PV backfeed + existing load breaker sizing

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GoldDigger

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I was wondering if anyone was going to comment on that post :huh:

No wrong assumptions on my part. PV L2 is zero amperes and degrees to the utility. What you marked in red is via POCO transformer... but it is same current that originated from PV L1. It then goes through the load then back to PV L1 via the neutral. There is no additive neutral current.
Try listing your phase angles again please. The return current on the neutral as sourced by PV on L1 will be in the same direction as the return current on the neutral from the load on L2. So they do add. All the way back to the POCO transformer, but not into the primary side of the POCO transformer. If you really have a center tapped winding on one core, there will be no current on the primary side. It is true that the power flows from PV to load, but that does not mean that the current does the same thing!.
(The normal current you show at zero degrees on L1 is of zero amplitude, so I assume that you just drew it for reference.)

If you have 30A going into the transformer secondary at L1 and 30A going into the transformer secondary at L2, you cannot just get rid of them both by showing them going to the primary. :)
voltage (phase) polarities and then in addition say that there is another 180 phase shift because of the direction of the arrow.

The load current at L2 is at 180 degrees to the voltage at L2, not to the voltage at L1.

At the point in the cycle where L1 is positive with respect to the neutral, current will flow toward the transformer through L1 and will flow toward the transformer through L2 (which is negative at that point in time) from the load. That leaves double the current to come back through the neutral.
How I wish I could draw animated pictures. <sigh>

If you look at the two halves of the center tapped secondary as two separate windings on an autotransformer (or an isolation transformer for clarity), you will see that the current will be flowing in the same direction relative to the midpoint in both windings. That allows the magnetic fields of the two windings to cancel in the core (not counting magnetization current for now.)

PS: I can see how you try to justify your error, and I sympathize, but you are still wrong.
 
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ggunn

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ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
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Electrical Engineer - Photovoltaic Systems
Okay, as I said, partly semantics. In actual electrical distribution, where we don't have superconductors, the inverter causes a voltage rise at its own terminals. Better?

In a non-superconductor, if the voltage isn't higher at the source than at the load, then power isn't flowing along that circuit. True?



I didn't refer to the impedance of the loads themselves, but rather the "impedance between [loads] and the inverter", by which I actually meant the impedance between the inverter and the parallel connections to loads. I think that in the real world (no superconductors), more impedance between the inverter and a parallel connection also logically entails that the parallel connection is physically farther away along the wire path. But if you prefer to think that that's just a coincidence given the existence of superconductors, I won't argue the point further.

To go back to your statement that "An inverter does not power loads closest to it preferentially;" I still think this is absolutely demonstrably false. Okay, it's not a matter of the inverter's 'preference'; but as a matter of physics, an inverter does power the loads that are connected in parallel closest to it along the wire path, in succession.

I could prove this by putting a series of directional watt meters in between an inverter, a bunch of parallel connections to loads, and the utility. If the inverter is producing power, then you will see power flow from the inverter towards the utility in the meters that are nearest the inverter along the wire path. You will never see a meter showing power coming from the utility if that meter is closer to the inverter than a meter that shows power coming from the inverter. Or vice versa.

This would be true even with superconductors.



Thanks for the lesson, but we're not having this discussion because these concepts are new to me. :cool:

One more thing: I wouldn't describe it as "massively parallel." For any inverter, there will only be one parallel connection in the system that is powered by both inverter and the next source out. That's very 'locally parallel', IMO.

Trying again... Grid.jpg

Your inverter is part of the massively parallel circuit which includes the grid generator and all the loads on the grid. The current through your meter is determined by Kirchoff's Law which states that the sum of all currents entering and leaving a node must be zero. If all your loads are off, then Im = Ii and you are supplying power to the grid. If IL exceeds Ii, then Im is negative and you are consuming power from the grid. That's the "net" in net metering.

Your inverter is a current source with its voltage clamped at Vg, and Ii does not depend on the size of your loads. Your inverter does not preferentially feed your loads; it feeds the whole circuit. Note that this circuit functions just fine with ideal conductors (R=0) and Vg as a constant all along the top rail.
 
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BillK-AZ

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Location
Mesa Arizona
That's the "net" in net metering.

No, net metering is generally an energy term, not power as in this discussion. Generally refers to utility billing practice of crediting customer backfeed to the grid to offset future use.
 

jaggedben

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...It's because of where the loads are topologically in the circuit and where you put the meter.

Ahem. Right. That's what I've been saying all along. The inverter powers the loads that are closest to it. That is, the closest parallel connections along the conductor path. The closest topologically, if you prefer.

I may have been off base with bringing in impedance as having a causal relation to the direction of power flow. (I'd say now that voltage drop is a phenomena which happens to exactly parallel power flow in resistive conductors. Caused by the same thing, with parallel results, but not caused by each other.) So I learned something about that in this discussion, which is good, and my thanks to you and others for helping me better grasp that. But let's not confuse anyone else by suggesting that an inverter doesn't power the loads that have the closest connections to it. Ordinary people will not be able to make sense of net-metering in any other way. Besides which, it's just true, under any sensible definitions of power and energy.

Your inverter is a current source with its voltage clamped at Vg and Ii does not depend on the size of your loads. Your inverter does not preferentially feed your loads; it feeds the whole circuit.

You are engaging in a strawman here. I never suggested that the size or ownership of the loads has anything to do with anything, or that the inverter has a 'preference' for where its power goes. In fact I've disclaimed all of that already over my last few posts.

As for whether the inverter "feeds" the whole circuit, that's a nifty semantic construction that avoids the real question of where the inverter's power goes in a given moment. Sure, it 'feeds the whole circuit'. But its power still goes through the closest connections on that circuit first. Or to put it simply to an net-metering customer, it powers all the building loads before it sends power to the grid.
 
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ggunn

PE (Electrical), NABCEP certified
Location
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Ahem. Right. That's what I've been saying all along. The inverter powers the loads that are closest to it. That is, the closest parallel connections along the conductor path. The closest topologically, if you prefer.

I may have been off base with bringing in impedance as having a causal relation to the direction of power flow. (I'd say now that voltage drop is a phenomena which happens to exactly parallel power flow in resistive conductors. Caused by the same thing, with parallel results, but not caused by each other.) So I learned something about that in this discussion, which is good, and my thanks to you and others for helping me better grasp that. But let's not confuse anyone else by suggesting that an inverter doesn't power the loads that have the closest connections to it. Ordinary people will not be able to make sense of net-metering in any other way. Besides which, it's just true, under any sensible definitions of power and energy.



You are engaging in a strawman here. I never suggested that the size or ownership of the loads has anything to do with anything, or that the inverter has a 'preference' for where its power goes. In fact I've disclaimed all of that already over my last few posts.

As for whether the inverter "feeds" the whole circuit, that's a nifty semantic construction that avoids the real question of where the inverter's power goes in a given moment. Sure, it 'feeds the whole circuit'. But its power still goes through the closest connections on that circuit first. Or to put it simply to an net-metering customer, it powers all the building loads before it sends power to the grid.

Look at it however it makes sense to you, but if all the loads in your house are off, your inverter feeds the loads on the grid. In a parallel circuit where the specific electrons come from which power a specific load is irrelevant. They are dumped on the "positive" bus and consumed by the loads. Your inverter is just one source among many.

But whatever; the circuit I showed was accurate though idealized. Where the current goes is determined by Kirchoff's Laws; the rest is semantics.

Peace,
Gordon
 

ggunn

PE (Electrical), NABCEP certified
Location
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Look at it however it makes sense to you, but if all the loads in your house are off, your inverter feeds the loads on the grid. In a parallel circuit where the specific electrons come from which power a specific load is irrelevant. They are dumped on the "positive" bus and consumed by the loads. Your inverter is just one source among many.

But whatever; the circuit I showed was accurate though idealized. Where the current goes is determined by Kirchoff's Laws; the rest is semantics.

Peace,
Gordon

And just one more note - I can agree with what you said for some meanings of the word "closer". You could have a PV inverter wired into a panel next to it that has a load 500' away. When the grid is up your inverter is connected to your next door neighbor's loads as well, but that pump or whatever which is 500' away is topologically "closer" to your inverter than your neighbor's loads, even though it is physically farther away. It's not a matter of distance but where the nodes and the meter fall in the circuit.
 

tallgirl

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True, but the power would flow even if the differential were not there.

Sorry, I've bitten my tongue completely in half not responding to this one. Unless one of my cats clicked "Reply" while I wasn't looking ...

Electrons only flow from higher potentials to lower potentials. The inverter might try leading the grid, but unless there is a negative voltage gradient, the inverter is NOT going to supply any current to anything. Electrons don't "just flow" because there are lots of them waiting to explore the rest of the electric grid.

Now to see a doctor and have my tongue sewn back together.
 

jaggedben

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In a parallel circuit where the specific electrons come from which power a specific load is irrelevant. They are dumped on the "positive" bus and consumed by the loads. Your inverter is just one source among many.

I think we all know, especially since this is AC power, that the electrons can't be traced back to a particular source.

But whatever; the circuit I showed was accurate though idealized. Where the current goes is determined by Kirchoff's Laws; the rest is semantics.

I actually never talked about current. I talked about power (and energy). I don't think that its merely semantic to talk about which source powers which loads. If it was merely semantic then net-metering wouldn't work.

And just one more note - I can agree with what you said for some meanings of the word "closer". You could have a PV inverter wired into a panel next to it that has a load 500' away. When the grid is up your inverter is connected to your next door neighbor's loads as well, but that pump or whatever which is 500' away is topologically "closer" to your inverter than your neighbor's loads, even though it is physically farther away. It's not a matter of distance but where the nodes and the meter fall in the circuit.

I think I was pretty clear in my last few posts that 'closer' referred to the location of the parallel connection to the load. I'm fine with "topologically closer" as shorthand for that. I totally agree that the distance (or impedance) from the parallel connection point to the load is irrelevant.

Electrons only flow from higher potentials to lower potentials. The inverter might try leading the grid, but unless there is a negative voltage gradient, the inverter is NOT going to supply any current to anything. Electrons don't "just flow" because there are lots of them waiting to explore the rest of the electric grid.

That's what I thought I had learned elsewhere, and then ggunn practically convinced me it wasn't true. I guess that inside my brain the jury is still out.
 

Smart $

Esteemed Member
Location
Ohio
...

PS: I can see how you try to justify your error, and I sympathize, but you are still wrong.
After re-evaluating, I was wrong. I depicted current through L2-part winding backwards. It is actually 30A@0? because it is in the same direction as normal L1-part winding, i.e. normal L2 to N is same as normal N to L1.
 

GoldDigger

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I actually never talked about current. I talked about power (and energy). I don't think that its merely semantic to talk about which source powers which loads. If it was merely semantic then net-metering wouldn't work.
Although I agree with your conclusions as well as your power meter based analysis, I still say that net metering does not care directly about which source powers which loads. It cares only about the NET power flow through the meter. It could in principle (although indistinguishable from the other situation in practice) be the case that the grid is powering the local loads and the GT is powering (some of) the grid loads. As long as the meter measures and calculates net power flow (one single number with an algebraic sign to indicate direction) it does not matter whether it is seeing (+3) + (-1) or just +2.

"It is not just a question of semantics, it is also about the words we use."
--- S. Colbert

From the circuit analysis and the meter readings you can conclusively say which direction the net power flow is at any point, but you cannot say as comfortably what power is going where.

One of the useful principles of Electricity and Magnetism is the formal principle of linearity. That says that if two charges each independently produce a particular electric or magnetic field, then the actual observed field value will be the sum of the two. (Not that this does NOT imply that all impedances are linear, just that in the absence of non-linear devices the resulting fields add linearly.)
 

tallgirl

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That's what I thought I had learned elsewhere, and then ggunn practically convinced me it wasn't true. I guess that inside my brain the jury is still out.

Don't worry -- there are some laws of physics which say that electrons don't flow from lower to higher potentials for no particularly good reason. Even in a superconductor, sooner or later there has to be a voltage drop to keep them moving because there are some other laws of physics which say so.
 

tallgirl

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Great White North
Occupation
Controls Systems firmware engineer
Although I agree with your conclusions as well as your power meter based analysis, I still say that net metering does not care directly about which source powers which loads. It cares only about the NET power flow through the meter. It could in principle (although indistinguishable from the other situation in practice) be the case that the grid is powering the local loads and the GT is powering (some of) the grid loads. As long as the meter measures and calculates net power flow (one single number with an algebraic sign to indicate direction) it does not matter whether it is seeing (+3) + (-1) or just +2.

Unless the inverter has the same current and voltage phase as the load, the actual answer is that both the grid and the inverter are powering the load.

This is why utilities are upset about PV -- they sell us real power, we sell them real power, but somewhere along the line, someone better be providing VARs because there are non-unity power factor loads out there which are consuming VARs that the PoCO doesn't charge us for.
 

jaggedben

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"It is not just a question of semantics, it is also about the words we use."
--- S. Colbert

Yup.

Again, I'm going to go back to the ordinary net-metering customer, many of whom do not have a head for the difference between a series and parallel circuit. Much simpler to explain that the power from the inverter goes to the loads inside their house first. It matters how this is explained to them so they get less confused instead of more confused.

Beyond that, when I look at a dictionary definition of 'power', I just can't really imagine how 'net power flow' is a real thing different from 'power flow', if we're talking about real power. I'd make that analogy that the speed of an object is not a mathematical result of calculating its 'net forward speed minus its net backward speed'. It is only going in one direction. It's an Occam's Razor sort of thing here; the simple explanation is the correct one.

Tallgirl makes a good point about power factor, but even in that case, I would argue that the 'net' calculation that would result in a reading of zero between two nodes is apparent power minus reactive power. Real power flow is still actually zero in both directions, and a proper meter reads that.
 

ggunn

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Sorry, I've bitten my tongue completely in half not responding to this one. Unless one of my cats clicked "Reply" while I wasn't looking ...

Electrons only flow from higher potentials to lower potentials. The inverter might try leading the grid, but unless there is a negative voltage gradient, the inverter is NOT going to supply any current to anything. Electrons don't "just flow" because there are lots of them waiting to explore the rest of the electric grid.

Now to see a doctor and have my tongue sewn back together.
Sorry about your tongue, but an inverter would still supply current to the grid if the resistance in the conductors were zero ohms. In the real world, of course, we don't yet have superconducting cables at reasonable temperatures, and there is a voltage gradient from one end of the conductor to the other, but it is a chicken/egg problem; does the gradient exist because there is current flowing, or does current flow because the gradient exists? The answer is "yes".

This is second semester EE stuff. Admittedly, it's been a long time since then for me, but ideal current sources do not depend on a voltage gradient to supply current to a circuit. No real world voltage or current source is ideally one or the other, but GT PV inverters are current sources to a first approximation.

Granted, the difference may be semantic or so deep in theory as to have no significance in the real world. We know how GT inverters behave well enough to design them into systems. We know that they are going to source the power that they are given by the PV modules irrespective of the local loads, and if the power has nowhere to go, they will shut down. The rest is just arguing about the number of angels that can dance on the head of a pin.
 

GoldDigger

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Don't worry -- there are some laws of physics which say that electrons don't flow from lower to higher potentials for no particularly good reason. Even in a superconductor, sooner or later there has to be a voltage drop to keep them moving because there are some other laws of physics which say so.

Actually, that is just completely wrong. In a superconductor there is literally zero resistance, so once the electrons have been accelerated (by an electric field resulting from a changing magnetic field, for example), they will continue to remain in motion until acted upon by some other force. Period.

Electrons do not start moving (accelerate) in any particular direction for no particularly good reason. But once moving, they will happily keep on moving in that same direction for a very good reason, without any external forces applied (as by an electric field and a resulting potential difference.)
 

tallgirl

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Actually, that is just completely wrong. In a superconductor there is literally zero resistance, so once the electrons have been accelerated (by an electric field resulting from a changing magnetic field, for example), they will continue to remain in motion until acted upon by some other force. Period.

Electrons do not start moving (accelerate) in any particular direction for no particularly good reason. But once moving, they will happily keep on moving in that same direction for a very good reason, without any external forces applied (as by an electric field and a resulting potential difference.)

Pay careful attention to what you wrote -- "without any external forces applied". Because a change in voltage potential ("higher voltage") is an external force. Changing potential from a lower to higher value without an external force violates conservation of energy. Which isn't just a good idea, it's the law.

An amp is a coulomb per second (6.24 x 10^18 electrons). A change of one volt would make that a watt-second (Joule), and the total number of Joule's in that system is a constant. Meaning, it doesn't happen unless the energy comes from somewhere else (like, the conductors suck heat from the environment, instead of giving it off as resistance normally causes them to do ...)
 

ggunn

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Electrical Engineer - Photovoltaic Systems
Pay careful attention to what you wrote -- "without any external forces applied". Because a change in voltage potential ("higher voltage") is an external force. Changing potential from a lower to higher value without an external force violates conservation of energy. Which isn't just a good idea, it's the law.

An amp is a coulomb per second (6.24 x 10^18 electrons). A change of one volt would make that a watt-second (Joule), and the total number of Joule's in that system is a constant. Meaning, it doesn't happen unless the energy comes from somewhere else (like, the conductors suck heat from the environment, instead of giving it off as resistance normally causes them to do ...)
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.
 

jaggedben

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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.

Well, actually, in order to design systems we need to know about voltage rise/drop in order to properly size inverter output conductors. But you know that.
 

ggunn

PE (Electrical), NABCEP certified
Location
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Electrical Engineer - Photovoltaic Systems
Well, actually, in order to design systems we need to know about voltage rise/drop in order to properly size inverter output conductors. But you know that.
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?
 
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