Power factor and VA vs Watts

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jghrist said:
Actually, I should have said for a circuit with resistance an reactance. You will find that I??R peaks at the same time as current, not at the same time as voltage as on Post #229. If you insist on treating real power as a variable with time, then I think plotting I??R would be the thing to plot.

Personally, I consider real power to be the average of instantaneous power, so it really has no meaning except when considered for a particular period of time like a full cycle. I think plotting it as a function of time has no utility.
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Allow me to pick a very small nit. The term "average" here is a bit of a misnomer. You can't really average a time varying quantity over time; calculating an average involves adding a bunch of quantitative values and then dividing by the number of terms, but in the case of a continuous function there are an infinite number of points. That leads to infinity terms in both the divisor and dividend. What we have to do is calculate the integral of the function over the period between t1 and t2.

As you were... :D
 
rattus said:
Seems to me that at any instant when the source is delivering energy, that is positive work. When the source is absorbing energy, that is negative work. Therefore it seems appropriate to use the WATT as the unit of power.
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I have no problem with this when considering the complete system, i.e. inclusive of the POCO gennies, for example. The issue of separation is for at the load end only.

Real and reactive powers are added vectorially (although they are not vectors) in the power triangle. Doesn't seem right to add them algebraically. And too, the usage of the VA and VAR is with steady state (average) values. I see no benefit in separating out reactive power in the p(t) formula. That being said, you may use any units you wish if it makes things clearer for you.
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Using steady-state values is fine for actual calculations... this is why everyone is saying there is no advantage. From this perspective, I agree. However, plotting of steady-state values does not provide a true depiction of occurence. We refer to AC voltage and current by their RMS values. Is it truly representitive to depict these as a horizontal line on a graph?
 
jghrist said:
Actually, I should have said for a circuit with resistance an reactance. You will find that I??R peaks at the same time as current, not at the same time as voltage as on Post #229. If you insist on treating real power as a variable with time, then I think plotting I??R would be the thing to plot.
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The problem with doing that is it does not take into account impedance varies with time. If one factor's in z as the resistance it equals v/i, and will shift the plotted curve to align with vi. I?R becomes i?z... i?(v/i)... i(v).

Personally, I consider real power to be the average of instantaneous power, so it really has no meaning except when considered for a particular period of time like a full cycle. I think plotting it as a function of time has no utility.
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See my reply to rattus a couple posts back.
 
ggunn said:
Allow me to pick a very small nit. The term "average" here is a bit of a misnomer. You can't really average a time varying quantity over time; calculating an average involves adding a bunch of quantitative values and then dividing by the number of terms, but in the case of a continuous function there are an infinite number of points. That leads to infinity terms in both the divisor and dividend. What we have to do is calculate the integral of the function over the period between t1 and t2.

As you were... :D
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Moot point.
The infinitesimal calculus makes the leap from the number of discrete terms to the continuous infinite number of points.
It is still average.
:grin:
 
Besoeker said:
But it isn't. And you can't seem to grasp that.
One Volt and one Amp occurring at the same instant is one Watt.
Just that. One Watt. Real or reactive is totally meaningless in this context.
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Whether it is one watt or not is irrelevant for the discussion. You are too concerned with terminology. The terminology I'm using is used by conventional electrical power engineering. In other words, if VA and VAR's are truly are a measure of watts, so be it. This still does not mean the instantaneous values of "watts" cannot be separated into "real" and "reactive" portions. This is what you are not grasping.
 
Besoeker said:
Moot point.
The infinitesimal calculus makes the leap from the number of discrete terms to the continuous infinite number of points.
It is still average.
:grin:
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I said it was a very small nit, didn't I? :grin:
 
Smart $ said:
The problem with doing that is it does not take into account impedance varies with time. If one factor's in z as the resistance it equals v/i, and will shift the plotted curve to align with vi. I?R becomes i?z... i?(v/i)... i(v).
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Impedance varies with time? Real power is what is dissipated in the resistor, I??R.
 
jghrist said:
Impedance varies with time?
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Hmmm... I technically misstated that. Though impedance is a steady-state value, its equivalent R value varies with time where reactive components are involved. For example, apply a DC voltage to a [discharged] capacitor. Initially its resistive-equivalent ohmage is near zero and through time increases exponentially to the point where the resistive-equivalent ohmage is near infinity. Applying an AC voltage to a capacitor just changes the rate at which the resistive-equivalent ohmage changes. An inductor functions just the opposite.

Real power is what is dissipated in the resistor, I??R.
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My earlier reply was a misunderstanding in your premise, since you mentioned a resistive and reactive circuit. In order to plot I?R (not I?Z), you'd need to separate out the resistive portion of the current. The resistive portion of the current aligns with the voltage. Separating out the resistive portion of the current is at the root of this discussion.
 
Smart $ said:
... We refer to AC voltage and current by their RMS values. Is it truly representitive to depict these as a horizontal line on a graph?
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Yes. "RMS" has a specific definition.

Vrms = (1/T ∫ (v(t)^2) dt)^.5 , integrated over the interval of one period (T). And that definition is a number - invariant over the one period.

For a non-varing load - that is a flat line.

cf
 
Smart $ said:
.. This still does not mean the instantaneous values of "watts" cannot be separated into "real" and "reactive" portions. This is what you are not grasping.
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Bes is grasping this just fine. The issue is again a definition - this time "instantaneous"

For the sinusouids we deal with:

The instantaneous voltage at any specific time is v(t) = A sin(ωt)
And that is a number. It has no real or reactive portion.

cf
 
For what its worth:

For what its worth:

Besoeker said:
It can't.
........
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Beg to differ here.

p = vi = [VmIm/2][cos(theta) - cos(2wt)cos(theta) + sin(2wt)sin(theta)]

The sum of the first two terms yields the instantaneous real power, and the thrid term yields the instantaneous reactive power.
 
rattus said:
Beg to differ here.

p = vi = [VmIm/2][cos(theta) - cos(2wt)cos(theta) + sin(2wt)sin(theta)]

The sum of the first two terms yields the instantaneous real power, and the thrid term yields the instantaneous reactive power.
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There is no phase displacement between instantaneous values.
 
Smart $ said:
Hmmm... I technically misstated that. Though impedance is a steady-state value, its equivalent R value varies with time where reactive components are involved. For example, apply a DC voltage to a [discharged] capacitor. Initially its resistive-equivalent ohmage is near zero and through time increases exponentially to the point where the resistive-equivalent ohmage is near infinity. Applying an AC voltage to a capacitor just changes the rate at which the resistive-equivalent ohmage changes. An inductor functions just the opposite.
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Now you have me confused. I was about to admit that indeed Z can be thought of to vary with time because X = L?di/dt. I'm afraid that I don't follow your argument about equivalent R value and "resistance-equivalent ohmage."
My earlier reply was a misunderstanding in your premise, since you mentioned a resistive and reactive circuit. In order to plot I?R (not I?Z), you'd need to separate out the resistive portion of the current. The resistive portion of the current aligns with the voltage. Separating out the resistive portion of the current is at the root of this discussion.
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I can't agree. The heat dissipated in a resistor depends on the total current flowing through the resistor. It gets a little tricky talking about heat dissipation because this is I??R?t. It takes time to produce heat. I guess I??R would be the rate of heat production or some such thing. To think of this in a physical manner and keep the concept of real power being a function of time, I guess you could consider that the rate of heat production falls to zero when the electrons vibrating back and forth change direction so they are not moving. This is really stretching things, however. Heat is produced by the vibration of the electrons. All the more reason for me to consider the concept of sinusoidally varying real power as bogus.
 
Smart $ said:
Hmmm... I technically misstated that. Though impedance is a steady-state value, its equivalent R value varies with time where reactive components are involved. For example, apply a DC voltage to a [discharged] capacitor. Initially its resistive-equivalent ohmage is near zero and through time increases exponentially to the point where the resistive-equivalent ohmage is near infinity. Applying an AC voltage to a capacitor just changes the rate at which the resistive-equivalent ohmage changes. An inductor functions just the opposite.
Click to expand...

Smart, "resistive-equivalent ohmage" is a meaningless term which has no place in circuit analysis!
 
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