PG&E transformer upgrade: why?

[wheisenburg]

What happens if you have 100A of backfeed at 240v and 200A consumed on one L1? 100A two pole OCPD

That would be 100A from Utility on L1
100A to Utility on L2
200A to Utility on N

First of all you look at the net power running on each hot leg. If you are have -100 on both L1 and L2 that results in 0 amps on the neutral. The -means the current is flowing out of the house against the voltage so you are exporting power. If you then add a +200 amp load on L1, the current flow on L1 reverses for a net of +100 amps on L1. This load is not balanced so the entire 200 amps will need to return on the neutral. This doesn't have anything to do with L1 and L2 adding together. If you put a 200 amp 120 volt single phase load on a system, it will have a 200 amp return current on the neutral. What does get added together is power that is produced locally with power that is imported. The power flow then becomes this. 12,000 watts exported on L2. 12,000 watts imported on L1. You are consuming 24,000 watts on L1. So the power produced locally on L1 never get exported. It is consumed locally and never flows out of the house. The neutral is at zero volts so it is just a return path for the current. Watts = volts * amps, so 0 * x = 0. There is no power flowing over the neutral.

So it looks like this:

Net Power Flow:

L2 = -100 *120 = -12,000 watts (exported power on L2)
L1 = (200 - 100) * 120 = +12,000 watts (imported power on L1)

Net Exported is 0 watts.

Power Production:

L1 & L2 = -100 * 240 = -24,000 watts.

Power Consumed:

L1 = +12,000 watts produced locally +12,000 watts imported = +24,000 watts.

So then we get down to the following. If you are back feeding on the supply side of your main breaker the 200 amp load will trip the breaker. If you are back feeding into a breaker in the main panel, then by code, that breaker would be limited to about 20% of the panel rating.

Now if you installed this in such a way that you allowing the system to accept unlimited locally produced power and then to also pull 200% of the panel rating out of a single leg, then yes you could overload the neutral wire. Such an installation would not be code compliant. The bus bars in the panel are generally only rated for a 20% rating over the panel rating. So you would also be overloading the one leg bus bar and the neutral bus too. By code a 200 amp panel would allow about 40 amps of back feed into a breaker installed in the panel. You can go bigger, but only by lowering the rating of the main breaker such that the combined input current would not exceed the rating of the bus bar. This is why a solar installation is required to either be installed on the supply side of the main panel or to tie into a breaker in the main panel that limits the current that is back fed into the main panel.

So if the question is can an improperly designed system overload the neutral and/or the panel bus bars without tripping the main breakers, then the answer is yes. This is why there are code provisions that limit the way multiple sources of current are allowed to be be combined together.

 

[TheHappyNomads]

Hope your experience means PG&E doesn't charge me. I'm my own electrician, so I'm cheap but slow.

What did you pay for parts and labor?
"Service lines" - is that anything on utility side of meter? PG&E's questionnaire asked about design and trenching on their side.

I've spent $100 for meter socket, $200 for 200A Square-D main breaker only panel. Some amounts previously for 200A and 125A main breaker load centers (prices have doubled in recent years). Need to get a wiring box and conduit to put in basement, connect to those boxes. Maybe 50' of 3/0 wire (Lowes price will be around $350)

I believe it was around $3500 for everything. Parts were minimal as he was able to reuse the conduit up to the pull but he ran 2 new ground rods and all new ground wire, 200a panel. It was actually the most affordable quote I got, dude did a really clean job on it too.

 

[zanydroid]

Now if you installed this in such a way that you allowing the system to accept unlimited locally produced power and then to also pull 200% of the panel rating out of a single leg, then yes you could overload the neutral wire. Such an installation would not be code compliant. The bus bars in the panel are generally only rated for a 20% rating over the panel rating. So you would also be overloading the one leg bus bar and the neutral bus too. By code a 200 amp panel would allow about 40 amps of back feed into a breaker installed in the panel. You can go bigger, but only by lowering the rating of the main breaker such that the combined input current would not exceed the rating of the bus bar. This is why a solar installation is required to either be installed on the supply side of the main panel or to tie into a breaker in the main panel that limits the current that is back fed into the main panel.

This is not true, I have a code compliant install with way more than 20%. Because I use Hawaiian Tie In / 705.12 (B) (3). This was approved twice by AHJ and POCO, for my original and for my expansion, and three entities looked at it closely (me, the designer drafting for me, and the code review firm hired by my AHJ)

1. The sum of the ampere ratings of all overcurrent devices on panelboards, both load and supply devices, excluding the rating of the overcurrent device protecting the busbar, shall not exceed the ampacity of the busbar. The rating of the overcurrent device protecting the busbar shall not exceed the rating of the busbar. Permanent warning labels shall be applied to distribution equipment displaying the following or equivalent wording:
   WARNING:
   EQUIPMENT FED BY MULTIPLE SOURCES. TOTAL
   RATING OF ALL OVERCURRENT DEVICES EXCLUDING
   MAIN SUPPLY OVERCURRENT DEVICE SHALL
   NOT EXCEED AMPACITY OF BUSBAR.
   The warning sign(s) or label(s) shall comply with 110.21(B).

Also you can do that with 120% rule with a 225A busbar panel if you derate to 100A main on 100A service as you're supposed to.
120% * 225 - 100 = 170A

That is why I picked the numbers the way I did in my original post, to correspond to the 100A service case, because I've done those calculations a lot for my two installs on my house.

As well, the neutral is overloaded already at 80% * 100A = 80A of solar. You can consume up to 180A on one leg without overloading the breaker on that leg (since it's net 100A in from utility). That has 180A neutral current.

This 180A is also well within the 225A limit of the busbar above. However, the neutral service conductor is very sad in this situation.

 

[altenergy20]

That's what YOU (and almost every body else) thinks. It is taught in school.
Question Everything.

Yes, "I am certain." So long as someone has a 120V load.

Analyze an auto-transformer. If your head doesn't explode first, you will come to the conclusion that center-tap ("Neutral") carries the sum of transformer's L1 and L2. NOT the difference, the imbalance.

You can figure this out with pencil and paper by applying Kirchhoff's Current Law or by simply reasoning on power flow. You can analyze it with SPICE. You can measure it with a clamp ammeter.

If we consumers only draw power from the grid, utility transformer is an isolation transformer. But if we backfeed power while anyone draws single-phase load, we use it as an auto-transformer (plus isolation for any power flowing to or from primary side.)

He thinks like he's thinking 3 phases. For the load side of the transformer, current on the common wire - the neutral is the the sum of two parallel circuits.

 

[Hedges]

So then we get down to the following. If you are back feeding on the supply side of your main breaker the 200 amp load will trip the breaker. If you are back feeding into a breaker in the main panel, then by code, that breaker would be limited to about 20% of the panel rating.

Now if you installed this in such a way that you allowing the system to accept unlimited locally produced power and then to also pull 200% of the panel rating out of a single leg, then yes you could overload the neutral wire. Such an installation would not be code compliant.

So if the question is can an improperly designed system overload the neutral and/or the panel bus bars without tripping the main breakers, then the answer is yes. This is why there are code provisions that limit the way multiple sources of current are allowed to be be combined together.

For a single consumer's electrical panel, I agree it has to violate code by using "200%" rather than 120% rule.

But I can do it to the utility grid overhead wires. Load them single phase 120V to 200% of rated current by having many neighbors each load their panel reasonably.

Then backfeed 100% of rated current at 240V by having many neighbors each use PV compliant to 120% rule.

He thinks like he's thinking 3 phases. For the load side of the transformer, current on the common wire - the neutral is the the sum of two parallel circuits.

For 3-phase, I'm less clear on what we could do to neutral. I think it is still 200% of L1, because L2 and L3 are vectors with 120% angles, each supply 0.5x the current in phase with L1.

 

[Shimmy]

Anybody know if utility fuses the primary or the secondary?

Varies by location and size and type of transformer. Secondaries usually unfused below ~100kVA shared pole mount and just one side of the primary for single phase. Pad mounts are usually dedicated and unfused secondaries. Underground vaults I have seen fuse everything. Three phase transformers usually fuse all three primaries.

 

[Shimmy]

But I can do it to the utility grid overhead wires. Load them single phase 120V to 200% of rated current by having many neighbors each load their panel reasonably.

For clarity, you are saying loading an unbalanced 120V (all "L1-N"), correct?

 

[Hedges]

Yes, all loads unbalanced, all sources balanced so utility transformer provides all neutral current.

 

[altenergy21]

Don't listen to your lying head.
It is easy to think the transformer should do something, and this took me a while to wrap my brain around.
Just apply Kirchhoff's Current Law. Or consider power flow and calculate amps based on volts and watts.

Seeing the schematic with your eyes makes it so much easier:

View attachment 175356

The key is that current will only flow in one winding of the transformer if there is current flowing opposite direction in other winding.
Apply a load to the neutral, then current is allowed to flow in the directions shown.

I believe your circuit is only partial representation of reality. You treated it as a an ideal case.
Reality is the B field from the primary coil "pushes" against the one on the secondary. This is significant because the setup is a transformer.

Kirchhoff's circuit laws - Wikipedia

en.wikipedia.org

 

[Hedges]

Transformer coupling is the reason current flows in opposite directions through the two coils of auto-transformer.
Without load connected to center tap "neutral", inductance of the coils would keep current near zero.

I think an ideal transformer is fine for understanding the circuit. The things that differ for a real implementation don't matter for this. They are resistive loss, hysteresis, saturation.

Do you see any error in my claim that center tap current equals sum of L1 and L2 current?
Any more details to understand its behavior?

 

[altenergy21]

Transformer coupling is the reason current flows in opposite directions through the two coils of auto-transformer.
Without load connected to center tap "neutral", inductance of the coils would keep current near zero.

I think an ideal transformer is fine for understanding the circuit. The things that differ for a real implementation don't matter for this. They are resistive loss, hysteresis, saturation.

Do you see any error in my claim that center tap current equals sum of L1 and L2 current?
Any more details to understand its behavior?

I(center)=I(L1) + I(L2) + I(B)

Without I(B), I(B) = 0, you're right.

Being "pushed" by B field from primary coil means secondary coil loses energy.

E=VIt - Energy of an electric circuit.

The energy losses equals voltage and current losses. This this tells you that you need to account for I(B). I(B) is significant because your circuit is a transformer.

 

[Hedges]

How "significant"?
I expect a couple or few percent energy loss.
I haven't checked how much imbalance there can be between I(L1) and I(L2), which are ideally equal and opposite.

Operating a power transformer at rated voltage, I find its no-load current to be several percent of full load. Operating it at 1/2 of rated voltage it dropped to about 1/8 as much, so I prefer that when powered by inverter not grid.