Power factor and VA vs Watts

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Smart $

Esteemed Member
Location
Ohio
Multiply the voltage by the current and plot the result.
That gives you power.
Yes, that is the simple HOW.

But I was hoping to see Ham' come up with the "proofing" answer, since he went to such degree for proofing the plot of power (when pf=1) based on Pavg. The exercise, per se, was more a question of proofing the relationship of scalar and vector approaches using S, P, and Q (hint).
 

Smart $

Esteemed Member
Location
Ohio
There is nothing disguised about it, and it's called "parody". Look it up. :grin:
Okay. Here it is, from Merriam-Webster Collegiate? Dictionary
1 par?o?dy
Pronunciation: 'per-ə-dē, 'pa-rə-
Function: noun
Inflected Form: plural -dies
Etymology: Latin parodia, from Greek parōidia, from para- + aidein to singular ― more at ODE
Date: 1598

1 : a literary or musical work in which the style of an author or work is closely imitated for comic effect or in ridicule
2 : a feeble or ridiculous imitation
synonyms see CARICATURE
?pa?rod?ic \pə-'r?-dik, pa-\ adjective
?par?o?dis?tic \ˌper-ə-'dis-tik, 'pa-rə-\ adjective

2 parody
Function: transitive verb
Inflected Form: -died ; -dy?ing
Date: circa 1745

1 : to compose a parody on <parody a poem>
2 : to imitate in the manner of a parody
I'm not even going to bother with elaborating.

And that's "led", not "lead", mr. maven of the English language, or are you using a different meaning for "lead"? ;)
It just happens to be a simple typo, using the wrong tense... you know, "led" being the past and past part of the transistive verb "lead".

PS: You still have not defined the "RMS domain" for me; are you ever going to get around to that?
Considering all your heckling, what do you think?
 

Hameedulla-Ekhlas

Senior Member
Location
AFG
Yes, that is the simple HOW.

But I was hoping to see Ham' come up with the "proofing" answer, since he went to such degree for proofing the plot of power (when pf=1) based on Pavg. The exercise, per se, was more a question of proofing the relationship of scalar and vector approaches using S, P, and Q (hint).
I have not gone anywhere I am coming :grin:

by they way, I was just mentioned that 0i = 0v it means that it is only real instantanous power. If you take the integral of real instantenous power in a period of time, ofcourse you will get only real power without reactive part and the p.f will be equal to 1. I dont know why too much discussion going on this issue, try to come on conclusion instead of debate.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Considering all your heckling, what do you think?
Convenient. All I've been doing is trying to get you to substantiate your claims. You can't. This time it's because I'm heckling you, you poor thing. :grin:

But ain't the internet grand? To win an argument, you don't have to be correct or even have a modicum of a clue about the pseudoscientific mumbo-jumbo you are spouting, all you have to do is dodge questions and outlast your opponent.

OK, you win. It's my own fault for smacking the tarbaby (that's another literary reference, BTW, not a racial slur) back on page 1.

Insert last word here =>
 

Smart $

Esteemed Member
Location
Ohio
Convenient. All I've been doing is trying to get you to substantiate your claims. You can't. This time it's because I'm heckling you, you poor thing. :grin:

But ain't the internet grand? To win an argument, you don't have to be correct or even have a modicum of a clue about the pseudoscientific mumbo-jumbo you are spouting, all you have to do is dodge questions and outlast your opponent.

OK, you win. It's my own fault for smacking the tarbaby (that's another literary reference, BTW, not a racial slur) back on page 1.
From my perspective I've substantiated my claims... and it's not my problem if you fail to recognize such. I could get into the details, but that would just add fuel what you are errantly calling an argument. I could care less what you consider the outcome.
 

Smart $

Esteemed Member
Location
Ohio
When you install receptacles do you favor ground up or ground down? ;)
Being a moderator, you don't have to ask this question to close the thread :D

...but since you asked, I favor ground up.

Dare I ask, did I substantiate that enough for all to understand? ;)
 
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Smart $

Esteemed Member
Location
Ohio
:grin:

No, you provided nothing.

You never said why you like ground up. :grin:
Ohh! Darn... another typo (omission type this time) :roll:

Actually substantiation is three-fold:
  1. Nema configuration diagram depicts it as such.
  2. Experience has taught me extension cord plugs are better retained in the receptacle when subject to moderate pulling tension, especially in receptacles mounted at above knee height from the floor or it's equivalent.
  3. As a result of #2, grounding prongs are broken off less.
 

Besoeker

Senior Member
Location
UK
Yes... but the question as quoted was asked in the context of the discussion at that point, as explained.
OK.
In practical terms here's what you can do:
Take the voltage as v*sin(wt) the current as i*sin (wt-phi), phi being the displacement between voltage and current for lagging PF. Multiply them to get power as a function of time.
With just pencil and paper and a simple scientific calculator you can work out values and plot them on paper.
I generally use a spreadsheet to display waveforms simply to give a better presentation but the fundamental calculations are the same.

Here is 0.7 PF
 

Smart $

Esteemed Member
Location
Ohio
OK.
In practical terms here's what you can do:
Take the voltage as v*sin(wt) the current as i*sin (wt-phi), phi being the displacement between voltage and current for lagging PF. Multiply them to get power as a function of time.
With just pencil and paper and a simple scientific calculator you can work out values and plot them on paper.
I generally use a spreadsheet to display waveforms simply to give a better presentation but the fundamental calculations are the same.

Here is 0.7 PF
Got some D?j? Vu going on here :D

Anyway, you're still branching off from the context of my earlier question... but I'll go with the flow and try to direct it back...

You say your graph indicates a 0.7 PF. AFAICT, the green line indicates what p(t), which is apparent power in instantaneous form... correct?

If so, what is the real power? ...reactive power?
 

gar

Senior Member
100420-2238 EST

Besoeker:

I disagree with calling your green curve a plot of instantaneous power and therefore the use of the word power. What is plotted in green is the instantaneous volt-ampere curve.

The instantaneous power curve is K (1-cos 2wt) for a sine wave voltage curve, is never negative, and is only zero at the voltage zero crossings. This I discussed in posts way long ago.

The word power all by itself is the rate of doing work, the resistive component.

.
 

Besoeker

Senior Member
Location
UK
100420-2238 EST

Besoeker:

I disagree with calling your green curve a plot of instantaneous power and therefore the use of the word power. What is plotted in green is the instantaneous volt-ampere curve.
Power is instantaneous. A rate.


The instantaneous power curve is K (1-cos 2wt) for a sine wave voltage curve, is never negative, and is only zero at the voltage zero crossings. This I discussed in posts way long ago.
Your (1-cos 2wt) is applicable for just unity power factor for which specific case power is never negative. Introduce a phase shift and you then have this relationship:

2sinAsinB = cos(A-B) - cos(A+B)

This yields positive, negative and zero values.
 

gar

Senior Member
100421-1526 EST

Besoeker:

Power is the rate of doing work. It equals volt-amperes when the load is resistive. Reactive current is not doing any work and therefore does not contribute to power.

.
 
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