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With a given two-pole motor, with all else being equal, if the pole count is increased to 6, the nominal speed will be reduced to 1/3. Can it be freewheeled to the same speed as the original 2-pole motor?

The maximum speed of the bearings would naturally be of concern, as well as the higher centrifugal forces. Would motors of the same model range be using the same bearings? Would the rotor be produced with the same process, thus allowing the same centrifugal forces?

Is it normally possible to understand this maximum mechanical limit speed from the motor's nameplate or specification sheet?

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  • \$\begingroup\$ I’m concerned about the voltage, and the eddy current losses will be higher too I think \$\endgroup\$
    – Bryan
    Commented Jan 24, 2023 at 1:58
  • \$\begingroup\$ The back EMF of the PMSM increases linearly with speed. If you run it faster than rated speed, the back EMF will also be higher than rated. This can lead to control problems (unable to maintain zero torque unless field weakening is supported) and overvoltage problems (if the back EMF is > the rated voltage of the DC Link capacitors). Many motors list a mechanical maximum speed as well as a rated speed. If you plan to over-speed the motor, search for one that lists a maximum speed in addition to the rated speed. \$\endgroup\$
    – user57037
    Commented Jan 24, 2023 at 1:58

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With a given 2-pole motor, with all else being equal, if the pole count is increased to 6, the nominal speed will be reduced to 1/3.

Actually, this does not really follow. The pole count can be increased without changing the nominal speed. The VFD can simply run at a higher frequency to compensate.

But can it be freewheeled to the same speed as the original 2-pole motor?

There are two components to this. The first is the mechanical safe speed. I have seen a maximum speed listed in the spec sheet for some motors which is higher than the rated speed. If this is important to you, you should select a motor that lists this.

The second component is electrical. When you spin a PMSM faster than its rated speed, the motor will act as a generator and produce a voltage higher than its rated voltage. As long as the VFD can handle the higher voltage, this shouldn't result in a catastrophe. However, the VFD also may have problems controlling torque at over rated speed. Even if it will not actually be destroyed, it may produce undesirable mechanical drag at high speed.

The reason for this unwanted mechanical drag is that the generated voltage will be rectified by the VFD output switches and increase the voltage of the DC-link bus. Some VFDs contain power dissipation elements to help control the DC-link bus voltage. This can lead to continuous current flowing from the motor into the power dissipation element. This continuous power loss will manifest as a drag force on the rotor.

If the VFD has no way to dissipate power directly from the DC-link, then the link voltage will rise until it matches the generator voltage of the motor. As long as the generated voltage does not exceed the maximum DC-link voltage it should be OK. If the voltage gets too high (motor spins too fast) at some point it may cause a catastrophic failure of the DC-link capacitors or switching elements.

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