Posted by firstname.lastname@example.org on April 18, 2011, 2:48 pm
Absolutely agree. And this is probably as good a place as any to
post an analysis of the model Wilkins provided a while back. I
bet most of those on the other side of this issue have not
bothered to do any analysis of it. If you do, it shows exactly
what you are saying above:
1 1 Vg 1
1 1 1 ohm
------R1-- Rw----------Rg-----------Rg--------Rw ---
! ! =
! ! =
PV V1 Rload 100
ohms V2 240V
array ! ! =
! ! =
Here's a simple model of what we have. V1 and R1 are a
Thevinin model of the PV array. R1 is the internal resistance
of the PV array. Rw is the resistance of the
wire connecting the array to the grid. Rg is the resistance
of the grid wire connecting the load two blocks away. That
load is driven by another power source in parallel, modeled
by V2 and R2, it;s connecting wire Rw. I hope we can
all agree that this is a valid model for the discussion at
Step 1: The sun is not shining, the PV is supplying no
current. What is the voltage at the grid connection point
Vg? We know the only current flowing is from V2. That
current is 240/(100+1+1+1) = 240/103= 2.33 amps. That
current flows through the Rload resistor producing a
voltage of 100 X 2.33 = 233 volts across the load.
know zero current is flowing in the left side of the circuit,
the current through R1, R2, Rw must be zero. The only
way for that to occur is for Vg to be equal to 233 volts. The
grid voltage is now 233 volts. The voltage at V1 must also
be 233 volts.
Step 2: The sun comes out and the PV is going to put it's
power on the grid. The only way for current to start flowing
through V1 is for V1 to INCREASE above 233 volts. As it
does, the voltage on the grid will slowly increase too. How
do we know this? Let's look at the simplest case, where
the PV array can supply half the total current. Writing the
Kirchoff equation for the right side of the loop we have:
240 = i2 x (R2 + Rw + Rg) + (i1 + i2) x Rload
240 = i2 x 3 + ( i1 + i2) x 100
240 = 100 x i1 + 103 x i2
Where i2 is the current flowing from source V2 and i1
is the current flowing from the PV array. We
also know that each source is supplying half the
current, so i1 = i2. Substituting, we have
240 = 100 x i1 + 103 x i1
240 = 203 i1
or i = 1.182 amps.
We know that twice that is flowing through the load, so the
voltage across the load is 100 x 1.182 x 2 = 236.4 volts
Before we brought V1 online it was 233 volts. Now it is
3.4 volts HIGHER. Intuitively this makes sense, because
the voltage there must RISE enough to reduce the current
that was flowing from the other power source V2.
Now that we have the voltage across the load, and know
that I1 is 1.182 amps we can calculate the voltages at the
grid, Vg and what V1 must be:
Vg = 236.4 + i1 x Rg = 236.4 + 1.182 = 237.6 volts
V1 = 236.4 + i1 x (Rg + Rw+ R1) = 236.4 + 1.182 x 3 = 239.9 volts
All the voltages along the way have increased, the voltage at
the load, the voltage at the grid, and the voltage at the PV
Posted by g on April 15, 2011, 5:12 pm
On 15/04/2011 05:47, email@example.com wrote:
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" :)
Posted by firstname.lastname@example.org on April 15, 2011, 12:33 pm
This is what he said:
"If they are unequal in capacity, then yes that (one will be charging
other) will eventually happen. "
Which of course is total nonsense and I hope you agree.
Posted by Home Guy on April 15, 2011, 1:14 pm
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
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.
Capacity is the "quantity of electrons" - which by itself tells you
nothing of the potential (voltage).
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
Posted by email@example.com on April 15, 2011, 2:14 pm
Total nonsense and such a basic failure at elementary circuit basics
that it discredits just about anything else you have to say. The
of the array IS the voltage of the grid at the point it's connected.
can it be anything else, unless you want to include the resistance of
the wire used to make the connection, which is immaterial for the
Take a look at the dual voltage source circuit diagram that Jim
supplied a couple posts back. It's example #1.
It's a simple diagram of two ideal voltage
sources with series resistors connected to a load. That serves as a
basic model for two batteries or two generators or two PV arrays, etc
connected to a load. The example gives the full equations for what
would be two batteries of differing voltages and internal resistance
connected in parallel to a load. Change the voltages so that they
are equal, make them 20V. Solve those equations and you'll find
that BOTH sources are supplying current to the load. The
voltage on the "grid", ie across the load resistor is just one value.
One source is not at a higher value to "push" current.
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