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SteveB wrote:

Wattage devided by the voltage (120) is amps. Or just add up the wattage. 20 amps x 120v = 2400 watts. Lights are easy since the wattage is written on each bulb.

Dear Steve, the formula is

volts times amps = watts

So, you can say that the 110 volt circuit, times 20 amps = 2200 watts.

Hint: Try using flourescent, or compact fluorescent bulbs. They use a lot less power.

At a lower power factor... watts = 0.4 x volts x amps, for some undercounter fluorescents.

Nick

Sure... 0.4 in the case above.

Yes.

You am wrong. W = PFxVxA.

Nick

Ignorance and belligerance are an unfortunate combination.

No... In the case above, the real power is 15W = 0.4x120Vx0.3125A, but "w=va" = 120Vx0.3125A = 37.5 VA overestimates it by a factor of 2.5.

Yes.

That's a good idea, in most cases.

Nick

QUESTION about this "power factor":

(it has been***so long*** since I understood any
of that stuff, that I've forgotten all but a
few words describing it.)

With DC, the pf is 1.0?

With AC, I'm not sure what it is.

Of course there's the "rms" stuff, trying to get an average value of a sine-wave.

The pf, I recall from***ages*** (decades) ago, had something to
do with the voltage and current "waves" getting out of sync
with each other, due to a coil or a capacitor (one shifting
in one direction, one in the other).

I recall something about having to use trig to get the pf, maybe it was the sine or cosine of the degrees of lead or lag?

If so, then since those functions range between plus and minus one, then maybe the pf I dimly recall is the reciprocal of yours?

Anyway -- here's my question:

How do you get a substantial pf for a fluorscent (sp?) light? (Huge electric motor, I understand how.)

And a pf for an incandescent light, that would involve no phase shift at all?

Obvously, I could use some mental fill-in!

Thanks!

David

Yes.

With AC, it's the sum of instantaneous products of current and voltage over a cycle, divided by the product of their average values. It's 1.0 if current is proportional to voltage.

It's also the number you see when you push the PF button on a Kill-a-Watt meter. I did that with a "0.1A 120V" Little Giant fountain pump the size of a golfball and saw 0.42. The meter said the pump used 5 watts.

Nick

What's the phase shift of a switching power supply? :-)

Nick

Thanks, all!

Now, where (on net) can I learn more, esp about that paragraph above?

(Yeah, I can google, but I'm not quite sure for exactly what, and***maybe*** one of you already have
a good site in mind.)

David

....

Yes, I now recall that.

Indeed!

.... ....

The harmonics being like the terms in a (***very*** dimly recalled) fourier series
or transform that sum up to approximate the chopping-caused square or
impulse or whatever waveform it is?

As I asked in my other reply, do you know offhand any good sites that cover this stuff (with drawings, too)?

The above formula isn't very useful in this case.

So, what is?

Jeez, on the surface, that sounds like the***opposite***, maybe,
of what one would expect???

Maybe you could say some more?

(I knew way back then that going for a double-E would be way over my ability!)

Wow... Maybe you could say a bit more about that!

(This stuff is NOT simple!)

Thanks to all for a fascinating look into the dark.

David

#### Site Timeline

- posted on August 10, 2005, 2:50 am

Wattage devided by the voltage (120) is amps. Or just add up the wattage. 20 amps x 120v = 2400 watts. Lights are easy since the wattage is written on each bulb.

- posted on August 11, 2005, 3:02 pm

volts times amps = watts

So, you can say that the 110 volt circuit, times 20 amps = 2200 watts.

Hint: Try using flourescent, or compact fluorescent bulbs. They use a lot less power.

--

Christopher A. Young

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Christopher A. Young

Learn more about Jesus

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- posted on August 11, 2005, 11:46 am

At a lower power factor... watts = 0.4 x volts x amps, for some undercounter fluorescents.

Nick

- posted on August 12, 2005, 3:30 am

snipped-for-privacy@ece.villanova.edu wrote:

Do fluorescent lights have a power factor? would like to know. But, that said, if the undercounter light fixture say 15W, then that is how much is used regardless of power factor and w=va still applies. or am I wrong and the labels are incorrect.

Do fluorescent lights have a power factor? would like to know. But, that said, if the undercounter light fixture say 15W, then that is how much is used regardless of power factor and w=va still applies. or am I wrong and the labels are incorrect.

- posted on August 12, 2005, 6:09 am

Sure... 0.4 in the case above.

Yes.

You am wrong. W = PFxVxA.

Nick

- posted on August 13, 2005, 4:30 am

snipped-for-privacy@ece.villanova.edu wrote:

Did you go to embassy doublespeak school? If you did, congratulations in your studies. You certainly confused me. You answered the 2nd question yes and the third saying I am wrong. Those answers are inconsistent. The subject was totaling up the wattage or amps. If the fixtures says 15 watts and it has a power factor, the powerfactor is already applied. If you total it up with other appliances you don't apply the power factor again.

I think I'll just stick with totaling up the wattage of fixtures/appliances.

Did you go to embassy doublespeak school? If you did, congratulations in your studies. You certainly confused me. You answered the 2nd question yes and the third saying I am wrong. Those answers are inconsistent. The subject was totaling up the wattage or amps. If the fixtures says 15 watts and it has a power factor, the powerfactor is already applied. If you total it up with other appliances you don't apply the power factor again.

I think I'll just stick with totaling up the wattage of fixtures/appliances.

- posted on August 13, 2005, 3:41 am

Ignorance and belligerance are an unfortunate combination.

No... In the case above, the real power is 15W = 0.4x120Vx0.3125A, but "w=va" = 120Vx0.3125A = 37.5 VA overestimates it by a factor of 2.5.

Yes.

That's a good idea, in most cases.

Nick

- posted on September 11, 2005, 6:08 am

QUESTION about this "power factor":

(it has been

With DC, the pf is 1.0?

With AC, I'm not sure what it is.

Of course there's the "rms" stuff, trying to get an average value of a sine-wave.

The pf, I recall from

I recall something about having to use trig to get the pf, maybe it was the sine or cosine of the degrees of lead or lag?

If so, then since those functions range between plus and minus one, then maybe the pf I dimly recall is the reciprocal of yours?

Anyway -- here's my question:

How do you get a substantial pf for a fluorscent (sp?) light? (Huge electric motor, I understand how.)

And a pf for an incandescent light, that would involve no phase shift at all?

Obvously, I could use some mental fill-in!

Thanks!

David

- posted on September 11, 2005, 8:43 am

Yes.

With AC, it's the sum of instantaneous products of current and voltage over a cycle, divided by the product of their average values. It's 1.0 if current is proportional to voltage.

It's also the number you see when you push the PF button on a Kill-a-Watt meter. I did that with a "0.1A 120V" Little Giant fountain pump the size of a golfball and saw 0.42. The meter said the pump used 5 watts.

Nick

- posted on September 11, 2005, 7:58 am

"With AC, it's the sum of instantaneous products of current and voltage

over a cycle, divided by the product of their average values. It's 1.0 if current is proportional to voltage. "

David's memory is correct. For AC, its the cosine of the phase angle between the voltage and current. When there is zero phase shift, the power factor is 1.0

over a cycle, divided by the product of their average values. It's 1.0 if current is proportional to voltage. "

David's memory is correct. For AC, its the cosine of the phase angle between the voltage and current. When there is zero phase shift, the power factor is 1.0

- posted on September 11, 2005, 12:01 pm

What's the phase shift of a switching power supply? :-)

Nick

- posted on September 11, 2005, 3:39 pm

On Sun, 11 Sep 2005 03:58:33 -0700, trader4 wrote:

As Nick Pine eludes to, this isn't a good definition because it doesn't take into account the harmonic content of the waveforms. It works for purely sinusoidal voltage abd current though. Power factor is more appropriately the Power divided by RMS Volts * RMS Amps, or PF == P/VA. The power can always be found by averaging the instantaneous V*A over the cycle, as nicksanspam wrote above.

As Nick Pine eludes to, this isn't a good definition because it doesn't take into account the harmonic content of the waveforms. It works for purely sinusoidal voltage abd current though. Power factor is more appropriately the Power divided by RMS Volts * RMS Amps, or PF == P/VA. The power can always be found by averaging the instantaneous V*A over the cycle, as nicksanspam wrote above.

--

Keith

Keith

- posted on September 12, 2005, 9:13 am

Thanks, all!

Now, where (on net) can I learn more, esp about that paragraph above?

(Yeah, I can google, but I'm not quite sure for exactly what, and

David

- posted on September 11, 2005, 3:49 pm

On Sun, 11 Sep 2005 02:08:43 +0000, David Combs wrote:

PF is kinda meangless with DC, since P=VA.

Not average. When Volts is multiplied by amps (power) there is a squared term in there. RMS == Root Means Square (root of the mean square), so when you calculate power, the squared term is taken into account. Average voltage isn't all that useful; the average for sine wave is zero. ;-)

Or a non-linear load adding harmonic content to the waveforms.

Cosine of the angle between them, but that really only works for pure sine waves.

The fluorescent is a non-linear load. It chops the current waveform, adding harmonics to it. The above formula isn't very useful in this case.

A "huge electric motor" should have a PF very close to one if it's heavily loaded. If it's creating no mechanical work, then it will have a worse PF.

The phase shift will be small, but it will only conduct current during part of the cycle, chpping the current waveform. Since the current isn't being used during the entire waveform, the power uring this part of the cycle isn't useful.

PF is kinda meangless with DC, since P=VA.

Not average. When Volts is multiplied by amps (power) there is a squared term in there. RMS == Root Means Square (root of the mean square), so when you calculate power, the squared term is taken into account. Average voltage isn't all that useful; the average for sine wave is zero. ;-)

Or a non-linear load adding harmonic content to the waveforms.

Cosine of the angle between them, but that really only works for pure sine waves.

The fluorescent is a non-linear load. It chops the current waveform, adding harmonics to it. The above formula isn't very useful in this case.

A "huge electric motor" should have a PF very close to one if it's heavily loaded. If it's creating no mechanical work, then it will have a worse PF.

The phase shift will be small, but it will only conduct current during part of the cycle, chpping the current waveform. Since the current isn't being used during the entire waveform, the power uring this part of the cycle isn't useful.

--

Keith

Keith

- posted on September 12, 2005, 9:30 am

....

Yes, I now recall that.

Indeed!

.... ....

The harmonics being like the terms in a (

As I asked in my other reply, do you know offhand any good sites that cover this stuff (with drawings, too)?

The above formula isn't very useful in this case.

So, what is?

Jeez, on the surface, that sounds like the

Maybe you could say some more?

(I knew way back then that going for a double-E would be way over my ability!)

Wow... Maybe you could say a bit more about that!

(This stuff is NOT simple!)

Thanks to all for a fascinating look into the dark.

David

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