Neutral Current Calculation

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I think what is missing here is the actual maximum load current. We can talk buy/sell current all we want but ultimately the inverter can be producing more current than what is in shortage or excess. All that current would flow on the neutral.
 
080712-0945 EST

There are several undefined assumptions in the question. These are:

(1) The inverters are in perfect synchronization with the power company.
(2) The waveforms are sinusoidal. There are no harmonics.
(3) The loads are resistive.
(4) The unbalanced current in the power company neutral is the component resulting from the different distribution of current from the inverters. In this case the 10 A is correct. But note the power company neutral might be 25 A, 10 A from the inverter unbalance, and 15 A from a load unbalance.

Nothing was said about the output current from each inverter. The question does not state that each inverter is current limited at the same value.

It is probably more appropriate to consider each source as a constant current supply instead of voltage. Certainly at least for the inverters.

The convention for the direction of positive current flow is positive to negative in the circuit external to the source of energy. Using the assumptions 1, 2, and 3 above you can draw the circuit with batteries for the voltage sources and simplify your thought process,

Note: In the case of an inductor with a steady-state current of I and just before opening a switch the current inside the inductor is flowing from the + end to the - end. In this case the inductor is a load. The direction and magnitude of current thru an inductor can not change instantaneously. Just after opening of the switch the current is unchanged, this means the voltage polarity across the inductor has to reverse instantaneously and the inductor becomes the energy source.

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niffur said:
I think what is missing here is the actual maximum load current. We can talk buy/sell current all we want but ultimately the inverter can be producing more current than what is in shortage or excess. All that current would flow on the neutral.
Click to expand...

This is a 120/240 setup -- dual inverters. In which case neutral current is the sum of the maximum load plus the maximum inverter output. Using a worked example, I can "buy" 60A (inverter passthru limit) on L1 and "sell" up to 30A (inverter output limit) on L2 and N is then 90A. The OCPD's on L1 and L2 are both 60A (passthru is greater than inverter output) and there is no OCPD on N.

That's why I asked for the code section -- I don't recall seeing any code requirements for neutral sizing different than maximum current on L1 or L2.
 
niffur said:
I think what is missing here is the actual maximum load current. We can talk buy/sell current all we want but ultimately the inverter can be producing more current than what is in shortage or excess. All that current would flow on the neutral.
Click to expand...

But Tallgirl's statement was concerning the wiring which is coming from the utility transformer. If there is 5 amps sell on one leg and five amps buy on the other leg, there is 10 amps total on the neutral going into the centertap, regardless of the actual connected load and regardless of how much current the inverter is putting out.
 
crossman said:
The attachment is not working for me.
Click to expand...

Hmmm. Works for me, but the board didn't let me edit the post I'd attached it to, so I'm wondering what's up with that. Here it is again --

attachment.php


Oh, and really quick -- if you want to see it in real time, it's http://julie-world.mooo.com/JavaMate.html Click on "Total".

Software has bugs, and it's limited to a small number of connections, so don't camp on it.
 
tallgirl said:
I'm not sure his diagram is either useful or correct. I've been trying to make sense of it since I saw it and it seems to include an incorrect assumption, as well as just plain being wrong.
Click to expand...

As a schematic, Smart's diagram looks okay to me.

tallgirl said:
The setup for a standard grid-interactive inverter is that the voltage sources are on the "same side" of the load
Click to expand...

The placement of the components in a schematic do not represent the actual physical location of the components. A schematic is a simplistic diagram of what is connected to what.

tallgirl said:
power from the PoCo passes through the inverter on the way to the load.
Click to expand...

I'm not sure this statement is correct. Maybe it is true of some systems, but the ones I have seen, the inverter is simply a second source and is connected in parallel with the utility, of course, OCP is inserted.
 
tallgirl said:
This is a 120/240 setup -- dual inverters. In which case neutral current is the sum of the maximum load plus the maximum inverter output.
Click to expand...

I don't think this statement is accurate. I think your math is correct though, but it doesn't match what you said.

Say my maximum connected load is 40 amps on each leg. And my inverter is producing a maximum 40 amps on both legs. Seems to me that the inverter is providing all the power to the load in this case and the neutral from the utility will have zero amps.

But if we add max load + max inverter output, we get 80 amps.

I am definitely no expert in all this. My point with being contrary is to force you guys to teach me something.
 
crossman said:
But Tallgirl's statement was concerning the wiring which is coming from the utility transformer. If there is 5 amps sell on one leg and five amps buy on the other leg, there is 10 amps total on the neutral going into the centertap, regardless of the actual connected load and regardless of how much current the inverter is putting out.
Click to expand...

I see - I thought we were sizing the neutral for the load.

So I would say for the original 5A buy + 5A sell question the actual current would be zero since we're dealing with opposing currents. What should be used for sizing would be something different, I'm sure.
 
Niffur, the problem here is that the currents are not opposing. One is a "buy" and one is a "sell". This negates the old rule of "the neutral carries the imbalance of the two currents".

Certainly, if both currents are "buy", then the two five amp currents results in zero on the neutral. But a "sell" situation on one leg means this current is going backwards and you would add, not subtract.

Disclaimer: I am taking a simplistic approach and using all positive currents. The old "imbalance" formula will work if we use "negative" currents.
 
crossman said:
Niffur, the problem here is that the currents are not opposing. One is a "buy" and one is a "sell". This negates the old rule of "the neutral carries the imbalance of the two currents".

Certainly, if both currents are "buy", then the two five amp currents results in zero on the neutral. But a "sell" situation on one leg means this current is going backwards and you would add, not subtract.

Disclaimer: I am taking a simplistic approach and using all positive currents. The old "imbalance" formula will work if we use "negative" currents.
Click to expand...

Yes, I see it now :)
 
080712-1158 EST

Make this a DC circuit. Two voltage sources (power company). Two current sources, 30 A each (inverters). One resistive load draws 35 A and the other 25A. Serparate this into two circuits. Draw this and you will see that both have 5 A flowing toward the neutral point. When you connect the neutral points back together and combine the neutral wires you then have 5 + 5 = 10 A.

This analysis fails for some AC and non-resistive loads. You can still do the two separate circuits, but then you need to add the instantaneous neutral currents and calculate the RMS value.

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Hey Tallgirl,

I want to make sure that you understand that I understand what you are saying. I never thought about it before....

With a PV system and inverter, under certain circumstances, we could overload the neutral. This could happen with a fairly large inverter and when the only loads turned on in the house were all from the same utility leg, in other words, maximum unbalanced load.

Edit: Now, upon further reflection, since the neutral is going to be sized to carry the maximum unbalanced load already, there is no way to overload the neutral even in the worst case scenario. That's the way i see it.
 
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080712-1555 EST

crossman:

Without inverters I can oveload the neutral if the design is marginal.

I will design a load that draws a short narrow current pulse. For the load on one side of center tap the pulse is turned on just before 90 and 270 deg, and the load on the other side of center tap is turned on just after 90 and 270 deg. Now we have two independent current pulses in the neutral every half cycle.

Is this probable? No.

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Without an inverter, and with a 120/240 single phase system, it is going to be impossible to overload the neutral of a code-compliant installation by any type of switching the loads on and off during certain periods in the sine wave.

Now, on a three-phase wye, I am totally in agreement with you.

Back to the 120/240 single phase service with inverter.... I am sticking with the point that it is impossible to overload the neutral of a code-compliant installation no matter what loads are on or off.

If I am incorrect, I am certainly open for an education. Convince me I am wrong with diagrams, theory, etc.
 
crossman said:
Without an inverter, and with a 120/240 single phase system, it is going to be impossible to overload the neutral of a code-compliant installation by any type of switching the loads on and off during certain periods in the sine wave.

Now, on a three-phase wye, I am totally in agreement with you.

Back to the 120/240 single phase service with inverter.... I am sticking with the point that it is impossible to overload the neutral of a code-compliant installation no matter what loads are on or off.

If I am incorrect, I am certainly open for an education. Convince me I am wrong with diagrams, theory, etc.
Click to expand...

That's why I asked the question about which section covered neutral sizing in this situation.
 
For the utility service neutral, see 220.61. It is actually going to require doing a load calculation based on the provisions of Article 220 first. Then we can come up with the required size for the neutral.

For the photovoltaic system / inverter conductors, see 690.8.
 
080712-2149 EST

crossman:

I am not sure how your comment on code compliant fits into the description I provided.

From hot line 1 to neutral I have a load that draws current from 80 to 85 degrees with a peak value of I1. Assume it to be a rectangular pulse. This dissipates I1^2*Rneutral*5/360 watts.

From hot line 2 to neutral I create an identical pulse but delayed to occur between 95 to 100 degrees and of opposite polarity. Clearly these do not overlap and therefore do not cancel. This second pulse dissipates the same power in the neutral as the first pulse.

Thus, if we assume the neutral wire size is the same as either hot line, then the power dissipated in the neutral is double that in either hot line. Note: power dissipation has nothing to do with the polarity of the current when the different currents do not overlap.

This is something I could create. Why? I have no idea for a use other than to prove that in a center tapped single phase application that I can produce a higher neutral current than the current in either hot line.

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After further thought, I agree with the theoretical aspects of your statement. The neutral could certainly carry twice as much curent as the line conductors.

But I am wondering that if you built this equipment and got a UL listing on it, would the nameplate current be based strictly on the RMS value of the line current, or would the nameplate current be based more on the peak of the spike?

The RMS value of the large spike could actually be rather small. This would lead to small line and neutral conductors in the installation. And the neutral may be overloaded. But if UL required a nameplate current based more on the "spikey" nature of the load, then we would use bigger wire and we still wouldn't overload the neutral.

I hope that makes sense. It is all semi-educated speculation on my part.:smile:
 
Back to the original thought of the OP:

After having pondered this a good bit more, I still hold the conclusion that, considering normal residential loads, there is no way to overload a properly sized service neutral conductor regardless of the loading. There just aren't any scenarios where the service neutral carries more current than the unbalanced load.

And since the service neutral has to be sized based on the maximum unbalanced load, we're okay.
 
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