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Power Factor correction

  1. May 25, 2013 #1
    The problem statement

    Using the nomogram of FIGURE 10, calculate the size of
    capacitor bank to give a final power factor of 0.95 for a single phase,
    230 V, 50 Hz lighting installation which has a load current of 17.5 A and
    operates at 0.8 pf.

    The nonogram gives a value k = 0.44

    Relevant equations

    Qc = Pk

    Xc = 1/2pifC

    P = VI



    The attempt at a solution

    I have come up with two ways but i think the first is wrong as the current though the capasitor (If in parrallel) will not be the same as the load current.

    Attempt 1

    P = 230 x 17.5 = 4025W

    Qc = 4.03 x 0.44 = 1.77 KVAr

    P = I^2 x Xc

    Xc = P/I^2 = 1.77 x 1000 x 17.5^2 = 5.783Ω

    so C would come out at 550μF (as I said I think this is wrong)

    Attempt 2

    Power still = 4025W and Qc = 1.77 KVAr

    but P = V^2/Xc so Xc = V^2/P

    Xc = 230^2 x 1.77 x 1000 = 18.38Ω

    so C would work out at 173μF

    Just a bit of guidance would be appreciated.
     
  2. jcsd
  3. May 25, 2013 #2

    gneill

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    Staff: Mentor

    Hi topcat123, Welcome to Physics Forums.

    While I'm not that familiar with the usual methodologies used in Power Systems (like the nomogram you mention), I can suggest a more generic approach.

    Since you are given the current magnitude and the power factor you can write the current phasor (complex or polar form). Knowing the voltage too, you can write the impedance or admittance of the load (Y = 1/Z = I/V).

    Recalling that admittances add in parallel, you can find the admittance (of a capacitor) to add in order to bring the pf from 0.8 to 0.95; It's just a matter of reducing the imaginary component of the total admittance to the desired value, and capacitors have negative imaginary admittance...
     
  4. May 26, 2013 #3

    rude man

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    Homework Helper
    Gold Member

    Ahem .... :smile:
     
  5. May 26, 2013 #4

    gneill

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    Hah! Well caught. Dunno why that brain cell fired :smile:
     
  6. May 27, 2013 #5
    So draw a phase diagram for 0.8 pf.
    Work out the restive current.
    Overlay the new 0.95 pf and work out the change in the reactive component.
    from that i con work out the capacitor size.

    Doing it this way i get 123μF

    before i got 173μF
     
  7. May 27, 2013 #6

    gneill

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    Staff: Mentor

    Can you show more of your work?

    I think it would be better to work with admittances, since they add in parallel (presumably the capacitor bank will be placed in parallel with the load). I'm seeing a smaller value of capacitance for the result.
     
  8. May 27, 2013 #7
    Ok

    Z=V/I = 230/17.5 = 13.14Ω

    cos θ = 0.8 so R = 0.8 x 13.14 = 10.512Ω

    Xl = √Z^2 - R^2 = 62.16Ω

    New pf = 0.95 = cos θ
    θ = 18.19°
    R tan 18.19° = Xl-Xc =3.455Ω

    There for Xc = 58.71Ω

    C=1/2πfXc = 54.22μF
     
  9. May 27, 2013 #8

    gneill

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    Check your calculation of Xl. 62.16Ω looks too large.

    Remember that impedances in parallel don't add algebraically. Admittances (1/Z) add.
     
  10. May 27, 2013 #9
    yes Xl should be 7.88Ω forgot to √it.

    So Xl-Xc = 3.455Ω
    (1/3.455)-(1/7.88) = 1/Xc

    Xc = 6.152Ω

    C=1/2πfXc = 1/(2π x 50 x 6.152) = 0.000517F

    Is this better?
     
  11. May 27, 2013 #10

    gneill

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    Staff: Mentor

    One problem, while it is true that admittance is the inverse of impedance, Y = 1/Z, it is not true that the imaginary part of the admittance is equal to the inverse of the imaginary part of the impedance. That is, Im(Y) ≠ 1/Im(Z).

    Start by finding Y = 1/Z. Then work with its components; you want to reduce its polar angle in the same fashion that you're trying to do for the impedance.
     
  12. May 27, 2013 #11
    Z = V/I = 230/17.5 = 13.14Ω

    Y = 1/Z = 0.076

    G = Y cosθ = 0.076 x 0.8 = 0.061

    Bl =√z^2 - G^2 = 0.076^2 - 0.061^2 = 0.045

    new pf 0.95

    cosθ = 0.95 θ = 18.19°

    Bl-Bc = G tan 18.19° = 0.020

    Bc = 0.025

    Xc = 1/Bc = 40Ω

    which makes C = 79.5μF

    Hope i have it this time...
     
  13. May 27, 2013 #12

    gneill

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    That looks MUCH better! I think if you hang on to a few more digits of accuracy for intermediate values your final result will be closer to 81.6 μF.

    Cheers!
     
  14. May 27, 2013 #13
    Thanks for the help gneill.
     
  15. May 27, 2013 #14

    gneill

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    Staff: Mentor

    No worries, I'm glad to help!
     
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