50Hz/60Hz Problem

Hello all,

Not strictly a PLC question, but it does relate to controls. I have a system I am currently installing in Canada. When we tested the system in Germany we had no issues. Now, upon supplying power to the system, I have an issue with a reversing contactor tripping out. The contactor is made by Phoenix Contact and has the following part number:
2297031 ELR W3- 24DC/500AC- 2I


When looking this up, I saw the contactor was rated for 0.83 kW to 2.23 kW, with a range of 0.18 to 2.4 Amps. The motor is rated at 0.80 kW.

The trick is, in Germany we had a 50Hz supply, where here we have a 60Hz supply.

I believe in Germany the motor was running at the power limit of the reversing contactor and as such there was not an issue (the motor powers a small conveyor belt).

But the trick is, since we are working on a 60Hz Power supply in Canada as opposed to 50Hz, the motor's power has dropped, and as such is beyond the operating limits of the reversing contactor.

I have tried to search for equations to prove my theory (I'm a programmer, not an electrician) but cannot seem to find any.

Does anybody have some help for me regarding the voltage/amps/frequency/power relationship?

Thanks in advance...

boswellw said:

...
(the motor powers a small conveyor belt).

But the trick is, since we are working on a 60Hz Power supply in Canada as opposed to 50Hz, the motor's power has dropped, and as such is beyond the operating limits of the reversing contactor...

Click to expand...

The problem isn't power. At 60 Hz, the motor is trying to spin 20% faster. Your problems are related to that.

And a 20% increase in speed correlates to a 20% decrease in power. Having a reverse contactor that is rated at 0.83kW with a motor at 0.80kW @50Hz will most certainly make a difference. I asked for power/voltage/current/frequency relationships and equations if I recall?

Boswell
Keith is correct. The motor is turning approx 20% faster hence is working 20% harder and drawing 20% more current. If the starter was fully loaded then it will be 20% over with the increased frequency.

Not exactly what you are looking for but some useful info anyway.

http://www.patchn.com/index.php?option=com_content&task=view&id=43&Itemid=74

boswellw said:

And a 20% increase in speed correlates to a 20% decrease in power. Having a reverse contactor that is rated at 0.83kW with a motor at 0.80kW @50Hz will most certainly make a difference. I asked for power/voltage/current/frequency relationships and equations if I recall?

Click to expand...

Not quite.
Things rotating faster tend to use more energy..................
But if you can show me how to rotate 20% faster & use LESS KW I'd appreciate it.

hello,

you may try this link. it has the basic formulas, easy & clear.

http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm

Took a look at the spec sheet on that contactor, it's max rated voltage is 550V, which might also be a problem since most industrial 3 phase in canada is 575V-600V.

Looking through the certifications I also didn't see it as CSA approved, the inspector might not like that. (Although it is CUL listed)

boswellw said:

I asked for power/voltage/current/frequency relationships and equations if I recall?

Click to expand...

Sorry for the short answer. You did say you were a programmer, not electrician.
The way I see it, math won't help you because the problem is mechanical, not electrical. The contactor is properly sized. The overload was set in Germany. The setting is now wrong for Canada. If the system can run 20% faster without a problem, then the overload setting needs to be increased because current draw is higher. We can't know what that value is without looking at the machine, i.e., the load from the conveyor.
I see a motor with gearbox driving a belt. If the load is now beyond the rating of the motor, then we need a different gear ratio. If faster belt speed is okay, and the load is light, then the current increase is probably small. Just increase the overload setting a little.
Without looking at the actual load on the motor, we are in the dark.
Edit: Chistoff84 raises another issue. What is the voltage?

I got an idea it may or may not work, what about implmenting a VFD ??

CroCop said:

Not quite.
Things rotating faster tend to use more energy..................
But if you can show me how to rotate 20% faster & use LESS KW I'd appreciate it.

Click to expand...

Some say less HP some say more HP for the same unit.

HP = 746 watt
HP = (torque X RPM) / 5252

Motor at 60 HZ is running 20% faster than at 50.
Conveyer is constant torque load.
Assuming you are not changing loading of conveyer then as you speed it up HP thus Kw must go up.

You need to check the overload setup on your unit and find what they are set to trip at. Do NOT change it so far they are doing a fine job of protecting motor.
Check your line amperage and make sure you are not overloading motor both in foward and in reverse.

Once you know motor amps you can then decide if your contactor is undersized.

Dan Bentler

Dan, could he not use a VFD?

texasmaverick30 said:

Dan, could he not use a VFD?

Click to expand...

One small thing I forgot - with direct on line starting and reversing -
the motor CANNOT
reverse everything
WITHOUT letting everything come to complete stop. Crash reverses are very hard on contactors, motors and gearboxes.

I would think a VFD may be a good idea
BUT he has to know his motor can handle the load. Right now all he knows is he is tripping overloads but not WHY they are tripping.

Once he knows that then
if overloading motor either get a gearbox with greater reduction that will let him reduce torque at motor (slower conveyer speed though) or get a larger motor and associated set of controls - at which stage a VFD may simplify a lot of things for him including reversing.

If not overloading motor then he has to ensure he has right overloads they are set correct and then he can worry about having the right approval tags on the contactor.

Dan Bentler