Power Factor & kWH?

Sounds like you have a lot of time on your hands.

Here is a link to some useful charts:

The power company puts power factor correction equipment on their lines.

I cant answer you but a meter that is fun and usefull is a Kill-A-Watt apx 25$, it lets you plug in an apliance and measures very accuratly time, watt, amp, power factor, Kwh used and the hours, V. Hz and more, you will easily be able to audit usage of anything 120v you plug into it and find the hogs to show the hogs. It is also very good on low draw equipment on standby, such as the tv off. It measures over a 100 hr period. You would be suprised how older things can cost 1$ a month un used but newer "energy star" rated can cost 1 penny to keep plugged in a month, It helped me get my electric to 14-20 a month from 50. It also made me get a new frige that uses 1/6th the power. They have been independantly tested very accurate, somethimes Radio shack has them.

No, kWH = kilowatts multiplied by time in hours.

Residential power meters measure watts, not reactive VA. Power factor problems are paid for by the utility, not you.

In this case, I'm talking a 480V 3 phase commercial meter. What is the power factor charges on the bill? . Aside from the normal dials to read the kWh used, does the power factor get interpreted into it? The meter has a needle indicator for power factor also and stays pretty much in the same place.

Fluorescent lights and large electric motors receive part of their power by shifting the voltage and the current slightly out of phase. A residential meter ignores this and usually the effect is small for a household. A commercial meter measures power factor as well as kilowatts and the customer is charged accordingly.

try

These charges are not uniform and often vary greatly from utility to utility.

or 2nd - Contact the billing specialist at your utility company and request a copy of the tariffs. If they have the ability to explain it to you, in addition, consider yourself lucky.

Beachcomber

This was done a few years ago. I'll have to dig out some bills ot see how well it worked. There were three or four capacitors at different locations in the plant.. This is allegedly better than one big one at the 800A main panel.

The tariffs are easy enough to find. Explanation is a whole other scenario.

OK, then you apparently do get charged a penalty for reactive power use.

The physics of reactive power requires calculus to understand.

In simple terms, a motor or other inductive load can draw a portion of costly current and generating capacity from the utility, beyond the power realized by the device, even though you are not realizing that power in your facility. A simple power meter does not measure that loss, so a factor is measured by a more complicated meter to charge you for reactive power (what the utility sent to you) instead of real power (what you actually used). This is reasonable, and should encourage you to fix your installation to properly reduce the reactive power component.

You have to match the applied reactive power correction to the different times of day when your motor loads are creating a low power factor. Otherwise, an overcorrection (too much capacitive reactance) is just as bad, if not worse and may make your line voltage levels fluctuate all over the place.

It matters not if it is done at different locations or the main panel (assuming the main panel feeds the entire plant). A good main panel installation will have matched banks of capacitors that can be added in stages to correct the power factor. This can be automatic, timer-driven, or manually controlled.

It all depends on the load and mostly the motor load for that part. Is your plant idle at night, weekends, holidays? Are all motors running continously or do you have a lot of start-stop operations? You may have a base load (say of pumps and air blowers) that are on 24 hours a day, hence there might be the need for a certain base value of power factor correction.

Beachcomber

"The physics of reactive power requires calculus to understand. "

It's really pretty simple to understand the basics. Instantaneous power is always voltage times current. With an AC circuit and a purely resistive load, the voltage and current are always in phase with each other. Place a graph of voltage over a graph of current and they line up perfectly. So simply multiplying RMS Voltage times RMS Current gives power.

With a load that has capacitance or inductance in addition to resistance (eg a motor), the voltage and current are no longer in phase. Place a graph of one over the other and they appear shifted. So when voltage is at it's peak, current is not, hence the power consumed will be less. How much less depends on how far out of phase voltage and current are. With a purely capacitive load or a purely inductive load, the power will be zero.

A little trig is sufficient, IMO.

Nick

.... : : Sounds like you have a lot of time on your hands. Yeah, may be: It happens when one is suddenly disabled, thrown out of work because of it, housebound and not allowed to drive or even do the checking ;-( any longer. : : Here is a link to some useful charts: :

Thanks; looks like a decent deal, actually. I'll likely try that. Still leaves me wondering how it can do that though .

Thanks again & Regards

Below is part of what I was wondering about: Doesn't that mean, in reality, that, unless the PF is corrected, that the customer is the one with the advantage? As in, "free" power? IE of shifted waveforms is going to be less than in-phase IE over time. Therefore, the correction equipment is to "correct" the numbers so the power company isn't delivering power it isn't charging for? If so, why would anyone voluntarily install a capacitor system?

Pop

"Below is part of what I was wondering about: Doesn't that mean, in reality, that, unless the PF is corrected, that the customer is the one with the advantage?"

I guess that depends on exactly what the electric power meter installed at your home is measuring. I don't know, but I would guess that for a home it's not sophisticated and is likely only measuring RMS amps and basing it on that? If that's true, then the billing advantage is to the power company. But I would think the cost delta due to the power factor issue is pretty small in the typical home.

No. They measure real watts.

Nick

No, it is just wasted in the transmission lines and/or in reduced plant generating capacity. The point of the reactive meter is to get the culprit customer to pay for it.

A purely reactive load would be consuming all kinds of current down the line, and be loading the utility power plant, but show zero watts on a real power meter.

The answer to that question is.....

If it were a perfect world full of honorable people (particularly at the utilities) then .... if every homeowner installed equipment to make their home's power factors as close to unity as practically possible (Probably that could just be a capacitor right across each significantly sized motor.), the utility would have less line losses and be able to pass the savings back to their customers in the form of lower rates.

Remember, I said "perfect world"...

Jeff (Another old phart EE who's forgotten more than he learned, but one thing I'll never forget are the words of that Brit professor who tought our "rotating machinery" lab who said, "You boys will never become good electrical engineers until you learn to "take" a shock.")