Hello lynlynch,

A very useful post!

GA to you sir.

bb

Another GA from me. A very clear and well laid out explantion.

Thank you lyn lynch

It's rare to see a well written technical report that is clearly written and in terms that can be understood by everyone without paronizing the lay person or putting advanced tech types to sleep.

Nicely done and informative.

L.J.

I hope you folks realize that lyn lynch copied and pasted the FAQ entry from another Electrical Engineering forum. That FAQ was written by a friend of mine.

The typical method to reference someone else's work like that would be yo post a link to the site, not copy it. But at least lyn lynch did not attempt to conceal the fact that this was a copy-paste.

As you are probably aware, ideally you should supply the 50Hz motor with (380 x 60/50) = around 460V 60 Hz to maintain the design V/Hz ratio to use the same motor.

I suspect the cost of the step-up transformer you are suggesting may be similar to a new correct voltage 60 Hz motor. This would be best matched with a new pulley and belts to get the compressor speed back to 50 Hz design speed (if that is a problem - if not see below).

Be aware that if you are not changing the pulley sizes, then the power requirement to drive the compressor ~ 6/5 time faster will increase (e.g. if square law this would require 1.44 times the power at 50 Hz speed) for the extra speed, and hence may require a larger power drive motor,

Hi,

Just to summarise what everybody has said:

Yes you can operate the 380v 50Hz motor from a 460v 60Hz supply by using the step up transformer. The V/f ratio will be correct and hence the magnetic flux in the motor will be according to the designed flux.

The speed of the motor will increase by 20% and the torque the motor produce reduced by 20%. You therefore need to change the pulleys as you suggested, if not the motor will be operating under an overload condition.

The reduction in torque produced by the motor will be cancelled by the 20% increase in your pulley ratio (pulley ratio need to increase by 20% to keep the compressor speed constant).

By changing the pulleys and keeping the compressor speed constant, the absorbed power from the compressor will not increase at all. At the same time the torque available on the compressor pulley will increase by 20%, hence the same torque will be available to the compressor as if supplied by a 50Hz motor (with the smaller pulley ratio.

Regards

Thanks for the short answer.

Thanks to you all.

I frequently run 220 volt 3 phase motors from single phase 220 volts using solid state inverters. I need to vary the speed of the motor carefully for manufacturing process involving optics and, knowing I can do so safely with 3 phase motors is what drives this application.

The speed is always controlled by varying the output frequency. This type of controller for a horsepower environment such as yours would likely be costly but the technology does exist to run mismatched motors.

L.J.

That's correct, the inverter will convert the single phase input, to a controlable 3 phase output which controll the speed of the motor to the set output frequency.

It must be said though that this technology is only available on small units (up to 3 Hp in SA) because of the high current drawn on the single phase supply side.

Correct me if I am wrong but I believe a step up or a step down transformer will only change the voltage. It will not change the Frequency.

Yes you are correct. The reason that the XFRMR is mentioned is not to change the frequency it is to change the voltage up or down, thus keeping the V/Hz ratio close to the same.

50Hz/60Hz=0.833 so

380V/0.833=456V

If a 380V 50Hz motor is to be run at 60Hz you should step the voltage up to 456V to keep the Voltage Frequency ratio the same. Luckily 460V is close enough and it is a standard US voltage to boot.

In summary here are the two best options:

1. Connect directly to 460 vac and the run the mechanical system at a 60/50 over-speed. This the best (least cost) choice if the mechanical system is capable of the 20% speed up.

2. Add as solid state inverter which would connect to 460V and drive the motor at its voltage and frequency rating. Inverters have down in cost and this approach would not overstress the mechanical system or require belts and pulleys to be changed. (Asuming the compressor has belts and pulleys.)

Apart from internal electro-mechanical forces, the dominant load on the motor is caused by the air compressor.

The output of an air compressor is proportional to the speed - which in turn is proportional to the horse-power needed to drive it.

The rated output of a motor is governed by (1) the maximum available magnetic field strength, where an increase of current in the field windings will increase field strength until the iron core saturates, and then no more torque is delivered to the shaft, and then (2) the heat that the motor can run at before the windings burn out.

Going to 60Hz means an increase of 20% power. Everything be equal it will be safe to assume your motor has been matched to the compressor for use at 50Hz and thus will be over-loaded at 60 Hz. But check with the manufacturers.

It could be found your motor will deliver the increased power, but note that the compressor also has a maximum output largely governed by the limitations of the cooling arrangements. Which if not adequate will put an enormous additional load on the downstream after-coolers, filters, dryers etc.

On balance the simplest solution is to change the pulleys.

Here you state that the power goes up by 20% using 60 Hz. Can I assume you mean the motor power? Earlier above it was stated that the torque will go down by 20% which would make the power the same regardless of the frequency.

As it was pointed out in the original question that the pulley would of course be changed to account for the higher speed and there would not be a higher load demand from the driven machinery.

Does the motor output power increase with higher frequency or does it stay the same? Thanks.

The delivered motor power must remain constant to remain within the motor design criteria. A 30Hp motor at 50Hz is designed and manufactured for 30 Hp and will be overloaded if suddenly asked to produce 36 Hp. There will be an increase in running current resulting in overheating of the motor and even insulation failure (depending on the service factor of the motor). Motors manufactured according to Nema specifications have a service factor of 1.2, which means the motor can produce continually 20% more power (maybe our friends from the Americas will confirm this statement). Motors manufactured according to IEC specifications has a SF of 1, which means the rated power on the motor rating plate is what you can get out of the motor, without destroying it.

The electric motor will sacrifice himself to try and deliver what ever you are asking from him, and thats why we need to protect him from destruction. You therefore need to set the overload protection to 30Hp even at 60 cycles.

The absorbed power from the compressor will (depending what type of load curve the compressor has) increase if operating at 20% higher speed (without changing the pulleys). Different applications has different type of speed (rpm) versus torque (Nm) curves and this will have an influence on the increase in absorbed power from the compressor, eg.:

Constant torque: absorbed torque remains constant with a increase of speed (rpm). Typical example: Conveyor belt

Parabolic torque: required torque increase with the root 3 (T³) with an increase of speed. Typical : centrifical pumps and fans

Constant power: Torque will decrease to the same ratio as the increase in speed. Typical: Mine winders

Regards

A

Basically a motor has a power rating that must not be exceeded.

What happens is a pro-rata increase in the load caused by the compressor running faster. Therefore in order not to overload the motor the pulley needs to be changed to get back to the original compressor speed.

If the ratio is not changed, then the motor will try to deliver the power needed by the compressor. At full compressor output the motor will struggle to keep up speed, and to do this it will pull more current. The motor overheats because it is trying to deliver more power than it is designed for. Motor-starters with overload protection and winding heater sensors usually stop the motor to protect it from this sort of thing.

It is like riding a bike. Your legs turn the pedals at much the same speed. You change gear to match the pedal speed to the wheel speed for the load conditions.

In bottom gear you will not be able to pedal fast enough to keep up with the bike going downhill, and in top gear you will not be able to deliver the torque at low pedal speed when going slowly uphill.

Your body has a maximum power output that delivers a wheel speed that is matched to the road conditions.