# Power factor calculations?

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

1. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Last edited: Oct 5, 2011
2. ### crutschowWell-Known MemberMost Helpful Member

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I believe that gives the average value of V, not the RMS.

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5. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Yep, something looks wrongish:

integral (sin(t) from t=0 to t=pi) is http://www.wolframalpha.com/input/?i=integral+(sin(t)+from+t=0+to+t=pi)
=2

and rms (sin(t) from t=0 to t=pi) is http://www.wolframalpha.com/input/?i=rms+(sin(t)+from+t=0+to+t=pi)
= 1/sqrt(2)=0.707

So, I guess you have to do something a little fancy: http://www.wolframalpha.com/input/?i=rms

Last edited: Oct 5, 2011
6. ### RobertRossNew Member

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Hello guys,

>So, I guess you have to do something a little fancy: http://www.wolframalpha.com/input/?i=rms

Yes, it seems like the same thing as this (just in lamens terms):

http://www.analytictech.com/mb313/rootmean.htm

HOWEVER, something still doesn't make sence. First of all, please take a quick look at the graph in my attachment below:

If we go by the method explained in my latter link above, then let's assume that Voltage at 100% will be 100V peak. Now, let's do this:

Step#1 (square all values of the top part of the AC wave):
V @ 10% = 10 Volts ² = 100
V @ 20% = 20 Volts ² = 400
V @ 30% = 30 Volts ² = 900
V @ 40% = 40 Volts ² = 1600
V @ 50% = 50 Volts ² = 2500
V @ 60% = 60 Volts ² = 3600
V @ 70% = 70 Volts ² = 4900
V @ 80% = 80 Volts ² = 6400
V @ 90% = 90 Volts ² = 8100
V @ 100% = 100 Volts ² = 10000
V @ 90% = 90 Volts ² = 8100
V @ 80% = 80 Volts ² = 6400V
V @ 70% = 70 Volts ² = 4900V
V @ 60% = 60 Volts ² = 3600V
V @ 50% = 50 Volts ² = 2500V
V @ 40% = 40 Volts ² = 1600V
V @ 30% = 30 Volts ² = 900V
V @ 20% = 20 Volts ² = 400V
V @ 10% = 10 Volts ² = 100V

Step#2 (Sum all values):
Sum of all values is: 67000

Step #3(Do the average of all these values):
Average: 67000/19 slices = 3526

Step#4 (Do the square root of the average):
RMS = √3526 = 59.38 Volts ?????

59.38 volts is *NOT* the rms value of 100V peak????

RMS is supposed to be 70.7 Volts ?????

Can someone explaine this discrepentie. Confused!

r

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• ###### Sinewave.png
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7. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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I've attached a xls file in ZIP form that does the RMS calculations for a 120 V sine wave. Ignore the 155*. It's 169.68*sin(t) where t is in degrees. The increment is every degree.

AVE: 14395.6512 SQRT(AVE) 119.9818786

#### Attached Files:

• ###### RMS calc 120 v.xls.zip
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<bump>

9. ### RCinFLAWell-Known Member

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You calculated the rms value for a triangular wave.

For sinewave you have to take the sine of 100v peak for each of the evenly time spaced increments and it needs to be over a complete repeating segment.

Last edited: Oct 6, 2011
10. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Sorry I think I confused you with one of my examples.

In the spreadsheet, I used 1 to 360. I should have used 0 to 259; but the answer is the same because it's a complete cycle.

11. ### RobertRossNew Member

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Hello KISS,

Sorry for the delay of my reply... super busy here....

Ahh! I see what you did. Basically you are taking the sine of every degrees in rads
and multiplying it by the peak voltage. So it is up to me to read in the Peak voltage which is 169V and then plug it in to the formula for every degree:

AVE =169.68*SIN((PI()/180)*AX)

where x is every degree.

After I have done this for every degree, I square every value like you have in column C. I then Add up coloumn C and divide it by 360 and square root the result. I repeat this for current and I finally have my RMS V and I .... right?

Now one question though! I have 2 attachements below. One of a clean wave form and one of a noisy shaped wave form. The example you showed me in the .xls file will do for the clean wave form. But can it also apply to the noisy shaped wave form?

I appreciate your help... its nice of you.

Regards
Ross

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• ###### NoisyWaveForm.png
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12. ### RobertRossNew Member

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RCinFLA,

>You calculated the rms value for a triangular wave.

Hummm! I did didn't I!!!!

Okay, back to the drawing board.... thanks

r

13. ### RobertRossNew Member

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Hi KISS,

>In the spreadsheet, I used 1 to 360. I should have used 0 to 259; but the answer is the same because it's a complete cycle.

Now you really confused me LOL
Why 259... what's the 259.... where did you get the 259 from ????

I have to step out... Be back in a few hours or tomorrow morning!

Have yourself a good evening!

ross

14. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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OOPS: 359 !!. Since sine is defined at 0, I should have used zero and one sampling point less then than the last zero crossing of the periodic waveform. Include both a positive and negative half cycle.

X-10 and other devices communicate on/near the zero crossings.

You don;t READ the peak voltage. The sin() function goes for +1 to -1, I had to multiply that by the 0-peak value to SIMULATE a 120 V sine wave. What's necessary in the design is to capture a full cycle. i.e. time the zero crossings.

You probably should not assume exactly 60 Hz, 1/60 seconds. You should really be using the leading zero cross value of zero up until the value before the next+1 zero crossing. The important part: is is the waveform periodic?

The power computation is valid. Suppose a spike was in the AC line. It could be missed. I know for many AC voltage instruments, the crest factor http://en.wikipedia.org/wiki/Crest_factor enters into the ability to measure the signal correctly. Fluke and another company in Japan used a thermal bridge sensor which did a really good job of measuring AC signals from DC to 20 MHz.

Last edited: Oct 6, 2011
15. ### RatchitWell-Known Member

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RobertRoss,

Sorry I did not get around to posting in this thread sooner, but I was on vacation for the last couple of days.

OK.

Why are you getting wrapped around the axle with impedance? You just want the three components of the power triangle, don't you?

Yes, basically you have the right idea.

In reviewing this thread I saw some terms that are off key. First of all, there is no such thing as "RMS power". All power is average power. There is RMS voltage and current, but no RMS power.

You can get the average true or active power of any waveshape if you can sense and process/compute the values fast enough. If you are not evaluating a sinusoidal wave, then you are dealing with a multifrequency entity, and the question of apparent power, reactive power, and power factor becomes ambiguous or difficult. But continue on and pretend you have a good sinusoidal wave.

So all you have to do is multiply each instantaneous current and voltage value, sum them up, and divide by the period or multiple period of the waveform. Make sure you don't calculate with fractions of a period. You can get the phase by discerning with good programming, the difference in time between the zero crossing of the voltage and current when both are at either their positive or negative slopes. Divide that number by the period, and convert into degrees or radians to get the phase. With the power and the phase angle you can complete the power triangle.

Ratch

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16. ### RobertRossNew Member

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Hello KISS,

>The sin() function goes for +1 to -1,
Yes.

>had to multiply that by the 0-peak value
What's a "0-peak value" ?

>to SIMULATE a 120 V sine wave.
Yes, that's okay. But you are assuming the peak voltage is 169.68 V to simulate a method to calculate the 120V RMS value as you proved in your xls file. Which is good. But suppose in a real situation of an AC circuit, the peak might not be 169.68V ... let's say its 173.57 or 164.55... this could happen. So, in your RMS method where you do:

@1st degree =169.68*SIN((PI()/180)*A2)

isn't 169.68 supposed to be for example either 173.57 or 164.55 depending on what the AC line voltage is that day!!!

>What's necessary in the design is to capture a full cycle. i.e. time the zero crossings
Yes, to figure out phase angle between I and V... no?

A little shady on the 169.68 V stuff though as I explained above.

r

Last edited: Oct 7, 2011
17. ### RobertRossNew Member

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>Why are you getting wrapped around the axle with impedance? You just want the three components of the power triangle, don't you?

Yes, but I prefer getting the power triangle through the instantaneous V and I. The reason is let's suppose one day I don't only want the power triangle! Suppose one day I need the instantaneous I and V to do something else.... its just a precaution.

<
You can get the average true or active power of any waveshape if you can sense and process/compute the values fast enough. If you are not evaluating a sinusoidal wave, then you are dealing with a multifrequency entity, and the question of apparent power, reactive power, and power factor becomes ambiguous or difficult. But continue on and pretend you have a good sinusoidal wave.
>

okay.

>So all you have to do is multiply each instantaneous current and voltage value, sum them up, and divide by the period
Quick question, this gives us the apparent power ... right?

r

18. ### RatchitWell-Known Member

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RobertRoss,

The method I outlined does give you the power triangle by processing the instantaneous V and I values over a single or multiple period. You cannot find the power triangle by just one instantaneous value each of V and I. Whatever else you want to do with the instantaneous values of V and I, can be done with no conflict with the afore power triangle method. Is there a problem with that?

Wrong, it gives you the average true or active power, which is the power dissipated by the resistive components of the circuit. If you want apparent power, divide the true power by the cos of the phase angle.

Ratch

19. ### RobertRossNew Member

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Hi Ratch,

God I am having a hard time with this. I apologize.

Okay, let me conclude one thing at a time with you.

First, doing the product for each instataneous values for V and I, and then summing them up and dividing this result by the period is true power. Okay, I did more reading and true power is defined as work done per time by the resistive components in an AC circuit.

Now, going back to my innitial example, if we would multiply the 120V by the 1.41A, this will give us 169 va's... right! However, the 120 and the 1.14 are the values that a meter (VOM) would read ... in other words the "rms" voltage and current right?

So then, in my example the peak voltage is (120/0.707)= 169 volts and that the peak current is (1.41/0.707)=1.99A, then, what is the units of the porduct of(169x1.99)= 336.8 called? "336.8 peak va" ???

r

Last edited: Oct 7, 2011
20. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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One word: PERIODIC PERIODIC PERIODIC
Synonym: One complete cycle. All positive values of V and all negative values of V

What you need to measure:
RMS of v(t) over one cycle --> Vrms
RMS of i(t) over one cycle ----Irms
RMS of: v(t) * i(t) over over one cycle; S, I believe

Calculate from above:

Vrms * Irms is Reactive Power Q in VAR, I believe
PF = P/S
True power = Q * PF

and thus Apparent Power S in units of VA, Reactive Power in VAR and True Power in Watts

So, there you have it. There is no mention of the peak value of anything. PERIODIC PERIODIC PERIODIC

Now, since you want the power over another interval, average the powers over x line cycles, so the numbers don't get out of hand.

You could try to do the calcs over 5 minutes or whatever interval you wanted, but the numbers could get way out of hand.

What's always more fun is to do your interval such that your always including the last 5 minutes of numbers. That's what you want to do, not 5 min and then the next 5 min and the next 5 min. You want, 1 cycle (display), 2 cycles (display), n cycles (display), n-1 to n+1 cycles (display) etc.

Initially while the meter is trying to average over 5 minutes, it can't so it averages on what it has. Then it gets 5 minutes and can display a 5 min average so it does. Then the average starts to loose the first cycle and then the first 2 and then the first 3, but the interval is still 5 minutes.

0-Peak is Zero to Peak. p-p or peak to peak is 2 * the Zero to Peak value for a sine wave

If you want the zero to +peak value or the 0 to -peak value or the peak-peak value, you have to get them separate.
e.g. Peak detector or you pick them off from the A/D converter numbers.

Last edited: Oct 7, 2011
21. ### RatchitWell-Known Member

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RobertRoss,

My name is not God, nor am I God. Many folks have a hard time understanding this subject, even those who have been associated with this field for many years.

So far, so good.

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

Yes, but you need to know the phase relationship between V and I to get the complex or apparent power.

Call it nonsense. In a reactive circuit, the peak voltage and current occur at different times, so it does not make sense to multiply them together without regard to their phase relationship.

Ratch