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

N

narke

Jan 1, 1970
0
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.
 
G

Guest

Jan 1, 1970
0
narke said:
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?
------------------
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
--------------------------
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?
-----------
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.
Finally, for these AC concepts, is there a good text book in amazon.com?

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
 
N

narke

Jan 1, 1970
0
------------------
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
--------------------------

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?

-----------
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.

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

Guest

Jan 1, 1970
0
narke said:
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?
---------------
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.----------
And, for the prove, I don't understand. Can I find it in what you
mentioned Schaum's outline?
---------
it may not be but is in any general EE text. Calculus is used.
Do you agree what I mentioned: apparent energy should be a sum of real
energy and reactive energy?
----------
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
 
N

narke

Jan 1, 1970
0
---------------
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

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.
 
G

Guest

Jan 1, 1970
0
narke said:
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.

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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