rotor and stator together and the rotor is virtually pulled into step with the speed of revolving flux (i. e. , synchronous speed). (iii) If now the external prime mover driving the rotor is removed, the rotor will continue to rotate at synchronous speed in the clockwise direction because the rotor poles are magnetically locked up with the stator poles. It is due to 295 this magnetic interlocking between stator and rotor poles that a synchronous motor runs at the speed of revolving flux i. e. , synchronous speed. Fig. (11. 3) 11. Making Synchronous Motor Self-Starting A synchronous motor cannot start by itself. In order to make the motor self-starting, a squirrel cage winding (also called damper winding) is provided on the rotor. The damper winding consists of copper bars embedded in the pole faces of the salient poles of the rotor as shown in Fig. (11. 4). The bars are short-circuited at the ends to form in effect a partial Fig. (11. 4) squirrel cage winding. The damper winding serves to start the motor. (i) To start with, 3-phase supply is given to the stator winding while the rotor field winding is left unenergized.
The rotating stator field induces currents in the damper or squirrel cage winding and the motor starts as an induction motor. (ii) As the motor approaches the synchronous speed, the rotor is excited with direct current. Now the resulting poles on the rotor face poles of opposite polarity on the stator and a strong magnetic attraction is set up between them. The rotor poles lock in with the poles of rotating flux. Consequently, the rotor revolves at the same speed as the stator field i. e. , at synchronous speed. iii) Because the bars of squirrel cage portion of the rotor now rotate at the same speed as the rotating stator field, these bars do not cut any flux and, therefore, have no induced currents in them. Hence squirrel cage portion of the rotor is, in effect, removed from the operation of the motor. 296 It may be emphasized here that due to magnetic interlocking between the stator and rotor poles, a synchronous motor can only run at synchronous speed. At any other speed, this magnetic interlocking (i. e. , rotor poles facing opposite polarity stator poles) ceases and the average torque becomes zero.
Consequently, the motor comes to a halt with a severe disturbance on the line. Note: It is important to excite the rotor with direct current at the right moment. For example, if the d. c. excitation is applied when N-pole of the stator faces Npole of the rotor, the resulting magnetic repulsion will produce a violent mechanical shock. The motor will immediately slow down and the circuit breakers will trip. In practice, starters for synchronous motors arc designed to detect the precise moment when excitation should be applied. 11. 5 Equivalent Circuit Unlike the induction motor, the synchronous motor is connected to two electrical systems; a d. . source at the rotor terminals and an a. c. system at the stator terminals. 1. Under normal conditions of synchronous motor operation, no voltage is induced in the rotor by the stator field because the rotor winding is rotating at the same speed as the stator field. Only the impressed direct current is present in the rotor winding and ohmic resistance of this winding is the only opposition to it as shown in Fig. (11. 5 (i)). 2. In the stator winding, two effects are to be considered, the effect of stator field on the stator winding and the effect of the rotor field cutting the stator conductors at synchronous speed.
Fig. (11. 5) (i) The effect of stator field on the stator (or armature) conductors is accounted for by including an inductive reactance in the armature winding. This is called synchronous reactance Xs. A resistance Ra must be considered to be in series with this reactance to account for the copper losses in the stator or armature winding as shown in Fig. (11. 5 (i)). This 297 resistance combines with synchronous reactance and gives the synchronous impedance of the machine. (ii) The second effect is that a voltage is generated in the stator winding by the synchronously-revolving field of the rotor as shown in Fig. 11. 5 (i)). This generated e. m. f. EB is known as back e. m. f. and opposes the stator voltage V. The magnitude of Eb depends upon rotor speed and rotor flux ? per pole. Since rotor speed is constant; the value of Eb depends upon the rotor flux per pole i. e. exciting rotor current If. Fig. (11. 5 (i)) shows the schematic diagram for one phase of a star-connected synchronous motor while Fig. (11. 5 (ii)) shows its equivalent circuit. Referring to the equivalent circuit in Fig. (11. 5 (ii)). Net voltage/phase in stator winding is Er = V ? Eb Armature current/phase, I a = where 2 Zs = R 2 + X s a hasor difference Er Zs This equivalent circuit helps considerably in understanding the operation of a synchronous motor. A synchronous motor is said to be normally excited if the field excitation is such that Eb = V. If the field excitation is such that Eb < V, the motor is said to be under-excited. The motor is said to be over-excited if the field excitation is such that Eb > V. As we shall see, for both normal and under excitation, the motor has lagging power factor. However, for over-excitation, the motor has leading power factor.
Note: In a synchronous motor, the value of Xs is 10 to 100 times greater than Ra. Consequently, we can neglect Ra unless we are interested in efficiency or heating effects. 11. 6 Motor on Load In d. c. motors and induction motors, an addition of load causes the motor speed to decrease. The decrease in speed reduces the counter e. m. f. enough so that additional current is drawn from the source to carry the increased load at a reduced speed. This action cannot take place in a synchronous motor because it runs at a constant speed (i. e. , synchronous speed) at all loads.
What happens when we apply mechanical load to a synchronous motor? The rotor poles fall slightly behind the stator poles while continuing to run at 298 synchronous speed. The angular displacement between stator and rotor poles (called torque angle ? ) causes the phase of back e. m. f. Eb to change w. r. t. supply voltage V. This increases the net e. m. f. Er in the stator winding. Consequently, stator current Ia ( = Er/Zs) increases to carry the load. Fig. (11. 6) The following points may be noted in synchronous motor operation: (i) A synchronous motor runs at synchronous speed at all loads.
It meets the increased load not by a decrease in speed but by the relative shift between stator and rotor poles i. e. , by the adjustment of torque angle ?. (ii) If the load on the motor increases, the torque angle a also increases (i. e. , rotor poles lag behind the stator poles by a greater angle) but the motor continues to run at synchronous speed. The increase in torque angle ? causes a greater phase shift of back e. m. f. Eb w. r. t. supply voltage V. This increases the net voltage Er in the stator winding. Consequently, armature current Ia (= Er/Zs) increases to meet the load demand. iii) If the load on the motor decreases, the torque angle ? also decreases. This causes a smaller phase shift of Eb w. r. t. V. Consequently, the net voltage Er in the stator winding decreases and so does the armature current Ia (= Er/Zs). 11. 7 Pull-Out Torque There is a limit to the mechanical load that can be applied to a synchronous motor. As the load increases, the torque angle ? also increases so that a stage is reached when the rotor is pulled out of synchronism and the motor comes to a standstill. This load torque at which the motor pulls out of synchronism is called pull—out or breakdown torque.
Its value varies from 1. 5 to 3. 5 times the full— load torque. When a synchronous motor pulls out of synchronism, there is a major disturbance on the line and the circuit breakers immediately trip. This protects the motor because both squirrel cage and stator winding heat up rapidly when the machine ceases to run at synchronous speed. 299 11. 8 Motor Phasor Diagram Consider an under-excited ^tar-connected synchronous motor (Eb < V) supplied with fixed excitation i. e. , back e. m. f. Eb is constantLet V = supply voltage/phase Eb = back e. m. f. /phase Zs = synchronous impedance/phase (i) Motor on no load
When the motor is on no load, the torque angle ? is small as shown in Fig. (11. 7 (i)). Consequently, back e. m. f. Eb lags behind the supply voltage V by a small angle ? as shown in the phasor diagram in Fig. (11. 7 (iii)). The net voltage/phase in the stator winding, is Er. Armature current/phase, Ia = Er/Zs The armature current Ia lags behind Er by ? = tan-1 Xs/Ra. Since Xs >> Ra, Ia lags Er by nearly 90°. The phase angle between V and Ia is ? so that motor power factor is cos ?. Input power/phase = V Ia cos ? Fig. (11. 7) Thus at no load, the motor takes a small power VIa cos ? phase from the supply to meet the no-load losses while it continues to run at synchronous speed. (ii) Motor on load When load is applied to the motor, the torque angle a increases as shown in Fig. (11. 8 (i)). This causes Eb (its magnitude is constant as excitation is fixed) to lag behind V by a greater angle as shown in the phasor diagram in Fig. (11. 8 (ii)). The net voltage/phase Er in the stator winding increases. Consequently, the motor draws more armature current Ia (=Er/Zs) to meet the applied load. Again Ia lags Er by about 90° since Xs >> Ra. The power factor of the motor is cos ?. 300 Input power/phase, Pi = V Ia cos ?
Mechanical power developed by motor/phase Pm = Eb ? Ia ? cosine of angle between Eb and Ia = Eb Ia cos(? ? ? ) Fig. (11. 8) 11. 9 Effect of Changing Field Excitation at Constant Load In a d. c. motor, the armature current Ia is determined by dividing the difference between V and Eb by the armature resistance Ra. Similarly, in a synchronous motor, the stator current (Ia) is determined by dividing voltage-phasor resultant (Er) between V and Eb by the synchronous impedance Zs. One of the most important features of a synchronous motor is that by changing the field excitation, it can be made to operate from lagging to eading power factor. Consider a synchronous motor having a fixed supply voltage and driving a constant mechanical load. Since the mechanical load as well as the speed is constant, the power input to the motor (=3 VIa cos ? ) is also constant. This means that the in-phase component Ia cos ? drawn from the supply will remain constant. If the field excitation is changed, back e. m. f Eb also changes. This results in the change of phase position of Ia w. r. t. V and hence the power factor cos ? of the motor changes. Fig. (11. 9) shows the phasor diagram of the synchronous motor for different values of field excitation.
Note that extremities of current phasor Ia lie on the straight line AB. (i) Under excitation The motor is said to be under-excited if the field excitation is such that Eb < V. Under such conditions, the current Ia lags behind V so that motor power factor is lagging as shown in Fig. (11. 9 (i)). This can be easily explained. Since Eb < V, the net voltage Er is decreased and turns clockwise. As angle ? (= 90°) between Er and Ia is constant, therefore, phasor Ia also turns clockwise i. e. , current Ia lags behind the supply voltage. Consequently, the motor has a lagging power factor. 301 ii) Normal excitation The motor is said to be normally excited if the field excitation is such that Eb = V. This is shown in Fig. (11. 9 (ii)). Note that the effect of increasing excitation (i. e. , increasing Eb) is to turn the phasor Er and hence Ia in the anti-clockwise direction i. e. , Ia phasor has come closer to phasor V. Therefore, p. f. increases though still lagging. Since input power (=3 V Ia cos ? ) is unchanged, the stator current Ia must decrease with increase in p. f. Fig. (11. 9) Suppose the field excitation is increased until the current Ia is in phase with the applied voltage V, making the p. . of the synchronous motor unity [See Fig. (11. 9 (iii))]. For a given load, at unity p. f. the resultant Er and, therefore, Ia are minimum. (iii) Over excitation The motor is said to be overexcited if the field excitation is such that Eb > V. Under-such conditions, current Ia leads V and the motor power factor is leading as shown in Fig. (11. 9 (iv)). Note that Er and hence Ia further turn anti-clockwise from the normal excitation position. Consequently, Ia leads V. From the above discussion, it is concluded that if the synchronous motor is under-excited, it has a lagging power factor.
As the excitation is increased, the power factor improves till it becomes unity at normal excitation. Under such conditions, the current drawn from the supply is minimum. If the excitation is further increased (i. e. , over excitation), the motor power factor becomes leading. Note. The armature current (Ia) is minimum at unity p. f and increases as the power factor becomes poor, either leading or lagging. 302 11. 10 Phasor Diagrams With Different Excitations Fig. (11. 10) shows the phasor diagrams for different field excitations at constant load. Fig. (11. 10 (i)) shows the phasor diagram for normal excitation (Eb = V), whereas Fig. 11. 10 (ii)) shows the phasor diagram for under-excitation. In both cases, the motor has lagging power factor. Fig. (11. 10 (iii)) shows the phasor diagram when field excitation is adjusted for unity p. f. operation. Under this condition, the resultant voltage Er and, therefore, the stator current Ia are minimum. When the motor is overexcited, it has leading power factor as shown in Fig. (11. 10 (iv)). The following points may be remembered: (i) For a given load, the power factor is governed by the field excitation; a weak field produces the lagging armature current and a strong field produces a leading armature current. ii) The armature current (Ia) is minimum at unity p. f and increases as the p. f. becomes less either leading or lagging. Fig. (11. 10) 11. 11 Power Relations Consider an under-excited star-connected synchronous motor driving a mechanical load. Fig. (11. 11 (i)) shows the equivalent circuit for one phase, while Fig. (11. 11 (ii)) shows the phasor diagram. Fig. (11. 11) 303 (i) (ii) Input power/phase, Pi = V Ia cos ? Mechanical power developed by the motor/phase, Pm = Eb ? Ia ? cosine of angle between Eb and Ia = Eb Ia cos(? ? ? ) Armature Cu loss/phase = I 2 R a = Pi ? Pm a Output power/phasor, Pout = Pm ?
Iron, friction and excitation loss. (iii) (iv) Fig. (11. 12) shows the power flow diagram of the synchronous motor. Fig. (11. 12) 11. 12 Motor Torque Gross torque, Tg = 9. 55 where Pm N-m Ns Pm = Gross motor output in watts = Eb Ia cos(? ? ? ) Ns = Synchronous speed in r. p. m. Shaft torque, Tsh = 9. 55 Pout N-m Ns It may be seen that torque is directly proportional to the mechanical power because rotor speed (i. e. , Ns) is fixed. 11. 13 Mechanical Power Developed By Motor (Armature resistance neglected) Fig. (11. 13) shows the phasor diagram of an under-excited synchronous motor driving a mechanical load.
Since armature resistance Ra is assumed zero. tan? = Xs/Ra = ? and hence ? = 90°. Input power/phase = V Ia cos ? Fig. (11. 13) 304 Since Ra is assumed zero, stator Cu loss (I 2 R a ) will be zero. Hence input power a is equal to the mechanical power Pm developed by the motor. Mech. power developed/ phase, Pm = V Ia cos ? Referring to the phasor diagram in Fig. (11. 13), (i) AB = E r cos ? = I a X s cos ? Also AB = E b sin ? ? E b sin ? = I a X s cos ? or I a cos ? = E b sin ? Xs Substituting the value of Ia cos ? in exp. (i) above, Pm = = V Eb Xs VEb Xs per phase for 3-phase
It is clear from the above relation that mechanical power increases with torque angle (in electrical degrees) and its maximum value is reached when ? = 90° (electrical). Pmax = V Eb Xs per phase Under this condition, the poles of the rotor will be mid-way between N and S poles of the stator. 11. 14 Power Factor of Synchronous Motors In an induction motor, only one winding (i. e. , stator winding) produces the necessary flux in the machine. The stator winding must draw reactive power from the supply to set up the flux. Consequently, induction motor must operate at lagging power factor.
But in a synchronous motor, there are two possible sources of excitation; alternating current in the stator or direct current in the rotor. The required flux may be produced either by stator or rotor or both. (i) If the rotor exciting current is of such magnitude that it produces all the required flux, then no magnetizing current or reactive power is needed in the stator. As a result, the motor will operate at unity power factor. 305 (ii) If the rotor exciting current is less (i. e. , motor is under-excited), the deficit in flux is made up by the stator. Consequently, the motor draws reactive power to provide for the remaining flux.
Hence motor will operate at a lagging power factor. (iii) If the rotor exciting current is greater (i. e. , motor is over-excited), the excess flux must be counterbalanced in the stator. Now the stator, instead of absorbing reactive power, actually delivers reactive power to the 3-phase line. The motor then behaves like a source of reactive power, as if it were a capacitor. In other words, the motor operates at a leading power factor. To sum up, a synchronous motor absorbs reactive power when it is underexcited and delivers reactive power to source when it is over-excited. 11. 15 Synchronous Condenser
A synchronous motor takes a leading current when over-excited and, therefore, behaves as a capacitor. An over-excited synchronous motor running on no-load in known as synchronous condenser. When such a machine is connected in parallel with induction motors or other devices that operate at low lagging power factor, the leading kVAR supplied by the synchronous condenser partly neutralizes the lagging reactive kVAR of the loads. Consequently, the power factor of the system is improved. Fig. (11. 14) shows the power factor improvement by synchronous condenser method. The 3 ? ? load takes current IL at low lagging power factor cos ?
L. The synchronous condenser takes a current Im which leads the voltage by an angle ? m. The resultant current I is the vector sum of Im and IL and lags behind the voltage by an angle ?. It is clear that ? is less than ? L so that cos ? is greater than cos ? L. Thus the power factor is increased from cos ? L to cos ?. Synchronous condensers are generally used at major bulk supply substations for power factor improvement. Advantages (i) By varying the field excitation, the magnitude of current drawn by the motor can be changed by any amount. This helps in achieving stepless control of power factor. ii) The motor windings have high thermal stability to short circuit currents. (iii) The faults can be removed easily. 306 Fig. (11. 14) Disadvantages (i) (ii) (iii) (iv) There are considerable losses in the motor. The maintenance cost is high. It produces noise. Except in sizes above 500 RVA, the cost is greater than that of static capacitors of the same rating. (v) As a synchronous motor has no self-starting torque, then-fore, an auxiliary equipment has to be provided for this purpose. 11. 16 Applications of Synchronous Motors (i) Synchronous motors are particularly attractive for low speeds (< 300 r. . m. ) because the power factor can always be adjusted to unity and efficiency is high. (ii) Overexcited synchronous motors can be used to improve the power factor of a plant while carrying their rated loads. (iii) They are used to improve the voltage regulation of transmission lines. (iv) High-power electronic converters generating very low frequencies enable us to run synchronous motors at ultra-low speeds. Thus huge motors in the 10 MW range drive crushers, rotary kilns and variable-speed ball mills. 307 11. 17 Comparison of Synchronous and Induction Motors S. Particular No. 1.
Speed 2. 3. 4. Power factor Excitation Economy 3-phase Induction Motor Remains constant (i. e. , Ns) from Decreases with load. no-load to full-load. Can be made to operate from Operates at lagging lagging to leading power factor. power factor. Requires d. c. excitation at the No excitation for the rotor. rotor. Economical fcr speeds below Economical for 300 r. p. m. speeds above 600 r. p. m. Self-starting No self-starting torque. Auxiliary means have to be provided for starting. Complicated Simple More less Synchronous Motor 5. Self-starting 6. 7. Construction Starting torque 308