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About The Test Standard For Energy-saver For Lightly Loaded Motor


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#1 yhg

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Posted 04 November 2007 - 01:11 PM

dear marke:

Now a lot of energy-savers for lightly loaded motor and electricity distribution networks, but many energy-saver are not so good , and product quality is low, so we must regulating the makets of energy-saver,and protect the consumer's benefits. so we must issue the national standards for these energy-savers,and these energy-savers must be tesed in lab before it propagation in the maket.
so I want to ask if there are test standard for these energy-savers in your country?

thank you
best regards

#2 marke

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Posted 04 November 2007 - 05:56 PM

Hello yhg

Welcome to the forum.

I am not aware of any specific standards that apply to energy savers for induction motors as such, rather they are normally just covered by the motor controller standards such as IEC60947.4 (I think that is correct!!) This does not include any tests for quality of components etc, it is more of a safety standard. Neither does it refer to/measure the ability to save energy.

I believe that if there were independent tests done on the ability to save energy to the manufacturers claims, all (or most) would fail due to unrealistic claims except under extreme circumstances. - There is no magic bullet on these devices due to the efficiencies of induction motors except for small single phase motors and motors that are well overfluxed. (motors that do not meet modern standards.)

Best regards,

#3 Brigitlc

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Posted 22 April 2008 - 08:15 AM

QUOTE (yhg @ Nov 4 2007, 09:11 PM) <{POST_SNAPBACK}>
dear marke:

Now a lot of energy-savers for lightly loaded motor and electricity distribution networks, but many energy-saver are not so good , and product quality is low, so we must regulating the makets of energy-saver,and protect the consumer's benefits. so we must issue the national standards for these energy-savers,and these energy-savers must be tesed in lab before it propagation in the maket.
so I want to ask if there are test standard for these energy-savers in your country?

thank you
best regards


I cannot think of any specific standard besides IEC60947.4 either.

Sometimes the quality of those products are low because the unrealistic claims of those manufacturers, that's true. But I think some people picking the wrong model or have little knowledge in this field, is also a huge problem.

By the way, this forum is pretty nice! wink.gif

#4 Sami

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Posted 12 August 2009 - 06:56 AM

QUOTE (Brigitlc @ Apr 22 2008, 11:15 AM) <{POST_SNAPBACK}>
I cannot think of any specific standard besides IEC60947.4 either.

Sometimes the quality of those products are low because the unrealistic claims of those manufacturers, that's true. But I think some people picking the wrong model or have little knowledge in this field, is also a huge problem.

By the way, this forum is pretty nice! wink.gif



Dears

Greetings.

as stated, really this forum is very nice and fruitful.

Could we make an equation or formula which helps in Selecting the best application for an energy saving product?
in other words, when we want to install an Energy saving device in a factory,for example,we have to consider different factors in order to select the best application (Motor)to install the power saving device on it,so we can get the best savings ever!

We may consider the following :Max KW Load (Measured) compared to the Motor max KW rating, how much Time the motor is under no load?,Motor age & Manufacturer,Ampient Temp and the Motor size,Kw rating, (whetjer it is a big motor or small motor in KWs)


What do You think ?

regards
sami

#5 marke

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Posted 12 August 2009 - 08:25 AM

Hello sami

Welcome to the forum.

Unfortunately, there is no magic formula that I know or that will give you an immediate answer.
If you have not already done so, have a look at http://www.lmphotonics.com/energy.htm This may answer some questions for you.
The energy that you can save with the Nola type controllers, is limited to a portion of the iron loss of the motor. The iron loss is independent of load and the maximum energy saving is at minimum load.
To make an appreciable saving on a motor, you need to have a motor with a high iron loss that is running at very lightly loaded conditions for a long time.
If you look at the power factor of the motor while it is running at light load, and the power factor is less than 0.4 and the current is reasonably high at this low power factor, then you may have a candidate. Large motors tend to be very efficient and so have a low iron loss per KW motor rating. It is not uncommon for the iron loss on large motors to be less than 3% of the motor rating so there is little energy that can be saved even at open shaft conditions.
Small motors have a much higher iron loss per KW and there is more potential to save energy at light load.

Find small motors operating with a power factor of less than 0.4 that are running very hot and you may get a result.

Best regards,
Mark.

#6 Sami

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Posted 13 August 2009 - 12:41 PM

QUOTE (marke @ Aug 12 2009, 11:25 AM) <{POST_SNAPBACK}>
Hello sami

Welcome to the forum.

Unfortunately, there is no magic formula that I know or that will give you an immediate answer.
If you have not already done so, have a look at http://www.lmphotonics.com/energy.htm This may answer some questions for you.
The energy that you can save with the Nola type controllers, is limited to a portion of the iron loss of the motor. The iron loss is independent of load and the maximum energy saving is at minimum load.
To make an appreciable saving on a motor, you need to have a motor with a high iron loss that is running at very lightly loaded conditions for a long time.
If you look at the power factor of the motor while it is running at light load, and the power factor is less than 0.4 and the current is reasonably high at this low power factor, then you may have a candidate. Large motors tend to be very efficient and so have a low iron loss per KW motor rating. It is not uncommon for the iron loss on large motors to be less than 3% of the motor rating so there is little energy that can be saved even at open shaft conditions.
Small motors have a much higher iron loss per KW and there is more potential to save energy at light load.

Find small motors operating with a power factor of less than 0.4 that are running very hot and you may get a result.

Best regards,
Mark.


Dear Mark,

Thanks alot for your usefull reply.

what is the range for small Motors.? i.e. :could we consider a 22KW ,for example, as a small motor ?.this is as per what you stated : (small motors shall give you much savings),

More Over, what about the other factors : Ambient Temp, OEM Manufcr.... are they have any effects to gain much savings?

Also, in order to know the (Kw\h) consumption , what is the best way :to measure it or by calculations?iand how to do it accurately in both cases.

Thanks & regards

Sami

#7 marke

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Posted 13 August 2009 - 07:16 PM

Hello Sami

1. Small motors. Have a look at a motor data sheet and you will find that the full load efficiency of motors is high for larger motors and only drops when the motors are quite small. The best candidates are those with a low efficiency and those where the half load efficiency is significantly lower than the full load efficiency. This does depend on the range of motors that you are looking at, but in many cases, I would think that you would be looking below 7.5KW.

2. Ambient temperature will not have any major effect. It comes down to the design of the motor. If it is overfluxed due to design or operating on a high voltage relative to the design voltage, you may achieve results provided that the load is low enough.

3. Measurement. You must use a proper three phase metering device, preferably use a power board type KWHr meter to give a good indication. DO NOT RELY ON CALCULATIONS!!

Best regards,
Mark.

#8 Sami

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Posted 15 August 2009 - 07:50 AM

QUOTE (marke @ Aug 13 2009, 10:16 PM) <{POST_SNAPBACK}>
Hello Sami

1. Small motors. Have a look at a motor data sheet and you will find that the full load efficiency of motors is high for larger motors and only drops when the motors are quite small. The best candidates are those with a low efficiency and those where the half load efficiency is significantly lower than the full load efficiency. This does depend on the range of motors that you are looking at, but in many cases, I would think that you would be looking below 7.5KW.

2. Ambient temperature will not have any major effect. It comes down to the design of the motor. If it is overfluxed due to design or operating on a high voltage relative to the design voltage, you may achieve results provided that the load is low enough.

3. Measurement. You must use a proper three phase metering device, preferably use a power board type KWHr meter to give a good indication. DO NOT RELY ON CALCULATIONS!!

Best regards,
Mark.


hello mark

as you stated , Large motors have high effeciency ,so no much savings is expected here.On the other hand, using very small motors (below 7.5kw) may give you high percentage of saving even above 50%..but still 50% of 7.5kw is veryy littel .Accordingly, you will not get big savings.but if we use a middel sized motor (37kw) we could get 15 % savings which is relatively good saving. for example: 15% of 37 kw motor is much much better than 50% of 7.5 kw motor

So i think the middel sized motors make a compromise and could result in good savings.... also because the large sized motor (above 100kw)could not result in any saving at all!!


thanks & regards

Sami

#9 marke

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Posted 15 August 2009 - 08:36 AM

Hello sami

I do not believe that you will get 15% of 37KW savings with a 37KW motor.
A 37KW motor will have a high efficiency. I am looking at the data for a 37KW 4 pole motor and it has an efficiency of 93%, so the total losses are 7%. Therefore, the iron losses are in the order of 2.5% of 37KW and at very light load, we may save as much as 50% of the iron loss which equals 1.25% of 37KW or about 420 watts.

If we look at a 5.5KW motor, we find that the efficiency is of the order of 90%, so at very light load, we could save perhaps 2.5% of 5.5KW or about 140 watts.

I do not believe that you will save 15% of 37KW with a 37KW motor as this is considerably higher than the iron loss on the motor, ifact it is about double the total losses of the motor!!
You can only save a portion of the iron loss of the motor and then, only at very light load.

Best regards,
Mark.

#10 Sami

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Posted 16 August 2009 - 03:20 PM

QUOTE (marke @ Aug 15 2009, 11:36 AM) <{POST_SNAPBACK}>
Hello sami

I do not believe that you will get 15% of 37KW savings with a 37KW motor.
A 37KW motor will have a high efficiency. I am looking at the data for a 37KW 4 pole motor and it has an efficiency of 93%, so the total losses are 7%. Therefore, the iron losses are in the order of 2.5% of 37KW and at very light load, we may save as much as 50% of the iron loss which equals 1.25% of 37KW or about 420 watts.

If we look at a 5.5KW motor, we find that the efficiency is of the order of 90%, so at very light load, we could save perhaps 2.5% of 5.5KW or about 140 watts.

I do not believe that you will save 15% of 37KW with a 37KW motor as this is considerably higher than the iron loss on the motor, ifact it is about double the total losses of the motor!!
You can only save a portion of the iron loss of the motor and then, only at very light load.

Best regards,
Mark.

hi mark

the 93% efficiency of 37 kw motor is at full load! and so the total losses are 7% but at full load. so when the load decreses the savings increase.because always the efficiency is maximum at full load and minimum at low load.

#11 marke

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Posted 16 August 2009 - 07:08 PM

Hello sami

QUOTE
the 93% efficiency of 37 kw motor is at full load! and so the total losses are 7% but at full load. so when the load decreses the savings increase.because always the efficiency is maximum at full load and minimum at low load.


Exactly, but your increased savings are of a lower figure. Many of the sales people selling this type of product neglect to tell you that although the percentage saving at light load is much higher, the power consumed is much lower. 50% savings of 3.5% is the same as 1.75% savings of 100%.
You can only save a portion of the energy that is wasted, and at light load, the primary energy loss is iron loss. As the load increases, the copper loss increases with the square of the current and at full load, the highest loss is typically copper loss.

The iron loss does not change with load, therefore we can look at the full load efficiency and estimate what the iron loss will be, and we know that under open shaft conditions we may save as much as 50% of this.
So, if a motor has a full load efficiency of 93%, then the iron loss is going to be less than 50% of the full load losses, so less than 3.5% of the motor rated power. At best, under very light load, we may save as much as 50% of the iron loss, so the true KW energy saving is less than 1.75% of the motor rating.

If the motor has a full load efficiency of 60%, then the iron loss may be in the order of 15% of the motor rating and you may save 7.5% of the motor rating under ideal conditions.

Be careful when looking at the quoted percentage savings. 50% of nothing, is still nothing.

Take our 37KW motor operating at a shaft power of 5%, (perhaps a clutch powered punch press)
Our iron loss is in the order of say (100 - 93)/2 (assuming a high iron loss example)
= 3.5% = 1.3KW
At a shaft load of 5%, our load power = 1.85KW
Now apply the energy saver and save 50% of the iron loss.
Total load input is now 1.85 + 0.65 = 2.5KW
Before the energy saver was applied, total power input was 1.85 + 1.3 = 3.15KW
so energy saving is 21% or 0.65KW
21% sounds much better than 650 watts, so the perce3ntage is always quoted.
The example above is not strictly correct as the savings will be less due to the additional losses associated with copper loss , windage loss and frictional losses. Plus, the iron loss reduction would probably be less than 50%.

Go back to actual watts loss each time and you will see what I am saying, however be very careful using quoted watts (KW) saved as the quoted figures are not always true. I have seen many cases where the quoted savings can only be true if the motor is less than 50% efficient!! and as one sales person tried to tell me, the iron loss increases as the load reduces so that a 37KW motor draws 37KW under all load conditions!!

Best regards,
Mark.

#12 yuri

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Posted 17 August 2009 - 05:39 PM

Hello Mark.

Honestly, I dont know anything about those p. savers. They may be much more costly than I think or are not very reliable (prone to breakdowns). So, not an argument, just a thought.

I understand it so that power saved on an 37kW engine would be about 650 W - independent of load. But still, for instance, on a fan in an air handling unit working unstopably, savings may be 650*24*365 = 5694 kW. At the rate 0.10 EC per 1 kW = 569 Euros a year. In our days also not bad ?

Best regards.

#13 marke

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Posted 17 August 2009 - 06:59 PM

Hello Yuri

QUOTE
power saved on an 37kW engine would be about 650 W - independent of load

No, the maximum power that can be saved is a function of the iron loss.
The iron loss is independent of load.
The percentage of the iron loss that can be saved is very dependent on load. Maximum watts saving at no load and the watts saved reduces rapidly as the load is increased.
On an air handling unit, the load would probably never be low enough to save any energy, infact because these devices actually use energy, there could easily be a loss situation!!

Best regards,
Mark.




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