Average power factor

[BrownOut]

My electricity comes out of the ground like water and natural gas. The telephone and cable TV/internet wires too.

My sewer goes into the ground. Maybe the ground converts my sewer waste into electricity and all the other utilities.

LOL!

That's it! We've finally found the unifying theory of Physics!

 

[The Electrician]

Hello again,


Ok, so you now got the same result i did several posts back?

I got the same result for the resultant PF, but since you didn't give a numerical value for the resultant magnitude, I can't say about that.

In post #18, you said:

"There still seems to be a problem with your calculations however, and that
is that it is very hard to believe that you can have two currents, one
that is 42.33 and another that is 4.52 where they are out of phase with
each other and still end up with a current and i quote:
"46.849 A with a PF of .362096".

As a sanity check, you can see that 42.33 plus 4.52 equals 46.85, which is
0.001 amps different than your answer, and that could only happen if the
two where in phase with each other. This leads me to believe very strongly
that you made a mistake in your calculation and you should review that."

So what numerical result do you get for the RMS value of the resultant?

 

[MrAl]

Hello again,


I was of course more concerned with the power factor, but for the
magnitude i got close to 46.62 which is close to what you got
the second time around.

 

[The Electrician]

So you've resolved these doubts you expressed?

"As a sanity check, you can see that 42.33 plus 4.52 equals 46.85, which is
0.001 amps different than your answer, and that could only happen if the
two where in phase with each other. This leads me to believe very strongly
that you made a mistake in your calculation and you should review that."

 

[MrAl]

Hello again,

What are you talking about, resolved those doubts?
What doubts?

You corrected me (again) and (again) posted the wrong answer,
then tell me i might be resolving some 'doubts' ? Geeze

I use two completely different techniques to find an answer to most
questions just in case a mistake is made using the first one.
How many do you use?

 

[The Electrician]

In which post # was it where I "...(again) posted the wrong answer..."?

Which answer in that particular post was it that was wrong?

What is the correct answer and how should it be calculated?

 

[MrAl]

Hello again,


I dont understand your question when you corrected your own error already
so how could you not know where it was?

In any case, if you want to sum up your method here that might be nice.
Maybe present a procedure for calculating the power factor and related
information? Just an idea if you feel like doing it.

 

[The Electrician]

Hello again,

I dont understand your question when you corrected your own error already
so how could you not know where it was?

You said that I "...(again) posted the wrong answer...". Isn't that saying that I posted the wrong answer more than once? You know, the "again" part? I was asking where the second post of the wrong answer occurred.

In any case, if you want to sum up your method here that might be nice.
Maybe present a procedure for calculating the power factor and related
information? Just an idea if you feel like doing it.

I explained it in one sentence in post #17, "...simply calculate the components of each current, add those up and the calculate the power factor of that vector sum.", and went through the calculations for your example in detail in post #19.

But I'll give the procedure I used in general terms.

Assume that the line voltage is Vline, and a number of loads are measured with a number of measured apparent powers, P1 through Pn. The load currents are I1 through In; these are the total currents associated with each load. These can be determined from the apparent powers and the line voltage if not measured directly. The power factors associated with each load are given by PF1 through PFn.

Decompose each load current into its component real and reactive parts. Take the sign of the reactive part to be minus if it's leading (capacitive).

Symbolically we can show the components of the n currents as:

I1, PF1 = I1_real ± I1_reactive
I2, PF2 = I2_real ± I2_reactive
I3, PF3 = I3_real ± I3_reactive
.
.
.
In, PFn = In_real ± In_reactive

Proceed with the decomposition.

Given a current Ik, calculate the phase angle, Φk, as Φk = arccos(PFk). Then the components are given by:

Ik_real = Ik*cos(Φk)

Ik_reactive = ±Ik*sin(Φk) adding the minus sign if it's a leading load.

Continue this process for k=1 to n, until all the load currents have been decomposed into their components, then add up all the real components into a total real component and all the reactive components into a total reactive component.

Note that since the reactive components can be either positive or negative, the total reactive component can have either sign.

Itot = ∑Ij_real + ∑Ij_reactive, where j goes from 1 to n, and the final result is:

Itot = Itot_real ± Itot_reactive

The RMS value of Itot is Itot_rms = SQRT(Itot_real^2 + Itot_reactive^2), and its power factor is PFtot = cos(arctan(Itot_reactive/Itot_real)) or PFtot = Itot_real/Itot_rms, with the power factor described as leading if the sign of Itot_reactive is negative.