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Reactive Power Apparent Power Issue

by azaharak
Tags: apparent power, reactive power
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azaharak
#1
Aug8-11, 01:23 PM
P: 152
This is not a HW question.

Assume you have a sinusoidally varying voltage, such that current lags by some phase angle phi

Then you can write the power as a function of time as

p(t)=i(t)v(t)=2*Vrms*Irms*cos(omega*t)*cos(omega*t-phi)

You can break this up via trig identity into the following

Eq1: p(t)=2*Vrms*Irms*( [1+cos(2*omega*t)]*cos(phi) + sin(2*omega*t)*sin(phi) )

If you average over 1 period, then you obtain that the average REAL power is

P=Vrms*Irms*cos(phi) This is average real power

The peak real power would be twice this value since (take t=0 in Eq1:)
real power is associated with cos(phi) term.

The average reactive power is zero, it bounces back and forth between inductive and capacitive loads (B and E fields).

But the peak reactive power is defined as Q=Irms*Vrm*sin(phi).

HERE IS THE QUESTion.


I see often that the Apparent power S = P + i*Q. Unfortunately it is never throuroughly defined. Why would would be adding the Average value of real power and Peak value of Reactice power and call it the apparent power.


Am I misunderstanding this? Shouldn't we be adding the PEAK real power and PEAK reactive power and call this the apparent power.


Also, other relations are P=(I^2)R Q=(I^2)X S=(I^2)Z





THANK YOU!!



Am i correct in saying that the reason why it is defined this way is so that "you" could say that the phase angle is given by

Phi = arctan(Q/P).

I'm just a little weired out by combining the Average Real power with Peak Reactive power.

Thanks
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tiny-tim
#2
Aug8-11, 04:42 PM
Sci Advisor
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tiny-tim's Avatar
P: 26,148
hi azaharak!
Quote Quote by azaharak View Post
HERE IS THE QUESTion.

I see often that the Apparent power S = P + i*Q.
no, that's the complex power

the apparent power is VrmsIrms it's voltage times current
Unfortunately it is never throuroughly defined. Why would would be adding the Average value of real power and Peak value of Reactice power and call it the apparent power.
so far as i know, neither complex power nor apparent power have any precise physical significance

(but you need to know them because they come up in exam questions! )

from the pf library on impedance
Power:

Power = work per time = voltage times charge per time = voltage times current:

[tex]P = VI =\ V_{max}I_{max}\cos(\omega t + \phi/2)\cos(\omega t - \phi/2)[/tex]
[tex]=\ V_{max}I_{max}(\cos\phi + \cos2\omega t)/2[/tex]
[tex]=\ V_{rms}I_{rms}(\cos\phi + \cos2\omega t)[/tex]
[tex]=\ (V_{rms}^2/|Z|)(\cos\phi + \cos2\omega t)[/tex]

So (instantaneous) power is the constant part, [itex]P_{av} = V_{rms}I_{rms}\cos\phi[/itex] (the average power), plus a component varying with double the circuit frequency, [itex]V_{rms}I_{rms}\cos2\omega t[/itex] (so a graph of the whole power is a sine wave shifted by a ratio [itex]\cos\phi[/itex] above the x-axis).

Apparent power, reactive power, and complex power, are convenient mathematical definitions with no precise physical significance: apparent power is the (constant) product of the r.m.s. voltage and current, [itex]S=V_{r.m.s.}I_{r.m.s.}[/itex]: it is what we would expect the average power to be if we knew nothing about reactance!

Similarly reactive power is defined as [itex]Q=S\sin\phi=P_{av}\tan\phi[/itex], and complex power is defined as [itex]Se^{j\phi}=P_{av}+jQ[/itex].


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