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Homework Help: KVAR needed to correct PF to 0.95

  1. Oct 5, 2015 #1
    Hello All,

    I need to know if i have done this correctly (its been some time):


    The 3-phase load at a bus is 9000 kW at .89 lagging PF. What is the total reactive power supplied by a 3-phase capacitor bank that will increase the PF to .95 lagging?


    9000kW/0.95 = 9473.68kVA

    kVAR = sqrt( (9473.68kVA)^2 - (9000kW)^2 )
    kVAR = 2958.14

    Thanks in advance.
  2. jcsd
  3. Oct 5, 2015 #2


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    Yes until now it's correct, but:
    ( from PF = 0.89 ).
  4. Oct 7, 2015 #3
    I think the way i did this is correct because it is asking for the total kVAR needed to achieve 95% PF, i think you are implying it says what is the additional kVAR needed to achieve 95% which would be difference of kVAR at 95% and kVAR at 89%.

  5. Oct 7, 2015 #4


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    Gold Member

    You must calculate the amount of KVAr's consumed by PF = 0.89. Now the capacitor will compensate/produce an amount of kVAr's, so that the new PF becomes 0.95.
    A capacitor is consuming negative reactive power, thus producing reactive power, and thereby increasing an inductive PF.

    Thoughts: I'm not in doubt about that. :wink:
  6. Oct 7, 2015 #5
    Ok so then if i understand correctly the following should be true:

    kVAR's at 89% = 4610
    kVAR's at 95% = 2958

    so kVAR's supplied by capacitor bank should be 4610-2958 = 1652 kVAR

    This would be true because the capacitor bank will effectively cancel out some kVAR to raise the power factor.
  7. Oct 7, 2015 #6


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    Yes, that's absolutely correct.
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