Balancing a on neutral

[sparkyway]

Got to settle this issue with my contractor friend...Using a 208 volt 3 phase system...will the lighting neutral load balance out if (for example) I run a black phase (to carry 9 amps) and a red phase (13 amps) with a common neutral. There isn't a need to include a third phase in the conduit for load concerns. I appreciate the technical insights in this issue and I apologize if its a repeat question ( I did check ). Thanks

 

[iwire]

Got to settle this issue with my contractor friend...Using a 208 volt 3 phase system...will the lighting neutral load balance out if (for example) I run a black phase (to carry 9 amps) and a red phase (13 amps) with a common neutral.

It's safe and legal but in your example the neutral would not 'balance out' to 0 amps. Without doing the calculations I would think you would have around 13 amps on the neutral.

If it was a single phase system you would only have 4 amps on the neutral.

 

[sparkyway]

Thanks! I just couldn't see how not including the third phase in the conduit would make a difference in balancing the neutral's load. I appreciate the help and fast answer very much.

 

[brian john]

13 is close enough- 11.5 something amps

 

[iwire]

Thanks! I just couldn't see how not including the third phase in the conduit would make a difference in balancing the neutrals load.

Oh but it does.

If you had the following

A = 5 amps

B = 5 amps

C = 5 amps

The neutral would have 0 amps.

Turn of any single phase

A = 5 amps

B = 0 amps

C = 5 amps

The neutral would have now have between 5 and 10 amps on it.

In my opinion, if your running multi wire branch circuits from 3 phase panels it makes much more sense to run and use all three phases.

 

[sparkyway]

I totally agree... but my friend seemed to consider this a violation of safe electrical procedures... I disagreed considering info stated here and the fact that each phase is 120 degrees out of phase with one another. The earth will continue spinning and the sun just might rise another day...had to smile a bit. Sure, it'd be "better" to figure in the third phase but not "necessary or crucial."

 

[brian john]

I cannot locate my link for the formula but basically it i

(AxA+BxB+CxC)-(AxB+BxC+CxA) and take the square root of this number.

 

[LarryFine]

I just couldn't see how not including the third phase in the conduit would make a difference in balancing the neutral's load.

Here's a simple way to picture it:

Start with a fully-balanced 3-phase MWBC. There's zero neutral current, right? Now, reduce the load on any one phase by one amp.

The neutral current increases by that one reduced phase amp. Reduce the phase current by 5a, and the neutral current becomes 5a.

If you eliminate one phase entirely, as in your case, the neutral current equals the current in the remaining phases, if they're equal.

 

[LarryFine]

Turn of any single phase

A = 5 amps

B = 0 amps

C = 5 amps

The neutral would have now have between 5 and 10 amps on it.

As I see it, the neutral current would be exactly 5a.

 

[hurk27]

.5 x the sum of the remaning current.

A = 8
B = 12
C = 0

8 + 12 = 20 x .5 = 10 amps on the neutral

Is this right?

 

[480sparky]

Got to settle this issue with my contractor friend...Using a 208 volt 3 phase system...will the lighting neutral load balance out if (for example) I run a black phase (to carry 9 amps) and a red phase (13 amps) with a common neutral. There isn't a need to include a third phase in the conduit for load concerns. I appreciate the technical insights in this issue and I apologize if its a repeat question ( I did check ). Thanks

11.5325625947 amps.

..............
Turn of any single phase

A = 5 amps

B = 0 amps

C = 5 amps

The neutral would have now have between 5 and 10 amps on it.......

5 amps.

The formula is : SqrRt of (Ai?+Bi?+Ci?-AiBi-AiCi-BiCi)

 

[480sparky]

I finally located the graphic I have of the formula:

Where:

Ai = Current on A phase
Bi = Current on B phase
Ci = Current on C phase
Ni = Current on Neutral

 

[hurk27]

11.5325625947 amps.

The formula is : SqrRt of (Ai?+Bi?+Ci?-AiBi-AiCi-BiCi)

Where did the .5325625947 amps come from? thats more current then you had to start with?

with the 9 and 13 amp loads in the OP on two phases of a Y system I came up with 11 amps.

(9+13)x.5 = 11

Or has it just been to long ago for me:roll:

 

[480sparky]

Where did the .5325625947 amps come from? thats more current then you had to start with?

with the 9 and 13 amp loads in the OP on two phases of a Y system I came up with 11 amps.

(9+13)x.5 = 11

Or has it just been to long ago for me:roll:

Where did 0.5 come from?

I use the formula, not just a multiplier.

 

[charlie b]

.5 x the sum of the remaning current. . .

Is this right?

No it is not. It might sometimes give you an answer that is close. But the correct formula has been given in this thread by several members.

 

[sparkyway]

According to this diagram, if I use A phase at 5 amps and B phase at 13 amps...my neutral load would be 12.449899 amps. Thanks for the assists everyone...the sun did rise this morning!! Have a great day people!!

 

[sparkyway]

Must have hit the wrong button... I keep getting 11.35+ amps so we're good to go still.

 

[Besoeker]

Must have hit the wrong button... I keep getting 11.35+ amps so we're good to go still.

I get 11.53A with that formula.
But note that this formula is valid only if the load on each of the phases has the same power factor.

 

[gar]

090912-1216 EST

To see what is happening draw a vector diagram.

Draw a vertical line 9 units long. Using conventional notation this is an angle of 90 deg. From the top end of the first vector draw a second vector 13 units long to the right at an angle of -30 deg. Note 0 deg is horizontal to the right. Also note that the right hand end of the 13 unit vector is not as low as the bottom end of the 9 unit vector.

Draw a construction line horizontally from the lower right end of the 13 unit vector to the point of intersection with the 9 unit vector.

The sine of 30 deg is 0.5 exactly, and the sine of 60 deg is approximately 0.86602540 (1/2 the sq-root of 3).

The base line of a right triangle of hypotenuse 13 units and a 30 deg angle is 6.5 units exactly. Thus, the intersection point on the first vector is 6.5 units from the top end. That leaves 2.5 units below the intersection point.

Now a second right triangle can be constructed with sides of 2.5 and 13*0.866 = 11.258. The hypotenuse of this triangle is the desired neutral current. The solution is (2.5^2 + 11.258^2)^0.5 = 11.533 . The same result as Besoeker.

.

 

[sparkyway]

I agree with you and Besoeker...I made an excel spreadsheet with the formulas and I get the same. My old outdated calculator is now in a much happier place. Now I won't need to travel far for a quick accurate answer either. Thanks all.