Maker Pro
Maker Pro

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

For once I agree with Harry. We can forget about generators and
distributions systems. Just take two 12V batteries and connect
them in parallel to a 12ohm resistor. Under Homeguy's
theories, I don;t know what he thinks would happen. 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.

Oh, but according to Homeguy, that doesn't work because batteries are VOLTAGE
sources. said:
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.

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

David Nebenzahl

Jan 1, 1970
0
I think he gets it directly from Ohms Law. V=IR.

Or, I = V/R

If V, the voltage is zero, then I, the current must be zero. Or, in
other words, current will only flow if there is a difference in
voltage.

But that's not Ohm's Law (the statement "current will flow only if there
is a difference in voltage"). Actually, that is a *tautology* (look it
up). In other words, that's the very definition of current, which
requires a potential difference (voltage > 0) to flow. Ohm's law didn't
establish that, because it was already established by the time he came
along.

You've correctly stated Ohm's Law, but that's not what it says. Strictly
speaking, what Ohm determined was that the current flowing in a circuit
is proportional to the voltage and inversely proportional to the
resistance--but only for certain resistors. Specifically, his carefully
calibrated metal resistances, at a certain temperature. So "Ohm's
law"--what he determined experimentally and published--is only this:

I = E / R

and that only at fixed temperature. Turns out "Ohm's law" does *not*
hold for a lot of things that look like resistances in the real world
(for example, any humble tungsten filament fails to observe it).

But that's going waaaaay deeper into it than we need to 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*?)
 
H

Home Guy

Jan 1, 1970
0
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?

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.

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.
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.
 
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?


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!
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 shit
flow uphill. ...though you have been pumping enough of the latter here.
If they are unequal in capacity, then yes that will eventually happen.

Wrong!
 
H

Home Guy

Jan 1, 1970
0
"[email protected]" 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. Motors won't turn faster, lights won't
really burn brighter. They will just give off a little more heat thanks
to the extra current the panels are supplying to the grid.

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

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.
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.

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 shit
coming on-line if they have a choice.
 
H

Home Guy

Jan 1, 1970
0
What a dumbass!

You're the dumbass!

Read the following:

=============
SMUD wanted to investigate what effect reverse power flow from exporting
PV systems would have on service and substation voltage regulation.
Figure 1 shows the Home-to-Substation voltage difference (top) and solar
irradiance (bottom) on a clear, cool day, Saturday, March 7, 2009,
representative of a day with relatively low load and high local PV
penetration. Penetration is defined as the amount of PV output divided
by the load at a particular point in time.

At night the substation voltage ranged between 0.4 V to 0.7 V higher
than the home voltage. This is representative of a typical circuit with
voltage drops through the line and transformer impedances. During
daylight hours, this reversed and the home voltage rose as high as 0.7
V, or 0.6%, greater than the substation voltage.
============

Did you read the last sentence dumbass?

The differential between substation and home voltage went from -.7 to
+.7 volts - a difference of almost 1.5 volts due to the effect of the
PV panels injecting current into the grid.

If, according to you, there was no such phenomena of an increase in
local grid voltage caused by PV panels, then there should be no basis
for the point of this IEEE research paper I posted yesterday.

But engineers know that there will be a voltage increase because of
these panels, and the excercise now is to figure out how much PV power
can come on-line before the substation becomes unable to properly
regulate it's output voltage levels.

==============
Since the PV penetration levels were relatively low, there were no
adverse effects on voltage regulation. It was possible to see the
effects of the PV systems on the voltage at the individual homes and the
distribution transformers.
==============

There will be more PV-equipped homes coming on-line in that project and
it's not yet known if in total their operation will cause poorly
controlled or unstable grid voltage.
They will share in proportion to their capacity.

Capacity is the "quantity of electrons" - which by itself tells you
nothing of the potential (voltage).
Electricity, water, nor shit flow uphill.

And height is exactly equivalent to voltage potential.

So PV panels can't push current into the grid unless the invertors raise
their voltage higher than the grid voltage. Just matching the grid
voltage gets you to the point where your current flow is ZERO. Every
millivolt you adjust your output voltage higher than the grid voltage
means some small increment in current outflow from your panels. Since
you can't store the energy coming from the panels via battery bank, then
it's in your best interest to always maximize the amount of current
you're injecting into the grid up to the full potential of the panel's
output capacity. That means raising the output voltage as high as you
need to so that every milliamp is squeezed out of them and onto the
grid.
 
H

Home Guy

Jan 1, 1970
0
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.

===================
The negative sign for I1 means that the direction of current flow
initially chosen was wrong, but never the less still valid. In fact, the
20v battery is charging the 10v battery.
==================
 
H

Home Guy

Jan 1, 1970
0
Take a look at the dual voltage source circuit diagram that Jim
Wilkins supplied a couple posts back. It's example #1.

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

I did, you dumbass #2.

Go and read my last post.
It's a simple diagram of two ideal voltage sources with
series resistors connected to a load.

And note what happens when the voltage sources have unequal voltages.

And note that we are not talking about batteries here in the case of a
municipal power grid and a PV system.
And I'll bet if you do the equations with the voltage sources
at the same value, you'll find that twice as much current
flows from the voltage source with the 10 ohm resistor as
the one with the 20 ohm resistor.

If that diagram shows reverse current flow because Battery 1 has a lower
voltage than Battery 2 (and current I1 is negative), then at what point
does current I1 become zero? What would the voltage of battery 1 have
to be for current I1 to be zero?
 
B

bud--

Jan 1, 1970
0
Impedance is the vector sum of resistance and reactance measured in
Ohms not a percentage.
Regulation of a transformer mau be measured in percentage terms.
What exactly are yo on about?

Use of "per-unit" values for voltage, current, impedance, ... is common
in the electric power field. It makes calculations easier, particularly
as the system gets more extensive. One of the "per-unit" values is "%
impedance".

A utility transformer is likely to be rated in "% impedance".

Daestrom has written about this. Looks like mII (or whoever) is familiar
with it. That leaves you. If you knew as much as you think you know you
would be familiar with % impedance.

You really need to go for some instruction. These things can't be worked
out by lying on your bed and thinking about it.
 
B

bud--

Jan 1, 1970
0
"[email protected]" unnecessarily full-quoted:

(Which is why I didn't try.)
So why not run a 120V motor with 240 volts then?
Duh...


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.

A motor running at a constant RPM creates a fixed amount of mechanical
power for a given load. RPM of induction motors is not very sensitive to
voltage. The electrical power used is tied to the mechanical power
consumed. Raising the voltage a little lowers the current a little.
 
G

g

Jan 1, 1970
0
So, do you agree that under the condition of two identical
fully charged batteries at exactly 12V, connected in
parallel to a load, the current will flow from both
batteries through the load? I hope you do.

Your example, while it is correct, have little bearing on real life
problems. The grid can in no circumstances be looked at as identical to
a PV array voltage converted to AC voltage.

Two identical batteries have identical inner resistance, that is why
your example works.

As for Homeguy
he apparently believes one has to be at a higher voltage
to "push" current.

He is correct. Why? Because he wants to "push" current into the grid.
With your 2 batteries at the exact same voltage, how do you get current
flowing from the "home battery" to the "grid battery"?

I have yet to hear him explain how the
batteries then decide which one it will be and how they
will change their voltage to obtain the allegedly necessary
"push" to get the current flowing.

He does not have to explain that, it is self explanatory when one
understand that there are no perfect conductors with zero resistance in
a power distribution system.

To get power into the grid from a local generated power the voltage has
to be higher. HOW much higher depends on the impedance in the systems.
 
G

g

Jan 1, 1970
0
That
circuit is EXACTLY the model for the dual battery
example I brought up.

No it is not. It is an example with two different voltages.

It is a perfect example of how to supply power to the grid, resulting in
higher voltage at the "home-owners end" :)
 
M

Mho

Jan 1, 1970
0
hmmmmm. worked for a power company? Perhaps not an electrical "power "
company?
A weight lifter?
Meter Reader?

Sounds like a complete misnomer. Correctly that should be "energy" company.


All teasing aside, m II is quite correct in asking about percent impedance.
Have a look at any distribution or large power transformer and you will see
a rating called "%Imp."

I have seen pole transformers down to 1.2%Imp and up to 13.8%Imp.

This is where the electrical energy distribution companies get concerned
about losses. (see I even remembered my own comment. LOL)

Worked on a 2MVA Grounding Bank with a 89 %Imp. rating. (maybe it was 93???
Can't remember now)

As an aside... (utility guys will appreciate this one)
In the old days, before equipment was invented, we used to prove polarity of
CTs with a 6v lantern battery and an old Model T voltmeter coupled with a
variable resistor in series to avoid bashing the needle too hard. We would
generate positive and negative going pulse on the primary of the CT and see
the needle deflection on the secondary of the CT winding.

Trouble was this high impedance bitch would not pass any current from the 6V
lantern battery through the primary winding and through the CT primary
(bushing type). SO we get two 6V lantern batteries in series and try again
to no avail. hmmmmm...

Well we get talking about it and while still holding the leads on the
primary (14kV) bushings my partner begins to yell that the needle was now
deflecting upwards, so I pull it off!...Well! I drew about a 10" long arc
(yeah bare hands..no grid connection) and friggin' nearly fell off the
transformer, some 12-15' high, to the ground. WHEW!!!

Back we went to the 6V lantern battery, waited a little longer for this
impedance haystack to react and wore the 30kV gloves. Sitting on top was a
little more uncomfortable too...LOL



Here is your link with an oversimplified explanation.
http://www.transformerworld.co.uk/impedance.htm


---------------
"harry" wrote in message
Link?
I used to work for a power company. Never once heard of Impedance
being expressed as a percentage.
Transformers are rated in Kva.
There may be a continuous KVA and a higher one for a specified period.
Regulation is sometimes expressed as a %. ie how the output voltage
varies between no load and full load conditions.
Here's my link:-
http://en.wikipedia.org/wiki/Impedance_(electrical)
Now, where's yours?
 
G

g

Jan 1, 1970
0

I am not sure what you think I am wrong about here. Could you please
indicate which of my above statements are wrong?

The topic is " Feeding solar power back into municipal grid". How are
the voltage levels when there is a flow into the grid from the PV array?

Maybe I missed a topic change somewhere in the previous posts, which is
easy to do since a few posters here does not bother to trim posts.

Go back to the example in the link, as I suggested to
Homeguy. You have two ideal voltage sources with different voltages
connected through two resistors of different values, one 20 ohms,
the other 10 ohms. That is a basic model of a battery, where the
10 and 20 ohm resistors represent the different internal resistances
of the batteries. So they are batteries with identica voltage, but
different internal resistances.

I did read the example, and yes, I understand Kirchoff's Law.

But you seem to be contradicting yourself, you state above the batteries
have different voltages, then you state they are batteries with
identical voltages. Maybe I am not understanding what you mean.

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.

I have read all of Homeguy's messages, and I may have missed it, but
does he not claim the need to have a higher voltage to push current into
the GRID, not the LOAD.

I never said current flows from one battery to the other. It's
apparently
Homeguy and now you who are hung up on that for reasons unknown.

The reason of my "hangup" is the topic: Feeding solar power back into
municipal grid.

Your example of batteries with equal voltage might be interesting, but
irrelevant to the topic.
 
H

Home Guy

Jan 1, 1970
0
Home said:
:

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.
 
"[email protected]" unnecessarily full-quoted:


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.
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.
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 shit
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.
 
H

Home Guy

Jan 1, 1970
0
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?
 
S

Smitty Two

Jan 1, 1970
0
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.
 
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.
 
Top