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Iron loss = copper loss energy saving


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#1 zac

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Posted 10 May 2003 - 09:44 PM

Hello,

I am in touch with a Japanese person claiming he had invented a power saving device, which can considerably save energy. The savings he is talking about are at the magnitude of 25% up to 50% and even more of the total energy consumed by the motor (3 phase AC motors).

According to the manufacturer, the core technology is based on the physical concept of "iron loss = copper loss". That is, he regulates the voltage and frequency to the motor in such a way that keeps it working at a point in which the two losses, copper and iron, are equal. At this point, the total energy saving is greatest.

The machine had already been sold to a large number of well known companies in Japan, and apparently there is a market to it.

At this point I would like to note, that I am totally aware of the fact that this sounds like a complete bull, yet I have personally tested one machine, performing the measurements myself, and was stunned to find that indeed there was a 30% reduction in the motor power consumption. Also, the energy saving device improved the PF from 0.79 to 0.89.

The measurements were done using a sophisticated instrument, measuring the 3 phases. The saving I got was in KW of course (e.g. not in KVARs). The motor speed was maintained at the original value (1500 RPM), as well as the pressure built by the pump the motor was driving.

Another thing I noticed was that the motor vibrations considerably decreased, as well as its operating noise and its surface temperature.

The motor used in the test is 400 Volt, 60HP, driving a hydraulic pump for cooling water. Loading conditions are constant (the system is in uniform state uniform flow).

The motor initially was not regulated in any way - it was directly connected to the plant power supply.

I intend to conduct yet another test soon, with a smaller motor (30HP) to try and validate my earlier findings again.

Originally I was very skeptical, and the truth is that I am still, but the test that I conducted stands as proof. What worries me the most is the fact that I know that even regular (non high-eff) motors are quite efficient, with efficiencies of 90% and higher, and the fact that only at very light loads does the motor efficiency sharply drop.

Is there a catch? is there a sting here? am I missing something?

Would be very happy to get any comments and light shedding on this.

-Zac.

#2 marke

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Posted 11 May 2003 - 12:27 AM

Hello Zac

Welcome to the forum.
The major losses in and induction motor are the iron and copper loss, and under reduced load conditions, it is possible to reduce the iron loss of the motor. This I believe is a well accepted fact, but the big question is how much KW does this amount to and what is the cost of saving it. - what are the real returns and payback periods. The other problem is that there have been so many marketing campaigns based on this type of technology that are misleading to say the least, that there is 1) a lot of skepticism in regard to whether the technology actually works at all, and 2) an unreasonable expectation as to what should be able to be achieved. I have seen many claims and "certified" results showing savings that just can not be achieved without making the motor operate at greater than 100% efficiency!! I have heard claims of "save 50% at half load" which is crazy on a motor that is operating at 85% efficiency.

I believe that the first thing to do, is to study the characteristics of motors and learn from that just what the opportunities are to save energy. Remember, you can only save what is being wasted.

The iron loss of the motor is dependant on the flux in the iron. If we reduce the flux in the iron, we reduce the iron loss, but we also reduce the torque capacity. One way to reduce the iron loss without reducing the torque, is to increase the amount of iron, i.e. to increase the frame size, but that of course results in a more expensive and larger motor. The commercial demands today are for a smaller size and lower cost, so the frame size is minimised rather than maximised. The iron loss becomes a function of the turns, the stator voltage and the iron itself. If we operate the motor from a fixed supply voltage, the iron loss is essentially independent of load. (I have had one supplier of energy saving devices arguing adamantly that the iron loss increases dramatically with reducing load!!) If we reduce the applied voltage, the iron loss will reduce.

The copper loss of the motor is dependant on the current squared and the resistance of the stator winding. As the motor load reduces, the current will reduce down to a minimum of the magnetising current of the motor. As the current reduces, the copper loss reduces.
At rated load, the copper loss and the iron loss are of the same order of magnitude, (approximately equal) with the copper loss typically a little higher than the iron loss. If we analyse the current flowing in the stator, we have two major current components, the load current and the magnetising current. If at any load we reduce the applied voltage, the magnetising current will drop, but the load current will rise. The load current is directly proportional to the shaft load divided by the voltage. If the load current is higher than the magnetising current, we will see an increase in total current and an increase in copper loss. If the load current is significantly less than the magnetising current, then we can see a reduction in copper loss.

So, at light loads, a reduction in the applied voltage will yield a reduction in the energy consumed (wasted) by the motor, but the important points to note here are:

  • the energy reduction applies only to the energy wasted by the motor
  • the shaft load must remain the same
  • the energy wasted by the motor is very small relative to the motor rating except in very small single phase machines
  • the energy saving is greatest under open shaft conditions

      As the shaft load increases, the actual energy saved reduces and can in some cases increase.

      An indication of the maximum energy that can be saved is to accept that only a portion of the iron loss can be saved, the iron loss is essentially constant and will generally be less than half the full load losses. If a 50KW motor has a full load efficiency of 85%, then the iron loss would be less than 50kw x 0.15 / 2 = 3.75kw The open shaft energy saved may be as high as 2 KW, but by the time the machine is operating with a shaft load of 10KW, this saving would be significantly reduced and may be below 500W. Very few machines operate below 25% rated load for any significant period of time.

      To get a payback, you need a small motor, operating 24/7, under almost open shaft conditions for the majority of the time.

      Any tests, are best done using a three phase rotating disk KWHr meter. Other instruments can be fooled by chopped waveforms etc.

      Best regards,

#3 zac

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Posted 11 May 2003 - 12:34 PM

Hello Mark,

Thank you for your profound explanations. I totally agree with you about the fact that the energy saving limit is the energy being wasted or lost, it is simply common sense. Yet, my objective measurements clearly show a major saving of about 30%, which I cannot explain.

During the test we made sure that the loading conditions of the motor were preserved. The motor surely was not working open shafted, since it was building pressure (by means of a hydraulic pump) of approximately 10atm. The only thing that I cannot tell is at which point was the motor working, e.g. how far from the BEP was it on the motor graph. I can only say that the speed of the motor was the same as the plate reads. In other words, I do not know the efficiency the motor was working at, so it might have originally been lossy, yet I do not imagine that it was working in an extemely low efficiency, since the whole system was designed and constructed by a knowledgable engineering contractor.

I agree that the measuring equipment might be misleaded by non standard waveforms, such as trimmed waves or even wierder forms, but I can clearly say that this is not our case in this test, since the measurement was taken from the input of the energy saving device, and not on the output going to the motor. On the input to the saver device the waveform should basically be sinusoidal.

One question I have is, did you hear about the "iron loss equals copper loss" concept? this is the point that the manufacturer claims its saver device is bringing the motor to work in. Schematically, when iron loss is greater than copper loss then the voltage is reduced, and vice versa, when the copper loss is greater than the iron loss then the voltage is increased. This way, iteratively, these two loss components are equalized at a certain point, supposedly the best working point, energy saving wise.

Have you heard about this "iron loss = copper loss" approach before? is there a profound technical basis for it?

Best regards,

#4 marke

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Posted 12 May 2003 - 09:03 AM

Hello zac

The bottom line is that you can only save a portion of the energy that is being wasted. No one can validly argue this, but some do try!
I would be interested to know what the actual KW drawn by the motor was when you made the 30% saving, i.e. 30% of how much.
The current through the energy saving device will be the same on the input and the output, unless it is an inverter, in which case the current waveform will be different. With a chopper type energy saver, or with an inverter, the current waveform on the input is non sinusoidal and can yeild false readings with some instrumentation.
If you apply a variable voltage to an induction motor, as you reduce the voltage, the iron loss will reduce, and the copper loss will increase, except where the load current is significantly below the magnetising current at which time, the copper loss will also reduce then begin to rise again. If you tune the voltage for the minimum losses, that will be about where the copper loss and iron loss are equal. This is part of the design criteria used in optimising motor designs.

Best regards,

#5 zac

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Posted 12 May 2003 - 09:45 PM

Hello Mark,

Yes, I believe that we both agree on the maximal achievable saving as the energy being wasted. This is common sense.

Here are the results that were measured. Each value is given in the following order: Rated || Motor_only || Motor+saver

Power(KW): 45 || 42 || 29
Voltage (V): 380 || 415 || 415
Current(A): 87.5 || 78 || 50
Speed(RPM): 1475 || 1470 || 1460
Power-Coeff: 0.84 || 0.79 || 0.89


Regarding the chopped waves etc - as I mentioned in the earlier entry, we have performed the measurements upstream, on the input to the saving machine, where the waves are sinusoidial. The waveforms that go upstream to the motor might indeed be chopped or whatever, but since this is not where we measured it does not really interest us here.

Regards,
-Zac

#6 marke

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Posted 13 May 2003 - 10:12 AM

Hello zac

Yes I agree that being on the input, the meter will see non distorted voltage, but the current waveform will be distorted on both the input and the output. It is the current waveform that can upset measuring equipment.

What is the full load, three quarter load and half load efficiency of this motor?

Best regards,
Mark.

#7 zac

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Posted 19 May 2003 - 03:59 PM

Hi Mark,

If the current waveform might have been distorted in the input to the device than I can understand how this can fool the measuring device.

In the test that we plan on conducting soon we will therefore use the good old watt hour meter, which you also recommended in your article about energy saving devices.

I am sorry but I do not have the motor efficienciy values that you asked me. Where do I find these values? on the motor plate? in its documentation? this is an old motor (10 years or so) so I am not sure that the documentation can be found. Perhaps I could estimate by finding a similar motor, or looking on the internet. Why do you want to know these?

Best regards,
-Zac

#8 marke

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Posted 19 May 2003 - 07:24 PM

Hello zac

You will find the efficiency at varying loads on the data sheet for the motor. The efficiency at a particular load will show you the amount of energy that is being wasted by the motor at that load. In n umerous situations where I have been quoted very high KW savings at part load, I have found that the claimed KW saved is higher than the motor wasted at that load. If we know the efficiency, then we can get a reality check on the "measured" results. Maybe a multiplier set wrong or something.
I think that sometimes people read the meter incorrectly because they expect to see a number when the actual number is one tenth of that.

Good luck,
Best regards,

#9 karthi

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Posted 22 June 2003 - 06:55 AM

sir,
I am student.I am doing my project in energy saving in induction motor.
we loaded the 5H.P machine at half full load.
the supply to the motor is given through the generator.
The generator output voltage is varied by using 3 phase variac.
The frequency of the generator is varied by varying the speed of the prime mover.
i thing in this method the current waveform may not distorted
Is method of finding the energy saving in the induction motor is correct.
I also agree that for this test we have to go for a alternator and a prime mover.
My question is this way of proceeding is correct .

#10 zac

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Posted 23 June 2003 - 05:55 PM

Hi Marke,

Thanks for all the info so far. I dropped by to close this thread by telling the end of the story.

As it turned up, the initial measurements of the speed of the motor were wrong. When we performed the tests again, we found out that the speed of the motor was actually dramatically reduced (about 20% less in RPM). This of course means that the whole idea of the "saver" saving energy while conserving other output parameters the same is not true, and further means that this device is simply an inverter with a fancy name.

Too bad for me, yet better that we found this before going into investment...

Thanks for all the help,
-Zac.




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