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Capacitance power factor question

  1. Mar 30, 2012 #1
    1. The problem statement, all variables and given/known data

    A 50 KW load operates from a 60 Hz 10KV rms line with a power factor of 60% lagging. Determine the capacitance that must be placed in parallel with the load to achieve a 90% lagging power factor.


    3. The attempt at a solution



    for 60% p.f:-

    θ1 = Cos-1(0.6) = 53.13 degrees

    Q1 = 50,000*tan(53.13) = 66.667 kVAR

    for 90% p.f:-

    θ2 = Cos-1(0.9) = 25.84 degrees

    Q2 = 50,000*tan(25.84) = 24.22 kVAR



    Qcapacitor = Q2 - Q1 = -42.45 kVAR



    Xc = VRMS /Qc = (104)2 / -42450 = -2356 Ω


    ω = 2.∏.f = 2.∏.60 = 377

    C = 1 / ω.Xc = 1 / 377*2356 = 1.126 μF



    can anyone check if i'm on the right path here please?
     
    Last edited: Mar 30, 2012
  2. jcsd
  3. Apr 2, 2012 #2
    anyone?
     
  4. Apr 2, 2012 #3

    NascentOxygen

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    Staff: Mentor

    I posted a response to this about 6 hours ago, saying that's the answer I got.

    Not sure where my reply went. https://www.physicsforums.com/images/icons/icon9.gif [Broken]
     
    Last edited by a moderator: May 5, 2017
  5. Apr 2, 2012 #4
    Ok thats good. thank you
     
  6. Sep 19, 2013 #5
    Was this answer correct as i am getting a different value..?
     
  7. Sep 20, 2013 #6
    yes it is correct.
     
  8. Aug 6, 2014 #7
    message deleted
     
    Last edited: Aug 6, 2014
  9. Apr 30, 2017 #8

    why isn't it - 2356 in the final formula ?
     
  10. Apr 30, 2017 #9

    gneill

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    Staff: Mentor

    oxon88 calculated the signed reactance of the capacitor to be -2356 Ω. The minus sign reflects the fact that it is capacitive reactance. He then used the formula for the magnitude of the reactance, which is unsigned, to find the value of the capacitor.

    An alternative approach might have been to use complex impedance which retains the sign throughout, but that would achieve the same end result.
     
  11. Apr 30, 2017 #10
    Thanks for the reply i now understand
     
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