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Need some help -- Power Factor Correction for a Motor

  1. Dec 26, 2014 #1
    < Mentor Note -- thread moved to HH from the technical engineering forums, so no HH Template is shown >

    Hi there. Im wondering if someone can help me.. Below is a question and answer, I need help with IV . Where does the 0.51 and 0.28 come from? Ive tried cos and sin and thitha but doesnt give me correct number
     

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    Last edited by a moderator: Dec 26, 2014
  2. jcsd
  3. Dec 26, 2014 #2

    phinds

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    Seriously? You ask people for help by posting a tiny unreadable image of something
     
  4. Dec 26, 2014 #3
    Cant you enlarge it by opening the actual picture? Sorry but im new here..
     
  5. Dec 26, 2014 #4

    phinds

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    The "enlargement" is tiny

    The issue is not about this forum it's about the size of the image you posted.
     
  6. Dec 26, 2014 #5
    A 12.5 kW motor is to be powered from a 440 V, 50 Hz supply. Data from

    factory testing indicates the motor has an operating efficiency of 85% and

    that the power factor is 0.86 lagging. The power utility feeding the factory

    requires that the motor’s power factor is 0.96 lagging or better.


    (i) Explain how the addition of a capacitor to the motor would give the required power factor.

    (ii) Determine the motor current in the original configuration.

    (iii) Determine the supply current with the capacitor in place.

    (iv) Determine the current flowing in the capacitor.

    (v) Calculate the value of the capacitance which must be added to give the

    desired performance

    (vi) Determine the ‘VAr’ rating of the additional component.
     
  7. Dec 26, 2014 #6
    (ii) Efficiency of motor = Power output/power input,

    ð 0.85 = 12500/power input

    ð power input = 14706 W


    Power input = VImcosФ = (440)(I)(0.86)

    ð Im = 14706/(440)(0.78) =

    ð Im = 42.85 A

    The motor current Im is 42.85 A


    (iii) The current in the parallel capacitor leads the supply voltage by 90°.

    The phasor sum of the original current Im and the capacitor current Ic must be such that the supply current Is lags by 0.96 or better.


    Resistive component of motor current = Im cosФ = (42.85)(0.86) = 36.85 A


    Resistive component of new supply current = Is cosФ’ = Is (0.96) = 0.96 Is A


    As resistive component of current does not change => 36.85 = 0.96 Is

    => New supply current, Is = 38.39 A


    (iv) Reactive component of motor current = Im sinФ = (42.85)(0.51) = 21.87 A


    Reactive component of new supply current = Is sinФ’ = (38.39)(0.28) = 10.75 A


    As reactive component of current in capacitor is balance of current => Ic = 21.87 – 10.75

    => Capacitor current, Ic = 11.12 A


    (v) Capacitor current, Ic = V/Xc => Xc = V/Ic


    ð 1/(2πfC) = V/Ic

    ð C = Ic /(2πfV) = 11.12/(2π(50)(440))

    ð C = 1.26 x 10-4 = 80.44 μF



    (vi) The var rating of the capacitor = VIc = (440)(11.12) = 4.89 kvar
     
  8. Dec 26, 2014 #7
    Old PF is cosФ = 0.86 = arccos(0.86) = 30.68° so the sin of 30.68° degrees is sin (30.68°) = 0.51
    Since the new PF must be equal to cosФ = 0.96 = arccos(0.96) = 16.26° so the sin of 16.26° is 0.28.
     
    Last edited: Dec 26, 2014
  9. Dec 26, 2014 #8
    Thank you
     
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