PV backfeed + existing load breaker sizing

[jaggedben]

Absolutely, although what "properly" means is a bit of a moving target and economics is a player. I can reduce Vd as much as you want; how much money do you want to spend on wire?

Enough to keep the inverter from tripping off line due to exceeding its proscribed voltage limits.

 

[tallgirl]

Enough to keep the inverter from tripping off line due to exceeding its proscribed voltage limits.

+1.

 

[tallgirl]

Nevertheless, electrons continue to move through a superconductor without the exertion of an outside force once they are set in motion because of the self same conservation of energy.

When you get way down into the real nitty gritty of moving carriers in a conductor (or a semiconductor, especially) a lot of counterintuitive issues come up in the math. Quantum mechanics is nothing but bizarre at the micro micro level. I took courses in QM in engineering school but I will not pretend that I understood all of it.

As long as we accept the fact that a GT inverter connected to the grid puts out what it is capable of irrespective of the local loads (i.e., behaves as a current source with its output voltage clamped by the grid), then we can design systems. Whether or not the inverter has to raise its voltage above the line voltage in order to push current into the grid is not really relevant.

Even in QM, Conservation of Energy holds -- and especially in QM. There are some interesting things to do with probability which might cause some random electron to behave microscopically in a way that seems to violate macroscopic rules (see Hawking Radiation, for example ...), but when it comes to the macroscopic world of real conductors (meaning, not the world of "first, assume a frictionless pulley ..."), absent an external force, electrons flow from higher to lower potentials. Some may flow within regions of the same potential, but I'm not seeing a lot of service laterals made from high temperature superconductors.

 

[ggunn]

Even in QM, Conservation of Energy holds -- and especially in QM. There are some interesting things to do with probability which might cause some random electron to behave microscopically in a way that seems to violate macroscopic rules (see Hawking Radiation, for example ...), but when it comes to the macroscopic world of real conductors (meaning, not the world of "first, assume a frictionless pulley ..."), absent an external force, electrons flow from higher to lower potentials. Some may flow within regions of the same potential, but I'm not seeing a lot of service laterals made from high temperature superconductors.

I'm not sure what you are trying to say, but if it is that if there were real world high temperature superconductors and one connected a GT inverter to a service with them, the inverter wouldn't deliver current to the grid because there would be no voltage rise in the conductors, then we are just going to have to agree to disagree.

 

[tallgirl]

I'm not sure what you are trying to say, but if it is that if there were real world high temperature superconductors and one connected a GT inverter to a service with them, the inverter wouldn't deliver current to the grid because there would be no voltage rise in the conductors, then we are just going to have to agree to disagree.

We can agree to disagree all you want, but Conservation of Energy really does require that sooner or later, there's a voltage drop or those electrons really just aren't going anywhere. It can be at the other end of that hypothetical superconductor, but it most certainly is =somewhere=.

A current source is just a model -- you can't, for example, connect an ideal current source to an open circuit. The output voltage would have to be infinite to overcome the infinite resistance of the open circuit and have current flow. Likewise, a voltage source is just a model and it can't be connected across a short circuit -- it would have to supply infinite current across the 0 voltage drop of the short circuit. In both of those instances, the appropriate series and parallel resistances must be included so that the ideal source behaves like a real-world source. See the Norton and Thevenin equivalent circuits for more info.

To make this relevant to PV, a PV cell is a current source, where the current is created by photons striking the junction. It's the internal equivalent series and parallel resistances which create the I-V curve.

 

[ggunn]

To make this relevant to PV, a PV cell is a current source, where the current is created by photons striking the junction. It's the internal equivalent series and parallel resistances which create the I-V curve.

Yes, I know, and likewise, to a first approximation a GT PV inverter is also a current source, as I have been saying all along. As you say, current sources and voltage sources are simply models we use as analogies to explain what we see happening, just as is the notion that electrons cannot move except in the presence of an electric field. Whether the voltage gradient in a real conductor is the cause or the result of the movement of electrons is an argument bordering on the philosophical. It doesn't make any difference in the real world; it does what it does and we know how to deal with it. But one fact remains: the lower the resistance of the conductors, the less the voltage drop, and the amount of current from the inverter does not depend on the magnitude of that voltage drop. The voltage drop could be zero and the inverter would still function just fine. Yes, I know that in the real world as we know it so far, the voltage drop isn't ever zero. That doesn't mean that it can't ever be.

On another subforum there was a protracted argument about where the energy flow is when current flows in a conductor. One position was that all of it travels outside the conductor. It did not invalidate all the math we use to calculate current densities and ampacities in conductors, it's just a different model, a different way of analogizing the observable. It is neither right nor wrong because it's just a model.

And no, I am not trying to reignite that argument.

 

[jayzbond007]

Hi All,
I appreciate the replies to this post, but I still would like a simple answer to my original post. I've read through the posts, but it's too technical and don't understand. I am just looking for a simple answer on sizing the breaker at the main panel.

Thanks,


======================================================================
Hi,


I am designing a PV array that will have (2) strings of modules, which will require 20A dpb each and consolidate into a load center. This will tie back to the main service panel (MSP). Since the MSP is full, I will move (1) 40A dpb circuits to the load center to make room and replace that breaker in the MSP with a breaker size that can accomodate both the PV and the existing 40A circuit. My question is, how do you size that? Since the PV is a backfeed and the existing 40A is a load, do I take my 40A for the PV + the existing 40A and use an 80A dpb?


Any help is greatly appreciated.


Thanks!

======================================================================

 

[shortcircuit2]

Hi All,
I appreciate the replies to this post, but I still would like a simple answer to my original post. I've read through the posts, but it's too technical and don't understand. I am just looking for a simple answer on sizing the breaker at the main panel.

Thanks,


======================================================================
Hi,


I am designing a PV array that will have (2) strings of modules, which will require 20A dpb each and consolidate into a load center. This will tie back to the main service panel (MSP). Since the MSP is full, I will move (1) 40A dpb circuits to the load center to make room and replace that breaker in the MSP with a breaker size that can accomodate both the PV and the existing 40A circuit. My question is, how do you size that? Since the PV is a backfeed and the existing 40A is a load, do I take my 40A for the PV + the existing 40A and use an 80A dpb?


Any help is greatly appreciated.


Thanks!

======================================================================

Oh sure go ahead and hijack the thread!

I'll take a shot...feed a 100 amp sub-panel with an 80amp breaker and feeder from the MSP and good to go.

 

[ggunn]

Hi All,
I appreciate the replies to this post, but I still would like a simple answer to my original post. I've read through the posts, but it's too technical and don't understand.

I apologize for my part in the hijack. Sometimes I just cannot help myself.

 

[GoldDigger]

Oh sure go ahead and hijack the thread!

I'll take a shot...feed a 100 amp sub-panel with an 80amp breaker and feeder from the MSP and good to go.

If you put in an 80 A breaker, some inspectors will require you to size the feeder wires for 120A .
My opinion is that you can use a 40A breaker and then size the feeder for 80A, just to avoid possible hassles.

 

[jaggedben]

If you put in an 80 A breaker, some inspectors will require you to size the feeder wires for 120A .
My opinion is that you can use a 40A breaker and then size the feeder for 80A, just to avoid possible hassles.

I concur, except to add that the 120% rule allows the feeder to be sized for only 67A.