Instantaneous 3-phase power:

[rattus]

Re: Instantaneous 3-phase power:

Crossman,

An instantaneous value refer to a value at some point in time. This point is not a millisecond wide, not a microsecond, not a picosecond. It is infinitetesimally small.

It does not matter how many phases we are considering, this point in time is measured relative to some arbitrary starting time, t-zero.

There can be no ambiguity when discussing time.

 

[steve66]

Re: Instantaneous 3-phase power:

rattus:

I don't think you have read a word i have posted.

You posted;

If Vp = 78V, then dv/dt = 0 and

i(t) = (0)*C = 0, not 78/Xc

dv/dt=0??

Since when did we start talking about DC But no amount of reason will convince you that you are missing some understanding of what you are talking about.

Editied because my smiley face didn't work

Steve

[ January 25, 2005, 09:15 PM: Message edited by: steve66 ]

 

[rattus]

Re: Instantaneous 3-phase power:

Steve,

Yes, I have read your posts, and I am trying to set you straight.

At the peak of a sinusoidal voltage, dv/dt changes from positive to negative or vice-versa. At that instant, dv/dt = 0. This is fundamental. You must understand that fact and understand elementary calculus to understand my example.

As for understanding, I proved my case. Did you?

Why don't you give us an example from a textbook and cite the source? You have not done that. Until you do that, your arguments are nothing more than your misunderstandings.

Now, you can convert time domain voltages and currents to RMS and convert the reactive components to reactances, but you cannot mix them as my example shows.

Just because v(t) is sinusoidal does not mean you can use reactances to compute i(t).

[ January 25, 2005, 11:54 PM: Message edited by: rattus ]

 

[crossman]

Re: Instantaneous 3-phase power:

There can be no ambiguity when discussing time.

Rattus, I agree, about time, there is no ambiguity thre. What I was missing is that the GIVEN information of "3 phase power" defines the voltage and current at every instant in time and then a lot of calculations and assumptions and facts can be used in the proof.

My mistake for not realizing that obvious fact, I guess I realized it actually but let some bad thinking of mine get in the way.

 

[crossman]

Re: Instantaneous 3-phase power:

But I do have a question:

Is it possible for voltage and current to be, say, 37 degrees out of phase at a single infinitismally small instant?

And there is what has been snagging me all along. I am going to answer this question with NO.

Current can flow only one of two ways in a wire. It is either going the same way that the source voltage is pushing it, or it is going backward from the direction the source would have it go.

When voltage and current are both positive or both negative, the power produced is positive. When voltage and current are of opposite polarities, the power is negative.

Power Factor is based on a combination of all the positive powers and negative powers over time. It is not an instantaneous quantity. I think this holds true for reactance and VARs also.

 

[rattus]

Re: Instantaneous 3-phase power:

Crossman,

Yes, it is not only possible for the phase angle to be 37 deg. for an instant, but it would be 37 deg. at all times. If you know the phase angle, you can compute the magnitude and sign of the current at any point in time.

Think of it this way. The phase angle can be considered as a time delay. In 3-phase work, that delay is 120 deg. which is 1/3 of a cycle. At 60 Hz the voltage sinusoids are separated by about 5.5 milliseconds. In a balanced system, the currents would also be separated by the same amoutn plus they could be delayed from the voltage waves if reactance is part of the balanced load.

If the loads are reactive and unbalanced, the currents would be separated by different amounts of time.

Remember this:

In making calculations we consider the impedances to be constant. Then if we apply AC voltages of constant amplitude and phase, the amplitudes and phases of the currents will be constant at all times.

The reason we speak of "instants" is that AC values change in the smallest time increment you can imagine. So, instantaneous voltages are defined at some infintely small slice of time.

[ January 26, 2005, 04:18 PM: Message edited by: rattus ]

 

[rattus]

Re: Instantaneous 3-phase power:

Crossman,

To be strictly correct, energy--not power--flows back and forth in a reactive load. You are right, PF is a constant determined by the lead or lag of the load current, say 37 deg. It is not used in instantaneous equations.

Likewise, reactance is a constant at a given frequency and cannot be used in instantaneous equations. Let me provide another example:

Consider a voltage,

v(t) = Vp(sin(wt))

applied across an inductor, L.

Now consider the instant when wt = 90 deg.

v(t) = Vp

but since the current lags by 90 deg.,

i(t) = 0, zilch, nada, nothing, 0^2,

THIS IS FUNDAMENTAL AND CANNOT BE DISPUTED!

but,

Vp/wL > 0 at this juncture.

Both expressions cannot be right.

 

[rattus]

Re: Instantaneous 3-phase power:

edited out duplicate post

[ January 27, 2005, 12:06 PM: Message edited by: rattus ]

 

[crossman]

Re: Instantaneous 3-phase power:

(1) The current through an inductor is zero when the source voltage is peaking at 90 degrees.

(2) When viewed over a period of time, the source voltage and current are 90 degrees out of phase in an inductor.

But you can't use statement (2) to prove statment (1) unless you prove statement (2) first.

Is there anything about inductive reactance that mathematically proves the voltage and current are 90 degrees out of phase?

Not that I know of. So we take it on faith that there is a 90 degree phase angle?

It can be said that inductive reactance is actually a vector with a 90 degree phase angle. Again, we are accepting that 90 degree part in faith.

Does inductive reactance itself cause a 90 degree phase shift? Does inductive reactance actually even exist in reality? Or is it just a convenient mathematical tool that we use to deal with the math of sine waves over time?

In actuality, reactance produces a CEMF which has a negative peak when the source voltage has a positive peak. And it is these two voltages interacting at the instant that the peak is reached that causes the current to be zero. It has nothing to do with impedance or reactance measured in ohms. Ohms used with anything but resistance is an artificial mathematical construct that does not exist in reality and CANNOT be used in any instantaneous calculations.

So for this reason, using anything other than voltage and current in proofs that calculate power at distinct instants of time is incorrect.

[ January 26, 2005, 11:31 PM: Message edited by: crossman ]

 

[crossman]

Re: Instantaneous 3-phase power:

Darnit rattus! You still have me spinning.

Your most recent post confirms some of what I have been saying all along.... You can't use RMS values in the instantaneous calculations.

I certainly enjoyed your original proof and learned from it.

Where I disagree with you is here:

p(t) = (v1^2 + v2^2 + v3^2)/R

I don't think this works for every instant in time. You are relying on the assumption that True Power is consumed only in a resistor. Yes, I have seen that statement and that definition in the textbooks before. Yes, I use it all the time in AC theory class. Yes, I even teach that. Yes, it is hugely convenient. But it is also a huge simplification designed to make the math easier when dealing with RMS values over time.

Perhaps our difference on this boils down to a differing view of the definition of a Watt. But your view of the "watt" is an RMS watt. And we can't use RMS quantities in equations involving each instant in time, which I think you have stated yourself.

One reason I say your use of a Watt is RMS is that your Watt is actually a vector that can be used in conjunction with the RMS vectors of VARs and RMS Volt-Amps. I am talking about instantaneous watts. You are talking about RMS watts. There is a dictionary definition of the Watt that is based on energy. This is the real scientific definition of a Watt. And it certainly takes real energy to charge a capacitor. And this real energy can most sertainly be expressed in Watts.

Another notion: In the first quadrant of a sine wave voltage, the voltage is actually a rising DC voltage. And if we connect a capacitor to this rising DC voltage, we will get a decreasing current that falls to zero once we stop increasing the source voltage. Is there a phase angle between voltage and current in this DC circuit? Does current x voltage in this DC circuit give you watts? Remember, it is all DC in the first quadrant. Can a DC circuit where both the source and the current are positive have a power factor less than 100%?

Another thought: Considering our rising DC source voltage and decreasing current through the capacitor... I could take a variable resistor and cause the exact same current to flow. It could start very high, I would start increasing the resistance, and could cause the current to fall to zero.

Two exact graphs of voltage and current... one in a resistor, the other in a capacitor. Exactly the same voltage and current characteristics.

1) Could the source tell any difference?

2)Did the source expend a different type of energy to make the current flow?

3) Does the fact that in one case the current is converted to heat, and in the other case it is converted to an electrostatic field make a difference?

4) Is one using watts but the other isn't?

5) If the power is not the same in each case, what formula would account for the difference?

Very good questions.

[ January 27, 2005, 01:59 AM: Message edited by: crossman ]

 

[rattus]

Re: Instantaneous 3-phase power:

Crossman, you are making this too complicated. Reactive phase leads and lags have been proven mathematically a gazillion times and need no further proof.

Reactances and impedances are expressed in Ohms, but only a pure resistance can be used in time domain analysis. My example proves that.

The CEMF appears in inductors, but not in capacitors which are also reactive. But rather than explain the zero current this way, you should understand that,

e = -L(di/dt)

Then the CEMF, which is always equal and opposite to the applied voltage in a pure inductance, reaches a peak when |di/dt| is maximum, and this is at zero crossing. That is the proper explanation. You can prove this to yourself by taking the derivatives of a sine or cosine wave at zero crossing.


.

 

[crossman]

Re: Instantaneous 3-phase power:

Holy cow! Check this out!

Here is the equation for the impedance of an AC sine wave LCR circuit at any instant in time!

(Using the same notation that Charlie B used, this is cut and paste from his given identities except the last term, Z(t) is mine)

________________________________________________
Terms
VaM Maximum value of line-to-neutral voltage on Phase A.
VM Maximum value of line-to-neutral voltage. It is assumed that the loads are balanced, so that VM = VaM

Va(t) Voltage from Phase A to neutral, as a function of time ?t.?

IaM Maximum value of phase current on Phase A.
IM Maximum value of phase current It is assumed that the loads are balanced, so that IM = IaM

Ia(t) Current on Phase A, as a function of time ?t.?

w Frequency in radians per second. Equal to 2pi times the frequency in cycles per second.

u Phase angle between voltage and current on Phase A

Let Va(t) = VM cos (wt)

Let Ia(t) = IM cos (wt - u)

Za(t) = Impedance of phase A to neutral as a function of time.
_________________________________________________

Za(t) = VM cos(wt)/IM cos(wt-u)

Now if some of you math gurus want to convert that to another form that has some meaning, please do.

Would the VM/IM reduce to ZM and just exactly what would ZM mean? Would that be the Impedance of A-phase divided by .707? (impedance of A-phase based on the vector addition methods and ohm's law depending on whether a series or parallel LCR load) I think it is!

so let's say you want to do Charlie B's proof but I only give you the voltage and impedance. You can still do his proof!

And I think this shows that you cannot ignore the reactive components in the instantaneous values.

 

[crossman]

Re: Instantaneous 3-phase power:

Originally posted by rattus:

The CEMF appears in inductors, but not in capacitors which are also reactive.
.

Capacitors do have a CEMF. The plates are charged opposite of the source and this is definitely a counter voltage that is 180 degrees out of phase with the source voltage.

I have never seen the mathematical proofs of the out-of-phase conditions of inductors and capacitors. I am assuming that said mathematical proofs do not involve the RMS values of reactance measured in Ohms.

I do assume they involve the dt stuff that you posted.

I don't think I am making it too hard. I am thinking that I am getting down to the nitty-gritty of it all. And I like it! I'm talking about the nitty gritty that all the previous geniuses had to fathom before they discovered all the non-real mathematical constructs that make the math so easy for us today. Effective voltage is one example. It is a mathematical entity to make the math easy. It doesn't really exist. The problems arise when we start thinking that these things are real... and we start using them in formulas that cannot accomodate them.

 

[rattus]

Re: Instantaneous 3-phase power:

Crossman,

Where I disagree with you is here:

p(t) = (v1^2 + v2^2 + v3^2)/R

I don't think this works for every instant in time. You are relying on the assumption that True Power is consumed only in a resistor. Yes, I have seen that statement and that definition in the textbooks before. Yes, I use it all the time in AC theory class. Yes, I even teach that. Yes, it is hugely convenient. But it is also a huge simplification designed to make the math easier when dealing with RMS values over time.

*Crossman, it is not an assumption, it is a fact. Energy is converted to heat in a resistor, energy is converted to mechanical energy in a motor, energy is merely stored in a capacitor or inductor.

You are beating a dead horse!

In a resistor:

p(t) = v(t)i(t) = i(t)^2*R = v(t)^2/R

If that is not true, then my proof is invalid.

p(t) (in this case) tells us the rate of energy conversion at any instant. This is real power.

Add an inductor in parallel and energy flows back and forth between the source and the inductor. This "inductive power" if you will, can be expressed in Watts, and is easily computed, but I don't want to do that before noon on a rainy day.

This point is that this inductive power has no effect on the resistive power!

Think about two identical AC sources, one connected to an inductor, the other to a resistor. No energy is converted in the inductor. Energy is converted in the resistor. Now tie the sources in parallel and tie the loads in parallel. The same currents flow in the loads. No change, same, same, ditto, same as before.

Now a Watt is a Watt is a Watt! We are talking instantaneous power here, but we most often are speaking of average power. The Watt is the established unit of power and has no phase angle.

The Watt is not a unit of energy, the Joule is the established unit of energy.

Pavg = Vrms*Irms*PF, no angles here.

"RMS Power" is not defined. RMS applies to voltages and currents. Do not bring them into a discussion of functions of time.

You cannot arbitrarily challenge the established definitions

Pavg = Vrms*Irms*PF, no angles here.

And the counter emf term applies only to inductors and motors, not caps. The mechanism is quite different in a cap.

Also, if you do not understand the "dt stuff", you cannot fully understand AC circuits. But, that doesn't make you a bad person.

And, one more time, impedance is NOT a function of time!

I will let someone else comment on the rest of your post.

[ January 27, 2005, 01:21 PM: Message edited by: rattus ]

 

[steve66]

Re: Instantaneous 3-phase power:

As for understanding, I proved my case. Did you?

The only thing you have proved is that you are very bad at math.


You posted:

Likewise, reactance is a constant at a given frequency and cannot be used in instantaneous equations. Let me provide another example:

Consider a voltage,

v(t) = Vp(sin(wt))

applied across an inductor, L.

Now consider the instant when wt = 90 deg.

v(t) = Vp

but since the current lags by 90 deg.,

i(t) = 0, zilch, nada, nothing, 0^2,

THIS IS FUNDAMENTAL AND CANNOT BE DISPUTED!

but,

Vp/wL > 0 at this juncture.

Both expressions cannot be right.

If you didn't just throw magnitudes in for phasors your last equation would be:

Vp@90/wL@90 which does equal zero

Thanks for proving my point.

 

[rattus]

Re: Instantaneous 3-phase power:

Steve,

One more time, you are injecting vectors into instantaneous equations. This indicates that you have your wires crossed.

You call them phasors which are the polar representations of vector quantities, that is:

Vp[(sin(wt) + jcos(wt)] = Vp*exp(jwt)

Complex numbers are not used in time domain analysis, so phasors have no application here.

v(t) which equals Vp at 90 degrees has no phase angle. wL has no phase angle.

Furthermore,

Vp@90/|Z|@90 = (Vp/|Z|)@0 which is not zero and is not a function of time! This indicates that you only the foggiest notion of what you are talking about.

Crossman has finally realized that you cannot use impedance in instantaneous equations. How many times must you be hit with a 2x4 before you understand that?

Instantaneous values are functions of time, period. They can be positive, negative, or zero. Phase angles are embedded in the arguments of the trig functions, period. RMS values and vector analysis are an outgrowth of time domain analysis. They are related but separate methodologies.

You still have not cited a reference for any of your examples. The challenge still stands. Until you do that, I must assume that you can't because there are no such references.

[ January 27, 2005, 02:19 PM: Message edited by: rattus ]

 

[charlie b]

Re: Instantaneous 3-phase power:

I have been out of town for two days. I return to find this thread half way on its way to an infinite discussion. I can?t wait for the other half.

It?s going to take me a while to digest the new stuff, and decide if I agree or disagree with anything to a high enough degree to add any more comments myself. But I will address one statement (it was stated twice in my absence, in two different ways):

Originally posted by rattus: Charlie B., I don't recall using Z in my proof. I would not know how.

Charlie, I did not use V^2/Z in my proof, maybe in an example, but not in the proof. Just pretend that the concept of RMS, reactance, etc. have not been invented. One can still solve problems in the time domain with calculus and no mention of X or Z.

Yes you did. You did it in your third post on the first page of this thread:

We can represent the load as a parallel combination of resistance and reactance, then all we have to do is prove that the sum of the squares, (normalized voltage):

4) vn^2 = v1^2 + v2^2 +v3^2 = K

Your use of the phrase ?normalized voltage? is the point at which you declared no interest in actual values of the maximum voltage, and therefore no interest in the actual constant value that power will attain. You set peak voltage at a value of ?1.? I have already agreed that this is a valid approach. That is, it you can show that a constant value is attained, and if you don?t care what that constant is, you can safely ignore the actual values of maximum voltage.

But your equation #4 has the value of Z built into it. You don?t see the ?Z?? Look again. Look only at the left and right sides of the first equation mark. Ignore, for the moment, that you are also setting both sides equal to a constant ?K.? You can safely ignore that, because you are declaring it to be the desired end point of your proof. You don?t use that second equation in your proof, but rather are driving your proof towards that equation.

Now, when does the square of one voltage (expressed as a function of time) equal the sum of the squares of three other voltages (each having been expressed as a function of time)? The answer is, ?never.? More to the point, when does power equal the square of any voltage? That answer is also, ?never.? Power can be the square of a voltage that is then divided by an impedance. If you don't care about constants, and wish to "normalize" things to eliminate the constants, that is OK. But the constants are still there. They may be numerically set to "1," but they keep their units of measure, and they are still there.

Your equation #4, as written, is meaningless. It can only have meaning when you write out the several equations that precede it, then point out the constants that are present in these ?background? equations, state that you don?t need to know their actual values, bring them all over to the left side of equation #4, and set them collectively equal to ?1,? so that they don?t obscure the simplicity of the remainder of your proof.

Here are the equations that you left out, the ones that explicitly show the ?Z? value that you did not explicitly show, the ones you would have needed to show, to make your proof ?rigorous?:
</font>

• <font size="2" face="Verdana, Helvetica, sans-serif">Power, as a function of time, in Phase A, can be expressed as pa(t) = va(t) x ia(t)</font>

<font size="2" face="Verdana, Helvetica, sans-serif"></font>

• <font size="2" face="Verdana, Helvetica, sans-serif">Power, as a function of time, in Phase B, can be expressed as pb(t) = vb(t) x ib(t)</font>

<font size="2" face="Verdana, Helvetica, sans-serif"></font>

• <font size="2" face="Verdana, Helvetica, sans-serif">Power, as a function of time, in Phase C, can be expressed as pc(t) = vc(t) x ic(t)</font>

<font size="2" face="Verdana, Helvetica, sans-serif"></font>

• <font size="2" face="Verdana, Helvetica, sans-serif">Total Power, as a function of time, can be expressed as pt(t) = pa(t) + pb(t)+pc(t).</font>

<font size="2" face="Verdana, Helvetica, sans-serif"></font>

• <font size="2" face="Verdana, Helvetica, sans-serif">Current, as a function of time, in Phase A, can be expressed as ia(t) = va(t)/Z</font>

<font size="2" face="Verdana, Helvetica, sans-serif"></font>

• <font size="2" face="Verdana, Helvetica, sans-serif">Current, as a function of time, in Phase B, can be expressed as ib(t) = vb(t)/Z</font>

<font size="2" face="Verdana, Helvetica, sans-serif"></font>

• <font size="2" face="Verdana, Helvetica, sans-serif">Current, as a function of time, in Phase C, can be expressed as ic(t) = vc(t)/Z</font>

<font size="2" face="Verdana, Helvetica, sans-serif">Therefore,
</font>

• <font size="2" face="Verdana, Helvetica, sans-serif">Total Power, as a function of time, can be expressed as pt(t) = va(t) times va(t)/Z + vb(t) times vb(t)/Z +vc(t) times vc(t)/Z.</font>

<font size="2" face="Verdana, Helvetica, sans-serif"></font>

• <font size="2" face="Verdana, Helvetica, sans-serif">Thus, pt(t) = va(t)^2/Z + vb(t)^2/Z +vc(t)^2/Z.</font>

<font size="2" face="Verdana, Helvetica, sans-serif">Multiplying both sides of this equation by Z, and you get
</font>

• <font size="2" face="Verdana, Helvetica, sans-serif">Total power, having been multiplied by a constant I no longer care about, equals
  va(t)^2 + vb(t)^2 +vc(t)^2.</font>

<font size="2" face="Verdana, Helvetica, sans-serif">This is the essence of your equation #4. The rest of your proof goes on to show that this expression is a constant.

So, yes, you did use ?Z? in your equation #4.

Charlie B. insists that I did not ignore the reactive current. My proof has no equation for current anywhere, resistive or otherwise. That is about is ignorant as one can be!

Charlie B. insists that the values of X, Z, etc, were swallowed up by my method. No Charlie, they were never considered.

Please note that I have just proven you to have been wrong. You did use current, although you failed to show us the equations that contain current, and that you converted into equations that only show voltage. You jumped in with ?power equals voltage squared divided by impedance,? and that inherently converts an expression containing current into an equation that only shows voltage. And you did swallow up the value of ?Z,? in the manner I have just described.

Also please note that if you choose now to assert that your equation #4 was not derived in this manner, then I would reassert that your equation #4 can have no other meaning. Your proof would then fall apart completely.

Having finished that evaluation of your methods of mathematical analysis, I will now state that I take great exception to your tone and to your personal remark. I think you need to reconsider your method of expressing yourself. We are professionals here, and your statements to myself and to several others were condescending and clearly unprofessional.

 

[steve66]

Re: Instantaneous 3-phase power:

(Vp/|Z|)@0 = (Vp/|Z|) sin 0 = 0

I suppose you will now try to argue that 0 not = 0. :roll:

...which is not a function of time

Of course it is not a function of time. You wanted to evaluate the function of time at one point to disprove it. Let me refresh your memory - you said:

Now consider the instant when wt = 90 deg.

You also posted:

You still have not cited a reference for any of your examples.

Any very basic 1st semester circuits textbook. For example: "Electrical Engineering REference Manual", John A. Camara, P.E., 6th edition, page 27-11:

Ohms Law: Ohms law for AC circuits with linear circuit elements is similar to Ohms law for DC circuits (footnote #13). The difference is that all the terms are represented as phasors. (footnote #14). V = IZ

Edit: Just for the record, I posted this before I saw Charlie's reply.

[ January 27, 2005, 05:24 PM: Message edited by: steve66 ]

 

[crossman]

Re: Instantaneous 3-phase power:

It would be terrific if we could all get together sometime and spend a few hours on this. I know I would learn a great deal from all of you.

charlie b, I am directing this toward you.

pt(t) = va(t)^2/Z + vb(t)^2/Z +vc(t)^2/Z

So if the proof that Rattus gave was applied to a parallel RL circuit, then the proof did indeed include the reactive branch current along with the resistive branch current?

But I am just not seeing how a constant impedance can be used to solve for instantaneous values.

How would you define impedance mathematically?

(Stop me when I am wrong)

1) It could be defined using pythagorean's theorem in a series LCR "impedance triangle" from the resistance and reactance vectors. This would be an RMS definition and is a vector with constant angle and magnitude.

2) Could it be defined as voltage divided by current? Yes, and in fact, in parallel LCR circuits, we do this. We use pythagorean's theorem to combine the RMS vector branch currents, then we divide the RMS source voltage by the calculated total current to get the impedance. This is still all using RMS values.

3) How would you define impedance for an instant in time? Would it still be voltage divided by current? More precisely, the instantaneous voltage divided by the instantaneous current?

If I told you we had an instantaneous source voltage of 66 volts and the instantaneous current was 2 amps then the impedance at that instant is 33 ohms?

[ January 27, 2005, 05:09 PM: Message edited by: crossman ]

 

[rattus]

Re: Instantaneous 3-phase power:

Charlie B.,

First: let me apologize if I offended anyone. I did not intend to be offensive, only emphatic.

Second: the term vn(t)^2 is an equivalent voltage, that is the sum of the squares. Bearing this in mind,

p(t) = vn(t)^2/R,

then since R is constant, we need to prove only that vn(t)^2 is constant to prove the point which I am trying to do as concisely as possible.

Third: The load presents an impedance, but since I choose to make the load a parallel combination, an admittance if you will, I can ignore the reactive portion. That is the real power (resistive) is given by,

p(t) = v(t)i(t) = i(t)^2*R = v(t)^2/R

Reactive voltage current relationships are described with calculus, not reactance or impedance.

Fourth: In time domain analysis, voltages and currents are scalar functions of time. Phase angles are embedded in the trigonometric arguments. There is no accommodation for reactance or impedance.

Fifth: If you disagree with the previous statement, I would respectfully request that you cite a reference that proves me wrong.

Rattus

[ January 27, 2005, 10:47 PM: Message edited by: rattus ]