Single Fault Condition - Motor Test

[Roland chung]

Hi all,

I have two questions about the motor test. I will appreciate any opinion you can provide.

1) One of the locking test preconditions stated in clause 13.2.8 is equipment has one or more motors with a locked rotor torque smaller than the full load torque.

I have never seen a motor with locked rotor torque smaller than the full load torque. If the full load torque is larger, it would draw higher current. That means full load test is a worse case. Why locking test is still required for this case?

2) Regarding the motor running overload test (13.2.13.3 b), the standard requires the load (current) to be increased in appropriate steps, the supply voltage to be maintained at its original value.

Normally, the voltage will decrease when the load increased. Why supply voltage should be maintained during the test?

Regards,

Roland

[Peter Selvey]

For 13.2.8, just from a quick look at some web literature it seems that motor's maximum available torque can be larger than the locked rotor torque, but to avoid damage the torque at rated load is usually lower than the locked rotor torque. As such the provision would not apply for most motors.

If the locked torque is less than the full load torque, there is a higher possibility of locking. A locked rotor can still get hotter even if there is lower current, because there can be less cooling (rotor is not moving, or there could be a cooling fan attached).

For 13.2.13.3 b), it seems kind of obvious, the voltage should be maintained, but it's not really that critical either for overload tests. Variations within 1% would not be a concern. The same applies when overloading a transformer, in principle you keep the voltage constant but slowly increase the current load. A fall of more than 1% typically only occurs if the change in current load is large (e.g 5A), or long cables/ smalll autotransformers with low current rating are used.

[Roland chung]

2) Regarding the motor running overload test (13.2.13.3 b), the standard requires the load (current) to be increased in appropriate steps, the supply voltage to be maintained at its original value.

I found the provision for motor running overload test specified in IEC 60335 is nearly the same as the requirement in IEC 60601. But IEC 60335 is clear to indicate the current through the windings is increased by 10 % steps.

There are two opinions on understanding the 10 % steps:

Assuming the rated current of motor is 5.0 A:

1) 10 % of the rated current --> step 1............5.0 A
step 2............5.5 A
step 3............6.0 A
step 4............6.5 A
step 5............7.0 A
...

2) 10 % of the current of last step --> step 1............5.0 A
step 2............5.5 A
step 3............6.05 A
step 4............6.66 A
step 5............7.32 A
...

Two loaded methods produce two test results. Which one is correct?

Please kindly advise it.

Thanks,
Roland

[Peter Selvey]

The results should not be that much different assuming a reliable protection method is used.

There are three temperatures to consider:
1) temperature of the winding
2) temperature of the protective device (e.g. thermal cut-out)
3) measured temperature (e.g. thermocouple, rise of resistance)

For (1) a larger step size should produce a higher temperature, because of the time delay to heat up and trigger the thermal cut out. But, with small steps of 10% it will not make much difference which method is used (10% of starting or 10% of last current).

For (2) the operating point should be the same regardless of the step size, provided it is a reliable device.

For (3) it depends on how well the measurements are made. It is actually very difficult to make accurate measurements unless you are using on-line resistance measurement. If you use thermocouple the result can be very different depending on the step size, because it is usually even more delayed than the thermal cut out. If you use periodic resistance measurement, then the result will depend on when the resistance measurement is made. If it is made after each step change is stable, then yes the result will be very different; it will be random as to which step point ended up being closest to tripping point of the thermal cut-out. Although relatively small, steps of 10% in current are 21% in I-squared-R of the winding. So, the random factor, rather than the step size could be the key problem. If you are using periodic resistance measurements, small steps are recommended.

However, at the end of the day, usually a thermal cut out with a ~20degC margin to the limit is used, so the test method does not affect the pass/fail result.

[Roland chung]

it will be random as to which step point ended up being closest to tripping point of the thermal cut-out. Although relatively small, steps of 10% in current are 21% in I-squared-R of the winding. So, the random factor, rather than the step size could be the key problem. If you are using periodic resistance measurements, small steps are recommended.

I could not understand very well why small steps are recommended. The larger the step size, the larger the I-squared-R.

[Peter Selvey]

Consider a test to find out the highest temperature before a thermal cut out (TCO) opens in an oven.

Reference Test:
gradually raising temperature until TCO trips
Result 133C

Step Test 1:
Steps of 33C
Result 132C

Step Test 2:
Steps of 20C
Result 120C

Step Test 3:
Steps of 10C
Result 130C

You can see that the largest step got the best result. But actually it's just chance (random) as to which step gets closest to the trip point.

[Roland chung]

I don't think the current step size would result in the same temperature rise in each step. The measured temperature is also not simply the product of the steps and temperature rise in each step.

I know you just gave an example. It sounds too good to be ture.

[Roland chung]

If you use thermocouple the result can be very different depending on the step size, because it is usually even more delayed than the thermal cut out.

The thermocouple is a main tool for measuring temperature in the test house. Although the measurement is not that "real" time, the short delay (response of thermocouple) would not cause very different result.

This is of course just my thinking. Please correct me if I am wrong.

Other question: if there is a larger test current and the measured temperature obtained before operating of the thermal cut-out exceeds the limit (because of the time delay to heat up and trigger the thermal cut out), is the result pass or failure?