No, you missed my answer - right above.
Yes, I know that's what you stated. But you're wrong. You're having a very hard time grasping this simple concept: the 80% rule DOES NOT APPLY in this situation.
20A x 120V / 100W = 24 bulbs."TURTLE" wrote
I would, but I am afraid that I might hit a light bulb.
Steve
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
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.
This is Turtle.
I see you would load up a break to it's max. amps and if the people in the house happen to turn on all 24 light bulbs. It will blow the breaker in about 2 or 3 hours and then call you to ask why did my break throw. You maybe able to do this by NEC but I will not load anything up to the max. what so ever.
Now what I did up above by just putting 20 light bulbs on a 20 amp breaker is Nothing wrong with doing it. You had said it was wrong well i tell you it is nothing wrong with it if i wanted to do it. Now you can explain any wrong as NEC goes please explain it to me. Let me explain to you what right and wrong is. Right is you can do it. Wrong is you Can't do it. Now take these two words and explain wrong as you say above here.
TURTLE
This is Turtle.
I would just go ahead and Punt on First 10 .
TURTLE
Please read more carefully, Turtle. I never said that putting only 20 100W bulbs on a 20A breaker was wrong -- I said you were wrong to claim that 19 bulbs was the maximum permitted because of the 80% rule.
I've explained it several times already, but you're not paying attention. Let's try again. The 80% rule applies to continuous loads. Residential lighting is not a "continuous load" as defined in the NEC. Therefore the 80% rule does not apply to residential lighting circuits.
Yep, and by that definition putting twenty-four 100W light bulbs on a 20A 120V circuit is right. So is twenty bulbs. Or five bulbs. Or one.
There's nothing wrong with loading a circuit to less than its capacity. What's wrong is your understanding of the capacity of a 20A circuit when used for residential lighting, and your understanding of the 80% rule.
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
"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
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
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.
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.
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.
.... ....
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
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