Electrical Motor Load

[RyanM]

I've been left puzzled by what someone has told me, and I'm hoping someone can help me out with an explaination.

If a 3-Phase motor has a name plate with: 415v, 20A in Star, and it was running with nothing attached to it, I.e no shaft etc. Will it be running at 20A?

Just someone today told me it wouldn't as it didn't have a load on it, and the only way you would measure 20A would be with a full load.
To me this sounds like rubbish, as how can you say what a full load on the motor will be? It could be rotation a fan blade made from paper, or on the other extreme, a huge conveyor belt!!??

Any education for me would be great from you guys.

 

[ricicle]

The current rating on the plate is it's Full Load Current (FLC) which is what it will draw when it is working at maximum power. (Ignoring efficiency and power factor ) An unloaded motor will draw less than FLC due to it not doing the work for which it was rated for.

If the motor draws more than it's rated current then it is being overloaded and will overheat/burn out. Protection should be in place for everything but the smallest motors.

 

[RyanM]

Thanks for the reply.

So what current draw would you expect from the above motor if it was just sat in a work shop running, with nothing attached to it?

This may sound stupid, but could the plates not show the un-loaded current, and full load current?



Cheers

 

[ricicle]

I have never seen a motor with loaded/unloaded current ratings on it.

The only time is when there are different voltages or different configurations specified.

 

[ban-all-sheds]

This may sound stupid, but could the plates not show the un-loaded current, and full load current?

Why would you want to know the no-load current?

What would you do with that information?

 

[Spark123]

Thanks for the reply.

So what current draw would you expect from the above motor if it was just sat in a work shop running, with nothing attached to it?

This may sound stupid, but could the plates not show the un-loaded current, and full load current?



Cheers

They could if they wanted to, but as it is less than that of the FLC then why does it matter? A motor running on no load i.e. with nothing attached is pretty pointless anyway.
Start up current is a bit more important than that.

 

[EFLImpudence]

It would be like asking the horse-power of your car while ticking over. ???

You could turn it off.

 

[SimonH2]

As all the above, the "no load" current isn't a very useful thing to put on a rating plate.

The figures you give suggest a fairly large motor - around 18-20hp. Especially when idle, it will present a highly inductive load and you may be surprised to find it takes considerable current - I'd guess possibly several amps. But this will be at a very low power factor so the actual power taken will be quite low.

As said, from the installation POV, the main things that matter are startup current (and duration) and FLC (plus duty cycle, start frequency, load characteristics, starter type). All the switchgear and cabling will need to be capable of taking the startup current - the duration of which will vary with the load. Separately, there will need to be overload protection which won't trip under the short term startup current - unless the nature of the load precludes overloading. And of course all the switchgear and cabling will need to be able to take the long term running current.
Designing this is far more involved than designing for other (eg domestic) loads. In practice, I suspect most equipment designers simply avoid the issue by over-specifying the supply requirements so that no-one has to think too hard

 

[RyanM]

Thanks for the replies.

I'll explain the situation a little better, as I probably didn't originally.

At work there is a spare motor, with gearbox attached to it which is for use in a freezer. This is just sat on a pallet until needed, but recently the manufacturer recommend we 'turn over' the motor/gearbox to keep it in good condition, but as it only has mineral oil in the gearbox, we should just run in for approx. 5 seconds.

The conversation I heard between the two 'engineers' at work was they both agreed to make this easier one of them should build a simple control panel to be able to start/stop the motor quite easily.
But, one of them said as it had no load on it, they wouldn't need to use a 20A overload due to it not having the full load on it......but then he couldn't say which overload should be used! The other guy said it should be built to spec as if it was going to run under full load, even though it never will while it is on the pallet.

I agree with the 2nd suggestion that it should be build with correctly sized overload, wiring etc, but cannot see the theory behind the first suggestion of building a panel which is under-specc'd for the motor?

I queried this, and he said the motor will never get anywhere near full load current, so there is no point buying new parts in to make the new panel, and use bits we have off the shelf. But which ones could you pick? Obviously a 18A over load would probably be fine, but I still don't agree that it should be used?

Is there some rules/regs on this?

I might be missing something totally obvious here, but I really don't know so thought I would ask you guys


Thanks

 

[ricicle]

If the motor/gearbox is standing alone then there would be no harm in using a smaller overload than required. As the manufacturers recommend only a 5 second turnover then there is little chance the overload will operate within that time (unless perhaps you have a locked rotor but then you will have bigger things to concern you))

 

[RyanM]

How would you spec it though?

As like you said, something smaller would work, but if doing the job properly, how would you know what to put in there?

For example a 1A overload wouldn't work....I know that is the extreme, but I don't understand how you could spec it correctly.

 

[SimonH2]

In this situation, the start current will be the deciding factor. For a 5 second run, it will barely have got going before you switch it off again !

So for the overload, whatever you have lying around. You mentioned 18A - well you aren't going to trip that in 5 seconds.

However, the initial start current will be the same as it would be loaded - though it will quickly drop off, much quicker than if it was loaded. So the switchgear MUST be rated for that size of motor irrespective of load. Ie, if you used (say) a 5A rated relay to start it, then you'd very quickly find the relay melted or it's contacts welded shut.
It's out of my area, but you'd need a contactor with the appropriate AC(x) rating for the size of motor. IIRC AC(x) ratings are the size of attached motor the gear is designed to start/run, with "x" being a number related to operating mode - eg single starts or repeated starts.
So in short, you'll need a contactor rated to start a 20A motor, but the overload can be smaller - and in reality for that use you don't need one, just some fuses will do.

So both your colleagues are correct !

You will also find that you'll need a decent supply - for the same reason given for the switchgear, you'll find it will trip MCBs that aren't considerably above 20A in rating unless they are type C or even D. But I'm not qualified to say what you will need there.

 

[IJWS15]

They have recommended you do this to re-distribute the oil in the gearbox which will all have drained to the sump - best thing to do is put a spanner or a wrench on the output shaft and turn it over by hand.

You should do that first anyway, full speed with no oil could damage the gearbox.