PF Correction, Average Power Assumption

Discussion in 'General Electronics Chat' started by oogieoogieful, Dec 2, 2013.

  1. oogieoogieful

    Thread Starter New Member

    Dec 2, 2013
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    In the Volume 2, Chapter 11 there is an example problem showing how to perform power correction:

    http://www.allaboutcircuits.com/vol_2/chpt_11/4.html

    My question is about the part where average power (the text calls it true power) is assumed to stay fixed when capacitive reactive power is added to inductive reactive power,

    "This correction, of course, will not change the amount of true power consumed by the load, but it will result in a substantial reduction of apparent power, and of the total current drawn from the 240 Volt source."

    How can we assume the correction will not change the amount of average or true power?
     
  2. JDT

    Well-Known Member

    Feb 12, 2009
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    Because the True Power is dissipated in the resistive part of the load. The inductive part does not dissipate power.

    The capacitance that you add to improve the power factor also does not dissipate power. Therefore the true power remains unchanged.

    (Provided the PFC capacitor is perfect, of course. Real capacitors will dissipate some power but this should be a very small proportion.)

    The power factor correction reduces the supply current to a minimum (if the PF = 1). This reduces power dissipated in the wires and transformers conducting the current to the load. The power dissipated in the load remains unchanged.
     
  3. Ramussons

    Active Member

    May 3, 2013
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    In the example indicated above, there will be NO change in the readings of the meters, whether or not the capacitor is connected the way it is :mad:

    The capacitor should be Across the Load for the Meters to show what the example claims.

    Ramesh
     
  4. oogieoogieful

    Thread Starter New Member

    Dec 2, 2013
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    When one adds capacitors in parallel, it changes both resistance and reactance of the impedance. Equivalent Impedance for components in parallel is

    Z = ((1/R+jX)+(1/-jX))^-1

    (This formula still treats the capacitor as if it were ideal, i.e. no resistance.)


    If the capacitor were added in series, it would not change the resistive part of the
    load, only the reactance part.

    Z = (R+jX) + (-jX).

    It appears that adding capacitors in parallel does dissipate power, the next question is how much?
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    In the truly general case, you can't assume that adding a reactive component won't change the real power. But in this case you can. The reason is that the capacitor is added in parallel to the source and therefore has no effect on the voltage applied to the load which then means that the load sees nothing different at all -- is still draws the same current at the same phase angle. That current can be viewed as the superposition of two currents, one that goes to the resistive part of the load and one that goes to the reactive part. The only thing the capacitor does is make it so that the current that is going to the reactive part is shuttled back and forth between the load and the capacitor while the source provides the current that goes to the resistive part of the load.

    If you were to include line resistance between the source and the capacitor, then you would see the addition of the capacitor result in a change in the real power dissipated in the load. The effect here is that, by reducing the current drawn from the source, there is less of a voltage drop in the line resistance and a corresponding increase in voltage appearing across the load.
     
  6. oogieoogieful

    Thread Starter New Member

    Dec 2, 2013
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    If the load has any resistance (not equal to 0 ohms) the voltage across it will change when capacitors in parallel are added. (To be clear, the set-up I am describing does not take into account a line resistance.)

    Given a load with an impedance Z = R + jX and capacitive impedance Z = -jX

    The new impedance is Z = r +jX.

    If R is Resistance before capacitor is added and r is resistance after capacitor is added in parallel, does R = r?

    One must first find equivalent impedance of load and capacitor. For components in parallel:

    Z = ((1/(R+jx)+(1/-jX))^-1 = ((R+jX)(-jX))/((R+jX)+(-jx)) = r +jX

    The only way the R = r is if the R = 0.

    Is the author claiming that R = 0 for the load in the diagram? I doubt it.

    The assumption must be something like the difference between R and r does not change the voltage across the load enough to matter.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    The problem is that you are answering a question about oranges by analyzing apples.

    The claim is that the average power dissipated in the load is unchanged, not that the resistive component of the load impedance is unchanged.
     
  8. oogieoogieful

    Thread Starter New Member

    Dec 2, 2013
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    My post responded to your claim, "The reason [one can assume that adding a reactive component wont change the true power] is that the capacitor is added in parallel to the source and therefore has no effect on the voltage applied to the load which then means that the load sees nothing different at all -- is still draws the same current at the same phase angle."

    I am showing how the voltage across the load does change when capacitors are added in parallel, since the impedance changes. According to Ohm's Law, the load cant draw the same current at the same phase angle.

    That is why I think analyzing the impedance is relevant.

    I am not claiming it is changing voltage of the supply, but the load. Perhaps this is where my reasoning goes wrong?
     
  9. WBahn

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    Mar 31, 2012
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    If the voltage across the load changes, then that must mean that the voltage across the supply changes, right? After all, they are in parallel. But how can the voltage across an ideal voltage source change?

    Give me an example of a simple circuit, configured like this one, in which the voltage across the load changes (and, no, you can't make the load impedance zero).
     
  10. oogieoogieful

    Thread Starter New Member

    Dec 2, 2013
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    Okay, I do see how I was writing about the voltage supply wrong. But I think I am right that the load draws neither the same current nor a current at the same phase angle. Isn't that the whole point of power correction?
     
  11. WBahn

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    Mar 31, 2012
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    If I tell you that the voltage across a load with a particular impedance is Vbob, can you tell me the current, both magnitude and phase, through that load? If I then make some changes on my side of a wall and tell you that the voltage is now Vfred, can you tell me the new current? What will the new current be relative to the old current if Vfred and Vbob are the same?
     
  12. oogieoogieful

    Thread Starter New Member

    Dec 2, 2013
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    Yes, I would apply Ohm's Law, since the voltages is constant, the current depends on the impedance.

    If the impedance does not change, the new current is the same.

    If the impedance does change, the new current will be different.

    Furthermore, the change in impedance depends on how the circuit was changed, namely were component(s) removed/added in parallel or in series.
     
    Last edited: Dec 5, 2013
  13. WBahn

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    Mar 31, 2012
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    But the impedance of the load did not change, hence the current in it does not change.

    The component that was added was a capacitor added in parallel to the load.

    The load sees NO change. The source DOES see a change. From the source's viewpoint, it sees a net load the has a different impedance than the original load. But the question wasn't about the current that the source delivers, it was about the average power dissipated by the load. If the load sees no change in the source source powering it, then the power dissipated in it doesn't change.

    Do the math! Figure out the average power dissipated in a generic load having an impedance of Z=R+jX. Then put a capacitor in parallel with it and figure out the average power dissipated in that same load. Then compare them.
     
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