AC Power: S^2 = P^2 + Q^2 ?

If waveform is purely sinusoidal, it was said the S^2 = P^2 + Q^2, where S is apparent power, P is real power and Q is reactive power.
In this formula, I have two questions. Firstly, S, P, Q here are what kind of power, average power? rms power (rms U * rms I)? Instantaneous power (U * I from moment to moment)? After clarify the definition, can anyone show a prove?
The second question is that I think, for energy, apparent energy should be a sum of real energy and reactive energy. I learned this from Andrew's article: http://blogs.sun.com/agabriel/entry/ac_power_power_factor_explained Am I right on the concept? If so, do you see the energy relationship between apparent/active/reactive is different with power?
Finally, for these AC concepts, is there a good text book in amazon.com?
Thanks in advance.
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------------------ Average power IS the product of rms U * rms I times power factor where power factor is the cosine of the phase angle between voltage and current P average= Urms* Irms cos (phase angle) Q average =- Urms* Irms sin(phase angle) (sign+ if I leads U ) S =P+jQ =Urms(conjugate of Irms) both expressed as complex numbers In terms of instantaneous power p(t) = Umax*cos(wt) and I(t)=Imax*cos(wt+a) where w is radian frequency and a is the phase shift of current with respect to voltage Integrate this over a cycle to get two terms - a constant term P and a double frequency term which does not contribute to average power but energy shuttled back and forth from the source and inductive/capacitive load-represented by represented by Q Since it is convenient to use rms voltage and current, it follows that we can express P and Q in terms of rms values and the phase angle of voltage with respect to current --------------------------

----------- Instantaneously, there is only power-the product of instantaneous voltage and current- averaging this over a cycle results in a constant P When we use rms U and I we are using a specific form of averaging. In terms or rms values, then it was recognized long ago that |V|*|I| magnitudes (S) don't represent real power ( i.e this represents the magnitude of apparent power) and that only the component of current that is in phase with voltage is what is real (non-zero average) power, The component of current 90 degrees out of phase with the voltage represents the effect of energy being shuttled back and forth.

There are many books available - look for electrical "circuit" theory. There are on-line references and cheap off line references include Schaum's Outlines Amazon has this and other references
This may help
http://fourier.eng.hmc.edu/e84/lectures/ch3/node9.html
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It seems you know the things, thank you for the explaination. But, I am so fool and I have not get clear the point. Could you directly answer the question: In S^2 = P^2 + Q^2, S,P,Q are rms powers or instantaneous powers?
And, for the prove, I don't understand. Can I find it in what you mentioned Schaum's outline?

Do you agree what I mentioned: apparent energy should be a sum of real energy and reactive energy?

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--------------- They are neither. They are found from a true average of instantaneous current and voltage over a period but can be expressed in terms of the product of rms voltage and current. The term "rms power" has been used but is wrong. It is better to use "true power" Rms power is a useful mode for considering heating of motors with varying loads but that is a different animal than what you are considering.---------- >

--------- it may not be but is in any general EE text. Calculus is used.

---------- As you have written "S^2 = P^2 + Q^2" This is a sum of squares as per Pythagorus' theorem for right triangles. true power and reactive are , in a phasor or power triangle, at right angles. True power P is the actual power involved with actual net energy transfer while reactive represents energy that is shuttled back and forth between source and load- average net energy is 0. Apparent power is simply what would appear if you measured the voltage and current and multiplied them together as if they were DC. Think of a right triangle with base =P , upright =Q and hypotenuse=S
check this reference http://en.wikipedia.org/wiki/Electric_power
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After several days of thinking, and studying of the integration process applied on the instantaneous power, I think I begin to understand. Thank you very much. Before that, what was always confusing me the that form of the S^2 = P^2 + Q^2 seems breaks the law of the conservation of energy, since I consider S as kind of power that supplied from the source, but it is not, rather it is merely V multiples I and is a common factor found in both items that express the P.
I also hope you can point me to a good EE textbook, could you? Thanks.

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wrote:

The textbooks that I have are old- by about 20 years- but the one that has gone on for decades with revisions is Fitzgerald, Higginbotham, etc "basic electrical engineering" It's a bit thin on basic development but is pretty solid overall. See what texts various universities are using for their basic EE courses. You should be able to do this on -line Note that knowledge of calculus and differential equations is a must.
As for the specific case of S,P,Q consider a voltage v(t)=Vm*cos(wt) and current i(t)=Im*cos(wt-a) where Vm and Im are maximum values and a is the phase shift of current with respect to voltage lagging for inductive loads. In terms of phasor representation using rms values V=( Vm/2) at angle 0 (reference) and I=Im/2 at angle -a with respect to V These frequency domain representations useful for sinusoidal steady state carry the essential magnitude and phase information and allow the use (as complex numbers) of all the circuit theory developed for steady state DC -extending it to AC. Note -rms values are mathematical constructs but we can make meters that measure in these terms. What we have in reality is the time varying quantities. Now we can consider instantaneous power p(t)=v(t)*i(t) =(Vm*Im)*cos(wt)*cos(wt-a) =(Vm*Im)*[cos(wt)*cos(wt)*cos(a) +cos(wt)*sin(wt)*sin(a)] =(Vm*Im/2)*[(1+cos(2wt))*cos(a) +sin(2wt)] Without formal integration over a cycle it can be seen that the 2wt terms average 0 leaving the term Pave=(Vm*Im/2)cos(a) =V*I*cos(a) is related to the net energy per cycle that is converted to some other form of energy. The term (Vm*Im/2)sin(a) represents an "average rate energy shuttled back and forth from source to inductance or capacitance (or between them - magnetic and electric field energy-similar to potential and kinetic energy in a mechanical system with masses and springs) Energy stored for part of the cycle and returned for the rest of the cycle. This affects the magnitude of the current for a given real power This term can be represented by Q In phasor notation S=P+jQ and S^2= P^2+Q^2
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