Not sure what you are trying to say, but you must agree that the phase lead or lag between voltage and current must be taken into account in the p(t) equation? And, you must agree that the presence of reactive components affects the value of p(t) at any instant?
Power factor and VA vs Watts
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Not sure what you are trying to say, but you must agree that the phase lead or lag between voltage and current must be taken into account in the p(t) equation? And, you must agree that the presence of reactive components affects the value of p(t) at any instant?
ggunn
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I think where some of the confusion is coming in is in the difference between a static and a dynamic analysis. In a dynamic analysis, every point on a curve in a timeline has two properties: value and slope, i.e., where it is and where it is going. In a static analysis, there is only one property of a point: value. You can look at it either way, and neither is right or wrong.
In a static analysis, phase angle is not a factor; there is only what the current is and what the voltage is at that point in time. If at the point in time you have chosen the current is zero and the voltage is not (or vice versa), it looks as if the impedance of the system has gone infinite (or gone to zero). From a statics POV, that is valid. But be careful; there are points in time where the current is positive and the voltage is negative (or vice versa); in those cases the impedance appears to have gone negative, and that is just as valid.
Of course, applying a static analysis to a dynamic system presents its own set of problems, and one might argue that although it's valid within its own frame of reference, it may not be useful.
I'll hang up and listen... :grin:
Cold Fusion
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Start by reading your posts 490 and 497.
Following your reasoning, if p is a scalar, then so is i. A scalar has no phase angle.
In your words, "No other conclusion is possible."
No, I am not going to find you a text reference on basic math to prove this conclusion.
cf
You are confusing phasors and vectors. Vectors must include a directional angle. But phase angles do not indicate direction; they indicate a time lead or lag. But, this argument is pointless, so let's cut to the heart of the matter.
Do you claim that the expressions,
v(t) = Vm*sin(wt) and
i(t) = Im*sin(wt + phi)
are invalid and cannot be used in the equation for p(t)??
That is what I am hearing, and I am not deaf yet.
The point is that the power at any instant is just the product of the current and the voltage at that instant in time. It doesn't matter how they got to those values or where they are going.
The truth of your first statement is obvious. Now the fine details matter to me and others who wish to understand this in depth. Since you didn't clearly answer my question, I will answer it myself:
Clearly both the reactive and resistive components affect the value of i(t), and since
p(t) = v(t)i(t),
the answer must be yes.
wasasparky
Senior Member
I think it was changed to 500.
Evidently NOT
I always thought so but you, and others, seemed to be finding disagreement with it in earlier posts.
Anyway, on the basis of your agreement with my point, you can surely see that the graphical data I presented in posts #141 and 156 are correct while that posted in #186 cannot be.
It is that simple from your perspective only because you choose to limit the issue to such. However, even from my perspective it is still simple, and not as difficult a leap as you make it out to be. It's not really much different than determining how total power is divvied up among combination loads.
As I see it, regarding the latter mentioned (mine), 3 of 5 sinusoisal waveforms depict the same information as the former two (yours).
All three are pictured below in the order you mentioned them...
I see your two graphs depicting different loads. The first appears to be an inductive load while the second appears to be an ideal capacitor. Now suppose we took these two loads and connected them in parallel (to the same supply, as it appears voltage is the same in both graphs). Wouldn't the total current and power (using your terminology) change? What about current and power to each load? Can the current and power to each load be distinguished from the power to the combined load?
Cold Fusion
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No
However I am heartened to see you use the notation v(t) instead of just "v". I was sure I knew what you meant, but for a purist such as your self, it made it look like you could be confusing vectors/phasors with time domain equations.
Note to smart: "time domain" has a definition.
cf
Cold Fusion
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I think you already know all of this. You're the one that said you have the same educational background I do. So I don't know what you are looking for.
Tell you what - I'll give non-rigorous answers, you fill in the gaps. Then you reciprocate with an answer to post 477. If you don't, then by default y.... :roll:
So here are your answers:
Here is the model and calculation for inductors: (this is all in vectors, not time based equations)
First some terminology definitions:
Let this symbol, "<" mean "phase angle"
Bold (V)are vectors (phasors for rattus if you like). Un-bolded (V) are magnitudes.
The model for an inductor is a resistance, R, in series with an inductance, L. We will drive the circuit with a sinusoid, V, or V<0 (magnitude V(rms), phase angle 0), frequency w
Impedance is Z = R + jwL
convert Impedance to polar:
Z = (R^2 + (wL)^2)^.5 < (arctan(wL/R))
Z = (R^2 + (wL)^2)^.5 and <phase angle = β = (arctan(wL/R)
I = V/Z = (V <0)/(Z < β)
I = V/Z <(- β)
Power (average) dissipated in the resistor is:
P(res) = I^2 * R = (V/Z)^2 *R
And if you wish to assign a phase angle to P, use <(-2β), referenced to V
Since you are insistent that you understand all of this classical engineering education nonsense (paraphrasing your words), I figure your only reason to ask is to find some nits to pick. Being the nice guy that I am, I put fifteen minutes into this strictly for your enjoyment. So enjoy.
cf
Hameedulla-Ekhlas
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As I see it, regarding the latter mentioned (mine), 3 of 5 sinusoisal waveforms depict the same information as the former two
Smart$:
In your graph, you have shown, Real power, Reactive Power, Average Power, and Apparent power. Where is the intantaneous power and I think there is a mistake between instantaneous power and apparent power. Apparent power is magnitude. Isnt it?
I didn't. There is no leap to be made.
Sorry, no intent of nits to pick... but as a consolation, consider me entertainedI think you already know all of this. You're the one that said you have the same educational background I do. So I don't know what you are looking for.
Tell you what - I'll give non-rigorous answers, you fill in the gaps. Then you reciprocate with an answer to post 477. If you don't, then by default y.... :roll:
So here are your answers:
Here is the model and calculation for inductors: (this is all in vectors, not time based equations)
First some terminology definitions:
Let this symbol, "<" mean "phase angle"
Bold (V)are vectors (phasors for rattus if you like). Un-bolded (V) are magnitudes.
The model for an inductor is a resistance, R, in series with an inductance, L. We will drive the circuit with a sinusoid, V, or V<0 (magnitude V(rms), phase angle 0), frequency w
Impedance is Z = R + jwL
convert Impedance to polar:
Z = (R^2 + (wL)^2)^.5 < (arctan(wL/R))
Z = (R^2 + (wL)^2)^.5 and <phase angle = β = (arctan(wL/R)
I = V/Z = (V <0)/(Z < β)
I = V/Z <(- β)
Power (average) dissipated in the resistor is:
P(res) = I^2 * R = (V/Z)^2 *R
And if you wish to assign a phase angle to P, use <(-2β), referenced to V
Since you are insistent that you understand all of this classical engineering education nonsense (paraphrasing your words), I figure your only reason to ask is to find some nits to pick. Being the nice guy that I am, I put fifteen minutes into this strictly for your enjoyment. So enjoy.
cf
That said, I don't see what gaps you are referring to. Your equations are explanatory enough, of the math model, from my perspective. Yet the math model doesn't really answer the questions posed with any certainty. All it indicates is that a real inductor may exhibit in part some resistance. It does not model the ideal inductor not having a resistive property. Ideal reactive components have perfect conductive and insulative (aka dielectric) properties, i.e. zero and infinite resistance respectively. Because real components' properties are less than perfect, resistance is encountered when energized. The resistance property of these components converts electrical energy to heat.
As for post #477...
...What fallacy? Your post stems from a discussion between rattus and Besoeker "discussing" phase shift of the the current, and somehow you brought into the discussion the topic of instantaneous power having, or is it not having a phase angle. From there you conclude a solution is being sought.
I was of the impression this discussion was to have a better understanding of the concepts involved. In short, theoretical. Hell, we can't even agree on the theoretical aspects of the topic and you're asking for a real world example of using instantaneous power to obtain a solution. I've never stated the purpose was to obtain some solution to a real world problem. I've actually stated several times, in several ways, my disposition is not to propose, develop, suggest, et cetera a new math model, but rather to provide a graphical representation of how power factor, watts, and volt-amperes (and other associated terminology) are related.
Just as you are educated enough, as are many others participating and lurking, to visual the relationships through equations, there are probably more lurkers (or there was) that have no idea what the math is all about... yet they may understand a pictorial representation, or perhaps grasp a little better understanding. The math just scares them off and they learn nothing.
And FWIW, I renounce the idea we have the same educational background. I am knowledgeable enough to know that even if we went to the same school and obtained the same degree there would be variations in the courses taken, knowledge retained, continuing education, and so on.
Apparent power is instantaneous power but in absolute form, i.e all values are measured using the lowest amplitude value as the baseline (zero). The single value used for apparent power in calculations is a magnitude equal to the average absolute instantaneous power. For a sinusoidal waveform, that'd be one-half the range of instantaneous power: p(t)ppk/2, or p(t)pk – p(t)avg.
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Well I can certainly give you credit for being persistent
And on that note, you also consistently ignore portions of my replies... or in their entirety. :roll::roll::roll:
Instantaneous power is just power. In Watts.
One Volt and one Amp at the same instant is one Watt.
It doesn't matter if the circuit is pure resistance or has reactive components. It is still just 1W.
Even if it was a DC circuit, it would still be 1W.
Suppose you take a very short segment of the waveform I posted in #156. Say 1us at 45deg. At that point you have 1V, 1A, 1W.
It doesn't matter how it got there. It is 1W. Nothing apparent about it.
Correct for both.
The current and voltage to each load is as previously posted.
The combined load:
ggunn
PE (Electrical), NABCEP certified
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- Austin, TX, USA
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- Consulting Electrical Engineer - Photovoltaic Systems
You guys are really arguing apples and oranges. What you say is correct from a statics standpoint, but what the other guys are talking about is from a dynamics perspective, and never the twain shall meet unless you are talking about a DC circuit.
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