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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. --------------------

wrote in message

wrote:

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.

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.

http://www.solarpanelsplus.com/solar-inverters/How-Solar-Inverters-Work-With-Solar-Panels.pdf

http://www.solarpanelsplus.com/solar-inverters/How-Solar-Inverters-Work-With -Solar-Panels.pdf

David,

my suggestion to you is to google the term "current source". Most sources you are familiar with like batteries and generators are more like voltage sources. A current source is in a way the opposite concept and you need to think about it for a while.) A grid tie inverter emulates a current source. It puts out at its terminals (within limits) whatever voltage is needed to cause the desired current to flow.

They all follow Ohms law I=E/R. With a voltage source, V is fixed and I varies with R. With a current source, I is fixed and V varies with R.

In a normal grid tie situation, the amount the V has to vary is very small probably 1 or 2 volts at the most. That is the detail of how it controls the current. The voltage will rise or fall as required such that the desired current flows.

You can Google the design of current sources for more detail.

Mark

Google "Norton and Thevinin Equivalency" or "Norton Thevinin Circuits".

You'll see that a resistive circuit can be modeled as either a current source or a voltage source together with a resistor and that they are interchangeable and equivalent in how they behave.

On pg 76 it mentions a DSP

http://en.wikipedia.org/wiki/Digital_signal_processor

It can take DC voltage and build a sin wave using PWM (Pulse Width Modulation)

http://en.wikipedia.org/wiki/Pulse-width_modulation

As shown on pg 74 that is smoothed and fed into a transformer. A feedback circuit measures the current flow and adjusts the voltage of the constructed wave as needed. Not sure what you want here. If you want the exact details about how each part works google is your friend. 30 years ago I was working with drive panels for large industrial DC motors that did all this to feed current back into the lines when doing regenerative braking. When they want to slow down they don't worry about sharing a load. They know there are loads out there and they just pump the power into the line. At that time it was all controlled using analog circuits.

Picture two water tanks connected with a loop of pipe underneath. Put a ball in the connecting pipe. The two tanks are connected in parallel. The ball in the pipe doesn't go anywhere because it has equal pressure on both sides. Now pour some water into one of the tanks to raise its level (voltage). There will now be more pressure applied to the ball by that tank and the water will flow into the other tank, pushing the ball in that direction.

Attach a single drain pipe to the middle of the connector pipe above. If the tanks start at the same level and the resistance to flow in the pipes is the same then water will flow out of both pipes at the same rate. If one tank is higher than the other it is pushing harder and more water will flow out of it than the lower tank.

The power company has many stations and wants them to all contribute their share. We OTOH have spent big $$$ on our PV system and if it produces 5KW we want that whole 5KW to go to some load so that we get paid for it. We are not interested in playing nice and sharing a load if it means we don't get to contribute the full 5KW. So we raise the voltage just enough to flow the current we need to. That will result in less current flowing from the grid because the load is only going to accept so much flow and the PV system is taking more than its share.

Yes there will be flow at 119, 120 or 121, but from what source? Like I said, we aren't interested in playing nice. We are entitled to pump the power we produced into the line and get paid for it. If the power company has to reduce their production a bit to keep the voltage from climbing too high so be it.

But... but... but... EVERY car insurance company is cheaper than its competitors and every alkaline cell is the longest lasting... LOL

Seriously, the power stations are not higher than each other, but every one is higher than the grid, imagine three power stations, or 3 generators in the same station all putting out exactly 48KV, the substation/transformer they are connected to is getting slightly less, say 47,990 volts. If they then need to bring a fourth generator online they use a syncroscope to adjust the prime mover to sync up the frequency and phase, adjust the field to match the voltage IE 47,990, close the breakers, and slowly increase the field and prime mover so that the load is shared and now all 4 are at 48KV.

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 ! ! ! ! ! ! ---------------------------------------------------------------------------

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) = 240/103= 2.33 amps. That current flows through the Rload resistor producing a voltage of 100 X 2.33 = 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 = i2 x (R2 + Rw + Rg) + (i1 + i2) x Rload

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

240 = 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 = i2. Substituting, we have

240 = 100 x i1 + 103 x i1

240 = 203 i1

or i = 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 = 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 = 236.4 + i1 x Rg = 236.4 + 1.182 = 237.6 volts

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

#### Site Timeline

- posted on April 19, 2011, 2:13 am

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. --------------------

wrote in message

wrote:

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.

- posted on April 19, 2011, 2:18 am

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.

- posted on April 19, 2011, 3:00 am

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. ----------------

wrote in message wrote:

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.

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. ----------------

wrote in message wrote:

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.

- posted on April 19, 2011, 7:00 pm

On 4/18/2011 7:02 PM snipped-for-privacy@optonline.net spake thus:

Sorry, it's still a hopeless hash.

I really do want to follow your example, but I can't until I can see your circuit diagram properly. This ASCII-art thing clearly isn't working; any chance you can post a picture somewhere? Then I'll be able to follow along.

Sorry, it's still a hopeless hash.

I really do want to follow your example, but I can't until I can see your circuit diagram properly. This ASCII-art thing clearly isn't working; any chance you can post a picture somewhere? Then I'll be able to follow along.

--

The current state of literacy in our advanced civilization:

yo

wassup

The current state of literacy in our advanced civilization:

yo

wassup

Click to see the full signature.

- posted on April 19, 2011, 7:33 pm

You need to change your
"fixed spacing" font to one that is "fixed spacing"

I can even read it perfectly in your reply.

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

"David Nebenzahl" wrote in message wrote:

Sorry, it's still a hopeless hash.

I really do want to follow your example, but I can't until I can see your circuit diagram properly. This ASCII-art thing clearly isn't working; any chance you can post a picture somewhere? Then I'll be able to follow along.

I can even read it perfectly in your reply.

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

"David Nebenzahl" wrote in message wrote:

Sorry, it's still a hopeless hash.

I really do want to follow your example, but I can't until I can see your circuit diagram properly. This ASCII-art thing clearly isn't working; any chance you can post a picture somewhere? Then I'll be able to follow along.

--

The current state of literacy in our advanced civilization:

yo

wassup

The current state of literacy in our advanced civilization:

yo

wassup

Click to see the full signature.

- posted on April 19, 2011, 8:12 pm

On 4/19/2011 12:33 PM Mho spake thus:

Look, my display font IS monospaced. Like, duh.

It's still all garbled. Line wrapping and all that. Your news client (Microsoft Windows Live Mail 15.4.3508.1109) probably behaves differently from mine (Thunderbird).

Look, my display font IS monospaced. Like, duh.

It's still all garbled. Line wrapping and all that. Your news client (Microsoft Windows Live Mail 15.4.3508.1109) probably behaves differently from mine (Thunderbird).

--

The current state of literacy in our advanced civilization:

yo

wassup

The current state of literacy in our advanced civilization:

yo

wassup

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- posted on April 19, 2011, 11:31 pm

It shows up perfect when I look at it in David's replies too.
How about I just describe how to draw it on a piece of
paper which should be easy. It's very similar to that
example circuit Wilkins provided a while back

Left side is the PV array voltage source V1, which we will vary.

Right hand side is fixed voltage source V2= 240volts, representing the power company.

Now go across the top part of the circuit between V1 and V2 left to rightand draw a string of resistors connected in series that connects the two.

In order.....

R1 is the internal resistance of the PV array Rw is the resistance of the wire from the PV at the house to the grid Rg is the resistance of the grid as it goes down the block to the load Rg again is the resistance of the next piece of grid as it goes another block Rw is the resistance of the wire from the grid to the power company source R2 is the internal resistance of the power company source

For simplicity all these are 1 ohm each and per above connected look like this:

----R1-----Rw----Rg-----------Rg----Rw---R2----

Label the point between Rw and Rg on the left Vg. This is the connection point measurement voltage at the grid, where the PV array wire from the house attaches to the grid.

Connect the bottom of V1 to the bottom of V2

Add a load, Rload=100 ohms, one end connected between the two Rg resistors at the top and the other end connected to the bottom path between V1 and V2 that you drew in previous step

Here it is with the vertical connections left out:

--- R1 -- Rw ---Rg------Rg----Rw----R2----

V1 Rload=100 V2=240v

________________________________________

Remember Vg is the voltage between Rw and Rg on the left.

Now you have amodel of two power sources driving a load. Follow the math I posted previously.

Left side is the PV array voltage source V1, which we will vary.

Right hand side is fixed voltage source V2= 240volts, representing the power company.

Now go across the top part of the circuit between V1 and V2 left to rightand draw a string of resistors connected in series that connects the two.

In order.....

R1 is the internal resistance of the PV array Rw is the resistance of the wire from the PV at the house to the grid Rg is the resistance of the grid as it goes down the block to the load Rg again is the resistance of the next piece of grid as it goes another block Rw is the resistance of the wire from the grid to the power company source R2 is the internal resistance of the power company source

For simplicity all these are 1 ohm each and per above connected look like this:

----R1-----Rw----Rg-----------Rg----Rw---R2----

Label the point between Rw and Rg on the left Vg. This is the connection point measurement voltage at the grid, where the PV array wire from the house attaches to the grid.

Connect the bottom of V1 to the bottom of V2

Add a load, Rload=100 ohms, one end connected between the two Rg resistors at the top and the other end connected to the bottom path between V1 and V2 that you drew in previous step

Here it is with the vertical connections left out:

--- R1 -- Rw ---Rg------Rg----Rw----R2----

V1 Rload=100 V2=240v

________________________________________

Remember Vg is the voltage between Rw and Rg on the left.

Now you have amodel of two power sources driving a load. Follow the math I posted previously.

- posted on April 18, 2011, 9:39 pm

http://www.solarpanelsplus.com/solar-inverters/How-Solar-Inverters-Work-With-Solar-Panels.pdf

- posted on April 18, 2011, 10:56 pm

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

http://www.solarpanelsplus.com/solar-inverters/How-Solar-Inverters-Work-With-Solar-Panels.pdf

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.

http://www.solarpanelsplus.com/solar-inverters/How-Solar-Inverters-Work-With-Solar-Panels.pdf

Y'know, that's the

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.

--

The current state of literacy in our advanced civilization:

yo

wassup

The current state of literacy in our advanced civilization:

yo

wassup

Click to see the full signature.

- posted on April 18, 2011, 11:11 pm

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

------------------------ "David Nebenzahl" wrote in message

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.

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

------------------------ "David Nebenzahl" wrote in message

Y'know, that's the

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.

- posted on April 19, 2011, 3:07 am

http://www.solarpanelsplus.com/solar-inverters/How-Solar-Inverters-Work-With -Solar-Panels.pdf

David,

my suggestion to you is to google the term "current source". Most sources you are familiar with like batteries and generators are more like voltage sources. A current source is in a way the opposite concept and you need to think about it for a while.) A grid tie inverter emulates a current source. It puts out at its terminals (within limits) whatever voltage is needed to cause the desired current to flow.

They all follow Ohms law I=E/R. With a voltage source, V is fixed and I varies with R. With a current source, I is fixed and V varies with R.

In a normal grid tie situation, the amount the V has to vary is very small probably 1 or 2 volts at the most. That is the detail of how it controls the current. The voltage will rise or fall as required such that the desired current flows.

You can Google the design of current sources for more detail.

Mark

- posted on April 19, 2011, 11:56 am

Google "Norton and Thevinin Equivalency" or "Norton Thevinin Circuits".

You'll see that a resistive circuit can be modeled as either a current source or a voltage source together with a resistor and that they are interchangeable and equivalent in how they behave.

- posted on April 19, 2011, 10:49 pm

On Tue, 19 Apr 2011 04:56:03 -0700 (PDT), " snipped-for-privacy@optonline.net"

Bingo! It doesn't matter which you model the inverter as, the physics doesn't change.

Bingo! It doesn't matter which you model the inverter as, the physics doesn't change.

- posted on April 19, 2011, 11:36 pm

On Apr 19, 6:49 pm, " snipped-for-privacy@att.bizzzzzzzzzzzz"

Yes, we'e on the same page. They are interchangeable, but I think Mark had a good idea in bringing up the idea of the current source, as conceptually it's a bit easier to understand. It appears to have got Smitty at least partly convinced.

Yes, we'e on the same page. They are interchangeable, but I think Mark had a good idea in bringing up the idea of the current source, as conceptually it's a bit easier to understand. It appears to have got Smitty at least partly convinced.

- posted on April 19, 2011, 11:38 am

On pg 76 it mentions a DSP

http://en.wikipedia.org/wiki/Digital_signal_processor

It can take DC voltage and build a sin wave using PWM (Pulse Width Modulation)

http://en.wikipedia.org/wiki/Pulse-width_modulation

As shown on pg 74 that is smoothed and fed into a transformer. A feedback circuit measures the current flow and adjusts the voltage of the constructed wave as needed. Not sure what you want here. If you want the exact details about how each part works google is your friend. 30 years ago I was working with drive panels for large industrial DC motors that did all this to feed current back into the lines when doing regenerative braking. When they want to slow down they don't worry about sharing a load. They know there are loads out there and they just pump the power into the line. At that time it was all controlled using analog circuits.

- posted on April 18, 2011, 4:16 pm

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.

--------------------- "David Nebenzahl" wrote in message

It***sounds***--and I'm sure you'll correct me if I'm wrong--as if you're
agreeing with me, and with Smitty, and others when we say that it is
***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.

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.

--------------------- "David Nebenzahl" wrote in message

It

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.

--

- posted on April 18, 2011, 3:15 am

Picture two water tanks connected with a loop of pipe underneath. Put a ball in the connecting pipe. The two tanks are connected in parallel. The ball in the pipe doesn't go anywhere because it has equal pressure on both sides. Now pour some water into one of the tanks to raise its level (voltage). There will now be more pressure applied to the ball by that tank and the water will flow into the other tank, pushing the ball in that direction.

Attach a single drain pipe to the middle of the connector pipe above. If the tanks start at the same level and the resistance to flow in the pipes is the same then water will flow out of both pipes at the same rate. If one tank is higher than the other it is pushing harder and more water will flow out of it than the lower tank.

The power company has many stations and wants them to all contribute their share. We OTOH have spent big $$$ on our PV system and if it produces 5KW we want that whole 5KW to go to some load so that we get paid for it. We are not interested in playing nice and sharing a load if it means we don't get to contribute the full 5KW. So we raise the voltage just enough to flow the current we need to. That will result in less current flowing from the grid because the load is only going to accept so much flow and the PV system is taking more than its share.

Yes there will be flow at 119, 120 or 121, but from what source? Like I said, we aren't interested in playing nice. We are entitled to pump the power we produced into the line and get paid for it. If the power company has to reduce their production a bit to keep the voltage from climbing too high so be it.

- posted on April 18, 2011, 3:32 am

But... but... but... EVERY car insurance company is cheaper than its competitors and every alkaline cell is the longest lasting... LOL

Seriously, the power stations are not higher than each other, but every one is higher than the grid, imagine three power stations, or 3 generators in the same station all putting out exactly 48KV, the substation/transformer they are connected to is getting slightly less, say 47,990 volts. If they then need to bring a fourth generator online they use a syncroscope to adjust the prime mover to sync up the frequency and phase, adjust the field to match the voltage IE 47,990, close the breakers, and slowly increase the field and prime mover so that the load is shared and now all 4 are at 48KV.

- posted on April 18, 2011, 12:53 pm

To Smitty

You can't have two voltage sources EXACTLY in parallel, its like dividing by 0. There is always some small resistance between voltage sources or the sources themselves have resistance.

The sources have to be higher then the grid, not higher then eaach other. Lets say the grid is some point in the middle of a square and it is at 120.0. Lets say there are 4 sources feeding that center point from the 4 corners. Each feed has a small resistance between it and the "grid point". Each one can be at 120.1 for example and power will flow from each to the grid point. If feed point one is a big nuclear plant it might be at 120.3 and your small solar plant at point 2 might be at 120.01. But each will feed power TO the grid point load. Each plant adjusts itself to the right slight increse in voltage to feed the amount of power it has avaialbe.

(This is an oversimplification of what really happens. What really happens also has a lot to do with frequency and phase.)

If the load is at the grid point and is at 120.0, then any source under 120.0 will pull power and not source it., but the load as a reistor will pull current no matter what voltage is on it. A resistor can never geneate power. But a motor connected to a grid can pull power if you load the shaft or it can push power if you hook an engine to the shaft and drive it.

Mark

You can't have two voltage sources EXACTLY in parallel, its like dividing by 0. There is always some small resistance between voltage sources or the sources themselves have resistance.

The sources have to be higher then the grid, not higher then eaach other. Lets say the grid is some point in the middle of a square and it is at 120.0. Lets say there are 4 sources feeding that center point from the 4 corners. Each feed has a small resistance between it and the "grid point". Each one can be at 120.1 for example and power will flow from each to the grid point. If feed point one is a big nuclear plant it might be at 120.3 and your small solar plant at point 2 might be at 120.01. But each will feed power TO the grid point load. Each plant adjusts itself to the right slight increse in voltage to feed the amount of power it has avaialbe.

(This is an oversimplification of what really happens. What really happens also has a lot to do with frequency and phase.)

If the load is at the grid point and is at 120.0, then any source under 120.0 will pull power and not source it., but the load as a reistor will pull current no matter what voltage is on it. A resistor can never geneate power. But a motor connected to a grid can pull power if you load the shaft or it can push power if you hook an engine to the shaft and drive it.

Mark

- posted on April 18, 2011, 2:48 pm

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 ! ! ! ! ! ! ---------------------------------------------------------------------------

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) = 240/103= 2.33 amps. That current flows through the Rload resistor producing a voltage of 100 X 2.33 = 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 = i2 x (R2 + Rw + Rg) + (i1 + i2) x Rload

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

240 = 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 = i2. Substituting, we have

240 = 100 x i1 + 103 x i1

240 = 203 i1

or i = 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 = 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 = 236.4 + i1 x Rg = 236.4 + 1.182 = 237.6 volts

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

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