1. Why is the rotational speed of the motor freely changeable?
Motor rotation speed unit: r/min The number of rotations per minute can also be expressed as rpm. For example: 2-pole motor 50Hz 3000 [r/min] 4-pole motor 50Hz 1500 [r/min]
Conclusion: The rotational speed of the motor is proportional to the frequency
The rotational speed of an inductive AC motor (hereinafter referred to simply as a motor) is approximately determined by the number of poles and frequency of the motor. The number of poles of the motor is fixed by the working principle of the motor. Since the pole value is not a continuous value (a multiple of 2, for example, the number of poles is 2, 4, 6), it is generally uncomfortable and the speed of the motor is adjusted by changing the value.
In addition, the frequency can be supplied to the motor after being adjusted outside the motor, so that the rotational speed of the motor can be freely controlled.
Therefore, the inverter for the purpose of controlling the frequency is the preferred device for the motor speed control device.
n = 60f/p
n: synchronization speed
f: power frequency
p: motor pole pairs
Conclusion: Changing frequency and voltage is the optimal motor control method
If only the frequency is changed without changing the voltage, the frequency will decrease and the motor will be over-voltage (overexcitation), causing the motor to be burned out. Therefore, the inverter must change the voltage at the same time while changing the frequency. When the output frequency is above the rated frequency, the voltage cannot continue to increase, and the maximum can only be equal to the rated voltage of the motor.
For example, in order to reduce the rotational speed of the motor by half, change the output frequency of the inverter from 50Hz to 25Hz, then the output voltage of the inverter needs to be changed from 400V to about 200V.
2. What is the output torque when the motor’s rotational speed (frequency) changes?
The starting torque and maximum torque when the inverter is driven are less than that of the direct-frequency power supply. When the motor is powered by the commercial power supply, the starting and acceleration shocks are large, and when the inverter is used for power supply, these impacts are weaker. A direct start of the power frequency produces a large starting and starting current. When the inverter is used, the output voltage and frequency of the inverter are gradually added to the motor, so the starting current and impact of the motor are smaller. Generally, the torque produced by the motor is reduced as the frequency decreases (the speed decreases). The reduced actual data is given in some of the drive manuals.
By using a flux vector controlled inverter, the torque of the motor at low speeds is improved, and even in the low speed range, the motor can output sufficient torque.
3. When the inverter is adjusted to a frequency greater than 50Hz, the output torque of the motor will decrease.
The usual motor is designed and manufactured at a voltage of 50 Hz, and its rated torque is also given within this voltage range. Therefore, the speed regulation below the rated frequency is called constant torque speed regulation. (T=Te, P<=Pe) When the inverter output frequency is greater than 50Hz, the torque generated by the motor should be linearly inversely proportional to the frequency. decline. When the motor is running at a frequency greater than 50 Hz, the size of the motor load must be considered to prevent the motor from outputting insufficient torque.
For example, the torque generated by the motor at 100 Hz is reduced to approximately 1/2 of the torque at 50 Hz.
Therefore, the speed regulation above the rated frequency is called constant power speed regulation. (P=Ue*Ie)
4. Application of inverter above 50Hz
As you know, the rated voltage and current rating of a particular motor are constant. If the inverter and motor are rated: 15kW/380V/30A, the motor can work above 50Hz. When the speed is 50Hz, the output voltage of the inverter is 380V, and the current is 30A. At this time, if the output frequency is increased to 60Hz, the maximum output voltage of the inverter can only be 380V/30A. Obviously, the output power is unchanged. So we call it constant power speed regulation.
What is the torque situation at this time?
Because P = wT (w: angular velocity, T: torque). Because P does not change, w increases, so the torque will decrease accordingly.
We can also look at another angle:
The stator voltage of the motor U = E + I*R (I is the current, R is the electronic resistance, and E is the induced potential)
It can be seen that when U, I are unchanged, E does not change.
And E = k*f*X, (k: constant, f: frequency, X: flux), so when f is from 50–>60Hz, X will decrease accordingly.
For the motor, T = K * I * X, (K: constant, I: current, X: flux), so the torque T will decrease as the flux X decreases.
Meanwhile, when it is less than 50 Hz, since I*R is small, when U/f=E/f is constant, the magnetic flux (X) is constant. The torque T is proportional to the current. This is why the inverter is usually used. Overcurrent capability to describe its overload (torque) capability. Also known as constant torque regulation (rated current is constant –> maximum torque is constant)
Conclusion: When the inverter output frequency increases from above 50Hz, the output torque of the motor will decrease.
5. Other factors related to output torque
The heat and heat dissipation capacity determine the output current capability of the inverter, which affects the output torque capability of the inverter. Carrier frequency: Generally, the rated current of the inverter is the highest carrier frequency, and the value of continuous output can be guaranteed at the highest ambient temperature. When the carrier frequency is reduced, the current of the motor will not be affected. However, the heat of the components will decrease. Ambient temperature: It is not like increasing the protection current value of the inverter because it detects that the ambient temperature is low.
Altitude: The altitude increases, which has an effect on heat dissipation and insulation performance. Generally, it can be ignored below 1000m. Above 5% per 1000m can be used.