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Power factor calculations?

Discussion in 'General Electronics Chat' started by RobertRoss, Oct 4, 2011.

  1. Ratchit

    Ratchit Well-Known Member

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    RobertRoss,

    Yes, you can do it that way, but your dividing by √2 to get the RMS value relies on the wave being sinusoidal. As others have pointed out, if nonsinusoidal, then that method does not give you the RMS value anymore. Also I²Z does not work either on a nonsinusoidal wave because something like a rectangular or triangle wave will have multifrequency components of different amplitudes. So what is the impedance "Z" of a component driven by a multifrequency wave? Have you ever seen anyone calculate or define it?

    When finding the average, the units of time are unimportant because they all get cancelled by the division. I could have used sin(2*pi*60*t) and divided up the intervals by (1/60*(1/360) and gotten the same answer, with more computation. However, the degrees represent a time unit of something, so it works for the average. The important thing is to make the intervals even.

    Wrong, the method of finding the true power from average the the VI product works for whatever shape the wave is as long as you have enough points to represent the wave.

    Yes, as long as it is sinusoidal. If not sinusoidal, then use the method I outlined.

    Now, the spreadsheet I sent you contains an error. The result is correct, but there is something wrong with the pair values at each point. Can you find the error?

    Ratch
     
  2. RobertRoss

    RobertRoss New Member

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    Hi Ratch,

    >However, the degrees represent a time unit of something, so it works for the average.
    So this is an average... not an RMS... right?


    >Now, the spreadsheet I sent you contains an error. The result is correct, but there is something wrong with the pair values at each point. Can you find >the error?

    The only thing I can think of is that the magnitude of the impedance (85....) is a constant , and shouldn't be!!! I am nit sure though.
    Other than that I see no other problem.... I guess you do huh!

    Get back I am curious.

    r
     
  3. Ratchit

    Ratchit Well-Known Member

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    RobertRoss,

    As I said in my first post in this thread, #54, all power is average. No such thing as RMS power. RMS is not involved in the method I suggested. You read the values of the instantaneous V or I and find the product. No such thing as a RMS instantaneous value, either. However, when computing RMS for V or I, RMS also involves an average.

    The impedance in a linear circuit is determined by the components and the frequency. Therefore in your example, it will always be constant unless you change the above.

    Indeed, I do. Look at the equation for the current. See something wrong?

    Ratch
     
  4. dave

    Dave New Member

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  5. RobertRoss

    RobertRoss New Member

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    Okay, here are a few things that may appear odd to me... but this is pretty much between a guess and an affirmation.

    #1) After 315 degrees, you keep on adding the degrees to 45.... and beyond this point you are the sin of greater degrees than 360. It still should give you the correct answer though. For example, the last one would be:

    360 + 45 = 405..... Convert deg to rads = 1.125 rads x 2PI....... In rads mode.... sin(7.06858) = 0.707
    then... (0.707 * 169)/85 = 1.40589.... that seems still pretty close!

    #2) Yous say only the current formula... well, both formula's 120V constant should still alternat from 120 to (-120)....

    #3) Mayby should be like this... but I don't think much difference, unless division done before multiplication!!!!
    =(120*SQRT(2)*SIN(RADIANS(A9+45.1517071234)))/85.07838132

    Sorry, Ratch, I don't see anything else....

    ross
     
    Last edited: Oct 9, 2011
  6. Ratchit

    Ratchit Well-Known Member

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    RobertRoss,

    The example you gave is for an series inductive circuit. Therefore the current lags the voltage. In the formula for the current, I have the current leading the voltage as it would in a capacitive circuit. Since the reactance is the same for both, the result for the true power is the same. Change the formula to =(120*SQRT(2)*SIN(RADIANS(A9-45.1517071234)))/85.07838132 to show a lag of approximately 45°. The negative VI product will then show up at the beginning of the period instead of the end, but the result will be the same.

    Ratch
     
  7. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    1. V & I sinusoidal. Special case. f not always 60Hz. Must know lead/lag info too for your way.

    2. 0 to 259 step 1 is 360 equal spaced data points.

    3. Any periodic shape of V or I as the reference. i.e. Calc good for one cycle
     
  8. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    If you change Ratch's input Lead/lag signs should change.
     
  9. RobertRoss

    RobertRoss New Member

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    Hi KISS,

    >1. V & I sinusoidal. Special case. f not always 60Hz. Must know lead/lag info too for your way.
    can't lead lag be determined by which the degree of the voltage and current in respect to the zero crossing. I mean by closely monitoring the voltage and cureent for the first 45 degrees of the wave form, we should be able to tell if V leads/lags current ... no?
     
  10. RobertRoss

    RobertRoss New Member

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    >If you change Ratch's input Lead/lag signs should change.

    understood.
     
  11. RobertRoss

    RobertRoss New Member

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    /////////kjhgkjhgk
     
    Last edited: Oct 9, 2011
  12. Ratchit

    Ratchit Well-Known Member

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    KISS,

    To whom are you addressing the above comments? It would be nice if you put a name at the top of your post. Whose way are you talking about?

    Ratch
     
    Last edited: Oct 9, 2011
  13. RobertRoss

    RobertRoss New Member

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    Ratch,

    >The example you gave is for an series inductive circuit. Therefore the current lags the voltage. In the formula ....

    aaaaaahahaharrrhr !

    I was looking, looking... and I knew it was some little tricky detail like this.

    One more thing. This is okay to measure an AC circuit... assuming its a 120VAC (Hot plus neutral) system . Suppose we were to measure a 240 VAC circuit... (Please view attachment) ...you know like the one that powers your hot water tank in your home. I know that in this case the circuit would be purely resistive... but let's suppose it was for other loads which would also have an inductive component as well. Given this type of circuit... since the voltages of one hot is 120VAC and the other hot is 120VAC but 180 degrees apart... can we do the same calculations on one of the 120VAC and doulbe the readings for the other 120VAC side.

    For example in our sample we get 120 watts of true power.... if it were a 240 circuit though.... it would obviously be 240 watts.... is this correct?


    Thanks Ratch.
     

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    Last edited: Oct 9, 2011
  14. RobertRoss

    RobertRoss New Member

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    No comments
     
    Last edited: Oct 9, 2011
  15. Ratchit

    Ratchit Well-Known Member

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    RobertRoss,

    Since there are two separate power sources connected to a common ground on a 240 circuit, I would think that you would need two sets of VI measuring transducers. One for each hot line, and both referenced to neutral or ground. The total power would be the sum of the two.

    Ratch
     
  16. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Ratch: 86 reply to 84. Using cell phone. Posts smaller. Sorry.
     
  17. RobertRoss

    RobertRoss New Member

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    Ratch and KISS,

    thanks for all your help... this was very insightfull for me and is pretty much what I needed to know to continue what I am doing. Its been great... til the next time.

    Thanks all.

    ross
     
    Last edited: Oct 9, 2011

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