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Power factor correction

  1. Apr 3, 2014 #1
    1. The problem statement, all variables and given/known data

    A small industry operates from 220 volts supplied by a utility. The small
    industry represents a load to the utility that represents 22,000 watts and a power factor of 0.8.Determine the value of a capacitor placed at main voltage panel to correct the circuit to unity power factor. The supply frequency is 60 Hz.

    2. Relevant equations

    cosø = 0.8
    Xc = V/Ic
    3. The attempt at a solution

    I was able to get 80A as a value for I using P=VIcosø but it does not have an imaginary component which is what I need to solve the problem. Can someone please tell me how I am supposed to get the imaginary component?
  2. jcsd
  3. Apr 4, 2014 #2


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    Staff: Mentor

    Hi kaalad, Welcome to Physics Forums.

    Check your current calculation --- I don't see 80 A as the magnitude of the current.

    It's helpful to know about power triangles, current triangles, and impedance triangles for AC calculations. It's also worth knowing that the the phase angle associated with the power is the same as that for the impedance (i.e. the power and impedance triangles are similar triangles).

    If you know the magnitude of the current and the power factor then you can fill in the adjacent and opposite sides of the current triangle and hence write the current as a complex number (phasor). From there to the impedance is straightforward given that you known the source voltage.
  4. Apr 4, 2014 #3
    Thanks for replying :)

    I did over the calculations and got 100 A as the current.

    I also drew a current triangle and managed to get IL as 80 A and Ic as 60 A.
    This is where I realised that what I actually need (well what i think I need) is the phasor current of IL.

    What I learnt was that for the unity power factor, the imaginary component of the total current IT should be 0. And IT = IL + Ic

    Therefore, it seems that I may need to find the phasor current of IL as well and use the opposite of its imaginary part as Ic. For instance, if IL = 3 + j5, Ic would be equal to -j5 for a unity power factor.

    The value of Ic would then be substituted into the equation Xc = V/Ic.
    Xc would then be put into C = 1/(2∏fXC) to get the value of the required capacitance.

    Sorry I didn't put all this in the original question but I was a bit confused as to what was needed and I was a bit sleepy :redface:

    But in summary what I would like help with now is how to find the phasor current of IL. Of course, if you are aware of another way on how to solve the question I am open to suggestions.
  5. Apr 4, 2014 #4


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    Staff: Mentor

    Can you show your calculation? Which current value is that? The magnitude of the total current phasor or that of the real component only???

    Again, you should show your calculations so we can tell what it is you're calculating. Is IL the total load current? How do you arrive at Ic with the power triangle? (Assuming Ic is supposed to be the capacitor current???)

    You can find the magnitude of the total current from the real power, the source voltage, and power factor as you stated in the first post. What value do you get for the magnitude of the current? Show your calculation.

    A current triangle can then be created with this current value on the hypotenuse and using the power factor value to find the other side lengths: real current, imaginary current). That gives you the current phasor in rectangular form.

    With the source voltage and current phasor you can determine the impedance, or more handy, the admittance....
  6. Apr 5, 2014 #5
    Sorry! That's the total current value. BUT while I was typing it here i realized I did something wrong...again :frown: so what I put here is my new value.
    I used I = P/Vcosø
    = 22000/(22 x 0.8)
    = 1250 A

    So here are all the calculations based on this value (1250 A). Sorry again x.x

    (I'm not sure how to draw a triangle here so I'll put it in words)
    The Current triangle:
    The hypotenuse is the total current, I = 1250 A
    The angle was cos-1(0.8) = 36.9
    The opposite side is the capacitive current, Ic
    The adjacent side is the inductive current, IL

    IL = cos(36.9) x 1250 = 0.8 x 1250 = 1000
    Ic = sin(36.9) x 1250 = 0.6 x 1250 = 750

    ∴I = 1000 + j750 A

    For unity power factor, we would want I = 1000
    ∴ Ic of the required capacitor would have to be -j750

    So substituting |Ic| = 750 into |Xc| = |V|/|Ic|gives:
    |Xc| = 220/750 = 0.2933

    Then substituting |Xc| into C = 1/(2∏fXc) gives:
    C = 1/(2∏ x 60 x 0.2933)
    = 0.00904 F
    = 904 μF

    So the required capacitance is 904 μF.

    Hopefully I made everything clear this time :/
    Is this correct?
  7. Apr 5, 2014 #6


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    Staff: Mentor

    Right, except that the source voltage is 220 V, not 22 V. That means that rest of your current values (below) will be 10x too high. No worries, it's just a scaling issue that doesn't affect the basic procedure.

    Okay, bit of a problem there. The opposite side represents the REACTIVE current, that is, the imaginary portion of the current phasor, while the adjacent side represents the REAL current. There's no capacitor involved in the circuit yet.

    Yup, just divide that current by 10.

    And, thanks to the one slip above being just a scaling issue which cancels out in the calculation, you've arrived at a correct value for the capacitance :smile:

    Just fix your voltage value as mentioned above and scale the currents accordingly and you'll be okay.
  8. Apr 5, 2014 #7
    Whew! Thank you soooo much for your help and your patience!!! :smile:
  9. Apr 16, 2014 #8
    Just chiming in with another question...
    Looking at the power triangle, if you adjust the power factor on a main unit generator, does it cause the reactive power and apparent power to change, while maintaining the real power?
  10. Apr 16, 2014 #9


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    All of the quantities are interrelated. Changing the phase angle of the source will change the "allocation" of the power available to real or imaginary components of the power in the load. Usually the method of changing the power factor of the source involves altering the load in some way (even if it's adding a "load" at the source side of a distribution network).
  11. Apr 16, 2014 #10
    Well for the problem I am looking at, I must choose one that is maintained the same when changing the PF. So I imagined it was the real power that stayed the same. Because changing the angle, it has nothing to do with the adjacent leg of the angle.
  12. Apr 16, 2014 #11


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    Staff: Mentor

    Sure. But you would then choose the changes made accordingly. For example, you might have to change the source voltage along with the power factor.
  13. Apr 17, 2014 #12
    Here is what I used for the power triangle. I am not familiar at all with electric systems.

    The main page showing power triangle, apparent/real/reactive power:

    Using the side bar on the left to get to the "calculating power factor" page, I feel like my original answer is wrong. On this page it says:
    So if at PF=1, its a horizontal line, and if at PF=0 its a vertical line, that means it is changing the true (horizontal) and reactive (vertical) lines. However, the apparent power (hypotenuse) changes with both of these changes. So I dont know how I'm supposed to pick a side that will remain the same when changing the PF..:confused:
  14. Apr 17, 2014 #13


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    Staff: Mentor

    Why do you need to pick a side?

    But supposing that you want to keep the real power the same despite changing the power factor, you may find that you have to adjust something else, such as the source voltage to make it so.
  15. Apr 17, 2014 #14
    Because thats what the problem asks for. Out of the 3 "sides" (or powers), they ask by changing PF, which two sides change, which one remains the same.
  16. Apr 17, 2014 #15


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    Staff: Mentor

    The power factor can be changed by adjusting the resistive or reactive parts of the load, or both. In "real world" practice the adjustment of power factor is usually done so as to maintain the real power to the load as a constant and reduce the reactive power (which produces no work and is thus wasteful for the generator to have to produce). The practical way to accomplish this is to place an additional purely reactive load in parallel with the original load in order to at least partially cancel out the reactance of the load. This brings the power factor closer to 1. So I suppose in this sense the "side" that you want to remain constant in the power triangle is the real power.
  17. Apr 17, 2014 #16
    Thanks! that's what I was thinking too.
  18. Apr 17, 2014 #17


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    Gold Member

    Here's my take:
    220 volts
    22,000 watts
    PF= .8

    22,000/ .8 = 27,500 VA

    Using pythagrean theroem I can find my VARS, or the vertical part of the triangle.
    Vars= 16,500

    I know that V^2/R= Power

    In this case I'll substitute in VARS for power and 1/jwc for R.

    V^2/VARS = jwc

    220^2/16,500= j377C

    Ignore the J, solve for C. (The J flipped into denominator will just make it minus 90 degrees which we already know)
    C= 7.78 mF.

    Sound right....or did I screw that up?

    Two more things. Correcting to perfect power factor (unity or 1) creates harmful harmonics in factories....a .95 or .9 would be better in reality.
    Also, the beauty of power factor correction is that it does NOT affect the voltage or current thru your motors or inductive loads.
    If you had a motor with 220 volts and 125 amps at .8 power factor.....It will still be 220 volts and 125amps at the motor WITH your power factor correction of 1. This is quite convenient since a motor will not run properly if you change the voltage or amperage!
    Last edited: Apr 17, 2014
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