Calculating Unbalanced Neutral Current in 3p-4w system

[stevee]

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?

 

[roger]

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

 

[480sparky]

Two write it out a bit better, it's

√((Ia?+Ib?+Ib?)-(Ia*Ib)-(Ib*Ic)-(Ia*Ic))

 

[roger]

Two write it out a bit better, it's

√((Ia?+Ib?+Ib?)-(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

 

[480sparky]

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.

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.

 

[roger]

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

 

[480sparky]

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 Ib? in the first set of parenthesis twice. It should be:


√((Ia?+Ib?+Ic?)-(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

 

[roger]

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

 

[480sparky]

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?

 

[roger]

Edit to add: I just noticed my first formula was wrong. I had Ib? 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

 

[480sparky]

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

I'm nothing like you, Roger. I gotta go back and edit two or three times after I click the "Preview Post" button. Even then, I miss something that I end up seeing once it's posted. Then the ten-minute clock is ticking, and I didn't catch that one quick enough.......

Can you fix me up with a private 24-hour edit button?

 

[roger]

I'm nothing like you, Roger. I gotta go back and edit two or three times after I click the "Preview Post" button. Even then, I miss something that I end up seeing once it's posted. Then the ten-minute clock is ticking, and I didn't catch that one quick enough.......

Believe it or not, that is exactly like me

Can you fix me up with a private 24-hour edit button?

No, but anytime you really need something edited, like something left out of advice that could possibly leave someone in a dangerous situation, you know you can let one of the Mods know and they can edit for you.

BTW, I live by the same rules as everyone else when it comes to editing my posts.

Roger

 

[gar]

081212-1406 EST

Above the assumption has been made that phase angle of the load to its source voltage was the same for each load.

I would approach the problem by drawing a vector diagram and calculating the resultant which makes it easy for loads with non-equal phase angles.

.

 

[roger]

081212-1406 EST

Above the assumption has been made that phase angle of the load to its source voltage was the same for each load.

I would approach the problem by drawing a vector diagram and calculating the resultant which makes it easy for loads with non-equal phase angles.

.

Yeah, most electricians, electrician apprentices, laborers, landscapers, pool boys, Hooters waitresses, fishermen, and the overall world population would too, I don't know what I was thinking.

Roger