Feeding solar power back into municipal grid: Issues andfinger-pointing

[g]

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.

If you take the voltage drops into account, the voltage at the inverter
has to be higher than the voltage at the grid connection point.

This is a real world scenario.

 

[no spam]

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

Pigs-arse you were "teasing". You were sucked in
by the GymmyBob troll.. another of a few hundred
of his 'victims'. Deservedly so on this occasion.. you
prove a clueless dolt.

Get yourself educated on "how to" in reading news, rube.
netscape.public.mozilla.general

Then go check out your 'victor' :-/
"Gymy Bob" <[email protected]>/http://preview.tinyurl.com/3auj4kc
"John P. Bengi" <[email protected]>/http://preview.tinyurl.com/3zgxjxa
"Solar Flare" <[email protected]>/http://preview.tinyurl.com/42ytgkq
"Pizzza Girl" <[email protected]>/http://preview.tinyurl.com/3p979nm
"Janice" <[email protected]>/http://preview.tinyurl.com/3bctgwr
"Joesepi" <[email protected]>/http://preview.tinyurl.com/3qkwmxr

... mind how you go. Don't want you
bleeding all over the place:-D

 

[Mho]

Especially since you answered your own questions in your response.

 

[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.

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

"Mark" wrote in message

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]

Perfect logic and based on real thinking.

Bonus points...select a vowel.
-----------------

"Smitty Two" wrote in message

I've chewed this over a bit, and I still don't like it, and here are my
reasons:

1: Voltage sources in parallel do not push *against* one another.

2: If no voltage source can join the grid without being at a higher than
grid potential, then every contributing power station would have to be
at a higher potential than every other one, and that's impossible.

3: While voltage might *push*, it's the load that it said to *pull* the
current. If there's a demand, current will flow whether the supply
voltage is 119, 120 or 121.


Where is my thinking flawed?

 

[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.

---------------------
"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.


--

 

[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.

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

"g" wrote in message If you take the voltage drops into account, the voltage at the inverter
has to be higher than the voltage at the grid connection point.

This is a real world scenario.

 

[Mho]

Sorry, Your inconsideration, using all caps and messing with header info is
intolerable, m II, no spam, g

<PLONK>
-----------------------

"The Ghost in The Machine" wrote in message
DOESNT YOUR HUSBAND MAKE YOU HAPPY.....?

I SEE YOU ALSO HAVE A PERVERTED SEMEN FETISH, YOU DEVIANT!!

PLEASE LEAVE OUR USENET GROUPS AND PERSUE YOUR TWISTED FANCY IN SOME
DATING SERVICE FOR UNHAPPLIY WED PERSONS LIKE YOURSELF, OR TAKE IT
ELSEWHERE FELLER, WE'RE NOT IN THE MOOD FOR YOUR MANBOY CRAVING
SHENANIGANS.

PATECUM

 

[g]

Pigs-arse you were "teasing". You were sucked in
by the GymmyBob troll.. another of a few hundred
of his 'victims'. Deservedly so on this occasion.. you
prove a clueless dolt.

Get yourself educated on "how to" in reading news, rube.
netscape.public.mozilla.general

Then go check out your 'victor' :-/

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]

The inverter output is higher -internally- by the V=IR drop between it
and the grid.

In this case I is the independent variable, the array's output, and V
is whatever it takes to make I pass through R to get to the grid
voltage.

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.


--
The current state of literacy in our advanced civilization:

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wassup
nuttin
wan2 hang
k
where
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[David Nebenzahl]

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.


--
The current state of literacy in our advanced civilization:

yo
wassup
nuttin
wan2 hang
k
where
here
k
l8tr
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[David Nebenzahl]

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.

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?)


--
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*?)

 

[[email protected]]

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

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

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

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



------------------------
"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.

 

[[email protected]]

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?)

Make sure it's in a fixed space font.

 

[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.
--------------------

wrote in message


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]

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

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.

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

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

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.

 

[MarkK]

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.

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

 

[Smitty Two]

A grid tie inverter emulates a current
source.

Good lord. That's what all this is about? 9,000 rambling posts full of
arithmetic and gibberish and it all comes down to that? Why didn't you
speak up earlier? Well, maybe you did, I confess I've gotten bogged down
with all the long-windedness that's permeated the thread and I've
deleted a few posts without reading them.

Now I feel like we're getting somewhere, or at least, I might be. All
that talk of inverters putting out an extra volt to compensate for the
resistance of the wire from the house to the pole made no sense at all.
A current source, that's something I can comprehend.

 

[David Nebenzahl]

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
nuttin
wan2 hang
k
where
here
k
l8tr
by

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