VFD w/ Motor w/ Pump High Amps so pick bigger motor

[sg000]

Hi everyone. First post here. First off, please forgive me, i'm no EE, I just play one in my spare time because I have the best background in the company and the genes (father was an EE). I'm MechE by education.

This situation has a couple of questions I really could use your help, and my information sources are not really helping.

Situation: 10 HP WEG Motor on an Omron VFD (460V, 3 phase) driving a gear pump. 22 Hz, 12.9 A, 667 RPM, 152V, 3.4 KW power draw. We're at the top of the amp limit for the motor already at only 1/3 the speed. Its just gets worse as we go faster.

Initial impression: We need a bigger motor. Ok, I can live with that, if thats the right answer. I just don't want to go bigger if I don't have to and there's a simple answer in how the motor is controlled by the VFD.

Second impression: Some setting in the drive is causing this? We are running at constant torque mode on the drive, I think. I've been reading, and every time constant and variable torque comes up, pumps are instructed to be run in variable. Problem is, that doesn't make much sense to help my situation because it will output less torque, less voltage and raise the amps at this speed, not lower the amps. Am I WAY off here?

Alternate issue: I asked local supplier for motor recommendations for a 15hp, and I have a choice between a 10:1VT/10:1CT and 20:1VT/10:1CT. Thats what lead to my research on constant/variable torque. My problem is, I have no idea which one is better for me!

Any help would be appreciated! Thanks!

--Scott

 

[Jraef]

Embedding points in your post:

Situation: 10 HP WEG Motor on an Omron VFD (460V, 3 phase) driving a gear pump. 22 Hz, 12.9 A, 667 RPM, 152V, 3.4 KW power draw. We're at the top of the amp limit for the motor already at only 1/3 the speed. Its just gets worse as we go faster.

So you are at 37% speed (22Hz/60Hz), and 46% power (3.4kW out of 7.46kW, which is 10HP), but you are at only 26.5% voltage (152/460). That indicates that someone has likely messed with the V/Hz pattern, either deliberately or unintentionally. Basically at the simplest level, your V/Hz ratio should be constant, at 7.67V/Hz, the ratio that your motor was designed for (460/60). So at 22Hz, the voltage should have been 168V, but it is less, about 6.9V/Hz or 90% of what it should be. That means that at this ratio, your peak torque is at about 81% of what the motor is capable of, and in a given load, that means the motor runs at higher slip and pulls more current than it needs to to attempt to keep up with the demand put on it.

Initial impression: We need a bigger motor. Ok, I can live with that, if thats the right answer. I just don't want to go bigger if I don't have to and there's a simple answer in how the motor is controlled by the VFD.

Do not jump to that conclusion, at least not yet.

Second impression: Some setting in the drive is causing this? We are running at constant torque mode on the drive, I think. I've been reading, and every time constant and variable torque comes up, pumps are instructed to be run in variable. Problem is, that doesn't make much sense to help my situation because it will output less torque, less voltage and raise the amps at this speed, not lower the amps. Am I WAY off here?

You have the right idea, just the wrong conclusion.
Pumps being "variable torque" is a commonly made mistake, in that not ALL pumps are variable torque, only CENTRIFUGAL pumps are. A gear pump is a Positive Displacement (PD) pump, they are definitely NOT centrifugal, so therefor they are NOT "variable torque", they are "constant torque". That might possibly explain the V/Hz issue. VFDs have an "energy savings" trick for use WHEN you have a VT / centrifugal load. What they do is tweak the V/Hz ratio at a rate of V/Hz squared, which LOWERS the available torque from the motor as speed is reduced, ostensibly because IF it is a centrifugal machine, the torque is not needed (the definition of "variable torque). This saves a tiny bit extra energy by reducing the magnetizing current in the motor windings. But... this can be a BIG problem if the load is NOT a centrifugal machine... i.e. a gear pump!

Alternate issue: I asked local supplier for motor recommendations for a 15hp, and I have a choice between a 10:1VT/10:1CT and 20:1VT/10:1CT. Thats what lead to my research on constant/variable torque. My problem is, I have no idea which one is better for me!

Any help would be appreciated! Thanks!

--Scott

Don't jump to that solution just yet.

Get out your manual, look for the V/Hz pattern function, make sure it is set for Constant Torque. Do not turn on "Torque Boost" or anything that sounds like that, you want Plain Jane CT settings. If you think someone may have started messing with settings and programming trying to figure out what was going on, there will be a parameter that will return everything to the factory default setting, you may have to do that. BUT BEFORE YOU DO, write down EVERY setting that is not already at the factory default. If you have the software and a way to talk to the drive with a PC, that is easier, otherwise it is a process of scrolling through maybe a couple of hundred parameters. So Check to see if it is just the V/Hz setting first, if it is, fix just that before trying anything else.

Possibility #2: your drive is set up to be in "Sensorless Vector Control" (SVC) operating mode, which is common for gear pumps, but nobody performed what is called an "Autotune" procedure that has the drive configure itself specifically for that motor. It must be performed every time a motor is changed / commissioned or the drive will not work correctly, or the drive must be set for standard "V/Hz" (aka Scalar) Mode instead of SVC.

Possibility #3, albeit more remote: You have a dual voltage motor (230/460V) and someone mistakenly hooked up the motor in the LOW voltage pattern, i.e. 230V, but now you are giving is a reference of 460V, so TWICE the V/Hz than it needs, it is saturating and drawing current needlessly, which is doing nothing more than heating up the windings.

Possibility #3A, even more remote... I noticed you said it was a Weg motor. Sometimes, because Weg comes from Brazil, they fulfill back orders with motors designed for use IN Brazil, where they use 690Y400V motors a lot. You can use a 400V 50Hz motor at 460V 60Hz, the V/Hz ratio is close enough. But where that screws people up is that the Weg connection diagram just says "Low" and "High" for voltage. People here are used to "High" being 460V, when in fact for THAT motor, it would be "Low"! I have seen that dozens of times.

 

[nhee2]

Great response by JRaef. Another example of what a great tool this forum is.

 

[qcroanoke]

Great response by JRaef. Another example of what a great tool this forum is.

I enjoy reading Jraef's answers. Even if I have no idea what he is talking about.......
His knowledge of drives is amazing

 

[Jraef]

I enjoy reading Jraef's answers. Even if I have no idea what he is talking about.......
His knowledge of drives is amazing

psssst... I have no idea what I'm talking about either... but thanks.

 

[augie47]

I enjoy reading Jraef's answers. Even if I have no idea what he is talking about.......
His knowledge of drives is amazing

His posts could be used in a dictionary as an example of "awe"
(unlike many of mine that are awwww ).
Said it before and agree with others, he is a valuable as gold here.

 

[sg000]

First, some BIG BIG thanks to the speed and detail of your response. I should have started with this forum 2 weeks ago when we first exhibited this issue. I'm happy to see it come with a number of supporters!

A gear pump is a Positive Displacement (PD) pump, they are definitely NOT centrifugal, so therefor they are NOT "variable torque", they are "constant torque".

Thanks for the lesson...I almost didn't include that it was a gear pump. More info is always better!

make sure it is set for Constant Torque. Do not turn on "Torque Boost" or anything that sounds like that,

Yup, checked again and again that we are constant torque. I've pasted a screenshot of the relevant settings, all default (yes, the PC software is awesome!). So, doesn't look like that is the culprit.

http://screencast.com/t/nkGVT1tqD0FR

Possibility #2: your drive is set up to be in "Sensorless Vector Control" (SVC) operating mode

Nope, not set for that. Fortunately the consultant who set up the drive originally didn't do a lot of fancy adjustments, it is mostly defaults.

Possibility #3, albeit more remote: You have a dual voltage motor (230/460V) and someone mistakenly hooked up the motor in the LOW voltage pattern

First thing we checked and double checked.

But....

Possibility #3A, even more remote... I noticed you said it was a Weg motor. Sometimes, because Weg comes from Brazil, they fulfill back orders with motors designed for use IN Brazil, where they use 690Y400V motors a lot. You can use a 400V 50Hz motor at 460V 60Hz, the V/Hz ratio is close enough. But where that screws people up is that the Weg connection diagram just says "Low" and "High" for voltage. People here are used to "High" being 460V, when in fact for THAT motor, it would be "Low"! I have seen that dozens of times.

Didn't know about this. Will look into it. Though we did a No-Load test up to full speed (1800 RPM) and current was just where it should have been (5A if I recall correctly).

I should note we're using the drive output readings for the current values, not a meter on the wires into the motor.

I'm still curious about my query with the 10:1VT and 20:1VT from the motor supplier. What does that really mean? Is it relevant to the situation? Its not something shown on the basic datasheets, so its new to me.

Thanks Again

 

[Besoeker]

Embedding points in your post:

So you are at 37% speed (22Hz/60Hz), and 46% power (3.4kW out of 7.46kW, which is 10HP), but you are at only 26.5% voltage (152/460). That indicates that someone has likely messed with the V/Hz pattern, either deliberately or unintentionally. Basically at the simplest level, your V/Hz ratio should be constant, at 7.67V/Hz, the ratio that your motor was designed for (460/60). So at 22Hz, the voltage should have been 168V, but it is less, about 6.9V/Hz or 90% of what it should be. That means that at this ratio, your peak torque is at about 81% of what the motor is capable of, and in a given load, that means the motor runs at higher slip and pulls more current than it needs to to attempt to keep up with the demand put on it.

Yes, but it's still going to be 46% power at 37% speed even with nominal V/f ratio.

 

[Jraef]

Yes, but it's still going to be 46% power at 37% speed even with nominal V/f ratio.

Yes, but my thinking was that the added power represented the wasted flux current in the motor. Remember, the drive will only measure and display the total power, not what it consisted of.

 

[Jraef]

...

I should note we're using the drive output readings for the current values, not a meter on the wires into the motor.

I'm still curious about my query with the 10:1VT and 20:1VT from the motor supplier. What does that really mean? Is it relevant to the situation? Its not something shown on the basic datasheets, so its new to me.

Thanks Again

1. The VFD reading will be more accurate than most meters, unless the price tag of the meter had 4 or more digits to the left of the decimal point.

2. In a VT load, the load expressed upon the motor drops at the cube of the speed change. So what they are saying is that the motor can survive a 20:1 speed ratio if the load is VT, because at 5% speed (1/20), the load on that motor will be so low that any loss of cooling capacity will be offset by the extremely low power being used. In the 10:1 VT version, the motor has less cooling capability. On a CT version they are the same, but the fact that the VT version at 20:1 means it has better cooling capability (likely TENV as opposed to TEFC), I would use that one.

 

[Jraef]

Yes, but my thinking was that the added power represented the wasted flux current in the motor. Remember, the drive will only measure and display the total power, not what it consisted of.

Oh wait, I see my mistake now. On one had I was postulating that the motor was being under fluxed by having a VT setting, then on the other had assuming that the extra power consupmtion shown was the result of being OVER fluxed. Can't have it both ways...

Oh flux... :slaphead:

 

[qcroanoke]

Oh wait, I see my mistake now. On one had I was postulating that the motor was being under fluxed by having a VT setting, then on the other had assuming that the extra power consupmtion shown was the result of being OVER fluxed. Can't have it both ways...

Oh flux... :slaphead:

See that it doesn't happen again....

 

[Besoeker]

Oh wait, I see my mistake now. On one had I was postulating that the motor was being under fluxed by having a VT setting, then on the other had assuming that the extra power consupmtion shown was the result of being OVER fluxed. Can't have it both ways...

Oh flux... :slaphead:

Happens to us all.
The 3.4kW was the input power. I don't know what the overall drive/motor efficiency is - but take a punt at 88%.
So shaft power would be around 3 kW.
Or about about 40% of the 10 HP rating - one day you guys will will get with the programme.......you don't need both kW and HP - power is power.....
At 22/60 or 37% speed.

Even if the load is truly constant torque, and few really are, 10 HP isn't enough for the application.
In short, I think the OP is right. He needs a bigger motor.

 

[Jraef]

...you don't need both kW and HP - power is power.....

So explain to me how it is that I can have a 5.5kW motor running at full load, and it is drawing 6.3kW then?

Even "you guys" have to "qualify" the difference between MECHANICAL kW and ELECTRICAL kW by calling them (respectively) "shaft" kW vs "absorbed" kW right?. So we just use HP as the Mechanical unit, kW as the electrical unit. In my mind, that is less confusing.

But I didn't want to turn this into a debate, we can agree to disagree.

And by the way, we originally got these terms from YOU GUYS...

I too am coming to the conclusion of not enough motor, barring a possible defect of something somewhere, or the oddball Weg nameplate connectivity issue.

 

[Besoeker]

So explain to me how it is that I can have a 5.5kW motor running at full load, and it is drawing 6.3kW then?

Efficiency.

Even "you guys" have to "qualify" the difference between MECHANICAL kW and ELECTRICAL kW by calling them (respectively) "shaft" kW vs "absorbed" kW right?.

Nope. Power is power. Mechanical or electrical. Different units are not required. So why complicate the issue?

And by the way, we originally got these terms from YOU GUYS...

Indeed. And not moved on.

Slightly more seriously, I grew up with Imperial units. I know what an inch and a yard look like. And that a person weighing 200 pounds is a pretty big chap.
It's still ingrained in my brain that a horespower is 33,000 ft-lbf/minute. Never use it.
But, for engineering, SI is just so much easier.

I too am coming to the conclusion of not enough motor, barring a possible defect of something somewhere, or the oddball Weg nameplate connectivity issue.

Difficult to conclude otherwise on the basis of the information presented.

 

[sg000]

Efficiency.
Slightly more seriously, I grew up with Imperial units. I know what an inch and a yard look like. And that a person weighing 200 pounds is a pretty big chap.
It's still ingrained in my brain that a horespower is 33,000 ft-lbf/minute. Never use it.
But, for engineering, SI is just so much easier.

Agreed...the other day I was asked how many bar was my 300 psi. was in my presentation (20.6 bar for those keeping track). He had no sense what 1 psi was, and I've never had a good sense of bars.

Proof its better in engineering: In school, they mixed imperial problems into our problem sets, so I would convert them to SI, do the calculations, then convert the final answer back into imperial. lbm, lbf, slugs, ugh! :blink:

I too am coming to the conclusion of not enough motor, barring a possible defect of something somewhere, or the oddball Weg nameplate connectivity issue.

I think I was able to prove the "not enough motor" yesterday.
15 Hp motor: At 667 RPM, I had 12.1A, 3.3 kW, 155-160V, 23 Hz on a 15hp motor
vs. the 10 hp motor: 22 Hz, 12.9 A, 667 RPM, 152V, 3.4 KW power draw.

The result is much better efficiency, and my system the pump is driving runs better too, my output pressure was MUCH better. I'm wondering if the 667 RPM from the 10hp was even real...never thought to put a tach on it to check. Might have been dragging.

Thanks again for your help guys. Even if the result is just too small of a motor, I did learn a lot about what I should be looking for and how to set the drive up that I didn't know before. :thumbsup::thumbsup::thumbsup::thumbsup::thumbsup::thumbsup::thumbsup:

 

[Besoeker]

Proof its better in engineering: In school, they mixed imperial problems into our problem sets, so I would convert them to SI, do the calculations, then convert the final answer back into imperial. lbm, lbf, slugs, ugh! :blink:

You make the point very well - two lots of conversions when none would be required if it was all SI.
And you're stuck with SI for volts, amps, watts, ohms, etc.

Anyway, I'm glad you found the thread helpful.