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

[The Phantom]

"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.

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

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

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.

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

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.

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

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.

 

[The Phantom]

OK, good example: the product E*I is integrated over time to
get

power = lim (s=> infinity){ (1/s) integral_0^s { sin(t)*(sin(t) +
sin(3t)) dt}

and we note that sin(t) * sin(t) is 1/2(1+sin(2t)) and therefore has
a nonoscillatory part = 1/2 plus an oscillatory part. In the limit
of large 's',
the oscillatory part does nothing but oscillate, but the '1/2' part
grows.
Similarly, sin(t)*sin(3t) is a purely oscillatory function, with
components
at sin(2t) and sin(4t), but nothing that will grow with time.
Thus when we take the limit of large 's',

power = 1/2 = 0.500

is the result. Now, RMS value of E is 1/sqrt(2), and of I, 1, so the

VAR =Erms * Irms = 1/sqrt(2) = 0.707

and the power factor is power/VAR = .707

Ok, but you said in an earlier post, "The correct way to measure I is not
just RMS, but filtered for exactly the frequency and phase of E."

Can you give an example where doing that would give a result different than
the result gotten by just taking the RMS value of I?

 

[Eeyore]

As measured with true RMS meters, right?

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

If watts are higher than VA, the power factor is greater than 1, since power
factor is defined as watts/VA.

According to http://en.wikipedia.org/wiki/Power_factor, power factor is never
greater than 1.

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

Can you give a specific example of non-sinusoidal waveforms where the power
factor would be greater than 1?

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

Graham

 

[Eeyore]

If you measure the voltage and current with a true-RMS (AC+DC) meter, I
don't think there is any condition where true power is greater than
apparent power.

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

Graham

 

[The Phantom]

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

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.

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

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 (http://mathworld.wolfram.com/SchwarzsInequality.html)
guarantees it.

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

http://www.eng-tips.com/viewthread.cfm?qid=193052&page=1

http://kmh.ync.ac.kr/PcNcMicro/glossary/VA.HTM

http://www.lmphotonics.com/faq/faq_qanda.php?id=29

Plus many more if you search on:

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

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

Quite the contrary. This is an example of loads that don't have a high
power factor. For example, this site:
http://power-solutions.com/watts-va.php
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.

 

[John Fields]

Cite?

---
Not necessary; see your next sentence, below,
---

leading me to believe that that's what you

I'm very tolerant of nice people, even
black/gay/dumb/hill-billy/liberal ones.

---
Even??? And "Nice", of course, being defined by you.
---

I'm not as tolerant of mean,
profanity-hurling creeps.

---
ISTR a profane shadow puppet, so I guess you're talking about all
the mean, profanity-hurling creeps but you.
---

Of course I can decide what he's for. So can you.

---
Sorry, I don't have my God garb on right now, so I can't.
---

Phil is so obnoxious and so obscene and so rude that he has excluded
himself from civility.

---
Ah, I see...

_You_ exclude him from what you call "civility" because you disagree
with his rhetoric and his delivery, and that gives you the right to
berate him with impunity?
---

Read his posts.

---
I do
---

And you don't?

---
Nope.

I guess what you and I call "tolerance" are two very different
things.
---

and then generate insulting caricatures of his being a

I like women. And I'm making fun of Phyllis; so at least he serves a
purpose, to amuse me.

---
Then, instead of trying to trivialize him by casting him in the role
of a woman, you should be thanking him for the amusement he
provides.

In truth, though, what with your stated abhorrence of his demeanor,
it sounds more like you're amusing yourself at his expense.

 

[John Fields]

Who cares?

 

[whit3rd]

Ok, but you said in an earlier post, "The correct way to measure I is not
just RMS, but filtered for exactly the frequency and  phase of E."

Can you give an example where doing that would give a result different than
the result gotten by just taking the RMS value of I?

This example shows the effect clearly: the part of I that was sin(3t)
contributed nothing to the integral of E* I, because it was at a
different
frequency and thus uncorrelated to the E function (which was sin(t)).
That 'cos theta' or 'power factor' number is also called the
correlation
of the two functions... sorry, I'll try not to get too out-of-hand
with
the multiple nomenclature.

So, the integral selected exactly ONE of the Fourier components of the
I function, effectively filtering out all but one exact frequency. It
also
selected the sin(t) phase, not the cos(t) phase, of the I function.
The actual contributing part of the I function was exactly the same
frequency and phase as E, and if one had filtered I for that
frequency and phase, the power would have been exactly
equal to Erms* I[filtered, rms] ... and this is exactly the result
that one gets from a power meter that is really doing the multiply
and integrate operation.

 

[The Phantom]

Ok, but you said in an earlier post, "The correct way to measure I is not
just RMS, but filtered for exactly the frequency and  phase of E."

Can you give an example where doing that would give a result different than
the result gotten by just taking the RMS value of I?

This example shows the effect clearly: the part of I that was sin(3t)
contributed nothing to the integral of E* I, because it was at a
different
frequency and thus uncorrelated to the E function (which was sin(t)).
That 'cos theta' or 'power factor' number is also called the
correlation
of the two functions... sorry, I'll try not to get too out-of-hand
with
the multiple nomenclature.

So, the integral selected exactly ONE of the Fourier components of the
I function, effectively filtering out all but one exact frequency. It
also
selected the sin(t) phase, not the cos(t) phase, of the I function.
The actual contributing part of the I function was exactly the same
frequency and phase as E, and if one had filtered I for that
frequency and phase, the power would have been exactly
equal to Erms* I[filtered, rms] ... and this is exactly the result
that one gets from a power meter that is really doing the multiply
and integrate operation.

All right, I see what you're doing with this mathematically defined
situation.

Suppose you make measurements with a true wattmeter, true RMS ammeter, and
true RMS voltmeter. How do you apply your technique "The correct way to
measure I is not just RMS, but filtered for exactly the frequency and phase
of E." to quantities that aren't known mathematically, but are measured?

 

[whit3rd]

Suppose you make measurements with a true wattmeter, true RMS ammeter, and
true RMS voltmeter.  How do you apply your technique "The correct way to
measure I is not just RMS, but filtered for exactly the frequency and phase
of E." to quantities that aren't known mathematically, but are measured?

Huh? It's not so much 'how' as 'why' that confuses me.
If you measure power, it hardly matters that RMS measurements
after filtering would also give the power result. You already HAVE
that result. You also have enough info to calculate the power
factor (for utilities billing schemes that require this).

What benefit can be expected from redoing the same measurement
with another apparatus?

 

[John Larkin]

Profanity? Cite, please. The strongest word I recall using is "crap."

But why are you chasing me around like a crazed pekinese nipping at my
ankles?


John

 

[The Phantom]

Huh? It's not so much 'how' as 'why' that confuses me.
If you measure power, it hardly matters that RMS measurements
after filtering would also give the power result. You already HAVE
that result. You also have enough info to calculate the power
factor (for utilities billing schemes that require this).

What benefit can be expected from redoing the same measurement
with another apparatus?

Your very first post in this thread included one line from John Larkin and
the following response:
-------------------------------------------------------------
">
But it IS the formula of interest when you have a pure sinewave
for 'E', which is nearly correct for mains power.
The correct way to measure I is not just RMS, but filtered for
exactly the frequency and phase of E. So two RMS meters
won't tell you the power factor any more than two average-meters
would."
---------------------------------------------------------------

It seemed to me that your sentence:

"The correct way to measure I is not just RMS, but filtered for
exactly the frequency and phase of E."

was referring to the "I" in John's line, but your subsequent posts have
persuaded me that I was mistaken.

 

[John Fields]

Profanity? Cite, please. The strongest word I recall using is "crap."

 

[John Larkin]

Haven't a clue what you mean. Please explain.

You keep attacking, or at least trying to. Most weird.

John

 

[John Fields]

Haven't a clue what you mean. Please explain.

---
Well, I'm not going to waste any of my life tracking down your links
since it should suffice to say that you should recall a picture you
posted with the shadow of an uplifted middle finger prominent.

Surely you realize that's a profane gesture equivalent in meaning to
"**** you" and belies your claim that you're not in the same boat
with the rest of the "profanity-hurling creeps."

 

[John Larkin]

---
Well, I'm not going to waste any of my life tracking down your links
since it should suffice to say that you should recall a picture you
posted with the shadow of an uplifted middle finger prominent.

Surely you realize that's a profane gesture equivalent in meaning to
"**** you" and belies your claim that you're not in the same boat
with the rest of the "profanity-hurling creeps."

Well, as long as you're having fun, that's all that matters.

John

 

[HC]

While I don't have anything new to add, I'd like to cover this
issue at the most elementary level, just for the exercise...



Just to be pedantic, the procedure of using an average-reading
AC meter for current, then for voltage, ignores all polarity.
So, the reading can indicate that the power grid is powering
the appliance, OR that the appliance is powering the grid,
you need more info to determine which. This uncertainty
as to power direction also applies instant-to-instant, so
it is easy to see that the net power could be alternately
positive and negative.

Power forward from grid to appliance yields 'power factor' = 1.
Power backward from appliance to grid yields 'power factor' = -1
Power alternating can get a 'power factor' anywhere in the range
of (+1, -1)




But it IS the formula of interest when you have a pure sinewave
for 'E', which is nearly correct for mains power.
The correct way to measure I is not just RMS, but filtered for
exactly the frequency and phase of E. So two RMS meters
won't tell you the power factor any more than two average-meters
would. You need a power meter like the one the power company
bolted to your house, OR its electronic equivalent, that actually
measures E times I first, then integrates up all the timeslices
(the multiply/accumulate function here is KEY for digital signal
processors, because it shows up in LOTS of real problems).

The 'cos theta' is a number in the range (+1, -1) which is also
called the power factor, and it is the missing piece in any
of the schemes that measure E and I separately.

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

 

[John Fields]

Well, as long as you're having fun, that's all that matters.

 

[HC]

Hey, all, I'm trying to check something and I've come up with a number
and I'm not sure I've done my math right. What I want to do is figure
out how much a device is costing me to operate in electricity
charges. I'm not sure I'm doing the wattage calculations correctly
(I'm not well-versed with AC calculations and have only light hobbyist
experience with DC), and I'm not sure I'm going from the wattage to
the kW hours correctly.

I have a device that runs on 120v AC. I put a DMM in line with one of
the two conductors in the power cord and I measure 16.56 mA (my DMM
has a setting for measuring AC amps and that's what I used). I then
used the same DMM to measure AC voltage at the receptacle and I
measured 124.8 VAC. I'm not sure if this DMM does true RMS (as I see
that some manufacturers advertise that their meters measure "true
RMS") and the manual does not say yes or no.

Now, I take the amps times voltage to get watts so: 0.01656 x 124.8
and I get 2.066688 watts. I get kWh by 2.066688 x 24 hours / 1000
(the device is on constantly) and I get roughly 0.0496 kWh per day.
Say the billing cycle is for 31 days so that's 0.0496 x 31 = 1.5376
kWh each month and all I have to do is multiply that by my cost per
kWh and I should be good (for simplicity, say it costs me 10 cents per
kWh it's 15 cents a month).

Did I do all that right or am I mucking up the math or the wattage
calculation for AC?

Thanks in advance.

--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 (http://en.wikipedia.org/wiki/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

 

[Phil Allison]

"HC" = some demented fool with a multimeter.

Thank you all for your replies.

** What - even my one ?

There have been a variety of useful
responses that have helped me.

** Nothing will help a smug idiot like you.

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.

** What the **** takes you any time to do that ?

BTW:

YOU either have or have NOT got a suitable test instrument for measuring the
standby power consumption of a TV set.

No ordinary DMM has that capability.

The whole idea is a fucking waste of everybody's time.

SO BLOODY PISS OFF - TROLL.




........ Phil