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Can anyone explain power factor to me?

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  • 22-04-2013 1:22pm
    #1
    Registered Users Posts: 1,977 ✭✭✭


    I've taken a quick google but there are many new terms being introduced like wattless power etc.

    I currently have a device running at about 1Kw (I am using a measuring tool on the socket).

    The power factor shows as about 94. What does this mean in layman terms/ or just to get me started. I have a basic understanding of electricity - voltage, amps, watts...


    Thanks.


Comments

  • Registered Users Posts: 378 ✭✭frankmul


    Power factor is the ratio between the KW (Kilowatts) and the KVA (Kilo voltamperes) drawn by an electrical load where the KW is the actual load power and the KVA is the apparent load power. It is a measure of how effectively the current is being converted into useful work output

    kW = V x A x PF
    KVA = V x A

    Hope this might help!


  • Registered Users Posts: 163 ✭✭BrianDug


    Just to echo what frankmul said,

    If you have something plugged in for instance thats drawing 1kW at a p.f of 0.94 this means that the TOTAL power the device is taking from the supply is 1kW/0.94 = 1063W (1.063kW).

    Think of kW as useful power, kVar as power thats been wasted (in very simple terms) and kVA as the TOTAL power been drawn from the main supply.

    Mentally picture it as right angle triangle, kW=adjacent, kVAr=opposite kVA=hypotenuse, at the phase angle/p.f = cos-1(0.94) = 19.94 degrees

    So in in your case kW=1kW, kVAr=0.036kVAr, kVA = 1.063kVA

    kVA = sqrt[(kW)^2 + (kVAr)^2]
    kVAr = sqrt[(kVA)^2 - (kW)^2]
    p.f = kW/kVA

    The electricity companys do not charge domestic customers by kVA but by kW. For commercial customers the billing process is more complicated but they penalise you for low power factor and penalise you per kVAr if memory serves me correct.

    The problem is if you are running an inefficient system thats wasting a lot of power, the network operator(E.S.B) still have to supply you with the power which means that all there equipment will have to be bigger to handle the bigger loads even, but if you had a good power factor then the total power you are pulling will be less and they can reduce the size of their equipment (in simple terms).

    Hope this helps.


  • Closed Accounts Posts: 13,422 ✭✭✭✭Bruthal


    euser1984 wrote: »
    I've taken a quick google but there are many new terms being introduced like wattless power etc.

    I currently have a device running at about 1Kw (I am using a measuring tool on the socket).

    The power factor shows as about 94. What does this mean in layman terms/ or just to get me started. I have a basic understanding of electricity - voltage, amps, watts...


    Thanks.

    A power factor of .94 means 94% of the current drawn is doing useful work. So the useful work is the kw, which is the amps x volts x pf.
    KVA is volts by amps, because kva is what's needed to power the load.

    For example, if a motor is 1kw, the amps required would be 4.35, but if the pf is .5, the motor would take 8.7 amps. Only 4.35 amps is doing work, so it is a 1kw motor. But a power supply to power it would need to provide the 8.7 amps. So its taking 2kva from the supply.

    So as said, pf is the ratio of kw to kva and is actually kw/kva.


  • Registered Users Posts: 10,952 ✭✭✭✭Stoner


    euser1984

    Three people have taken the time to answer your question, there is a thank you button at the end of their posts, its good practice to use it, otherwise we don't know if you got the answer you wanted or if time was wasted addressing the issue that you needed help with.


  • Closed Accounts Posts: 4,372 ✭✭✭im invisible


    I think the power factor has gone to your head....


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  • Registered Users Posts: 5,756 ✭✭✭el diablo


    beer-mug-and-power-factor_zps95aeecdf.jpg

    We're all in this psy-op together.🤨



  • Registered Users Posts: 378 ✭✭frankmul


    el diablo wrote: »
    beer-mug-and-power-factor_zps95aeecdf.jpg

    I might use that if you don't mind


  • Registered Users Posts: 1,977 ✭✭✭euser1984


    Stoner wrote: »
    euser1984

    Three people have taken the time to answer your question, there is a thank you button at the end of their posts, its good practice to use it, otherwise we don't know if you got the answer you wanted or if time was wasted addressing the issue that you needed help with.

    Sorry man. I've only really got the time to come back to the thread now and concentrate on it properly.

    I do get it now. Thanks very much for your replies and the beer analogy is really good, "very funny"!


  • Registered Users Posts: 4,394 ✭✭✭Pac1Man


    A bit late to the party here but an analogy which made me understand it was the horse pulling the canal boat. The green line shows the required power (kW), the red line shows the actual power (kVA) and the black line is effectively the 'wasted' power (kVAr).

    The greater the angle = the greater the power factor = the harder the horse will have to work to achieve the same results.

    Disclaimer: Rope was attached to the tail purely for illustrative purposes. Horse may or may not be to scale.

    PowerFactor.jpg


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