Power factor and VA vs Watts

[Smart $]

As RMS values.
Now think about what RMS means.

No thought necessary. I know what RMS means (pun intended ).

From the current and voltage given by Rattus, about 0.707VAr

Now you are definitely comparing apples to oranges

Regarding both quoted statements, using RMS def's and equations, the values for S, P, and Q are all absolute averages, not their average effect in the circuit.

 

[Besoeker]

Bes, must be a man of few words because this is the first time I have heard him explain his position, and this is a no-brainer. BTW, I said essentially the same thing a few posts back.

Is this what you are agreeing with?

If you need the passage of time to work out the the real and reactive components you can't plot instantaneous values.

 

[Besoeker]

RMS def's and equations, the values for S, P, and Q are all absolute averages.

Absolute and average seem like mutually exclusive terms.

 

[Cold Fusion]

Absolute and average seem like mutually exclusive terms.

I didn't get that either. But that would only be one of many things I did not get.:roll:

cf

 

[gar]

100505-1431 EST

Try average of the absolute value.

.

 

[rattus]

Is this what you are agreeing with?

If you need the passage of time to work out the the real and reactive components you can't plot instantaneous values.

No, I don't agree with that statement because a plot involves a series of instantaneous values over time. If you plot these values over one period, you should be able to work out the nature and values of the equivalent circuit.

Now from a single value of vi, you might see p(t) < 0 which could indicate a reactive component, or it could indicate negative real power. Although you could, I think you would not plot a single point.

 

[Smart $]

Absolute and average seem like mutually exclusive terms.

I didn't get that either. But that would only be one of many things I did not get.:roll:

cf

100505-1431 EST

Try average of the absolute value.

.

Its definitely not the average of absolute values...

As I see it, the absolute average of a sinusoidal curve on a cartesian plot is the value equal one half the y-axial range. Where y = cos(x), it would be equal to [(y_max-y_min)/2], or [1-(-1)]/2 = 1, whereas the average would be zero, and the avaerage of the absolute values would be 0.636618, where the curve has an max' amplitude of 1 (speaking of a VAr curve originallly centered on zero with a range of 2).

Another way to look at it is to offset the entire curve (all values) so the lowest amplitude equals zero and then average the offset instantaneous values: y = avg(cos(x) +1), where +1 represents the offset, but it's not really part of a solution equation.

Is there some other name for this? The closest I can fathom is the modulus of the complex waveform... but I'm don't believe calling it the complex modulus (or just the modulus, for short) is the appropriate terminology either.

 

[Smart $]

... Although you could, I think you would not plot a single point.

Technically, you would not be able to see it

Though we generally represent it as a visible spec.

 

[rattus]

Technically, you would not be able to see it

Though we generally represent it as a visible spec.

The point is that the point in question is an infinitesimal point in the center of the visible point. Get the point?

 

[Smart $]

The point is that the point in question is an infinitesimal point in the center of the visible point. Get the point?

A blind man could see your point

 

[Cold Fusion]

I went to post this in the new thread, but realized it actually belonged here.

For all of the militant, repetative posting insisting that instantaneous power (what you call the "vi product") has an "offset", it just does not. No matter how you doll it up, that "offset" appears to be a phase angle between the instananeous values calculated from v(t) and i(t) - and that is a problem. But it turns out to be a rather minor problem.

Here's my translation:
You (plural - but not "all-ya'll") decree there is a phase shift between the sinusoids V and I. Then you use the phase angle to calculate instantaneous i(t1) from the value v(t1). Now because you have decreed there is a phase shift, then you say the instantaneous power has a phase angle.

Now why is that concept patently useless?
Because of the circular reasoning. First you decreed there is a phase shift between V and I. Using that phase shift, you calculate an instantaneous power and decree that it is into a reactive load because of the phase shift. Uhh... okay, you want to start with a phase shifted V - I and then calculate the load is reactive. Wow, now that is some real engineering reasoning - or just patently useless. I'll let you choose.

cf

 

[rattus]

I went to post this in the new thread, but realized it actually belonged here.

For all of the militant, repetative posting insisting that instantaneous power (what you call the "vi product") has an "offset", it just does not. No matter how you doll it up, that "offset" appears to be a phase angle between the instananeous values calculated from v(t) and i(t) - and that is a problem. But it turns out to be a rather minor problem.

Here's my translation:
You (plural - but not "all-ya'll") decree there is a phase shift between the sinusoids V and I. Then you use the phase angle to calculate instantaneous i(t1) from the value v(t1). Now because you have decreed there is a phase shift, then you say the instantaneous power has a phase angle.

Now why is that concept patently useless?
Because of the circular reasoning. First you decreed there is a phase shift between V and I. Using that phase shift, you calculate an instantaneous power and decree that it is into a reactive load because of the phase shift. Uhh... okay, you want to start with a phase shifted V - I and then calculate the load is reactive. Wow, now that is some real engineering reasoning - or just patently useless. I'll let you choose.

cf

cf, just what do you call the lead or lag between v(t) and i(t), or maybe you are saying there is zero lead or lag?

I have provided a reference which demonstrates the use of a phase angle with i(t). Do you not believe that reference?

 

[Smart $]

I went to post this in the new thread, but realized it actually belonged here.

For all of the militant, repetative posting insisting that instantaneous power (what you call the "vi product") has an "offset", it just does not. No matter how you doll it up, that "offset" appears to be a phase angle between the instananeous values calculated from v(t) and i(t) - and that is a problem. But it turns out to be a rather minor problem.

Here's my translation:
You (plural - but not "all-ya'll") decree there is a phase shift between the sinusoids V and I. Then you use the phase angle to calculate instantaneous i(t1) from the value v(t1). Now because you have decreed there is a phase shift, then you say the instantaneous power has a phase angle.

Now why is that concept patently useless?
Because of the circular reasoning. First you decreed there is a phase shift between V and I. Using that phase shift, you calculate an instantaneous power and decree that it is into a reactive load because of the phase shift. Uhh... okay, you want to start with a phase shifted V - I and then calculate the load is reactive. Wow, now that is some real engineering reasoning - or just patently useless. I'll let you choose.

cf

For a cartesian plot of power, the offset is y-axial, not x-axial... and that simply means the waveform (i.e. or its mean value) is not centered on the x axis. No one is saying or calculating a "phase shift" in the mathematical description of power. The phase shift "theta" is only used to calculate i(t) respective of v(t). This is necessary to compensate for the manner in which we conventionally quantify the two continuous measures of AC voltage and current. If we had actual correlated voltage and current measures through time, we would not have to use any phase shift at all in the calculations There is no circular reasoning going on that I can fathom.

 

[weressl]

I went to post this in the new thread, but realized it actually belonged here.

For all of the militant, repetative posting insisting that instantaneous power (what you call the "vi product") has an "offset", it just does not. No matter how you doll it up, that "offset" appears to be a phase angle between the instananeous values calculated from v(t) and i(t) - and that is a problem. But it turns out to be a rather minor problem.

Here's my translation:
You (plural - but not "all-ya'll") decree there is a phase shift between the sinusoids V and I. Then you use the phase angle to calculate instantaneous i(t1) from the value v(t1). Now because you have decreed there is a phase shift, then you say the instantaneous power has a phase angle.

Now why is that concept patently useless?
Because of the circular reasoning. First you decreed there is a phase shift between V and I. Using that phase shift, you calculate an instantaneous power and decree that it is into a reactive load because of the phase shift. Uhh... okay, you want to start with a phase shifted V - I and then calculate the load is reactive. Wow, now that is some real engineering reasoning - or just patently useless. I'll let you choose.

cf

Could it be just simply stated that the phase shift represent time?

It is representative of the time difference between the I and V maximum values.

 

[weressl]

cf, just what do you call the lead or lag between v(t) and i(t), or maybe you are saying there is zero lead or lag?

I have provided a reference which demonstrates the use of a phase angle with i(t). Do you not believe that reference?

The lead or lag is representative of TIME difference and therefore not INSTANTANEOUS. Instantaneous means without time. So if you were capable of measuring the V and I in the same instant there is no lead or lag to consider, the measurement itself would give you the actual value. Instantaneous is a theorethical value and can not be directly measured because EVERY measurement involves time averaging. At 60Hz the actual instantaneous voltage and current goes from 0 to max in 1/240 of a second, max to 0 in the next 1/240 sec. 0 to -max in the next 1/240 second and so on, and 'instantaneous' measuring device would not be able to 'follow' the values, display it and you be able to read it.

With analog meters this was not a problem as the magnetic force would of needed to move a mass and the acceleration takes time. (Thanks Isac).

 

[rattus]

The lead or lag is representative of TIME difference and therefore not INSTANTANEOUS. Instantaneous means without time. So if you were capable of measuring the V and I in the same instant there is no lead or lag to consider, the measurement itself would give you the actual value. Instantaneous is a theorethical value and can not be directly measured because EVERY measurement involves time averaging. At 60Hz the actual instantaneous voltage and current goes from 0 to max in 1/240 of a second, max to 0 in the next 1/240 sec. 0 to -max in the next 1/240 second and so on, and 'instantaneous' measuring device would not be able to 'follow' the values, display it and you be able to read it.

With analog meters this was not a problem as the magnetic force would of needed to move a mass and the acceleration takes time. (Thanks Isac).

Now Laszlo, we know that it takes time to take a sample, but if we can do it in a microsecond say, that is for all intents and purposes instantaneous if we are measuring a 60hz AC quantity. Let's not get too theoretical.

Of course lead or lag can be expressed in time, however we express the trig arguments in radians--wt that is. And, wt is just a number, no units at all. We do not write sin(t)!


Now consider the waves,

v(t) = sin(wt) = 0 at t = 0
i(t) = sin(wt + pi/4) = sqrt(2)/2 at t= 0

Clearly, i(t) leads v(t) by pi/4 just like I @ pi/4 leads V @ 0 by pi/4 radians.

 

[Cold Fusion]

cf, just what do you call the lead or lag between v(t) and i(t), or maybe you are saying there is zero lead or lag? ...

I don't know what you are asking for. I didn't say anything about or refer to "lead or lag".

...I have provided a reference which demonstrates the use of a phase angle with i(t). Do you not believe that reference?

This is an example of the reptative, insistent postings. Saying this louder and more often does not make it any closer to truth or usefulness. As already discussed and answered, this reference you quote is a 1950's out-of-print, unavailable text. I believe I have no clue as to the context.

cf

 

[Cold Fusion]

...Now consider the waves,

v(t) = sin(wt) = 0 at t = 0
i(t) = sin(wt + pi/4) = sqrt(2)/2 at t= 0

Clearly, i(t) leads v(t) by pi/4 just like I @ pi/4 leads V @ 0 by pi/4 radians. (emphasis added by cf)

Let's go make two circuit measurments at t=0, v(t0) = 0, i(t0)= .707. From those two measurements it clearly is not clear that i(t) leads v(t).

Rather, i(t) leads v(t) because: (drum roll)

You decreed it did (see the emphasis in your quote)- not because of any calculation you did.

You want "i(t) leads v(t)", it is okay with me. However, that does not impart any additional information other than: Thou hath decreed ...

cf

 

[Smart $]

Let's go make two circuit measurments at t=0, v(t0) = 0, i(t0)= .707. From those two measurements it clearly is not clear that i(t) leads v(t).

Rather, i(t) leads v(t) because: (drum roll)

You decreed it did (see the emphasis in your quote)- not because of any calculation you did.

You want "i(t) leads v(t)", it is okay with me. However, that does not impart any additional information other than: Thou hath decreed ...

cf

Taking a measure of v and i at time t is pointless in power engineering, for p does not exist where Δt=0. Yes, you can call it the vi product. And you can assign the unit measure volt-ampere to it. However, by implicit definition, you cannot call it power (p).

Instantaneous measures (note plural) is not restricted to only the measures at a single point in time (an instant). It is usually a record of measures over a period of time... usually one complete cycle or multiples thereof. As for current leading or lagging the voltage, we ascertain the relationship from our data set. We are not multiplying voltage value at one instant with current value of another. We are just adjusting the time references... synchronizing waveforms.

Anyway, I have no idea who even brought up the idea of limiting ourselves to instantaneous measures of a single point in time, but that is simply not the case. I've never even run across anyone (other than here) that would confine their defintion of instantaneous to a single point in time when discussing power engineering. Additionally, we basically use the term "instantaneous" to distinguish between DC-equivalent, single-value measures from measures that continually varying through expiration of time. Summarily, doing so is within the bounds of conventional concepts and calculations.

 

[Cold Fusion]

Quote from smart:
Taking a measure of v and i at time t is pointless in power engineering,
Generally, Yes -cf

for p does not exist where Δt=0.
That's the part that is just not true. p is a rate. And by definition a rate is an instantaneous measure - unless you want to use RMS values (for V and I) and get an average power over one period.-cf

Yes, you can call it the vi product. And you can assign the unit measure volt-ampere to it. However, by implicit definition, you cannot call it power (p).
I'm not going to use Noetic Science definitions. I'm going to stick with IEEE 100 concepts. Yes, it is power - cf

Anyway, I have no idea who even brought up the idea of limiting ourselves to instantaneous measures of a single point in time, but that is simply not the case.
As I recall, that was you - remember back to your insistent discussions of impedance varying from zero to infinity depending on what part of the V-I waveform the cycle was in. - cf

I've never even run across anyone (other than here) that would confine their defintion of instantaneous to a single point in time when discussing power engineering.
This one completely baffles me. Yes it is rare that one uses concepts on instantaneous power in power engineering. However there is no definition of "instantaneous" other than being confined to a single point in time. As Bes has said over and over, (paraphrased), "that's what instantaneous means - at a single point in time." -cf

Additionally, we basically use the term "instantaneous" to distinguish between DC-equivalent, single-value measures from measures that continually varying through expiration of time. Summarily, doing so is within the bounds of conventional concepts and calculations.
Huh? -cf