A/C Motor Voltage, Overload Question

[RD12]

I'm a bit confused reading through some textbook material regarding under/over voltage when dealing with A/C induction motors. What I believe I understand is this:

---A given motor operating at a lower voltage than rated on the nameplate will cause the motor to overheat and burn up from the high amps passing through it.

---A given motor that is operating under a heavier load than rated on the nameplate (in Horsepower) will cause the motor to overheat and burn up from the high amps passing through it.


So my question is this.......If an induction motor has a given impedance/resistance (in ohms) then wouldn't it burn up from a higher voltage than rated on the nameplate as well? (Due to I/R)

Looking for some guidance!

 

[gar]

180831-2361 EDT

RD12:

A motor, AC or DC, is not a constant impedance.

To a large extent a motor can be viewed as two separate power components, losses and power delivered to the shaft load.

The biggest power component at full load is output power. If motor RPM is approximately constant, then shaft power output is approximately constant. Which means motor input power is approximately constant.

If motor power input is approximately constant as voltage is changed, then input current will vary inversely with input voltage.

.

 

[ActionDave]

The short answer is a motor has to spin or it will burn up. Too low voltage and it can't spin and it burns up. Too high voltage it burns up before it starts to spin.

The key to getting this is separating impedance from resistance, how they are similar and how they are different.

Your gut is right on about this. Motors are a strange animal, but if you get them then you get a whole lot of electrical stuff.

 

[GoldDigger]

Another way to look at it is that the motor winding current is the result of three factors in addition to the applied voltage:
1. The winding resistance
2. The winding inductance, assuming the iron core does not saturate.
3. The counter-EMF developed as the result of the rotation.

When the applied voltage is too high the second factor fails to limit the current.

When the applied voltage is too low or the load is too high the motor speed does not reach it's operating range and the third factor fails to limit the current.


Sent from my XT1585 using Tapatalk

 

[retirede]

Do a search for induction motor curves.
You will find that (at rated HP) slight over-voltage results in lower current and can actually be beneficial. But at some level of over voltage, the current will begin to rise and result in overheating.

There are many other factors that influence motor heating such as voltage imbalance. A slight imbalance is much more detrimental to the motor than slight over voltage.

 

[Besoeker]

I'm a bit confused reading through some textbook material regarding under/over voltage when dealing with A/C induction motors. What I believe I understand is this:


So my question is this.......If an induction motor has a given impedance/resistance (in ohms) then wouldn't it burn up from a higher voltage than rated on the nameplate as well? (Due to I/R)

Looking for some guidance!

Depends on how much higher or lower. There normally a tolerance, say ±10%. Reduced voltage increases the slip, the difference between synchronous speed and operating speed, causing the rotor current increases. Reduce the voltage to the point where the motor won't get over the hump in the speed torque curve and will sit at several times full load current.

If the voltage is too high, the stator field may saturate, again resulting in several times full load current.

If either is allowed to persist that will certainly cause the motor to fail but if the overload protection is correctly set it should trip before that happens.

 

[gar]

180901-0921 EDT

RD12:

In my post numbered #2 I made much use of the word approximate. So here I am changing the motor to a synchronous motor for AC or a perfectly compensated compounded DC motor. Both of these motors will have a constant output speed independent of load up to an overload point.

The full rated continuous load power capability of a motor will be much less than this breakdown overload point. Some of the motor power losses are moderately constant (hysteresis and eddy current), and others are roughly I^2*R (resistive losses), but in a moderate size motor with good efficiency the total losses may be less than 5 %. Thus, at full rated load the variations in losses are not large compared to the power output to the motor shaft. Note that with constant output RPM that windage loss will be a constant. Thus, these losses can be lumped into load power.

The very large power component is motor shaft load and since speed is constant this is constant. So power input to the motor is Ploss + Pload. Ignoring Ploss the input power to the motor is Pload. Thus, we have Pconstant = v * i, or i = P/v (the inverse relationship).

It should be noted that magnetic core saturation is not as big a problem in a motor as in a transformer because of the much larger air gap in the magnetic path of a motor vs a transformer.

.

 

[Jraef]

...
---A given motor operating at full or near full load with a lower voltage than rated on the nameplate will cause the motor to overheat and burn up from the high amps passing through it.
...

This part is a little light on detail, I added the qualifier. What really takes place is that as the voltage is reduced, the motor full load torque is reduced proportionately and more importantly, the motor PEAK torque is reduced at the SQUARE of the voltage reduction. Full load torque is the value used to determine the maximum safe operating point of the motor at the desired speed. If the load on the motor is significantly lower than what the motor was designed for proportionate to the reduction in voltage, it will not overheat. If you think about it, we do this on purpose when we use a “reduced voltage starter” of any type; soft starter, Wye-delta, autotransformer etc. in using those however, you have to know that the load does not NEED that extra torque to accelerate before you hit the overload trip point. But because Peak torque is what the motor uses to RE-accelerate the running motor after a step change in load, if that step change takes place while under that low voltage condition, it may fail to re-accelerate or stall, causing an overload.