Power Factor

[The_Wallock]

Hi all

Thought this would be the best place for these questions, saves clogging up the main boards.

Could someone explain, what is Power Factor and what is the correct way to calculate it ?

[Ohmslaw]

Hi there.


Power Factor = True power/Apparent power.


In every electrical circuit there will be - TRUE power, APPARENT power, and REACTIVE power.

Basically VA is Apparant power this is volts supplied x amps used.


W is True power. this is amps used x voltage developed acrross the purely resistive element of the load. I.E (W=VxI)

VAR (volts amps reactive) is Amps used x Volts across purely reactive element of the
load

Reactive power can be re-intoduced to the system usually using PFCC's (capacitors) to improve the power factor.

The Power Factor is W/VA or VAR/W

Now in practice, you will always have resistance and reactance in a circuit (unless it is something like a pure resistor) (heater).

so the PF will always be less than 1.

Around 0.8 is normal.

If you were using a lot of inductive loads such as motors (in a factory) then your PF would decrease and you would be wasting a lot of the supplied VA (which is what you are paying for!), in which case you can improve the PF by adding capacitance to the circuit

What this means is, even for computers, unless they are the ONLY thing on the supply, the VA (and the PF) of the total load will be in***enced by EVERYTHING on the circuit, this side of the electricity meter.

So lets say you have a PF of 0.8 , this gives you a ratio of 3:4:5 of VAR:W:VA

So for every 4W you have 5VA

As an example divide the load by 4 and multiply by 5 for VA @ 0.8 PF

And, the important thing is, your electric bill is worked out on the VA supplied, not the Watts used.


hope this helps.

[Central Scrutinizer]

Hi Mr ohmslaw

when i saw the post, i thought, "there is a subject i can talk about!"

and then i read your reply and thought "jeez, thats pretty much what i would say, and how i would say it" (i taught this subject for many years)

and then i thought "no thats TOO close" and got abit of deja vue.....

so i did a quick google search of a username i used to use on another forum.....and found MY original reply to a similar question.....

"basically VA is apparant power this is volts supplied times amps used


W is true power. this is amps used times voltage developed acrross the purely resistive element of the load

VAR (volts amps reactive) is amps used times V across purely reactive element of the
load

The power factor is W/VA or VAR/W

now in practice, you will malways have resistance and reactance in a circuit (unless it is something like a pure resistor (heater))

so the PF will always be less than 1.

Around 0.8 is normal.

If you were using a lot of inductive loads such as motors (in a factory) then your PF would decrease and you would be wasting a lot of the supplied VA (which is what you are paying for!), in which case you can improve the PF by adding capacitance to the circuit

What this means is, even for computers, unless they are the ONLY thing on the supply, the VA (and the PF) of the total load will be in***enced by EVERYTHING on the circuit, this side of the electricity meter.

So lets say you have a PF of 0.8 , this gives you a ratio of 3:4:5 of VAR:W:VA

so for every 4W you have 5VA

so your 6.2KW divided by 4 = 1550

times 5 for VA = 7750 VA = 7.75KVA @ 0.8 PF

And, the important thing is, your electric bill is worked out on the VA supplied, not the Watts used

Or of course, i could just be making this up from stuff i dreamt..........its been known..."

nice to see you took the trouble to change some of my calculations, but ruddy hell old chap!!!!!!!!!!!

still, i suppose the guy has got his answer - but i came here anonomously, and i STILL cant escape myself!


hope this helps

[The_Wallock]

Well who ever wrote it, I appreciatte the help, I am gonna print it out and go through it with me other notes, who do I send the copyright fees too ?

cheers chaps

[Central Scrutinizer]

well i will leave the fees to you - I know who wrote it - and so does Mr Ohmslaw

Might be interesting if you study and asborb my original hypothesis regarding true and apparant power (which i formulated and started teaching in the mid-90's) - to the point where you can ask further questions on it

then lets see who answers!!!!

[Ohmslaw]

Wind your neck in trembley one, I knew it was you. I was tempting you to see if you'd reveal yourself and what better way than with a question on power factor!!

How's things at Access??? Do you hear from the others much...Doc, Paul etc.

Cheers.

P.S I owe you a bow!!!!

[Central Scrutinizer]

hahaha

got me!!!!!

yeah we are going great

i would ask how you knew it was me, but seeing as i spout the same old twoddle wherever i land, i suppose it was only a matter of time!

do i know you by any other names........

[The_Wallock]

Ahh, I think I have stumbled in to sparks re-untied

Guys, can you take a look at this & advise me if I am on the right track:

230 V supplied to a 16amp motor = 230V x 16A = 3680VA or 3.68KVA

If our motor uses 14A and reduces the flow of current to 220V the Watts will = 3080W or 3.08KW

So 3.08KW / 3.68KVA = a PF of 0.84

Am I on the right track or have I missed something ?

[Ohmslaw]

mmmmm on a second look it dont seem quite right.

CS will be along to clarify I'm sure.


Told you.

[Central Scrutinizer]

Ahh, I think I have stumbled in to sparks re-untied

Guys, can you take a look at this & advise me if I am on the right track:

230 V supplied to a 16amp motor = 230V x 16A = 3680VA or 3.68KVA

If our motor uses 14A and reduces the flow of current to 220V the Watts will = 3080W or 3.08KW

So 3.08KW / 3.68KVA = a PF of 0.84

Am I on the right track or have I missed something ?

yes, you have missed something

how do you know the motor is a 16 A motor?

what you seem to be suggesting is that by reducing the V to 220V the motor will draw less current (14A), which is not really how it works, and is not really about power factor

say if i was supplying a motor from a generator (or a number of motors from any power supply)

if the motors are running at a PF of 0.8 for example then for every 4 kw i use (real power) then i will draw 5 kva from the supply (the apparant power, which i pay for)

so lets say you have a factory using lots of inductive loads (motors, ***orescent lights etc), then overall you will be paying for a lot of power you dont use.

Now, there is a pneumonic we use which is CIVIL

Capacitance - Current - Voltage - Current - Inductance (because inductance is measured in Henry's which is L)

what this means is that in capacitive circuits the current leads the voltage, and in inductive cirucits the current lags the voltage (or out of phase)

so, theoretically if you supplied a circuit which is equally capacitive and inductive, then the current will neither lead or lag the voltage, and will be in phase

think of it as capacitance cancelling out inductance

so in the factory, it is common practice to have PF correction with large banks of capacitors which automatically switch in and out to counteract the PF loses caused by the inductive loads. the resultant is that the PF moves more towards unity (1), and real power (kw) and apparant power (kva) will move to a point where they match

in real money, it means you pay for what you get, rather than paying for what is being lost, plus what you get

to really understand this you have to start going into vectors, and starts getting a bit hairy!