- Thread starter CraigM
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- San Francisco Bay Area, CA, USA

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- Electrical Engineer

A motor is designed to provide a certain amount of torque at a given speed:

HP = torque x RPM / 5250.

The design of the windings is based on providing rated torque at a ratio of Voltage and Frequency or V/Hz. So on a 460V 60Hz motor, the design is going to produce rated torque at 60Hz based on 460/60 or a V/Hz ratio of 7.67:1 If you apply a lower frequency to a motor designed for a higher one, it will not only spin slower, but it will produce less HP and the windings will be over excited because the ratio is wrong. So 460V 50Hz will give you a V/Hz ratio of 9.2:1 meaning that you are applying too much voltage for that frequency and saturating the windings, increasing the thermal losses in the motor. If on the other hand you had a motor designed for 380V 50Hz, guess what... The V/Hz ratio is 380/50 = 7.6:1, essentially the same as a 460/60 motor design!

So back to your more likely situation of having a 460V 60Hz designed motor and applying 480V 50Hz. The motor will turn 17% slower (50/60) so your HP capacity will drop by that much as well. If your load on the motor does not change, AND the motor was sized very close to the load requirements, you may end up overloading the motor. Current would increase at least 17% because the motor is producing less torque and slip in increasing. However if the motor was over sized by 20% (a common design technique), then you may be OK.

BUT... even though you may get away with it from a loading standpoint, you still may have trouble. If you are using a 60Hz designed motor and applying 50Hz at the same voltage, the motor may overheat even without being overloaded because of the motor being over saturated. If you reduced the voltage to 380 though, you would again be back in the motor's design specs. In THAT case, your current draw would then depend on the load, but assume it will be roughly the same as if it were at 460/60.

- Location
- Ann Arbor, Michigan

CraigM and ta50214:

If we assume 90% efficiency for a motor, then the electrical input power, Pinput = Poutput/0.9 .

This has nothing to do with motor size, output RPM, frequency, power output, and voltage given we know the efficiency. However, all these factors influence efficiency.

If we were speaking of a DC motor, then

Power input = Vin * Iin.

where

Power in watts

Voltage in volts

Current in amperes

Power output = FT-#/minute . One HP = 33,000 ft-#/minute. So application of 1 HP for 1 minute will raise 1# 33,000 ft. Or 33,000# can be raised 1 ft in 1 minute.

Rotational mechanical Power = Torque * RPM / 5252

where

Power in horsepower, one HP = 746 watts

Torque in #-ft

RPM in revolutions per minute

Power input to an AC motor where both voltage and current are sine waves is

power input = Vin * Iin * PowerFactor

.

- Location
- Ann Arbor, Michigan

CraigM:

If you have studied my last post and have some understanding of my comments, then consider your original question.

If there are two different motors, one 50 Hz and the other 60 Hz, such that both are loaded to 10 HP on the output, have the same efficiency, the same power factor, and input voltage, then the input currents should be essentially the same.

If you have a 50 Hz induction or synchronous motor, apply the same HP load on the shaft, meaning a higher torque when on 50 Hz than 60 Hz to maintain constant output HP, and the same input voltage, then you can expect the current to be different because both efficiency and power factor will change.

You may need to refine your question to get at the real issue of what you want to know.

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