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

[David L. Jones]

On Dec 30, 12:48 am, NoS...@daqarta.com (Bob Masta) wrote:

On Sat, 29 Dec 2007 00:08:08 -0800 (PST), HC <hboo...@gte.net> wrote:
On Dec 28, 11:54 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
Your math is right for a simple (ie, resistive) load, but might be off
by, say, 2:1 in either direction, depending on the actual load
waveform and/or power factor.

John

Hey, John, the more I've been reading the more I think my math and
measurements have been badly wrong because they're rooted in DC
methodology which seems to be quite different from the AC world (or
can be for equipment/circuits that are not 100% resistive). So, as
you say, I could be quite wrong by just simply not knowing the
intricacies of AC power flow. I found a formula for calculating AC
current consumption on single-phase which is P = V x I x cosine Theta
which is great except that Theta is the "power factor angle" of the
equipment which I don't currently know. I approached this problem
thinking like I do about DC and I've found that it is wrong and the
problem is considerably more complex. I'm going to do a lot more
reading about this and "reboot" my whole test.

Check out the "Kill-A-Watt", it does *exactly* what you want.
I bought mine on-line for about $20 (with free shipping!) for
uses like this. I tested 2 different TV models of the same size
(27") and similar age and found that one used nearly 16 watts
when off, while the other used less than 8.

The Kill-A-Watt shows RMS volts, amps, watts, KWH,
power factor, line frequency, and probably some others
I've forgotten. You plug the Kill-A-Watt into the wall, then
plug the thing you want to test into the Kill-A-Watt, and
push a button to select what you want to read.

I've wanted a watt-meter for *years* and always
figured I'd have to bite the bullet and build a crude
one. But the Kill-A-Watt does more than anything
I'd ever have built... and for only $20 !!! This is by
far my favorite toy of the year.

(PS: I have no affiliation with Kill-A-Watt, just
a delighted customer.)

Best regards,

Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!

Sadly, the Australian market still does not have a low cost consumer
power meter like that one.
*sigh*
One was released recently, but the supplier removed it from sale
because it had accuracy issues on the low end of the scale.

Dave.

 

[The Phantom]

On Sat, 29 Dec 2007 00:08:08 -0800 (PST), HC <hboothe@gte.net> wrote:

Your math is right for a simple (ie, resistive) load, but might be off
by, say, 2:1 in either direction, depending on the actual load
waveform and/or power factor.

John

Hey, John, the more I've been reading the more I think my math and
measurements have been badly wrong because they're rooted in DC
methodology which seems to be quite different from the AC world (or
can be for equipment/circuits that are not 100% resistive). So, as
you say, I could be quite wrong by just simply not knowing the
intricacies of AC power flow. I found a formula for calculating AC
current consumption on single-phase which is P = V x I x cosine Theta
which is great except that Theta is the "power factor angle" of the
equipment which I don't currently know. I approached this problem
thinking like I do about DC and I've found that it is wrong and the
problem is considerably more complex. I'm going to do a lot more
reading about this and "reboot" my whole test.

Thank you for your reply.

I checked my 20 inch Sony and it was drawing 2 watts when turned off, but 5 VA
(volt-amperes, the product of separately measured volts and amps; the same thing
you did).

The power was measured with a Yokogawa electromechanical wattmeter specifically
designed to measure power when the power factor is low.

These two measurements mean that the power factor was about .4.

What you should know is that the measurement you made is an upper bound for the
true power. For example, if the power factor (which you don't know) is .33333,
then the true power consumption is 1/3 of what you measured and you are paying
less than 15 cents a month for that TV's idle power; more like 5 cents. Do you
really need to know more than that? If so, then get a Kill-a-watt, but be aware
that the Kill-a-watt will have inadequate resolution for really accurate
measurements at such low powers.

 

[Phil Allison]

"The Phantom" hn

I checked my 20 inch Sony and it was drawing 2 watts when turned off, but
5 VA
(volt-amperes, the product of separately measured volts and amps; the same
thing
you did).

** Like hell the trolling, scumbag cretin did.

You need a " true rms" amp meter to measure VA.





........ Phil

 

[amdx]

"David L. Jones" <altzone@gmail.com> wrote in message
news:225f59f5-6244-4384-b702-1aa6afbe5640@b40g2000prf.googlegroups.com...

On Dec 30, 12:48 am, NoS...@daqarta.com (Bob Masta) wrote:
On Sat, 29 Dec 2007 00:08:08 -0800 (PST), HC <hboo...@gte.net> wrote:
On Dec 28, 11:54 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
Your math is right for a simple (ie, resistive) load, but might be off
by, say, 2:1 in either direction, depending on the actual load
waveform and/or power factor.

John

Hey, John, the more I've been reading the more I think my math and
measurements have been badly wrong because they're rooted in DC
methodology which seems to be quite different from the AC world (or
can be for equipment/circuits that are not 100% resistive). So, as
you say, I could be quite wrong by just simply not knowing the
intricacies of AC power flow. I found a formula for calculating AC
current consumption on single-phase which is P = V x I x cosine Theta
which is great except that Theta is the "power factor angle" of the
equipment which I don't currently know. I approached this problem
thinking like I do about DC and I've found that it is wrong and the
problem is considerably more complex. I'm going to do a lot more
reading about this and "reboot" my whole test.

Check out the "Kill-A-Watt", it does *exactly* what you want.
I bought mine on-line for about $20 (with free shipping!) for
uses like this. I tested 2 different TV models of the same size
(27") and similar age and found that one used nearly 16 watts
when off, while the other used less than 8.

The Kill-A-Watt shows RMS volts, amps, watts, KWH,
power factor, line frequency, and probably some others
I've forgotten. You plug the Kill-A-Watt into the wall, then
plug the thing you want to test into the Kill-A-Watt, and
push a button to select what you want to read.

I've wanted a watt-meter for *years* and always
figured I'd have to bite the bullet and build a crude
one. But the Kill-A-Watt does more than anything
I'd ever have built... and for only $20 !!! This is by
far my favorite toy of the year.

(PS: I have no affiliation with Kill-A-Watt, just
a delighted customer.)

Best regards,

Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!

Sadly, the Australian market still does not have a low cost consumer
power meter like that one.
*sigh*
One was released recently, but the supplier removed it from sale
because it had accuracy issues on the low end of the scale.

Dave.
How about as a work around for the poor low end accuracy;

Put three 100 watt light bulbs on as a load, measure
the power then add the low power load he wants to measure and note the
difference. Would the power meter have the resolution to do this?
Mike
Mike

 

[John Larkin]

On 30 Dec 2007 03:38:02 -0600, The Phantom <phantom@aol.com> wrote:

On Sat, 29 Dec 2007 00:08:08 -0800 (PST), HC <hboothe@gte.net> wrote:




Your math is right for a simple (ie, resistive) load, but might be off
by, say, 2:1 in either direction, depending on the actual load
waveform and/or power factor.

John

Hey, John, the more I've been reading the more I think my math and
measurements have been badly wrong because they're rooted in DC
methodology which seems to be quite different from the AC world (or
can be for equipment/circuits that are not 100% resistive). So, as
you say, I could be quite wrong by just simply not knowing the
intricacies of AC power flow. I found a formula for calculating AC
current consumption on single-phase which is P = V x I x cosine Theta
which is great except that Theta is the "power factor angle" of the
equipment which I don't currently know. I approached this problem
thinking like I do about DC and I've found that it is wrong and the
problem is considerably more complex. I'm going to do a lot more
reading about this and "reboot" my whole test.

Thank you for your reply.


I checked my 20 inch Sony and it was drawing 2 watts when turned off, but 5 VA
(volt-amperes, the product of separately measured volts and amps; the same thing
you did).

The power was measured with a Yokogawa electromechanical wattmeter specifically
designed to measure power when the power factor is low.

These two measurements mean that the power factor was about .4.

What you should know is that the measurement you made is an upper bound for the
true power. For example, if the power factor (which you don't know) is .33333,
then the true power consumption is 1/3 of what you measured and you are paying
less than 15 cents a month for that TV's idle power; more like 5 cents.

If the gadget has a transformer first thing, yes. But if it's a
switcher, with a rectifier and filter cap first, and he uses a
not-true-RMS meter, the error could go in the opposite direction.

"E I cos theta" is sort of meaningless for radical waveforms.

John

 

[David L. Jones]

On Dec 31, 3:08 am, "amdx" <a...@knology.net> wrote:

"David L. Jones" <altz...@gmail.com> wrote in messagenews:225f59f5-6244-4384-b702-1aa6afbe5640@b40g2000prf.googlegroups.com...

On Dec 30, 12:48 am, NoS...@daqarta.com (Bob Masta) wrote:
On Sat, 29 Dec 2007 00:08:08 -0800 (PST), HC <hboo...@gte.net> wrote:
On Dec 28, 11:54 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
Your math is right for a simple (ie, resistive) load, but might be off
by, say, 2:1 in either direction, depending on the actual load
waveform and/or power factor.

John

Hey, John, the more I've been reading the more I think my math and
measurements have been badly wrong because they're rooted in DC
methodology which seems to be quite different from the AC world (or
can be for equipment/circuits that are not 100% resistive). So, as
you say, I could be quite wrong by just simply not knowing the
intricacies of AC power flow. I found a formula for calculating AC
current consumption on single-phase which is P = V x I x cosine Theta
which is great except that Theta is the "power factor angle" of the
equipment which I don't currently know. I approached this problem
thinking like I do about DC and I've found that it is wrong and the
problem is considerably more complex. I'm going to do a lot more
reading about this and "reboot" my whole test.

Check out the "Kill-A-Watt", it does *exactly* what you want.
I bought mine on-line for about $20 (with free shipping!) for
uses like this. I tested 2 different TV models of the same size
(27") and similar age and found that one used nearly 16 watts
when off, while the other used less than 8.

The Kill-A-Watt shows RMS volts, amps, watts, KWH,
power factor, line frequency, and probably some others
I've forgotten. You plug the Kill-A-Watt into the wall, then
plug the thing you want to test into the Kill-A-Watt, and
push a button to select what you want to read.

I've wanted a watt-meter for *years* and always
figured I'd have to bite the bullet and build a crude
one. But the Kill-A-Watt does more than anything
I'd ever have built... and for only $20 !!! This is by
far my favorite toy of the year.

(PS: I have no affiliation with Kill-A-Watt, just
a delighted customer.)

Best regards,

Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!

Sadly, the Australian market still does not have a low cost consumer
power meter like that one.
*sigh*
One was released recently, but the supplier removed it from sale
because it had accuracy issues on the low end of the scale.

Dave.

How about as a work around for the poor low end accuracy;
Put three 100 watt light bulbs on as a load, measure
the power then add the low power load he wants to measure and note the
difference. Would the power meter have the resolution to do this?

Your average punter is not going to want to do this, and even I
couldn't be bothered. Also, that is not a solution for long term
consumption monitoring which is what these things are good at also.
I've got a silicon chip power meter kit (which I got ridiculously
cheap), and that is pretty good, but I wouldn't mind a one or two more
cheap commercial ones. They are dime-a-dozen in the US, but not here.

Dave.

 

[David L. Jones]

On Dec 30, 5:06 pm, "David L. Jones" <altz...@gmail.com> wrote:

On Dec 30, 12:48 am, NoS...@daqarta.com (Bob Masta) wrote:



On Sat, 29 Dec 2007 00:08:08 -0800 (PST), HC <hboo...@gte.net> wrote:
On Dec 28, 11:54 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
Your math is right for a simple (ie, resistive) load, but might be off
by, say, 2:1 in either direction, depending on the actual load
waveform and/or power factor.

John

Hey, John, the more I've been reading the more I think my math and
measurements have been badly wrong because they're rooted in DC
methodology which seems to be quite different from the AC world (or
can be for equipment/circuits that are not 100% resistive). So, as
you say, I could be quite wrong by just simply not knowing the
intricacies of AC power flow. I found a formula for calculating AC
current consumption on single-phase which is P = V x I x cosine Theta
which is great except that Theta is the "power factor angle" of the
equipment which I don't currently know. I approached this problem
thinking like I do about DC and I've found that it is wrong and the
problem is considerably more complex. I'm going to do a lot more
reading about this and "reboot" my whole test.

Check out the "Kill-A-Watt", it does *exactly* what you want.
I bought mine on-line for about $20 (with free shipping!) for
uses like this. I tested 2 different TV models of the same size
(27") and similar age and found that one used nearly 16 watts
when off, while the other used less than 8.

The Kill-A-Watt shows RMS volts, amps, watts, KWH,
power factor, line frequency, and probably some others
I've forgotten. You plug the Kill-A-Watt into the wall, then
plug the thing you want to test into the Kill-A-Watt, and
push a button to select what you want to read.

I've wanted a watt-meter for *years* and always
figured I'd have to bite the bullet and build a crude
one. But the Kill-A-Watt does more than anything
I'd ever have built... and for only $20 !!! This is by
far my favorite toy of the year.

(PS: I have no affiliation with Kill-A-Watt, just
a delighted customer.)

Best regards,

Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!

Sadly, the Australian market still does not have a low cost consumer
power meter like that one.
*sigh*
One was released recently, but the supplier removed it from sale
because it had accuracy issues on the low end of the scale.

Dave.

For those playing along at home, this is the new power meter in
question available on the Australian market:
http://cgi.ebay.com.au/Save-Electricity-Mains-Power-Monitor-Meter-Outlet_W0QQitemZ320198707286QQihZ011QQcategoryZ79253QQssPageNameZWDVWQQrdZ1QQcmdZViewItem

The ATA who was selling the unit (not via eBay like the link above)
initially had this to say in the advert:

"Note: Some of the readouts on the display on this meter may be hard
to read if your eyesight isn't so good, although the main readout uses
around 9mm high digits. Also, meter accuracy measuring small values,
below 10 watts, is not great, so it may not display accurate reading
for small loads. "

But then removed it from sale for the following reason:

"Low cost digital power meter possible accuracy issues:
Our new low cost power meter has been a very popular item, however
some customers have reported accuracy issues with the meter. Our
initial testing showed it to work quite well, but it seems the
accuracy is variable, with some meters working well, and others being
inaccurate.

Because of this problem, we have decided to no longer stock the meter.
We are instead looking for alternatives and expect to have a
replacement meter available in a couple of months at most.

If you have recently purchased a low-cost energy meter from ATA and
have doubts about its accuracy, or have any other problems with the
meter, you are welcome to return it for a refund."

Seems like eBay sellers are still happy to supply it, and probably
haven't heard about any problems with it.

Dave.

 

[HC]

On Dec 29, 7:48 am, NoS...@daqarta.com (Bob Masta) wrote:

On Sat, 29 Dec 2007 00:08:08 -0800 (PST), HC <hboo...@gte.net> wrote:
On Dec 28, 11:54 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
Your math is right for a simple (ie, resistive) load, but might be off
by, say, 2:1 in either direction, depending on the actual load
waveform and/or power factor.

John

Hey, John, the more I've been reading the more I think my math and
measurements have been badly wrong because they're rooted in DC
methodology which seems to be quite different from the AC world (or
can be for equipment/circuits that are not 100% resistive). So, as
you say, I could be quite wrong by just simply not knowing the
intricacies of AC power flow. I found a formula for calculating AC
current consumption on single-phase which is P = V x I x cosine Theta
which is great except that Theta is the "power factor angle" of the
equipment which I don't currently know. I approached this problem
thinking like I do about DC and I've found that it is wrong and the
problem is considerably more complex. I'm going to do a lot more
reading about this and "reboot" my whole test.

Check out the "Kill-A-Watt", it does *exactly* what you want.
I bought mine on-line for about $20 (with free shipping!) for
uses like this. I tested 2 different TV models of the same size
(27") and similar age and found that one used nearly 16 watts
when off, while the other used less than 8.

The Kill-A-Watt shows RMS volts, amps, watts, KWH,
power factor, line frequency, and probably some others
I've forgotten. You plug the Kill-A-Watt into the wall, then
plug the thing you want to test into the Kill-A-Watt, and
push a button to select what you want to read.

I've wanted a watt-meter for *years* and always
figured I'd have to bite the bullet and build a crude
one. But the Kill-A-Watt does more than anything
I'd ever have built... and for only $20 !!! This is by
far my favorite toy of the year.

(PS: I have no affiliation with Kill-A-Watt, just
a delighted customer.)

Best regards,

Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!

Cool, thanks, Bob, I'll check that out.

--HC

 

[The Phantom]

On Sun, 30 Dec 2007 08:44:51 -0800, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On 30 Dec 2007 03:38:02 -0600, The Phantom <phantom@aol.com> wrote:

On Sat, 29 Dec 2007 00:08:08 -0800 (PST), HC <hboothe@gte.net> wrote:




Your math is right for a simple (ie, resistive) load, but might be off
by, say, 2:1 in either direction, depending on the actual load
waveform and/or power factor.

John

Hey, John, the more I've been reading the more I think my math and
measurements have been badly wrong because they're rooted in DC
methodology which seems to be quite different from the AC world (or
can be for equipment/circuits that are not 100% resistive). So, as
you say, I could be quite wrong by just simply not knowing the
intricacies of AC power flow. I found a formula for calculating AC
current consumption on single-phase which is P = V x I x cosine Theta
which is great except that Theta is the "power factor angle" of the
equipment which I don't currently know. I approached this problem
thinking like I do about DC and I've found that it is wrong and the
problem is considerably more complex. I'm going to do a lot more
reading about this and "reboot" my whole test.

Thank you for your reply.


I checked my 20 inch Sony and it was drawing 2 watts when turned off, but 5 VA
(volt-amperes, the product of separately measured volts and amps; the same thing
you did).

The power was measured with a Yokogawa electromechanical wattmeter specifically
designed to measure power when the power factor is low.

These two measurements mean that the power factor was about .4.

What you should know is that the measurement you made is an upper bound for the
true power. For example, if the power factor (which you don't know) is .33333,
then the true power consumption is 1/3 of what you measured and you are paying
less than 15 cents a month for that TV's idle power; more like 5 cents.

If the gadget has a transformer first thing, yes. But if it's a
switcher, with a rectifier and filter cap first, and he uses a
not-true-RMS meter, the error could go in the opposite direction.

Tim Wescott in the first response to the OP's question mentioned the need for a
correct RMS measurement. I expected that the issue was settled then and my
comments assumed as much.

"E I cos theta" is sort of meaningless for radical waveforms.

John

 

[whit3rd]

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...

On Dec 31, 4:26 am, The Phantom <phan...@aol.com> wrote:

mentioned the need for a
correct RMS measurement.  

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)

"E I cos theta" is sort of meaningless for radical waveforms.

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.

 

[Rich Grise]

On Fri, 28 Dec 2007 18:03:19 -0800, HC wrote:

On Dec 28, 6:28 pm, Don Bowey <dbo...@comcast.net> wrote:
....... Phil

Wow, so you're this group's neighborhood troll. Nice to meet you.
Hey, when you grow up your dicky might get bigger than it is now. Be
patient. In the meantime, find another way to compensate for it.

Ignore what he says at your peril.

Don, I appreciate your comment and I will look for more information to
try to get a better understanding of my question and the answers I've
received, including Phil's. I just didn't appreciate being spoken to
harshly for asking a simple question and being maligned for leaving out
details that, in my ignorance, are perhaps important.

No, just ignore the trolls, of which Phalluson is one of the worst.

This is sci.electronics.basics, where no question is unwelcome -
rudeness like his (and a few others) is severely frowned upon.

Cheers!
Rich

 

[Phil Allison]

"Rich Grise"

No, just ignore the trolls,

** Especially Rich Grise.

A monumental fuckwit, a smug turd, an asshole and a net cop jerk off.

This is sci.electronics.basics, where no question is unwelcome -

** The bone-headed moron thinks he OWNS the newsgroup and his bizarre
opinions are unchallenagble.

He could not be more wrong.

About everything.

Piss the fool off, or killfile him.




......... Phil

 

[Phil Allison]

"The Phantom"

Tim Wescott in the first response to the OP's question mentioned the need
for a
correct RMS measurement. I expected that the issue was settled then and
my
comments assumed as much.

** What a spectacularly STUPID assumption to make.

But only one of a great many.




......... Phil

 

[John Larkin]

On Tue, 01 Jan 2008 00:05:10 GMT, Rich Grise <rich@example.net> wrote:

On Fri, 28 Dec 2007 18:03:19 -0800, HC wrote:
On Dec 28, 6:28 pm, Don Bowey <dbo...@comcast.net> wrote:
....... Phil

Wow, so you're this group's neighborhood troll. Nice to meet you.
Hey, when you grow up your dicky might get bigger than it is now. Be
patient. In the meantime, find another way to compensate for it.

Ignore what he says at your peril.

Don, I appreciate your comment and I will look for more information to
try to get a better understanding of my question and the answers I've
received, including Phil's. I just didn't appreciate being spoken to
harshly for asking a simple question and being maligned for leaving out
details that, in my ignorance, are perhaps important.

No, just ignore the trolls, of which Phalluson is one of the worst.

This is sci.electronics.basics, where no question is unwelcome -
rudeness like his (and a few others) is severely frowned upon.

Cheers!
Rich

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.

John

 

[Phil Allison]

"John Larkin"


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

Those are only his good points.




....... Phil

 

[John Larkin]

On Tue, 1 Jan 2008 16:11:32 +1100, "Phil Allison"
<philallison@tpg.com.au> wrote:

"John Larkin"


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

Those are only his good points.

Well, now we know what sorts of things you admire.

John

 

[Phil Allison]

"John Larkin"


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

Scum of the planet.




...... Phil

 

[Phil Allison]

<stratarse46@yahoo.com>


** Drop dead - pork brain.





....... Phil

 

On Dec 31, 9:45 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:

"John Larkin"

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

 Scum of the planet.

..... Phil

And a Happy New Year to you as well

GG

 

[The Phantom]

On Mon, 31 Dec 2007 14:48:26 -0800 (PST), whit3rd <whit3rd@gmail.com> wrote:

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...

On Dec 31, 4:26 am, The Phantom <phan...@aol.com> wrote:

mentioned the need for a
correct RMS measurement.  

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)

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

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

"E I cos theta" is sort of meaningless for radical waveforms.

But it IS the formula of interest when you have a pure sinewave
for 'E', which is nearly correct for mains power.

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

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

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

So two RMS meters
won't tell you the power factor any more than two average-meters
would.

I don't think anybody said that two RMS meters (alone) would tell you the power
factor, did they?

You need a power meter like the one the power company
bolted to your house,

Actually, the meter bolted to your house is an energy meter, not a power meter.

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.