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

[[email protected]]

<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.

 

[g]

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.

 

[Home Guy]

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.

 

[[email protected]]

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.

 

[Mho]

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.

If your impedance source is lower than the grid's at that point you will
supply the majority of the current. If your impedance source is higher (to
the load) then you will supply less than the grid.

One very simple rule. Two supplies in parallel output the same voltage. When
you measure each of them you will measure across the same two points for
both measurements.

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

"g" wrote in message

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.

 

[Mho]

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

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

Sources share the load between themselves according to the impedance between
and including the source and the load. If ht e grid were a perfect conductor
and had zero impedance no co-gen source could work.

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

wrote in message

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/dccircuits/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.

 

[g]

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?

 

[David Nebenzahl]

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*?)

 

[[email protected]]

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.

 

[David Nebenzahl]

On 4/17/2011 2:20 PM [email protected] spake thus:

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

Look at it this way. Suppose the utility pole at my house ia at
120.000V. I hook up a power source, be it a generator, PV, whatever
on the end of the line inside my house. I place exactly 120.000V on
my end of that line. The wire from my power source to the pole has
some small resistance, R. With 120.000V on one end of a resistor and
120.000V on the other end, by ohm's law, how much current will flow?
Zero.

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

Let's change the example just a little. We'll have a large AC source
(call it "the grid"), and a smaller AC source (the solar system's
inverter), connected *in parallel*, and then some resistance (the total
load of "the grid") after all that, completing the circuit.

In this example, "the grid" and the PV inverter are operating at
*exactly* the same voltage.

Let me argue this negatively, and see what you say to it:
If you're saying that this will not work (i.e., that the PV inverter
cannot contribute any power to the circuit because it isn't at a higher
voltage), then *no* circuit where you have two power sources in parallel
could ever work under the same circumstances.

No electric vehicle would ever work, because the battery cells in them
are (pretty close to) exactly the same voltage, so how would each tiny
cell (tiny in comparison to the total power of its siblings) ever be
able to "push" electricity into the circuit?

That elementary electronics tutorial (Kirchoff's Law, etc.) explains
everything you need to know here. And it doesn't require any higher voltage.

I look forward to your reply.


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

yo
wassup
nuttin
wan2 hang
k
where
here
k
l8tr
by

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

 

[[email protected]]

On 4/17/2011 2:20 PM [email protected] 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.

 

[MarkK]

test

 

[Smitty Two]

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?

 

[[email protected]]

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.

 

[Smitty Two]

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


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.


You assume wire has zero resistance. Bad assumption.


That is "said" has little bearing on physics.


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.

 

[[email protected]]

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.

 

[Daniel who wants to know]

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?

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

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

 

[David Nebenzahl]

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

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.


--
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]]

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".)

No, the voltage will STILL be higher if you're supplying current to the grid.
Wires have resistance. Current sources don't go against physics and really
are the same thing as voltage sources (Norton/Thevenin duality). Physics
doesn't lie.

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.

You're throwing around the pseudo-science. If you don't like the treatment,
you can easily leave.

 

[g]

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.

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