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What happens if an induction motor given with 3 phase supply is forced to rotate with synchronous speed in

a) same direction

b) opposite direction

Please help me with the above problem. I think in the first case, the current induced in rotor will be 0. So, no losses in rotor.

In second case, slip will be 2.

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  • \$\begingroup\$ Is this a homework question? What research have you done? \$\endgroup\$ – LMS May 12 '18 at 16:25
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    \$\begingroup\$ In case a), yes, it seems that the rotor current would be zero (or very low). Stator current will not be zero, however. In case b) I think it is safe to say that if a motor is forced to operate in this condition for an extended period, where the rotor is moving in opposite direction from magnetic field, the stator current will be very large and the motor will overheat. Motors in industrial use are often reversed, on-the-fly, but in this case, the rotor stops and reverses fairly quickly. \$\endgroup\$ – mkeith May 12 '18 at 16:27
  • \$\begingroup\$ @LMS I am preparing for interviews. So, such questions can be asked. \$\endgroup\$ – Nikhil Kashyap May 12 '18 at 17:50
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In case a), you are correct.The slip is zero so no current is induced in the rotor. The stator current is the magnetizing current. The power drawn from the source will be the stator iron losses. This condition is almost the same as no-load operation, but driving the rotor to exactly synchronous speed will mean that the the mechanical losses, friction and aerodynamic drag (windage) will be supplied by the external drive.

Case b) is called plug braking, the motor brakes the load with both the braking energy and high rotor slip losses dissipated in the rotor. The stator current will be somewhat higher than the motor's locked rotor current.

Re case b) effect on motor:

The image below shows torque power and current vs. slip for a particular induction motor. At slip = 2, the motor's braking torque is about 66% of rated torque and the braking power (P = -5.3kW) is about 66% of rated motor power. The copper losses in the rotor are about 10.8 kW, making the total power dissipated in the rotor 16.1 kW, about 65 times the power dissipated during normal full load operation. The copper losses in the stator at s=2 are about 22 kW, about 44 times the power dissipated during normal full load operation. When the motor is started at full voltage, the power dissipated in the rotor and stator is initially (S=1, locked rotor condition) about 39 times tor normal power dissipation in each. Either locked-rotor or S=2 operation will cause the motor to fail due to internal heating rather quickly. For S=2, the heating is more severe, particularly in the rotor.

enter image description here

Image and motor data from Fitzgerald, Kingsley, Umans Electric Machinery 4th ed.

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  • \$\begingroup\$ in case b, will the windings burn if kept in such situation is maintained for a long period? And also the torque will be low in such case ? \$\endgroup\$ – Nikhil Kashyap May 13 '18 at 3:42
  • \$\begingroup\$ See addition to answer. \$\endgroup\$ – Charles Cowie May 13 '18 at 17:18

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