Posted by *krw@att.bizzzzzzzzzzzz* on April 17, 2011, 8:49 pm

wrote:

*>On 4/15/2011 8:53 PM krw@att.bizzzzzzzzzzzz spake thus:*

*>> On Fri, 15 Apr 2011 18:45:07 -0700, Smitty Two*

*>> *

*>>> Home guy, try this wikipedia article for starters:*

*>>> *

*>>> http://en.wikipedia.org/wiki/Grid_tie_inverter *

*>>> *

*>>> Here is an excerpt from that page:*

*>>> *

*>>> "Inverters take DC power and invert it to AC power so it can be fed*

*>>> into the electric utility company grid. The grid tie inverter must *

*>>> synchronize its frequency with that of the grid (e.g. 50 or 60 Hz)*

*>>> using a local oscillator and limit the voltage to no higher than*

*>>> the grid voltage."*

*>>> *

*>>> Repeat: *no higher than the grid voltage!**

*>>> *

*>>> I realize wikipedia has its detractors, but it is peer reviewed and*

*>>> if that statement were as blatantly inaccurate as you believe, it*

*>>> would have been amended by now.*

*>>> *

*>>> I'm going to put forth an analogy, and welcome feedback on it. It*

*>>> may or may not be an accurate analogy, but this is the way I look*

*>>> at it: The grid is a big freeway. Picture 6 lanes in one direction,*

*>>> with all the cars moving along at 60 mph. The speed represents*

*>>> voltage. The number of cars represents amps.*

*>>> *

*>>> Now, you're going to add your little PV supply to it, so you cruise*

*>>> down the onramp and merge into traffic. You match the speed*

*>>> (voltage) of 60. But, you've added some current to the grid. Not a*

*>>> big percentage, but some. You don't have to go 61 mph to get on the*

*>>> freeway, in fact, it would be disruptive to do so.*

*>> *

*>> No, you can't get on the "freeway" unless you're going faster than 60. If*

*>> you're going slower, they're actively pushing you off.*

*>Whoa. Where do you get that (using Smitty's analogy)? Sure, you can't be *

*>going slower; that's self-evident. You say you have to be going faster. *

Yes.

*>But you say nothing about going (approximately) *the same speed*, which *

*>is what Smitty's example was saying.*

"Approximately" means a little faster, or slower. Slower does *not* work.

(And is what ever article I've ever

*>read about inverters, grid interties, etc., has said. (*None* of them *

*>say "the voltage of the contributing system has to be slightly higher *

*>than the grid in order to feed current into it". None of 'em.)*

Of course they don't say that. Physics does.

*>(Which, by the way, is eggs-ackley the same thing I've been saying here ...)*

Whatever you've been saying doesn't change physics.

Posted by *trader4@optonline.net* on April 17, 2011, 9:20 pm

*> On 4/15/2011 8:53 PM k...@att.bizzzzzzzzzzzz spake thus:*

*> > On Fri, 15 Apr 2011 18:45:07 -0700, Smitty Two*

*> >> Home guy, try this wikipedia article for starters:*

*> >>http://en.wikipedia.org/wiki/Grid_tie_inverter *

*> >> Here is an excerpt from that page:*

*> >> "Inverters take DC power and invert it to AC power so it can be fed*

*> >> into the electric utility company grid. The grid tie inverter must*

*> >> synchronize its frequency with that of the grid (e.g. 50 or 60 Hz)*

*> >> using a local oscillator and limit the voltage to no higher than*

*> >> the grid voltage."*

*> >> Repeat: *no higher than the grid voltage!**

*> >> I realize wikipedia has its detractors, but it is peer reviewed and*

*> >> if that statement were as blatantly inaccurate as you believe, it*

*> >> would have been amended by now.*

*> >> I'm going to put forth an analogy, and welcome feedback on it. It*

*> >> may or may not be an accurate analogy, but this is the way I look*

*> >> at it: The grid is a big freeway. Picture 6 lanes in one direction,*

*> >> with all the cars moving along at 60 mph. The speed represents*

*> >> voltage. The number of cars represents amps.*

*> >> Now, you're going to add your little PV supply to it, so you cruise*

*> >> down the onramp and merge into traffic. You match the speed*

*> >> (voltage) of 60. But, you've added some current to the grid. Not a*

*> >> big percentage, but some. You don't have to go 61 mph to get on the*

*> >> freeway, in fact, it would be disruptive to do so.*

*> > No, you can't get on the "freeway" unless you're going faster than 60. If*

*> > you're going slower, they're actively pushing you off.*

*> Whoa. Where do you get that (using Smitty's analogy)? Sure, you can't be*

*> going slower; that's self-evident. You say you have to be going faster.*

*> But you say nothing about going (approximately) *the same speed*, which*

*> is what Smitty's example was saying. (And is what ever article I've ever*

*> read about inverters, grid interties, etc., has said. (*None* of them*

*> say "the voltage of the contributing system has to be slightly higher*

*> than the grid in order to feed current into it". None of 'em.)*

I was on your side of this issue before I went back and looked at the

circuit

model for two power sources driving a load that Jim Wilkins provided.

I

made a post today in reply to Mho where I went through the math.

Sadly, Mho didn't even bother to go through the detailed circuit

analsysis

I provided. If you look for the post, I encourage you to review the

analysis

and the math and see what you conclude. The

conclusion I came to is HomeGuy and KRW are right. To get power onto

the grid, the additional source coming online has to be slightly

higher that

the voltage on the grid.

Look at it this way. Suppose the utility pole at my house ia at

120.000V.

I hook up a power source, be it a generator, PV, whatever on the end

of the line inside my house. I place exactly 120.000V on my end of

that

line. The wire from my power source to the pole has some small

resistance, R. With 120.000V on one

end of a resistor and 120.000V on the other end, by ohm's law, how

much current will flow? Zero.

How do I get current to flow? At least one of three things must

happen:

1 - I raise the voltage on my end of the wire slightly.

2 - The load increases on the rest of the system which in turn lowers

the voltage at the utility pole outside my house

3 - Whatever else is driving the load reduced it's voltage slightly,

which

in turn lowers the voltage at the utility pole by house.

Those are the only choices to get a potential difference across the

line

connecting the source in my house to the grid and only then can

current flow and power from my house make it onto the grid. And with

option 1, my raising the voltage slightly on the house end, in turn

must

raise the voltage at the utility pole slightly as well. In other

words, I've

raised the voltage of the grid at the utility pole at my house. It's

a very

small amount, but it's real.

Posted by *David Nebenzahl* on April 17, 2011, 9:56 pm

On 4/17/2011 2:20 PM trader4@optonline.net spake thus:

[cut to the chase, i.e., trader4's example:]

*> Look at it this way. Suppose the utility pole at my house ia at *

*> 120.000V. I hook up a power source, be it a generator, PV, whatever*

*> on the end of the line inside my house. I place exactly 120.000V on*

*> my end of that line. The wire from my power source to the pole has*

*> some small resistance, R. With 120.000V on one end of a resistor and*

*> 120.000V on the other end, by ohm's law, how much current will flow?*

*> Zero.*

All I can say is, good example, but wrong conclusion.

Let's change the example just a little. We'll have a large AC source

(call it "the grid"), and a smaller AC source (the solar system's

inverter), connected *in parallel*, and then some resistance (the total

load of "the grid") after all that, completing the circuit.

In this example, "the grid" and the PV inverter are operating at

*exactly* the same voltage.

Let me argue this negatively, and see what you say to it:

If you're saying that this will not work (i.e., that the PV inverter

cannot contribute any power to the circuit because it isn't at a higher

voltage), then *no* circuit where you have two power sources in parallel

could ever work under the same circumstances.

No electric vehicle would ever work, because the battery cells in them

are (pretty close to) exactly the same voltage, so how would each tiny

cell (tiny in comparison to the total power of its siblings) ever be

able to "push" electricity into the circuit?

That elementary electronics tutorial (Kirchoff's Law, etc.) explains

everything you need to know here. And it doesn't require any higher voltage.

I look forward to your reply.

--

The current state of literacy in our advanced civilization:

yo

wassup

nuttin

wan2 hang

k

where

here

k

l8tr

by

- from Usenet (what's *that*?)

Posted by *krw@att.bizzzzzzzzzzzz* on April 17, 2011, 10:31 pm

wrote:

*>On 4/17/2011 2:20 PM trader4@optonline.net spake thus:*

*>[cut to the chase, i.e., trader4's example:]*

*>> Look at it this way. Suppose the utility pole at my house ia at *

*>> 120.000V. I hook up a power source, be it a generator, PV, whatever*

*>> on the end of the line inside my house. I place exactly 120.000V on*

*>> my end of that line. The wire from my power source to the pole has*

*>> some small resistance, R. With 120.000V on one end of a resistor and*

*>> 120.000V on the other end, by ohm's law, how much current will flow?*

*>> Zero.*

*>All I can say is, good example, but wrong conclusion.*

*>Let's change the example just a little. We'll have a large AC source *

*>(call it "the grid"), and a smaller AC source (the solar system's *

*>inverter), connected *in parallel*, and then some resistance (the total *

*>load of "the grid") after all that, completing the circuit.*

*>In this example, "the grid" and the PV inverter are operating at *

*>*exactly* the same voltage.*

*>Let me argue this negatively, and see what you say to it:*

*>If you're saying that this will not work (i.e., that the PV inverter *

*>cannot contribute any power to the circuit because it isn't at a higher *

*>voltage), then *no* circuit where you have two power sources in parallel *

*>could ever work under the same circumstances.*

*>No electric vehicle would ever work, because the battery cells in them *

*>are (pretty close to) exactly the same voltage, so how would each tiny *

*>cell (tiny in comparison to the total power of its siblings) ever be *

*>able to "push" electricity into the circuit?*

Simple: The motor is at a lower potential.

*>That elementary electronics tutorial (Kirchoff's Law, etc.) explains *

*>everything you need to know here. And it doesn't require any higher voltage.*

Elementary, sure, but you're still not getting it.

*>I look forward to your reply.*

Posted by *trader4@optonline.net* on April 17, 2011, 10:33 pm

*> On 4/17/2011 2:20 PM trad...@optonline.net spake thus:*

*> [cut to the chase, i.e., trader4's example:]*

*> > Look at it this way. Suppose the utility pole at my house ia at*

*> > 120.000V. I hook up a power source, be it a generator, PV, whatever*

*> > on the end of the line inside my house. I place exactly 120.000V on*

*> > my end of that line. The wire from my power source to the pole has*

*> > some small resistance, R. With 120.000V on one end of a resistor and*

*> > 120.000V on the other end, by ohm's law, how much current will flow?*

*> > Zero.*

*> All I can say is, good example, but wrong conclusion.*

But what exactly is wrong with this example, which closely conforms to

what we have been discussing? We have one end of a power wire

connected to the PV array in my house. The other end is connected

100 feet away to the grid, ie the utility pole by my house. The sun

is

behind an eclipse, the array isn't generating any power. The voltage

at the utility pole is 120.000V. The sun comes out. To put power on

the grid, current must flow from my house, through that wire, to the

utility pole. The wire has some small resistance, R. According

to physics in my world, the only way to get current flowing in that

wire is to have the end in my house at a HIGHER voltage than it

is at the pole. It doesn't have to be a lot higher and it actually

will be only slightly higher. But without that difference, tell me

how could current ever flow?

The instant it does flow, I'm now pushing current out onto the grid.

Assuming the rest of the grid stays the same, ie the load is fixed,

the other power sources don't change, that means that the

voltage at the pole now increase slightly as well. Net result is

the voltage in my house is now say 120.1V, the voltage at the pole is

now 120.05V and I have in fact raised the voltage of the grid.

*> Let's change the example just a little. We'll have a large AC source*

*> (call it "the grid"), and a smaller AC source (the solar system's*

*> inverter), connected *in parallel*, and then some resistance (the total*

*> load of "the grid") after all that, completing the circuit.*

*> In this example, "the grid" and the PV inverter are operating at*

*> *exactly* the same voltage.*

Only if you assume the connection between the two is zero ohms,

ie a perfect conductor. That would be akin to declaring the line

connecting my house to the pole to be a perfect conductor and

changing the model from what it is in the real world. The model

that corresponds to what we have and also to your new proposed

example is essentially the circuit

example that Wilkins provided. The two resistors R1 and R2

represent the internal impedance of the two power sources.

If you want to model the grid connecting them, then you could

add two resistors, one after R1, the other after R2 to model

the resistance of the grid between each of the power sources

and the load. It doesn't change the analysis.

*> Let me argue this negatively, and see what you say to it:*

*> If you're saying that this will not work (i.e., that the PV inverter*

*> cannot contribute any power to the circuit because it isn't at a higher*

*> voltage), then *no* circuit where you have two power sources in parallel*

*> could ever work under the same circumstances.*

*> No electric vehicle would ever work, because the battery cells in them*

*> are (pretty close to) exactly the same voltage, so how would each tiny*

*> cell (tiny in comparison to the total power of its siblings) ever be*

*> able to "push" electricity into the circuit?*

You can model a two cell battery with that circuit diagram as well.

Let's

leave V2 at 20V, making it a 20V cell with an internal resistance of

20

ohms modeled by R2. It's companion cell is V1 with internal

resistance

of 10 ohms. Do the math and you'll find that with V1 at 13.2 volts,

no

current flows through the V1 half of the circuit. All the load

current

comes from V2. Start increasing V1 and only then does current

flow through V1 and through the load. The consequences of that

are then that the voltage at the load increases, which in turn

decrease the current flowing from V2.

In your real battery with cells connected in parallel, R1 and R2, the

internal resistances, would be very close or equal. And V1 would

be close in value to V2, both would be supplying about half the power.

But it doesn't change the application of Kirchoff's Law or the

conclusions.

*> That elementary electronics tutorial (Kirchoff's Law, etc.) explains*

*> everything you need to know here. And it doesn't require any higher voltage.*

*> I look forward to your reply.*

Did you look at the detailed analysis I provided in an earlier post

where

I went through the analysis of that circuit example? Here it is

repeated.

Take a look at that circuit and go through it step by step. It is a

model

of two power sources driving a load.

http://www.electronics-tutorials.ws/dccircuits/dcp_4.html

Voltage source V1 and R1 represent a simple battery,

with R1 being the internal resistance of the battery.

Same with V2 and R2. For our purposes a suitable

model for a PV array and another power source on

the other end of the distribution lines.

Let's leave V2 as is at 20V, supplying all

the current to the load, with no current coming from V1.

You then have a simple series circuit

consisting of resistors R2 and R3 connected to voltage

source V2. A current of 20/(40+20) = .33A is flowing,

which is I2 in the drawing. The only way for no power to be

flowing through the other half of the circuit encompassing V1 is

if V1 is at the exact same potential as the load resistor. With

.33A flowing through the load, R3, you have R3 at 13.2 V.

That means V1 must be 13.2 volts. With V1 at 13.2 volts

the voltage across R1 is zero and no current flows.

So, everything is in balance. V1=13.2V, I1=0, the voltage

on the load is 13.2 volts, and I2= .33A is flowing from V2

through R2, R3.

Now, if we want V1 to start supplying part of the power, what

has to happen? V1 has to increase ABOVE 13.2 volts. And

when it does, the voltage across the load resistor R3 will

also increase. As that happens, current will start to flow now

from V1 through the load resistor and at the same time the current

from V2 will decrease slightly, as the potential drop across

R2 is decreased slightly.

The net result of this is that the voltage across the load

has increased. Current I1 is now flowing from V1, I2 from V2

is now slightly less and the combined currents of I1 and I2

which together are I3 has increased slightly.

The other ways to get V1 to supply power

would be for either the load resistor R3 to decrease in

value or for voltage source V2 to decrease.

If you want to more closely model the situation, we could

add two more resistors to model the distribution line

resistance. A resistor could be added after R1 and after

R2. But if you go through the analysis, it doesn't change

the basics of the above analysis. For source V1 to supply

power, the voltage on it's portion of the distribution system

and across the load must increase.

>On 4/15/2011 8:53 PM krw@att.bizzzzzzzzzzzz spake thus:>> On Fri, 15 Apr 2011 18:45:07 -0700, Smitty Two>>>>> Home guy, try this wikipedia article for starters:>>>>>> http://en.wikipedia.org/wiki/Grid_tie_inverter>>>>>> Here is an excerpt from that page:>>>>>> "Inverters take DC power and invert it to AC power so it can be fed>>> into the electric utility company grid. The grid tie inverter must>>> synchronize its frequency with that of the grid (e.g. 50 or 60 Hz)>>> using a local oscillator and limit the voltage to no higher than>>> the grid voltage.">>>>>> Repeat: *no higher than the grid voltage!*>>>>>> I realize wikipedia has its detractors, but it is peer reviewed and>>> if that statement were as blatantly inaccurate as you believe, it>>> would have been amended by now.>>>>>> I'm going to put forth an analogy, and welcome feedback on it. It>>> may or may not be an accurate analogy, but this is the way I look>>> at it: The grid is a big freeway. Picture 6 lanes in one direction,>>> with all the cars moving along at 60 mph. The speed represents>>> voltage. The number of cars represents amps.>>>>>> Now, you're going to add your little PV supply to it, so you cruise>>> down the onramp and merge into traffic. You match the speed>>> (voltage) of 60. But, you've added some current to the grid. Not a>>> big percentage, but some. You don't have to go 61 mph to get on the>>> freeway, in fact, it would be disruptive to do so.>>>> No, you can't get on the "freeway" unless you're going faster than 60. If>> you're going slower, they're actively pushing you off.>Whoa. Where do you get that (using Smitty's analogy)? Sure, you can't be>going slower; that's self-evident. You say you have to be going faster.Yes.

>But you say nothing about going (approximately) *the same speed*, which>is what Smitty's example was saying.