Instantaneous 3-phase power:

steve66 · Jan 21, 2005

Re: Instantaneous 3-phase power:

Rattus:

Sorry, but in your post of your proof, I didn't see any equations with power in them. They all seem to only have voltages.

Then you posted:

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

You are using resistance, and ignoring any reactance and phase angle. But I will assume that you mean v1, v2, and v3 to be a function of time.

If I may suggest something that would send any mathmatician into an uproar: You could substitute R with Z where Z is a vector that has a magnitude and a angle. Now we are dividing phasors by vectors

Since Z isn't a function of time, we can replace it with a constant (vector). Now it is pretty obvious that if the sum of the squares of the voltages are constant, the function is constant.

I still agree with Charlie that this isn't a rigorous proof. I think it would have any mathmatician pounding their head on the table in disbelief. But I think it does point to the fact that the function is constant.

Steve

rattus · Jan 21, 2005

Re: Instantaneous 3-phase power:

Steve,

You need to study my proof more carefully.

1) We are proving that the instantaneous power, p(t), in a balanced 3-phase system is constant, therefore, the proof is presented in terms of instantaneous values, not RMS values.

2) Vector analysis cannot be used with instantaneous voltages and currents. The phase angles of v1, v2, and v3 are included in their definitions.

3) The load can be represented as a parallel combination of resistance, R, and either inductance or capacitance. Then p(t) is simply the sum of the squares of v1, v2, and v3 divided by R. The reactive components have no effect on power. Furthermore, since R is a constant, we can drop it from our proof and show that the sum of the squares of v1, v2, and v3 is constant and that proves that p(t) is constant.

Please go back and tell me precisely what I am doing wrong. Looks pretty good to me.

[ January 21, 2005, 10:57 AM: Message edited by: rattus ]

steve66 · Jan 21, 2005

Re: Instantaneous 3-phase power:

Rattus:

Since we are talking about instantenous power (I think we all agree to that) you can not leave out the reactance component. The reactance causes a power that flows back and forth, and effects the instantenous power. Since instantenous power is what we are concerned about, the reactance has to stay in the equation.

Tehcnically, we are using phasor analysis and not vecotr analysis. Vectors only have an angle and a magnitude. Phasors have a magnitude, and an angle that varies with time. (They rotate at the same frequency as the applied voltage or current). But if we look at one single instant in time, in effect the math for phasors reduces to vectors.

Steve

crossman · Jan 21, 2005

Re: Instantaneous 3-phase power:

I certainly have been agreeing with Rattus up to this point and my thinking was that his logic is right on target. For a purely resistive load, it works. But now the issue of reactance and instantaneous values has my head spinning as I ponder the implications...

It seems to me that the proof is going to have to be based on the P=IxEx1.73 direction like was done in Charlie B's proof. You have to multiply the current and voltage sine waves at every point. You cannot use Z for instantaneous computations because Z is in actuality an RMS quantity. Does that sound right?

I mean, the 2 x Pi x f in reactance is converting the inductance or capacitance into an RMS value is it not?

Do I have a clue here? This is the first time I have ever pushed my elementary knowledge of the subject this far.

rattus · Jan 21, 2005

Re: Instantaneous 3-phase power:

Steve and Crossman, here are some points to consider:

* Since we are trying to prove that p(t) is constant, we use instantaneous values in the proof, that is, time domain analysis. There is no need to do anything else.

* Vector analysis and/or frequency domain analysis cannot be mixed with time domain analysis. Impedance with a phase angle has no meaning; RMS values have no meaning; average values have no meaning; power factor has no meaning. One would have to use calculus to deal with the reactive components in time domain analysis.

* The load on each of the three phases can be expressed as the parallel combination of a resistor, R, and a reactive element.

* The power taken by each load is:

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

Where v(t) is the instantaneous voltage across R. This is fundamental.

* The reactive components in parallel with R draw current but have no effect on the power. We are talking about power in KW, not apparent power in KVA. This is also fundamental.

* We do not need an absolute value for p(t); we need only to prove that it is constant, therefore we can normalize the voltages and even set R to unity to simply the math. We need not include sqrt(3) either.

steve66 · Jan 21, 2005

Re: Instantaneous 3-phase power:

My head is spinning also. I don't think impedence is a "RMS" value, but its value does depend on the shape of the wave. But we are assuming a sine wave input. I think you can use 2*pi*f and the angle of the impedence (+90 or -90) and use vecotr math to combine it with R. I'm not sure if the result is a vector or a phasor with a frequency of 0.

Steve

rattus · Jan 21, 2005

Re: Instantaneous 3-phase power:

Steve,

Impedance only has meaning for sinusoidal, RMS voltages and currents, that is:

Irms = Vrms/Z

Each of these quantities has a phase angle, and:

Pavg = Vrms x Irms x PF,

But this is not instantaneous power.

You are going off on a tangent when you introduce impedance into a time domain problem.

oliver100 · Jan 21, 2005

Re: Instantaneous 3-phase power:

It was proven beyond reasonable doubt that the taken power is constant (provided that the system is balanced and the break torque is constant as well). However, the subject of efficiency was not discussed - 3 phase motors versus single phase motors.

As far as I remember my small electrical machines classes, we were discussing symmetrical components analysis. The method was first presented Dr .C. L. Fortescue in 1918. Do I correctly remember, that the magnetic field in single phase motors is dissolved on three components - positive rotating field, negative rotating field and a 0 sequence field? The only useful field is the positive rotating field, thus the efficiency of the single phase motor is quite low.

What is your take?

jtester · Jan 21, 2005

Re: Instantaneous 3-phase power:

rattus
One of my power books says the following, Consider an impedance element Z at an angle 'a', v(t)=Vm coswt, w=omega,
the instantaneous current is i(t)=Im cos(wt-a) where Im=Vm/Zmagnitude.
Then P(t)=VmIm/[cosa+cos(2wt-a)].
There are a couple of steps shown in the book that I left out, to get to the final equation, but clearly the phase angle of the impedance, 'a', is related to the instantaneous power.
Jim T

steve66 · Jan 21, 2005

Re: Instantaneous 3-phase power:

Rattus:

You posted:

Impedance only has meaning for sinusoidal, RMS voltages and currents, that is:

and:

You are going off on a tangent when you introduce impedance into a time domain problem.

Sorry, but you are incorrect on both statements. Impedence is a phasor, which is a function of time. In phasor notation ohms law is V=IZ where all three are phasors (and V and I vary with time). Here are a couple of footnotes I found regarding this:

1. The term "vector addition" is not technically correct since phasors, with the exception of impedance and admittance phasors, rotate with time. The methods, however, are identical.

2. In general, phasors are considered to rotate with time. A secondary definition of a phasor is any quantity that is a complex number. As a result, the impedance is also considered a phasor even though it does not change with time.

I have stated at least three times that my math is not 100% rigorous. The two footnotes above were part of that. And I found one more caveat:

For all circuit elements at the same frequency, the circuit must be in steady state conditions.

So if we can't assume steady state, I think that only Charlie's proof holds. (that explains the two pages

Steve

rattus · Jan 21, 2005

Re: Instantaneous 3-phase power:

Originally posted by jtester:
rattus
One of my power books says the following, Consider an impedance element Z at an angle 'a', v(t)=Vm coswt, w=omega,
the instantaneous current is i(t)=Im cos(wt-a) where Im=Vm/Zmagnitude.
Then P(t)=VmIm/[cosa+cos(2wt-a)].
Jim T

Jtester, we have a mix of apples and oranges here. Impedance, a vector quantity, is only used to determine Im which is then plugged into the equation for instantaneous current. In my case, I chose a parallel RLC load so I would not need to worry about the current.

You must agree that the instantaneous power into a parallel RLC load is v(t)^2/R, and although the LC portion draws reactive current if has no effect on the power, and for this proof, we do not care about its magnitude and phase.

The point that you guys are missing is that this is a time domain analysis, therefore reactance cannot be used, only resistance, capacitance, and inductance. In a series RLC loop, the voltage is iR +L(di/dt)+int(idt)/C. There are reactive components, but no reactances!

rattus · Jan 21, 2005

Re: Instantaneous 3-phase power:

Steve, you said that impedance is a function of time. You know better than that!

As for introducing phasors into this discussion, I fail to see the connection with my proof. We don't need rotating vectors to manipulate some simple equations and trig functions.

As I told Jtester, the reactive portion of the load current is of no concern to my proof.

rattus · Jan 22, 2005

Re: Instantaneous 3-phase power:

How about an example to clear this thing up?

Consider a load of 10 Ohms @ 45 deg, and 120V @ 0 deg applied.

This leads to a current of 12A @ -45 deg or 8.5A real and 8.5A inductive.

The load can be represented as either a series RL combo or a parallel RL combo.

The tendency is to represent the load as 7.07 Ohms resistive in series with 7.07 Ohms inductive.

I choose to represent the load as 14.14 Ohms resistive and 14.14 Ohms inductive in parallel.

The average power then can be calculated in three ways:

1) Pavg = 12A x 12A x 7.07 Ohms = 1018 W (I^2 x R)

2) Pavg = 120V x 12A x cos(45) = 1018 W (VxIxPF)

3) Pavg = 120V x 120V/14 .14 Ohms = 1018 W (V^2/R)

We could use any of these equations to compute p(t) by writing equations for i(t), but this makes the proof unduly complicated.

Note that in eqn 3, there is no need to know anything about the current or the reactive elements of the load. In other words, I have recognized a difficulty and avoided it. The only angles we need consider are the phase angles of the 3-phase voltages. Then for each phase the instantaneous power is:

4) p(t) = ((Vp(sin(u + phi)) ^2)/R

Where u = 2 x pi x f x t and phi is the phase angle.

If we show that p1(t) + p2(t) + p3(t) is constant, then the proof is complete. It is that simple!

charlie b · Jan 22, 2005

Re: Instantaneous 3-phase power:

Originally posted by rattus:Please go back and tell me precisely what I am doing wrong. Looks pretty good to me.

As you wish.

You started off OK, when you essentially eliminated all of the constant multipliers (e.g., peak voltage, a value of 1/2, and a few others), and just went about showing that the final result was a constant. That was a reasonable and valid track. However, you did make one significant error in your mathematical manipulations.

The error came in when you made the substitution of the trigonometric identity (your equation number 5) into your equation number 4. You made the same error twice. The first time is in converting the second term of #4 into the second and third terms of #7. The second time is in converting the third term of #4 into the fourth and fifth terms of #7.

Here is the error: You took this form:

e = ab + cd

Then squared both sides, to get

e^2 = (ab + cd)^2

But then you did not correctly square the right hand side. You had the following:

e^2 = a^2b^2 + c^2d^2

When you should have gotten the following (note the term that appears in bold):

e^2 = a^2b^2 + c^2d^2 + 2 (abcd)

Here is how it showed up in your proof:

Consider the second term of #4 (and ignore the constant Vp): v2^2.

From #2, v2^2 would be equal to sin^2 (u + 120)

Substituting trig identity #5 gives you v2^2 = (sin(u)cos(120) + cos(u)sin(120)) ^2

When you square the right hand side, you should get three terms. You only showed two of them. Your equation #7 should have read:

7) vn^2 = sin^2(u) + { sin^2(u)cos^2(120) + cos^2(u)sin^2(120) + 2 sin(u)sin(120)cos(u)cos(120)} + { sin^2(u)cos^2(-120) + cos^2(u)sin^2(-120) + 2 sin(u)sin(-120)cos(u)cos(-120)

Now that I have typed out this response, I see that the extra two terms will cancel each other out. So your proof was very close to being correct. A little extra "rigorous" treatment, and you would have had it.

g1matrix · Jan 22, 2005

Re: Instantaneous 3-phase power:

Charlie:

Just a question if you have VmIm/2 is that not the same as (Vm/(Sq.Rt.2))*(Im/(Sq.Rt.2)) ? The first part is what you get after invoking the Product identity. Is this not already rms ?

then by adding this result to the Sum identity you wind up with the complete P(t) expression.
So what youre getting in the cosine terms is the power Pr(t) and in the Sine terms Px(t)

Pr(t) the energy flow into the circuit and Px(t) the energy borrowed and returned by the circuit

The term (using rms), VI cos(theta)cos2(wt+ang.V)
has a frequency twice that of the source, accounts for the sinusoidal variation in the absorption of power by the resistive portion of the load. Since the average value of this function is zero.
But it was said that the instantaneous power in a balanced 3 phase is constant then how is this accounted for ?

rattus;

Also you could use P(t) = (Veff^2*R)/((Z)^2),

or (14400*7.07)/(100) = 1018 watts

gary

rattus · Jan 22, 2005

Re: Instantaneous 3-phase power:

Charlie B.,

You lost me. Just where do I need to show more rigor? I did not wade through all the manipulations if that is what you mean. If there is something else please tell me.

After all this exchange, I hope all agree that impedance, reactance, RMS, PF, etc. cannot be used in time domain equations. However, they can be used to determine phase angles and magnitudes for v(t) and i(t) equations.

charlie b · Jan 23, 2005

Re: Instantaneous 3-phase power:

Originally posted by rattus:Charlie B., You lost me. Just where do I need to show more rigor? I did not wade through all the manipulations if that is what you mean. If there is something else please tell me.

You had tried to make a case that the premise ?power is constant in a balanced 3-phase system? can be proven easily with simple math. Then you showed simple math. But your simple math left out the complicated steps, without which your ?proof? would have been invalid. You took a different path than I. You first converted current to an equivalent voltage (given knowledge of the reactance), and therefore had to use different trig identities. We came to the same conclusion. But your math, had you showed all the steps (as I did), would have been every bit as complicated as mine.

In other words, it?s OK not to show the rigorous math. But if you just show a few simple steps, and don?t show the rigor, you don?t get to state that the proof can be done without rigorous math.

After all this exchange, I hope all agree that impedance, reactance, RMS, PF, etc. cannot be used in time domain equations. However, they can be used to determine phase angles and magnitudes for v(t) and i(t) equations.

Now you are not being consistent, and I do not agree with your statement. The expression v(t) is a time domain equation. Point in fact, I did use RMS as part of my proof, and I did that proof in time domain. Another point in fact, you did use both impedance and reactance in your proof, and you did yours in time domain as well. So long as you are using the trig identities, you are in the time domain. It is just that you swallowed the inductance and reactance in amongst all the other unnamed constants, since you only needed to show that the answer was a constant (without caring what constant it was).

charlie b · Jan 23, 2005

Re: Instantaneous 3-phase power:

Originally posted by g1matrix: Just a question if you have VmIm/2 is that not the same as (Vm/(Sq.Rt.2))*(Im/(Sq.Rt.2)) ? The first part is what you get after invoking the Product identity. Is this not already rms ?

Yes. And I used that fact in the following part of my proof:

Originally posted by charlie b: Then P(t) = = (3) ? (VM IM) cos (u)

= (3) (VM / sqrt 2 ) (IM / sqrt 2) cos (u)

Now take note that (VM / sqrt 2 ) is the same as VRMS, and that (IM / sqrt 2) is the same as IRMS.

P(t) = (3) (VRMS) (IRMS) cos (u)

Originally posted by g1matrix: Since the average value of this function is zero.
But it was said that the instantaneous power in a balanced 3 phase is constant then how is this accounted for ?

As I said earlier, average power means nothing in this discussion. The mission was to prove that instantaneous power is constant. I did that in the following step of my proof (I have added bold text to the second and third lines):

Originally posted by charlie b:
Then P(t) = (3) ? (VM IM) cos (u)
+
? (VM IM) cos (u) { cos (2wt) + cos (2wt - 240) + cos (2wt - 480)}
+
? (VM IM) sin (u) { sin (2wt) + sin (2wt - 240) + sin (2wt - 480)}

Note that the second and third terms are at all times equal to zero (because the system is balanced).

Then P(t) = = (3) ? (VM IM) cos (u)

I was apparently not clear in my intentions here. I?ll say it another way, by looking at the second line (what I called the ?second term?) of this equation.

Yes, cos (2wt) has an average value of zero. That means nothing.

Yes, cos (2wt - 240) has an average value of zero. That means nothing.

Yes, cos (2wt - 480) has an average value of zero. That means nothing.

What matters is that if you add these three terms, the result is zero at every instant in time.

That is how it is accounted for: By adding three cosine curves, each spaced 120 degrees apart from the other two. That sum adds to zero, and what you are left with is a constant (the time variable has dropped out).

crossman · Jan 24, 2005

Re: Instantaneous 3-phase power:

I am trying to follow all of the above, but I have a question:

I realize that over time, the true power (watts) used in an inductor is zero. But when we are talking instantaneous power, how should a person look at it? Does it take true power (watts) to build up the magnetic field around the inductor? Then as the magnetic field collapses, it is putting this true power (watts) back to the source?

When you draw the sine wave graphs of current and voltage in an inductor, then multiply the voltage and current at each instant in time, you get a graph of power that is sometimes positive and sometimes negative. This positive and negative power cancels out over time, hence leaving the power in the inductor equal to zero.

So is the instantaneous power of an inductor considered as wattage?

charlie b · Jan 24, 2005

Re: Instantaneous 3-phase power:

There are three separate units of power: Watts, VA, and VAR. They can all be reduced to the same fundamental units of measure: (kilograms) x (meters ? squared) / (seconds ? cubed). But they must be treated separately. The power that is given to an inductor in one half of a cycle, and the power that is given back by the inductor in the other half of a cycle, must be expressed in terms of ?Volt-Amps-Reactive? (VAR). You cannot use the word ?watts? to describe this power. The reason has to do with the way you add ?watts? to ?VAR.? The sum of ?3 watts? plus ?4 VAR? is not ?7 anything.? It is ?5 VA.? The sum of ?3 watts? minus ?4 VAR? is not ?negative 1 anything.? It is also ?5 VA.? You treat the Watts and the VA as being two sides of a right triangle (i.e., the two sides with the 90 degree angle between them), and calculate the sum using the Pythagorean Theorem: (3)^2 + (4)^2 = (VA)^2, so VA = 5.

Instantaneous 3-phase power:

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