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Help with power factor calculations, how to find KW from KVA and KVAR

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    Help with power factor calculations, how to find KW from KVA and KVAR

    A single line feeds 2 separate loads. The first load is 1200kw at .80 pf. The second load is 800kva and 300 kvar. What is the total load on the single line feeder?

    I have been racking my brain trying to figure out this question that someone else said was on the Level 4 exam. My answer is 1702kW, can anyone let me know if I am right? Here is how I figured it out:

    Load 1: Easy, just correct kW for power factor. 1200 * 0.8 = 960kW

    Load 2: This is where it got tricky but this is how I was able to calculate kW from kVA and kVAR, I think:

    (800kVA)2 = 640000
    (300kVAR)2 = 90000
    640000 - 90000 = 550000
    Square root (550000) = 742kW

    Load 1 (960kW) + Load 2 (742kW) = 1702kW

    Just looking to know if this is correct. Thanks in advance.
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    kva kvar kw calculation

    The equation is: KW = KVA x pf, therefore KVA = KW/pf
    1200 kW = KVA x 0.8 ..... KVA = 1200/0.8 = 1500 KVA,
    In other words, the 1200 kW, with pf 0.8, needs to have available 1500 KVA.

    For the second question we have:
    (KVA)2 = (KVAR)2 + (KW)2 .... (800)2 = (300)2 + (KW)2 (This is you got it right)

    Observe that you can use the above equation to calculate the KVAR for the first question:
    (1500)2 = (KVAR)2 + (1200)2
    KVAR = Square root ( 2250000 - 1440000) .... KVAR = Square root(810000) = 900
    Load: 1500 + 800 = 2300 KVA
    1200 + 742 = 1942 KW
    900 + 300 = 1200 KVAR
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    how to find kvar

    doh! thats what i get for not double checking my work. thanks for clearing this up for me!

    Quote Originally Posted by joaogemal View Post
    The equation is: KW = KVA x pf, therefore KVA = KW/pf
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    kva to kw

    No calculation is necessary for load 1, you are given the kW.

    Quote Originally Posted by veracon0700 View Post
    Load 1: Easy, just correct kW for power factor. 1200 * 0.8 = 960kW
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    kvar kw

    i found this power factor triangle to be helpful in understanding these kind of calculations
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