Energy Savers for Induction Motors 3

[Marke]

I was recently challenged

http://www.lmphotonics.com/forum/viewthread.php?tid=23&pid=170#pid170

over an article that I wrote in relation to Energy Savers for induction motors. (

http://www.lmphotonics.com/energy.htm

) The challenger contested that my article was unduly negative and some interesting dialog followed, some in public and some in private. To prove my bias, I was presented with a set of "certified" test results that showed considerable savings were made on a test motor. When I looked at the results, I discovered that the losses in the 15KW motor were 3.5KW at full load, 2.5KW at half load and a whopping 7.9KW at no load. I questioned the no load losses and it was suggested that "what is printed in Theory & what actually happens in practice experiments can be a whole lot different".
I have asked for suggestions as to why this sudden increase in losses occurs with induction motors and I have not had a reply. I have been unable to find any evidence of this myself, and in this case, I would expect that the true losses would be in the order of 2KW.

Does any one have any theory, evidence or experience to support this contention?

I believe that the losses at full load are primarily iron loss and copper loss and both are of the same order of magnitude. As the load is reduced, the iron loss stays essentially constant, while the copper loss reduces with the current squared. This certainly is supported by my experience in the field. Mark Empson

http://www.lmphotonics.com

 

[cbarn24050]

Hi, an interesting article which does seem a bit negative, it seems you have selected a few bad examples to illestate your point.There are a number of benefits from using these softstarters, apart from the energy saving aspect, which you fail to mention at all. Also your pricing seems a bit out, a 7.5kW unit should be around $250 not the $3K+ you mentioned. Harmonic legislation will probably make these units illegal soon anyway. The latest craze is to use an inverter for enery saving, this is a great improvement, and coupled with new "green" tax laws these units have a faster payback time.

 

[electricpete]

What can possibly cause losses to increase as load decreases? I can only come up with two ideas:

#1 - Core losses. As load increases there is less voltage drop across the series leakage reactances. Clearly this will increase the flux in the rotor iron, which would tend to increase rotor losses. Although there will be a competing effect that the frequency of the rotor field decreases (as we decrease load) which tends to decrease the rotor losses.

What about stator core losses? I have to think a little bit about the meaning of the equivalent circuit parameters to decide whether portions of stator core are exposed to higher flux. What do you think?

#2 - Friction/windage. For a high-slip motor (Nema design D), there might be a substantial change in speed between full-load and no-load. This will increase the friction and windage losses. For a NEMA B it would be less.... in that case we are assured the change in speed is less than 5%. I don't know if there is a formula relating friction losses to speed (friction ~ speed^2?).

None of the above was very reasonable. Here is I think a more likely explanation for that data. The data gatherer simply measured the current and assumed a constant power factor. At no-load the current decreased to about half of full load so he calculated about half power. I would ask him to tell you the power factor as function of load to see if he measured it (if he accurately measured power input he must have measured power factor).

 

[electricpete]

I still think measurement error (not measuring power factor) is most likely cause.

But, trying to keep an open mind, I can imagine one more possibility. Perhaps if this motor is supplied from an electronic drive, the characteristics of the voltage may change with load (not so difficult to imagine). A change in the high frequency content of the voltage might influence the core losses dramatically. (It's a stretch)

 

[electricpete]

One more related item... a change in waveform can introduce measurement errors also. (still in the not-very-likely category).

 

[electricpete]

One thing that should not be obscured by my discussion of remotely-possibly explanations is the most-likely scenario: This person simply assumed power factor was constant for all power levels.

 

[Marke]

cbarn24050 Thank you for your reply. The article is not about soft starters, just the energy savers. I appreciate that a number of soft starter manufacturers include this functionality as an added feature. My comments are really aimed at the marketing people making sweeping claims based on tests on very small single phase motors. The examples quoted are from manufacturers datasheets and are typical of the claims that I have seen.
Soft starters have many other advantages. I have spent the last 23 years working on soft starter designs, so don't worry, I do believe in them!!

electricpete You sort of confirm my theories in that it is hard to come up with an explanation. I was informed that this is the norm for induction motors and was why there were good savings to be made, and my estimates were unacceptable.

I will keep looking, perhaps someone can give me a good explanation, till then, from what I have seen with induction motors, I remain a sceptic!!
Mark Empson

http://www.lmphotonics.com

 

[electricpete]

I will open up my mind to the possibility that your source may be correct. Here is an interesting excercize that I was surprised at.

IEEE 738-1995 (IEEE Recommended Practice for Energy Management of Industrial Facilities) figure 5-12 and MG10-1994 give “typical” efficiency vs load curves for 1hp, 10hp, 100hp motors. I read (guessimate) the values for efficiency from those curves at 25%/50%/75%/100% load as follows:

efficiency
100hp 10hp 1hp
100% load: 84 79 62
75% load: 81 74 53
50% load: 77 65 45
25% load: 70 55 33

The I compute losses using the formula
Losses (%)=output (1/eff - 1)

100hp 10hp 1hp
100% load: 19.0 26.6 61.3
75% load: 17.6 26.4 66.5
50% load: 14.9 26.9 61.1
25% load: 10.7 20.5 50.8

I’ll admit I was floored to see no significant increase in losses with increasing power as we might expect from the I^2*R component. Also I was surprised at the high losses on the 1HP. And your 50% data point (2.5kw losses=17%) and 100% data point (3.5kw=23%) falls between the 10hp and 100hp results as expected (if anything, the 50% losses of 2.5kw reported represents even a little lower than would be predicted by the above data, indicating better efficiency and more of a decreasing loss with decreasing load than shown above).

But that NEMA data is still not even close to consistent with what your source reported for no-load losses.

First you’ll notice that computed losses do in fact stay roughly constant or decrease as load decreses toward zero. Second you’ll notice that your motor should fall between 10 hp and 100hp (closer to 10hp). If we generously place your 15hp motor in the 10hp category, we still see the losses should not exceed approx 27% at any load which would be at most 3.75kw for your 15kw motor. (we have been generous also in not following the trend of decreasing losses with decreasing load).

One thing missing is a zero-power data point. The efficiency figure cannot go to zero load (undefined efficiency) and in my curve stops at 25%. But with the shape of the curves at the known data points and physical reasoning there is no reason I would suspect a sudden jump in losses going from 25% load to 0% load.

So the NEMA/IEEE data seems to contradict your source’s observation of >8kw losses at zero power and increasing losses with decreasing load.

The only points that might have lead me toward incorrect conclusion:
- perhaps newer motors behave differently than the older motors in this pre-1994 data (but I would expect them to be more efficient at all loads)
- perhaps I am mistaken in assuming there is no abrupt change in the curve between 25% and 0%?
- perhaps this is a large single-phase motor which might have some special characteristics?

I still say that it should not be unreasonable/insulting for you to ask your source for his power factor readings. The data seems fairly consistent with the pattern expected with an assumption of constant power factor vs load. In the meantime if I get time I’ll see if I can come up with a more quantitive demonstration of that hypothesis.

 

[electricpete]

I will clarify that "losses (%)" is intended to represent losses in percent of rated output power. ie for the 15kw motor, 1.5kw losses at a given power level would be 10% losses.

 

[electricpete]

If I go back and re-read the 1hp 75% load point I see it's closer to 55% which gives losses of 61.5% which gets rid of the anomaly in the curves.

Now a pattern becomes evident:
Losses are approx constant over the range of 50%-100% for these 1hp and 10hp typical motors. (smaller range of flat losses as horsepower increases)

But we know the I^2*R components increase substantially between 50% to 100%. Not quite quadruple but surely at least double for a high-speed motor with low exciting current.. So.... is the I^2R so small in comparison to other losses that it's increase doesn't show up? Or is there another loss component which significantly decreases as load increases to counteract the increasing I^2*R?

Perhaps whatever phenomenon is responsible for my inability to explain this nema data is responsible for our inability to explain Mark's friend's test data.

 

[electricpete]

Here you will find data sheet for a 20hp 400v motor:

http://www.reliance.com/pdf/pdf/aced/PD16-G7473.pdf

At the no-load point they have 9.9A, 4% power factor
gives real power input = 9.9*sqrt(3)*400*0.04=0.3kw if I have done my math right. (although that does sound terribly low).

At the 1/4-load point they have 11.6A, 51% power factor
gives real power input = 11.6*sqrt(3)*400*0.51=4.1kw total input. Subtracting out 3.75kw input gives ~ 0.4kw. This is an order of magnitude lower than suggested by the NEMA.

Just to check the 100% point they have 28.1, 83% power factor, 92.6% efficiency gives real power OUTPUT = 28.1*sqrt(3)*400*0.83=*0.926=15KW (confirms the math and voltage etc).

I got to that motor data sheet by selecting motor parameters at this page:

http://www.reliance.com/cgi-bin/mtrquery.pl#modelchar

Draw your own conclusions. I think perhaps these are simply quite a bit more efficient than the older NEMA data. The published full-load efficiency of 93% confirms that (they had only 84% for 100hp motor!).

So really this new data about more efficient modern motors pushes your clients claims further out of whack.

Note in particular there is no dramatic change in losses between 25% and 0%

(by the way don't forget to ask for the power factor data).

 

[Marke]

Hello Electricpete

This one has really got to you hasn't it!!

With small motors, the magnetising current can be very high, sometimes as high as 70% of the full load current. Under those circumstances, the I2R will change very little between full load and no load, so the info above is quite as expected and easily explained. On very large motors, the magnetising current is in the order of 20% so I would expect to see a much larger variation between losses at full load and no load due to copper loss variations.

Additionally, the leakage reactance is higher on small motors than large motors so the voltage regulation across the magnetising component of the equivilent circuit would not be as good as with a larger motor. There fore I would expect to see a small increase in iron loss at no load, but certainly not a multiple of 3!! An increase of 10% - 20% on small motors would be acceptable in my book.

I prefer to use the equivilent circuit model to explain the losses and when you look at this, there is no way that I can see for such a massive increase in iron loss!!

Unfortunately, I asked too many questions of the other party and he is now ignoring me completely, so no further information is available.
Best regards,
Mark. Mark Empson

http://www.lmphotonics.com

 

[electricpete]

Yes, it is an interesting one for me. I was coming to the same conclusions as you have mentioned, although not yet 100% confident in them.

I stumbled upon the fact that small motors have much higher percentage of no-load losses through an interesting side-track.... The shape of the efficiency vs load curve can be used to estimate the relative no-load and load losses, as follows:

Assume the losses are of the from
L=NL+LL * P^2
where L=total losses, NL=no-load losses, LL=load losses at full-load, P=output power from 0..1

divide by P:
L/P = NLL/P + LL *P

Max efficiency occurs at the minimum of the quantity L/P which is found by taking derivative of L/P with respect to P and setting it to zero:

0 = -NLL/P^2 + LL
Pbep = sqrt(NLL/LL) where Pbep = "best efficiency point"... peak of the efficiency vs power curve.

LL>NLL => Pbep<1 (np) <=> peak eff within op range

NLL>LL => Pbep >1 (np) <=> peak eff above op range

Looking at the NEMA efficiency curve for larger motors indicates that peak efficiency occurs perhaps around 1.0 or slightly. Indicates these large motors have LL>NLL

Looking at the NEMA efficiency curve for smaller 1hp motors indicates that efficiency is still increasing at the point where P=1. Pbep>1 indicates these small motors have NLL>LL.

I am still thinking about the stator core losses. I understand your thought process that increased current causes increased voltage drop across stator leakage reactance and reduced voltage applied to magnetizing branch. I have heard people imply before that voltage across magnetizing branch is directly related to flux.

But, I struggle a little with the physical understanding of that. Let us say that we operate the motor at no-load so the rotor is an open circuit. The exciting current flows through series combination of L1 and Xm. It is really the series combination of L1 and Xm that represent the magnetizing inductance in this case.

If we add load and allow rotor current to flow, then some of the flux does not couple the rotor which leads to the need to separate L1 and Xm. But the same voltage is still applied across the series combination of L1 and Xm. We distinguish them to separate the flux which is not coupled to the rotor, but does that in any way reduce the stator flux? i.e. the &quot;voltage drop accross L1&quot; that we propose is reducing our voltage available at Xm is STILL creating stator flux in L1... it just isn't linked to the rotor.

I'll think some more. Any thoughts?

 

[electricpete]

my discussion of comparing NLL and LL neglects the effect of magnetizing current, but near full load where the peak occurs, the error is not severe.

 

[electricpete]

Regarding the shape of the efficiency vs power curve and resulting inferences about LL vs NLL. The inductive current does introduce some error so the results should be considered qualitatviely, not quantitatively accurate. (It is not exactly true that LL=NLL at the point where efficiency curve peaks... but fairly close).


I am still thinking that perhaps physical view of the equivalent circuit does not support the view that stator core losses decrease with load. The stator flux will need to satisfy Vapplied=N dPhi/dt, regardless of secondary load. That seems to apply that total stator flux does not change with load (although local distribution may change).

It is true that the flux linking the rotor decreases. That is why we need to separate the total primary reactance into X1 (not linked to rotor) and Xm (linked to rotor). But from my perspective that does not provide a mechanism for decreasing stator flux. Maybe I am missing something?

As for rotor flux, it will decrease, which would tend to decrease core losses with load. BUT, the rotor frequency increases with load which would tend to increase rotor core losses with load (core loss proportional to f^2).

 

[electricpete]

I think maybe I have the answer to my question about effect of load on flux. (just thinking... not sure).

There are at least two different (series) contributors tof the stator leakage reactance: stator toothtop leakage reactance and end-turn leakage reactance.

I believe the stator toothtop leakage reactance acts the way I predicted.... current flow through it does not reduce flux to stator. That is because the associated flux does magnetize the stator. It bypasses the rotor but not the stator core.

I believe the stator endturn leakage reactance acts the way you predicted.... current flow through it does reduce flux to stator. That is because the associated flux does not magnetize the stator. It bypasses both the stator core and rotor core. Equivalently imagine that the stator was energized by a very long pair of leads representing the leakage reactance. As current increases a voltage drop is created which reduced voltage available for flux production in core.

So bottom line is that there is some reduction in flux due to increase in load. But I don't think it is as large as we would predict if we considered the entire stator leakage reactance X1 to create a voltage drop which does not produce flux in the core (the portion associated with tooththop leakage reactance does produce flux in the core).

At this point the discussion wanders pretty far from the original point. I'm just trying to learn a little in the process.

 

[electricpete]

Just for completeness, I believe that in the case of a transformer, virtually all of the leakage reactance acts the way you described, which is similar to the motor-end-turn leakage reactance in that the associated voltage drop reduces the flux in the core.

For the core form transformer with core insdie of LV winding inside of HV winding, tany flux which does not link both windings must be outside of the LV winding (mostly in the space between windings). If it is outside of the lv winding it is outside of the core. Any associated current contributes to voltage drop without magnetizing the core.

In contrast for the stator tooth top portion of motor winding leakage reactance, the majority of the flux path is in the stator core. The current associated with this leakage reactance creates a voltage drop which limits rotor voltage, but also contributes to magnetization of the core.

 

[electricpete]

Just for completeness, I believe that in the case of a transformer, virtually all of the leakage reactance acts the way you described, which is similar to the motor-end-turn leakage reactance in that the associated voltage drop reduces the flux in the core.

For the core form transformer with core insdie of LV winding inside of HV winding, tany flux which does not link both windings must be outside of the LV winding (mostly in the space between windings). If it is outside of the lv winding it is outside of the core. Any associated current contributes to voltage drop without magnetizing the core.

In contrast for the stator tooth top portion of motor winding leakage reactance, the majority of the flux path is in the stator core. The current associated with this leakage reactance creates a voltage drop which limits rotor flux, but also contributes to magnetization of the stator core.

 

[busbar]

Maybe this has been adequately covered in other threads { thread238-21978 thread237-19632 } but does the validity of the NASA/Frank Nola/power-factor controller and related claims bear discussion here?

 

[electricpete]

I have another question to throw in... why is it that smaller motors tend to have a higher fraction of no-load losses than large motors?