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Apparent and reactive power

  1. Aug 17, 2011 #1
    hi dudes...
    i m confuse about apparent, reactive and real power, please explain.........any help wll be appreciated..thanks in advance
     
    Last edited by a moderator: Aug 17, 2011
  2. jcsd
  3. Aug 17, 2011 #2

    sophiecentaur

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    I'm not sure that "reactive power' has a meaning. Power refers to the rate of energy transfer and that works for AC or DC.
    Reactance is a quantity that is used for AC circuits and no net energy transfer is involved. Energy is stored, temporarily, in a capacitor or an inductor during parts of the AC cycle but comes out in another part of the cycle. In the light of that, I would say that you can't actually have reactive power.

    The power transferred in an AC circuit will be V I cos(θ) , where θ is the phase angle between V and I. Cos(θ) is the 'power factor'. For a power factor less than 1, there will be higher voltage or current for a given power. This will introduce higher losses or involve embarrassingly higher voltages than the equipment can handle - but I wouldn't have said that was actually reactive power.

    I guess that V I would be what you refer to as "apparent power"

    In a resonant circuit, you can have reactive energy which is built up in the reactive components as energy is supplied at the right frequency.

    Does that make sense?
     
  4. Aug 18, 2011 #3
    The apparent power is a complex number, S=P+jQ. I'd think of it as active power is the energy that feeds the load. Reactive power cant be really be used to feed a load, though it stresses components and transmission lines. Therefore you have to dimension everything according to the apparent power. I dont know if i'm right thats just the way I think about it. Search for the power triangle on google and then it might be easier to understand.
     
  5. Aug 18, 2011 #4

    sophiecentaur

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    But isn't the definition of power rate of energy transfer? Energy in the reactive bits is not transferred anywhere. I think the two terms "apparent" and "reactive" are not very helpful and don't actually need to be used. 'VI' is a useful term, which is used everywhere and that would, presumably, be 'apparent power', I guess. This isn't worth losing sleep over as it's only a fringe term - possibly introduced by someone who wanted to 'help' students in some way. If it doesn't help then don't use it.
     
  6. Aug 18, 2011 #5
    I like the idea of looking at reactive power as energy that is simply stored and released with no net effect. It results in additional movement of current because energy is being stored, but no actual work is done other than expanding and collapsing a magnetic and/or electric field. That's why a power supply with a rating for 1000VA may only be able to supply 600W to a purely resistive load. That's how I've always looked at it and hopefully my understanding is sound.

    I believe the term apparent is being used as a synonym for real power, the real component of the complex power vector, where the imaginary axis is the reactive component.
     
    Last edited: Aug 18, 2011
  7. Aug 19, 2011 #6
    There is a BIG difference between reactive, active and apparent power and it is why the terms were invented for AC circuits. One must be careful not mixing them with instantaneous power. In AC circuits, power is a function of time [itex]t[/itex] given by

    [tex] p(t)=e(t)i(t)[/tex]
    where [itex]e(t)[/itex] is the instantaneous voltage and [itex]i(t)[/itex] is the instantaneous current. Note that in DC, everything is constant, so instantaneous power is just power.

    Now let us examine 3 basic cases for this instantaneous power:

    The Resistor

    Let us assume we have a resistor of linear resistance [itex]R[/itex] connected to a voltage source [itex]e(t)=E\cos{\omega t}[/itex] (i.e. a sinusoidal waveform of magnitude [itex]E[/itex] and angular frequency [itex]\omega[/itex]). Since the current in the resistor is in phase with the voltage, we get some current [itex]i(t)=I\cos{\omega t}[/itex] in the resistor. From the definition of [itex]p(t)[/itex] above, we get
    [tex] p(t)=e(t)i(t)[/tex]
    [tex] =E\cos{\omega t}\cdot I\cos{\omega t}[/tex]
    [tex] =EI\cos^2{\omega t}[/tex]
    [tex] =\frac{1}{2}EI(1+\cos{2\omega t})[/tex]
    (the last step included the identity [itex]2\cos^2{\omega t}=1+\cos{2\omega t}[/itex])
    Hence, we can express the instantaneous power of the resistor as 2 parts: a first part which is independent of time (mean value) and a second part which oscillates at twice the frequency. The active power [itex]P[/itex] is defined as the mean value of the instantaneous power
    [tex]P=\frac{1}{2}EI[/tex]
    Notice here the presence of the [itex] \frac{1}{2}[/itex], because [itex]E[/itex] and [itex]I[/itex] are the magnitudes of the cosines, not the RMS value. This is where the definition of RMS kicks in, since we know the power in DC (instantaneous or mean value, all the same) is
    [tex]P=EI[/tex]
    and we know for sinusoidal waveforms the RMS values are [itex]E_{RMS}=\frac{E}{\sqrt{2}}[/itex] and [itex]I_{RMS}=\frac{I}{\sqrt{2}}[/itex]. Hence, we get
    [tex]P=E_{RMS}I_{RMS}[/tex]
    for the sinusoidal waveforms of the resistor. We see the equivalence of the AC so-called active power with the power in DC. The RMS values in AC represent the values needed to perform the same amount of work in DC with a resistor. We can express this active power by using Ohm's law in other forms for the resistor
    [tex]P=E_{RMS}I_{RMS}=RI_{RMS}^2=\frac{E_{RMS}^2}{R}[/tex]


    The Inductor

    This case is a bit trickier. Let us assume we have the same voltage source [itex]e(t)=E\cos{\omega t}[/itex] applied to an inductor of inductance [itex]L[/itex]. The current of an inductor is lagging by 90 degrees its voltage, so the current is [itex]i(t)=I\sin{\omega t}[/itex]
    From the definition of [itex]p(t)[/itex] above, we get
    [tex] p(t)=e(t)i(t)[/tex]
    [tex] =E\cos{\omega t}\cdot I\sin{\omega t}[/tex]
    [tex] =\frac{1}{2}EI(\sin{2\omega t})[/tex]
    (the last step included the identity [itex]2\cos{\omega t}\cdot \sin{\omega t}=\sin{2\omega t}[/itex])
    Here, we see a big difference from the case of the resistor. The mean value of instantaneous power is 0. Which means that energy is transferred back and forth within a cycle (actually twice within a period of [itex]e(t)[/itex]).

    Even though the mean instantaneous power is 0, that doesn't mean we don't have current in the source necessary for the inductance (in practice, this current has to flow in the power lines, cables, etc.). Hence, we need a measure of this fictitious necessary power that draws AC current from the source. In a similar fashion as in the resistor [itex](p(t)=P+P\cos{2\omega t})[/itex], we can define this reactive power [itex]Q[/itex] as [itex]p(t)=Q\sin{2\omega t}[/itex], where [itex]Q[/itex] will be again given by
    [tex]Q=\frac{1}{2}EI[/tex]
    or
    [tex]Q=E_{RMS}I_{RMS}[/tex]
    These formulas are, of course, for the inductor. Then again, we could replace the last equation by using Ohm's law (in complex form)
    [tex]Q=E_{RMS}I_{RMS}=L\omega I_{RMS}^2=\frac{E_{RMS}^2}{L\omega}[/tex]

    The impedance

    Since the capacitor case is relatively simple, let us now look at the impedance case, where we get an arbitrary phase difference [itex]\theta[/itex] between voltage and current waveforms. Again, let us assume we have the same voltage source [itex]e(t)=E\cos{\omega t}[/itex]. The current of the impedance will be of the form [itex]i(t)=I\cos{(\omega t-\theta)}[/itex]. So we will get for the instantaneous power
    [tex] p(t)=e(t)i(t)[/tex]
    [tex] =E\cos{\omega t}\cdot I\cos{(\omega t-\theta)}[/tex]
    [tex] =EI\cos{\omega t}(\cos{\omega t}\cos{\theta}+\sin{\omega t}\sin{\theta})[/tex]
    [tex] =EI(\cos^2{\omega t}\cos{\theta}+\cos{\omega t}\sin{\omega t}\sin{\theta})[/tex]
    with the 2 identities mentioned earlier, we finally get
    [tex] p(t)=\frac{1}{2}EI\cos{\theta}(1+\cos{2\omega t})+\frac{1}{2}EI\sin{\theta}(\sin{2\omega t})[/tex]
    [tex] =P(1+\cos{2\omega t})+Q(\sin{2\omega t})[/tex]

    We can see that in the general case (impedance), the active power in the last equation is now [itex]\frac{1}{2}EI\cos{\theta}[/itex] and the reactive power is now [itex]\frac{1}{2}EI\sin{\theta}[/itex]. Since both powers do affect the current that must be drawn from the AC source, there was also a name given for that power, which is the apparent power. Following the complex notation for the impedance, we can formulate this power as a complex power, which real part comes from active power [itex]P[/itex] and the imaginary part comes from [itex]Q[/itex]
    [tex] \vec{S}=P+jQ[/tex]
    [tex] S=\frac{1}{2}EI=E_{RMS}I_{RMS}[/tex]

    **From an AC point of view, this is all very important. In order not to mix these powers, each one has a different unit (even though all equivalent to the SI unit W for power). Active power is in W, reactive power is in var (lower-case), which stands for volt-ampere reactive and apparent power is in VA (volt-ampere). The active power is the power that creates work (lift charges, heating, etc.). The reactive power is stored and restored power (in the form of electromagnetic field). The apparent power is the result of both powers, it is what you will measure in the end, by multiplying the RMS voltage and current. Hence, this is why you usually have ratings for transformers in VA, because depending on what the impedance (connected to it) is, you must not have more than a certain amount of current (RMS) in the conductors (in order for them not to overheat) and a certain amount of voltage (RMS) on the ends of the conductors (in order for the dielectric to withstand this voltage). Of course, there is always a margin for these (steady-state) values, but one has to understand that if there was a transformer rated 10W, that would mean I could connect 10W and whatever reactive power to it (such as 1Mvar). But the currents drawn from that load would melt the conductors right away... This is why the distinction in AC is important.

    Sorry for the long post, perhaps only the last part is necessary.

    M.
     
  8. Aug 22, 2011 #7
    Nice explanation, thanks for that.
     
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