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Feeding solar power back into municipal grid: Issues and finger-pointing - Page 29

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Posted by g on April 16, 2011, 5:17 am
 
On 15/04/2011 16:56, Home Guy wrote:


Well, in a simplified scenario, (Grid, load, PV array) there will be
"load sharing" between the grid and the PV array. With the voltage
pretty much equal. I say pretty much equal, because there are line
resistance losses to take into account.


I think there is some misunderstanding about the concepts here.

trader4 talks about 2 batteries supplying a load. The example is good
for calculating load sharing between 2 voltage sources with load
resistor and line resistance.

The way I see it is that the LOAD with this battery analogy is the house
load. One battery models the PV array, the other the grid.

One thing should be clear: To get power into the grid at all, the house
load must be lower than what the PV array can supply. If you remove the
house load, then all the available PV array current will flow into the
grid, with the inverter at a higher voltage.

To get back to that Kirchoff's Law example, if we remove the 40 ohms
resistor (the load), there are basically 2 voltage sources opposing each
other. When these voltages are equal, no current flows.

To allow current to flow into the grid, the PV array voltage has to be
higher, whether there is a house load or not.





Posted by Mho on April 17, 2011, 3:22 pm
 
You guys need to get over your basic understanding of electricity.

When you connect to a grid the voltage is always **EXACTLY*** the same as
the connect point. They are in parallel.

If your impedance source is lower than the grid's at that point you will
supply the majority of the current. If your impedance source is higher (to
the load) then you will supply less than the grid.

One very simple rule. Two supplies in parallel output the same voltage. When
you measure each of them you will measure across the same two points for
both measurements.

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


On 15/04/2011 16:56, Home Guy wrote:


Well, in a simplified scenario, (Grid, load, PV array) there will be
"load sharing" between the grid and the PV array. With the voltage
pretty much equal. I say pretty much equal, because there are line
resistance losses to take into account.


I think there is some misunderstanding about the concepts here.

trader4 talks about 2 batteries supplying a load. The example is good
for calculating load sharing between 2 voltage sources with load
resistor and line resistance.

The way I see it is that the LOAD with this battery analogy is the house
load. One battery models the PV array, the other the grid.

One thing should be clear: To get power into the grid at all, the house
load must be lower than what the PV array can supply. If you remove the
house load, then all the available PV array current will flow into the
grid, with the inverter at a higher voltage.

To get back to that Kirchoff's Law example, if we remove the 40 ohms
resistor (the load), there are basically 2 voltage sources opposing each
other. When these voltages are equal, no current flows.

To allow current to flow into the grid, the PV array voltage has to be
higher, whether there is a house load or not.





Posted by trader4@optonline.net on April 17, 2011, 4:56 pm
 
That was where I was coming from too.  However, having gone back
and revisited that basic circuit diagram of two voltage source driving
a load, I have come around to where I see Homeguy's point that if a
new
source wants to deliver power onto the grid, it can raise the voltage
at the load.  Let's say I have a solar array that is covered up and
its sunny outside.  It's connected via distribution lines to a load
that
is a block away.   Another power source is located a similar
distance away from the load on the other side.  In other words,
a case like the circuit example Jim Wilkens provided.


When I uncover the PV array, for it's additional X KW of
power to make it to the load down the block, at least one
of 3 things needs to happen:

1 - The PV raises the voltage on it's end of the distribution system
slightly.

2 - the load increases

3 - the power source on the other end of the distribution system
lowers it's voltage.

Here's the circuit example again that Jim provided:

http://www.electronics-tutorials.ws/dccircuits/dcp_4.html

It's example one.

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.

I think this is what Homeguy has been saying all along.  So,
I've come full circle here and now agree with him on that
issue.   I still disagree that the slight increase in voltage
in a distribution system means that the power is wasted.
The issue there is how the loads respond to being given
121V instead of 120V.   HG claims that except for resistance
heaters, that energy goes to waste.  And I say there he
is wrong, but that topic is being covered in another part
of this thread.






Posted by Mho on April 17, 2011, 5:09 pm
 All that "load increases" is a bunch of baloney!

A fixed load is just that....***FIXED***

Sources share the load between themselves according to the impedance between
and including the source and the load. If ht e grid were a perfect conductor
and had zero impedance no co-gen source could work.

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

wrote in message


That was where I was coming from too.  However, having gone back
and revisited that basic circuit diagram of two voltage source driving
a load, I have come around to where I see Homeguy's point that if a
new
source wants to deliver power onto the grid, it can raise the voltage
at the load.  Let's say I have a solar array that is covered up and
its sunny outside.  It's connected via distribution lines to a load
that
is a block away.   Another power source is located a similar
distance away from the load on the other side.  In other words,
a case like the circuit example Jim Wilkens provided.


When I uncover the PV array, for it's additional X KW of
power to make it to the load down the block, at least one
of 3 things needs to happen:

1 - The PV raises the voltage on it's end of the distribution system
slightly.

2 - the load increases

3 - the power source on the other end of the distribution system
lowers it's voltage.

Here's the circuit example again that Jim provided:

http://www.electronics-tutorials.ws/dccircuits/dcp_4.html

It's example one.

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.  V1.2V,  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.

I think this is what Homeguy has been saying all along.  So,
I've come full circle here and now agree with him on that
issue.   I still disagree that the slight increase in voltage
in a distribution system means that the power is wasted.
The issue there is how the loads respond to being given
121V instead of 120V.   HG claims that except for resistance
heaters, that energy goes to waste.  And I say there he
is wrong, but that topic is being covered in another part
of this thread.





Posted by g on April 17, 2011, 8:17 pm
 On 17/04/2011 10:09, Mho wrote:

Define a fixed load.

As a teaser, say I buy a 2 kW heater to heat my office in my home on
those cold winter nights when I am reading posts in
alt.energy.homepower. Is that 2 kW heater a fixed load?


 > If ht e grid were a

I disagree. But I have been known to be wrong :)

Could you state why it does not work?

Is it possible to hook up any power source to such a grid?
If the answer is yes, why is co-gen not possible?


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