Defining Average Power
- Thread starter dnem
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Average poweris only meaningful if defined as the time average over an integral number of cycles of the instantaneous power.
If by average voltage you mean the arithmetic mean of the absolute value of the voltage, then your first equation is incorrect.
Your relations to peak value are only valid for a pure sine wave.
The RMS "average" was defined so it has the useful property that for any periodic waveform the second equation is true. (And you can argue that it is also useful, with a slightly different definition involving limits, for a non-periodic waveform.)
FWIW, you can average the power over one cycle. That is a period of time. Instantaneous is a single point within that period.
Carultch
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The definition of the average value of a function in general, is you "add up" the output value throughout an interval of the input variable (time in this case). By "add up", I mean the concept of integration in Calculus, which means the area under the curve as you graph the function.
Since neither voltage nor current is a complete picture of the power on its own, it is incorrect to simply average volts and amps and multiply the two together, expecting a meaningful output. This is why we have the RMS concept, which means you square the value as you add it up, then divide by the time interval, and square root it. For a sine wave, RMS value is sqrt(2)*peak value. RMS is a fancy kind of average, that allows you to multiply RMS voltage and RMS current, to get the time average of power. That is, when both waveforms are synchronized in phase and have exactly the same shape.
If you average the absolute value of voltage over a complete AC cycle, you do indeed get 0.637*peak voltage. But multiplying this with average current, doesn't get you any meaningful figure on power.
Yes, a complete AC sine wave was my intent
It might not be a meaningful figure but Im trying to get an understanding of these relationships and the text books say RMS Voltages times RMS Amps equals Average Power. So Im assuming that there is no such thing as RMS Power.
Lets try a math approach:
1 peak volt x 1 peak amp = 1 peak watt
.707 rms volt x .707 rms amp = .707 x .707 = .499849 watt
So does this mean that the average (or rms) of 1 watt is .5 watt ? Is this the Average Watt.? If yes, then the Average Watt is neither peak x .637 nor peak x .707
In phase and same shape.? So if there is reactance in the circuit, I understand that Pythagoreans Theorem tells you how to add horizontal true resistive with vertical reactive to get diagonal Apparent (Applied/Source) Power. But RMS reactive volts x RMS reactive amps should still = reactive VARs. And RMS resistive volts x RMS resistive amps should still = Watts.
When youre considering a real life circuit (that has both reactance and resistance), you will have Apparent (Applied/Source) Power, you will have True (Active/Work) Power, and you will have Reactive (Wasted Circulating) Power. You should be able to calculate both peak and RMS numbers for each power type even with a phase shift. The phase shift increases the reactive component and increases the theta angle and makes the triangle taller, but RMS is still .707 of Peak
You can't just use the real or reactive components
S = P + jQ = V I*
V and I is phasor or complex form and * is the conjugate
assume
V = 1/45 deg = 0.71 + 0.71j
I = 1/45 deg = 0.71 + 0.71j (* = 1/-45 = 0.71 - 0.71j)
S = 1/45 x 1/-45 = 1/0 = 1 + 0j va
or
S = (0.71 + 0.71j) (0.71 - 0.71j) = 1/2 - 1/2j + 1/2j +1/2 = 1 + 0j = 1/0 va
so
P = 1 watt
Q = 0 var
Using only the real V and I components: 0.71 x 0.71 = 1/2 watt NE P = 1 watt
Reactive V and I: 0.71j x -0.71j = 1/2 watt, not 0 var
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Carultch
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The average power in this example would be 0.5 Watts. The instantaneous peak power would be 1 Watt. Both answers assume that both volts and amps are in phase and pure sinusoids in shape.
Technically, if you average a pure AC sinusoidal waveform over one cycle, you get 0.
Carultch
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Technically, if you average a pure AC sinusoidal waveform over one cycle, you get 0.
True indeed. Hence it is really the absolute value that you average to get 0.637*peak voltage. The rectified waveform with an ideal diode full wave rectifier.
Carultch
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RMS Volts * RMS Amps is by definition the apparent power. The kVA.
Instantaneous Volts * Instantaneous Amps, integrated over a complete cycle, divided by period, is the real power. The time average of the instantaneous power.
In the special case of pure sinusoids for both Volts & Amps, you can construct the Pythagorean theorem as described. kW^2 + kVAr^2 = kVA^2. We defined kVA, we defined kW, now you can determine kVAr.
Non-sinusoids, i.e. harmonic distortion, make the relationship much more complicated. RMS will no longer be 1/sqrt(2) of peak. This is why I specified same shape.
RMS being 0.707 of peak applies to a single sinusoidal wave, centered upon zero. The power as a function of time is not centered upon zero. The voltage and current are, for a pure waveform of exclusively AC.
That whole thing just shot right over my head
Verbose comes to mind.
I can think of other ways to describe it.
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That doesn't make the average power zero though.Technically, if you average a pure AC sinusoidal waveform over one cycle, you get 0.
Yes you can. For power all you need is the real component.
That's why it's called real.
Yes let's please limit the scope to a beautifully smooth and arcing sinewave
1 peak volt x 1 peak amp = 1 peak watt
.707 rms volt x .707 rms amp = .707 x .707 = .499849 watt
And so a 1 peak watt is .5 average watt
.637 average volt x .637 average amp = .637 x .637 = .405769 watt
Does this .405769 value have a label.?
Carultch
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Observe that the 0.707 factor is meant to be exactly 1/sqrt(2). This means when you multiply both factors together, it isn't approximately 1/2, it is exactly 1/2. So exactly 1 peak Watt is exactly 0.5 Watts average.
Nope. That is usually an irrelevant concept, with no practical or conceptual application of which I am aware.
The value of 0.637*Vpeak is the average function value for the ideal rectified (as in positive only) waveform, but multiplying this value for current and voltage together has no connection to any statistical measure of the power waveform.
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Technically, if you average a pure AC sinusoidal waveform over one cycle, you get 0.
But we were talking about power, and with AC voltage and current sine waves in phase, the power delivered will always be positive (or negative).
If you look at the power waveform, its somewhat like a sine wave, but not quite. Its a little more triangular.
Here is an example, but the current and voltage aren't quite in phase. In your example with 1 V and 1A, the power waveform would start at zero, and go to 1 W peak.
https://openenergymonitor.org/emon/sites/default/files/reactive.jpg
If you can picture the power waveform going from 0 to 1W peak, then picture cutting the waveform in half at 0.5 W. Then flip the top peaks upside down, and you will find they have just enough area to fill in the spaces between the peaks under 0.5 watts. So there is enough area under the curve to completely fill in the graph up to 0.5 W.
I'm a little rusty on this, so someone correct me if I'm wrong, but in this case, the RMS power of 0.5 W is also the average power of 0.5 watts.
That's the whole point of using RMS for voltages and currents, we can use them to give us the average power for AC sine waves, which is also the RMS power.
Here is an example, but the current and voltage aren't quite in phase. In your example with 1 V and 1A, the power waveform would start at zero, and go to 1 W peak.
https://openenergymonitor.org/emon/sites/default/files/reactive.jpg
If you can picture the power waveform going from 0 to 1W peak, then picture cutting the waveform in half at 0.5 W. Then flip the top peaks upside down, and you will find they have just enough area to fill in the spaces between the peaks under 0.5 watts. So there is enough area under the curve to completely fill in the graph up to 0.5 W.
I'm a little rusty on this, so someone correct me if I'm wrong, but in this case, the RMS power of 0.5 W is also the average power of 0.5 watts.
That's the whole point of using RMS for voltages and currents, we can use them to give us the average power for AC sine waves, which is also the RMS power.
thats a good visual image
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