V/f motor

The motor will run slower. In some cases, this could cause the motor to overheat, but this can only be determined from the motor manufacturer's data.

We ship motors all over the world, when we get a 50Hz order, the motor manufacturer will take the next size up motor, and simply change the nameplate to derate the motor's HP if it's to be used in 50hz service. Most large electric motor manufactures do it this way.

It will most likely work as long as the service factor of the motor is high enough... like say... 1.5... As long as the motor is not already underpowered.

Thank you for your fast answer. I knew that the motor would run slowly. My doubt is how much the motor current will increase, considering that we are talking about a fridge, will it resist the overcurrent?

Thank you in advance

The motor will pull as much current as it needs. I would guess that the manufacturer spec'ed in a motor that would just do the job, which is fairly common. No extra headroom. But I could not give an answer in confidence unless I knew what the service factor of the motor is. If you have a 1.5 service factor, you can run the motor half again as hard as it was designed to run without immediate failure. The overall lifespan of the motor will be less if you are operating the motor above the 1.0 service factor range.

I suspected it was so!

Thank you for your time.

Cheers from Argentina

"The motor will run slower. In some cases, this could cause the motor to overheat, but this can only be determined from the motor manufacturer's data."

I am not familiar with the concept, but why would it overheat if its running slower?

A spinning motor also acts as a generator, creating a voltage which opposes the applied voltage, commonly called back emf. When the motor slows down, the back emf drops allowing more current to flow and thus more resistive heating in the windings.

"Since the magnetic field (proportional to V/f is just a bit higher 2 Vs 2.2, the current is likely to go down a bit). The 60Hz motor at 50Hz would have expected 100V to maintain the flux level but it gets 110V so the current will go down by about 10%."

Could you explain the significance of the V/f ratio? From what you have typed here, I understand it the ratio of Voltage to Frequency, so the expected voltage would be 2 times 50 Hz = 100V. But the supplied voltage is 110 V which is 10% greater than 100V. But why would the current go down? Wouldn't it be proportional to the increase in voltage? V = IR? Just wondering.

I personally believe that generally speaking the motor current will increase due to the lower frequency. By how much? No idea.....

The lower voltage will cause the current to drop slightly.

I have no idea what the final value will be but I bet there are some good posts somewhere on the web for that, like here for example:-

http://www.engineersedge.com/wwwboard/posts/6717.html

The flux density is proportional to the

V/f ratio directly provides flux density of the magnetic circuit.

The torque is proportional to current x flux

As the suplied voltage increases the flux increases. Assuming the same torque demanded by the load, the current hence will go down. (here it is not a simple IR since the demand is by the compressor which will demand the rotational torque).

The applications (the loads) will decide how a system will behave under variable frequency.

Of course as AG pointed out the lowering of frequency will demand higher magnetising current (and it is associated in fact with a bit of higher magnetc circuit losses too). As will the efficiency of the motor. It will no more be working near BEP (Best Efficiency point) . These are the con aspects as I mentioned, and may not be too significant damaging in 50 Vs 60 Hz.

Why anonymous?

You have had some good answers already here, but a simple test will allow you to make some extra judgement. Take an AC Ammeter and measure the current taken at the lower voltage/frequency and compare the current taken with the motor info plate, if the current is similar to the plate or only slightly higher, you should be OK......(lower is of course OK!)

Its not exactly 100% accurate, but it will give you a good idea.

It really depend on if the motor is working at "full stretch" on the original voltage/frequency or not. If it was, you may get problems with the lower voltage/frequency.....

WARNING!

If you are NOT electrically trained, get someone who is to do it for you!!!

Thank you all for your contribution. In my opinion when frecuency goes down current will increase due to the motors impedance goes down. The motor will overheat because the current will be higher.

Than you for your help

The current will be affected by the type of load imposed to the motor and the magnetic saturation.

For this application, a compressor will reduce the power demand due to the speed reduction. If it is a centrifugal compressor the power could drop as far as 58%. That mean it will run up to 72% more time to make the same cooling work, but the active current will drop.

The magnetizing current will go up because 110 volts are 10% over a constant V/HZ ratio.

The final total current is the vectorial addition of active current plus reactive current.

For this case the higher probability is a total current drop, but the operating temperature must be monitored.

The temperature will depend on the losses and cooling system. The mechanical friction will drop with speed and stator and rotor winding losses will be reduced by the current reduction (I^2*R). The core losses will be similar since the lower frequency will compensate for the higher saturation(10%). The flow of refrigerant over the motor will be reduced around the same 58% because of speed reduction.

Note that an induction motor IS NOT a constant impedance device. Impedance changes with saturation and shaft slip from synchronous rpm.