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Feeding solar power back into municipal grid: Issues and finger-pointing - Page 18

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Posted by krw@att.bizzzzzzzzzzzz on April 17, 2011, 8:49 pm

"Approximately" means a little faster, or slower.  Slower does *not* work.

(And is what ever article I've ever

Of course they don't say that.  Physics does.

Whatever you've been saying doesn't change physics.

Posted by trader4@optonline.net on April 17, 2011, 9:20 pm

I was on your side of this issue before I went back and looked at the
model for two power sources driving a load that Jim Wilkins provided.
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
I provided.   If you look for the post, I encourage you to review the
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
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
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

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,
in turn lowers the voltage at the utility pole by house.

Those are the only choices to get a potential difference across the
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
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.

Posted by David Nebenzahl on April 17, 2011, 9:56 pm
 On 4/17/2011 2:20 PM trader4@optonline.net spake thus:

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

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:

   wan2 hang

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

Posted by krw@att.bizzzzzzzzzzzz on April 17, 2011, 10:31 pm

Simple: The motor is at a lower potential.

Elementary, sure, but you're still not getting it.

Posted by trader4@optonline.net on April 17, 2011, 10:33 pm
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
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.

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.

You can model a two cell battery with that circuit diagram as well.
leave V2 at 20V, making it a 20V cell with an internal resistance of
ohms modeled by R2.   It's companion cell is V1 with internal
of 10 ohms.  Do the math and you'll find that with V1 at 13.2 volts,
current flows through the V1 half of the circuit.  All the load
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

Did you look at the detailed analysis I provided in an earlier post
I went through the analysis of that circuit example?   Here it is
Take a look at that circuit and go through it step by step.  It is a
of two power sources driving a load.


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.

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