Mohammed Ahmed
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Voltage controller of induction motor
- Thread starter Mohammed Ahmed
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Help Support Professional Engineer & PE Exam Forum:
I believe the idea is that the general circuit diagram of an induction motor is equivalent or very similar to a transformer. So if the stator (primary) voltage is reduced, then the rotor (secondary) voltage is also reduced.
Torque is proportional to power (Tm = 9.549 * Pout / n). To maintain the same torque and power with reduced voltage, the rotor current must be increased (P = V * I * pf).
Torque is proportional to power (Tm = 9.549 * Pout / n). To maintain the same torque and power with reduced voltage, the rotor current must be increased (P = V * I * pf).
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RedRaider2020
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Torque ∝ V^2 * Slip/ R
Since we are not changing what the motor is driving the required Torque and Resistance will be constant.
I ∝ V * Slip / R
When the voltage is decreased the slip will change and can be calculated with this ratio
Vold^2 * oldSlip = Vnew^2 * newSlip
newSlip = oldSlip (Vold^2/Vnew^2) = oldSlip (1^2/0.5^2) = oldSlip * 4
I ∝ V * Slip / R = 0.5 * 4 / R = 2 / R
To add to what Akyip said in order to maintain the same torque the power will have to increase because the motor speed will decrease in response to the slip increasing. If the Voltage is half and the current is double they will cancel each other out in the power equation P = V * I * pf but since the power factor will increase the power will also increase.
The equivalent circuit of a motor and a transformer look the same. A difference in the construction of each is that the windings of a transformer use an iron core to transfer flux while the motor has to transfer its flux over an air gap which is much more difficult. Therefore the magnetizing current in a motor is much larger than that of a transformer. This large magnetizing current makes the efficiency and power factor much smaller, possibly around 60 percent vs a transformers which is possibly around 99 percent. The magnetizing current can be 70 to 90 percent of the full load current. Magnetizing current is Im=V1/jXM. A change in voltage V1 will make the magnetizing current smaller while I2 is increasing making the real power greater and increasing the power factor.
Since we are not changing what the motor is driving the required Torque and Resistance will be constant.
I ∝ V * Slip / R
When the voltage is decreased the slip will change and can be calculated with this ratio
Vold^2 * oldSlip = Vnew^2 * newSlip
newSlip = oldSlip (Vold^2/Vnew^2) = oldSlip (1^2/0.5^2) = oldSlip * 4
I ∝ V * Slip / R = 0.5 * 4 / R = 2 / R
To add to what Akyip said in order to maintain the same torque the power will have to increase because the motor speed will decrease in response to the slip increasing. If the Voltage is half and the current is double they will cancel each other out in the power equation P = V * I * pf but since the power factor will increase the power will also increase.
The equivalent circuit of a motor and a transformer look the same. A difference in the construction of each is that the windings of a transformer use an iron core to transfer flux while the motor has to transfer its flux over an air gap which is much more difficult. Therefore the magnetizing current in a motor is much larger than that of a transformer. This large magnetizing current makes the efficiency and power factor much smaller, possibly around 60 percent vs a transformers which is possibly around 99 percent. The magnetizing current can be 70 to 90 percent of the full load current. Magnetizing current is Im=V1/jXM. A change in voltage V1 will make the magnetizing current smaller while I2 is increasing making the real power greater and increasing the power factor.
