Posted by *trader4@optonline.net* on April 18, 2011, 2:48 pm

*> To Smitty*

*> > I've chewed this over a bit, and I still don't like it, and here are my*

*> > reasons:*

*> > 1: Voltage sources in parallel do not push *against* one another.*

*> You can't have two voltage sources EXACTLY in parallel, its like*

*> dividing by 0. There is always some small resistance between voltage*

*> sources or the sources themselves have resistance.*

*> > 2: If no voltage source can join the grid without being at a higher than*

*> > grid potential, then every contributingpowerstation would have to be*

*> > at a higher potential than every other one, and that's impossible.*

*> The sources have to be higher then the grid, not higher then eaach*

*> other. Lets say the grid is some point in the middle of a square and*

*> it is at 120.0. Lets say there are 4 sources feeding that center*

*> point from the 4 corners. Each feed has a small resistance between it*

*> and the "grid point". Each one can be at 120.1 for example and power*

*> will flow from each to the grid point. If feed point one is a big*

*> nuclear plant it might be at 120.3 and your small solar plant at point*

*> 2 might be at 120.01. But each will feed power TO the grid point*

*> load. Each plant adjusts itself to the right slight increse in voltage*

*> to feed the amount of power it has avaialbe.*

*> (This is an oversimplification of what really happens. What really*

*> happens also has a lot to do with frequency and phase.)*

*> > 3: While voltage might *push*, it's the load that it said to *pull* the*

*> > current. If there's a demand, current will flow whether the supply*

*> > voltage is 119, 120 or 121.*

*> If the load is at the grid point and is at 120.0, then any source*

*> under 120.0 will pull power and not source it., but the load as a*

*> reistor will pull current no matter what voltage is on it. A resistor*

*> can never geneate power. But a motor connected to a grid can pull*

*> power if you load the shaft or it can push power if you hook an engine*

*> to the shaft and drive it.*

*> Mark*

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 ---

R2------

! ! =

!

! ! =

!

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

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 *g* on April 15, 2011, 5:12 pm

On 15/04/2011 05:47, trader4@optonline.net wrote:

*>>*

*>> How to properly analyze the*

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

*>> Look at Example 1.*

*>>*

*>> jsw*

*> That*

*> circuit is EXACTLY the model for the dual battery*

*> example I brought up.*

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 *trader4@optonline.net* on April 15, 2011, 12:33 pm

*> On Apr 15, 5:46am, "k...@att.bizzzzzzzzzzzz"*

*> > >"k...@att.bizzzzzzzzzzzz" wrote:*

*> > >> > We can forget about generators and distributions systems.*

*> > >> > Just take two 12V batteries and connect them in parallel*

*> > >There's your problem right there.*

*> > >It's usually not a good idea to connect batteries together in parallel,*

*> > >unless they are exactly of the same type, age, condition, etc. If you*

*> > >get a weak cell in one of the batteries it will turn into a load.*

*> > >And what happens when they are not exactly at the same voltage before*

*> > >being connected together? How do you insure that you always get current*

*> > >flowing out of both of them?*

*> > >> > But clearly he thinks if we put a second AC power source on*

*> > >> > a distributions system, it has to be at a higher voltage to*

*> > >> > "push" current out.*

*> > >The IEEE paper I posted earlier today shows exactly that - that PV*

*> > >systems raise local grid voltage and the utility company must compensate*

*> > >by reducing primary supply voltage to down-regulate the secondary*

*> > >voltage coming from the distribution transformer.*

*> > What a dumbass!*

*> > >> > So, what happens with the two batteries? Under the*

*> > >> > laws of physics the rest of us use the voltage would*

*> > >> > remain at 12 volts and BOTH batteries would be supplying*

*> > >> > part of the 1 AMP flowing through the resistors.*

*> > >Take a 12 V car battery and wire it up in parallel with 8 AAA batteries*

*> > >connected in series. Then connect a load and tell me how much current*

*> > >the AAA batteries will supply vs their what their potential current*

*> > >supply could be if they were connected to their own isolated load.*

*> > They will share in proportion to their capacity. Electricity, water, nor shit*

*> > flow uphill. ...though you have been pumping enough of the latter here.*

*> > >> Not on Homeguy's and Vaughn's planet. One of the batteries*

*> > >> will be charging the other.*

*> > >If they are unequal in capacity, then yes that will eventually happen.*

*> > Wrong!- Hide quoted text -*

*> > - Show quoted text -*

*> He said unequal voltage. In which case he's right.*

*> Capacity is not voltage.- Hide quoted text -*

*> - Show quoted text -*

This is what he said:

"If they are unequal in capacity, then yes that (one will be charging

the

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

"krw@att.bizzzzzzzzzzzz" wrote:

*> >> > But clearly he thinks if we put a second AC power source on*

*> >> > a distributions system, it has to be at a higher voltage to*

*> >> > "push" current out.*

*> >*

*> > The IEEE paper I posted earlier today shows exactly that - that PV*

*> > systems raise local grid voltage and the utility company must*

*> > compensate by reducing primary supply voltage to down-regulate the*

*> > secondary voltage coming from the distribution transformer.*

*> *

*> What a dumbass!*

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

distribution transformers.

==============

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.

*> > Take a 12 V car battery and wire it up in parallel with 8 AAA*

*> > batteries connected in series. Then connect a load and tell me*

*> > how much current the AAA batteries will supply vs their what*

*> > their potential current supply could be if they were connected*

*> > to their own isolated load.*

*> *

*> They will share in proportion to their capacity.*

Capacity is the "quantity of electrons" - which by itself tells you

nothing of the potential (voltage).

*> Electricity, water, nor shit flow uphill. *

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

grid.

Posted by *trader4@optonline.net* on April 15, 2011, 2:14 pm

*> "k...@att.bizzzzzzzzzzzz" wrote:*

*> > >> > But clearly he thinks if we put a second AC power source on*

*> > >> > a distributions system, it has to be at a higher voltage to*

*> > >> > "push" current out.*

*> > > The IEEE paper I posted earlier today shows exactly that - that PV*

*> > > systems raise local grid voltage and the utility company must*

*> > > compensate by reducing primary supply voltage to down-regulate the*

*> > > secondary voltage coming from the distribution transformer.*

*> > What a dumbass!*

*> 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*

*> distribution transformers.*

*> ==============*

*> 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.*

*> > > Take a 12 V car battery and wire it up in parallel with 8 AAA*

*> > > batteries connected in series. Then connect a load and tell me*

*> > > how much current the AAA batteries will supply vs their what*

*> > > their potential current supply could be if they were connected*

*> > > to their own isolated load.*

*> > They will share in proportion to their capacity.*

*> Capacity is the "quantity of electrons" - which by itself tells you*

*> nothing of the potential (voltage).*

*> > Electricity, water, nor shit flow uphill.*

*> 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.*

Total nonsense and such a basic failure at elementary circuit basics

that it discredits just about anything else you have to say. The

voltage

of the array IS the voltage of the grid at the point it's connected.

How

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

discussion.

>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*

*> grid.*

Take a look at the dual voltage source circuit diagram that Jim

Wilkins

supplied a couple posts back. It's example #1.

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

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

ohm resistor.

> To Smitty> > I've chewed this over a bit, and I still don't like it, and here are my> > reasons:> > 1: Voltage sources in parallel do not push *against* one another.> You can't have two voltage sources EXACTLY in parallel, its like> dividing by 0. There is always some small resistance between voltage> sources or the sources themselves have resistance.> > 2: If no voltage source can join the grid without being at a higher than> > grid potential, then every contributingpowerstation would have to be> > at a higher potential than every other one, and that's impossible.> The sources have to be higher then the grid, not higher then eaach> other. Lets say the grid is some point in the middle of a square and> it is at 120.0. Lets say there are 4 sources feeding that center> point from the 4 corners. Each feed has a small resistance between it> and the "grid point". Each one can be at 120.1 for example and power> will flow from each to the grid point. If feed point one is a big> nuclear plant it might be at 120.3 and your small solar plant at point> 2 might be at 120.01. But each will feed power TO the grid point> load. Each plant adjusts itself to the right slight increse in voltage> to feed the amount of power it has avaialbe.> (This is an oversimplification of what really happens. What really> happens also has a lot to do with frequency and phase.)> > 3: While voltage might *push*, it's the load that it said to *pull* the> > current. If there's a demand, current will flow whether the supply> > voltage is 119, 120 or 121.> If the load is at the grid point and is at 120.0, then any source> under 120.0 will pull power and not source it., but the load as a> reistor will pull current no matter what voltage is on it. A resistor> can never geneate power. But a motor connected to a grid can pull> power if you load the shaft or it can push power if you hook an engine> to the shaft and drive it.> Mark