For a synchronous motor, the maximum value of torque developed at an angle of
Right Answer is:
For a synchronous motor, the maximum value of torque is developed at an angle of 90 degrees.
When a synchronous motor operates under load it draws active power from the line. The active power absorbs by the motor depends on the supply voltage E and excitation Voltage Eo.
Therefore power is given as
P = ( E × Eo/Xs)sinδ.
P = mechanical power of the motor per phase
Xs = synchronous reactance per phase
δ = Torque angle between Eo and E.
This equation shows that the mechanical power increase with the torque angle and its maximum value is reached when δ = 90°.
Therefore peak power per phase is given as
Pmax = EoE/Xs.
The maximum torque occurs at a torque angle of 90 electrical degrees. This is called the pull-out torque, which indicates the maximum value of torque that a synchronous motor can develop without pulling out of synchronism. In general, its value varies from 1.25 to 3.5 times the full-load torque. In actual practice, however, the motor will never operate at a torque angle close to 90 electrical degrees because the stator current will be many times its rated value at this operation.
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