Power factor and VA vs Watts

[Hameedulla-Ekhlas]

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$:
Regarding to pf = 1 for instantaneous power is already stated by Fryze and I have mentioned many times 0v = 0i for you

"Fryze verified that the active power factor (f) reaches maximum ( f = 1 ) if and only if the instantaneous current is proportional to the instantaneous voltage, otherwise power factor ( f < 1 ). "

Please do not ask me, who is Fryze.

 

[Hameedulla-Ekhlas]

Got some D?j? Vu going on here

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?

no, wrong,
apparent power is the magnitude of complex power
we can call it instantaneous complex power which contains both.

 

[Hameedulla-Ekhlas]

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) (hint).

Ok, you need the graph for p.f = 1 see below

When the power factor is unity, the voltage and current are in phase and sin0=0. Hence, the reactive power is zero. In this case, the instantaneous power is never negative. This case is illustrated by the current,voltage and power waveforms in attached photo. the power curve never dips below the exis and there is no exchange of energy between the source and the circuit. At the other extreme, when the power factor is zero, the voltage and current are 90 degree out of phase and sin0 = 1. Now the reactive power is a maximum and the average power is zero. In this case, the instantaneous power is positive over half a cycle (of the voltage) and negative over the other half. All the energy delivered by the source over half a cycle is returned to the source by the circuit over the other half.

It is clear, then that the reactive power is a measure of the exchange energy between the source and the circuit without being used by the circuit. Although none of this exchanged energy is dissipated by or stored in the circuit and it is retured unused to the source, nevertheless it is temporarily made available to the circuit by the source.

 

[LarryFine]

Please do not ask me, who is Fryze.

That's the guy who first cut potatoes into strips and cooked them in oil.

grin: Sorry. I couldn't resist.)​

 

[Hameedulla-Ekhlas]

That's the guy who first cut potatoes into strips and cooked them in oil.

grin: Sorry. I couldn't resist.)​

yeah, very good guess: grin:

 

[Smart $]

It indicates power. Nothing apparent about it.
Just power.

Okay, if you're going to be that way about it.

Let me restate the questions...

How much of that power is realized energy (i.e. converted to another form of energy, or does work [otherwise known as real power])? ...and how much of that power is unrealized energy (i.e. is neither converted nor does work [otherwise known as reactive power])?

Keep in mind that I may not have worded these questions 100% correct... but hopefully you can get the gist of them.

 

[Smart $]

Smart$:
Regarding to pf = 1 for instantaneous power is already stated by Fryze and I have mentioned many times 0v = 0i for you

"Fryze verified that the active power factor (f) reaches maximum ( f = 1 ) if and only if the instantaneous current is proportional to the instantaneous voltage, otherwise power factor ( f < 1 ). "

Please do not ask me, who is Fryze.

no, wrong,
apparent power is the magnitude of complex power
we can call it instantaneous complex power which contains both.

I'm using conventional names. The so-called "real power" is the only term that represents power. It equals volts times amperes where pf=1. The point of the discussion is what does "real power" equal when pf<1.

The other two names—"apparent power" and "reactive power"—are misnomers. Power, is the time rate of work. For "reactive power", current flowing back and forth doing no work in the process is not power. Yes, it is stored and released energy... but no work equals zero power. Since the so-called "apparent power" is a complex magnitude of "real power" and "reactive power", then it too is not power.

You are correct in that it is complex

FWIW, the only time a plot of v, i, and |v|?|i| [aka hereinbefore p(t)] indicates power accurately—when pf <1—is at V_+pk, V=0, and V_-pk. For all other instances of t, "reactive power" is affecting the magnitude of p(t).

 

[cschmid]

If it will help you respect me more, I won't debate it. Doesn't mean I agree with what you are saying though You'll have to let me know where you stand on this

In the meantime, can you at least agree that in a reactive circuit there is energy which goes into the load, is stored temporarily, then returns from the load, and in intervals? ...and that energy is transferred in the form of electrical current? If yes to the immediately preceding question, does this action represent utilized energy (i.e. not the capacity or potential to do work, but actual work)?

I know better then this...yes and no :grin:

 

[cschmid]

I have to add I disagree with the authors' implications in their figure (below)...

The authors state, "During the time interval corresponding to area "A", the source is delivering energy to the load,whereas during the time interval corresponding to area "B", a percentage of this amount of energy is being returned back to the source."

While their statement is not a complete fallacy, it erroneously implies that during the time interval corresponding to area "B" is the only time during which energy is being returned to the source. This implication is not true. At best, it is indicative only of some energy being returned to the source and is of the time interval of the leading half of energy being returned.

Referring to my latest chart below, my area "A" corresponds with the authors area "B". Pease note the overall hatched area with the line color cyan. The instantaneous magnitudes in this area are depicted by the "reactive power" waveform (dashed cyan). The time interval for both my areas "A" and "B" is the period which energy is being returned to the source, while the time interval corresponding to both my areas "C" and "D" is the period which energy is being stored by the load. There is only an instant between periods.

my what program are you using it is sweat..

 

[Smart $]

... [aka hereinbefore p(t)] ...

... p(t).

Regarding the notation p(t), I am only using it as I believe it to have been used in this discussion. It is fine for a resistive circuit. However, I find using it to represent volts times amperes of a reactive circuit to be misleading. Since volts times amperes in a reactive circuit does not represent power (at least not accurately within the strict definition of power), I would prefer a different notation be used...

 

[Smart $]

my what program are you using it is sweat..

See Post #298 in this discussion.

As an additional note to what I wrote there, I have been using the Canvas program for over ten years and am quite adept with it. While a good portion of its usage is intuitive, many of the nuances take experience to realize their full potential.

 

[cschmid]

Smart$:

If you need it, i will send you the free link and you can download for free its pdf. You dont need it to buy it.

I will take the link..:grin: I am cheap

 

[cschmid]

..as it unfolds, all of a sudden Smart $ just picks up his marbles and leaves the playground.:grin:

alternative ending:

..."When I use a word," Humpty Dumpty said in a rather a scornful tone, "it means just what I choose it to mean ? neither more nor less."!!:grin:

I really thought it was Willy Wonka who stated that but what do I know. :grin:

 

[cschmid]

See Post #298 in this discussion.

As an additional note to what I wrote there, I have been using the Canvas program for over ten years and am quite adept with it. While a good portion of its usage is intuitive, many of the nuances take experience to realize their full potential.

I got to that post several minutes after I posted that and I thank you as you are very proficient with it. Now do not let me interrupt you gentlemen as i am enjoying the lesson.

 

[Besoeker]

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.

Yes, power is rate of doing work.
And it also follows that work or energy is power times time.

Take a capacitor for PFC. It is entirely reactive. But work is done.
Current leads the voltage as shown in post #141.
For the first quarter cycle both current and voltage are positive and the capacitor is being charged ending up with 0.5*C*V^2 Joules of energy. Now energy is power times time. So we have power flow from the supply to the capacitor during that quarter cycle. In the next quarter, current is reversed and power flow is also reversed.
The mean value of power is zero but power does flow even in this entirely reactive circuit.

The waveforms I put in post #334 show power as a function of time over one cycle for 0.7 pf lag. To ge t0.7 lag the circuit has to be inductive. The inductive element stores energy (0.5*L*i^2) and returns it to the supply over the course of a cycle. That's why the green curve has both positive and negative parts. Make the power factor lower and the positive parts get shorter and the negative parts get longer until at zero pf they are equal and average power is then zero.

 

[Besoeker]

Okay, if you're going to be that way about it.

Let me restate the questions...

How much of that power is realized energy (i.e. converted to another form of energy, or does work [otherwise known as real power])?

I'd like to be able to explain better but your question isn't logical.
You need first to understand that power isn't energy.

One Volt times one Amp occurring at the same instant is one Watt.
Just that. Nothing more.
Can we go from there?

 

[Smart $]

Yes, power is rate of doing work.
And it also follows that work or energy is power times time.

Take a capacitor for PFC. It is entirely reactive. But work is done.
Current leads the voltage as shown in post #141.
For the first quarter cycle both current and voltage are positive and the capacitor is being charged ending up with 0.5*C*V^2 Joules of energy. Now energy is power times time. So we have power flow from the supply to the capacitor during that quarter cycle. In the next quarter, current is reversed and power flow is also reversed.
The mean value of power is zero but power does flow even in this entirely reactive circuit.

The waveforms I put in post #334 show power as a function of time over one cycle for 0.7 pf lag. To ge t0.7 lag the circuit has to be inductive. The inductive element stores energy (0.5*L*i^2) and returns it to the supply over the course of a cycle. That's why the green curve has both positive and negative parts. Make the power factor lower and the positive parts get shorter and the negative parts get longer until at zero pf they are equal and average power is then zero.

You are using the definitions of work and power ambiguously.

The stricter (physics) definition of power is the rate at which work is performed or energy is converted. Under this definition, a capacitor neither performs work nor converts energy (aside from losses considered insignificant to this discussion). I believe you are using a much looser definition of power, essentially the rate at which energy is transferred. Perhaps appropriate in a less technical discussion, but I do not recognize this loose definition as here.

 

[Smart $]

I'd like to be able to explain better but your question isn't logical.
You need first to understand that power isn't energy.

I understand power is not energy (and vice-versa). See post #357

One Volt times one Amp occurring at the same instant is one Watt.
Just that. Nothing more.
Can we go from there?

I think this is where we are having a major snafu...

One volt times one ampere occuring at the same instant is one volt-ampere. Until we include such conditions or parameters which indicate work is performed or energy converted at that same instant, it cannot be one watt.

 

[ggunn]

I'd like to be able to explain better but your question isn't logical.
You need first to understand that power isn't energy.

One Volt times one Amp occurring at the same instant is one Watt.
Just that. Nothing more.
Can we go from there?

Good luck with your quest. :grin:

 

[weressl]

...all this from the guy that said math is the proof

Exactly. Or do you not understand how it supports my previous statement, hand-in-glove?.