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Feeding solar power back into municipal grid: Issues and finger-pointing - Page 28

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Posted by trader4@optonline.net on April 15, 2011, 2:41 pm
 

Maybe it wasn't clear what I meant when I said:

"Just take two 12V batteries and connect
them in parallel to a 12ohm resistor.  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 resistor."

I didn't mean two batteries labled nominally as 12V that are
actually at different voltages because one is fully charged, the
other only partially charged.  We were trying to discuss a
simple a comparison as possible of using two power sources
on a circuit.

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

Posted by g on April 15, 2011, 5:03 pm
 
On 15/04/2011 07:41, trader4@optonline.net wrote:


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.



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"?



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.









Posted by trader4@optonline.net on April 15, 2011, 5:46 pm
 
Wrong.  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.
Very basic stuff.  The Kirchoff equations are right
 there.  All you have to do is change the voltages so that they
 are equal.

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 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.
If the voltage sources are equal, no current flows between the two,
which
is  the situation that is most desirable when powering
a load with two batteries in parallel. They just BOTH supply part of
the
 current to the load resistor.


As I said, go back and solve the equations for the case where the
voltages are equal and you will find that they BOTH supply current
to the load.  Because of the differing resistors which would represent
the internal resistance of the batteries, one supplies twice the
current of the other.  Capiche?



Imperfections do not render a model useless or change the facts.  If
you
want to model the transmission line, then insert some additional
resistors
after the 10 ohm and 20 ohm resistors to model the line.   Make a
model
that includes capacitance and inductance too, and make it an AC
circuit .    It changes
nothing with regard to the ridiculous requirement that in order to
supply
current, a source can't just be equal in voltageto another source
connected to the same load and that it has to be
higher.  If you have a model that shows Homeguy's planet, we'd like
to see it.  Until then, Jim's model is perfectly fine.


Posted by g on April 15, 2011, 8:10 pm
 On 15/04/2011 10:46, trader4@optonline.net wrote:

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.



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.



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.



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.





Posted by Home Guy on April 15, 2011, 11:56 pm
 "trader4@optonline.net" 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.

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?

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