partial motor load energy

[caporter]

A brand name motor drive manufacturer writes:

"Motor size is often increased by 1.2 to 1.5 times for load applications such as fans, blowers and pumps. Consequently, the motors in such applications are generally operating at 70% rated capacity when the rated output of the equipment is reached. Therefore, with even lighter load, efficiency drops as the motor is supplied with the rated voltage. This wastes energy and potentially causes damage to the motor itself as the unused energy is dissapated into the motor windings."

The last sentence doesn't sound right to me. Is there a mechanism for heat buildup in motors at low partial load?

 

[edison123]

You're right, the last sentence isn't.

Who is this "brand name" drive manufacturer ?

Muthu

http://www.edison.co.in

 

[electricpete]

A given motor has varying efficiency with load. Peak efficiency typically near 80% load (varies).

It was traditional wisdom for a long time that putting in a motor which was oversized would push you to the inefficient range below 80% which would waste energy.

However it's a little tricky to evaluate because the relationship to peak efficiency of the motor is not the only variable. Large motors tend to have higher efficiency accross the board. So a large motor operating off of its "best efficiency point" might still be more efficient than small motor operating on its best efficiency point.

We had a thread awhile back where someone plugged the numbers and showed it was not the case.

The only real way to begin to guess is tell us the horsepower of motors under consideration and loading level. Then we (or you) can do a comparison based on typical catalogue motors. But there is a big variation within those as well so the comparison is most valid when comparing data for the exact motors you are considering.

Needless to say there may be other variables besides efficiency. Initial cost is one and the larger motor will of course have higher intial cost.

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[edison123]

pete - reread the last sentence.

Muthu

http://www.edison.co.in

 

[electricpete]

I take from the context that heat buildup is not of interest for reasons of temperature, but only as it relates to energy useage.

To take it to an extreme, if I have a 12hp load and I want to power it with the least energy possible, should I use a 15hp motor or a 1500hp motor?

It's safe to say the 15hp motor will probably be more efficeint AT THAT particular load.

The reason is that the 1500hp carries a bigger burden of no-load losses. Those include core losses and I^2*R losses assocaited with magnetizing current component.

So there are cases where large motor wastes more energy than small motor at low load level.

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[electricpete]

This wastes energy and potentially causes damage to the motor itself as the unused energy is dissapated into the motor windings."

I agree with edison that particular sentence is nonsensical.


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[jraef]

It's a really common part of "energy saver" scam artists repertoire. They use the basic premise in E-pete's argument, but fail to mention the fact that as the motor is less loaded, efficiency becomes less and less of an issue. So they allow the reader to make their own implication that lower efficiency means more heat in the motor, rather than explain the truth that it's relative to begin with. I got into a fun argument with one of them at a trade show once when I heard him spewing this scam. I hit him with the math, pointing out that if my efficiency drops from 95% to 93% at 50% load, the overall energy loss in that motor is still FAR BELOW the rated capacity the motor was designed to handle, because at 50% load, it is already pulling 70% current (or thereabouts), so even if we add back in 2% for reduced efficiency, it's irrelevant. He countered with a "conspiracy" argument about my being part of the "electrical establishment who wants to prevent the truth from getting out to the public", the last defense of a weak mind...


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[waross]

A motor draws enough energy to drive the load and supply the motor losses. If the load is less, the motor draws less energy and you pay for less energy. If a 15 HP motor is driving a 12 HP load, it just draws enough energy to support the 12 HP load. There is no 3 HP lost as heat. (15 HP - 12 HP = 3 HP)

A brand name motor drive manufacturer writes:

Is this a well known brand? It is a common misconception of those who are uninformed concerning motors and electricity that a motor draws full current regardless of the load.
NOT TRUE.
There must be a misunderstanding here somewhere.

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

 

[edison123]

Going by their "theory", an unloaded motor can be used as a room heater.

Stupid burns.

Muthu

http://www.edison.co.in

 

[LionelHutz]

There must be a misunderstanding here somewhere.

No misunderstanding. I bet this paragraph is part of an energy saving feature description. It was either written by someone who blantantly lied or someone who has no clue, both of which should not be trusted to give any real info.

Most likely this is part of the description why a SCR based controller or soft-starter will save energy by reducing the motor voltage.

A brand name would be interesting to hear. I'm betting it's someone like Powerboss or Power Efficiency Corp.

 

[caporter]

You are right that it is part of a SCR power saving device. The brand name is Yaskawa.

 

[jraef]

Yaskawa has resorted to that scam? Sad...

Are you sure this is not part of an ad for a VFD? There is a legitimate issue with VFDs where you can save EXTRA energy and reduce motor heating on some types of loads when compared to other methods of flow control. But a VFD is not "SCR" based. I was not aware that as good a company as Yaskawa had sunk so low as to start pedaling the energy saver scam.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[jraef]

Google told me that this statement came from this paper.

It is, as I suspected, written about a VFD Energy Saver function. This is nothing like the SCR based energy saver scams. You should try to be more clear in your descriptions.

Thant said, I do think the Yasakawa technical writer took some liberty with the concept of "...potentially causes damage to the motor itself as the unused energy is dissapated into the motor windings." The fact that they failed to use spell check on the word "dissipated" is a clue that this was not written by an engineer.

The energy savings concept in VFDs is valid, this part of this statement is bogus.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
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[jraef]

Link error was not my fault, for some reason the URL format is incompatible with this site's TGML.

[URL unfurl="true"]http://www.yaskawa.com/site/dmdrive.nsf/link2/MHAL-7M6UUV/$file/AN.V1000.02.pdf[/URL]

That "link2... part seems to get messed up, even when I tried to encapsulate it as "code".



"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
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[LionelHutz]

Well, Yaskowa does VFD's so the description is not likely to be for a soft-starter.

This is broken too;
[ignore]
pdf file
[/ignore]
Which should create a highlighted "pdf file" link.

And the spot to paste the url of an attachment doesn't seem to work with that url either but we'll see when I submit this.

 

[LionelHutz]

Nope, the attachements showed up completely broken.

 

[waross]

I followed jraef's tracks and Googled it. The link in the google hit works.

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

 

[LionelHutz]

This is more or less the same as the NOLA circuit though. Lower the motor voltage and reduce the iron losses. If you reduce the iron losses you increase the efficiency of the motor.

The losses in a motor should go down as the load decreases. One of the losses is the copper resistance I^2xR losses in the stator windings. As you reduce the load you reduce the stator current. Less stator current produces less heating in the stator windings.

The output voltage waveform of a VFD will increase the losses of the motor or energy used by the motor. The energy savings due to reducing the voltage can be very small at times unless the motor is running below about 25% load. So, installing a VFD to reduce energy use with this voltage reduction feature could actually increase your energy use.

The common and blantant use of "energy savings" to sell a product need to always be considered in details. Mostly, I see VFD's as being useful for process control where you need a certain pressure or flow and for use in places where the efficiency of a process can be optimized due to poor initial component selection.

Basically, the marketplace has too much focus on simply reducing the cost to run the motor and not enough on reducing the cost to get the work done. Of course, it is easier to focus on how VFD's will reduce the energy going into a motor and much harder to show how it will make the whole process more efficient.

For example, if you are pumping water, you want to focus on reducing the energy cost per gallon of water pumped, not to focus on reducing the cost per hour to run the motor turning the pump. Lowering the speed of a water pump can easily reduce the efficiency of the whole pumping system which increases your cost per gallon of water pumped.