Power factor question

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Well most home owners do not have a meter to measure harmonics so the cost of to have one would be ?

I know they make a watt hour meter electronic commercially to adjust / measure the harmonic content of the service power and it gives calculated adjusted wattage used but the harmonic 3rd 4th ect ect have the negative effects on a normal watt meter used today generally on 99 percent of the homes today the old watt meter would actually benefit the bill as it would not measure the total wattage used by the chopping trashed ac wave passing thur that standard watt meter do you agree ?
 
Most of my discussions are for medium to large facilities. Home users do not have to worry from all of that, as the losses are mainly at the utility side (this is one reason why their charges are higher).

The wattage in harmonics is minimal (volts harmonics multiplied by current harmonics). It will not reduce your bill and neither increase it by more than 1%.

The major losses due to harmonics is in the network itself.
 
iwire said:
No one is charged based on harmonics, so even if the meter can record it nothing changes with the KWH consumed.

You might as well no right off the bat that in my opinion you are a snake oil salesman.
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You are welcome to follow this link to decide if I am a snake oil salesman or power quality expert. You may Google my name to see some more links.
 
PowerQualityDoctor said:
Most of my discussions are for medium to large facilities. Home users do not have to worry from all of that, as the losses are mainly at the utility side (this is one reason why their charges are higher).

The wattage in harmonics is minimal (volts harmonics multiplied by current harmonics). It will not reduce your bill and neither increase it by more than 1%.

The major losses due to harmonics is in the network itself.
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Well what if one had a way to inject harmonic content into the ac wave at meter and adjusted the total wattage measured or used but did not effect the equipment using the real power of the line by multipule frequency adjustment .

The opposite of measuring the harmonics but making harmonics in a safe way .
 
ohmhead said:
Well what if one had a way to inject harmonic content into the ac wave at meter and adjusted the total wattage measured or used but did not effect the equipment using the real power of the line by multipule frequency adjustment .

The opposite of measuring the harmonics but making harmonics in a safe way .
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Technically, this will reduce the consumption. The problem is that such injection will cost you more than your saving.
 
Well question if one sent you info on there system PF issues say i have a plant with MV motors 4160v 3 phase and a low voltage system included .

With also office branch low voltage normal system online .

We send you our data for your analysis .

We need info from you say do we test harmonics line to line or do we test line to neutral balanced or unbalanced loads what do you need for a good data program to compute PF correction in a industrial or commercial area .


Have you heard of frequency injection ?
 
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100214-0803 EST

iwire:

I believe there is a slight language problem in the discussion, but PowerQualityDoctor does have a point that deserves attention.

Suppose I have a substantial length of wire from the meter to an inductive load and that load has a power factor of 0.5, then the real power at the load is 50% of measured VA to the load. The current in the branch circuit is approximately double that required for the actual work being performed, and the power losses in the branch circuit are 4 times that required to do the actual work.

Consider a real example:

A 20 A load at a PF = 0.5 and 5% voltage drop on the branch circuit. I will use a 120 V circuit for some actual values.

5% of 120 = 6 V. Power lost in the branch circuit is 20*6 = 120 W. The actual usable load power is (120-6)*20*0.5 = 1140 W.

As an approximation assume the load power remains the same and there is 1/2 the branch circuit voltage drop by changing the load power factor to 1.0, then the load current is about 1140/117 = 9.75 A. The power lost in the branch circuit is about 3*9.75 = 29.25 W.

The cost of operating at PF = 0.5 is 1140 + 120 = 1260 W.
The cost of operating at PF = 1.0 is 1140 + 29.3 = 1169 W.
Thus a saving of 91/1169 = 0.0778 or about 7.8%.

Most power factors won't be this bad. So this would be a high estimate on possible savings. Voltage drop is a trade off of the cost of copper to energy saving, then there are the capacitor costs to consider. It gets very complex at this point as to whether the capital costs are worth it. If we gamble that there will be substantial inflation and increase in energy costs, then the capital investment may be worth while.

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Being an engineer for more than 20 years in power quality analysis, power quality correction and motor controls I think I have some knowledge that can contribute to the community. I am not familiar with most of the electricity issues, but I am an expert in these three. Since that I have created a free web site to assist people with power quality analysis. I do it as a hobby on my free time (sorry, wife).

As the only person who has two video interviews at Leonardo Energy website and reputable presenter at various IEEE and other conferences I think I deserve more than curses and threats by the forum moderator himself !!!

Sorry for those of you who didn't receive an answer. If you want, you can find me elsewhere.
 
gar said:
100214-0803 EST

iwire:

I believe there is a slight language problem in the discussion, but PowerQualityDoctor does have a point that deserves attention.

Suppose I have a substantial length of wire from the meter to an inductive load and that load has a power factor of 0.5, then the real power at the load is 50% of measured VA to the load. The current in the branch circuit is approximately double that required for the actual work being performed, and the power losses in the branch circuit are 4 times that required to do the actual work.

Consider a real example:

A 20 A load at a PF = 0.5 and 5% voltage drop on the branch circuit. I will use a 120 V circuit for some actual values.

5% of 120 = 6 V. Power lost in the branch circuit is 20*6 = 120 W. The actual usable load power is (120-6)*20*0.5 = 1140 W.

As an approximation assume the load power remains the same and there is 1/2 the branch circuit voltage drop by changing the load power factor to 1.0, then the load current is about 1140/117 = 9.75 A. The power lost in the branch circuit is about 3*9.75 = 29.25 W.

The cost of operating at PF = 0.5 is 1140 + 120 = 1260 W.
The cost of operating at PF = 1.0 is 1140 + 29.3 = 1169 W.
Thus a saving of 91/1169 = 0.0778 or about 7.8%.

Most power factors won't be this bad. So this would be a high estimate on possible savings. Voltage drop is a trade off of the cost of copper to energy saving, then there are the capacitor costs to consider. It gets very complex at this point as to whether the capital costs are worth it. If we gamble that there will be substantial inflation and increase in energy costs, then the capital investment may be worth while.

.
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Gar correct me if I am wrong but just because I may reduce losses in the lines does not mean I will see any reduction at the meter.

If my equipment runs just as well with voltage drop I really see no benefit in reducing loses in the conductors. In fact in some instances lowering the voltages to lightly loaded motors increases their efficiency.
 
PowerQualityDoctor said:
As the only person who has two video interviews at Leonardo Energy website and reputable presenter at various IEEE and other conferences I think I deserve more than curses and threats by the forum moderator himself !!!
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You where not cursed, I simply let you know my opinion.

As far as being threatened you must follow the forum rules just like anyone. We do not allow advertising for any purpose on the forum.
 
As my limited understanding in English "Snake Oil Salesman" in not a blessing...

You are welcome to view similar discussion to this post at http://forums.mikeholt.com/showthread.php?t=110653&highlight=.com and check the advertisements there. It includes a link to a company that some posts called their statements as "lies". Oh, and see some signatures over there referring to other companies.

And just to continue the debate:
iwire said:
Well I know you are sales person as I removed your advertising.

You have not answered two direct questions with any direct answers.


So far I am staying with my original assessment.
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It seems that you know things that I don't KNOW. You can guess, you can assume, but if you KNOW something which is wrong, what does it mean about you?
 
Well lets talk just PF Questions which i have about it and as your the engineer on PF . Iam here to learn more

So heres one resistance increases with frequency so a ampere of harmonic current above fundamental produces more of a loss than does an ampere of the fundamental frequency which produced it to start with this in mind a larger wire diameter has more resistance to this harmonic current at higher frequency 3rd 4th ---- without a capacitor used could one use a wire or conductor to eliminate such current inline distribution work we use tube conductors at high voltage correct skin effect ?

What if we used tube conductors with a outer shell of say a higher resistance material to limit the harmonic current flowing in the outer shell on the surface i know temp would rise on tube or conductor and increase resistance of conductor but what if it did not meaning one had a way to limit the temp rise and would it even be of any value to cause any heat as its reactive by example ?


Meaning a resistive filter inline for just harmonics or should i say HARM MONICS which is to me a bigg part of PF
 
iwire said:
But if I am not paying for those losses I save nothing by eliminating them.
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Exactly right! I agree if not penalized then there is no issue. From forever I've been tought PF means $$ other than that if they do not penalize from a customers standpoint there is no concern. There are people who would not know anything about PF if it were not due to the $$$.:)
 
I like Weressl's comment best:

To sum it up: there are Statistics, there are Damned Lies and there are Electric Power Saving Devices.
__________________
 
100214-1202 EST

iwire:

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

The input power to most motors is equal to the load power, torque * RPM * constant, plus internal losses. Speed does not change much on normal induction motors with small voltage changes, and not at all on synchronous motors. Thus, under constant load torque the work power into a motor does not change much with a small voltage change. Thus, in my sample calculation I kept the load power constant even though there was about a 3 V change with and without power factor correction at the motor.

Power factor correction at the motor, a capacitor, has little effect on what is going on inside the motor and its load.

What I showed in my sample calculation was a change in the power dissipated in the branch circuit wire to the motor. At PF = 0.5 this wire loss was 120 W for a load real power of 1140 W. By changing the power factor at the load to 1.0 while holding the real load power constant at 1140 W I reduced the losses to about 29 W in the branch circuit.

If this motor runs continuously for one year at $0.10 per KWH the wire losses cost 8760*0.12*0.10 = $105 vs 8760*0.029*0.10 = $25. Thus, per year of continuous operation there is a $80 saving. Ten years and the saving is $800. Will this pay for the PF correction and the lost use of the money for something else. Very likely, especially with the expected inflation to come. But as I said a PF of 0.5 is quite unlikely for a motor in normal use. My 1/2 HP single phase capacitor run lathe motor under no load has a PF of 0.54 . Most fairly continuous duty motors are run with fairly high loads, and have a higher power factor.

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