1. Why can the rotation speed of the motor be changed freely?
The unit of motor rotation speed: r/min the number of rotations per minute, can also be expressed as rpm.
Example: 2-pole motor 50Hz3000 [r/min]
4-pole motor 50Hz1500[r/min]
Conclusion: The rotational speed of the motor is divided into two parts of the frequency
The rotational speed of an induction AC motor (hereinafter referred to 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, e.g. the pole number is 2, 4, 6), it is generally not suitable to adjust the speed of the motor by changing the value.
In addition, the frequency can be adjusted outside the motor and then supplied to the motor, so that the rotation speed of the motor can be controlled freely.
Therefore, the inverter for the purpose of frequency control is the preferred equipment for motor speed control equipment.
n=60f/p
n: synchronization speed
f: Power frequency
p: Number of pole pairs of the motor
Conclusion: Changing the frequency and voltage is the best way to control the motor
If you only change the frequency without changing the voltage, the motor may burn out due to overvoltage (overexcitation) when the frequency drops. Therefore, it is necessary to change the voltage of the inverter at the same time as changing the frequency. When the output frequency is above the additional frequency, the voltage cannot be continuously added, and the highest voltage can only be equal to the additional voltage of the motor.
For example, if the output frequency of the inverter is changed from 50 Hz to 25 Hz in order to halve the rotation speed of the motor, the output voltage of the inverter needs to be changed from 400 V to about 200 V
2. When the rotation speed (frequency) of the motor is changed, what happens to its output torque?
The starting torque and the maximum torque when driven by the inverter are smaller than those driven by the direct industrial frequency power supply
The starting and acceleration impact of the motor is large when the power frequency power supply is used, and when the power supply is used by the inverter, these shocks are weaker. A large starting current will occur when the power frequency is directly started. 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.
In general, the torque generated by the motor decreases as the frequency decreases (speed decreases). The actual data of the reduction is explained in some inverter manuals.
The lack of torque in the motor at low speeds will be improved by using the inverter operated by magnetic flux vectoring, and the motor can output sufficient torque even at low speeds.
3. When the speed of the inverter is adjusted to a frequency greater than 50Hz, the output torque of the motor will decrease
The general motor is planned and manufactured according to the voltage of 50Hz, and its additional torque is also given in this voltage range. Therefore, the speed regulation under the extra frequency is called constant torque speed regulation. (T=Te,P<=Pe)
When the output frequency of the inverter is greater than the frequency of 50Hz, the torque of the motor decreases in a linear relationship inversely proportional to the frequency.
When the motor is running at a frequency speed greater than 50Hz, it is necessary to consider the size of the motor load to avoid the lack of motor output torque.
For example, the torque of a motor at 100 Hz drops to about 1/2 of that at 50 Hz.
Therefore, the speed regulation above the extra frequency is called constant power speed regulation. (P=Ue*Ie)
4. The use of the inverter above 50Hz
We know that for a particular motor, the extra voltage and extra current are constant.
For example, the additional value of the inverter and the motor is: 15kW/380V/30A, and the motor can work above 50Hz.
When the speed is 50Hz, the output voltage of the inverter is 380V, and the current is 30A. If the output frequency is increased to 60Hz, the maximum output voltage and current of the inverter can only be 380V/30A. Obviously, the output power remains unchanged.
What is the torque situation at this time?
Because P=wT (w: angular velocity, T: torque), because P does not change, w is added, so the torque will decrease accordingly.
Let's look at it from another point of view:
The stator voltage of the motor U=E+I*R (I is the current, R is the electronic resistance, E is the induced potential)
It can be seen that when U,I do not change, E does not change.
And E=k*f*X, (k: constant, f: frequency, X: magnetic flux), so when f is from 50 to > 60 Hz, X will decrease accordingly
For motors, T = K*I*X, (K: CONSTANT, I: CURRENT, X: MAGNETIC FLUX), SO THE TORQUE T WILL DECREASE WITH THE DECREASE OF MAGNETIC FLUX X.
Together, when it is less than 50Hz, because I*R is very small, the magnetic flux (X) is constant when U/f=E/f is constant. The torque T is proportional to the current. This is why the overcurrent ability of the inverter is generally used to describe its overload (torque) ability. It is also called constant torque speed regulation (the extra current does not change - the maximum torque of > does not change)
Conclusion: When the output frequency of the inverter is added from more than 50Hz, the output torque of the motor will decrease.
5. Other elements related to output torque
Heat generation and heat dissipation can determine the output current of the inverter, which affects the output torque of the inverter.
Carrier frequency: Generally, the additional current marked by the inverter is based on the highest carrier frequency, and the highest ambient temperature can ensure continuous output. The carrier frequency is lowered so that the current of the motor is not affected. However, the heat generation of the components will be reduced.
Ambient temperature: Just like the inverter protection current value will not be increased because the ambient temperature is relatively low.
Altitude: Altitude is added, which has an impact on heat dissipation and insulation performance. Generally, less than 1000m can be considered. More than 5% reduction per 1000 meters can be done.
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