Instantaneous 3-phase power:

[Ed MacLaren]

Re: Instantaneous 3-phase power:

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

I agree with Charlie that it is not.

True power (watts) is a measure of the rate at which energy is converted and no energy is converted if the inductor has no resistance.

In other words, a pure inductor can only store and release energy, and this process is what is called reactive power and measured in vars.

Ed

[ January 24, 2005, 12:24 PM: Message edited by: Ed MacLaren ]

 

[crossman]

Re: Instantaneous 3-phase power:

I understand all of the vectors, triangles, and pythagorean's theorem, but when doing the vectors and triangles we are talking about RMS values instead of instantaneous values, aren't we?

And in the proofs above, we are dealing with instantaneous values only.

For instantaneous values, if I have +4 amps of resistive current at this instant and I have +3 amps of inductive current at the same instant, then these currents will add up to 7 amps. For instantaneous values, the "90 degrees out-of-phase" condition of inductive reactance does not matter because we are talking about a specific instant in time, not 360 degrees of values being combined together vectorially to arrive at a single value.

I understand that when we have 4 amps (effective) resistive current and 3 amps (effective) inductive current, these two currents can be represented by sine waves that are 90 degrees out of phase. To find total current, we would add these two sine waves together and find the effective value by multiplying the new peak by .707. Since the sine waves do not peak at the same time, we can't just add 4 and 3 together. The simplest way to perform the addition of the sine waves is to use vectors that are 90 degrees apart and find the resultant. But this gives us an effective value, not an instantaneous value.

My question above is concerning the instantaneous values. I am not so sure that you can separate out the Watts, VARs, and VAs when we are talking about a single instantaneous value.

Charlie B, from my limited knowledge, it seems that your proof is the only correct one because it takes into account the total voltage and total current at each instant. The other proof that assumes the instantaneous currents of the reactive components can be ignored because they cancel over time seem wrong to me.

And I am still thinking that as far as instantaneous power is concerned, there is no difference between watts and vars and voltamps... because when speaking of instantaneous values, there are no phase angle differences. All the values are in phase at that very instant. (granted that the reactive current may subtract from the resistive current depending on what exact instant you choose)

Am I completely out of the ballpark?

 

[crossman]

Re: Instantaneous 3-phase power:

Ed, just read your reply. I am going to go on record as saying that the energy required to either charge a capacitor or "charge" an inductor is indeed watts. I could be totally wrong here.

Surely if I can charge a capacitor, then use it to light a lamp, the capacitor was storing wattage? And it took wattage to put that potential energy into the capacitor?

Now, when the capacitor gives the watts back to the source in an AC circuit, that is why we say that reactance has no wattage, only VARs... because the wattage coming in cancels out with the wattage going back out over time.

With reactance, there are actually 2 times per cycle that the power is positive and 2 times per cycle the power is negative.

 

[crossman]

Re: Instantaneous 3-phase power:

still thinking

okay, the power is being STORED not USED... does that make a difference? thinking.... thinking...

I hope you guys realize that I am seriously trying to understand this, I am not just being a pain in the rear...

 

[crossman]

Re: Instantaneous 3-phase power:

okay... if a source is charging a capacitor, or whether it is causing heat disapation in a resistor, does the source really know whether it is doing one or the other? Doesn't it take real energy to charge a capacitor just as it takes real energy to make heat in a resistor? Let's just look at the first quadrant of the voltage sine wave. In both the resistor and the capacitor, the voltage is positive and the current is positive. Yes, there is a difference in the current vs time, the resistive current rises as the voltage goes up to peak, with the capacitor, the current starts high and then falls to zero at peak, but does that make a difference if we are talking only about the first quadrant?

 

[steve66]

Re: Instantaneous 3-phase power:

rattus:

You are still using an equation for average power to start your proof Can't do that if we are trying to prove the instantenous power is constant.

Impedence is a function of time. In some cases, it is a constant function and does not vary with time. That would not be true for a motor that is starting, or has a varying load.

You also cannot ignore the current through the reactance in your proof.

Impedence and reactance can be used in time domain equations.

Steve

 

[steve66]

Re: Instantaneous 3-phase power:

Crossman:

I think you are very close to understanding this.
What you have said makes a lot of sense. I understand what you are saying about not having to use trig to add instantenous values. And I understand what you said about instantenous power all being the same.

But power can either be dissipated, or transfered to the load, or transfered to the source. That is why we still need to angles and trig to add instantenous power.

If you want to call all kinds of instantenous power the same thing, call it volt-amps (VA's), not watts.

Watts is power being disipated (it implies that some part of the voltage and current are in phase).

Volts-Amps-Reactive is power being stored (or returned to the source). It implies some part of the voltage and current are out of phase.

But the term VA doesn't imply any of the above. It would be the correct term for instantenous power.

Edited by Steve

[ January 24, 2005, 01:20 PM: Message edited by: steve66 ]

 

[Ed MacLaren]

Re: Instantaneous 3-phase power:

Crossman,
I'm thinking too, and you may be onto something there.
Hmmmm.
Ed

 

[crossman]

Re: Instantaneous 3-phase power:

From charlie b's proof:

Let Va(t) = VM cos (wt)
Vb(t) = VM cos (wt - 120)
Vc(t) = VM cos (wt - 240)

Let Ia(t) = IM cos (wt - u)
Ib(t) = IM cos (wt - u - 120)
Ic(t) = IM cos (wt - u - 240)


Then P(t) = Va(t) Ia(t) + Vb(t) Ib(t) + Vc(t) Ic(t)

Now my thinking:

The currents in the above equation include the reactive currents, right? Also there is no inclusion of power factor, right? This makes me think that the VARs, VoltAmps, and Wattage are all the same thing when speaking only of a precise instant in time.

 

[crossman]

Re: Instantaneous 3-phase power:

Steve66... read your edited post and I will agree on your point about calling it Voltamps instead of watts.

But I still have some concerns. Let's say that we have a wattmeter capable of recording a graph of the watts.

In the first quadrant of the graph for an AC source feeding a capacitor, will it register a reading?

 

[steve66]

Re: Instantaneous 3-phase power:

Crossman:

You are making my head spin again

Your equation for power does include all the instantenous powers. But you must call them VA's. By definition watts and VAR's are different.

I think you could probably have an instantenous power triange.

P(w)^2 + (P(l) - P(c))^2 = P(instantenous)

where P(w) is the instantenous real power, P(l) is the instantenous inductive power, and P(c) is the instantenous capactitive power.

Steve

 

[crossman]

Re: Instantaneous 3-phase power:

In a wire, at any given instant, there are only two possibilities for the direction of current flow. It can be going one way, or going the other. Call one direction positive and the other negative. Simple addition of these instantaneous currents gives us the instantaneous total current. The amount out of phase, etc, does not matter because there are only two possibilities. Current in a horizntal wire is either going to the left or going to the right.

And in this light, the source voltage at any given instant is either in phase with the total current or 180 degrees out of phase with the total current. This gives us either 1) power going to the circuit from the source, or 2) power going from the circuit to the source.

When averaged out over time, the average power of reactance is certainly zero. But there is real power there at any given instant (except where the power graph crosses the horizontal axis)

 

[crossman]

Re: Instantaneous 3-phase power:

BTW, my head is spinning also!

 

[crossman]

Re: Instantaneous 3-phase power:

Okay I have it.

According to "Delmar's Standard Textbook of Electricity":

before true power or watts can exist, there must be some type of energy change or conversion.

Now, when a capacitor is charged, the energy in the moving electrons is converted into an electrostatic field in the capacitor. These are definitely two different forms of energy. I say it takes wattage to make the transfer of the energy. Just like electrical energy can be dissipated as heat, it can also be "dissipated" as an electrostatic field.

 

[crossman]

Re: Instantaneous 3-phase power:

Any takers on the wattmeter/capacitor question above?

 

[charlie b]

Re: Instantaneous 3-phase power:

Originally posted by crossman:For instantaneous values, if I have +4 amps of resistive current at this instant and I have +3 amps of inductive current at the same instant, then these currents will add up to 7 amps

Am I completely out of the ballpark?

Yes, they will, but yes, you are. Sorry.

We need to be clear about the currents. We typically describe sinusoidal currents as a single number: the RMS value. We may also talk about a second number: the phase angle, or equivalently, the power factor. You cannot use the RMS value of ?resistive current,? to use your phrase, and add it to the RMS value of ?inductive current,? and get a meaningful result.

You are correct in saying that at any instant in time, the current contributions will add in the manner you describe (3 + 4 = 7). But neither the ?3? nor the ?4? nor the ?7? is constant. One will be on the way up, the other on the way down, and the sum can go up or down. If you want to describe how the sum goes up and down as time goes on, you need to add the trigonometric function of the first current to the trigonometric function of the second current, and take whatever comes. But it will not be a simple summation of the RMS values of the two functions.

 

[crossman]

Re: Instantaneous 3-phase power:

charlie b, I am thinking I am on the same page as you on your last post. I understand that there is a difference between the 3 RMS value and the 3 instantaneous value and these two examples I picked are not necessarily the same examples.

In your rigorous proof above, you are using only instantaneous values. Your trig identities related for voltage and current of each phase serve no other purpose except to define the value of current and value of voltage at each individual instant in time that we are considering.


(speaking of only one phase of the three phase, but it applies to all three)
For each individual instant in time, we have only voltage and only current. Current can only flow one way or the other. For that precise instant in time, it does not make one bit of difference whether the current is rising or falling, it makes no difference whether the voltage is rising or falling either. For that precise instant in time, there is actually no way to tell whether the current is going to be rising in the next instant or falling in the next instant, if you only know information about that exact instant.

I feel certain that if, at one precise instant in time, I allowed you to measure all voltage characteristics and all current characteristics of a current flowing in a wire, that you would not be able to discern whether that current was flowing in a resistive circuit or a reactive circuit or combination thereof.

[ January 24, 2005, 03:05 PM: Message edited by: crossman ]

 

[crossman]

Re: Instantaneous 3-phase power:

oops, my last statement is wrong. You would know that something was "funny" if at the instant we picked, the current was flowing backwards from the source voltage. But if it was flowing backwards, could you tell if it was inductive or capacitive? I don't think so.

My wattmeter/capacitor question still begs for an answer.

 

[rattus]

Re: Instantaneous 3-phase power:

Originally posted by steve66:
rattus:

You are still using an equation for average power to start your proof Can't do that if we are trying to prove the instantenous power is constant.

Steve, v(t)^2/R is instantaneous power. This is fundamental. Where do you get this notion that it is average power?

Impedence is a function of time. In some cases, it is a constant function and does not vary with time. That would not be true for a motor that is starting, or has a varying load.

Impedance is never written as a function of time. Impedance may change, resistance may change, but in calculations, impedance must be treated as constant. Furthermore, to characterize a motor as a simple impedance simply cannot be done.

You also cannot ignore the current through the reactance in your proof.

I did ignore the reactive current and Charlie B. agrees that is legit. There will be current through the parallel inductance or capacitance. That current does not flow through the resistance, therefore it has absolutely no effect on power! Apparent power in KVA yes, but no effect on real power in KW. This is fundamental. You are missing the point that I set up the problem in such a way that the reactive current could be ignored.

Impedence and reactance can be used in time domain equations.

Give me an example please?

[ January 24, 2005, 04:39 PM: Message edited by: rattus ]

 

[charlie b]

Re: Instantaneous 3-phase power:

You did not ignore the reactive current. You expressed VI as V^2/Z, and started your proof from there. It takes calculus to prove that I = V^2/Z, but once you have that, you can use that. You are no longer treating current in the mathematical steps. Any actual values of R, X, Z, PF, Vp, and Ip were swallowed up by your method of just proving that power is constant, without caring about the value of that constant.

I think we are not all on the same page, with regard to the phrase ?time domain.? My view is that any expression that has time as a variable, such as v(t) = Vp cos (wt), is being handled in the time domain. Is there another meaning of this phrase?