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Different way of calculating power factor from multiple sources

  1. Apr 4, 2014 #1
    Hi,

    I did alot of googling and tried to search every forum I could, but I didn't quite found what I was looking for. Sorry if this has been asked before, english isn't my first language so there's a chance that I used wrong words to search for this question.

    Also I'm not sure if you use the same type of symbols that we use in Finland for electrical engineering. I will try to explain them if I can =)

    Okay to the problem in hand. I am an electrical engineering student, I have studied for 2 years. We had many instances that we had to calculate power factors (symbol that I used is "cos(phi)" and apparent powers (symbol that I used is "S") that the main transformer "sees" in the secondary side.

    Here is a picture that I hope clarifies the problem:
    http://postimg.org/image/jozu0g543/

    Then the question, I have always calculated all the different active- ("P") and reactive ("Q") powers and taken the square of their sums power to get the apparent power ("S"). Equation S = sqrt((sum P)^2 + (sum Q)^2).

    And from that I have calculated the power factor. Is there a faster/smarter way of doing this? The way I have been using isn't a problem if you are using under 10 outputs but after that it is quite laborious!

    Thanks for your input, and sorry for the long "question", also for my English!
     
  2. jcsd
  3. Apr 4, 2014 #2
    I know that you can calculate the secondary apparent power from the transformer by
    S = sqrt((AP(op1) + AP(op2) + AP(op3) + AP(op4))^2 + (RP(op1) + RP(op2) + RP(op3) + RP(op4)^2),
    which is about 12,064 MVA. op = output, AP = active power and RP = reactive power.

    And from that you can calculate the power factor
    cos(phi) = (AP(op1) + AP(op2) + AP(op3) + AP(op4)) / S, which is about 0,9201

    But is there a better way to do this?
     
  4. Apr 5, 2014 #3

    jim hardy

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    Your English is quite good.


    For the loads, you're given all the RP(Real Powers) and all the power factors cos(ø) ?

    for each load, is its ø = cos-1(cos(ø)),
    ...where cos(ø) was given;
    and is S = RP / cos(ø) ?
    and is AP = S X sin(ø) ?


    so you could sum the real powers and the reactive powers , as you said,
    and ø = tan-1(∑reactive/∑real)),,,

    I can remember how to do it with my slide rule and a piece of paper, so it shouldn't be difficult with a calculator.
    power factor = cos(tan-1(∑reactive/∑real) which looks awful but is only a few keystrokes...

    But i don't know whether it would be fewer keystrokes than all those squares and square roots....

    try it and see which you prefer.
     
    Last edited: Apr 5, 2014
  5. Apr 5, 2014 #4

    jim hardy

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    hmmm

    S = RP/cos(ø)

    and

    AP = S X sin(ø)

    so does AP = RP X sin(ø)/cos(ø)

    or, AP = RP X tan(ø) ?
    be careful about programming that into a calculator because it'll blow up at zero pf...
     
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