OT, under rated motor/gearbox change?

Hello to all

We have a several small inverters that are under rated, 1.5 FLA on the name plate on the motor, it has a 10 to 1 ratio gearbox, running speed is 45hz, the drive is large enough to handle 5 amps

I need more, its currently pulling close to 3 amps (and ever so slowly burning up the motors), if I change the gearbox to a 20 to 1 will that drop my FLA to 1.5? Not concerned with the speed reduction, I have time up to 15 minutes between cycles

Just trying to save a buck or two, any other ideas? and get what we should of got to begin with

As always many thanks
Mark

If you can stand to have the same work done in twice the time then you will probably be ok. So if the motor is moving a conveyor, and it was taking 10 seconds to move the product a given distance at 3hp, you should be able to move the product the same distance in 20 seconds at 1.5hp. This neglects any startup or efficiency issues...

The efficiency of the 20:1 reducer may be different than the 10:1 reducer so it will probably not be exact.

Are you experiencing failing motors or just looking to the future? On very low duty cycle applications we regularly use motors that would be much to small for a continuous duty with great success. You can read more about this here:
http://www.motorsanddrives.com/cowern/motorterms9.html

You should have a motor rated for use on a drive or the motor will fail for other reasons...

ndzied1 said:

...Are you experiencing failing motors or just looking to the future?

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We have lost one, its a new application, we clamped the motor and saw that we were twice as hi as the FLA, I just assumed that they went cheap and undersized them. Learn something everyday, I will look into other things to cause the failure...

Great link!

Yes its a conveyor and I can afford for it to take twice as long, its 10ft and cycles once very 30min

Thanks Norm

Assuming this is continuous duty, raising the ratio to 20/1 would get the motor current down to near 1.5amps. But why not change the ratio to 25/1 and then crank the motors up to 90Hz. Your FLA should still be below 1.5 and the motor cooling will be much better. Just be sure the gearbox can stand the extra speed on the input shaft.

I wouldn't say that this is a direct relationship. Let's say right now on your 1.5 fla motor that 0.5A (just to have a number), is being used for losses (magnetizing current, friction, windage, etc...). If it is pulling 3A that means that 2.5A is being used to do the work. If the ratio doubles then in theory you should need about 1.25A to do the work. Add that two the 0.5A that is lost just because the motor is running and you have a draw of 1.75A

Now this is assuming that the torque / amp relationship remains linear above the rating of the motor. This might not be the case.

Having said all that personally I would try a bigger reduction like you suggested, if it works great, if not you need a bigger motor.

geniusintraining said:

Not concerned with the speed reduction, I have time up to 15 minutes between cycles

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If you don't care about the speed why the VFD?

Edit: missed the last three posts.

I like DickDV's suggestion the best.

Just trying to save a buck or two, any other ideas?

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Is the motor really hot to touch? If it is truely overloaded it will be. Sounds like you have alot of cooling time. If you want twice the torque and there is a chain or belt between the gearbox and the roller? Change the one on the roll to 1/2 its size.

If you can stand to have the same work done in twice the time then you will probably be ok. So if the motor is moving a conveyor, and it was taking 10 seconds to move the product a given distance at 3hp, you should be able to move the product the same distance in 20 seconds at 1.5hp. This neglects any startup or efficiency issues...

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This is true for hp, but if torque is the problem? It`s going to stay the same and you still will not be able to pull the load! Don`t confuse hp with torque.

geniusintraining said:

we clamped the motor and saw that we were twice as hi as the FLA

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You can't use a clamp on amp meter to the motor from a vfd. Wrong waveform. Read the current draw from the vfd display or it's analog output monitor.

And who forgot to set the motor FLA parameter in the vfd ?

A 3 amp motor load (3phase @ 460vac) is only 1.5 hp motor. That is a darn cheap motor. Why not just get the correct size motor and be done with it.

The posters following mine all make excellent additional observations, especially about trying to read amps with a clampon ammeter on a drive output. This reading cannot possibly be accurate and should be read from the drive.

allscott's comments about motor magnetizing amps is correct also resulting in less amp reduction that a linear derate would indicate. Even tho that is true, whatever the existing motor hp is (and there seems to be some question about that), it is going to be able to run this load as some slower speed; the only question is how slow.

Thomas Sullens point about changing sheaves or sprockets is good too except that the driving (not the driven) sprocket needs to be reduced in half to effectively double the ratio.

My point about running the motor faster is simply that, from 60 to 90Hz, the motor produces full hp. At 45Hz it only produces 3/4 of the nameplate value and the cooling is less as well.

And, yes, it definitely sounds like the drive is not programmed right. In fact, the overcurrent condition, if there really is one, just might be due to programming errors. Clearly, the motor is not being protected properly by the drive overload calculations and that can only be due to incorrect motor nameplate data being entered into the drive or just using the factory default values.

Sounds like its time to do some rethinking on this whole arrangement.

GIT said:

Yes its a conveyor and I can afford for it to take twice as long, its 10ft and cycles once very 30min

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Is it by any chance a conveyor that pulls material out of a bin for example ?
How fast is the conveyor belt moving (m/s) ?
What material is being conveyed ?
If it is a such a draw-off belt conveyor, then I know a few tricks.

theDave2 said:

You can't use a clamp on amp meter to the motor from a vfd. Wrong waveform. Read the current draw from the vfd display or it's analog output monitor.

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It makes perfect sense but I never realized it. Thanks Dave

Issue resolved

Up date...

Let me give a little more detail, this was commissioned by the OEM, we relocated this line from a plant closure up north, we had the OEM add several more material conveyors, they also abandoned many that were being used up there.

So the frequency drives were all ready in the cabinets, in my previous post I used some general numbers just to try and understand the concept of doubling the ratio to reduce the current draw if it would work or not (my bad)

Also now just got done talking to the night shift tech, he found (as others here alluded to) the freq was still set up for the old application and that one drive was set to 3 amps thus the reasoning it drove up to 3.2 amps and burnt up that one motor.

I thought the reasoning was due to engineering change, below is a sketch the pick and place drop a single product onto the conveyer (a direct drive, belt conveyor), now they are trying to drop 50ish, then indexing, I did not have the whole story before (again my bad)

Also the terminology that I used 'clamped' was just a term, its way to easy to look at the drive

So... this motor is .7kw running at .8 amps, 460 3ph with a rated FLA of .95, this is with only 5 pieces

My suggestion is going to be if they want to continue to run higher and higher weights (3amps), buy new larger motor/gear boxes to handle the new larger load if we loose another one, there is plenty of cool down time between cycles.

Thanks for everyone’s help next time I will have the story straight before I post... But I still learned from this thread so thanks again

GiT