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Can anybody help with this power question?

  1. Sep 29, 2011 #1
    Hi Guys,

    This is my first post here. I am preparing for the EIT exam in October. I came across this practice problem during my review.

    Question:
    The circuit shown has a power factor of 0.8. What size capacitor must be added to this circuit to correct the power factor to 0.9?

    Drawing1.jpg

    (A) 10 kVAR in series
    (B) 10 kVAR in parallel
    (C) 13 kVAR in series
    (D) 13 kVAR in parallel

    It is easily calculated that the needed reactive power is ~13kVAR. However,the questions is about how to connect it: series or parallel.

    I think 13kVAR shall be connected in parallel to ensure that the existing load still draws 50kW at 0.8pf (by maintaining the terminal voltage, given the load is a constant impedance). However, the answer provided is (C). I can see the point of choosing (C) if the load is a constant KVA load. But if this is true, (D) is also a a correct answer isn't it?

    Let me know what you guys think.
     
  2. jcsd
  3. Sep 30, 2011 #2

    uart

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    Yes, you are correct. I can't imagine why the supplied answer was "C", other than a typo.
     
  4. Sep 30, 2011 #3
    It could be argued that the inductance in the circuit causing the power factor reduction, is in series with the resistance so the capacitance that must be added must also be in series. Another reason is that adding capacitance in parallel, but not enough to bring the power factor to unity, will raise the impedance, not lower it, reducing the power to the load.
     
  5. Sep 30, 2011 #4

    uart

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    No, not really. When the supply is a voltage source (or at least approx a VS) then the load power is unchanged by the addition of a parallel capacitor. Yes the overall impedance is increased and the reactive power reduced, but that's kind of the point of PFC.
     
  6. Sep 30, 2011 #5
    After doing a little complex arithmetic, I must change what I posted earlier.

    Adding -j13.3 in parallel to a series 50 + j12.5 (for pf = 0.8) results in an impedance of 3.53 - j13.2 or a magnitude of 13.7 KVA with a pf of 0.26.

    So contrary to my earlier post the impedance goes down because the capacitance overwhelms the inductance. This means the KVA would increase, not decrease.
     
  7. Sep 30, 2011 #6

    uart

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    My calculations are a little different.

    For V=480 and P=50 kW at PF=0.8, I calculate Z = 2.949 + j 2.212 ohms for the initial load impedance.

    For Q=13 KVAR I calculate the required parallel capacitor impedance is Zc = -j 17.72 ohms.

    For the combined parallel impedance of the above I get, Zt = 3.716 + j 1.821 ohms.

    As you can see the magnitude of the impedance has increased and it's angle has decreased (such that cos(phi) is now approx 0.9).
     
  8. Sep 30, 2011 #7
    A reactive current is not necessarily the inductive - it may be capacitive as well. There is nothing said also whether the active power should stay the same.
    The reactive (supposedly inductive) power is now 12.5 kVar (50/80 X 100 = 62.5 - 50 = 12.5) slightly less than the capacitive power (from the possibly right C variant) of 13 kVar. Although the capacitor(s) in series with the load will bring its active power drastically down to about 1 tength of what it is now, but, maybe then, the resultant capacitive reactive power will yield the number 0.9?
     
    Last edited: Sep 30, 2011
  9. Sep 30, 2011 #8

    uart

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    No. You can't calculate the reactive power by just subtracting the real power from the apparent power like that.

    It's VAR = sqrt(VA^2 - P^2), which give 37.5 KVAR reactive power!
     
  10. Sep 30, 2011 #9
    Yes, it is vector sum. (For PF 0.9 the reactive power here should be about 21kVAR).
     
    Last edited: Sep 30, 2011
  11. Sep 30, 2011 #10
    For V=480 and P=50 kW at PF=0.8, I calculate Z = 2.949 - j 2.212 ohms for the initial load impedance. 480^2 / [50000 + j37500] please note change in sign of reactance.

    For Q=13 KVAR I calculate the required capacitor reactance is Zc = -j 17.72 ohms.

    For the combined parallel impedance of the above I get, Zt = 2.281 - j 2.304 ohms. Calculation was done in Excel using the formula IMDIV(IMPRODUCT(COMPLEX(2.949,-2.212),COMPLEX(0,-17.72)),IMSUM(COMPLEX(2.949,-2.212),COMPLEX(0,-17.72))).

    This impedance has a magnitude of 3.242 ohms for a total KVA of 71065.

    As you can see the magnitude of the impedance has decreased and it's angle has increased (such that cos(phi) is now approx 0.704).
     
  12. Oct 1, 2011 #11

    uart

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    OK, granted that the question didn't explicitly say that the initial 0.8 PF was lagging.

    However, given that the question say's that the power factor is to be corrected with the addition of a capacitor, is there some particular reason why you chose to assume that the initial PF was leading?
     
  13. Oct 1, 2011 #12
    I believe an inductive load does cause a lagging power factor.

    480^2 / (50000 + j37500) = 2.949 - j2.212 The sign of the imaginary part changes when divided into a positive real.
     
  14. Oct 1, 2011 #13

    uart

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    Yes exactly, that's why we need a +ive imaginary part of the impedance (eg 2.949 + j2.212) to give rise to a lagging current.

    Your load impedance of 2.949 - j2.212 corresponds to a leading current and hence a capacitive load.

    The actual load in this question however is clearly NOT capacitive. The question clearly tells us that the addition of more capacitance will correct the power factor, not make it worse.
     
  15. Oct 1, 2011 #14
    "C" is typo as has been mentioned.

    Reactive current now at 0.8 PF is 77A.
    Reactive current at 0.9 will be 50A.
    13 kVAR capacitor provides 27 A reactive.
    Connecting the cap in parallel will bring the reactive current to 50 A and the PF to 0.9.
     
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