Single Phase Inverters on 208 3 Phase

bellington said:
Sorry, but I am a little hard-headed, or persistent, if you think positively. Is it possible for someone to create a combination of waves that expresses what happens when output from A-B inverter and the output of B-C inverter combine in N-B?
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I think we already went through the math at the level of waveforms for computing V(A-B) as V(A-N) - V(B-N), yes? That was for V(A-N) and V(B-N) of equal magnitude and 120 degrees apart.

Well, for your question, which is the last 3 paragraphs of post #288, the current going into node B will be I(B-C) - I(A-B); the minus sign on I(A-B) is because that quantity is the current going out of node B to node A. And I(A-B) and I(B-C) are of equal magnitude and 120 degrees apart.

So the math is exactly the same as what we saw earlier for computing V(A-B).

Cheers, Wayne
 
wwhitney said:
I think we already went through the math at the level of waveforms for computing V(A-B) as V(A-N) - V(B-N), yes? That was for V(A-N) and V(B-N) of equal magnitude and 120 degrees apart.

Well, for your question, which is the last 3 paragraphs of post #288, the current going into node B will be I(B-C) - I(A-B); the minus sign on I(A-B) is because that quantity is the current going out of node B to node A. And I(A-B) and I(B-C) are of equal magnitude and 120 degrees apart.

So the math is exactly the same as what we saw earlier for computing V(A-B).

Cheers, Wayne
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Yes, I agree with that. But what I am looking for is something that represents the combination of the output of inverter A-B and inverter B-C as seen across B-N (or N-B) and in relationship to grid voltage. When I combine A-B and B-C, I get something 60 degrees between both of them and it really messes up the relationship with the grid voltage of N-B, (or B-N). So, I conclude that my math must be wrong, and can't see where your math combines the interaction of waveforms A-B and B-C through N-B.

I feel like I'm asking how to extract the sugar and cream out of caramel, and I'm getting responses that explains how the sugar and cream combine to make caramel.
 
bellington said:
Yes, I agree with that. But what I am looking for is something that represents the combination of the output of inverter A-B and inverter B-C as seen across B-N (or N-B) and in relationship to grid voltage. When I combine A-B and B-C, I get something 60 degrees between both of them and it really messes up the relationship with the grid voltage of N-B, (or B-N).
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You're miscombining because you forgot a minus sign (we've all made that mistake).

Current A-B and current B-C are 120 degrees apart. To combine at B, you need to flip Current A-B to its negative Current B-A. Mathematically that looks like a 180 degree phase difference. So Current B-A and Current B-C are only 60 degrees apart.

The resulting sum of Current B-A and Current B-C will be 30 degrees apart from each of the summands. That sum will also be in phase with the Voltage B-N.

In the language of post #288 (once you trust this nomenclature and know how to use it, it is much quicker than drawing waveforms or writing equations with sin(t), yet provides identical results),

I(A-B) = 5.77A ∠ 330
I(B-A) = 5.77A ∠ 150
I(B-C) = 5.77A ∠ 90
I(B-A) + I(B-C) = 10A ∠ 120
V(B-N) = 120V ∠ 120

The last two phase angles are the same (120 degrees).

Cheers, Wayne
 
wwhitney said:
You're miscombining because you forgot a minus sign (we've all made that mistake).

Current A-B and current B-C are 120 degrees apart. To combine at B, you need to flip Current A-B to its negative Current B-A. Mathematically that looks like a 180 degree phase difference. So Current B-A and Current B-C are only 60 degrees apart.

The resulting sum of Current B-A and Current B-C will be 30 degrees apart from each of the summands. That sum will also be in phase with the Voltage B-N.

In the language of post #288 (once you trust this nomenclature and know how to use it, it is much quicker than drawing waveforms or writing equations with sin(t), yet provides identical results),

I(A-B) = 5.77A ∠ 330
I(B-A) = 5.77A ∠ 150
I(B-C) = 5.77A ∠ 90
I(B-A) + I(B-C) = 10A ∠ 120
V(B-N) = 120V ∠ 120

The last two phase angles are the same (120 degrees).

Cheers, Wayne
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But back up to the voltages that produced the current.
 
bellington said:
But the voltage is the starting point and current the results. I'm asking about extracting the sugar and cream, not how they combine.
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I don't know what you mean by the above or what you want.

The primary source creates V(A-N), V(B-N), and V(C-N). The voltages V(A-B), V(B-C), and V(C-A) are a mathematical consequence of V(A-N), V(B-N), and V(C-N). One inverter sees V(A-B), creates its own internal voltage that matches V(A-B), and pushes out current I(A-B) in phase with V(A-B). Another inverter does the same on B-C. The resulting currents I(A-B) and I(B-C) combine at node B to make a current that is in phase with the starting voltage V(B-N) (as long as each inverter is pushing out the same magnitude current).

Basically there's some reversibility here. Combine A-N and B-N to get A-B; combine B-N and C-N to get B-C; combine A-B and B-C to get back B-N. For the correct notion of "combine", which notion is applicable here.

Cheers, Wayne
 
wwhitney said:
I don't know what you mean by the above or what you want.

The primary source creates V(A-N), V(B-N), and V(C-N). The voltages V(A-B), V(B-C), and V(C-A) are a mathematical consequence of V(A-N), V(B-N), and V(C-N). One inverter sees V(A-B), creates its own internal voltage that matches V(A-B), and pushes out current I(A-B) in phase with V(A-B). Another inverter does the same on B-C. The resulting currents I(A-B) and I(B-C) combine at node B to make a current that is in phase with the starting voltage V(B-N) (as long as each inverter is pushing out the same magnitude current).

Basically there's some reversibility here. Combine A-N and B-N to get A-B; combine B-N and C-N to get B-C; combine A-B and B-C to get back B-N. For the correct notion of "combine", which notion is applicable here.

Cheers, Wayne
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What do you get when you combine A-B and B-C?
 
bellington said:
Is A-B 30 degrees ahead of N-B and 30 degrees after A-N? Is A-B combined with B-C 60 degrees after A-B and 60 degrees before B-C?
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Maybe the diagram below will help you see the answers to these questions. The center point of the circle should be labeled N. The diagram is from https://electronics.stackexchange.c...a-wye-transformer-make-30-degrees-phase-shift which I haven't read; the discussion there may or may not be helpful.

You can draw an arrow from any of the labeled points to any of the other labeled points, and the direction of that arrow will be the phase shift of the associated voltage. The arrows V(A-N), V(B-N), V(C-N), and V(A-B) are already drawn in for you.

I like the convention that V(A-N) is 0 degrees phase shift and have used it in this thread. And that V(B-N) is 120 degrees. The drawing below matches those choices if you use compass direction conventions, where North (towards the top of the page) is 0 degrees, and East (to the right on the page) is 90 degrees.

Mathematicians usually use the opposite convention, which makes 0 degrees towards positive x, and 90 degrees towards positive y. So I would draw it differently. But either convention works as long as you are consistent.

Cheers, Wayne

538HM.jpg
 
bellington said:
Sorry, but I am a little hard-headed, or persistent, if you think positively. Is it possible for someone to create a combination of waves that expresses what happens when output from A-B inverter and the output of B-C inverter combine in N-B?
Click to expand...
At any one point in a circuit there is one and only one voltage waveform relative to any other point in a circuit, and a PV inverter matches the voltage waveform that was already there wherever it is connected. You can look at Vab, Vbc, Vca, Van, Vbn, and Vcn waveforms on a three phase service with inverter(s) connected to them or not; they will look the same either way.
 
ggunn said:
At any one point in a circuit there is one and only one voltage waveform relative to any other point in a circuit, and a PV inverter matches the voltage waveform that was already there wherever it is connected. You can look at Vab, Vbc, Vca, Van, Vbn, and Vcn waveforms on a three phase service with inverter(s) connected to them or not; they will look the same either way.
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A-B B-C B-N Dashed N-B Dotted.png
The black sine wave is voltage across A-B, and the red one is voltage across B-C. The dashed red one is B-N, and the dotted red is N-B. Does this accurately express those voltage relationship while focusing primarily on B-N?
 
bellington said:
The black sine wave is voltage across A-B, and the red one is voltage across B-C. The dashed red one is B-N, and the dotted red is N-B. Does this accurately express those voltage relationship
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Yes. If you added V(A-N) to your graph, and omitted V(B-C), your graph would visually show that V(A-B) = V(A-N) - V(B-N). On the level of sinewaves, this exactly corresponds to sind(x) - sind(x-120) = sqrt(3) * sind(x-330), where sind takes degrees as its input, i.e. sind(x) = sin(x*pi/180)

To relate your graph to the figure in my previous post, for any of the sinewaves, find its zero crossing (x-axis crossing, place where the value is zero) between 0 degrees and 360 degrees that is 90 degrees before the maximum peak. [The other zero crossing is 180 degrees later, or 90 degrees after the maximum peak.] That angle is the phase angle of the sine wave, and it corresponds to the value A in the functional form sin(x-A) (that is clearest for your dotted red line). In the figure from my previous post, it is the compass direction for the corresponding arrow. And the length of the arrow corresponds to the height of the peak on your graph.

If you do that for each of your sine waves, you find they correspond exactly with the figure in my previous post. And this procedure lets you calculate sums and differences of sine waves without needing a graphing calculator. To compute V(A-B), you can draw the arrow V(A-N), draw the arrow (B-N), and then draw the arrow from B to A. That arrow is V(A-B), in that its length and direction give you the amplitude and phase angle of V(A-B).

Cheers, Wayne
 
wwhitney said:
Yes. If you added V(A-N) to your graph, and omitted V(B-C), your graph would visually show that V(A-B) = V(A-N) - V(B-N). On the level of sinewaves, this exactly corresponds to sind(x) - sind(x-120) = sqrt(3) * sind(x-330), where sind takes degrees as its input, i.e. sind(x) = sin(x*pi/180)

To relate your graph to the figure in my previous post, for any of the sinewaves, find its zero crossing (x-axis crossing, place where the value is zero) between 0 degrees and 360 degrees that is 90 degrees before the maximum peak. [The other zero crossing is 180 degrees later, or 90 degrees after the maximum peak.] That angle is the phase angle of the sine wave, and it corresponds to the value A in the functional form sin(x-A) (that is clearest for your dotted red line). In the figure from my previous post, it is the compass direction for the corresponding arrow. And the length of the arrow corresponds to the height of the peak on your graph.

If you do that for each of your sine waves, you find they correspond exactly with the figure in my previous post. And this procedure lets you calculate sums and differences of sine waves without needing a graphing calculator. To compute V(A-B), you can draw the arrow V(A-N), draw the arrow (B-N), and then draw the arrow from B to A. That arrow is V(A-B), in that its length and direction give you the amplitude and phase angle of V(A-B).

Cheers, Wayne
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A-B added to B-C with B-N Extracted.png

Would the green line represent the combination of A-B and B-C as seen across B-N?
 
First, I want to applaud you for getting out the graphing calculator and really trying to understand this.
bellington said:
Would the green line represent the combination of A-B and B-C as seen across B-N?
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Not quite

First, remember that the _current_ is what gets combined, not voltage. So first you need to calculate the _current_ flowing on the AB and BC path. Then you add these two currents to get the current on the BN path.

You have two voltage waveforms for V A-B and V B-C. But since you are trying to calculate the current flowing at _B_ you probably want your two voltage waveforms to be V B-A and V B-C. One of your L-L voltage waveforms should be inverted.

Next your _current_ is proportional voltage, so for qualitative understanding you can assume a proportionality constant of 1 and just sum the voltage values to represent the _net_ current from the two legs going to (or from) terminal B. Just keep in mind that this is a short cut, for a quantitative result you'd need to calculate the currents first.

Now compare this calculated net current (I B-A + I B-C) to the sine wave representing V B-N.

These should be in phase.
 
Just for clarification: the voltage waveforms will look the same whether or not there are inverters involved. The magnitudes may change a little due to Vd effects but the timing will not change.
 
bellington said:
Would the green line represent the combination of A-B and B-C as seen across B-N?
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In your previous post you specified what you were graphing (voltages). Now you haven't specified, so I'm going to clarify your question and say it's about currents.

But the basic answer is no, as the green line you show would be I(B-C) +I(A-B) (for some common magnitude of those L-L currents, I think 120/sqrt(3) RMS). But to compute a quantity at B, you need both currents to be going to B, i.e. you want I(B-C) + I(B-A). Since I(B-A) = -I(A-B), that means you want I(B-C) - I(A-B).

Also, given the formula that you graphed, you've sort of assumed the answer already (although you assumed the wrong phase shift per the above). So change the green line formula to "red line formula - black line formula". The red line and black lines are V(B-C) and V(A-B), but when the load/source is power factor 1, the currents I(B-C) and I(A-B) will be in phase with the voltages, so that will still give you the correct phase relationship. And if you want to consider loads of 1 ohm, it would give you the correct magnitude as well.

What you will find is that the resulting green line is in phase with your dotted redline, V(B-N).

Cheers, Wayne
 
wwhitney said:
What you will find is that the resulting green line is in phase with your dotted redline, V(B-N).
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Algebraically, in terms of sinewaves, expressed with the sind function which takes degrees instead of radians as its inputs, that will demonstrate the following relationship (I dropped the 170 coefficient in all of your formulas):

(sind(x-120) - sind(x-240)) - (sind(x) - sind(x-120)) = 3 * sind(x-120).

For the algebraically minded, that follows quickly from collecting terms and the observation that sind(x) + sind(x-120) + sind(x-240) = 0

Cheers, Wayne
 
wwhitney said:
Algebraically, in terms of sinewaves, expressed with the sind function which takes degrees instead of radians as its inputs, that will demonstrate the following relationship (I dropped the 170 coefficient in all of your formulas):

(sind(x-120) - sind(x-240)) - (sind(x) - sind(x-120)) = 3 * sind(x-120).

For the algebraically minded, that follows quickly from collecting terms and the observation that sind(x) + sind(x-120) + sind(x-240) = 0

Cheers, Wayne
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Desmos does not like sind, or I don't know how to use it. But, I did take the point by point addition of A-B and B-C, determined the 60 degree phase shift between the two, used -30 to get the correct position, and added in the 294.449 peak of 208. Does that green line not represent the combination of A-B and B-C voltage waveforms?A-B B-C Sum.png
 
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