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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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    Quote Originally Posted by veracon0700 View Post
    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.
    I guess you want to come up with all three values for both feeders to get all parameters of the load. In the case of the first load 1200KW at .8 - 1200/.8 = 1500KVA. The angle is 36.869 degrees so you have (sine) 900KVAR. Load 1 is fully described as 1500KVA, 1200KW, and 900KVAR.

    Load 2 at 800KVA and 300KVAR. The angle is ARC SIN 300/800 = 22.02 degrees. Watts then is 741.619 KW. Just add them all up for the total of each parameter. The assumption is that the vars and watt flow is in the same direction. Alternatively you can use the Pythagorean theorem for load 2 - that is 800(squared) = 300(squared) + x(squared) and you get the same 741.619 (as a cross check). www.cttestset.com
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