On Fri, 15 Apr 2011 00:40:05 -0700 (PDT), harry wrote:

On Apr 15, 5:46*am, "
wrote:
On Fri, 15 Apr 2011 00:27:36 -0400, Home Guy wrote:
" wrote:

We can forget about generators and distributions systems.
Just take two 12V batteries and connect them in parallel

There's your problem right there.

It's usually not a good idea to connect batteries together in parallel,
unless they are exactly of the same type, age, condition, etc. *If you
get a weak cell in one of the batteries it will turn into a load.

And what happens when they are not exactly at the same voltage before
being connected together? *How do you insure that you always get current
flowing out of both of them?

But clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

The IEEE paper I posted earlier today shows exactly that - that PV
systems raise local grid voltage and the utility company must compensate
by reducing primary supply voltage to down-regulate the secondary
voltage coming from the distribution transformer.

What a dumbass!

So, what happens with the two batteries? * Under the
laws of physics the rest of us use the voltage would
remain at 12 volts and BOTH batteries would be supplying
part of the 1 AMP flowing through the resistors.

Take a 12 V car battery and wire it up in parallel with 8 AAA batteries
connected in series. *Then connect a load and tell me how much current
the AAA batteries will supply vs their what their potential current
supply could be if they were connected to their own isolated load.

They will share in proportion to their capacity. *Electricity, water, nor ****
flow uphill. *...though you have been pumping enough of the latter here.

Not on Homeguy's and Vaughn's planet. *One of the batteries
will be charging the other.

If they are unequal in capacity, then yes that will eventually happen.

Wrong!- Hide quoted text -

- Show quoted text -

He said unequal voltage. In which case he's right.
Capacity is not voltage.

If he really means unequal voltage then he's proving my point. He's a dumbass
(second point proven).

Home Guy wrote:

" wrote:

How to properly analyze the problem:
http://www.electronics-tutorials.ws/...its/dcp_4.html

Look at Example 1.

Thank you Jim. I was sitting here wishing I could draw a
simple circuit into the newsgroup that models what we are
talking about and shows what really happens. You've gone
one better and found a perfect example from an independent
and credible source.

You dumbass.

Look more closely at the current flow in battery 1.
It's NEGATIVE.

Where's trader4?

Where did you go, you coward?

You have no response to what I wrote above?

You disappeared from this tangent thread pretty fast, didn't you?

You absolutely loved that link posted above, showing Example 1 - where
you thought it was proving me wrong.

Go ahead and substitute 13.33333 volts for Battery 1 in that example and
tell me how much current it's supplying to the load.

On Fri, 15 Apr 2011 00:47:53 -0400, Home Guy wrote:

" unnecessarily full-quoted:

You are claiming that any electricity produced by PV arrays that
goes onto the local grid just gets wasted because putting it on
the grid raises the voltage a tiny amount. I think that's what
he meant by saying "it doesn't work". That is you're saying
that PV arrays that have net current flowing into the grid
don't work, because the energy somehow just gets dissapears.

I'm not saying that it dissapears.

I'm saying that if your local grid is sitting at 120V and your panels
come on and raise it to 121V, and if the utility company doesn't
down-regulate their side to bring the local grid back to 120V, then the
current that your panels are injecting is wasted. It's wasted because
all the linear loads on the grid that are designed for 120V will not
operate any better at 121 volts.

Wrong.

Motors won't turn faster,

Motors won't turn faster, but they will take less current. They're doing the
same work so will take (roughly) the same power to do it.

lights won't really burn brighter.

Wrong, not that the higher intensity is always useful.

They will just give off a little more heat thanks
to the extra current the panels are supplying to the grid.

Wrong.

But sure - electric heaters will get hotter. They're the only devices
on the grid that are intended to convert electrical energy into heat.

You're batting 1000.

There is SO much wrong in your analysis, that I don't know
where to begin. But here's a start. You claim that with
a slightly higher voltage, an AC motor in an HVAC
compressor won't turn any faster and hence the additional
power is wasted. What you've completely overlooked is
that power is P=VI, or power is voltage times current.
Give that motor an extra half a volt and I'll bet it's
current decreases by a corresponding amount.

So why not run a 120V motor with 240 volts then?

Put the windings in series and it'll run better.

AC Motors are not simple loads like a resistor, but they will still
"consume" power (V x I) as a function of their supply voltage.

Wrong. You're still batting 1000.

I'm claiming that there won't be a corresponding voltage
down-regulation at the level of the neighborhood
distribution transformer to make the effort worth while
for all stake holders.

As Bud said a while back, you're new analysis must be
devastating to all the power companies in the world.

All the power companies in the world are in the business of generating
electricity in the thousands of volts and sending it out over
high-tension wires. That's what they'd rather do if they weren't being
hamstrung by crazy ideas and new rules / laws made by politicians about
small-scale co-generation.

"Co-generation"?

Look at the microFIT program in Ontario. When the rules were changed to
allow local utilities to veto hookups based on "network capacity" or
"substation insufficiency", they were only too happy to start swinging
their veto left and right. They don't want to see this small-scale ****
coming on-line if they have a choice.

They have to *pay* for that energy, not to mention manage the complexity of
the mess and lose money at the same time. Of course they'll opt out, if given
the chance. It shouldn't be done, but certainly not for the reasons you
suggest.

" wrote:

Solve the equations for the case where the voltage sources are
at identical voltages and you will find that current flows from
BOTH sources. One is NOT at a higher potential than the
other, which is what Homeguy claims must exist for both
to provide current to the load.

So what you're saying is this:

Connect 2 batteries of the same voltage together in parallel to the same
load and each battey will supply half the current to the load.

So if I extrapolate that situation:

If my service voltage is currently sitting at 120.0 volts, and if I
adjust my PV invertor output voltage to 120.0 volts, and then connect my
PV output to my service connection, then my PV system will somehow
magically supply half the current to to the load (the load being my
house and all other houses sharing the same service line).

Wow. That sounds like a really good bargain. Just by matching the
power companies voltage at my service input, my PV system will supply
half the current - always! And it doesn't matter how many PV panels I
have!

Wow. Who knew it would be that simple and effective?

In article , Home Guy wrote:

Home Guy wrote:

snip

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.

Y'all are welcome to take shots at this, I'm curious whether it seems
like a good analogy or not. But either way, I think the wikipedia
article is a good starting point for those that want to understand it
without an EE degree or reading Kirchoff as a bedtime story.

On Apr 14, 12:23*am, "
wrote:
On Wed, 13 Apr 2011 20:46:43 -0700 (PDT), The Ghost in The Machine





wrote:
On Apr 13, 11:38 pm, "
wrote:
On Wed, 13 Apr 2011 21:33:08 -0400, "vaughn"
wrote:

wrote in message
news

E=IR, certainly *IS* Ohm's law. I and E are proportional. You can't increase
I without increasing E.
Wrong. You CAN increase I without increasing E. You have 3 variables in that
formula, not just 2.

Dumbass, it's a fixed circuit.

Get it? I suppose not.

Apparently not.

You've only proved that you're just as stupid as David.

SO, NOW WE ALL SHOULD RESORT TO ANGRY NAMECALLING & FAGGOTY RESPONSES
HAHN???

GROW UP FOOL, OR DROP IT, YOUR STALE INSIGNIFICANT DULLARD POSTINGS
ARENT HELPING ANY.

Speaking of dumbasses...

PAT ECUM

Here is Roy Queerjano, A.K.A. Pat E. Cum/

DANNNNNNNNT WRONG !!!!
YOU SHOULD SHUT UP DONKEY KELLY WE KNOW ALL ABOUT YOU AND MICHAEL
TERRELL'S GAYTARD BROKEN HEART OVER ROY Q...YOU FOOL....FYI I AM NOT
HIM ......YOU SYCOPHANTIC USENET ABUSER.
COLLECTING AND DELIVERING YOUR HATEFUL OPPRESSIVE RESENTFUL SOUL TO
HELL WILL BE A TREAT TO MANKIND.

PAT ECUM

On Apr 15, 3:40*am, harry wrote:
On Apr 15, 5:46*am, "





wrote:
On Fri, 15 Apr 2011 00:27:36 -0400, Home Guy wrote:
" wrote:

We can forget about generators and distributions systems.
Just take two 12V batteries and connect them in parallel

There's your problem right there.

It's usually not a good idea to connect batteries together in parallel,
unless they are exactly of the same type, age, condition, etc. *If you
get a weak cell in one of the batteries it will turn into a load.

And what happens when they are not exactly at the same voltage before
being connected together? *How do you insure that you always get current
flowing out of both of them?

But clearly he thinks if we put a second AC power source on
a distributions system, it has to be at a higher voltage to
"push" current out.

The IEEE paper I posted earlier today shows exactly that - that PV
systems raise local grid voltage and the utility company must compensate
by reducing primary supply voltage to down-regulate the secondary
voltage coming from the distribution transformer.

What a dumbass!

So, what happens with the two batteries? * Under the
laws of physics the rest of us use the voltage would
remain at 12 volts and BOTH batteries would be supplying
part of the 1 AMP flowing through the resistors.

Take a 12 V car battery and wire it up in parallel with 8 AAA batteries
connected in series. *Then connect a load and tell me how much current
the AAA batteries will supply vs their what their potential current
supply could be if they were connected to their own isolated load.

They will share in proportion to their capacity. *Electricity, water, nor ****
flow uphill. *...though you have been pumping enough of the latter here.

Not on Homeguy's and Vaughn's planet. *One of the batteries
will be charging the other.

If they are unequal in capacity, then yes that will eventually happen.

Wrong!- Hide quoted text -

- Show quoted text -

He said unequal voltage. In which case he's right.
Capacity is not voltage.

DON'T MIND KRW = KEITHTARD, HE AND HIS SUPPLIERS TERRROLL AND DONKEY
BOY ARE HAVING TROUBLE SCORING ROCK THESE DAYS....SO, TO STAY ON
TOPIC, HE IS JUST WIRED WRONG.

PATECUM

On Apr 15, 9:45*pm, Smitty Two wrote:
In article , Home Guy wrote:
Home Guy wrote:

snip

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.

Y'all are welcome to take shots at this, I'm curious whether it seems
like a good analogy or not. But either way, I think the wikipedia
article is a good starting point for those that want to understand it
without an EE degree or reading Kirchoff as a bedtime story.

NOT TOO BAD SMITTY TWO TWO...
BUT THIS GUYS SYSTEM ALL OF SUDDEN IS FACING AN ANGRY DRUNKEN MOM ON
MARIJUANA COMING AGAINST TRAFFIC..... THERE IS A REASON FOR NOT
LINKING IT UP INTO THE GRID.

PAT ECUM

On Fri, 15 Apr 2011 20:03:02 -0700 (PDT), The Ghost in The Machine
wrote:

On Apr 14, 12:23*am, "
wrote:
On Wed, 13 Apr 2011 20:46:43 -0700 (PDT), The Ghost in The Machine





wrote:
On Apr 13, 11:38 pm, "
wrote:
On Wed, 13 Apr 2011 21:33:08 -0400, "vaughn"
wrote:

wrote in message
news

E=IR, certainly *IS* Ohm's law. I and E are proportional. You can't increase
I without increasing E.
Wrong. You CAN increase I without increasing E. You have 3 variables in that
formula, not just 2.

Dumbass, it's a fixed circuit.

Get it? I suppose not.

Apparently not.

You've only proved that you're just as stupid as David.

SO, NOW WE ALL SHOULD RESORT TO ANGRY NAMECALLING & FAGGOTY RESPONSES
HAHN???

GROW UP FOOL, OR DROP IT, YOUR STALE INSIGNIFICANT DULLARD POSTINGS
ARENT HELPING ANY.

Speaking of dumbasses...

PAT ECUM

Here is Roy Queerjano, A.K.A. Pat E. Cum/

DANNNNNNNNT WRONG !!!!

"DANNNNNNNNNT"? That word isn't in my dictionary, Roy Queerjano?

YOU SHOULD SHUT UP DONKEY KELLY WE KNOW ALL ABOUT YOU AND MICHAEL
TERRELL'S GAYTARD BROKEN HEART OVER ROY Q...YOU FOOL....FYI I AM NOT
HIM ......YOU SYCOPHANTIC USENET ABUSER.

You're a liar, Roy Eat Cum.

COLLECTING AND DELIVERING YOUR HATEFUL OPPRESSIVE RESENTFUL SOUL TO
HELL WILL BE A TREAT TO MANKIND.

Go away, Roy.

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

In article , Home Guy wrote:

Home Guy wrote:

snip

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!*

That's good for a first order approximation. The fact is that water doesn't
run down hill and electricity does *not* flow from low potential to high. The
inverter's instantaneous voltage *must* be higher than the grid for current to
flow into the grid.

Think about turning a bicycle crank, one without a clutch makes the point
better. You're not doing any work unless you're applying pressure to the
pedals. If you do nothing it drives you.

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.

Y'all are welcome to take shots at this, I'm curious whether it seems
like a good analogy or not. But either way, I think the wikipedia
article is a good starting point for those that want to understand it
without an EE degree or reading Kirchoff as a bedtime story.

It's OK for a first order, but not for discussing the details people are
trying to get into here.

On 15/04/2011 16:56, Home Guy wrote:

So if I extrapolate that situation:

If my service voltage is currently sitting at 120.0 volts, and if I
adjust my PV invertor output voltage to 120.0 volts, and then connect my
PV output to my service connection, then my PV system will somehow
magically supply half the current to to the load (the load being my
house and all other houses sharing the same service line).

Well, in a simplified scenario, (Grid, load, PV array) there will be
"load sharing" between the grid and the PV array. With the voltage
pretty much equal. I say pretty much equal, because there are line
resistance losses to take into account.


Wow. That sounds like a really good bargain. Just by matching the
power companies voltage at my service input, my PV system will supply
half the current - always! And it doesn't matter how many PV panels I
have!


I think there is some misunderstanding about the concepts here.

trader4 talks about 2 batteries supplying a load. The example is good
for calculating load sharing between 2 voltage sources with load
resistor and line resistance.

The way I see it is that the LOAD with this battery analogy is the house
load. One battery models the PV array, the other the grid.

One thing should be clear: To get power into the grid at all, the house
load must be lower than what the PV array can supply. If you remove the
house load, then all the available PV array current will flow into the
grid, with the inverter at a higher voltage.

To get back to that Kirchoff's Law example, if we remove the 40 ohms
resistor (the load), there are basically 2 voltage sources opposing each
other. When these voltages are equal, no current flows.

To allow current to flow into the grid, the PV array voltage has to be
higher, whether there is a house load or not.

On Apr 16, 12:56*am, Home Guy wrote:
" wrote:
Solve the equations for the case where the voltage sources are
at identical voltages and you will find that current flows from
BOTH sources. *One is NOT at a higher potential than the
other, which is what Homeguy claims must exist for both
to provide current to the load.

So what you're saying is this:

Connect 2 batteries of the same voltage together in parallel to the same
load and each battey will supply half the current to the load.

So if I extrapolate that situation:

If my service voltage is currently sitting at 120.0 volts, and if I
adjust my PV invertor output voltage to 120.0 volts, and then connect my
PV output to my service connection, then my PV system will somehow
magically supply half the current to to the load (the load being my
house and all other houses sharing the same service line).

Wow. *That sounds like a really good bargain. *Just by matching the
power companies voltage at my service input, my PV system will supply
half the current - always! *And it doesn't matter how many PV panels I
have!

Wow. *Who knew it would be that simple and effective?

There are lots of people here who have wasted a lot of time here
trying to explain to you how it works. But it does.
Time for you to understand that you're too stupid to understand. Go
home, watch the TV. Something not too intellectual. Lie down. Drink
some beer. If you give up taxing your brain,the headache will go away.
There must be lot's out there you don't understand but don't worry
about it.

On Apr 16, 12:56*am, Home Guy wrote:
" wrote:
Solve the equations for the case where the voltage sources are
at identical voltages and you will find that current flows from
BOTH sources. *One is NOT at a higher potential than the
other, which is what Homeguy claims must exist for both
to provide current to the load.

So what you're saying is this:

Connect 2 batteries of the same voltage together in parallel to the same
load and each battey will supply half the current to the load.

So if I extrapolate that situation:

If my service voltage is currently sitting at 120.0 volts, and if I
adjust my PV invertor output voltage to 120.0 volts, and then connect my
PV output to my service connection, then my PV system will somehow
magically supply half the current to to the load (the load being my
house and all other houses sharing the same service line).

Wow. *That sounds like a really good bargain. *Just by matching the
power companies voltage at my service input, my PV system will supply
half the current - always! *And it doesn't matter how many PV panels I
have!

Wow. *Who knew it would be that simple and effective?

No. You match the voltage and then turn it up until the full load
current of your array is flowing in practical terms.
There. Do you understand that?

harry wrote:

So what you're saying is this:

Connect 2 batteries of the same voltage together in parallel to
the same load and each battey will supply half the current to
the load.

That sounds like a really good bargain. Just by matching the
power companies voltage at my service input, my PV system will
supply half the current - always!

No. You match the voltage and then turn it up until the full load
current of your array is flowing in practical terms.

There. Do you understand that?

That's exactly what I've been saying - that you "turn it up" (the
inverter's voltage output) to maximize the PV's current (I) supply into
the grid.

But everyone else (or most everyone else) is saying no - that simply
matching the grid voltage (as measured at your service connection) is
all that happens (and is all that needs to happen) for the entire PV
current (I) capacity of the PV system to be "injected" into the grid.

So now that we agree that PV systems need to raise the grid voltage if
they're going to "force" their maximal available supply capacity into
the grid, it's a moot or academic question as to what exactly their
supply situtation would be (how much current they'd supply into the
grid) if the invertors simply matched the grid voltage.

On Apr 16, 9:42*am, Home Guy wrote:
harry wrote:
So what you're saying is this:

Connect 2 batteries of the same voltage together in parallel to
the same load and each battey will supply half the current to
the load.

That sounds like a really good bargain. *Just by matching the
power companies voltage at my service input, my PV system will
supply half the current - always! *
No. You match the voltage and then turn it up until the full load
current of your array is flowing in practical terms.
There. Do you understand that?

That's exactly what I've been saying - that you "turn it up" (the
inverter's voltage output) to maximize the PV's current (I) supply into
the grid. *

But everyone else (or most everyone else) is saying no - that simply
matching the grid voltage (as measured at your service connection) is
all that happens (and is all that needs to happen) for the entire PV
current (I) capacity of the PV system to be "injected" into the grid.
...

You can only guess at what the inverter is forcing and sensing without
looking at the schematic and uP code or a technical explanation of it.
The Wiki type explanations are oversimplified for general readers,
engineers have better sources.

I was hired to decipher and troubleshoot several lead-acid and lithium
battery charging circuits after the designers quit, and found a couple
of different approaches in use. Generally they compared voltage and
current measurements to a model and used the result to pulse-width-
modulate the output control.

Power factor control is similar to the design issues of a grid-tie
inverter, with a large enough market to support custom ICs:
http://focus.ti.com/lit/an/slua144/slua144.pdf

jsw

On Apr 16, 2:42*pm, Home Guy wrote:
harry wrote:
So what you're saying is this:

Connect 2 batteries of the same voltage together in parallel to
the same load and each battey will supply half the current to
the load.

That sounds like a really good bargain. *Just by matching the
power companies voltage at my service input, my PV system will
supply half the current - always! *
No. You match the voltage and then turn it up until the full load
current of your array is flowing in practical terms.
There. Do you understand that?

That's exactly what I've been saying - that you "turn it up" (the
inverter's voltage output) to maximize the PV's current (I) supply into
the grid. *

But everyone else (or most everyone else) is saying no - that simply
matching the grid voltage (as measured at your service connection) is
all that happens (and is all that needs to happen) for the entire PV
current (I) capacity of the PV system to be "injected" into the grid.

So now that we agree that PV systems need to raise the grid voltage if
they're going to "force" their maximal available supply capacity into
the grid, it's a moot or academic question as to what exactly their
supply situtation would be (how much current they'd supply into the
grid) if the invertors simply matched the grid voltage.

The moment you connect the array to the grid, the voltages will be the
same.
The current could be flowing either way (except tby it's intrinsic
nature to the PV panels wont allow it Other non-PV generators will
allow it BTW)
So, you need a device to monitor the current and current direction.
And another device to"turn up" the voltage.
As it gets dark the PV gets to the point when it can't generate
sufficient volts to force current back into the grid. At which point,
the inverter disconnects it.
All this is done by the grid tie inverter automatically as you don't
want to be stood there all day.

On Apr 16, 3:36*pm, Jim Wilkins wrote:
On Apr 16, 9:42*am, Home Guy wrote:





harry wrote:
So what you're saying is this:

Connect 2 batteries of the same voltage together in parallel to
the same load and each battey will supply half the current to
the load.

That sounds like a really good bargain. *Just by matching the
power companies voltage at my service input, my PV system will
supply half the current - always! *
No. You match the voltage and then turn it up until the full load
current of your array is flowing in practical terms.
There. Do you understand that?

That's exactly what I've been saying - that you "turn it up" (the
inverter's voltage output) to maximize the PV's current (I) supply into
the grid. *

But everyone else (or most everyone else) is saying no - that simply
matching the grid voltage (as measured at your service connection) is
all that happens (and is all that needs to happen) for the entire PV
current (I) capacity of the PV system to be "injected" into the grid.
...

You can only guess at what the inverter is forcing and sensing without
looking at the schematic and uP code or a technical explanation of it.
The Wiki type explanations are oversimplified for general readers,
engineers have better sources.

I was hired to decipher and troubleshoot several lead-acid and lithium
battery charging circuits after the designers quit, and found a couple
of different approaches in use. Generally they compared voltage and
current measurements to a model and used the result to pulse-width-
modulate the output control.

Power factor control is similar to the design issues of a grid-tie
inverter, with a large enough market to support custom ICs:http://focus.ti.com/lit/an/slua144/slua144.pdf

*jsw- Hide quoted text -

- Show quoted text -

An unneccessary complication. Far easier and better to mount PF
correction capacitors on individual motors.

On Sat, 16 Apr 2011 11:29:27 -0700 (PDT), harry wrote:

On Apr 16, 3:36*pm, Jim Wilkins wrote:
On Apr 16, 9:42*am, Home Guy wrote:





harry wrote:
So what you're saying is this:

Connect 2 batteries of the same voltage together in parallel to
the same load and each battey will supply half the current to
the load.

That sounds like a really good bargain. *Just by matching the
power companies voltage at my service input, my PV system will
supply half the current - always! *
No. You match the voltage and then turn it up until the full load
current of your array is flowing in practical terms.
There. Do you understand that?

That's exactly what I've been saying - that you "turn it up" (the
inverter's voltage output) to maximize the PV's current (I) supply into
the grid. *

But everyone else (or most everyone else) is saying no - that simply
matching the grid voltage (as measured at your service connection) is
all that happens (and is all that needs to happen) for the entire PV
current (I) capacity of the PV system to be "injected" into the grid.
...

You can only guess at what the inverter is forcing and sensing without
looking at the schematic and uP code or a technical explanation of it.
The Wiki type explanations are oversimplified for general readers,
engineers have better sources.

I was hired to decipher and troubleshoot several lead-acid and lithium
battery charging circuits after the designers quit, and found a couple
of different approaches in use. Generally they compared voltage and
current measurements to a model and used the result to pulse-width-
modulate the output control.

Power factor control is similar to the design issues of a grid-tie
inverter, with a large enough market to support custom ICs:http://focus.ti.com/lit/an/slua144/slua144.pdf

*jsw- Hide quoted text -

- Show quoted text -

An unneccessary complication. Far easier and better to mount PF
correction capacitors on individual motors.

Motors aren't the only things that need PF correction. Capacitors aren't the
only, or often the, way of doing it.

On Apr 16, 2:29*pm, harry wrote:
...
Power factor control is similar to the design issues of a grid-tie
inverter, with a large enough market to support custom ICs:
http://focus.ti.com/lit/an/slua144/slua144.pdf

*jsw-
An unneccessary complication. Far easier and better to mount PF
correction capacitors on individual motors.-

An incorrect assumption. Those are for more complex loads like
switching power supplies with rectifier inputs.

jsw

On Apr 15, 7:46*pm, "
wrote:
On Fri, 15 Apr 2011 00:47:53 -0400, Home Guy wrote:
" unnecessarily full-quoted:

You are claiming that any electricity produced by PV arrays that
goes onto the local grid just gets wasted because putting it on
the grid raises the voltage a tiny amount. * I think that's what
he meant by saying "it doesn't work". * That is you're saying
that PV arrays that have net current flowing into the grid
don't work, because the energy somehow just gets dissapears.

I'm not saying that it dissapears.

Ok, subsitute the words "gets wasted and performs no useful
work". That is what you are saying.




I'm saying that if your local grid is sitting at 120V and your panels
come on and raise it to 121V, and if the utility company doesn't
down-regulate their side to bring the local grid back to 120V, then the
current that your panels are injecting is wasted. *It's wasted because
all the linear loads on the grid that are designed for 120V will not
operate any better at 121 volts.

Wrong.



That sure must be news to all the power companies that are paying
people for having PV arrays generate power to the grid.


Motors won't turn faster,

Motors won't turn faster, but they will take less current. *They're doing the
same work so will take (roughly) the same power to do it.

Actually, they do turn very slightly faster when supplied by a
slightly higher
voltage. Heres' a good reference that covers HVAC compressors:

http://www.hvactroubleshootingguides...ac-motors.html

Take a look at the graph, which shows slip, which is the variation
between
shaft speed and the sychronous magnetic field which is determined by
the
line freq. It shows that slip is also a function of voltage, that
when voltage
increases, RPMs increase slightly.




lights won't really burn brighter.

Wrong, not that the higher intensity is always useful.

Agreed. At higher voltage they do burn a tiny bit brighter. And that
would appear to be an example of what you could consider
wasted energy. Unless you want to factor in that in winter at
least, in some cases, it adds to the available heat.



They will just give off a little more heat thanks
to the extra current the panels are supplying to the grid.

Wrong.

But sure - electric heaters will get hotter. *They're the only devices
on the grid that are intended to convert electrical energy into heat. *

You're batting 1000.

There is SO much wrong in your analysis, that I don't know
where to begin. *But here's a start. *You claim that with
a slightly higher voltage, an AC motor in an HVAC
compressor won't turn any faster and hence the additional
power is wasted. * What you've completely overlooked is
that power is P=VI, or power is voltage times current.
Give that motor an extra half a volt and I'll bet it's
current decreases by a corresponding amount.

So why not run a 120V motor with 240 volts then?

Because 240V is out of the range of operation for a 120V
AC motor. Stick to the case at hand. We're talking about
running an AC motor at 120V or 121V. I say running it
at 121V means the current will be slightly less, resulting
in the motor operating at the same HP output, but at
slightly higher voltage and slightly lower current. And/or
part of the voltage increase will result in more power
being delivered by the motor to the AC comptressor.
You say what? The motor justs takes that extra volt and
turns it into pure heat? How does it know to do that?




Put the windings in series and it'll run better.



AC Motors are not simple loads like a resistor, but they will still
"consume" power (V x I) as a function of their supply voltage. *

Wrong. You're still batting 1000.

Well they do consume power in relation to their voltage
and current. Take a look at these formulas:

Look at the one "To find horsepower." Clearly I can get the same
HP output by raising the voltage slightly while the current gets
reduced. Also take a look at the previous graph, which clearly
shows that full load current DECREASES if you increase
voltage slightly.



I'm claiming that there won't be a corresponding voltage
down-regulation at the level of the neighborhood
distribution transformer to make the effort worth while
for all stake holders.


As Bud said a while back, you're new analysis must be
devastating to all the power companies in the world.

All the power companies in the world are in the business of generating
electricity in the thousands of volts and sending it out over
high-tension wires. *That's what they'd rather do if they weren't being
hamstrung by crazy ideas and new rules / laws made by politicians about
small-scale co-generation.

"Co-generation"?

Look at the microFIT program in Ontario. *When the rules were changed to
allow local utilities to veto hookups based on "network capacity" or
"substation insufficiency", they were only too happy to start swinging
their veto left and right. *They don't want to see this small-scale ****
coming on-line if they have a choice.

They have to *pay* for that energy, not to mention manage the complexity of
the mess and lose money at the same time. *Of course they'll opt out, if given
the chance. *It shouldn't be done, but certainly not for the reasons you
suggest.- Hide quoted text -


Again, you'd think that if most or all of that net energy that is put
onto the grid
by PV arrays is being wasted, we'd have heard about it from someone
long
before this.

On Apr 17, 11:22*am, "Mho" wrote:
You guys need to get over your basic understanding of electricity.

When you connect to a grid the voltage is always **EXACTLY*** the same as
the connect point. They are in parallel.

That was where I was coming from too. However, having gone back
and revisited that basic circuit diagram of two voltage source driving
a load, I have come around to where I see Homeguy's point that if a
new
source wants to deliver power onto the grid, it can raise the voltage
at the load. Let's say I have a solar array that is covered up and
its sunny outside. It's connected via distribution lines to a load
that
is a block away. Another power source is located a similar
distance away from the load on the other side. In other words,
a case like the circuit example Jim Wilkens provided.


When I uncover the PV array, for it's additional X KW of
power to make it to the load down the block, at least one
of 3 things needs to happen:

1 - The PV raises the voltage on it's end of the distribution system
slightly.

2 - the load increases

3 - the power source on the other end of the distribution system
lowers it's voltage.

Here's the circuit example again that Jim provided:

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

It's example one.

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.

I think this is what Homeguy has been saying all along. So,
I've come full circle here and now agree with him on that
issue. I still disagree that the slight increase in voltage
in a distribution system means that the power is wasted.
The issue there is how the loads respond to being given
121V instead of 120V. HG claims that except for resistance
heaters, that energy goes to waste. And I say there he
is wrong, but that topic is being covered in another part
of this thread.

On 17/04/2011 10:09, Mho wrote:
All that "load increases" is a bunch of baloney!

A fixed load is just that....***FIXED***

Define a fixed load.

As a teaser, say I buy a 2 kW heater to heat my office in my home on
those cold winter nights when I am reading posts in
alt.energy.homepower. Is that 2 kW heater a fixed load?


If ht e grid were a
perfect conductor and had zero impedance no co-gen source could work.


I disagree. But I have been known to be wrong

Could you state why it does not work?

Is it possible to hook up any power source to such a grid?
If the answer is yes, why is co-gen not possible?

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

On 4/15/2011 8:53 PM zzzzzzzzzz spake thus:

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

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.)

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


--
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*?)

On Sun, 17 Apr 2011 13:35:07 -0700, David Nebenzahl
wrote:

On 4/15/2011 8:53 PM zzzzzzzzzz spake thus:

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

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.

On Apr 17, 4:35*pm, David Nebenzahl wrote:
On 4/15/2011 8:53 PM spake thus:





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

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.

On Apr 17, 4:17*pm, g wrote:
On 17/04/2011 10:09, Mho wrote:

All that "load increases" is a bunch of baloney!

A fixed load is just that....***FIXED***

Define a fixed load.

You're wasting your time with Mho. He obviously isn't interested in
figuring out anything. I provided a detailed circuit analysis that
shows bringing an additonal power source online will result in
a slightly higher voltage at the grid and the load, provided
everything
else stays the same. The example was a FIXED load. I only
pointed out that one alternative to raising the grid voltage is that
the load could instead increase

His silly short response makes no sense at all.

On 4/17/2011 2:20 PM 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
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k
l8tr
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- from Usenet (what's *that*?)

On Sun, 17 Apr 2011 14:56:23 -0700, David Nebenzahl
wrote:

On 4/17/2011 2:20 PM 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.

On Apr 17, 5:56*pm, David Nebenzahl wrote:
On 4/17/2011 2:20 PM 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/...its/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.

test
"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

On Apr 16, 3:23*am, harry wrote:
On Apr 16, 12:56*am, Home Guy wrote:





" wrote:
Solve the equations for the case where the voltage sources are
at identical voltages and you will find that current flows from
BOTH sources. *One is NOT at a higher potential than the
other, which is what Homeguy claims must exist for both
to provide current to the load.

So what you're saying is this:

Connect 2 batteries of the same voltage together in parallel to the same
load and each battey will supply half the current to the load.

So if I extrapolate that situation:

If my service voltage is currently sitting at 120.0 volts, and if I
adjust my PV invertor output voltage to 120.0 volts, and then connect my
PV output to my service connection, then my PV system will somehow
magically supply half the current to to the load (the load being my
house and all other houses sharing the same service line).

Wow. *That sounds like a really good bargain. *Just by matching the
power companies voltage at my service input, my PV system will supply
half the current - always! *And it doesn't matter how many PV panels I
have!

Wow. *Who knew it would be that simple and effective?

No. You match the voltage and then turn it up until the full load
current of your array is flowing in practical terms.
There. Do you understand that?

DON't WORRY A FEW MORE HITS ON HIS CRACK PIPE AND K EITH RW WILL GET
IT DONE.
IN HIS OWN PUTRID MIND HE IS A VIRTUAL JACKASS OF ALL TRADES.

PAT ECUM

On Apr 15, 11:45*pm, "
wrote:
On Fri, 15 Apr 2011 20:03:02 -0700 (PDT), The Ghost in The Machine





wrote:
On Apr 14, 12:23 am, "
wrote:
On Wed, 13 Apr 2011 20:46:43 -0700 (PDT), The Ghost in The Machine

wrote:
On Apr 13, 11:38 pm, "
wrote:
On Wed, 13 Apr 2011 21:33:08 -0400, "vaughn"
wrote:

wrote in message
news

E=IR, certainly *IS* Ohm's law. I and E are proportional. You can't increase
I without increasing E.
Wrong. You CAN increase I without increasing E. You have 3 variables in that
formula, not just 2.

Dumbass, it's a fixed circuit.

Get it? I suppose not.

Apparently not.

You've only proved that you're just as stupid as David.

SO, NOW WE ALL SHOULD RESORT TO ANGRY NAMECALLING & FAGGOTY RESPONSES
HAHN???

GROW UP FOOL, OR DROP IT, YOUR STALE INSIGNIFICANT DULLARD POSTINGS
ARENT HELPING ANY.

Speaking of dumbasses...

PAT ECUM

Here is Roy Queerjano, A.K.A. Pat E. Cum/

DANNNNNNNNT WRONG !!!!

"DANNNNNNNNNT"? *That word isn't in my dictionary, Roy Queerjano?

YOU SHOULD SHUT UP DONKEY KELLY WE KNOW ALL ABOUT YOU AND MICHAEL
TERRELL'S *GAYTARD BROKEN HEART OVER ROY Q...YOU FOOL....FYI I AM NOT
HIM ......YOU SYCOPHANTIC USENET ABUSER.

You're a liar, Roy Eat Cum.

COLLECTING AND DELIVERING YOUR HATEFUL OPPRESSIVE RESENTFUL SOUL TO
HELL WILL BE A TREAT TO MANKIND.

Go away, Roy.

THIS ROY YOU KEEP BRINGING UP, HAVE YOU BEEN PERFORMING FELLATIO ON
HIM???
YOU MUST BE IN LOVE WITH HIM, YOU QUEER, SEEMS HE HAS GOT YOU ALL UP
IN A TIZZY, KIZZY.

I HOPE YOU TWO LIVE IN ONE OF THOSE LEGAL GAYTARD MARRIAGE STATES, ID
HATE TO HAVE YOU BACK IN HERE CRYING AND WHINING ABOUT HOW YOUR TWINKY
LITTLE HEART IS BROKEN CAUSE YOU CANT MARRY
HIM ....BWAHAHAHAJAHAHA !

AND IT'S MR. ECUM TO YOU FOOL.

In article ,
" wrote:

No, you can't get on the "freeway" unless you're going faster than 60.
If you're going slower (or the same speed as you say,)
they're actively pushing you off.

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?

On Sun, 17 Apr 2011 17:43:09 -0700 (PDT), The Ghost in The Machine
wrote:

On Apr 15, 11:45*pm, "
wrote:
On Fri, 15 Apr 2011 20:03:02 -0700 (PDT), The Ghost in The Machine





wrote:
On Apr 14, 12:23 am, "
wrote:
On Wed, 13 Apr 2011 20:46:43 -0700 (PDT), The Ghost in The Machine

wrote:
On Apr 13, 11:38 pm, "
wrote:
On Wed, 13 Apr 2011 21:33:08 -0400, "vaughn"
wrote:

wrote in message
news

E=IR, certainly *IS* Ohm's law. I and E are proportional. You can't increase
I without increasing E.
Wrong. You CAN increase I without increasing E. You have 3 variables in that
formula, not just 2.

Dumbass, it's a fixed circuit.

Get it? I suppose not.

Apparently not.

You've only proved that you're just as stupid as David.

SO, NOW WE ALL SHOULD RESORT TO ANGRY NAMECALLING & FAGGOTY RESPONSES
HAHN???

GROW UP FOOL, OR DROP IT, YOUR STALE INSIGNIFICANT DULLARD POSTINGS
ARENT HELPING ANY.

Speaking of dumbasses...

PAT ECUM

Here is Roy Queerjano, A.K.A. Pat E. Cum/

DANNNNNNNNT WRONG !!!!

"DANNNNNNNNNT"? *That word isn't in my dictionary, Roy Queerjano?

YOU SHOULD SHUT UP DONKEY KELLY WE KNOW ALL ABOUT YOU AND MICHAEL
TERRELL'S *GAYTARD BROKEN HEART OVER ROY Q...YOU FOOL....FYI I AM NOT
HIM ......YOU SYCOPHANTIC USENET ABUSER.

You're a liar, Roy Eat Cum.

COLLECTING AND DELIVERING YOUR HATEFUL OPPRESSIVE RESENTFUL SOUL TO
HELL WILL BE A TREAT TO MANKIND.

Go away, Roy.

THIS ROY YOU KEEP BRINGING UP, HAVE YOU BEEN PERFORMING FELLATIO ON
HIM???

No, Roy, you are *not* my type. Now go *away*!

YOU MUST BE IN LOVE WITH HIM, YOU QUEER, SEEMS HE HAS GOT YOU ALL UP
IN A TIZZY, KIZZY.

Isn't it cute when Roy Queerjano accuses others of being what he *is*.

I HOPE YOU TWO LIVE IN ONE OF THOSE LEGAL GAYTARD MARRIAGE STATES, ID
HATE TO HAVE YOU BACK IN HERE CRYING AND WHINING ABOUT HOW YOUR TWINKY
LITTLE HEART IS BROKEN CAUSE YOU CANT MARRY
HIM ....BWAHAHAHAJAHAHA !

Sorry, Pat Eats Cum, I'm already married - longer than you've been alive,
moron.

AND IT'S MR. ECUM TO YOU FOOL.

No it's *not*, Roy.

On Sun, 17 Apr 2011 18:15:20 -0700, Smitty Two
wrote:

In article ,
" wrote:

No, you can't get on the "freeway" unless you're going faster than 60.
If you're going slower (or the same speed as you say,)
they're actively pushing you off.

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

Again, physics doesn't care what you like and don't like. It is.

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

Well, I guess you could say that "currents" push against each other, but it
requires a difference in voltage to have a current. Think of the intersection
of two rivers.

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.

You assume wire has zero resistance. Bad assumption.

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.

That is "said" has little bearing on physics.

Where is my thinking flawed?

The biggest flaw is that resistance is not zero and you take what people "say"
too literally. Analogies are always flawed. That's why they're called
"analogies". ;-)

On Apr 17, 9:15*pm, Smitty Two wrote:
...
Where is my thinking flawed?

It makes more sense if you think of the inverter as forcing a constant
CURRENT and let the voltages be whatever the source (wire etc)
resistance makes them at that current.

The grid may or may not act like an infinite sink. The continual load
variations will probably swamp out any voltage measurement you might
make, so it's reasonable to consider it an infinite sink unless you
have a very large inverter. The GTI wants to dump all the current from
the array onto the line and will adapt itself to the line voltage,
whatever it may be.

If you connect a PV panel to a 12V battery the panel will source as
much current as the sunlight produces, at the voltage of the battery
even if the panel's open circuit voltage is above 20V. The battery
voltage will rise a little because of the IR drop in its internal
resistance.

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

jsw

In article ,
" wrote:

On Sun, 17 Apr 2011 18:15:20 -0700, Smitty Two
wrote:

In article ,
" wrote:

No, you can't get on the "freeway" unless you're going faster than 60.
If you're going slower (or the same speed as you say,)
they're actively pushing you off.

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

Again, physics doesn't care what you like and don't like. It is.

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

Well, I guess you could say that "currents" push against each other, but it
requires a difference in voltage to have a current. Think of the intersection
of two rivers.

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.

You assume wire has zero resistance. Bad assumption.

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.

That is "said" has little bearing on physics.

Where is my thinking flawed?

The biggest flaw is that resistance is not zero and you take what people "say"
too literally. Analogies are always flawed. That's why they're called
"analogies". ;-)

None of the things you just said mean anything. Saying it's "due to
physics" is meaningless. I know wire has resistance, so what? And I'm
sure you know that when I said "I don't like it" I meant "I don't agree
with it." And, uh, rivers don't fight against tributaries, last I
checked, but you shouldn't be using analogies if you don't think they
hold water. So to speak.

You are billed by how much current you draw. Let's say your normal
consumption is a steady 1 kw. Now you start generating 100 watts. So now
you're only drawing 900 watts from the rest of the grid, and that,
multiplied by hours, is what you get charged for.

OTOH, let's say you install some bigger panels, and you can generate
1500 watts. Now your net consumption is *minus* 500 watts, so you're
pumping 500 watts into the grid, and, hopefully, being compensated for
that by the power company. You're using 1 kw locally, and delivering the
rest where it's needed elsewhere.

I seriously fail to see where the resistance of the wire has anything to
do with it. Please don't tell me "that's just the way it is." If I'm
wrong, give me a reasonable and logical argument.

On Apr 17, 9:15*pm, Smitty Two wrote:
In article ,

" wrote:
No, you can't get on the "freeway" unless you're going faster than 60.
If you're going slower (or the same speed as you say,)
they're actively pushing you off.

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.

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.

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.

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.

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?

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.

On Sun, 17 Apr 2011 19:48:21 -0700, Smitty Two
wrote:

In article ,
" wrote:

On Sun, 17 Apr 2011 18:15:20 -0700, Smitty Two
wrote:

In article ,
" wrote:

No, you can't get on the "freeway" unless you're going faster than 60.
If you're going slower (or the same speed as you say,)
they're actively pushing you off.

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

Again, physics doesn't care what you like and don't like. It is.

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

Well, I guess you could say that "currents" push against each other, but it
requires a difference in voltage to have a current. Think of the intersection
of two rivers.

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.

You assume wire has zero resistance. Bad assumption.

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.

That is "said" has little bearing on physics.

Where is my thinking flawed?

The biggest flaw is that resistance is not zero and you take what people "say"
too literally. Analogies are always flawed. That's why they're called
"analogies". ;-)

None of the things you just said mean anything. Saying it's "due to
physics" is meaningless. I know wire has resistance, so what? And I'm
sure you know that when I said "I don't like it" I meant "I don't agree
with it."

Let me say it again, perhaps you'll catch on. Physics doesn't care what you
like. It is what it is.

And, uh, rivers don't fight against tributaries, last I
checked, but you shouldn't be using analogies if you don't think they
hold water. So to speak.

Because, like electricity, water always flows "down hill" - high to low.

You are billed by how much current you draw. Let's say your normal
consumption is a steady 1 kw. Now you start generating 100 watts. So now
you're only drawing 900 watts from the rest of the grid, and that,
multiplied by hours, is what you get charged for.

OK.

OTOH, let's say you install some bigger panels, and you can generate
1500 watts. Now your net consumption is *minus* 500 watts, so you're
pumping 500 watts into the grid, and, hopefully, being compensated for
that by the power company. You're using 1 kw locally, and delivering the
rest where it's needed elsewhere.

OK.

I seriously fail to see where the resistance of the wire has anything to
do with it. Please don't tell me "that's just the way it is." If I'm
wrong, give me a reasonable and logical argument.

Voltage is dropped across a resistance. Not all points in the grid go up
because your solar cells output more voltage because there is non-zero
resistance between all points. If you're generating electricity, your house
will be at a *higher* voltage than the pole. If the resistance of the wire
were zero this couldn't happen.