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Thread: PF Correction

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    PF Correction

    Hi,
    I have a 110, 14W energy saving globes installed in the building, single phase supply. I believe these globes have a PF of 0.06. Problem when the power fails and they are supplied by the inverter, the battery bank consisting of 4 X 12V X102APH batteries fails within two hours. I understand that the amperage rises to 11.66 due to the PF of the globes. I have calculated the a 1KVAR capacitor would be required to correct the PF to 0.9. Perhaps one of you who have more experience of PF correct could advise if my calculation is correct and where I may purchase the capacitor in Durban.
    Many thanks

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    Diamond Member AndyD's Avatar
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    I've come across issues in the past with large numbers of CFL's on a back-up inverter. I discovered the hard way just how nasty these things are when it comes to PF and harmonics and that includes the good brands, not just the no-name rubbish. Rule of thumb I work on is if the CFL is rated at 14watts, class it as triple its rated current so assume its load will be 42watts for inverter sizing purposes and for some cheaper CFL's even this figure may be too small. Also I can assure you the manufacturers lie like dogs when it comes to their published THD figures for CFL's. The best way to assess the load is by actual RMS current values tested on site.

    Problem with applying PFC is that if the lighting load is fluid and varied ie not all the lamps are running all of the time then your correction needs to be adaptive and for such a small load it's just not worth the outlay. Other option is that you apply correction to each individual circuit after the switch but this is often logistically awkward.

    I'm not sure how you got your figure but I get the reactive power of your total lamp load to be 2.2kVAR (At 11.66Amps I get your overall power factor to be 0.57 which is absolutely appalling but I can well believe it). I'd be looking at a 132uF correction capacitance but like I mentioned this is only correct for the full load, if the load varies you'll have to take an appropriate stab at a suitably lower correction figure and hope for the best.

    I think you're probably going to have to increase your battery power availability but good luck with the PFC.
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    jetpp1@absamail.coza (16-Aug-15)

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    Thanks AndyD, My mistake I should have stated that the PF of lamps is 0.6 not 0.06. As the load is constant i.e. stairway lighting I will install a capacitor to attempt to correct the suitation.

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    Hi AndyD. Please could you explain how you calculated the KVAR at 2.2. From my calculation(I may be incorrect) to improve the PF from 0.6 to 0.9. PF 0.6 Tan 53deg .13, = 1.3333, PF 0.09 Tan 25deg .50 = 0.4843 difference 0.849. 1.5KW X .849 = 1.301KVAR. I thank you for your comments.

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    Diamond Member AndyD's Avatar
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    I think the difference is coming in because you're correcting to cos φ = 0.9 where I was correcting to cos φ = 1.
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    jetpp1@absamail.coza (17-Aug-15)

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