Different way of calculating power factor from multiple sources

[Jookoo]

Hi,

I did a lot 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!

 

[Jookoo]

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?

 

[jim hardy]

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.

 

[jim hardy]

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...