Basic AC wattage question: am I doing my math right?

If you use other types of meters, then you haven't measured apparent power, have you? You've measured something else.

This is what John Larkin was getting at when he said, '"E I cos theta" is sort of meaningless for radical waveforms.'

But it is still useful if one adopts the modern view and decomposes the power factor into a "displacement power factor" and a "distortion power factor".

There was an extensive thread on this topic about 3 years ago:

"Power factor" is a definition, and it's somewhat ambiguous. It could mean...

(True power) / ((RMS volts) * (RMS amps))

or

Cos(E:I phase angle) where just the fundamental frequency components are considered

or

could be undefined for non-sinusoidal waveforms, as some older textbooks suggest.

There's not a lot of point arguing over definitions.

The first definition is probably the most sensible in most cases nowadays.

Again, some texts simply consider power factor to be undefined in unbalanced polyphase systems.

John

That may have been appropriate for the older times when those textbooks were current, but the present needs a definition for non-sinusoidal waveforms.

It's important that engineers everywhere mean the same thing when they use certain defined phrases and words such as ampere, volt and power factor, for obvious reasons. That makes it worthwhile to discuss definitions.

It is the one found in the rules promulgated by regulatory agencies, and defined by professional societies such as the IEEE. It's not just sensible, it's mandatory to use in certain scenarios.

The practicing engineers of that day may have found that a satisfactory state of affairs. Nowadays, there are working definitions in place to deal with unbalanced polyphase systems, and I doubt we will ever go back to saying it's undefined.

True RMS meters are only a requirement if the waveform is non-sinusoidal.

Try measuring a power supply without 'power factor correction' if you think that's true !

Most power supplies that don't have power factor correction.

Graham

Never measured a 'capacitor input filtered' power supply have you ?

Graham

As is the waveform associated with essentially all modern power supplies, including the OP's television. Even power factor corrected supplies don't have a perfectly sinusoidal current draw. One might find a .95 power factor acceptable in a particular circumstance, and the use of true RMS meters will be necessary to verify that. Using true RMS meters will always give the correct result whether the waveform is sinusoidal or not, whereas non-RMS meters will not always give a correct result. So, if you're measuring almost any household load other than perhaps an incandescent light bulb or a toaster, you'd better be using true RMS meters. And, of course, any meter used must have sufficient bandwidth for the measurement at hand.

I have and never got a power factor over 1.

If you got a power factor over 1, then you made a measurement error. Perhaps your instruments had insufficient bandwidth.

If power factor is defined as (true power)/(RMS volts*RMS current), then the power factor can never be greater than 1.

Schwarz's Inequality

guarantees it.

Other references to the fact that power factor can't be greater than 1:

Plus many more if you search on:

+"power factor"+"always less than 1"

Quite the contrary. This is an example of loads that don't have a high power factor. For example, this site:

gives the typical value of power factor for an uncorrected supply as .55 to .75.

The whole purpose of power factor correction is to raise the power factor as close to 1 as possible, not lower it from a value greater than 1 to just below 1.

In fact, a power factor greater than 1 would violate conservation of energy. It would mean that the actual power delivered to the load was greater than the product of volts and amps.

Thank you for your reply. The information you've provided helps to clarify what is going on and what makes it so complex. I had thought I would just "quickly" do this one thing (determining the power consumption of my TV at rest) but now I realize it's going to take a much deeper understanding of AC power flow than I have. Someday...when I have more time... :) In the meantime I'm going to stick to the basics and try to finish some other projects I have going so that I can focus on learning AC when the time comes.

Happy New Year.

--HC

Thank you all for your replies. There have been a variety of useful responses that have helped me. I found the link to Wikipedia about Power Factor

provided by The Phantom to be quite helpful. I am going to check my equipment to see what it states/claims it can do for AC readings of both voltage and current (see if it does RMS) and go from there. Thank you, Phantom, that helped a bunch.

Again, thanks everybody.

Happy New Year.

--HC

It might be well worth your while to spend $20 and get a Kill-a-watt. It can measure true watts, volt-amperes, power factor, amps, volts--the whole menu! The only problem it will have is that it won't have much resolution at very low power levels (below 1 or 2 watts), but at least it can give you an estimate even there. And, if it says 1 or 2 watts, then you can figure you're not paying much per month for that load.

I probably will, but I am still going to try to learn more about this so I can do it manually if for no other reason than that I would like to know how to do it, how it works, and why. Ya know? I want the answer, but I hate to give up on the knowledge of how it works.

Thank you.

--HC

Very much so. But even then the VA =/= the watts.

NOT for watts since the instantaneous power is the PRODUCT of the instantaneous volts and amps and NO meter measuring only one at a time can do that.

Graham

And I said:

"As measured with true RMS meters, right?"

believ "Using true RMS meters will always give the correct result"

I was referring to the measurement of VA since that was what the discussion had been about.

I never suggested that measuring voltage or amperage one at a time could give the instantaneous product of volts and amps.

An admirable attitude, and continued study will get you there. Good luck.

instantaneous

Not required if the waveform is sinusoidal.

Which formula ?

When the load current waveform is non-sinusodal, even using a true RMS meter will not help you get a true wattage figure. Hence whether you use a true RMS meter or not is irrelevant since with sinusoidal waveforms and a phase shift (a la classic power factor) the averaging meter will give equally accurate results.

Graham

always

instantaneous

You already said that the first time (and I've never denied it). I replied that since many loads draw non-sinusoidal current, by using RMS meters correct results for the VA part of the power factor formula will be obtained no matter whether the waveforms are sinusoidal or not.

Power factor = true power/(RMS volts * RMS current) which is the same as PF = watts/VA. And if VA is measured with true RMS meters, then this expression is correct no matter what the waveforms. If average responding meters are used, the correct VA is only given if the waveforms are all sinusoidal. So why not just use RMS meters and be done with it? Especially since nowadays so many pre-existing power supplies are uncorrected switchers, although regulatory agencies are now requiring PF correction.

will

As I have said several times now, the use of RMS meters is not intended to get a true wattage figure. It is intended to get a true VA figure. The VA number is needed to get the power factor.

As I said in another post, I used a wattmeter to get the true power. That, together with the VA figure derived from measurements with true RMS meters, allowed me to compute the power factor for the non-sinusoidal current drawn by my TV set.

classic

I've never said that average responding meters won't give accurate results, IF the waveforms are sinusoidal. But nowadays a substantial fraction of equipment loads are non-sinusoidal. Someday, when the regulations have been in effect for years, most switchers will be PF corrected, but that's not the case now. And accurate measurement of power factor requires the use of RMS responding meters to get an accurate value for VA.

You can beat the 1W resolution by doing a kWh measurement. At 2W, you'd need to wait about a day with the Kill A Watt to get to 0.05 kWh, from which you'd calculate the power to +/-20% accuracy (resolution of the Kill A Watt is 0.01 kWh).

Of course, the OP must be willing to go one day without turning on the TV.

Mark

I sold a few thousand multichannel energy loggers that utilities and researchers used for end-use load studies. One thing we instrumented was a chain of fast-food restaurants. Their deep-fat friers are resistive heaters with zero-crossing burst-mode triac temperature controllers. When the triac is on, PF is obviously 1. When it's off, there's no load so pf is undefined. So, what is net power factor? A number of "experts" could never decide on a definition.

John

This may not really help, if the basic resolution and zero offset errors compromise accuracy. The integral of a large number of bad readings will just give a bigger bad reading. If the system is analog, multiplier circuits are often non-linear at low levels and have offsets. A digital multiplier is limited by the resolution of the A/D components. When you are operating at two bits, you can have an error of -50% or +33%.

An alternative way to measure the actual power consumption would be to enclose the TV in a well-sealed styrofoam box, and measure the temperature rise above ambient to a point where it reaches a steady state. Then repeat the test with a resistor, and adjust the power until it produces the same temperature.

This will work with any waveform. But perhaps if there is an RF component that is radiating energy through the enclosure, it may draw more power than is measured by heat generation. The heat is simply wasted energy, assuming you don't want a heater. Along the same line of reasoning, if you use electric heat, and it is being used, the TV does not add anything to your utility bill. It may even reduce it if you have the TV under the thermostat and you don't change the setting.

Paul

On Fri, 04 Jan 2008 07:08:54 -0800, John Larkin

Remember this thread from 2004? You were a participant and you mentioned that same load:

I suppose one could ask, "What is the net power factor if I turn on an incandescent light bulb for 12 hours of the day, and leave it off for the other 12 hours?"

Some other interesting questions are "How many angels CAN dance on the head of a pin?" and "Just exactly what is the value of zero divided by zero?"

How about this for a method of MEASURING the net PF in any situation. A home (or business) already has an accumulating power meter, AKA a kilowatthour meter. Add another meter, an accumulating volt-ampere meter. Reset both meters to zero and then let them do their thing for a period of time for which it is wished to know the net PF. At the end of the time, take the ratio kilowatthours/kilovoltamperehours. That is the net PF for that interval of time. The meters won't be befuddled by the time when the power is off.

The power company would find that they never had to supply your deep-fat heaters with any VARs, so they would think your net PF was zero. The time when the PF is undefined is irrelevant and its contribution to the net PF has measure zero in the sense of Lebesgue integration, so don't even consider it. That's my "expert" opinion.