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

[Eeyore]

Now, I take the amps times voltage to get watts

Incorrect. Amps times Volts gives VA (volt-amperes) which is 'apparent
power'.

Finding true watts (real power) for a complex load is well .. complicated.
There may be phase angle issues between the volts and the amps (which make
the watts lower than the VA) and non-sinusoidal waveforms (which make the
watts HIGHER than the VA).

Fun eh ?

Graham

 

[The Phantom]

Incorrect. Amps times Volts gives VA (volt-amperes) which is 'apparent
power'.

As measured with true RMS meters, right?

Finding true watts (real power) for a complex load is well .. complicated.
There may be phase angle issues between the volts and the amps (which make
the watts lower than the VA) and non-sinusoidal waveforms (which make the
watts HIGHER than the VA).

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.

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

 

[Michael A. Terrell]

Phil loiters around the "basics" newsgroup looking for - in his
opinion - people who know less than he does, people who ask "basic"
questions, so he can abuse them and feel smart. He probably picks
fist-fights with six year old girls, too, so he can win.

He gave up fighting six year old girls. They kept beating the crap
out of him.

--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

 

[John Larkin]

And a Happy New Year to you as well

GG

Phil's great dream is to be elected Miss Congeniality Of Sci
Electronics Basics, to walk down that fabled runway in a white gown
and killer heels, tiara sparkling, tears of joy running down his face,
thousands of newbies cheering.

He's already stocking up on waterproof makeup.

John

 

[John Fields]

Phil's great dream is to be elected Miss Congeniality Of Sci
Electronics Basics, to walk down that fabled runway in a white gown
and killer heels, tiara sparkling, tears of joy running down his face,
thousands of newbies cheering.

He's already stocking up on waterproof makeup.

 

[whit3rd]

The usual definition of power factor doesn't admit of negative values.

That's a figment of the power industry sales agreement, in which
returning power to the grid gets credited in a separate arrangement
(and covered with a lot of safeguards against linemen being injured).
Mathematically, it's easiest to allow the full range. I'm a mathie.

But it is possible to have a pure sinewave for 'E' and a highly distorted
waveform for the current.


If the current waveform is highly distorted, how do you define the relative
phase between E and I?

THAT's the interesting question. You have to (as I indicated) ignore
all Fourier components of the I signal that do not share the frequency
of
E; the phase is then the phase of a single component of I that is
sinusoidal at the same frequency as E. _That_ phase is well
defined. For pure resistive, capacitive, and nonsaturating inductive
loads, and even for combinations of those, the current waveform
is not distorted, and the formula is easy to apply.

 

[The Phantom]

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.

Just to be sure I understand, are you saying that if a person measures the
voltage and current with two RMS meters, those two numbers, plus a wattmeter
measurement, can't be used to give power factor?

 

[John Larkin]

I can't help what you thought.

But gender? It's Phyllis I'm bashing, just for fun. Isn't that what
he's for?

John

 

[Phil Allison]

"Eeysore"

Finding true watts (real power) for a complex load is well .. complicated.
There may be phase angle issues between the volts and the amps (which make
the watts lower than the VA) and non-sinusoidal waveforms (which make the
watts HIGHER than the VA).

Fun eh ?

** Plus defies the law of conservation of energy.

All hell breaks loose then ......



....... Phil

 

[Paul E. Schoen]

As measured with true RMS meters, right?


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.

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

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. But if you use other types of meters, particularly with
capacitive coupling which removes the DC component, there might be such
effect.

Whenever there is asymmetrical distortion on a sine wave, phase angle
cannot be measured accurately, or in other words, it becomes undefined.
Most phase angle meters use zero crossings to determine the measurement
points, and thus a waveform with zero crossings not exactly spaced at 180
degrees will read differently when the polarity is reversed. Also when
there is crossover distortion, such that there is a flat spot at the zero
crossing, a phase angle meter cannot determine the point accurately.

Power can be measured accurately by using an analog multiplier circuit,
such as the AD534. I used this IC to design a DC wattmeter, which can also
be used on AC and distorted waveforms. The main problem was isolation from
the shunt that was used to measure current. A Hall effect sensor could have
been used, but an isolation amplifier did the trick.

You can also measure power using A/D techniques, especially if you use
simultaneous sampling converters. You must perform instantaneous products
of current and voltage, and then take an average of these sums for true
power.

Measuring power in three phase systems, especially with unbalanced and
non-linear loads, is even more complex. It is not so difficult in
star-connected systems where you can measure each leg and the voltages to
neutral, but a greater challenge in delta systems.

Paul

 

[whit3rd]

Just to be sure I understand, are you saying that if a person measures the
voltage and current with two RMS meters, those two numbers, plus a wattmeter
measurement, can't be used to give power factor?

No, if the wattmeter can run both forward and backward (some
utility-pole types cannot), it will register the RMS current times the
RMS voltage times the power factor. As long as the RMS current
and voltage aren't zero, you can divide and get the power factor.

In the case of complex waveforms, that (or an equivalent calculation)
is the only suitable way to determine the power factor. It's
what the big power-users' meters are based on.

 

[John Fields]

I can't help what you thought.

---
Well, that's not entirely true what with your earlier preaching of
tolerance being a virtue leading me to believe that that's what you
practiced.
---

But gender? It's Phyllis I'm bashing, just for fun. Isn't that what
he's for?

---
Nope. And exactly who the **** do you think you are to decide "what
he's for", you arrogant prick?

And bashing is fun? For bullies, maybe, but for the bashees,
hardly.

My point, though, was that since you obviously hold Phil in very low
esteem and then generate insulting caricatures of his being a
woman-like creature you're actually saying that women are like he is
and you're gender-bashing women.

 

[The Phantom]

That's a figment of the power industry sales agreement, in which
returning power to the grid gets credited in a separate arrangement
(and covered with a lot of safeguards against linemen being injured).
Mathematically, it's easiest to allow the full range. I'm a mathie.

I'm still not sure I understand your method. Let me give a concrete
example. Suppose you had a voltage source of e(t) = sin(t) and a
non-linear load which drew a current of i(t) = sin(t) + sin(3t), how would
you find the power factor? What would be your result for the power factor?

 

[John Larkin]

Cite?


leading me to believe that that's what you

practiced.

I'm very tolerant of nice people, even
black/gay/dumb/hill-billy/liberal ones. I'm not as tolerant of mean,
profanity-hurling creeps.


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

And bashing is fun? For bullies, maybe, but for the bashees,
hardly.

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

My point, though, was that since you obviously hold Phil in very low
esteem

And you don't?


and then generate insulting caricatures of his being a

woman-like creature you're actually saying that women are like he is
and you're gender-bashing women.

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

John

 

[The Phantom]

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. But if you use other types of meters, particularly with
capacitive coupling which removes the DC component, there might be such
effect.

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

Whenever there is asymmetrical distortion on a sine wave, phase angle
cannot be measured accurately, or in other words, it becomes undefined.

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".

http://en.wikipedia.org/wiki/Power_factor

http://en.wikipedia.org/wiki/Distortion_power_factor

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

http://groups.google.com/group/sci....8fa99d2?hl=en&lnk=st&q=&#doc_f16b87eafa50bca4

 

[Phil Allison]

""John Larkin"



** Larkin is a malicious, utterly autistic, criminal, pig arrogant Yank
cunthead.

Millions just like him make Americans the most hated people on earth.






....... Phil

 

[John Larkin]

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.'

"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.

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".

http://en.wikipedia.org/wiki/Power_factor

http://en.wikipedia.org/wiki/Distortion_power_factor

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

http://groups.google.com/group/sci....8fa99d2?hl=en&lnk=st&q=&#doc_f16b87eafa50bca4

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

John

 

[Don Bowey]

""John Larkin"



** Larkin is a malicious, utterly autistic, criminal, pig arrogant Yank
cunthead.

Millions just like him make Americans the most hated people on earth.

When did you and the donkey form the club?

 

[John Larkin]

""John Larkin"



** Larkin is a malicious, utterly autistic, criminal, pig arrogant Yank
cunthead.

Millions just like him make Americans the most hated people on earth.

Who cares?

John

 

[whit3rd]

  I'm still not sure I understand your method.  Let me give a concrete
example.  Suppose you had a voltage source of e(t) = sin(t) and a
non-linear load which drew a current of i(t) = sin(t) + sin(3t), how would
you find the power factor?  What would be your result for the power factor?

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