Power factor and VA vs Watts

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gar said:
100406-2031 EST

The basic definition of power factor is
PF = POWER/VA
where VA is based on RMS measurement of V and A. Although I have never seen a qualification that the voltage be a sine wave this may be a necessary useful assumption. There may be some logic to a requirement that the voltage be a sine wave. In most practical applications this may be approximately correct. On the other hand currents can be quite distorted.
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Do you mean a sine wave as opposed to other AC waveforms, or a sine wave vs. DC? For DC, none of this is relevant; a capacitor is an open and an inductor is a short.
 
steve66 said:
:-? It sounds like you are saying that the phase angle between the current and voltage doesn't depend on the impedance angle.

The phase angle between the voltage and current is determined by the impedance angle by ohms law: I=V/Z.

So for example, if the impedance angle is 45 degrees, we can see there will be a 45 degree shift between the voltage and current.

Steve
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I could have phrased that better, what I meant was phase angle and imoedance angle are 2 different things, different meansing, different applications.
 
100407-0905 EST

ggunn:

At steady state DC the power factor is always 1.0 . Yes I was referring to distorted or non-sinusoidal AC waveforms. There are many loads today that have peaked current waveforms, and looking at the AC line voltage distortion that exists today it is quite clear that this peaked type of load is significant.

.
 
gar said:
100407-0905 EST

ggunn:

At steady state DC the power factor is always 1.0 . Yes I was referring to distorted or non-sinusoidal AC waveforms. There are many loads today that have peaked current waveforms, and looking at the AC line voltage distortion that exists today it is quite clear that this peaked type of load is significant.

.
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Then yes, implicit in this discussion is that the voltage and current waveforms are sine waves. For AC waveforms that are not pure sine waves, the analysis gets more complex and must include a mathematical description of the waveform. A Fourier transform, maybe?
 
steve66 said:
:-? It sounds like you are saying that the phase angle between the current and voltage doesn't depend on the impedance angle.

The phase angle between the voltage and current is determined by the impedance angle by ohms law: I=V/Z.

So for example, if the impedance angle is 45 degrees, we can see there will be a 45 degree shift between the voltage and current.

Steve
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Ohm's law is for instantaneous values. The phases and the I=V/Z (where Z is complex number) is based on Ohm's law for sinusoidal waveforms. When harmonics get into the equation, it becomes correct for every harmony separately. This means I1=V1/Z1 (1 for first harmony), I2=V2/Z2 etc., but not necessarily I RMS = V RMS / Z.
 
PowerQualityDoctor said:
Ohm's law is for instantaneous values. The phases and the I=V/Z (where Z is complex number) is based on Ohm's law for sinusoidal waveforms. When harmonics get into the equation, it becomes correct for every harmony separately. This means I1=V1/Z1 (1 for first harmony), I2=V2/Z2 etc., but not necessarily I RMS = V RMS / Z.
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Yes, I think all that is a given, and I agree, except maybe for what the 1st sentance seems to imply. But I think you were just trying to emphasis that we have to be careful to stick with sine waves if we aren't dealing with instantenous values.

Steve
 
I think the clearest answer that I got was from Steve66. Does anyone disagree with what he said? I'll try not to ask anymore "dumb" questions...:roll:
 
zog said:
The angle being refered to has nothing to do with the relationship between voltage and current, that is a differtent topic and is called the phase angle. The angle I was refering to is called the impedance angle and is shown on a power triangle which show the relationship between true, reactive, and apparant power. (Watts being the x axis, vars being the y axis and VA being the hypotonuse).

The angle of this ?power triangle? graphically indicates the ratio between the amount of dissipated (or consumed) power and the amount of absorbed/returned power. It also happens to be the same angle as that of the circuit's impedance in polar form. When expressed as a fraction, this ratio between true power and apparent power is called the power factor for this circuit. Because true power and apparent power form the adjacent and hypotenuse sides of a right triangle, respectively, the power factor ratio is also equal to the cosine of that phase angle.
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You sure about this?

The angle between the voltage and current is the same as the angle real power and appearent power when it is expressed in a vectorial representation.
 
Hameedulla-Ekhlas said:
...and I think it is enough.
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Perhaps...

...but what I see missing from practically all explanations is the fact that reactance impedance is not steady state as it appears to be. That is, when doing the basic Xrcl component equations, one is left with an impression that impedance is constant in the instantaneous domain. It is not.

Attached is a simple pdf I whipped up with two pages giving a visual representation of what happens to impedance, with a sinusoidal voltage applied across an inductor's terminals. The first page is with an ideal inductor, and the second with a real inductor, such as a typical wire wound coil.
 
Smart $ said:
Perhaps...

...but what I see missing from practically all explanations is the fact that reactance impedance is not steady state as it appears to be. That is, when doing the basic Xrcl component equations, one is left with an impression that impedance is constant in the instantaneous domain. It is not.

Attached is a simple pdf I whipped up with two pages giving a visual representation of what happens to impedance, with a sinusoidal voltage applied across an inductor's terminals. The first page is with an ideal inductor, and the second with a real inductor, such as a typical wire wound coil.
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Have you checked this link wich I had given in page number 3
see it again please

http://www.rficdesign.com/power-factor-2
 
cyriousn said:
I think the clearest answer that I got was from Steve66. Does anyone disagree with what he said?
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I do not. Pretty basic, and straight to the point. All the calculation stuff isn't necessary to understand the concept

I'll try not to ask anymore "dumb" questions...:roll:
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Okay, but you have to promise to ask the "totally ignorant, other than dumb" questions first, before going on to the "not so dumb" questions... so I can keep up ;):D
 
Smart $ said:
Perhaps...

...but what I see missing from practically all explanations is the fact that reactance impedance is not steady state as it appears to be. That is, when doing the basic Xrcl component equations, one is left with an impression that impedance is constant in the instantaneous domain. It is not. ...
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That took me a while But maybe I understand your point.

Here's my translation: You are looking at the instantaneous current through, and voltage across, an inductor. You ar defining the inductor impedance, Z, using V/I, where V and I are instantaeous quanities. So, at the points in time where the voltage goes to Zero, there is still a finite current. So Z --> infinity. Is that your thinking?

cf
 
Cold Fusion said:
That took me a while But maybe I understand your point.

Here's my translation: You are looking at the instantaneous current through, and voltage across, an inductor. You ar defining the inductor impedance, Z, using V/I, where V and I are instantaeous quanities. So, at the points in time where the voltage goes to Zero, there is still a finite current. So Z --> infinity. Is that your thinking?

cf
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Well on my end it is past tense... ;)

And, it is actually at the current's zero crossings that impedance goes off the chart (a fold in the whole space-time contiuum, because it is both positive and negative infinity at the same time :D)

In a purely passive RCL circuit, Z is always 0 when V is 0, but I may be a non-zero value
 
jumper said:
This not a "dumb" question, I have a copyright to ask "dumb" questions and I ain't gonna allow you to use it.:grin:

I like these threads a lot and I am glad you started it. If we get Mivey in on this, it will really get fun. Where is he?
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I have called him and 2moro or the day after 2moro he will come.
 
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