Motor with Gearbox vs. High Efficiency High speed motor 4

[enverd]

I have a situation where I replaced an older motor that was running a process along with a gear box and the set up was controlled with an ABB VFD, we have replaced the entire set up with a new high efficiency high speed (5500RPM) motor but the new motor is still controlled by the existing ABB VFD.

Once everything was replaced and set up, we fired everything up and as the result we see the current used is by about 10% higher.

Why is that? Why would the current go up when we are using higher efficiency motor?

Thanks in advance for any replies.

Enver

 

[waross]

Where are you measuring the current and how are you measuring it.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[enverd]

Hi waross,

there is a PLC panel display at the unit we are driving that gives us all the information.
so I would say the load has a smart panel display that gives us that info.
new info has been compared to the old and we see 10% of current increase.

 

[waross]

Your VFD is outputting a variable voltage and frequency.
As an example; You may be running at 20% less voltage and 10% greater current. The actual efficiency may be further complicated by the relative power factors of the two motors.
10% current increase with a different motor running at a different gear ratio is meaningless by itself as an indication of efficiency.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[enverd]

Thanks waross,

I completely agree, wouldn't higher current usage mean more energy used?
I mean if we replace a motor and a gearbox (350HP, 480V, 3phase old and new) with new high speed motor to perform the same task and the new ends up using 10% more current wouldn't that translate in more energy used? And if so what was the point of making this change when our end result was supposed to be more energy efficient setup.

By the way, I am actually helping with this issue and do not have all the specifics, so essentially what I am trying to do is get my bases covered so that I have answers for most questions.

Thanks,

 

[waross]

See if the VFD displays Watts. Current alone is meaningless.

A 100 HP motor will draw about 100 Amps at 600 Volts.
A 50 HP motor will draw about 120 Amps at 240 Volts.
How much will the 100 HP motor save???
Which is more efficient???
You don't know what the effective voltage was to the old motor or for the new motor.
and
Maybe the salesman lied when he said that the new motor was high efficiency, but probably not.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[enverd]

Thanks,

Good point, I'll have a chance to look at some detailed information later this week.

Thanks for the input,

 

[mikekilroy]

If you bring back more information to share and compare, be sure to bring back also:

1) the base speed or number of poles for old and new motor
2) the speed each ran/runs in either hz from vfd display or rpm from hand tach
3) the old gearbox ratio (2:1?) and the new gearbox ratio (1:1?)
4) Include the old and the new motor part nos also in case anyone here has access to efficiency info on them

THEN the more meaningful comparisons can begin.

http://www.KilroyWasHere.com

 

[jraef]

No kidding. This is full of holes and HOPEFULLY, someone retained a qualified EE to investigate this rather than go by what some salesman told you.

One other important piece of information: What is the machine you have it connected to? A VERY VERY common mistake, one that is FREQUENTLY made by uniformed salesmen, is to think that speeding up a motor to make up for a gearbox speed increase is going to be OK. An AC motor can only be 2 poles, it cannot have less than that. 2 poles mean 3600RPM (synch). If you now have a motor that claims to be 5500RPM, and it cannot be any less than 2 poles so that means it is simply getting 92Hz (roughly) from the VFD. Once a motor goes over it's base speed however, the HP remains constant. So if they sold you on a motor that says it is 350HP at 5500RPM, that is not the same as having a motor that provides the same TORQUE as your 350HP 3600RPM motor! Do the torque calcs, you will see what I mean.

HP = Tq. x 5250 / RPM.
and therefore
Tq = HP x 5250 / RPM.

So:
350HP at 3600RPM = 510 ft. lbs. of torque
350HP at 5500RPM = 334 ft. lbs. of torque!

So for you to get the SAME torque as your original motor, 510 ft. lbs AT 5500RPM, that new motor would need to say it was at least 486HP at 5500RPM. Does it?

If it says it is still 350HP at 5500RPM, then you are getting only 65% of the torque you used to get. So if your old gearbox was anything less than a 1.54:1 ratio, you now have LESS shaft torque to work with. Less shaft torque means higher slip in the motor, higher slip mean higher current. So notwithstanding all of the other potential measurement issues mentioned above, this may be your problem, or a contributor to your problem, as well.


"Will work for (the memory of) salami"

 

[enverd]

Thanks jraef,

The reason why this is full of holes is because I am working with insufficient information. Unfortunately, this project has been lingering around for years and the Process and Mechanical Engineers have decided back in the day to do, what they saw as, one for one replacement. However when you compare the two setups, motor w/ gear box vs high speed motor is more like apples and oranges then anything else.

I posed the question in a hurry and without really thinking the entire situation through. I shouldn't be losing sleep over this because like you said this whole thing is just full of holes. Without having complete history and info we as EE's can't make solid recommendations.
Therefore I went back and did tell the engineers (of other disciplines) to get complete info on the old motor, gear box, setup, new motor, energy usage for old and new motors. Just then we can start to evaluate the entire situation.

In the meantime, to a certain extent I do agree with you but I also disagree on some items.
The new motor is a 350HP, 480V, 3phase, Inverter Duty, 5500 RPM Max, Constant HP. Its base speed is 3550 which makes it obviously a 2-pole motor with an approximate 1.4% slip. Because it is a constant HP the torque will get lower with the higher speed.
VFD's vary voltage with frequency in a V/Hz ratio so the if f is less the 60Hz the VFD will adjust the voltage as necessary. However if the higher speed is required then the voltage will tap out at 480V and the frequency will be increased to provide additional speed.
As a matter of fact they are currently operating the new motor and the entire process is working absolutely fine where the output frequency goes up to 130-150Hz and the RPM reach a max of 4500RPM, approx.
Obviously one needs to have the right motor, right VFD and the right controls programmed to have this setup.
The load they are driving is a WFI Still pump for a pharmaceutical process.

In any case, this whole thing seems to me a problem that is blown out of proportion. In reality there is no problem because the process works really well, they have a cleaner set up and just for kicks someone is questioning why the new motor uses 10% higher current without really telling me if the information is showing live feed vs max during the entire process or the actual average current usage. At the end of the day like waross said just having the information about current without knowing what my Watts are I can't tell them if the energy used is less or more (taking into account they haven't given me the Watts that the old setup used).

However, I am trying to get more info about the whole VFD, Motor, high frequency, high RPM setup and operation. I just can't find any good papers on it that will explain how that whole thing works.
I also heard that the higher the frequency, the motor operates on, the windings get smaller, can't say that made clear sense to me.

If anyone has some good papers, explanations, graphs I would love to get some more education on it.

Unfortunately as a consultant I am more into basis of design and management side on every project and I lose touch with the technical aspect so I have to do a lot of digging to get some answers.

Thanks again to all for the feedback and support.

 

[mikekilroy]

The new motor is a 350HP, 480V, 3phase, Inverter Duty, 5500 RPM Max......Its base speed is 3550

As a matter of fact they are currently operating the new motor and the entire process is working absolutely fine where the output frequency goes up to 130-150Hz and the RPM reach a max of 4500RPM, approx.

Glad to hear you went back and told them more real data needed to explain.

Consider NOT saying the above stuff to them though; it is too wrong: If motor is indeed 2 pole, then 5500rpm max means 90hz max output from vfd - this is VERY typical for max. There are few 350hp motors rated to 120hz and probably none from std mfgrs rated to 150hz. Along same line, 4500rpm & 150hz means a 4 pole motor running 2.5x base speed: again, I doubt they would have found such a motor in 350hp.

http://www.KilroyWasHere<dot>com

 

[enverd]

Hi Mike, Thanks for the feedback.

Below information is coming from the manufacturer. This is not something I came up with. It is actually information coming from the cut sheet.
The new motor is a 350HP, 480V, 3phase, Inverter Duty, 5500 RPM Max......Its base speed is 3550

This information is actual performance, something that "the field guys" have witnessed out in the field while operating the pump:
As a matter of fact they are currently operating the new motor and the entire process is working absolutely fine where the output frequency goes up to 130-150Hz and the RPM reach a max of 4500RPM, approx.

In the meantime while waiting for more information (if it every comes back to me) I contacted the manufacturer of the motor and this was his answer:
Essentially as the voltage goes from positive to negative, the current flow changes direction, and the associated magnet changes polarity. The higher the frequency, the faster the polarity changes.
The 3600 RPM is for 2 pole pairs changing at 60 hertz.
The formula is RPM = (Hz x 120) / poles
Therefore for 5500 RPM = (X hz x 120) / 4
Hz = 183

I agree with his 3600RPM for 2 poles sentence
I have no problem with the formula to calculate (manipulate) speed, etc..
How the heck do you tell me 3600RPM is for 2pole pairs and in the very last calculation you use a "4"? Doesn't make any sense. I mean is it a 2-pole motor or a 4-pole motor? Note: not asking for calculation just a silly question trying to figure out what these guys purchased

I know that vendor's, distributors, even most of manufacturer's engineers are tough to get answers from (for whatever reason), but I mean come on, you can do better than that.
Apparently he will get back to me with more information so I am anxious to see what else I can get out of him.

All the knowledge I have and info I was able to find confirms what you and other members already wrote in your responses and I completely agree with you all but in this case we seem to have a working example of this high speed motor that has some "new" characteristics that I would like to know more about.

It all seems to be confusing and until I get some more information I'm going to stay restless because I just have to know

 

[mikekilroy]

How the heck do you tell me 3600RPM is for 2pole pairs and in the very last calculation you use a "4"? Doesn't make any sense. I mean is it a 2-pole motor or a 4-pole motor? Note: not asking for calculation

I assume the salesman is mixing up the engineering facts. A 2 pole motor has "1 pole pair," a 4 pole motor has "2 pole pairs." Again, unless your customer has a very special (aka $ 35,000.00) motor, he does NOT have one capable of 5500rpm/150hz if 4 pole.

In our world of servos, we DO talk about "pole pairs" ; we generally have 8-32 poles so pole pairs makes sense to use. In the world of AC induction motors, I have never heard anyone refer to their motor as a "1 pole pair motor," but that does not mean there isn't somebody out there saying it.

Bottom line is the information you have is WRONG.

http://www.KilroyWasHere<dot>com

 

[mikekilroy]

YOU do the math: if 60hz gives 3600rpm, and speed is directly proportional to hz, then 90hz (90/60)= 5400rpm, 120hz=7200rpm & 150hz (150/60)=9000rpm. no magic here.

http://www.KilroyWasHere<dot>com

 

[LionelHutz]

Are you sure the motor is connected for 480V? If you connected a 240V motor to the VFD and programmed the VFD correctly then you could run the motor up to 120Hz while maintaining a constant torque output. This would let you get 700hp out of your 350HP motor.

 

[jraef]

Ok, so we know a little more, that helps. Most likely a WFI pump of that size is going to be centrifugal, I cannot imagine a PD pump like a gear pump or a progressive cavity pump being made to be able to stay clean enough for pharma Water For Injection processes. I've never seen one anyway.

So assuming it is centrifugal, it is entirely possible that the 350HP rating is so high specifically BECAUSE they intended it to be used at high pump shaft speeds, regardless of how that was attained. So knowing the torque you WANT at the high speed, you work backward to get a HP at a base speed that you need to purchase. I mention this just to get beyond the initial issue of possible mis-sizing. It may be intentional for all we know.

So now addressing only the higher current issue, there are other possibilities here. As you increase above motor base speed, like I said, the loss of torque can result in an increase in slip, which means more current draw. When the speed increase was done with a gearbox, the motor would have been at a fixed speed and rated slip, then the gearbox was increasing the shaft speed at the pump. One thing we dont know yet for sure (do we?) is the base motor speed of the original moror. My suspicion is that maybe what really happened here is that the original motor was 4 pole, to which they added a gearbox to speed it up to run the pump faster. Someone may have looked at that and said, "Why don't we just use a 2 pole motor and start off at a higher speed?" Seems reasonable to me. The problem would be that although you would be eliminating mechanical losses in the gearbox, i.e. friction and heat, you may at the same time be just shifting the looses to the motor now. If I determined I needed 350HP from a 4 pole motor to have enough shaft torque at the pump at a higher speed, I would need to replace that 350HP 4 pole motor with a 700HP 2 pole motor to get the same shaft torque at the pump! This is what's behind LionelHutz comment by the way, it's an old trick, although I've never seen it done at this power level. Without increasing the HP then, the faster motor has LESS shaft torque, therefore more slip and it draws more current. It might have been a LOT more were it not for the elimination of the losses in the gearbox, so all you are seeing is the current delta in the overall SYSTEM EFFICIENCY differences.

If however this WAS all still done correctly, and let's say the original design was a 4 pole 200HP motor that they replaced with a 2 pole 350HP motor (because of removing the gearbox losses), then the current difference may still be explained in other ways:

1) With the added mass of the gearbox, the 4 pole motor was running at closer to full load so it was at a better power factor. Now the 2 pole motor has less load on it, so the PF is technically worse. You will see higher CURRENT as a result, but the actual kW would more likely be LOWER. I find trying to explain PF issues to wrench turners to be less than a satisfying experience, I usually have to finish the conversations with someting like "Because that's the way it is in the electrical world". Good luck with that if this is the case. The infamous "Beer Mug" analog often helps, if for no other reason than to find something else to talk about with them...

2) All other issues are right with the world and all my above hypothesizing is completely in left field. But... They now added in the issue of running over base frequency. The current, if measured from the INPUT side of the VFD, is now slightly higher because you have added losse. Inside of the VFD, running at a higher output frequency adds switching losses that you did not have before. Its not much, but there would also be added heat in the motor, which is also a total throughput efficieny loss, which would also show up as a current increase.

"Will work for (the memory of) salami"

 

[ScottyUK]

A 240V 350HP motor would be fairly demanding in terms of current and cabling. In Europe around 350HP is where we would make the jump to 3.3kV or 6.6kV (equivalent would be 4.16kV in the US, I think).

The growing number of 690V drives in Europe would be amenable to a reconnecting a 400V delta machine to 690V star, with a corresponding 73% increase in base speed and power output.

 

[waross]

Further to Scotty's post; The existing VFD will probably not be able to supply the current that 350 HP at a 240 Volt connection would demand.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[ScottyUK]

Further to my own post, leave the 400V machine in delta and increase the frequency by 73% to operate at 690V. Brain in freewheel, fingers in top gear... [

 

[ozmosis]

An old colleague of mine wrote an interesting and useful piece a few years ago.
(

http://www.dpaonthenet.net/article/10084/Lies--damn-lies-and-VSD-efficiency-.aspx

)
He'd tried for many years, as I think he still does, to explain the myth that using a 'high efficiency class motor' along with a VFD does not equal savings as a result of the design of the motor. In fact, it can be quite the opposite.