1? / 3? Neutral Amp Question

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Re: 1? / 3? Neutral Amp Question

That's the equation I was looking for yesterday. I think we should stick with Wye configurations for this discussion.

When we use the equation for L-N loads, we are putting in phase currents. So if we put in 50A for A, and 50A for B, those two currents are assumed to be 120 deg apart (they have to be for the equation to work).

That is a completely different case than having a L-L load. In this case the 50A is not in phase with either of the currents in the above situation.

So in figuring the neutral current, you'd first have to eliminate all line-line loads from the phase totals, and then proceed with the calculation?
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You could do that knowing that L-L loads don't contribute any neutral current. Or you could convert your L-L loads into phase currents.

Steve
 
Re: 1? / 3? Neutral Amp Question

Steve, can you show the derivation of this formula? It must be an approximation since phase angles are not considered. Now if the phase currents must be separated by 120 degrees, then it is not applicable to the problem at hand.
 
Re: 1? / 3? Neutral Amp Question

Originally posted by rattus: Now open the C leg, then A = B, and the evaluation of the formula yields zero.
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No it doesn't. The formula yields "A."

I was going to post that same formula yesterday, when I first saw this thread. I was going to show how that formula proved that In = 0. Then I noticed that it did not prove that In = 0. That threw me, and I decided to go back to work, instead of trying to figure out what was wrong.

Originally posted by georgestolz: They're the amps each phase is pulling. If A = 50 amps, and B = 50 amps, C = 0 amps, you get 50 amps on the neutral. The beginning of the equation would be " I(n) = "
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That matches what I said above: The formula yields In = "A."
Originally posted by rattus:Steve, can you show the derivation of this formula?
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I would also like to see anyone show that derivation. I first saw the formula in the middle of one of Mike Holt's training videos on the NEC. I wrote it down, and I have used it from time to time. But I never could figure out how, why, and when it works, nor could I discover its limitations. If it does not apply to a DELTA, for example, then I could not tell you why.
 
Re: 1? / 3? Neutral Amp Question

I stand corrected. The formula does not yield 0 in this case. It is obviously an approximation with stringent conditions attached.

The formula is unusable with this problem.

You can look at the problem and see,

Ia = -Ib which means the currents are 180 degrees apart.

In = Ib -Ib = 0
 
Re: 1? / 3? Neutral Amp Question

Steve, can you show the derivation of this formula?
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I thought that would be an easy question, but I still don't have it after 10 min.

It must be an approximation since phase angles are not considered.
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Yes, it is somewhat of an approximation. It assumes that all the phase currents are shifted from the phase voltages by the same amount. That's usually a close approximation in power systems.

Now if the phase currents must be separated by 120 degrees, then it is not applicable to the problem at hand.
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Well, not directly applicable, but im still torn here. It seems like we should be able to convert the L-L current to phase currents, and then show the neutral current is 0. In effect, divide the load current into two other currents 120 deg out of phase. Then we should be able to get the equation to show a zero neutral current.

But, either my math is too rusty, or my reasoning is off, because I can't get that equation to give 0 for any combination of A & B with C=0.

Steve
 
Re: 1? / 3? Neutral Amp Question

Steve,

There is no way to get 120 degrees separation between Ia and Ib because they represent a single current with the magnitude and phase set by the one line to line load.

This formula simply is not applicable to this problem.
 
Re: 1? / 3? Neutral Amp Question

I think what we have been missing here is the initial setup of the problem. It started with a 3-phase panel, and postulated a line-to-line load that did not require a neutral wire. It said that a neutral wire was run to the appliance, "just for fun." But it did not state how, or even if, it was connected. So we need to talk about two different circumstances.

Case 1: Neutral is not connected at, and is not used by, the load.
Obviously there will be no neutral current, since the neutral wire is not connected. The voltage across the load will be the vector subtraction of Va ? Vb. That result has a magnitude of the square root of three times the magnitude of Va, and is at an angle of 30 degrees with respect to Va. The current in the load will be either in phase with that 30 degree angle voltage (if it is resistive) or out of phase (if it is not just resistive). The current coming from Phase A will be equal and opposite in phase from the current coming from Phase B.

The formula of (In^2) = (Ia^2 + Ib^2 + Ic^2) ? (IaIb + IaIc + IbIc) does not apply, because you are not looking at a 3-phase, 4-wire load. That formula gives you neutral current, and this case does not include a neutral.

Case 2: Neutral is connected at, and is used by, the load.
There will be current in the neutral. If Ia = Ib = 50 amps, then In = 50 amps also, as is predicted by the formula.

So whatever the origin and the basis of that formula, it seems clear to me that it only applies to a 3-phase, 4-wire load, and works even if the load in one phase is zero.
 
Re: 1? / 3? Neutral Amp Question

Yes, I have to agree with you Charlie. I was thinking of your second case.
Say you divided the load into two equal loads in series. I incorrectly assumed that the voltage at the point between the two loads would always be zero. So that would let you connect the neutral current at that point, and still get zero current.

Now I realize that since the phase voltages are out of phase, the voltage between the loads would not be zero, and connecting the neutral changes the circuit.

Steve
 
Re: 1? / 3? Neutral Amp Question

Charlie B.,

If Case 2 involves identical L-N loads, then you are correct. In = 50A.

But, L-L loads contribute nothing to In.

Agreed?
 
Re: 1? / 3? Neutral Amp Question

I have a somewhat related question: Tomorrow I will be pulling four [three phase Wye] circuits in 3/4" EMT. Circuits 14 and 24 are 12 gauge wire and circuits 39, 41 and 2,4 are 10 gauge wire. Would it be OK to create a 4 wire,3 phase circuit such as 24, 2 & 4 sharing the neutral? The phases, A B and C, would match as well as 14 [A], 39 and 41 [C phase]. But the amperage, as indicated by the different wire sizes, probably wouldn't.
The fate of 100' of white, 12 gauge wire rests in your hands. [Two 50' runs.]
~Peter
 
Re: 1? / 3? Neutral Amp Question

Originally posted by rattus:
There is no way to get 120 degrees separation between Ia and Ib because they represent a single current with the magnitude and phase set by the one line to line load.

This formula simply is not applicable to this problem.
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That clicks. :)

Peter: Could you guarantee that all circuits would be consuming power enough to knock a #10 ungrounded conductor supplying a load wouldn't overload a #12 neutral? Sounds pretty dicey to me. :)

[ September 22, 2005, 12:20 AM: Message edited by: georgestolz ]
 
Re: 1? / 3? Neutral Amp Question

Along those lines, a voltage drop question:

Picture a two-wire-circuit of #10 supplying a load 150' from the source, at 120 V, drawing 20 amps.

Now picture the neutral at #12.

The normal voltage drop equation I use is
Vd = 2 x R x I x D
Vd is Voltage Drop
2 for two identical conductors
R is Resistance
I is Amperage
D is Distance

Now I imagine this would change to
Vd = ( 2 x I x D ) x ( (R1 + R2) / 2)

Where R1 is the resistance of the one conductor of #10, and R2 is the resistance of one conductor of #12. Is that correct? I forget the formula for two resistors in series, as these would be, I think.

[ September 22, 2005, 12:31 AM: Message edited by: georgestolz ]
 
Re: 1? / 3? Neutral Amp Question

Originally posted by georgestolz:The normal voltage drop equation I use is
Vd = 2 x R x I x D
Vd is Voltage Drop, 2 for two identical conductors, R is Resistance, I is Amperage, D is Distance

Now I imagine this would change to Vd = ( 2 x I x D ) x ( (R1 + R2) / 2)

Where R1 is the resistance of the one conductor of #10, and R2 is the resistance of one conductor of #12. Is that correct? I forget the formula for two resistors in series, as these would be, I think.
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Yes it is correct. Two resistors in series give a total resistance equal to the sum of the two individual resistances.

But I prefer to think of the "2" in your "2 for two identical conductors" as representing "2 times the one-way distance "D." Another way to look at your proposed configuration is to note that you have one conductor going a one-way distance "D" and another conductor (with a different resistance) going a one-way distance "D." So my equation would be,
Vd = (1 x I x D ) x (R1) plus (1 x I x D ) x (R2)
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But this works out to the same equation as you showed. Essentially, what I would be doing is treating the two legs with separate voltage drop calculations, and then adding their results.
 
Re: 1? / 3? Neutral Amp Question

Originally posted by peter: The fate of 100' of white, 12 gauge wire rests in your hands. [Two 50' runs.]
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I think you need to buy 100' of white, 10 gauge wire.

What I think you are suggesting is that you have one circuit, slots 2 and 4, double-pole, 208 volt, 10 gauge wire, sharing a #12 neutral with a second circuit, slot 24, single pole, 120 volt, 12 gauge wire. You are also suggesting the same arrangement with the double-pole circuit, slots 39 and 41, and the single-pole circuit, slot 14.

My question is, "What is being supplied by the two double-pole circuits, and specifically, do they use the neutral?"

If the circuit in slots 2 and 4 does not need or use or even have connected a neutral wire, then the #12 you run with them is merely the neutral for circuit #24. But if that were the case, you would not have needed to ask the question.

So I am guessing that double-pole circuit in slots 2 and 4 (same for 39/41) does need, and use, a neutral. So the worst case would be for the portion of the load on slot 2 to be turn off, the portion of the load on slot 4 to be fully loaded (i.e., 30 amps on the #10 wire), and the circuit on slot 24 to be turned off. That would force the entire neutral current from slot 4 to go through the #12 neutral wire, and overload that neutral.

Your phase selection is OK. But your neutral wire has to be #10.

By the way, since I do not know the calculated load for these circuits, may I presume you have already taken into account the derating factor for 6 current-carrying conductors?
 
Re: 1? / 3? Neutral Amp Question

They won't let me have a copy of the prints but today I found that circuits 2 and 4 are 30 amp, 2 pole cold water heater. That is a 10-3 MC cable coming into the junction box and I get to decide whast runs to the panel through the conduit.
Circuits 39 and 41 are listed as "Washer/Dryer" and is again a 10-3 cable. Now it seems more likely that the #10 white is unnecessary and would make a fine neutral for the #12 clack [A ohase] circuit.
~Peter
 
Re: 1? / 3? Neutral Amp Question

I would call it balloon algebra its a bit inflated for a single phase two wire calculation. :D

This jokingly commit was about the original question you changed your question midstream.

And everyone using long math for a simple question seem like a joke to me.

But I'll get over it. ;)

But they are great graphics.

[ September 24, 2005, 04:39 PM: Message edited by: ronaldrc ]
 
Re: 1? / 3? Neutral Amp Question

Man, you guys are making my eyes cross and head ache -- would you please translate all of that engineering algebra and theory for calculating the wye neutral current into English? Have pity on a poor electrician who is trying to follow along with this discussion.

But... am I missing something here, or isn't that the whole point of 310.15(B)(4)(b)?

"In a 3-wire circuit consisting of two phase wires and the neutral of a 4-wire, 3-phase, wye-connected system, a common conductor carries approximately the same current as the line-to-neutral load currents of the other conductors and shall be counted when applying the provisions of 310.15(B)(2)(a)"

[ September 24, 2005, 03:44 AM: Message edited by: tx2step ]
 
Re: 1? / 3? Neutral Amp Question

Originally posted by tx2step:
Man, you guys are making my eyes cross and head ache -- would you please translate all of that engineering algebra and theory for calculating the wye neutral current into English?
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How about some doodling? :)

Let's see if I have this right:

:)

So, are my theories correct? I think I got it right.
 
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