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Power Factor Correction

  1. Apr 12, 2010 #1
    I've been working on a mathematical problem for a few days now. I'm not sure where my math is falling apart.

    I have an impedance connected in series with a source. In phasor notation, the impedance is 1.2649 at 71.565 degrees. In rectangular, this impedance is .4 + j1.2 . The power factor for this circuit is 0.32 lagging. I would like to make the circuit have a power factor of .98 leading, while retaining the original resistance (.4).

    One solution is to place a component in parallel that provides reactive compensation.

    I have attempted the solution in two ways. One is to convert the impedance into admittance, since parallel admittances add together (making the math simpler). The other way is to work through the math of two parallel impedances.

    When I try the problem via admittance values, I get a final answer that is a capacitor. This makes sense, as any additional real resistance will change the real power dissipation of the total circuit (my goal here is to make sure real power dissipation stays the same, even with power factor correction).

    When I try the problem via impedance values, my parallel component does not solve to the inverse value of the parallel admittance component I found in my first solution.

    In fact, when I calculate what the total impedance should be (load in parallel with component), my answer does not equal the inverse of the admittance.

    The total admittance I get from my first solution is .25 + j0.0508.

    The total impedance I get from my second solution is .4 - j0.0812.

    These two values are not inverse of each other. Why is that?
  2. jcsd
  3. Apr 12, 2010 #2
    The admittance corresponding to an impedance of 0.4+j1.2 ohms is 0.25-j0.75 mhos.

    Bob S
  4. Apr 12, 2010 #3
    Is this homework?

    Is your question what value of reactance do I add to a 0.32 lagging power factor to get a 0.98 leading power factor or is it why is the admittance value not the inverse of the impedance value?

    Instead of putting the reactance in parallel have you considered putting it in series?
  5. Apr 12, 2010 #4
    This is not homework, or I'd have posted it in the homework section.

    My question is not what value of reactance I need to place in parallel (I already have that answer). My question is: why do I not get the same answer when I evaluate the system as an impedance?

    I will post a visual reply (shortly) to make more sense of this.
  6. Apr 12, 2010 #5
    To answer that question we would need to see your calculations.
    Last edited: Apr 12, 2010
  7. Apr 12, 2010 #6
    I just finished drawing it up.

    Attached Files:

  8. Apr 12, 2010 #7
    You can see, I try to solve the problem two seperate ways: one via impedance, one via admittance. I am able to find the right-handed solution, but not the left. Why are the two values for total impedance/admittance inverse of each other? They should be, since I took the load impedance and admittance values and calculated what the necessary total values should be for the same power factor.
  9. Apr 12, 2010 #8
    When you add capacitance in parallel to an inductive impedance, you increase the value of R. In order to maintain R at the same value, you must add reactance in series. Your answer of 0.4 -j0.0812 is correct for a PF of 0.98. All you have to do is subtract the inductive reactance from that for the total capacitive reactance you need to add in series.
    Last edited: Apr 12, 2010
  10. Apr 12, 2010 #9
    I'm still not sure that answers my question. It seems to me that the answer should be the same, regardless of whether or not I choose to model the system as an admittance or an impedance. Why then, do I not get the exact (but inverse) results for what my system should look like at the correct power factor?

    Anyone else have any thoughts?
  11. Apr 12, 2010 #10
    The reciprocal of Z should equal Y and vice versa. The fact that they don't means your arithmetic is wrong. You don't show enough of your work to determine where the problem is. Can you explain in more detail how you get the reciprocal of ZL or YL? How did you calculate the value of ZC and YC? How do you combine ZC with ZL to get ZT? The same for YC, YL and YT.
  12. Apr 13, 2010 #11
    Gladly! Here are the mathematics. I've highlighted my conundrum.

    Note: Stating again, my problem is that those two values should be inverse of each other. The total impedance (Zt) (power factor corrected) should be the inverse of the total admittance (Yt). In the end, you're analyzing the same circuit, so why shouldn't they be inverse?


    Where am I going wrong with my math?
  13. Apr 13, 2010 #12
    Having a problem with the upload service. Will try again in a few.
  14. Apr 13, 2010 #13
    I had my file size too large. I've attached it as a .jpg. However, the quality might be too poor. So I've included a URL as well.

    http://i993.photobucket.com/albums/af55/khaldirth/PFProblem2.jpg [Broken]

    Attached Files:

    Last edited by a moderator: May 4, 2017
  15. Apr 13, 2010 #14
    Zl = 0.4 + j1.2

    For a PF of 0.98 and a real part of 0.4, the imaginary part must be 0.4 * sin(acos(.98))/-0.98 = -j0.08122.

    My way:
    The reactance that must be added in series to get that value is -j0.08122 - j1.2 = -j1.28122.

    Your way:
    To find Zc given Zl and Zt
    Zc = (Zl * Zt) / (Zl - Zt)

    Zl mag = sqrt(0.4^2 + 1.2^2) = 1.264911
    Zt mag = sqrt(0.4^2 + -0.08122^2) = 0.408163
    (Zl * Zt) mag = Zl mag * Zt mag = 0.51629

    Zl ang = atan(1.2 / 0.4) = 1.249046 radians
    Zt ang = atan(-0.08122 / 0.4) = -0.20033 radians
    (Zl * Zt) ang = Zl + Zt = 1.249 + -0.20033 = 1.0487 radians

    Zl - Zt = (0.4 +j1.2) - (0.4 - j0.08122) = 0.8 + j1.28122
    (Zl - Zt) mag = sqrt(0.8^2 + 1.28122^2) = 1.28122
    (Zl - Zt) ang = atan(1.28122 / 08) = -1.5708

    (Zl * Zt) / (Zl - Zt) mag = 0.51629 / 1.28122 = 0.402967
    (Zl * Zt) / (Zl - Zt) ang = 1.0487 - 1.5708 = -0.52209

    Zc real = 0.402967 * cos(-0.52209) = 0.349284
    Zc imag = 0.402967 * sin(-0.52209) = -0.200955

    There are two ways to get convert 0.4 + j1.2 to 0.4 - 0.08122. The first is to add 0 - j1.28122 in series. The second is to add 0.349284 - j0.200955 in parallel.
  16. Apr 14, 2010 #15
    Alright. So you say that the second way is to add an impedance of 0.349284 - j0.200955 in parallel to my load impedance.

    It is safe to say that this would be the same as adding an admittance of 1/(0.349284 - j0.200955) in parallel to my load admittance (.25 - j0.75)? The total admittance at this point would be (.25-j0.75) + (2.15099-j1.2375431), which equals 2.40099 + j.487543. This is not the same answer that I got in my admittance-based calculations (I determined that the final admittance would be .25 + j0.0508.
  17. Apr 14, 2010 #16
    Perhaps this touches on some concept of duality that I'm not getting?
  18. Apr 14, 2010 #17
    Yes it is safe to say that this would be the same as adding an admittance of 1/(0.349284 - j0.200955) in parallel to the load admittance (.25 - j0.75). However, 1/(0.349284 - j0.200955) is not 2.15099 - j1.2375431 but 2.15099 + j1.2375431 which when added to 0.25 - j0.75 equals 2.400998 + j0.487543. 1/(2.400998 + j0.487543) = 0.4 - j0.081223.
  19. Apr 14, 2010 #18
    Alright, I understand your math. I agree that you can place 0.349284 - j0.200955 in parallel with the load to get the desired impedance. What I don't understand is why you can also add an admittance of j0.801 in parallel and get the same circuit. You see in my right-hand calculations, the YC is j0.801 in parallel with the load admittance. Do you agree that this is the same as placing a 1/(j0.801) impedance in parallel with the load? This would be an impedance of -j1.2489. So my right side calculations show that you can add an impedance of -j1.2489 in parallel with the load. Your parallel impedance calculations also show that you can add 0.349284 - j0.200955 in parallel to the load. To me, that's two different circuits, and that's why I don't understand how both answers are right. Like I posted earlier, perhaps there's a concept of duality I am missing here?

    Also, thank you for entertaining the questions.
  20. Apr 14, 2010 #19
    No I don't agree. You cannot add a reactance to an impedance in parallel without affecting the value of the real part of the impedance. Complex arithmetic is not always intuitive. This is why I wanted to see your work, to see where the error was. It is also why I showed you my work, so you could follow the arithmetic.
  21. Apr 14, 2010 #20

    The Electrician

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    Gold Member

    The mistake you're making in all this is a false assumption that if you keep .25 as the real part of the final load expressed as an admittance, that when expressed as an impedance, its real part will be .4, which is your requirement.
  22. Apr 15, 2010 #21
    So, you're saying that if someone asked me to correct the power factor of a known load without changing the real impedance, I would not be able to do so by running the calculations via parallel admittance values (as I tried to do)?
  23. Apr 15, 2010 #22

    The Electrician

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    Gold Member

    Nope, that's not what I said.

    What I said is, if you want to do things on an admittance basis, you can't get the answer you want by keeping the real part of the admittance equal to .25, which is what you did so far.

    Your goal is to keep the real part of the impedance equal to .4, and that's what you have to do, whether you do things on an admittance or impedance basis.
  24. Apr 15, 2010 #23
    Okay. So how would you run the calculation for parallel admittances, such that the final result would give you an impedance with a real part of .4?
  25. Apr 15, 2010 #24
    Just as I showed you in post #14. As you know, the formula to calculate the parallel impedance given the two elements is Zt = (Z1 * Z2) / (Z1 + Z2). If you already know the resultant impedance and one of the two elements as in this case, the formula is Z2 = (Z1 * Zt) / (Z1 - Zt).
  26. Apr 15, 2010 #25
    I'm saying, via parallel impedances method Yt = Yc + YL. I'm asking this, because it seems to me that the solution to this problem is determined by the way you solve it. In other words, can you do the math (Yt = Yc + Yl, as seen in my attachment) in such a way that the result yields a real impedance of .4 with a pf of .98 leading?
    Last edited: Apr 15, 2010
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