Effect of Changing Excitation on Constant Load
As shown in Fig. 38.20 (a), suppose a synchronous motor is operating with normal excitation (Eb = V ) at unity p.f. with a given load. If Ra is negligible as compared to X S, then Ia lags ER by 90º and is in phase with V because p.f. is unity. The armature is drawing a power of V .Ia per phase which is enough to meet the mechanical load on the motor. Now, let us discuss the effect of decreasing or increasing the field excitation when the load applied to the motor remains constant.
(a) Excitation Decreased
As shown in Fig. 38.20 (b), suppose due to decrease in excitation, back e.m.f. is reduced to Eb1 at the same load angle a1. The resultant voltage ER1 causes a lagging armature current Ia1 to flow. Even though Ia1 is larger than Ia in magnitude it is incapable of producing necessary power V Ia for carrying the constant load because Ia1 cos f1 component is less than Ia so that V Ia1 cos f1 < V Ia.
Hence, it becomes necessary for load angle to increase from a1 to a2. It increases back e.m.f. from Eb1 to Eb2 which, in turn, increases resultant voltage from ER1 to ER2. Consequently, armature current increases to Ia2 whose in-phase component produces enough power (V Ia2 cos f2) to meet the constant load on the motor.
(b) Excitation Increased
The effect of increasing field excitation is shown in Fig. 38.20 (c) where increased Eb1 is shown at the original load angle a1. The resultant voltage ER1 causes a leading current Ia1 whose in-phase component is larger than Ia. Hence, armature develops more power than the load on the motor. Accordingly, load angle decreases from a1 to a2 which decreases resultant voltage from ER1 to ER2. Consequently, armature current decreases from Ia1 to Ia2 whose in-phase component Ia2 cos f2 = Ia. In that case, armature develops power sufficient to carry the constant load on the motor.
Hence, we find that variations in the excitation of a synchronous motor running with a given load produce variations in its load angle only.
Different Torques of a Synchronous Motor
Various torques associated with a synchro- nous motor are as follows:
1. starting torque
2. running torque
3. pull-in torque and
4. pull-out torque
(a) Starting Torque
It is the torque (or turning effort) developed by the motor when full voltage is applied to its stator (armature) winding. It is also sometimes called breakaway torque. Its value may be as low as 10% as in the case of centrifugal pumps and as high as 200 to 250% of full-load torque as in the case of loaded reciprocating two-cylinder compressors.
(b) Running Torque
As its name indicates, it is the torque developed by the motor under running conditions. It is
determined by the horse-power and speed of the driven machine. The peak horsepower determines the maximum torque that would be required by the driven machine. The motor must have a break- down or a maximum running torque greater than this value in order to avoid stalling.
(c) Pull-in Torque
A synchronous motor is started as induction motor till it runs 2 to 5% below the synchronous speed. Afterwards, excitation is switched on and the rotor pulls into step with the synchronously- rotating stator field. The amount of torque at which the motor will pull into step is called the pull-in torque.
The maximum torque which the motor can develop without pulling out of step or synchronism is called the pull-out torque.
Normally, when load on the motor is increased, its rotor progressively tends to fall back in phase by some angle (called load angle) behind the synchronously-revolving stator magnetic field though it keeps running synchronously. Motor develops maximum torque when its rotor is retarded by an angle of 90º (or in other words, it has shifted backward by a distance equal to half the distance between adjacent poles). Any further increase in load will cause the motor to pull out of step (or synchronism) and stop.
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