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Feeding solar power back into municipal grid: Issues and finger-pointing - Page 21

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Posted by trader4@optonline.net on April 18, 2011, 9:20 pm
 

Take a look at the simple circuit analysis I provided earlier today
that is
a few posts down in this thread.  I drew a model of the situation we
have been discussing which is very similar to the example circuit
Wilkins provided earlier.  And it shows how not only the internal
voltage
of the PV array must rise, but so too the voltage on the grid at the
point of connection.   It all follows directly from Krichoff's Law.





Posted by David Nebenzahl on April 18, 2011, 11:01 pm
 
On 4/18/2011 2:20 PM trader4@optonline.net spake thus:


The problem I have with that post is that the ASCII graphics you used
are all jumbled and I can't make sense of them (trouble with line
lengths, I think). Any chance you can redraw it in such a way that news
clients won't scramble it? (Maybe try drawing shorter lines with "hard
returns" at the end?)


--
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*?)

Posted by krw@att.bizzzzzzzzzzzz on April 19, 2011, 12:50 am
 wrote:


Make sure it's in a fixed space font.

Posted by trader4@optonline.net on April 19, 2011, 1:45 am
 
Yeah, I just looked at it and it came out looking all jumbled.  Here's
another try:



                 1       1      Vg    1          1
1           1  ohm
         ------R1--  Rw------Rg-------Rg--------Rw   --- R2------
         !                                     !                           =
               !
         !                                     !                           =
               !
 PV  V1                                   Rload
100                           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
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

Posted by trader4@optonline.net on April 19, 2011, 1:50 am
 wrote:
Another try a straightening out the drawing....



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