Reactive Power Apparent Power Issue

It is simply a mathematical representation convenient for calculations.In summary, the definitions of apparent power, reactive power, and complex power are purely mathematical and do not have a precise physical significance. They are simply convenient for calculations and are often used in exam questions.
  • #1
azaharak
152
0
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)ZTHANK 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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  • #2
hi azaharak! :smile:
azaharak said:
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! :wink:)

from the pf library on https://www.physicsforums.com/library.php?do=view_item&itemid=303"

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! :rolleyes:

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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1. What is the difference between reactive power and apparent power?

Reactive power is the portion of electrical power that flows back and forth between the source and load due to the presence of inductance and capacitance in the circuit. Apparent power, on the other hand, is the total power that is drawn from the source and is the combination of both real power and reactive power.

2. How does an issue with reactive power affect electrical systems?

An imbalance or issue with reactive power in an electrical system can cause a decrease in voltage, which can lead to equipment damage and power outages. It can also result in an increase in energy losses, causing higher electricity bills.

3. What are some common causes of reactive power issues?

Some common causes of reactive power issues include power factor imbalance, unbalanced three-phase loads, and the presence of harmonics in the electrical system. Faulty equipment and incorrect wiring can also contribute to reactive power problems.

4. How can reactive power issues be corrected?

To correct reactive power issues, power factor correction techniques can be used. This includes installing capacitors to offset the reactive power and improve the overall power factor of the system. It is also important to regularly monitor and maintain the power factor of the system to prevent future issues.

5. Can reactive power be used in a beneficial way?

Yes, reactive power can be beneficial in certain situations, such as inductive loads, where it is necessary to maintain the magnetic fields in motors and transformers. It can also be used for voltage regulation and to improve the power factor in industrial settings.

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