Neutral Current Calculation

Status
Not open for further replies.
tallgirl

tallgirl

Senior Member
Location
Glendale, WI
Occupation
Controls Systems firmware engineer
Apropos of nothing, on a 120/240 single phase system with one AC inverter on each leg, if one leg has a -5A "sell" current (the inverter is producing 5A more than being consumed) and the other leg has a 5A "buy" current (the loads on that leg are 5A more than the inverter is producing), the neutral current is 10A, right?
 
tallgirl said:
Apropos of nothing, on a 120/240 single phase system with one AC inverter on each leg, if one leg has a -5A "sell" current (the inverter is producing 5A more than being consumed) and the other leg has a 5A "buy" current (the loads on that leg are 5A more than the inverter is producing), the neutral current is 10A, right?
Click to expand...

How about a diagram showing the currents?
 
tallgirl said:
Apropos of nothing, on a 120/240 single phase system with one AC inverter on each leg, if one leg has a -5A "sell" current (the inverter is producing 5A more than being consumed) and the other leg has a 5A "buy" current (the loads on that leg are 5A more than the inverter is producing), the neutral current is 10A, right?
Click to expand...
Hi, Julie. Long time, no see.

I would think that the 5a will flow on the 'neutral' conductors between the two inverters, but this current should not affect the real shared-neutral circuit current.

Plus, the load currents on the two ungrounded conductors in the feeder will likely carry this 5a as part of the load currents. In any case, it sounds like unproductive current.
 
Tallgirl, I drew a diagram and "played" around with the numbers. Assuming you are talking about the service neutral between the inverter and the utility xfmr, then I agree with you. Ten Amps on neutral.
 
Yup!

Yup!

crossman said:
Tallgirl, I drew a diagram and "played" around with the numbers. Assuming you are talking about the service neutral between the inverter and the utility xfmr, then I agree with you. Ten Amps on neutral.
Click to expand...

Now that I understand the question, I agree with Crossman.
 
I thought Julie was talking about the outputs of the inverters.

Now I would like to see a schematic to respond intelligently.
 
Here is a diagram of what I am thinking:

Just to put some numbers in, I arbitrarily chose 10 amps as the output of the inverter, that is, 10 amps on each line conductor. (I am making an educated guess that the inverter pretty much maintains an equal current on both line conductors.)

To meet the requirements of 5 amps SELL on one leg to the utility and 5 amps BUY on the other leg of the utility, I placed a 5 amp load and a 15 amp load in the circuit as shown below.

The question is, what is the amperage at X, Y, and Z?
aaaPicture1.jpg


This is a very interesting question.

My answers are X = 10, Y = 0, and Z = 10.

I am by no means certain of my answers. It is fun to contemplate the directions of the currents and to figure out what is happening.

I hope this is similar to Tallgirl's scenario.
 
crossman said:
Tallgirl, I drew a diagram and "played" around with the numbers. Assuming you are talking about the service neutral between the inverter and the utility xfmr, then I agree with you. Ten Amps on neutral.
Click to expand...
I came up with something "different"...

buysell.gif
 
Smart, our diagrams are the same. Your use of the variables is more precise than my assumption of an inverter current. I'll get back to ya!:)

Edit: As I ponder... the drawing of the wires and loads is the same. I am not certain of your variables and current directions. I am assuming your drawing has the utility source on the left and the inverter on the right?
 
Last edited:
And on further inspection, your solution for the numbers looks good.

If we assume X = 10 and Y = 5 to give 10 amps on each leg of the output from the inverter, then your numbers match my solution.

15 amps on the top resistor, 5 amps on the bottom resistor, 10 amps on the neutral from the utility to the resistors, 0 amps on the neutral to the inverter.

But it seems some of your current directions are wrong. For instance, the very bottom wire from the inverter, the current needs to be flowing out of the terminal you have marked as negative.

Now, are we actuall correct on the numbers?:cool:
 
Last edited:
Here are my thoughts on current directions. This, of course, is for one half of the sine wave, for the other half, just reverse all the arrows.

The black leg from the utility is 5 amps SELL and the red wire from the utility is 5 amps BUY.
aaaPicture2.jpg
 
crossman said:
But it seems some of your current directions are wrong. For instance, the very bottom wire from the inverter, the current needs to be flowing out of the terminal you have marked as negative.
Click to expand...
I believe convention has current arrows in opposing direction to current flow... i.e. arrow points from positive to negative though charge flow is negative to positive.

(Feeling like the messenger right now.)
 
Doh! I didn't consider you were using conventional flow of positive to negative. 95% of my training was with electron flow and arrows from negative to positive.... it just seems standard to me. Sorry for the confusion. I think the Physics books still use conventional flow, but modern electrician's texts will use electron flow. I'm not sure what the modern engineering texts use.
 
Smart $ said:
I believe convention has current arrows in opposing direction to current flow... i.e. arrow points from positive to negative though charge flow is negative to positive.

(Feeling like the messenger right now.)
Click to expand...

Yes, but what does that have to do with the question at hand? It's AC current -- you going to erase and re-draw your arrows 60 times a second?
 
tallgirl said:
Yes, but what does that have to do with the question at hand? It's AC current -- you going to erase and re-draw your arrows 60 times a second?
Click to expand...
I didn't plan on a redraw... but I could generate an animated GIF to that effect :D
 
Smart $ said:
I didn't plan on a redraw... but I could generate an animated GIF to that effect :D
Click to expand...

You have too much free time on your hands. Here's how I handled it --

Code: 
				neutral = legAmps[0] - legAmps[1];
				if (neutral < 0)
					neutral = - neutral;

Absolute values are your friend.

I didn't bother with 3-phase because I'm not in the mood, and I don't have a 3-phase setup to verify the code on anyway ...

Oh -- you're a geek. I'm writing patent papers on a really neat electric meter. I'll have to remember to post the application number whenever it publishes. You should find it endlessly fascinating.
 
crossman said:
Smart, our diagrams are the same. Your use of the variables is more precise than my assumption of an inverter current. I'll get back to ya!:)

Edit: As I ponder... the drawing of the wires and loads is the same. I am not certain of your variables and current directions. I am assuming your drawing has the utility source on the left and the inverter on the right?
Click to expand...

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.

The setup for a standard grid-interactive inverter is that the voltage sources are on the "same side" of the load -- power from the PoCo passes through the inverter on the way to the load.

So, there is +5A from the transformer L1 to the first inverter input, and -5A from the transformer L2 to the second inverter input. Between the first inverter output and the load on its leg is +5A +/- any load on that leg, and so on.

The upshot of all this is that the service neutral current can exceed the current on either of the other two conductors. What's the code section that covers sizing conductors for this sort of thing?
 
Status
Not open for further replies.
Top