Motor Draws more amps at certain HZ

Hello everybody,

I am using TBwood drive to control a motor speed that drives an auger.
between 12-15 HZ motor drwas 10 amp-lower or higher than this frequency almost 5.5 amp.

any idea why this happens?

motor is 15 HP at 220 V FLA 15.5

Thank you

Your question is very hard to follow. Perhaps you could give your data in tabular form?

And check the nameplate on the motor again, no way is it 15HP with a FLA rating of 15.5 amps. Not at 220V.

In general though, if you are noticing higher currents at the lower end of the speed range, that is normal on a standard AC drive.

You may have a resonance at that frequency. The natural frequency of a part of the drive train lies there and may cause an increase in loading.

atatari

If you have a 3 phase 15 HP motor w/220vac supply your FLA should be around 42 Amps.

Rick has a good point. If your auger is "chugging" at lower speed the lack of a stable CEMF may cause higher than expected current draw. Are you experiencing a vibration of some sort that is accentuated at this speed?

Actually this is a good question for Dick. It would be interesting to see what he has to say....

Mike.

Sorry guys,

Motor was 2Hp ..so at that voltage ,FLA is around 7.8 amp.

but we found out that 12 Hz is the frequency that auger starts to turn and for some mechanical problem was drwaing high amp..but once it reached to 18-20 hz amp dropped.

so thanks for the replies ..its working now

Sounds like an open-loop Frequency drive. That's a LOT of slip.
You might want to check the 'Boost' settings, but I think I'd put a vector drive on that, probably with an encoder.

No... It happened again.

between cetrain frequencies 12-20 HZ ,motor draws more amp.
the only thing that has left is the motor.

I am going to change that tomorrow.

Torque boost is at 6% ...I didn't change that..has been like this for a long time.

It is open-loop ,but again it has been running like this for a long time.

I don know whats going on?

any ideas?

Now I will throw my 2 cents in

I would look for a mechanical problem...bearing seizing etc. At the lower frequencies there may not be enough torque to drive the auger if its under load and has a mechanical problem.

You said it has run this way along time so my suspicions would be to look closely at the mechanical, maybe disconnect motor from auger and turn by hand to see if it "feels" rough. Verify any and all bearings have been greased regularly, may even want to run auger with no load and grease while running then retest how it performs.

Let's see here. We've got a two hp motor nameplated 7.8amps at full load. And we've got a TB Woods V/Hz drive (ugh!--sorry, not fond of these at all!).

Assuming the drive is programmed for constant Volts per Hz from 60Hz down to near zero Hz, then we should see around 2 amps over the whole speed range 0-60Hz at no load and 7.8amps over the same load range at full rated load. I would double check that the drive is actually regulating the V/Hz ratio properly as everything else is dependent upon that. If it isn't and the parameters are set up properly, dump the drive and buy a new one. (May I suggest something other than TB Woods)

If everything in the above paragraph is in order, that pretty much leaves motor shaft torque as the cause of current variations. It is not uncommon, especially in screw conveyors for torque to vary all over at different speeds and with different consistencies of products going thru the conveyor. I would bet that the odd current behavior is actually due to odd torque behavior in the load.

One thing that seems to point to heavy low speed torque is the comment about it taking 12Hz at the input to the drive to get the motor to start turning. This is all wrong and points to either a motor that is way underexcited (low voltage) or break-away torques far above the rating of the motor. If 12amps at 3 or at most 4 Hz doesn't start the load moving, there are serious design problems with this system.

A 2hp motor at 1800rpm is rated for 6 ft-lbs continuous torque. At 3600rpm it would be 3 ft-lbs and at 1200rpm it would be 9 ft-lbs. You might just want to get a torque wrench on that conveyor and check the break-away torque. Sounds like it might be quite a bit higher than that. You can take the above torque numbers and increase them by 50% for short-term overloads if the drive is heavy-duty or constant torque rated.

This also would be a good time to say that top speed in the load should occur when the motor is at about 90Hz--not 60Hz. This improves motor cooling, available torque to the load at all but top speeds, and speed regulation under load changes. It doesn't hurt the motor at all. Don't use this rule at 1000hp or even at 500hp but certainly for four or more pole motors under 150hp, it's good power train design.

See.. Wadid I tell ya..

Dick, You should change your user name to Dr. DickDV!

atatari,

Let me summarize what you have said:

1. You have a 2 Hp, 220 volt AUGER motor that draws 10 Amperes in the range of 10-15 Hertz. The AUGER starts turning at 12 Hertz. So the 10 Amps is your "break-away" start-up current, the extra power required to get the material in an AUGER moving. This appears normal for a 2 HP. AUGER, as far as I am concerned.

2. At other speeds, it draws lower amps.

I do not see any difficulty with the above --so far. Now, if the current is increasing, getting larger than previously seen, as rsdoran, and elevmike pointd out, then you could have a mechanical problem, bad bearing, broken flight on the auger, frequency resonance, or your motor is going bad.

I have found that on an AUGER, if your motor has to run below 20 HZ for long periodsl, you need to either add an external cooling fan for the motor, or change out the gear reducer to a lower ratio so that the motor can run a little faster.

DickDV, note that he said AUGER, not conveyor. Yes, you can set the max frequency up to 90, IF the system designer over-built the equipment and put a large enough motor on so that the extra current at above-60 speed doesn't cause the drive to trip on current overload. About 1 out of 2 is the times that I have been able to do this. The last time, 70 was the fastest I could get at full load without tripping the overload.

Lancie, a couple of comments on your post. These are in a spirit of constructiveness, not personal criticism, if you please.

You observe that to start the auger, it takes 10 amps and 12 Hz. Now, look at that! 12 Hz and the motor is standing still. If the voltage is proper, the amps should be far higher. 12 Hz into a stationary motor is 20% slip. The motor nameplate would likely be not more than 4% slip to reach full load. Where should it be at 20% slip?!!! Buried in overload, at least. Clearly, something is way out of kilter!

Also, I don't understand your statement of overdesigning with a larger motor when using overspeed. In fact, the opposite is true, you can often get by with a slightly smaller motor. This is because, with the higher reduction ratio in the power train, the available motor torque is multiplied more so the load either has more available or the motor doesn't have to produce as much. The higher ratio also causes the motor to reach full rated horsepower at a slower load speed with full horsepower available from that point up to maximum speed. The power train doesn't need to be beefed up for the extra torque because system torque is limited by the nature of the load regardless of the availability of more torque. If it needs more torque at overspeed, it also would need it without overspeed. Actually, running the motor as I suggested is usually the cheapest way to get a nice improvement in performance. And, lower current is the expected result, not higher. I use this principle on applications every day with good results.

If you have questions, please ask. I'm trying to make this as clear as I can.

Hi All
I`m a little mixed up, but I would assume and you`ll know what happens when you do that. That if the drive said constant torque that would would mean o thru 60 Hz the torque would be the same.Although maybe not developing full Hp until about 15 through 60 Hz.After reaching what ever the synchronous speed of the motor is and running above this I would think the torque dropped off substantially if not you would be getting something for nothing? But on the other hand if torque is the product of the rotor current squared and your pulling more current a 12 hz , but not turning the rotor do we still have rotor current?
As for the 12 Hz set the drive so that it starts at 15 hz?
Now I`m really mixed up.

OK Thomas, let's go thru some motor basics when operated on inverters.

First, the term "constant torque" or "variable torque" when applied to inverters is a totally misleading term. It has nothing to do with torque but instead refers to how much short-term overload capacity the drive has. Variable torque drives usually have 10% overload capacity and constant torque drives usually have 50%. The same drive can be rated either way simply by derating the continuous current spec for the 50% overload rating.

Secpnd, an induction motor will generally develop its nameplate rated torque at nameplate voltage and current. Use the formula hp=TxRPM/5252 to calculate torque. For example, a motor nameplated 10hp, 460V, 16amps, 1780rpm, will develop about 30ft-lbs torque when fully loaded (that is 16 amps). Since motors are primarily inductive devices, if you reduce the frequency to the motor, you will also have to reduce the voltage in the same proportion to keep the current constant. That's what an inverter does when it is set up for constant volts per hz. So, at 60hz, the motor sees 460V, at 45hz it sees 345V, at 30hz 230V, and at 15hz 115V. Under those conditions the motor will draw the same current as when loaded at nameplate conditions. This means that an induction motor can be expected to produce constant torque from its nameplate base speed down to near zero speed if the V/hz ratio is held constant. Referring to the formula above, you can see that the hp drops linearly with speed from 10hp at nameplate speed down to zero at zero speed.

Third, when an induction motor is driven into its overspeed range by exciting it with a frequency over its nameplate, clearly you cannot continue to increase the voltage above the motor rating. So, above base speed, the voltage is held constant and the current that results is also constant. The motor input kw is therefore constant. It should come as no surprise then that the motor output kw or hp should also stay constant. Looking at the above hp formula again, you can see that, for hp to hold constant with increasing rpm, the torque must drop down in the same proportion as the rpm is increased. So, above nameplate base speed, the motor produces constant hp which results in the torque dropping as you go further and further into overspeed. This cannot go on indefinitely without the motor flying apart or, which usually happens first, the motor windings begin to misbehave due to the high frequencies and the motor hp actually starts to drop too. This varies some with different motor designs but, under 150hp, you can expect constant hp out to at least 90hz.

When applying an overspeed motor to a load, you must be sure that the load torque is always less than or equal to the available motor torque or you will not be able to drive thru that speed where the motor torque is less than load torque. If there is reduction in the power train, you must factor that into the calculation of torque too.

In summary, generally you can expect an induction motor on an inverter to develop constant torque at and below base speed and constant hp above base speed. All of this within reasonable limits.