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Feeding solar power back into municipal grid: Issues and finger-pointing - Page 20

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Posted by g on April 18, 2011, 4:29 am
 
On 17/04/2011 20:40, David Nebenzahl wrote:


If you take the voltage drops into account, the voltage at the inverter
has to be higher than the voltage at the grid connection point.

This is a real world scenario.


Posted by Mho on April 18, 2011, 4:21 pm
 
Sorry, totally incorrect assumption. I guess the real world isn't for you



Load current is shared depending on impedance of parallel source loops.

--------------------

If you take the voltage drops into account, the voltage at the inverter
has to be higher than the voltage at the grid connection point.

This is a real world scenario.


Posted by krw@att.bizzzzzzzzzzzz on April 18, 2011, 11:04 pm
 

You're wrong, 'g' is correct.  You can't even figure out how to use a
newsreader.


Word salad.



Posted by Jim Wilkins on April 18, 2011, 11:34 am
 
The inverter output is higher -internally- by the V=IR drop between it
and the grid.

In this case I is the independent variable, the array's output, and V
is whatever it takes to make I pass through R to get to the grid
voltage.

jsw

Posted by David Nebenzahl on April 18, 2011, 7:45 pm
 On 4/18/2011 4:34 AM Jim Wilkins spake thus:


OK, now we're getting somewhere.

At the risk of igniting another round of sniping here, how does that
work, exactly? I assume you're talking about the voltage drop between
the inverter and the point where it's tied to the external power lines
(= grid), correct? So since it can only "see" its own internal voltage,
how does the inverter even know what that voltage drop is? How does it
regulate its voltage so that it's equal to the grid voltage at the point
of connection?

Or is this somehow self-regulating, where the inverter simply "aims" at
what it calculates is the grid voltage, based on the current delivered
by the PV system, and the voltage self-stabilizes?

Gory details, please.


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