Power factor question

[mivey]

PowerQualityDoctor does make some valid points. I have also made some of the same arguments in some of the other threads.

Without a POCO pf penalty, I have found that for the correction to be economically feasible, the power factor has to low, the load has to be large, located a good distance from the meter, and run a lot of hours. That is just for a simple capacitor solution.

I'm not sure what PowerQualityDoctor's costs figures are or how efficient the devices are, but if they can do the job for less than a capacitor install, the economics would naturally be better. In some cases, even devices with a higher cost than capacitors would still show positive economics.

In all fairness, I think PowerQualityDoctor was clear in saying that this would not be economical for most residential customers.

Without a POCO pf penalty, I do not buy into solutions at the meter, but solutions at the load or farther out in the distribution system can make economic sense even then.

Yes you do pay for the losses between the meter and the load.

And I have posted info before showing these losses can be significant enough to make the correction worthwhile.

Add: Of course, a POCO pf penalty makes the economics even better and feasible for additional loads.

 

[iwire]

100214-1202 EST

iwire:

Yes you do pay for the losses between the meter and the load.

Yes the watse heat is paid for at the meter, but if my equipment is doing what it needs to do with the losses in the line I see no gain by reducing those losses in the lines.

A motor that is putting out it's rated HP may well draw more current at a reduced voltage but in most case reduced voltages to motors will result in less current and that means less power. Do you not feel the the reduced power consumption by the motors makes up for the losses in the lines?

I notice your figures do not take any of that into account, you only took into account what proves the point you where shooting for.

 

[ohmhead]

Well good PF = 120 volts x12 amps =1440va


BAD PF = 120 volts x12 x50%= 720 va




voltage drop on wire 3 volts x 12 =36 watts


So what your saying is if you have a good power factor your load uses full wattage at rated amps and the wire wattage goes down in watts .

But with a bad power factor the load uses less power to do the work and the wire uses some wattage . looks like a wash to me

 

[mivey]

...but in most case reduced voltages to motors will result in less current and that means less power...

If the motors are "over-productive", then it would make sense to step the output down at the motor. It can be reasonable to assume that you will try to operate efficiently for any given condition.

Maybe the increased motor output caused by the increased voltage makes the plant more efficient in other ways. In other words, Lucy & Ethel will be able to box more chocolates per shift.

 

[iwire]

Maybe the increased motor output caused by the increased voltage makes the plant more efficient in other ways. In other words, Lucy & Ethel will be able to box more chocolates per shift.

Possibly but IMO unlikely, besides, Lucy was already the limiting factor of production. :grin:

I know places that use large conveyors have controls that drop the voltage down when the full HP is not needed.

I also know that many grocery stores are intentionally reducing the voltage down from 120 to 114 in order to save KWH and increase the longevity of some equipment.

 

[gar]

100214-1529 EST

iwire:

Yes the watse heat is paid for at the meter, but if my equipment is doing what it needs to do with the losses in the line I see no gain by reducing those losses in the lines.

You pay for the losses in your branch circuit wiring. If you reduce those losses you will pay less. So the question is --- what is the payback on reducing those losses?

A motor that is putting out it's rated HP may well draw more current at a reduced voltage but in most case reduced voltages to motors will result in less current and that means less power. Do you not feel the the reduced power consumption by the motors makes up for the losses in the lines?

One has to have a lot more information to determine what happens when you increase the source resistance to a load relative to energy consumption.

I do not believe that well designed systems will reduce your energy consumption by not using power factor correction at the load.

The issue of running equipment on reduced voltage to reduce energy consumption is a whole different subject.




On a load with a low power factor when you compare the branch circuit losses with and without power factor correction at the load one can demonstrate some less energy consumed after power factor correction. Is it worth while to correct it is the real important question.

Yes I created conditions via assumptions to prove a point.

But I want to go back over my discussion on the assumption that the load is an induction motor doing a task. If a motor had no slip, then output RPM is synchronous speed and does not change with voltage. This is true of a synchronous motor until you reach breakaway torque. In an induction motor slip increases as load torque increases. However, the amount of this slip is a function of rotor resistance. Low rotor resistance produces little slip. High rotor resistance lots of slip. For a given rotor resistance the slip is approximately proportional to torque load until most of the way to breakaway.

If a motor is driving a constant power load, such as a pump, then at a fixed speed the torque does not change because of the constant power load. Since speed is controlled by frequency, slip (a modifier of speed) is determined by torque and only secondarily by voltage, this means speed does not vary very much with a low resistance rotor and small voltage changes to the motor and real power input to the motor for the load excluding motor losses is nearly constant with small voltage changes.

Experiments on my import drill-press. No load except friction and windage.

Volts -- Watts -- PF
100 ---- 293 ---- 0.6
110 ---- 312 ---- 0.52
120 ---- 344 ---- 0.49
130 ---- 375 ---- 0.43

Speed did not vary very much. With my crude tachometer consisting of a photocell, white paper-black tape, LED flashlight, and scope I saw no speed change. So I will guess less than 5% over this 30% voltage range.

The large power change, but still less than V^2 would predict, 1.27 to 1 as compared to 1.3^2 =1.69 to 1 would be much less if the motor had a full load constant torque load that would be large compared to varying motor losses.

A standard induction motor with a constant and near full load should show a current increase as voltage is lowered within a usable working range.

With this as background I think my assumptions to try to show the effect of power factor correction at the load on branch circuit losses is valid and informative.

Without getting into the details of what is happening to the load with a small change in supply voltage. Just consider the fact that if you reduce the supply current thru a circuit by 1/2, then you reduce the power dissipation in that circuit by 3/4 (the power dissipation is reduced to 1/4).

If a motor has been sized well to its load and the load is relatively constant, a refrigerator for example, then the power input to that motor for the load part should be relatively constant independent of supply voltage. However, the motor losses will increase with voltage. If the motor efficiency is good then load power should dominate over the motor losses. Efficiency will increase as motor size increases.

I hope most constant load applications are not designed such that current drops as line voltage drops.


The above is in response to your post #42.

I will comment on your other post later.

.

 

[iwire]

If a motor has been sized well to its load and the load is relatively constant, a refrigerator for example, then the power input to that motor for the load part should be relatively constant independent of supply voltage.

Supermarkets are reducing their KWH consumption by lowering the voltage supplied to the compressor racks. Apparently the compressor motors are over sized.

I will never attempt to dispute your engineering skills and your ability to calculate exceeds mine by a factor of at least 1000 but I do consider myself street smart and if there was decent KWH to be saved making a .85 PF become 0 PF we would be installing the equipment in the field.

 

[ohmhead]

Well i have the answer to power factor correction here it is

 

[mivey]

Supermarkets are reducing their KWH consumption by lowering the voltage supplied to the compressor racks. Apparently the compressor motors are over sized.

I will never attempt to dispute your engineering skills and your ability to calculate exceeds mine by a factor of at least 1000 but I do consider myself street smart and if there was decent KWH to be saved making a .85 PF become 0 PF we would be installing the equipment in the field.

A supermarket is not what I consider a target-rich environment for making PFC economically feasible on distribution losses alone.

I would question moving 85% to unity (that's 0.85 to 1.0). Correcting to unity would rarely make sense. On some numbers I have run, the pay-back curve makes a drastic slope change around the low to mid 90's. I would think you would need a power factor penalty or some real favorable load statistics to make it worth attacking an existing 85% pf.

As for PFC equipment, they are installing the equipment in the field. Just don't under-estimate the power of corporate stupidity. I'm sure you see many corporate decisions that just make you shake your head in disbelief.

 

[iwire]

A supermarket is not what I consider a target-rich environment for making PFC economically feasible on distribution losses alone.

That was not my point.

Just don't under-estimate the power of corporate stupidity. I'm sure you see many corporate decisions that just make you shake your head in disbelief.

No I cannot argue with that.

 

[mivey]

That was not my point.

I thought the point was (correct me if I butcher your position):
lower voltage = lower power consumption and that this lower power consumption is just as effective or better than reducing losses by adding PFC.

 

[iwire]

I thought the point was (correct me if I butcher your position):
lower voltage = lower power consumption and that this lower power consumption is just as effective or better than reducing losses by adding PFC.

Gar had said

If a motor has been sized well to its load and the load is relatively constant, a refrigerator for example, then the power input to that motor for the load part should be relatively constant independent of supply voltage.

My point was motors may not be well sized to the load as often as we might assume.

But basically my position is this, it can be shown on paper but what works on paper does not always pan out.

 

[mivey]

My point was motors may not be well sized to the load as often as we might assume.

And the loads are not always constant. The refrigeration I've seen varies quite a bit and would require dynamic correction except on the base refrigeration load units.

But basically my position is this, it can be shown on paper but what works on paper does not always pan out.

Agreed, but what comes out is only as good as what goes in.

 

[gar]

100214-2008 EST

iwire:

It is quite possible that lowering input voltage to a refrigeration compressor motor will lower input power. It will depend upon the motor design and how far into saturation the motor is driven. How old and inefficient the motor design is etc. I firmly believe that experimental evaluation is important backup for theoretical analysis.

Following are the results of some tests I ran after my previous post. This is on Slimline fixtures. All have the same model bulbs -- green end Phillips 60 W
Without looking for my light meter, just a qualitative reaction, I did not see a light change over this voltage range.

The following measurements were made with a Kill-A-Watt. Three different fixtures. They all date from the 60s but may have original or new ballasts. I have no electronic ballasts on Slimlines.

Volts -- Watts - VA --- PF --- Amps

100.7 --- 90 -- 101 -- 0.89 -- 1.01
110.5 --- 93 -- 103 -- 0.90 -- 0.93
120.5 --- 97 -- 107 -- 0.90 -- 0.89
130.4 -- 101 -- 116 -- 0.87 -- 0.89

100.3 -- 118 -- 119 -- 0.98 -- 1.19
110.0 -- 125 -- 127 -- 0.98 -- 1.12
120.3 -- 132 -- 134 -- 0.97 -- 1.12
130.3 -- 139 -- 143 -- 0.97 -- 1.09

100.3 -- 129 -- 129 -- 0.99 -- 1.29
110.7 -- 137 -- 138 -- 0.98 -- 1.25
120.2 -- 143 -- 144 -- 0.99 -- 1.21
129.9 -- 149 -- 153 -- 0.98 -- 1.18

On these lights I could possibly save 10% by operating at 100 V instead of normal line voltage of about 123-124. However, my current solution keep them off if not needed.

I have found my 60 year old GE light meter. On one fixture the reading was 24.5 foot-candles (just a hair below 25) at 130 V input and 25 foot-candles at 100 V.


An old computer

100.3 --- 38 --- 59 -- 0.65 -- 0.55
110.5 --- 40 --- 62 -- 0.63 -- 0.56
120.5 --- 40 --- 64 -- 0.62 -- 0.54
130.1 --- 41 --- 68 -- 0.62 -- 0.51

To avoid an uproar I do not want to play with the refrigerator now, and it is too cold in the garage to play with the freezers. That test will have to wait

.

 

[gar]

100213-2336 EST

Results from a cheapie 4 ft electronic ballast from Home Depot.

Tends to fail at 100 V I would consider 105 the lowest reliable voltage. Light intensity about the same as the Slimlines, but varies with voltage.

Volts -- Watts - VA --- PF --- Amps

104.9 --- 49 --- 55 -- 0.88 -- 0.54
110.7 --- 55 --- 66 -- 0.86 -- 0.59
120.7 --- 65 --- 84 -- 0.77 -- 0.69
130.6 --- 72 -- 102 -- 0.70 -- 0.78

I would have to do a more controlled experiment to determine if this was more or less efficient than the 8' Slimlines relative to light output vs power input. But its PF is clearly worse.

.

 

[mivey]

gar,

If you are feeling froggy, use the delta data you are getting and simulate an industrial load situation. Then compare the increase in power with the loss savings. However, I don't think that will prove much.

In iwire's case, the loss savings would still be recoverable because even if the voltage went up, the motor controls would just tap it back down. The end result would be the same power at the compressors, but less vars moving up and down the distribution system.

Why limit yourself to one type savings if the other is still a viable option?

 

[PowerQualityDoctor]

Following specific requests from ohmhead and mivey I have decided to comment on the last posts. I will totally ignore iwire, which his behavior questions if he deserves to be a moderator.

Voltage Changes
Most loads change their behavior when the input voltage changes. I have vast experience with two type of loads - discharge lights and asynchronous 3 phase motors. I work for a company that manufactures energy savers for both applications (not a salesmen, though) and did tens of measurements myself on actual loads. All my measurements are accurate and I put all my professional reputation on them, even in those cases where sales people tried to "convince" me to change them. You can see some of the results on my company website, but I am afraid to put a link so you will have to use Google (or see previous posts in which I was allowed by the Chief moderator to put links as they contribute to the technical discussion).

Lamps
For discharge lamps (fluorescent, metal-halide, HPS etc.) with electro-mechanical chokes, reducing the voltage by 10% after completing the startup process of 15 minutes (still in the allowed range) the consumption will be reduced by 20-30%. The light will also be somewhat reduced, usually less than the energy saving. Moreover, normally the light is more than required so it is saving only.

Motors
For induction motors - when the motor is loaded by more than 60% (by means of power, not current), reducing 25% of the voltage will reduce its consumption. The amount of reduction depends on the loading, motor size, motor efficiency and more parameters. The highest value I measured was 36.6% for 18kW (25HP) motor and much more for laborotary small motors. This idea is known since 1979. The problem that usually it creates harmonics which create other source of losses and it is not recommended. My company developed a unique technology that can do it without any harmonics at all.

If the motor is loaded by more than 80%, usually you can save if you increase the supplied voltage, but the saving is minimal of few percents only.

Field results: there are many applications where the load is changing and the motor is under partial load during significant amount of time. Some examples are granulators, grinders, escalators, mixers, conveyors, balers, hydraulic pumps, mud pumps, schredders, pelletizers and many more. In other cases the motors are oversized in order to withstand hard startup. According to the DoE, 40% or the motors in the US are loaded by 40% or less.

PF correction
I used to work in this business but changed position 2 years ago, so I assume I can add company names... I worked for Elspec, which manufactures the world's best power factor correction systems (as it was copied by many companies, it means not only I think this...)

PF correction can save you money even if no penalties occurs. However, in most cases if the PF is high enough (as mivey saied - 0.85 is acceptable, mid 90's is great value) the saving will be minimal. Note that sometimes there are indirect saving that can be very much, like saving on infrastructure or fixing voltage levels.

Distributed PF correction is complicated to implement and many times not cost effective unless the lines are very long and the PF is very low. It can be done, however, with partially loaded motors when reducing their supply voltage.

 

[iwire]

I will totally ignore iwire, which his behavior questions if he deserves to be a moderator.

I love you too.:grin:

Please understand we get about at least one person a week trying to sell mysterious power saving boxes so I am very quick to judge.

No one has banned you even though at least a couple of other mods have suggested it.

 

[PowerQualityDoctor]

I know. That's why I referred you to two other moderators with whom I had long discussion of what is allowed and what is not. I have followed what was agreed together and you still insisted to try and fight and not listen.

Yes, I am commercially involved with different electrical appliances companies, at part of them I worked and hold their stocks, some of them pay my salary, and some of them I just own stocks if I like their product. And yes, I would love that my posts will lead to increase sales.

But no, I am not a salesman, I am an engineer. I am the product specialists (which belongs to the marketing department). I do have more than 20 years of experience and am invited by reputable organization to lecture on various topics. I guess it is both my technical and presenting skills. My next lecture is next Monday about Power Quality monitoring at a training session of a society of electrical engineers, although I don't work in the PQ business for over 2 years. Just for the record, I do it for free. Also my website, whose link you deleted, is completly free. Why for free? My wife says that I am stupid. She is always right

 

[mivey]

...My company developed a unique technology that can do it without any harmonics at all...

Got a technical brief of the methodology or a patent reference?