Why would thermal overload trip when low voltage is supplied.

The background story goes like this. Our gas turbine has an output rating of 11,000 volts but we had a problem with the Automatic Voltage Regulator (AVR) and it had to be replaced. After replacement, the HMI indicated that the output voltage was 11,000V but the actual voltage on the bus was about 10,400V and this had a ripple effect on the whole system as the output of our 3.3kV transformer was 3.1kV and the output of the 440v transformer was 400v.

 

After a while, the thermal overload relays on the well-pump motors began to trip one after the other. This prompted us to troubleshoot, and we discovered that the supply voltage to the motor was below the nameplate rating.

 

Why would a low supply voltage trip the thermal-overload protection relay, normally a when a motor is overloaded it draws more current from the supply and when the motor tries to pull current higher than its full load amperage (FLA) rating the overload opens the circuit to prevent the motor from damage due to overheating. 


I will give two reasons why the thermal-overload relay tripped.

This is a little bit oversimplified but here it goes

1)     (1) A motor must draw a fixed amount of power from the supply (Power = VI) so when a motor is supplied with a voltage less than its nameplate rating it will compensate for the lower voltage supplied to it by drawing more current to meet the Power Threshold. If the current pulled starts to exceed the FLA of the motor, heat begins to build up. This mimics the events that occur in the event of an actual overload of the motor so and the Thermal overload relay operates to protect the motor.

2)   (2) Low voltage supplied to the motor causes the torque developed to be reduced, this reduced torque results in an increased slip and lower speed, the motor tries to reduce the slip by drawing more current from the supply and this trips the thermal-overload relay.

 Proof that Low voltage supplied to the motor causes the torque developed to be reduced.

  1. Torque equation: For an AC induction motor, the torque (T) is proportional to the square of the applied voltage (V):

             T ∝ V²

      2. Power Equation: The power supplied to the motor (PP) is given by:

P=V×I×cos(ϕ)
P = V \times I \times \cos(\phi)
      3. Torque-Power relationship: Torque is also related to power and speed

P = T * ω 

Where ω is the angular velocity (speed) in radians per second.

  • The torque (TT) developed by the motor is proportional to the square of the voltage (VV). Therefore, a reduction in voltage significantly impacts the torque.

TV2T \propto V^2

  • For example, if the voltage is reduced to 90% of its rated value: Tnew(0.9V)2=0.81V2T_{new} \propto (0.9V)^2 = 0.81V^2
  • This shows that the torque is reduced to 81% of its original value when the voltage is reduced to 90%.


 

 

 

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