Feeding solar power back into municipal grid: Issues and finger-pointing

[trader4]

Absolutely agree. And this is probably as good a place as any to post an analysis of the model Wilkins provided a while back. I bet most of those on the other side of this issue have not bothered to do any analysis of it. If you do, it shows exactly what you are saying above:

1 1 Vg 1 1 1 1 ohm ------R1-- Rw----------Rg-----------Rg--------Rw --- R2------ ! ! = ! ! ! = ! PV V1 Rload 100 ohms V2 240V array ! ! = ! ! ! = ! =20

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Here's a simple model of what we have. V1 and R1 are a Thevinin model of the PV array. R1 is the internal resistance of the PV array. Rw is the resistance of the wire connecting the array to the grid. Rg is the resistance of the grid wire connecting the load two blocks away. That load is driven by another power source in parallel, modeled by V2 and R2, it;s connecting wire Rw. I hope we can all agree that this is a valid model for the discussion at hand.

Step 1: The sun is not shining, the PV is supplying no current. What is the voltage at the grid connection point Vg? We know the only current flowing is from V2. That current is 240/(100+1+1+1) =3D 240/103=3D 2.33 amps. That current flows through the Rload resistor producing a voltage of 100 X 2.33 =3D 233 volts across the load. Since we know zero current is flowing in the left side of the circuit, the current through R1, R2, Rw must be zero. The only way for that to occur is for Vg to be equal to 233 volts. The grid voltage is now 233 volts. The voltage at V1 must also be 233 volts.

Step 2: The sun comes out and the PV is going to put it's power on the grid. The only way for current to start flowing through V1 is for V1 to INCREASE above 233 volts. As it does, the voltage on the grid will slowly increase too. How do we know this? Let's look at the simplest case, where the PV array can supply half the total current. Writing the Kirchoff equation for the right side of the loop we have:

240 =3D i2 x (R2 + Rw + Rg) + (i1 + i2) x Rload

240 =3D i2 x 3 + ( i1 + i2) x 100

240 =3D 100 x i1 + 103 x i2

Where i2 is the current flowing from source V2 and i1 is the current flowing from the PV array. We also know that each source is supplying half the current, so i1 =3D i2. Substituting, we have

240 =3D 100 x i1 + 103 x i1

240 =3D 203 i1

or i =3D 1.182 amps.

We know that twice that is flowing through the load, so the voltage across the load is 100 x 1.182 x 2 =3D 236.4 volts

Before we brought V1 online it was 233 volts. Now it is

3.4 volts HIGHER. Intuitively this makes sense, because the voltage there must RISE enough to reduce the current that was flowing from the other power source V2.

Now that we have the voltage across the load, and know that I1 is 1.182 amps we can calculate the voltages at the grid, Vg and what V1 must be:

Vg =3D 236.4 + i1 x Rg =3D 236.4 + 1.182 =3D 237.6 volts

V1 =3D 236.4 + i1 x (Rg + Rw+ R1) =3D 236.4 + 1.182 x 3 =3D 239.9 volts

All the voltages along the way have increased, the voltage at the load, the voltage at the grid, and the voltage at the PV array.

QED

[Mho]

Especially since you answered your own questions in your response.

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Well, I was just teasing him, did not really expect any answer.

[Mho]

Why would there be a local increase in the voltage?

The grid is filled with tapchangers and capacitors to adjust the voltage to a constant level, loads on a complex grid can shift load continuously to another place. A small PV source may increase the voltage up to the next house and would never be noticed.

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there is a VERY slight increase in local voltage.

If you want to push 5 kW back into the gird, the local voltage rises by the amount of voltage drop in the wires leading to the grid with 5 kW flowing through them. Its the same amount as it drops when 5 kW flows out.

For example, if the grid is 120.0 and your house is pulling 5 kW, then the local voltage at your house may drop to 119.9.

If your house pushes 5 kW into the grid the local voltage at your house may rise to 120.1.

The 5 kw is not wasted, the rest of the grid reduces its generation by that 5 kW to keep the grid at 120.0.

Another analogy is tandem bikes. If the back person pushes harder, the front person has to push less to go at the same speed. For synchronous AC motors and generators this is really a good analogy, they are all running at exactly the same speed and the PHASE slips ahead or behind slightly depending on which way the power flows. You can think of it as a bit of stretch in the bike chain one way or the other.

A lot of the engineering of power systems goes into how the load is shared among multiple sources.

But in any case, a 5 kW load or source is very small compared to the overall power flow in the grid.

Mark

[Mho]

Absolutely correct.

Two voltage sources in parallel are at the same voltage. Two exact voltage in parallel can supply the same load and split the load between them based on the impedance from source (including it's own internal impedance) to the load in the loop formed.

Yes the trolls are only trying to wreck another group. Sad from some mentally damaged types but it happens.

*not* required that the photovoltaic inverter supply a higher voltage in order to transfer current to the grid. (I take this from the last sentence in the next-to-last paragraph, where you say " ... will adapt itself to the line voltage, whatever it may be".)

The arguments against this, with all the pseudo-science being thrown around (most of it by the ones who are also slinging insults) are getting quite tiresome here.

[Mho]

Sorry, totally incorrect assumption. I guess the real world isn't for you

Load current is shared depending on impedance of parallel source loops.

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This is a real world scenario.

[g]

You apparently have spent a lot of time "checking out" people. Well, it is your time, waste it as you see fit. Better you than me.

Your tone of language though is one that tell much more about you than you understand. Maybe you and Mho can waste each others time checking each other out. For all I know you are one and the same.

Hopefully you will be so busy name-calling each other that the rest of us can forget about you. :)

[David Nebenzahl]

On 4/18/2011 4:34 AM Jim Wilkins spake thus:

OK, now we're getting somewhere.

At the risk of igniting another round of sniping here, how does that work, exactly? I assume you're talking about the voltage drop between the inverter and the point where it's tied to the external power lines (= grid), correct? So since it can only "see" its own internal voltage, how does the inverter even know what that voltage drop is? How does it regulate its voltage so that it's equal to the grid voltage at the point of connection?

Or is this somehow self-regulating, where the inverter simply "aims" at what it calculates is the grid voltage, based on the current delivered by the PV system, and the voltage self-stabilizes?

Gory details, please.

[trader4]

Take a look at the simple circuit analysis I provided earlier today that is a few posts down in this thread. I drew a model of the situation we have been discussing which is very similar to the example circuit Wilkins provided earlier. And it shows how not only the internal voltage of the PV array must rise, but so too the voltage on the grid at the point of connection. It all follows directly from Krichoff's Law.

[Bruce Richmond]

[David Nebenzahl]

On 4/18/2011 2:39 PM Bruce Richmond spake thus:

Y'know, that's the *second* time you've offered that document as a supposed answer to a question, and it doesn't contain any more relevant information to what I asked than it did the first time. It is chock-full of other interesting details, but it does *not* answer my question at all. The most they have to say is that a DSP is used to sense the line-side voltage and relay it to the intertie; however, they don't explain just how this all works in the detail I was asking for.

I invite you to point out specific sections that answer my question, as I have a copy of the PDF handy, if you think I'm mistaken.

[David Nebenzahl]

On 4/18/2011 2:20 PM [e-mail] spake thus:

The problem I have with that post is that the ASCII graphics you used are all jumbled and I can't make sense of them (trouble with line lengths, I think). Any chance you can redraw it in such a way that news clients won't scramble it? (Maybe try drawing shorter lines with "hard returns" at the end?)

[krw]

You're wrong, 'g' is correct. You can't even figure out how to use a newsreader.

Word salad.

[Mho]

The manufacturers are really the only ones that know. The rest of these here are jus guessing from logic and experience with similar equipment. IOW: it is a trade secret, mostly as to the exact details but..

The circuitry can adjust phase angle and open circuit voltage to the grid-ties point and sense the current that results. From this can tell the current magnitude and phase angle to maintain to deliver the quantity they desire. Tiny changes in phase angle of current drawn can tell them if the internal phasing to the grid phasing is getting "out-of-wack".

Voltage phasing is sensed when the grid-tie point (breaker) is open. Once in parallel (breaker closed) this is not possible as they are all the same voltage.

I am sure different manufacturers have different techniques.

Now if you are asking about the waveform synthesis you need somebody else. I know how to filter the crap out of a square wave but how the make a better sinewave (less distortion), I have only have ideas that I will not bore you with. I have repaired many of these things from 400kW 3 phase units down but never a proper sinewave unit. Maybe just better filtering?

Talk about the "cloud" now?...LOL

I invite you to point out specific sections that answer my question, as I have a copy of the PDF handy, if you think I'm mistaken.

[krw]

Make sure it's in a fixed space font.

[trader4]

Yeah, I just looked at it and it came out looking all jumbled. Here's another try:

1 1 Vg 1 1 1 1 ohm ------R1-- Rw------Rg-------Rg--------Rw --- R2------ ! ! = ! ! ! = ! PV V1 Rload 100 V2 240v array ! ! = ! ! ! = ! =20

------------------------------------------------------------------

Here's a simple model of what we have. V1 and R1 are a Thevinin model of the PV array. R1 is the internal resistance of the PV array. Rw is the resistance of the wire connecting the array to the grid. Rg is the resistance of the grid wire connecting the load two blocks away. That load is driven by another power source in parallel, modeled by V2 and R2, it;s connecting wire Rw. I hope we can all agree that this is a valid model for the discussion at hand.

Step 1: The sun is not shining, the PV is supplying no current. What is the voltage at the grid connection point Vg? We know the only current flowing is from V2. That current is 240/(100+1+1+1) =3D 240/103=3D 2.33 amps. That current flows through the Rload resistor producing a voltage of 100 X 2.33 =3D 233 volts across the load. Since we know zero current is flowing in the left side of the circuit, the current through R1, R2, Rw must be zero. The only way for that to occur is for Vg to be equal to 233 volts. The grid voltage is now 233 volts. The voltage at V1 must also be 233 volts.

Step 2: The sun comes out and the PV is going to put it's power on the grid. The only way for current to start flowing through V1 is for V1 to INCREASE above 233 volts. As it does, the voltage on the grid will slowly increase too. How do we know this? Let's look at the simplest case, where the PV array can supply half the total current. Writing the Kirchoff equation for the right side of the loop we have:

240 =3D i2 x (R2 + Rw + Rg) + (i1 + i2) x Rload

240 =3D i2 x 3 + ( i1 + i2) x 100

240 =3D 100 x i1 + 103 x i2

Where i2 is the current flowing from source V2 and i1 is the current flowing from the PV array. We also know that each source is supplying half the current, so i1 =3D i2. Substituting, we have

240 =3D 100 x i1 + 103 x i1

240 =3D 203 i1

or i =3D 1.182 amps.

We know that twice that is flowing through the load, so the voltage across the load is 100 x 1.182 x 2 =3D 236.4 volts

Before we brought V1 online it was 233 volts. Now it is

3.4 volts HIGHER. Intuitively this makes sense, because the voltage there must RISE enough to reduce the current that was flowing from the other power source V2.

Now that we have the voltage across the load, and know that I1 is 1.182 amps we can calculate the voltages at the grid, Vg and what V1 must be:

Vg =3D 236.4 + i1 x Rg =3D 236.4 + 1.182 =3D 237.6 volts

V1 =3D 236.4 + i1 x (Rg + Rw+ R1) =3D 236.4 + 1.182 x 3 =3D 239.9 volts

All the voltages along the way have increased, the voltage at the load, the voltage at the grid, and the voltage at the PV array.

QED

[trader4]

Another try a straightening out the drawing....

=A0 =A0 =A0 =A0 =A01 1 =A0 =A0 =A0 =A0 =A0 1 =A0ohm

=A0 =A0 =A0 =A0 =A0 ! =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 = =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0!

=A0 =A0 =A0 =A0 =A0 ! =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 = =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0!

=A0 =A0 Rload 100 =A0 =A0 =A0 =A0 =A0 =A0 =A0 240v=A0 =A0 V2

=A0 =A0 ! =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 = =A0 =A0 =A0 =A0 =A0!

=A0 =A0 =A0 =A0 =A0 ! =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 = =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0!

[trader4]

ote:

=A01 =A0 =A0 =A0 =A0 =A01 =A0 =A0 =A0 =A0 =A0 1 =A0 =A0 =A0 =A0 =A0 1 =A0oh= m

=A0 =A0 =A0 =A0 =A0 =A0 ! =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 = =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0!

=A0 =A0 =A0 =A0 =A0 =A0 ! =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 = =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0!

=A0 =A0 Rload 100 =A0 =A0 =A0 =A0 =A0 =A0 =A0 240v=A0 =A0 V2

=A0 =A0 ! =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 = =A0 =A0 =A0 =A0 =A0!

=A0 =A0 =A0 =A0 =A0 =A0 ! =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 = =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0!

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OK, got the drawing straightened out and it plus the analysis are above. Just to add some clarification, I said the only way for the PV array to start delivering power is for it to raise it's voltage, which in turn raises the grid voltage. That assumes that both the load and the other power source remain constant. As I stated in other posts, the other ways for the PV array to deliver power without it's voltage going up would be for the load to increase, ie Rload gets smaller, or for the other power source V2 to decrease in voltage.

[Mho]

Adding another source lowers the net impedance of the supply to the load.

This is simple network theorem. Even ohm's law can tell you the voltage requirements.

The voltage at the grid connection (assuming where Rg is) will not be the same as the V2 source.

In DC theorem what you are saying would be basically all true but in AC we have waveform phase angle and waveform distortion.

As an extreme example: consider a PV co-gen that is 180 degrees out of phase from the grid. Now we can have a 10 volt PV source hooked to a 240V grid and still supply current from it.

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OK, got the drawing straightened out and it plus the analysis are above. Just to add some clarification, I said the only way for the PV array to start delivering power is for it to raise it's voltage, which in turn raises the grid voltage. That assumes that both the load and the other power source remain constant. As I stated in other posts, the other ways for the PV array to deliver power without it's voltage going up would be for the load to increase, ie Rload gets smaller, or for the other power source V2 to decrease in voltage.

[trader4]

Well, duh! I think everyone here, on both sides of the discussion acknowledge that.

Not just basically, it is ALL exactly true with the equations to back it up. As for the complications of AC, there wouldn't appear to be much point in discussing that until there is agreement on what happens with a simple DC distribution system voltage example.

[Mho]

I guess I went too fast for you with the AC stuff.

The whole point is regarding AC connections and different rules apply. The DC basics are mostly valid no matter how much you want to disagree with something but still moot and established by the discussing people about 100 posts ago. Your ASCII schematic was nice.

I will let you disagree with yourself a little more for the next while.

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Well, duh! I think everyone here, on both sides of the discussion acknowledge that.

Not just basically, it is ALL exactly true with the equations to back it up. As for the complications of AC, there wouldn't appear to be much point in discussing that until there is agreement on what happens with a simple DC distribution system voltage example.