Posted by trader4@optonline.net on April 19, 2011, 2:02 am
wrote:
> Another try a straightening out the drawing....
> > 1 1 Vg 1 1 1 1 oh=
m
> > ------R1-- Rw------Rg-------Rg--------Rw --- R2------
> > ! ! =
!
> > ! ! =
!
> > PV V1 Rload 100 240v V2
> > 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
> > hand.
> > 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.
> > Since we
> > 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
> > array.
> > QED- Hide quoted text -
> - Show quoted text -
OK, got the drawing straightened out and it plus the analysis are
above. Just to add some clarification, I said the only way for
the PV array to start delivering power is for it to raise it's
voltage, which in turn raises the grid voltage. That assumes
that both the load and the other power source remain constant.
As I stated in other posts, the other ways for the PV array to
deliver power without it's voltage going up would be for the
load to increase, ie Rload gets smaller, or for the other power
source V2 to decrease in voltage.
Posted by Mho on April 19, 2011, 2:13 am
Adding another source lowers the net impedance of the supply to the load.
This is simple network theorem. Even ohm's law can tell you the voltage
requirements.
The voltage at the grid connection (assuming where Rg is) will not be the
same as the V2 source.
In DC theorem what you are saying would be basically all true but in AC we
have waveform phase angle and waveform distortion.
As an extreme example: consider a PV co-gen that is 180 degrees out of phase
from the grid. Now we can have a 10 volt PV source hooked to a 240V grid and
still supply current from it.
--------------------
wrote in message
wrote:
> > 1 1 Vg 1 1 1
> > 1 ohm
> > ------R1-- Rw------Rg-------Rg--------Rw --- R2------
> > ! !
> > !
> > ! !
> > !
> > PV V1 Rload 100 240v
> > V2
> > array ! !
> > !
> > ! !
> > !
>
> > ------------------------------------------------------------------
OK, got the drawing straightened out and it plus the analysis are
above. Just to add some clarification, I said the only way for
the PV array to start delivering power is for it to raise it's
voltage, which in turn raises the grid voltage. That assumes
that both the load and the other power source remain constant.
As I stated in other posts, the other ways for the PV array to
deliver power without it's voltage going up would be for the
load to increase, ie Rload gets smaller, or for the other power
source V2 to decrease in voltage.
Posted by trader4@optonline.net on April 19, 2011, 2:18 am
> Adding another source lowers the net impedance of the supply to the load.
> This is simple network theorem. Even ohm's law can tell you the voltage
> requirements.
> The voltage at the grid connection (assuming where Rg is) will not be the
> same as the V2 source.
Well, duh! I think everyone here, on both sides of the discussion
acknowledge that.
> In DC theorem what you are saying would be basically all true but in AC we
> have waveform phase angle and waveform distortion.
Not just basically, it is ALL exactly true with the equations to back
it up.
As for the complications of AC, there wouldn't appear to be much
point in discussing that until there is agreement on what happens
with a simple DC distribution system voltage example.
> As an extreme example: consider a PV co-gen that is 180 degrees out of phase
> from the grid. Now we can have a 10 volt PV source hooked to a 240V grid and
> still supply current from it.
> --------------------
Posted by Mho on April 19, 2011, 3:00 am
I guess I went too fast for you with the AC stuff.
The whole point is regarding AC connections and different rules apply. The
DC basics are mostly valid no matter how much you want to disagree with
something but still moot and established by the discussing people about 100
posts ago. Your ASCII schematic was nice.
I will let you disagree with yourself a little more for the next while.
----------------
wrote in message
> Adding another source lowers the net impedance of the supply to the load.
> This is simple network theorem. Even ohm's law can tell you the voltage
> requirements.
> The voltage at the grid connection (assuming where Rg is) will not be the
> same as the V2 source.
Well, duh! I think everyone here, on both sides of the discussion
acknowledge that.
> In DC theorem what you are saying would be basically all true but in AC we
> have waveform phase angle and waveform distortion.
Not just basically, it is ALL exactly true with the equations to back
it up.
As for the complications of AC, there wouldn't appear to be much
point in discussing that until there is agreement on what happens
with a simple DC distribution system voltage example.
> As an extreme example: consider a PV co-gen that is 180 degrees out of
> phase
> from the grid. Now we can have a 10 volt PV source hooked to a 240V grid
> and
> still supply current from it.
> --------------------
>
Posted by David Nebenzahl on April 19, 2011, 7:00 pm
On 4/18/2011 7:02 PM trader4@optonline.net spake thus:
> wrote:
>
>> Another try a straightening out the drawing....
>>
>>> 1 1 Vg 1 1 1 1
ohm
>>> ------R1-- Rw------Rg-------Rg--------Rw --- R2------
>>> ! !
!
>>> ! !
!
>>> PV V1 Rload 100 240v V2
>>> array ! !
!
>>> ! !
!
>>> ------------------------------------------------------------------
Sorry, it's still a hopeless hash.
I really do want to follow your example, but I can't until I can see
your circuit diagram properly. This ASCII-art thing clearly isn't
working; any chance you can post a picture somewhere? Then I'll be able
to follow along.
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> > 1 1 Vg 1 1 1 1 oh=
m
> > ------R1-- Rw------Rg-------Rg--------Rw --- R2------
> > ! ! =