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Thread: Calculating Unbalanced Neutral Current in 3p-4w system

  1. #1

    Calculating Unbalanced Neutral Current in 3p-4w system

    Is there a simple formula for determining the unbalanced neutral current in a 3p-4w system? I know how to calculate it if the line to neutral loads on two phases is equal and the line to neutral load on the third phase is different, but what if the line to neutral load is different on each phase?

    In other words, if I have a 3ph-4w 208/120v panelboard and connect three circuits, one on phase A, one on phase B, and one on phase C and the corresponding 120v loads are 10A on phase A, 8A on phase B, and 6A on phase C, what would my neutral current be from the unbalanced loads which are all 120 degrees out of phase?

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    Quote Originally Posted by stevee View Post
    Is there a simple formula for determining the unbalanced neutral current in a 3p-4w system? I know how to calculate it if the line to neutral loads on two phases is equal and the line to neutral load on the third phase is different, but what if the line to neutral load is different on each phase?

    In other words, if I have a 3ph-4w 208/120v panelboard and connect three circuits, one on phase A, one on phase B, and one on phase C and the corresponding 120v loads are 10A on phase A, 8A on phase B, and 6A on phase C, what would my neutral current be from the unbalanced loads which are all 120 degrees out of phase?
    SQRT (I^A + I^B + I^C) - (IA x IB) + (IB x IC) + (IC x IA)

    SQRT (10 x 10 + 8 x 8 + 6 x 6) - (10 x 8) + (8 x 6) + (6 x 10)

    SQRT (100 + 64 + 36) - (80) + (48) + (60)

    SQRT 200 - 188 = 12

    SQRT 12 = 3.46

    Roger
    Last edited by roger; 12-12-08 at 01:02 PM.
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    Two write it out a bit better, it's

    √((Ia²+Ib²+Ib²)-(Ia*Ib)-(Ib*Ic)-(Ia*Ic))

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    Quote Originally Posted by 480sparky View Post
    Two write it out a bit better, it's

    √((Ia²+Ib²+Ib²)-(Ia*Ib)-(Ib*Ic)-(Ia*Ic))
    Why do you think it's better? Might it have something to do with symbols or two?

    Actually, I think one could do it fine.


    Roger
    Last edited by roger; 12-12-08 at 01:19 PM.
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    Quote Originally Posted by roger View Post
    Why do you think it's better? Might it have something to do with symbols or two?

    Actually, I think one could do it fine.


    Roger
    I^A means what? I (amps) to the power of A (?)

    Also, you don't have enough parentheses nested.

    Quote Originally Posted by roger View Post
    SQRT (I^A + I^B + I^C) - (IA x IB) + (IB x IC) + (IC x IA)
    .....
    Roger
    Appears you take the square root of the first set of parantheses, then start subtracting the other three sets.
    Last edited by 480sparky; 12-12-08 at 01:23 PM.

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    Quote Originally Posted by 480sparky View Post
    I^A means what? I (amps) to the power of A (?)

    Also, you don't have enough parentheses nested.



    Appears you take the square root of the first set of parantheses, then start subtracting the other three sets.

    Oooops, good point should have been I^2A etc...., but I still don't know why it takes two to write it a bit better. :wink:

    Roger
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    Quote Originally Posted by roger View Post
    Oooops, good point should have been I^2A etc...., but I still don't know why it takes two to write it a bit better. :wink:

    Roger
    That's the problem with such a formula. Writing it out with common computer keyboard symbols makes it difficult to follow the hierarchy of the formula.

    You did a good job explaining it on your first post, though.

    Are you asking why it takes two sets of parentheses? To force the Square-Root function to be the last. The way you originally wrote it, it would actually work out this way:

    SQRT (I^A + I^B + I^C) - (IA x IB) + (IB x IC) + (IC x IA) means the SQRT function would only apply to the first set of parentheses. You'd end up with:

    SQRT (10*10 + 8*8 + 6*6) - (10*8) + (8*6) + (6*10)

    SQRT (100 + 64 + 36) - (80) + (48) + (60)

    SQRT (200) - (188)

    At this point, the SQRT function only would aply to the 200, so you end up with

    14.14 - 188 = -185.86.

    It's kinda of difficult to have a negative amperage flow. :smile:

    At least, this is how I remember it from school, back when dirt was invented.


    Edit to add: I just noticed my first formula was wrong. I had Ib² in the first set of parenthesis twice. It should be:


    √((Ia²+Ib²+Ic²)-(Ia*Ib)-(Ib*Ic)-(Ia*Ic))


    And to be hard on myself, I should also add the following variables explained:
    Ia = Amp flow on phase A
    Ib = Amp flow on phase B
    Ic = Amp flow on phase C
    Last edited by 480sparky; 12-12-08 at 01:45 PM.

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    Quote Originally Posted by 480sparky View Post
    Are you asking why it takes two sets of parentheses?
    No, I'm asking why you say Two write it out a bit better. Even though I was in a hurry and answering phone calls while I was posting, I think One (even if just myself) can write it out correctly.


    Hint, read the first sentence of your first post. :wink:

    Roger
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    Quote Originally Posted by roger View Post
    No, I'm asking why you say Two write it out a bit better. Even though I was in a hurry and answering phone calls while I was posting, I think One (even if just myself) can write it out correctly.


    Hint, read the first sentence of your first post. :wink:

    Roger
    OOOOOOOOOOoooooooooooo! Sea watt hapens wen theirs know SpellCheck?

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    Quote Originally Posted by 480sparky View Post
    Edit to add: I just noticed my first formula was wrong. I had Ib² in the first set of parenthesis twice. It should be:

    If you're like me and are doing anything else or are the least little bit preoccupied, things do get missed and typo's do happen.

    Roger
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