Single Phase Inverters on 208 3 Phase

[wwhitney]

I don't know what I should do to set up two sliders called C and S

Looks like you've already done it--whenever you type in a constant line like "R = 3", Desmos creates a little slider below it, and if you drag the dot left or right, it changes the value of the constant. It picks the initial bounds of the slider based on the initial value you assigned; you can see in your last screen grab that for R_B=3, it picked the bounds 1 and 5. Presumably there's a simple way to change the bounds yourself.

nor why I would slide constants.

Maybe it would help to think of it as adjusting parameters.

If you look at the last screen grab I posted, all the starting sinewaves were of the form A * sin(x+B). That's a two parameter family, with A the amplitude and B the phase angle. Then visually we could see that adding and subtracting them, and multiplying them by constants, the graphs all still look like sinewaves of the same frequency. In fact they are all of that form for different parameters A and B.

So if you understand how the various operations of addition, subtraction, and multiplication by a constant affect the parameters A and B, instead of working with the graphs directly, you can work with the parameters. Which is a lot less data, just two real numbers, rather than one real number for every value between 0 and 360 degrees.

It is also true that cos(x) is of the form A*sin(x+B). That means that C*cos(x) + D*sin(x) is another two parameter family of sinewaves; but it covers the identical collection of sinewaves. That is for each sinewave we can associate the parameters A, B, C, and D. [But not arbitrarily; if we choose A and B, that determines the sinewave, and thus C and D, and vice versa.]

When looking at sinewaves as vectors in the plane, we associate a sinewave with the point whose Cartesian coordinates are (C,D).

Cheers, Wayne

 

[wwhitney]

If we open all breakers, except our inverters, then coils A-B are in series. What my graph should be showing is an inside-out look at the interactions of the two currents AB and BC and the impact across and through coil B.

If you open all breakers, then you lose your voltage reference, and your inverters shut down, and you have no currents.

I'm missing why AC instantaneous values cannot also be added up directly.

They can, the sentence you quoted said "Instantaneous values add up directly". That's what you are doing with all your graphs, adding them up instantaneously for all value of a representative cycle.

Cheers, Wayne

 

[bellington]

If you open all breakers, then you lose your voltage reference, and your inverters shut down, and you have no currents.

Sorry, all breakers except for the inverters and the main breaker.

They can, the sentence you quoted said "Instantaneous values add up directly". That's what you are doing with all your graphs, adding them up instantaneous for all value of a representative cycle.

Cheers, Wayne

 

[wwhitney]

Sorry, all breakers except for the inverters and the main breaker.

Then see Carultch's comments about when the series paradigm applies.

If you have "conductor - some stuff in series - conductor," then it works. But the only connections to the "stuff in series" are at the ends. That doesn't apply to the subcircuit you are talking about.

Cheers, Wayne

 

[bellington]

Then see Carultch's comments about when the series paradigm applies.

If you have "conductor - some stuff in series - conductor," then it works. But the only connections to the "stuff in series" are at the ends. That doesn't apply to the subcircuit you are talking about.

Cheers, Wayne

If an inverter is the only thing connected to lines A and B of the transformer, there's only one path for the inverter to send current, through coil A and coil B, in series.

 

[wwhitney]

If an inverter is the only thing connected to lines A and B of the transformer, there's only one path for the inverter to send current, through coil A and coil B, in series.

OK, that's true.

But looking back at post #335, it appears you are trying to analyze a 3 ohm resistor across A-B along with a 3 ohm resistor across B-C. You calculate the series resistance R_T = 6 ohms, but nowhere in that arrangement do you have a pure series circuit. Since coil A connects at one end of the two resistors, coil B connects in the middle, and coil C connects at the other end, it doesn't fit the paradigm I described.

If this answer doesn't satisfy you, please draw out a circuit diagram of what you are modeling, and if you have further questions, post them with the diagram.

Cheers, Wayne

 

[Carultch]

I'm missing why AC instantaneous values cannot also be added up directly. I do agree that it becomes magnitudes easier with vectors, or waveforms.

Instantaneous AC values do add up directly.

What I meant by "AC values" in my earlier post, is AC amplitudes, RMS values of AC, or other AC data that gives the same information. These only add up directly in the special case of in-phase waveforms.

 

[bellington]

OK, that's true.

But looking back at post #335, it appears you are trying to analyze a 3 ohm resistor across A-B along with a 3 ohm resistor across B-C. You calculate the series resistance R_T = 6 ohms, but nowhere in that arrangement do you have a pure series circuit. Since coil A connects at one end of the two resistors, coil B connects in the middle, and coil C connects at the other end, it doesn't fit the paradigm I described.

If this answer doesn't satisfy you, please draw out a circuit diagram of what you are modeling, and if you have further questions, post them with the diagram.

Cheers, Wayne

This drawing would represent the waveforms of 335, the 3 Ω resistors represent the coils. I got a rough estimate of something similar to a resistance value for the inverters. That estimate of 43 kΩ, times 2 for 2 inverters in series with each other, but parallel with any L-L pair, I get an equivalent resistance 5.9995 Ω L-L. The inverters' resistance should be considered insignificant in relation to the voltages across and currents through the coils.

Effectively, the only current flowing out of one inverter will flow through the two connected coils in series and back into the other side of the inverter. Are there other paths that I have failed to consider?

 

[wwhitney]

This drawing would represent the waveforms of 335, the 3 Ω resistors represent the coils.

Nice drawing, but it's a bit odd to depict the transformer secondary coils as 3Ω resistors. Remember that the transformer coils define all the voltage relationships in your circuit, and the inverters only match the voltage they see. So your drawing is a bit deceptive in not showing the coils as voltage sources.

The current the inverters put out is not dependent on any external resistances, it depends only on the available DC power and any AC output current limit. The 43 kΩ resistors in parallel with them are neither here nor there; I guess if your inverter leaks 5 mA when off, that's an OK model, but it's fine to ignore them. So adding them in isn't of explanatory value.

Having said all that, in your circuit diagram, you can't analyze the loop consisting Inverter CA, Coil C, and Coil A in isolation. Because there are three other connections to that loop, from Inverter AB, Coil B, and Inverter BC.

Cheers, Wayne

 

[bellington]

These two circuits show what's in my head, which is, two inverters attempting to transfer power through the same coil of the secondary will produce less power out the primary than what they would across separate coils. Click the links, and click "Reset" for some reason, and compare voltage across and current through coil B in each simulation. Connected AB BC, B coil always has less RMS voltage and current than A, or C in this simulation:

http://tinyurl.com/2y7kw4e2

Connected AN BN, we get eliminate the subtracting result of the two inverters trying to use the same coil in this simulation:

http://tinyurl.com/2acz7lft

 

[ggunn]

These two circuits show what's in my head, which is, two inverters attempting to transfer power through the same coil of the secondary will produce less power out the primary than what they would across separate coils. Click the links, and click "Reset" for some reason, and compare voltage across and current through coil B in each simulation. Connected AB BC, B coil always has less RMS voltage and current than A, or C in this simulation:

http://tinyurl.com/2y7kw4e2

Connected AN BN, we get eliminate the subtracting result of the two inverters trying to use the same coil in this simulation:

http://tinyurl.com/2acz7lft

Are you still trying to say that the power from the the three single phase inverters will somehow not all (discounting minor losses that are in any transformer) get through the transformer? If so, you are still mistaken.

Also, as we all (most recently wwhitney) have been pointing out, the inverters are current sources, not voltage sources, and the voltage at a grid tied inverter is determined by the service voltage (as changed by the transformer in this case) plus the Vd of the conductors, not by the inverters.

 

[bellington]

I connected a wattmeter on this circuit running inverter AB on coils AB, and inverter BC running on coils BC. It gives 29.9 peak watts across and through each of the 3 load resistors, for a total of 89.7 total peak watts in the system:

http://tinyurl.com/2y8mg2la

On this circuit, I ran inverter A on coil A and inverter B on coil B. Peak power across and through the two corresponding resistors was 89.8 watts, for a total of 179.6 peak watts on this circuit:

http://tinyurl.com/25c74sl3

Why do these circuits show only half the power transferred with inverters connected L-L of what is transferred L-N?

 

[wwhitney]

These two circuits show what's in my head, which is, two inverters attempting to transfer power through the same coil of the secondary will produce less power out the primary than what they would across separate coils. Click the links, and click "Reset" for some reason, and compare voltage across and current through coil B in each simulation.

A few points:

- You can use the simulations if it helps you, but it's a lot more data to process than just thinking of the associated waveform vectors. Which as we've discussed multiple times represents the same information with just 2 real parameters. So personally I find the simulation to be more an obstacle than a benefit, although maybe it's nice to see once.

- Your schematic for the (2) delta connected inverters on the wye secondary is not correct. You have the first image below, but it should be like the second image below.

- I'm not sure whether with only (2) delta connected source on a wye secondary you will get the normal voltage behavior of the wye. Try closing the delta with a 3rd source. And remember, these grid following inverters only function on an existing voltage system, so as your model doesn't have that existing voltage system, if the simulation doesn't match expectations, it's a fault of your simulation.

- Looks like you have all your voltage sources set to 300 Vpp = 212 Vrms. If you want to match the voltages between the delta connected and the wye connected inverters, you'll need to set the delta connected inverters to 208 Vrms, and the wye connected inverters to 120 Vrms.

- Without any real loads in your simulation, the currents you have are not meaningful.

Cheers, Wayne

 

[ggunn]

Why do these circuits show only half the power transferred with inverters connected L-L of what is transferred L-N?

That's how it should be. With a L-N connection all the power goes through the connection to the line but in a L-L connection half the power goes through each conductor.

 

[bellington]

A few points:

- You can use the simulations if it helps you, but it's a lot more data to process than just thinking of the associated waveform vectors. Which as we've discussed multiple times represents the same information with just 2 real parameters. So personally I find the simulation to be more an obstacle than a benefit, although maybe it's nice to see once.

- Your schematic for the (2) delta connected inverters on the wye secondary is not correct. You have the first image below, but it should be like the second image below.

- I'm not sure whether with only (2) delta connected source on a wye secondary you will get the normal voltage behavior of the wye. Try closing the delta with a 3rd source. And remember, these grid following inverters only function on an existing voltage system, so as your model doesn't have that existing voltage system, if the simulation doesn't match expectations, it's a fault of your simulation.

- Looks like you have all your voltage sources set to 300 Vpp = 212 Vrms. If you want to match the voltages between the delta connected and the wye connected inverters, you'll need to set the delta connected inverters to 208 Vrms, and the wye connected inverters to 120 Vrms.

- Without any real loads in your simulation, the currents you have are not meaningful.

Cheers, Wayne

View attachment 2570352
View attachment 2570353

Yes, post 350 is wrong. The ones in 352 should be correct.

 

[wwhitney]

I connected a wattmeter on this circuit running inverter AB on coils AB . . .

OK, you've fixed your delta connected inverter schematic, good.

But you've got the sources in your delta connected schematic set to about 173 Vpp, while in your wye connected schematic they are still 300 Vpp. For a 208Y/120V system, your delta connected inverters should be set to 208 Vrms, and your wye connected inverters should be set to 120 Vrms.

If you do that, the power on your load(s) should match. If not, it's presumably the open delta issue I mentioned (I haven't thought about it enough to know if it is an issue), so add the 3rd source in your delta model and they will match.

In the wye model, you don't need a 3rd source, or even a second source. Each leg of your wye-wye transformer functions independently--if you notice, you can draw a loop around everything on one leg, and the only connection to the rest of the circuit is through one connection at one point, the neutral point. That means you can analyze it independently.

Cheers, Wayne

 

[wwhitney]

Yes, post 350 is wrong. The ones in 352 should be correct.

OK, but so far you have only dealt with points 2 and 5 of that post. Point 1 is philosophical, so it doesn't require a response, but points 3 and 4 remain to be dealt with. As I just reiterated in response to post #352.

Cheers, Wayne

 

[bellington]

A few points:

- You can use the simulations if it helps you, but it's a lot more data to process than just thinking of the associated waveform vectors. Which as we've discussed multiple times represents the same information with just 2 real parameters. So personally I find the simulation to be more an obstacle than a benefit, although maybe it's nice to see once.

- Your schematic for the (2) delta connected inverters on the wye secondary is not correct. You have the first image below, but it should be like the second image below.

- I'm not sure whether with only (2) delta connected source on a wye secondary you will get the normal voltage behavior of the wye. Try closing the delta with a 3rd source. And remember, these grid following inverters only function on an existing voltage system, so as your model doesn't have that existing voltage system, if the simulation doesn't match expectations, it's a fault of your simulation.

The first circuit in post 352 is now connected L-L and should be a correct wye configuration.

- Looks like you have all your voltage sources set to 300 Vpp = 212 Vrms. If you want to match the voltages between the delta connected and the wye connected inverters, you'll need to set the delta connected inverters to 208 Vrms, and the wye connected inverters to 120 Vrms.

Post 352, first circuit should be a wye, 300 was a rounding of 208 peak of 294.

- Without any real loads in your simulation, the currents you have are not meaningful.

The load is on the primary where I want to see the power transferred.

Cheers, Wayne

View attachment 2570352
View attachment 2570353

 

[wwhitney]

But you've got the sources in your delta connected schematic set to about 173 Vpp

OK, that comment is in error, as I see in your delta connected schematic you don't have a voltmeter across the source, just across the wye secondary coil. And with your approximation of 300 Vpp instead of 294 Vpp for 208 Vrms, you would expect 173 Vpp. So that's all good.

Still, on your wye connected schematic, you need to set your sources to 173 Vpp to match the voltage across the secondary coils to get a fair comparison. Right now your in your wye connected schematic, you have the secondary voltage sqrt(3) times the secondary voltage of your delta connected schematic, as your voltmeters show, and so you will get 3 times the power drop across your resistive loads. Which is exactly what you reported, 30W vs 90W.

Cheers, Wayne

 

[jaggedben]

I connected a wattmeter on this circuit running inverter AB on coils AB, and inverter BC running on coils BC. It gives 29.9 peak watts across and through each of the 3 load resistors, for a total of 89.7 total peak watts in the system:

http://tinyurl.com/2y8mg2la

On this circuit, I ran inverter A on coil A and inverter B on coil B. Peak power across and through the two corresponding resistors was 89.8 watts, for a total of 179.6 peak watts on this circuit:

http://tinyurl.com/25c74sl3

Why do these circuits show only half the power transferred with inverters connected L-L of what is transferred L-N?

Ggunn alluded to why already, but I'll take a stab at it too.

Of course if you put three 1K resistors across three 122V sources you are going to pull a different amount of power than if you put two 1k resistors across two 212V sources. That's basic Ohms law. So you can't then ask why the source sends a different amount of power; you chose that amount of load. It's not a reflection of the source capability or configuration. Your simulation clearly treats the inverters like voltage sources, and the load draws power accordingly. If you were to adjust your resistors so that they draw the same power for the voltage they are exposed to, your three phase simulation will deliver the same power.

I'm not sure if your simulation can handle a current source like a grid tied inverter. But regardless, for a realistic simulation you would have to have the grid as an additional source in parallel on the primary.