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Power factor calculations?

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

  1. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    How about?

    P (Watts) = V * I * PF always true
    PF =1 for a resistive load

    For AC loads where the current waveform is sinusoidal AND the current waveform is sinusoidal then P = Vrms*Irms*Cos(theta)

    Cos(theta) is the phase angle between voltage and current

    When the voltage waveform is sinusoidal and hence periodic and the current waveform is non-sinusoidal, but periodic (typical), you cannot measure the phase angle directly to determine the PF.
     
    Last edited: Oct 7, 2011
  2. Ratchit

    Ratchit Well-Known Member

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    KeepItSimpleStupid,


    Yes, but useless for RobertRoss, because he can calculate the instanteous products of the voltage and current and then find the average and the active power faster than calculating each RMS of the voltage and current and then finding the phase.

    Not a problem when finding the average of the product of instantaneous voltage and current values to get true power. PF not needed for that method.

    Ratch
     
  3. RobertRoss

    RobertRoss New Member

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    Hi Ratch,

    >No, I believe you want the apparent power, also called complex power, with the designated symbol S*. Its units are VA. S* = VI*, where I* is the >conjugate of I. Another relationship is S* = |I²|Z

    Wait a minute here:
    P = VI and 120 x 1.41 = 169

    Also:

    Apparent power = |I²|Z = (1.41² x 85) = 169

    Same thing??? So when we multiply V x I it gives us our va's which is apparent power!!!!!
    Why do you say "NO"
    Confused!

    Getting back to the example. I have drawn the Voltage and current wave forms in approximation to my example. I didn't draw _every_ degree... that would take too long.

    Please take a good look at the attachment below I have included in this post. Now let's put asside the fact that that a 120VAC main cannot just get connect to a micro-controller. We all know this. Obviously, the voltage and current will be attenuated by a conversion proccess. So let's set this asside for now. Let's just look at the AC theory.

    My micro controller will read the voltage sine wave *instantaneously*. So this means that it will see:
    @t1 = 2.96 v
    @t2 = 119.98 v
    @t3 = 167.1v
    @t4 = 169.25v
    @t5 = 167.1v
    @t6 = 119.98v
    @t7 = 2.96 v

    plus I will see aaaaaall the other voltage values between t1 and t2 and aaaaaall the voltage values between t2 and t3 etc.... So what I am saying is that for the sake of this sample, I just did seven time slices. But bear in mind I will be reading the voltage at every degree of the 360 degrees sine wave. So all positive and negative voltage values will be read into my micro controller and logged in a table. Hence instantaneous voltage. The same goes for the current wave form.

    Now what you have been explaining to me is to multiply every instantaneous values of VI at every time slice.
    >So all you have to do is multiply each instantaneous current and voltage value, sum them up, and divide by the period

    So let's do it:

    VI @ t1 = 2.96 x 1.057 = 3.128 watts
    VI @ t2 = 119.98 x 0.352 = 42.23 watts
    VI @ t3 = 167.10 x 0.352 = 58.81 watts
    VI @ t4 = 169.25 x 0.881 = 149.10 watts
    VI @ t5 = 167.10 x 1.233 = 206.03 watts
    VI @ t6 = 119.98 x 1.41 = 169.17 watts
    VI @ t7 = 2.96 x 1.233 = 3.649 watts

    So taking the absolute values of VI when in the negative part of the voltage sine wave, we get the same wattage values. So summing up all the wattages of the 7 time slices shown above and multiplying by 2 (to include the bottom part of the wave form) ... is:

    632.11 x 2 = 1264.23 watts

    Now we divide 1264.23 by the period of one cycle:

    1264.23/(1/60) = 75853 watts of true power ?????

    True power is supposed to be close to 120 watts???

    confused.
    Where have I gone wrong!
    Thanks for your help.
    r
     

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    Last edited: Oct 7, 2011
  4. dave

    Dave New Member

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  5. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Is that what he needs to do is take the average of v(t) * i(t) and not the RMS (v(t) * i (t) ) or are they the same?
     
  6. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Average is sum/n or 1264.23/7 or 180 VA . Not sure if that's close enough or not. Still seems off.
     
    Last edited: Oct 7, 2011
  7. RobertRoss

    RobertRoss New Member

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    >Average is sum/n or 1264.23/7 or 180 VA . Not sure if that's close enough or not.

    Well, I don't know, our example according to to the xls file... it should be closser to 120 watts????

    This is exhausting... ama going to lunch! :)

    r
     
  8. Ratchit

    Ratchit Well-Known Member

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    RobertRoss,

    ZT = 120/1.41A = 85.10 ohms

    The correct value of of the current is 120/(60+60.32J) = 1.41/_-45.15° amps. You have to include the phase of both the voltage and impedance to get the current. No phase makes it DC. So the apparent power is S* = VI* = 120(1.41/_45.15) = 169/_45.15 = |I²|Z = 1.41²*(60+60.32) = 169/_45.15° .

    I cannot bring up your attachment, so I cannot check your figures.

    Ratch
     
  9. Ratchit

    Ratchit Well-Known Member

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    KeepItSimpleStupid,

    An instantaneous value does not have a RMS value or an average value. It is the value of a DC meter measurement if the wave period was an hour, so the meter could follow it.

    Ratch
     
  10. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Converted DetailedWaveVI.png to jpg

    I could open it in Firefox just fine.
     

    Attached Files:

  11. RobertRoss

    RobertRoss New Member

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    >I cannot bring up your attachment, so I cannot check your figures.

    Try these!
     

    Attached Files:

  12. Ratchit

    Ratchit Well-Known Member

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    RobertRoss,

    Now I can bring up all the attachements you sent. However, it appears that the time intervals are not spaced evenly. Let me work on it after supper this evening with Excel, and see what I can do.

    Ratch
     
  13. RobertRoss

    RobertRoss New Member

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    >Let me work on it after supper this evening with Excel, and see what I can do.

    Okay, no problemo.... take your time... and "Bon appetit"

    ross
     
    Last edited: Oct 7, 2011
  14. RobertRoss

    RobertRoss New Member

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    Ratchit,

    >The correct value of of the current is 120/(60+60.32J) = 1.41/_-45.15° amps. You have to include the phase of both the voltage and impedance to get >the current. No phase makes it DC. So the apparent power is S* = VI* = 120(1.41/_45.15) = 169/_45.15 = |I²|Z = 1.41²*(60+60.32) = 169/_45.15° .

    Ratchit, whatever you do... just note that unfortunately, I am a little confused with phasor algebra... it quite goes back a long time ago. I am skimming my AC book as we speak and there is approx. 20 pages on the stuff... it will take me some time before I get the hang of this... atleast not for tonight.

    Thanks.
     
  15. RobertRoss

    RobertRoss New Member

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    Hello guys,

    169.28 is peak voltage and 1.41 was rms voltage which is wrong.... I would be reading peak current which really would be 1.994 A. So here is my examle again:

    I have hand drawn the Voltage and current wave forms in approximation to my example. I didn't draw every degree... that would take too long. So I took 7segments per 180 degrees. Therefore we have 14 time slices.

    Please take a good look at the attachment below I have included in this post. Now let's put asside the fact that that a 120VAC main cannot just get connect to a micro-controller. We all know this. Obviously, the voltage and current will be attenuated by a conversion proccess. So let's set this asside for now. Let's just look at the AC theory.

    My micro controller will read the peak voltage of an AC sine wave *instantaneously*. So this means that it will see:
    @t1 = 2.96 v
    @t2 = 119.98 v
    @t3 = 167.1v
    @t4 = 169.25v
    @t5 = 167.1v
    @t6 = 119.98v
    @t7 = 2.96 v

    plus I will see aaaaaall the other voltage values between t1 and t2 and aaaaaall the voltage values between t2 and t3 etc.... So for the sake of this sample, I just did seven time slices. But , again, bear in mind I will be reading the voltage at every degree of the 360 degrees sine wave. This means from 0 to + peak and from 0 to - peak. So all positive and negative voltage values will be read into my micro controller and logged in a table. Hence instantaneous voltage will be recorded. The same goes for the current wave form.

    Now what you have been explaining to me is to multiply every instantaneous values of VI at every time slice.
    >So all you have to do is multiply each instantaneous current and voltage value, sum them up, and divide by the period

    So let's do it:

    VI @ t1 = 2.96 x |-1.498| = 4.43watts
    VI @ t2 = 119.98 x |-0.498| = 59.75 watts
    VI @ t3 = 167.10 x 0.498 = 83.21 watts
    VI @ t4 = 169.25 x 1.246 = 210.88 watts
    VI @ t5 = 167.10 x 1.744 = 291.42 watts
    VI @ t6 = 119.98 x 1.994 = 239.24 watts
    VI @ t7 = 2.96 x 1.744 = 5.16 watts

    So summing up all the wattages of the 7 time slices shown above and multiplying by 2 (to include the bottom part of the wave form) ... is:

    894.09 x 2 = 1788.18 watts

    Now we divide 1788.18 by the period of one cycle:

    1788.18/(1/60) = 107290 watts of true power ?????
    IMPOSSIBLE!!!!!

    However when the sum of the powers are divided by the number of time slices we get a much more realistic answer which is:

    1788.18/14 = 127 watts of true power
    The slight imprecision could be due to my visual estimates of currents in my hand drawing.

    Confirm this at the following example (I used the same values) at:
    http://www.allaboutcircuits.com/vol_2/chpt_11/2.html

    Which still leaves me wondering. True power is "work / time" why does it not work with dividing by the period (1/60hz), but by the other token it works by dividing by the number of time slices... those 14 time slices are not "time" they are simply slices that I made up..... I think the latter way is for the V average!

    confused!!!!!

    r
     

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    Last edited: Oct 7, 2011
  16. Ratchit

    Ratchit Well-Known Member

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    RobertRoss,

    Ask the right questions, and it will be revealed.

    OK, I finished a Xcel spreadsheet of a 360° voltage sin wave of your problem. You can look at every value of every degree and see the formula for each cell. You can see the final value is in cell D363. You must compute through one complete period of either the voltage or the current, otherwise you don't get the negative values that a phase shift produces in the summation. Ask if you have any questions.

    Ratch
     

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  17. RobertRoss

    RobertRoss New Member

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    Thanks for your effort for both of you guys that took the time to do the xls spread sheet.

    Okay Ratch,

    Before I go on, can you explain to me why in the current's cell you are dividing by 85.07838132? I get the rest...

    r
     
  18. Ratchit

    Ratchit Well-Known Member

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    RobertRoss,

    85.07838132 is the magnitude of the impedance. It is divided into the phase shifted voltage to get the current at that point. It is what a dc ammeter would see if it were fast enough.

    Ratch
     
  19. RobertRoss

    RobertRoss New Member

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    Okay so its 120/1.41 = 85 ???

    is it? if not, either way I am out for tonight.... I will look at your xls carefully, and ask some questions tomorrow....

    Good night Ratchit... and thanks for your help

    ross
     
    Last edited: Oct 7, 2011
  20. Ratchit

    Ratchit Well-Known Member

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    RobertRoss,

    No, it is Z = R + Xl = 60 + j60.31857896 = 85.07838132/_45.1517071349°

    Ratch
     
  21. RobertRoss

    RobertRoss New Member

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    HI Ratch,

    Okay, I breifly went through the polar and rectangular coordinate stuff. Not as difficult as I thought. Probably need a little more practice to be comfortable at it though.

    ==========================
    85.07838132 at /_45.1517071349° is the magnitude of the impedance. It is divided into the phase shifted voltage to get the current at that point. It is what a dc ammeter would see if it were fast enough.
    ==========================

    Now that makes sence.

    So moving right along, you have created a spreadsheet actually simulating what my processor would read in terms of instantaneous voltages and currents. And we clearly see that multiplying every voltage by every current for every degree of the sine wave gave us its respective instantaneous power at every degree. You then summed every instantaneous power and divided this result by 360. Having done this, you have obtained "true power".

    I have 4 questions:

    Question #1) I am curious to know if this can also be done the way I innitially posted this thread.

    If my micro controller is able to read every instantaneous voltages and currents, then I would be able to detect the highest voltage value of the voltage's wave and the highest value of the current's wave .... right? So taking these two values and dividing them by 0.707 would give me their rms values respectively.... right? So let's put numbers to this in respect to your xls file.

    Highest Voltage detected = 169.0598 V
    Highest current detected = 1.994479 A

    RMS-V = 169.0598/√2 = 119.543 V
    RMS-I = 1.994479/√2 = 1.410309 A

    Then we can figure total impedence:
    ZT = 119.543/1.410309 = 84.76 ≈ 85 ohms

    Using ohms law we can then get apparent power by doing:

    va = I² * ZT ...... >>> as indicated here: http://www.allaboutcircuits.com/vol_2/chpt_11/2.html

    Therefore:

    169va = (1.410309)² * 85

    119.5watts = cos45° * 169

    So my question is... in terms of methods... is there a difference if I use your way (as done in your .xls) or my way as I did above? Or is it that your way can handle any wave shape and my can only handle well rounded wave shapes????

    Question #2)
    True power is work per a period of time. In your .xls file, you summed all the instantaneous powers, you then divided the result by 360°. Is degrees of a circle considered as "time". Isn't time the period of the wave sine (1/60) ... confused.

    Question #3) The method you have used will work on a sinuasoidal wave form only... right? If the AC wave I read is weirdly shape, then I would not be as precise with the True power calculations... right?

    Question #4)
    If my Ac circuit has very little or no reactive power, can I still use the above calculations?
    thanks
    ross
     
    Last edited: Oct 9, 2011

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