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Questions About Motor Performace and Voltage on 3 phase induction motors.

William_Turner

Plastic
Joined
Feb 13, 2015
I need some help. Please share your expertise. I'm like a gorilla banging on an engine with a bit of wood trying to get it to start. LOL.

Here's my problem. I have a nice 11kW spindle with ATC that I am / was / undecided to use on a build. It's 3 phase 380V.

My question is, what happens if I supply it with 220V 3 phase instead and use the VFD to limit the current to the motor?

I don't need 11kW of power. I am happy with much less. I have looked for the answer online, and all I've read is that if you try to run a motor at the same load with less voltage you will burn the motor out. Well, duh. But I want to run it at a reduced maximum power output. Why don't I just use a different spindle? Because affordable used ATC spindles are hard to get!

If you overload any asynchronous induction motor beyond it's rated load and don't have current protection in place would you not burn the motor out just the same?

What are the physical differences in the windings and components of let's say a 3 phase 220V 6.4kW motor and a 3 phase 380V 11kW motor?

Would the efficiency drop significantly? Would the max power output for the same current drop way more than 380/220 or 1.73? Has anyone done this? I'm grateful for any help.
 
I need some help. Please share your expertise. I'm like a gorilla banging on an engine with a bit of wood trying to get it to start. LOL.

Here's my problem. I have a nice 11kW spindle with ATC that I am / was / undecided to use on a build. It's 3 phase 380V.

My question is, what happens if I supply it with 220V 3 phase instead and use the VFD to limit the current to the motor?

I don't need 11kW of power. I am happy with much less. I have looked for the answer online, and all I've read is that if you try to run a motor at the same load with less voltage you will burn the motor out. Well, duh. But I want to run it at a reduced maximum power output. Why don't I just use a different spindle? Because affordable used ATC spindles are hard to get!

If you overload any asynchronous induction motor beyond it's rated load and don't have current protection in place would you not burn the motor out just the same?

What are the physical differences in the windings and components of let's say a 3 phase 220V 6.4kW motor and a 3 phase 380V 11kW motor?

Would the efficiency drop significantly? Would the max power output for the same current drop way more than 380/220 or 1.73? Has anyone done this? I'm grateful for any help.

ISTR there was a very similar challenge addressed within the last month or three, but with a much smaller motor. I also seem to recall the gurus did the math and figured out that with enough flexibility in the VFD, parameters could be set such that it could be made to work after a fashion.


"Flexible" VFD would not have been a rock-bottom sparse-features cheaply priced one.

For a a larger motor, those costs would be higher, yet.

Why would a transformer not make more sense? Is it because your proposed 3-P @ 230 VAC is only there at the generosity OF a VFD, and what you really have for service is 230 VAC single phase?

Not yet enough info on the table...
 
The issue of low voltage and burnout is because the ASSUMPTION is that you want the same power. You do not, so it is not an issue for you in the same way it is for some factory running a motor at full HP on a pug mill or similar tough load.

Obviously, the same power takes more current at a lower voltage, the power has to come in before it can be turned into shaft power. So the motor sucks more current at low voltage and same power than it does at rated voltage.

And, the "copper losses" in the motor (resistance losses, mostly) go as the square of current. So if the current goes up 10%, heating in the motor is about 21% more. With a large motor producing full power, that may overheat it over a bit of time, and cause it to fail. It is not likely to fail instantly, but it may in anything from hours to weeks, depending on the motor ratings and service factor. With some motors it will be fine at that overload, but 15% or 20% would fry it.

When you WANT significantly LESS power, there really is no problem with limiting current. So long as the motor goes at the same speed, which an induction motor will do well enough, and you do not allow any more than full load amps, you are just fine running it that way.

That's a big voltage reduction (43%), and will lead to a big power reduction. But if all you need is a couple kW of spindle power, there is nothing at all stopping you from doing that.

You will need to reset the motor rated voltage to your 240V, and set motor rated current to the normal nameplate rating. Frequency is not affected, I assume you want to run at normal speed. You can use the VFD to vary speed just as normal. You MAY want to use the low speed "boost" for the V/Hz curve, because you will be starting with a lower voltage, and motor resistance will be more of an issue.
 
Why would a transformer not make more sense? Is it because your proposed 3-P @ 230 VAC is only there at the generosity OF a VFD, and what you really have for service is 230 VAC single phase?

Firstly, Thank You for your input!

To answer your question, Yes. Exactly. So I would oversize the VFD input amps by 1.73 minimum, lets say double. And the single phase can't output the current required for an 11kW spindle, and I also don't need 11kW.

Also, I have read a few VFD user manuals, and on the ones I read you can only input one value for max Amps. So this would limit current (Amps) even in the constant v/f section of the power curve where voltage is lower. If I bump it up to 380 via a transformer, I have fewer Amps that I can use.

220V x 50A = 11KVA single phase
That gives 28.9A at 380V single phase assuming a 100% efficient ISO transformer.
Which equates to 28.9 / 1.73 = 16.7 Amps 3 phase out of the VFD assuming again 100% efficiency.

http://www.hsdusa.com/bo/allegati/Files/757_1423h0198_rev01_es988a_13kw5.pdf

If we look at this graph and the table below, we can see that around 26 Amps is required for full power. So if I limit it to 17 amps thats a huge difference, and it would be limited through the entire voltage range to 17 Amps if I read the manuals correctly?

Also, I don't want to buy the huge transformer if I don't have to.

Comparing the graph from HSD linked above, I figure that this would be the outcome:

Spindle Voltage Comparison.jpg

Does that look correct? It would have no effect in the Unaffected range? What would happen in the Affected range?

Also, 26A 3 phase x 1.73 = 44.98A single phase, well below 50A. The breaker on my 220V single is 60A, but I think I will limit my use to no more than 50A.

So this is my theory, please tell me what I'm wrong about as I've never actually done this.
 
I just posted a long post, but I received a message that a moderator must review it first. Is that because I posted a link and a picture?
 
And a big Thank You to you both for your help! You will see more info about my problem when the post I made is released......
 
An electric motor only draws power proportional to the work it is tasked to perform.

IF An 11Kw motor is doing 5Kw of work, it will draw 5Kw. There is no getting around that.
If your goal is to have the motor stall at the 5 Kw level, There are current limiting devices less sophisticated than a VFD.
 
I just posted a long post, but I received a message that a moderator must review it first. Is that because I posted a link and a picture?



No, its likely because you are new. I actually will not see it, The overall site moderator will, though. Patience..... And it showed up as I was typing this answer.
 
Does that look correct? It would have no effect in the Unaffected range? What would happen in the Affected range?

Also, 26A 3 phase x 1.73 = 44.98A single phase, well below 50A. The breaker on my 220V single is 60A, but I think I will limit my use to no more than 50A.

So this is my theory, please tell me what I'm wrong about as I've never actually done this.

Ah.... new info here..... Might have guessed.....

This is a high speed spindle motor. The VFD is really for running it HIGHER than the maximum 3450rpm or so for a 2 pole motor on 60Hz.

The curve shows the kw vs speed. The flat spot is, as you probably know, the area where the voltage is high enough to produce the needed motor current. First you have constant torque area, then constant HP area where torque reduces but speed goes up in proportion. As speed goes up farther, the current drops off in a way such that torque comes down faster than speed goes up, and power is reduced. The motor "runs out of volts" because the motor impedance is high.

When the voltage is reduced, the effect will be to move the curve lower in speed, so constant HP area is at a lower speed, and the whole curve becomes compressed. You have less voltage, but the motor impedance at any given speed is the same, so you run out of volts faster. The max HP will be reduced, and the whole curve is squeezed down.

It should not bother you as long as you do not want more power than is possible. The power is as the voltage reduction squared. So at 57% voltage, you cannot expect more than 0.57 x 0.57 x power, or about 1/3 of nameplate rating. 11kW will be reduced to a bit under 4 kW.
 
And it showed up as I was typing this answer.

I just tried to send you the following message in private:

Hi,

Thank You for your input on the thread I started.

You mentioned that my post showed up as you were typing. I see it wasn't posted however. May I ask, have I done something wrong? Was it the link or the picture? If I have inadvertently done something against the forum rules, I sincerely apologize. Please let me know what it was so that I don't repeat the mistake.

Thank You again for your help.

"JST has exceeded their stored private messages quota and cannot accept further messages until they clear some space."

LOL. I don't know what to say.
 
...I don't know what to say.

Maybe that was what got it intercepted, then?

Unless it is actually "Uncle JST" that Big Brother is watching?

:D

No fear.

Just try the original post that went walkabout again.

You'll have about 23 or so hours to edit or delete it if it comes up as a duplicate.
 
Why would a transformer not make more sense? Is it because your proposed 3-P @ 230 VAC is only there at the generosity OF a VFD, and what you really have for service is 230 VAC single phase?

Yes. I have single phase 220 / 230VAC. Yes, I plan to oversize the VFD to get my three phases.

I was thinking about a transformer.

220V x 50A = 11KVA single phase

380V x 28.9A = 11KVA single phase assuming 100% transformer efficiency

380V x 16.7A x 1.73 = 11KVA three phase assuming 100% VFD efficiency.

I have residential power up to 60A, but I don’t want to push beyond 50A maximum. The motor needs about 26A three phase to generate full power. 16.7A is quite a bit less than this. I read a few VFD user manuals, and if I interpreted correctly, there is only one input for current limiting across the entire RPM range? So if I use 380V I would be limited in current (Amps) even in the region where the voltage is lower in the constant v/f range of the power curve.

Here is a chart I put together to illustrate what I mean:

Spindle Voltage Comparison.jpg

What I don’t know is what will happen in the Affected range? Also, if my logic is sound. This is my first time doing this. Truth be told, I would be using the spindle at even less power than this most of the time, so I also wonder about efficiency.

26Amps 3 phase x 1.73 = 44.98A single phase, so I can get the full power at lower RPMs.
 
It's funny how I can post this message, which is text only, but when I tried to actually post about my problem, also with just text, it got rejected.
 
First time for everything..... I WAS supposed to approve it.... apparently I was the only one who COULD see it. In several years that was the first time of having that happen. all fixed, I think
 
I'll not venture an 'outcome' on this one. Rather a query-scenario:

- A VFD can shield your 50A breaker from motor starting load by ramping.

- A VFD's capacitor bank is on the uncontrolled side of the VFD. It presents an inrush load of its own. That load is greater when the VFD has had to be upsized to also operate off single-phase input.

- as an aside, if you went to a 10 HP Phase-Perfect, their smallest, it wants 55 A service. A P-P, however, does not provide soft-start / ramping. It could, however power a VFD that needed no de-rating nor oversized caps.

- so, too, could an RPC ahead of a 3-Phase-only VFD. If I were to do that, it would have two idlers. Start one, add the supplementary later. Again to try to better live within that meagre 50A breaker's comfort zone.

Every player in the room is straining, and to get to starvation rations, even if made to work.

So I ask.. just how hard is it to get the paltry 60A service upgraded to 100A or 200A?

Should this machine and its load be moved to a more favourable location?

Might it be better to trade-down to a less thirsty machine-tool?

I guess the 'real question' is simpler:

Why?
 
Thank you for your input. You bring up some interesting points. I’ve spent the last few several hours reading online trying to get to the bottom of everything you’ve mentioned.

As an aside, we are now talking on a different subject. I assume you have no objectionable things to say about running a 380V motor off of 220V so long as the current input is limited to the same as it would be at 380V? Effectively derating the motor along a portion of it’s speed range? Start up current aside. I’m wondering about efficiency and if there are any meaningful physical differences in the windings and components between a motor that is designed and manufactured to run at 220V along the blue line in my graph and a 380 designed for the pink line in my graph that would make this a bad idea?

A VFD can shield your 50A breaker from motor starting load by ramping.

Excellent! It’s a 60A breaker. I just wasn’t planning to push it over 50A.

- A VFD's capacitor bank is on the uncontrolled side of the VFD. It presents an inrush load of its own. That load is greater when the VFD has had to be upsized to also operate off single-phase input.

The user guides for Delta-A and ABB VFDs I looked at don’t even mention inrush current. The Delta-B mentions it and refers to Appendix B, accessories, for the use of an AC Input Reactor. Of course, I don’t know if the inrush they are referring to is for the capacitors or the VFD output (compressor with a short ramp up time?). The Yaskawa A1000 series user guide mentions fault codes and such for their inrush prevention circuit, which I assume is for the VFD’s capacitor bank as you mentioned. It’s interesting to note that they don’t see it as a special enough feature to mention it on the Specifications & Features page for the A1000 series.

I found this quote somewhere. I’m afraid if I post a link I may be banned:eek:. I don’t seem to have good luck posting links or pictures.

“VFDs have a capacitor charging inrush current that occur only when power is first applied to the VFD. All VFDs must limit that inrush to prevent damage to the diodes and DC bus capacitors. Various methods including combinations of methods are used to limit the charging current. VFDs are normally not energized and de-energized very often and the motor is normally prevented from starting until capacitor charging is complete. The peak charging current is very brief and the magnitude is partly determined by the source impedance. It is rarely a concern for the user and VFD manufacturers don't usually publish information about it. AC and/or DC inductance that is often built into VFDs to limit harmonic current also helps to limit charging current.”

So I guess it depends on the VFD. Something to think about for sure so thanks for bringing it up.:)

I’ve seen several posts from people on this forum who have oversized a VFD so that it can do the phase conversion for them. I’d really like to hear from someone who has done it. Can anyone please measure the current in Amps of their single phase input when power is applied to their VFD and compare that to the Input Current on the nameplate please? I’ve found lots of information in my searches from people explaining how to do the metrology, but no actual measurements. Does anyone know of any similar measurements that have already been posted online


Every player in the room is straining, and to get to starvation rations, even if made to work.

I disagree. If the 60A can handle the capacitors in the VFD when it is powered up (a momentary condition), I will have a nice ISO30 ATC spindle that I can run from home at up to 6.37 KW, which I won’t ever use. That Sir, is not starving. It just depends on the start up.

What I do want to take advantage of is the higher power available at lower RPMs that I would get with this spindle. So derating it in the higher RPM range is nothing I even care about so long as there are no ill effects. I’m concerned about what the efficiency will be.

So I ask.. just how hard is it to get the paltry 60A service upgraded to 100A or 200A?

I don’t know. I can look into it.

Might it be better to trade-down to a less thirsty machine-tool?

Yes. I actually have it up for trade. Some joker offered me $250 CAD for it. Still looking.

I was also looking at a 5.5kW ATC spindle that is available to me at a reasonable price. But the power curve was constant v/f all the way up to 18000 RPM, which would give me less HP than I want at lower RPMs, plus it was rated at 22 Amps, not much different a scenario that what I get by dropping the voltage on the one I already have. I still haven’t decided.

HSD makes some nice lower powered spindles with ATC that would suit my needs. They cost money that I don’t have. Have you looked at the Canadian dollar lately? Even Chinese knock offs with ATC cost over 3 grand US. LOL. If you know someone who wants to trade, please let me know, until then this thread is simply about whether or not I can use the very nice spindle I have at 220V effectively.

As always, thank you for your input!:Ithankyou:
 
I assume you have no objectionable things to say about running a 380V motor off of 220V

'Objectionable'? "Curious" rather. I don't understand why one would want to.

Nameplate calls for certain ration of power? Simplest thing is to provide it.

Never saw any more point in 'negotiating' with an electric motor than in trying to convince a horse to walk on but two legs because you only have two horseshoes.
 
Missing from this discussion is torque. When you reduce the voltage, you reduce the torque capability of the motor by the SQUARE of the voltage reduction. So giving a motor rated for 380V only 220V means that motor will be getting 58% voltage, ergo it will produce 33% of it's rated torque. So your 11kW motor becomes, essentially, a 3.63kW motor. You just need to understand the consequences.

One of those consequences is that every motor loses some of it's rated power because it must move the mass of it's own rotor. That's already factored into the kW rating of your motor at it's nameplate rating, but works against you in this scenario because you have the mass of an 11kW motor that is only capable of 3.63kW, so the extra motor mass will rob you of more capability just in trying to get out of it's own way. Bottom line, assume no more than 3kW of mechanical capability from it, while drawing current based on that 3.63kW (plus normal motor losses). If that's all you need, have at it.
 








 
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