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It is a situation where the load torque (Ta) suddenly decreases while rotating the shaft with a motor(PMSM) at a constant motor torque (Tm).

In my opinion, the rotational speed of the shaft (I-moment of inertia) and motor will be instantaneously increased by the torque difference, and then more power is consumed in proportion to the shaft speed, right?

Tm - Ta = I*wdot


before load decrease : Power = Tm * w 
After load decrease :  Power = Tm * (w + wdot * dt) 
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  • \$\begingroup\$ What do you understand by "at constant motor torque"? Is it a control system that keeps the torque constant? \$\endgroup\$ – AlexVB Mar 17 at 12:02
  • \$\begingroup\$ yes , control system keeps the torque \$\endgroup\$ – Lee Mar 17 at 12:02
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The rotor will accelerate over a very short period of time but after that it should remain in synchronous speed. In effect, the position of the rotor magnet relative to the rotating magnetic field will reduce by a few degrees of angle. That reduction reduces the forces that rotate the rotor and equilibrium then takes over.

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  • \$\begingroup\$ Is this the same result if you have a control system to maintain torque? \$\endgroup\$ – Lee Mar 17 at 12:07
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    \$\begingroup\$ If the control system attempts to produce a constant torque then rotor speed will increase to obtain that. "A" control system would have to increase the AC frequency to do so. If AC frequency remains constant, then what I said in my answer prevails. \$\endgroup\$ – Andy aka Mar 17 at 12:09
  • \$\begingroup\$ thank you Andy! \$\endgroup\$ – Lee Mar 17 at 12:10
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No load RPM is proportional only to voltage kRPM/V (& friction) while torque is proportional only to current and maximum torque starts at 0 RPM as BEMF reduces max torque available with applied voltage with rising speed.

So at your hypothetical test the torque will increase RPM to the new steady-state RPM according to the new lower load. Maximum power = Torque x RPM is often at or above 50% no load RPM depending on winding methods.

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