Consider an RC circuit which is supplied by a sine source
$$v(t)=V_m\cdot \sin(\omega t)$$
The current through the circuit will be
$$i(t)=I_m\cdot \sin(\omega t+\phi) \\
\phi=\arctan \left(\dfrac{1}{\omega RC} \right)$$
The instantaneous power equation is
$$p(t)=V_m\cdot I_m\cdot \sin(\omega t)\cdot\sin(\omega t+\phi)$$
Clearly from this image we can see that the power equation has some frequency component also.
The average of this instantaneous equation over one cycle will give us average power.
The real power is going to be the first term which, however, has a zero and a high frequency component in it and the reactive power however has a high frequency component only (by high frequency I mean the value of frequency is higher (i.e.) twice the supply frequency.)
Are real power, active power and average power the one and the same same?
If they are not, then we don't see text books mentioning this high frequency component in the real power/active power/reactive power.
They directly give the average value as zero frequency component. This reactive power that is mentioned here always has only a high frequency component here. In the case of power systems, if we are speaking about reactive power are we just speaking about the peak reactive load demands and quantifying it for load flows and in case of real power in power system will we be speaking only about average value of real power load demands and proceeding with the load flow?
I lose clarity here and I am sure that I am wrong at some basic point. I would be grateful if someone is able to clarify this.
Best Answer
I will be talking about the ideal cases where there are no harmonics, as I presume you're talking about, given your derivation.
The active power is the same as the real power. And any power is instantaneous by nature, i.e. at any instance of time it has a value. Averaging the instantaneous power results in an average, and this average value, like the instantaneous value, can come from any power.
Therefore: active power == real power and they refer to a specific type of power, while the average is the mathematical average performed on any quantity. No averaging means instantaneous.
We certainly do, since it's part of the very nature of the multiplication: there are two sines, multiplied, which gives the trigonometric equivalent of \$\cos(2\omega)\$. But the average of it is a fixed, non-oscillating value.
It seems you missed a few words ther, but even so, the part where "they" give only the average values is the part where only those matter for the counter, or for load-flow analysis. Remember that the counter performs an averaging in time. The result at the end comes out a fixed number.
You are misleading yourself by not continuing the derivation:
$$\begin{align} p(t)&=\dfrac{V_pI_p}{2}\{[1-\cos(2\omega t)]\cos(\theta)+\sin(2\omega t)\sin(\theta)\} \tag{1} \\ &=\dfrac{V_pI_p}{2}[\cos(\theta)-\cos(2\omega t)\cos(\theta)+\sin(2\omega t)\sin(\theta)] \\ &=\dfrac{V_pI_p}{2}[\cos(\theta)-\cos(2\omega t+\theta)] \tag{2} \\ &=\qquad{\bar p(t)}\qquad +\qquad{\tilde p(t)} \tag{3} \end{align}$$
Now you can see that there is a fixed value, \$\cos(\theta)\$, and an oscillating value at twice the frequency, which is naturally occuring when two sines are multiplied. The fixed value is nothing but the average. Since the cosine is an even function, the average is never negative, while the oscillating part never goes beyond twice the amplitude.
Remember that the total power, S, is made of both active, P, and reactive, Q, powers, and their relation is orthogonal: \$S=\sqrt{P^2+Q^2}\$. And S is calculated based on the RMS values of the voltage and current. This means that no matter what displacement exists, the RMS values will be divided by \$\sqrt2\$, and their multiplication will always be one half of the peak values. The instantaneous values will have a frequency twice the fundamental, and its peaks will never be more than twice the \$\bar S\$. For example, if V=3 and I=2, S=3 and the peak will never be above or below ±6. Here shown for an angle varying from 0 (blue) to π/[2,3,4,6 (red)]):
The load-flow assumes a behaviour in time, so the instantaneous values make little sense here. Therefore the only quantities of interest are magnitude and phase, which give the relevant average values.
As a final note, when talking about systems with harmoncs, the same reasoning as above can be used. For instantaneous values the displacement is now relevant only to one harmonic, and the total harmonic distortion (THD) takes place for the overall effect, while for a load-flow analysis, the same averages in time are done with the help of square-root sums of powers.