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!:p:p:p


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 :rolleyes:

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.:D

[Central Scrutinizer]

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

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!

[The_Wallock]

hmmm, I think I see.

So the Apparent power (KVA) would be a reading before any load and the real power (KW) would be a reading taken after the load.

The Inductance (lag of current (A)) is inherent in the supply of power from the generation of electricity and our capacitor's being able to hold a charge even out the lagging of current (A), sort of like a shock absorber to lessen the lagging effect from the supply. Although a PF of 1 (V & A in phase) would be nearly impossible to achieve, hence .8 being an acceptable achievable level.

The calculations are still a little hazy but i think I am grasping the basic concept.

My brain hurts

Thank you CS your patience is appreciatted.

[Besoeker]

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)
That's not actually correct.
You pay for electrical energy in kWh not kVAh.
Domestically, there is no penalty for poor power factor*
Many industrial plants have maximum (MVA) demand limits and exceeding these usually results in swingeing financial penalties.
*other than greater I2R losses in the conductors.

[Besoeker]

hmmm, I think I see.

So the Apparent power (KVA) would be a reading before any load and the real power (KW) would be a reading taken after the load.
Well, not really.
Assume you have a 230Vac single phase supply.
You won't have any current, kVA or kW until you connect a load to it.
Suppose you now connect a load that takes 10A.
The VA is simply the current times the voltage, 2300VA or 2.3 kVA in this particular case.
The power (W) is voltage times current times power factor.

Your 230Vac is alternating at a steady frequency much like the pendulum of a clock.
It swings one way, reaches a peak, stops, falls back to zero and swings the other way.
The current is also alternating at the same frequency. It will stay in step with the voltage but may not reach the peak of its swing at the same time. Power factor is a way of expressing the difference in that timing.
A picture is worth ..... well, a picture:

http://i36.photobucket.com/albums/e39/Besoeker/laggingpf03.jpg

I have shown the current lagging the voltage by 2.5ms.
That gives a lagging PF.
Enough for now.

[Central Scrutinizer]

That's not actually correct.
You pay for electrical energy in kWh not kVAh.
Domestically, there is no penalty for poor power factor*
Many industrial plants have maximum (MVA) demand limits and exceeding these usually results in swingeing financial penalties.
*other than greater I2R losses in the conductors.

whilst i agree with you, we are actually saying the same thing, I have not discussed domestically, and i was trying to simplify, but ultimately the user is responsible for the power factor, and yes whilst they will pay in kwh, their PF will require the electricity supplier to generate a larger kva (or Mva) to compensate, which of, course the user will pay for

I am pretty sure we are saying th same thing:D

[Besoeker]

whilst i agree with you, we are actually saying the same thing, I have not discussed domestically, and i was trying to simplify, but ultimately the user is responsible for the power factor, and yes whilst they will pay in kwh, their PF will require the electricity supplier to generate a larger kva (or Mva) to compensate, which of, course the user will pay for

I am pretty sure we are saying th same thing:D
The main down side of poor power factor is increased current. Most industrial plants I know of aim for correction to around 0.95 lag. The PFC capacitors to achieve this are often located local to point of use. In a typical sub or switchroom there will be a local 11kV/400V transformer. Corrected power factor on the LV side means that a lower kVA transformer can be used with the attendant cost savings on equipment, cabling, and installation costs. It's unusual for industrial users to be caught out exceeding their MD limit.

From the electricity suppliers' POV, the prime mover provides kW, not kVA.

[Central Scrutinizer]

From the electricity suppliers' POV, the prime mover provides kW, not kVA.

not quite sure what you mean by that

the prime mover is the power plant that drives the suppliers generator at source, i.e. steam turbine, so it does not provide any electricity, but rotational energy to drive the generator

and generators do not 'provide' kW's, you cannot choose what to provide - you generate a voltage (at a frequency dictated by the prime movers rpm), - when you connect a load to that voltage, you will draw a current - the composition of that load (its resistive and inductive elements, or even capacitive elements) will dictate how much of the generated power (in VA) you use as real or true power (in W) - it is the load (i.e. the user) that dictates the PF, not the supply (or supplier)

now hopefully we have rounded back to pretty much the same point that we started with, and hopefully we can agree that we are prety much where we should be - we are in great danger of straying from the tone of the original question, and, I can assure you, i have NO WISH to enter the kind of 'I can pi*s high than you' contest which ruins so many forums

and i say that in the nicest way possible!!!!

[Besoeker]

the prime mover is the power plant that drives the suppliers generator at source, i.e. steam turbine, so it does not provide any electricity, but rotational energy to drive the generator
The prime mover mechanical output in power terms is kW.
The driven alternator is generally rated in kVA.
The prime mover limits the kW. The alternator rating limits the kVA.
And no, it is not a p1ss1ng contest.
I just try to give just facts.

[The_Wallock]

OK, yep, I understand that unless we supply a load we will draw no KVA or KW's, (EMF present but not needed until we demand it), so no KVA / KW.

I know the basic calculations OHMS & Joules law.

I get the basic principle that current lag's Voltage.

It seems I have not quite grasped the principle of Inductance ?, so I would be grateful for a layman's description of Inductance.

Again many thanks for all your help & patience, it is quite frustrating when I feel this is all with in my ability to comprehend, but I can't quite get my head around it.

[Besoeker]

It seems I have not quite grasped the principle of Inductance ?, so I would be grateful for a layman's description of Inductance.

I'll try.
Typically, inductance is a property of any circuit that has a wound component. Like a motor, a transformer, a relay coil, even a run of cable.
If you apply a voltage to an inductive circuit, it takes time for the current to build up.
In an ac circuit that time equates to the lag between current and voltage.
It's a bit like trying to speed up and slow down a flywheel. The effort you put in takes time to show results.

[Central Scrutinizer]

OK, yep, I understand that unless we supply a load we will draw no KVA or KW's, (EMF present but not needed until we demand it), so no KVA / KW.

I know the basic calculations OHMS & Joules law.

I get the basic principle that current lag's Voltage.

It seems I have not quite grasped the principle of Inductance ?, so I would be grateful for a layman's description of Inductance.

Again many thanks for all your help & patience, it is quite frustrating when I feel this is all with in my ability to comprehend, but I can't quite get my head around it.

ok, in BASIC laymans terms, when you have relative movement between a conductor and a magnetic field, an emf (voltage) is induced, that is the basic principle by which generators and transformers work

now, consider applying a DC supply to a coil of wire. At the moment you connect, current and voltage will be at zero. As the voltage is connected, the current which is now being drawn by the coil rises (from zero)

as it does, it produces a magnetic field. so you have a magnetic field growing from zero, so effectively you have a 'moving' magnetic field, and a conductor.....and of course you have an induced emf....which is opposite to the force which created it, and the current produced by it opposes the current which originally created it (we call this back emf) (appreciate this is VERY laymans terms)

so....at the instant you connect the voltage, the current doesnt immediately rise to maximum, because the magnetic field it creates induces an emf which opposes it, so the current takes a bit longer, so the current is now behind (lagging) the voltage - the amount of 'opposition' the coil represnts to the current (number of windings etc) can be taken as its inductance. When the current reaches maximum, the magnetic field produced by it stops moving, there is no relative movement, no back emf, and there is effectively no inductance in the coil, the DC current will no longer 'see' it

of course, if we connect an ac sine wave to the coil, the voltage is constantly changing, from zero through the positive half cycle, through zero and the negative half cycle and back to zero.

So of course the current produced by it does the same, and the mangetic field produced by the curent does the same........so there is always a level of back emf or 'opposition' to the current flow, and we call this inductive reactance - which is effectively ac resistance - think of ac running through a 'pure resistor' (again laymans terms), such as a heating element, the current will have no opposition other than the heaters resisitance, wheras the same voltage and current running through an inductive circuit such as coil, motor, etc, will see an 'opposition' to the current flow

now go back to my origininal post - va (or kva) is the votage produced times the current drawn - in a purely resistive circuit, all of this voltage will make all of the current flow through the circuit, and W (or Kw) and VA (or kva) will be the same, and the power factor will be 1

of course, in reality, most loads have some level of resistance, and some level of inductance, which of course causes inductive reactance, or opposition to the ac current flow

so the more inductive a circuit, or load is, the more oppostion to current flow there will be

so, in a thoretical circuit, voltage produced times amps drawn is VA

amps drawn times voltage dropped accross the resistive element of the load is W, and amps drawn times voltage dropped accross the inductive (or reactive) element of the load is VAR (volts amps reactive)

if there was no inductive element to the circuit, pf would be 1, and the more inductive a circuit becomes (or is) the smaller the pf becomes, because more of the voltage is used just trying to overcome the inductive reactance produced in the circuit

with a ratio of 0.8 pf, a 25kva generator for example, will be producing 25kva of apparant power, of which you will only use 20kw of true power, and 15 KVAR of reactive power

obviously the less inductive your load(s) are, the less of what you produce you will lose

and as stated the capacitor, as a 'counter balance' to the circuits inductance, will tend bring the pf back towards 1, by counteracting the effect of the circuits inductance.

ok, hope that had helped, and as i have stressed, i have tried to keep it laymans - i guess if you wanted a theory book you would have bought/read one!!!!!